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Grade 6 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Rate problems with constant rates

Learning goals

  • Finds a unit rate and states its units.
  • Finds an amount or duration from a constant rate.
  • Compares rates expressed in a common unit.
  • Uses a constant rate in a multi-step calculation.
  • Distinguishes a constant-rate relationship from one with a fixed starting amount.

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Rate problems with constant rates 1
An elephant herd travels \(20\) miles in a day while moving at \(4\) miles per hour. How many hours does the herd travel?

Hints

- How far does the herd travel in one hour? - How many groups of \(4\) miles are in \(20\) miles? - Divide distance by rate to find time.

Solution

1. Use \(\text{time} = \text{distance} \div \text{rate}\). 2. Calculate \(20\,\text{mi} \div 4\,\text{mi/h} = 5\,\text{h}\).

Answer

The herd travels for \(5\) hours.
Rate problems with constant rates 2
A delivery van travels at a constant rate shown in the table. <table> <tr><th>Time (hours)</th><th>Distance (miles)</th></tr> <tr><td>\(2\)</td><td>\(90\)</td></tr> <tr><td>\(5\)</td><td>\(225\)</td></tr> </table> What is the unit rate, and what does it mean in this situation?

Hints

- A unit rate compares the distance with exactly one hour. - Use one row of the table to determine how much distance corresponds to one hour. - State both the numerical rate and its meaning in context.

Solution

1. Use either row to find the distance for one hour: \(90 \div 2 = 45\). 2. The unit rate is \(45\) miles per hour. 3. This means the van travels \(45\) miles during each hour when the rate stays constant.

Answer

\(45\,\text{mi/h}\). The van travels \(45\) miles each hour at this constant rate.
Rate problems with constant rates 3
A cyclist travels \(18\) miles in \(3\) hours at a steady speed. How far does the cyclist travel in one hour?

Hints

- Imagine the distance split evenly across three equal hours. - How much distance belongs to just one hour?

Solution

1. Divide the distance equally among the three hours. 2. In one hour, the cyclist travels \(18\div3=6\) miles.

Answer

\(6\) miles per hour
Rate problems with constant rates 4
A popcorn stand makes \(8\) bags of popcorn every minute at a steady rate. a) What does “\(8\) bags per minute” tell you? b) How many bags can it make in \(3\) minutes?

Hints

- Think about what the word “per” compares. - Does the rate describe one minute or all three minutes? - How do equal one-minute periods combine?

Solution

1. The rate means the stand makes \(8\) bags during each minute. 2. Three minutes contain three equal one-minute periods, so \(3\times8=24\) bags.

Answer

a) The stand makes \(8\) bags in each minute. b) \(24\) bags
Rate problems with constant rates 5
Four passes to a swimming pool cost \(\$14\). Each pass has the same price. How much does one pass cost? Explain what the unit price means.

Hints

- What does the given price cover? - How many equal price parts make the total? - What units belong with the answer?

Solution

1. Divide the total cost by the number of passes: \(14\div4=3.5\). 2. One pass costs \(\$3.50\); each additional pass costs the same amount.

Answer

\(\$3.50\) per pass; that is the price of one pass.
Rate problems with constant rates 6
A print shop's printer works at a steady rate. Fill in the missing number of pages and explain your method. <table><tr><th>Minutes</th><th>Pages printed</th></tr><tr><td>\(1\)</td><td>\(5\)</td></tr><tr><td>\(8\)</td><td>\(\square\)</td></tr></table>

Hints

- How does the number of pages change if the printer runs for more minutes at the same rate? - How could the one-minute row help you find the eight-minute total?

Solution

1. The printer makes \(5\) pages in one minute. 2. In eight minutes it makes \(8\times5=40\) pages.

Answer

\(40\) pages; at \(5\) pages per minute, \(8\times5=40\).
Rate problems with constant rates 7
How many miles does the hiker travel in one hour at the steady pace shown? Explain using a point on the graph.
Figure for problem 559340

Hints

- What does a point on the distance-time graph tell you? - Can you use a point that shows several hours to find distance in just one hour?

Solution

1. The line passes through \((2, 6)\): the hiker travels \(6\) miles in \(2\) hours. 2. The constant speed is \(6\div2=3\,\text{mi/h}\), or \(3\) miles in one hour.

Answer

\(3\,\text{mi/h}\); for example, the graph shows \(6\) miles in \(2\) hours, so one hour is half that distance.
Rate problems with constant rates 8
A student wraps one gift in \(3\) minutes. The student keeps the same pace. How many minutes will it take to wrap \(5\) gifts? Explain which quantity is measured “per gift.”

Hints

- What does one gift tell you about the time needed? - Does the number of gifts change the time for each gift at this pace? - Make sure the answer is in minutes.

Solution

1. The time rate is \(3\) minutes per gift. 2. Five gifts need \(5\times3=15\) minutes.

Answer

\(15\) minutes; the rate is \(3\) minutes per gift.
Rate problems with constant rates 9
At a farmers market, organic tomatoes cost \(\$4.80\) per pound. Mrs. Miller buys exactly \(2.25\) lb of tomatoes. What is the total cost?

Hints

- Use the price per pound and the number of pounds to find the total cost. - Keep track of decimal place value when multiplying. - Estimate first to check whether your answer is reasonable.

Solution

1. Multiply the unit price by the number of pounds: \(\$4.80/\text{lb}\times2.25\,\text{lb}\). 2. \(4.80\times2.25=10.80\). 3. The total cost is \(\$10.80\).

Answer

Mrs. Miller pays \(\$10.80\).
Rate problems with constant rates 10
Two cyclists compare training distances. Rider A completes \(14.5\) laps of a course that is \(2.25\) miles long. Rider B says, “I rode \(33\) miles today, so I went farther than you.” Use a calculation to determine whether Rider B is correct.

Hints

- Find Rider A's total distance first. - Then compare that distance with Rider B's distance. - Use place value carefully when multiplying decimals.

Solution

1. Find Rider A's distance: \(14.5\times2.25=32.625\) miles. 2. Compare the distances: \(33>32.625\). 3. Rider B is correct.

Answer

Rider B is correct. Rider A rode \(32.625\) miles, which is less than \(33\) miles.
Rate problems with constant rates 11
A grocery store has two orange specials. Deal A: \(15\) oranges for \(\$5.85\) Deal B: \(12\) oranges for \(\$4.80\) Which deal has the lower price per orange? Find both unit prices.

Hints

- Divide each total price by the number of oranges. - Use the same unit, dollars per orange, for both comparisons. - Compare the two unit prices.

Solution

1. Deal A: \(\$5.85\div15=\$0.39\) per orange. 2. Deal B: \(\$4.80\div12=\$0.40\) per orange. 3. Since \(\$0.39<\$0.40\), Deal A is less expensive per orange.

Answer

Deal A is less expensive at \(\$0.39\) per orange; Deal B costs \(\$0.40\) per orange.
Rate problems with constant rates 12
A class is making fruit punch for a school event. A recipe uses \(2.5\) gallons of juice for \(25\) people. How much juice is needed for \(60\) people? Write an equation to solve the problem.

Hints

- Find the amount of juice needed for one person. - Use a unit rate or equivalent ratio. - Check whether doubling the number of people doubles the amount of juice.

Solution

1. Find the amount per person: \(2.5\div 25=0.1\) gallon per person. 2. Let \(x\) be the gallons needed for \(60\) people. An equation is \(x=60\times 0.1\). 3. Solve: \(x=6\).

Answer

An equation is \(x=60\times 0.1\). The class needs \(6\) gallons of juice.
Rate problems with constant rates 13
A grocery store sells the same pasta in two package sizes. Package A contains \(12\,\text{oz}\) and costs \(\$1.80\). Package B contains \(20\,\text{oz}\) and costs \(\$2.80\). Find the price per ounce for each package. Which package is the better buy, and what is the difference in price per ounce?

Hints

- What does “price per ounce” mean? - Divide each price by the matching number of ounces. - Compare the unit prices, then subtract to find the difference.

Solution

1. Package A costs \(1.80\div 12=\$0.15\) per ounce. 2. Package B costs \(2.80\div 20=\$0.14\) per ounce. 3. Since \(0.14<0.15\), Package B is the better buy. 4. The difference is \(0.15-0.14=\$0.01\) per ounce.

Answer

Package A costs \(\$0.15\) per ounce. Package B costs \(\$0.14\) per ounce, so Package B is cheaper by \(\$0.01\) per ounce.
Rate problems with constant rates 14
A pancake recipe for \(4\) people uses \(2\) cups of flour and \(3\) eggs. a) How many cups of flour are needed for \(10\) people? b) If \(15\) eggs are used and all other ingredients are adjusted in the same ratio, how many people will the recipe serve?

Hints

- How much flour does the original recipe use per person? - In part b), compare the new number of eggs with the original number. - What stays in the same ratio when a recipe is scaled?

Solution

1. The flour per person is \(2\div4=0.5\) cup. 2. For \(10\) people, the flour needed is \(10\times0.5=5\) cups. 3. The egg quantity is multiplied by \(15\div3=5\). 4. The new number of servings is \(4\times5=20\).

Answer

a) \(5\) cups of flour b) \(20\) people
Rate problems with constant rates 15
A juice bar sells fresh juice in three cup sizes: - Small: \(8\,\text{fl oz}\) for \(\$2.20\) - Medium: \(12\,\text{fl oz}\) for \(\$3.00\) - Large: \(20\,\text{fl oz}\) for \(\$4.50\) a) Find the price per \(4\,\text{fl oz}\) for each cup size. b) Which cup size has the lowest price per \(4\,\text{fl oz}\)? c) If juice could be bought at that same rate, how much would \(36\,\text{fl oz}\) cost?

Hints

- Use the same comparison amount for all three cup sizes. - Count how many groups of \(4\,\text{fl oz}\) are in each cup. - After identifying the lowest rate, apply that rate to \(36\,\text{fl oz}\).

Solution

1. Small: \(8\,\text{fl oz}\) contains two groups of \(4\,\text{fl oz}\), so \(\$2.20 \div 2 = \$1.10\) per \(4\,\text{fl oz}\). 2. Medium: \(12\,\text{fl oz}\) contains three groups of \(4\,\text{fl oz}\), so \(\$3.00 \div 3 = \$1.00\) per \(4\,\text{fl oz}\). 3. Large: \(20\,\text{fl oz}\) contains five groups of \(4\,\text{fl oz}\), so \(\$4.50 \div 5 = \$0.90\) per \(4\,\text{fl oz}\). 4. The large-cup rate is lowest. 5. There are nine groups of \(4\,\text{fl oz}\) in \(36\,\text{fl oz}\), so \(9 \times \$0.90 = \$8.10\).

Answer

a) Small: \(\$1.10\) per \(4\,\text{fl oz}\); medium: \(\$1.00\) per \(4\,\text{fl oz}\); large: \(\$0.90\) per \(4\,\text{fl oz}\). b) The large cup has the lowest rate. c) \(\$8.10\)
Rate problems with constant rates 16
A clock’s minute hand moves at a constant rate. Find how many minutes pass while the hand sweeps through each angle. a) \(48^\circ\) b) \(162^\circ\) c) \(210^\circ\) d) \(540^\circ\)

Hints

- How many degrees does the minute hand move in one minute? - Divide each angle by the hand’s rate. - In part d, the angle is greater than one full revolution.

Solution

1. The minute hand sweeps \(360^\circ\) in \(60\) minutes, so its rate is \(360^\circ \div 60\,\text{minutes} = 6^\circ\) per minute. 2. For part a, \(48^\circ \div 6^\circ\text{ per minute} = 8\,\text{minutes}\). 3. For part b, \(162^\circ \div 6^\circ\text{ per minute} = 27\,\text{minutes}\). 4. For part c, \(210^\circ \div 6^\circ\text{ per minute} = 35\,\text{minutes}\). 5. For part d, \(540^\circ \div 6^\circ\text{ per minute} = 90\,\text{minutes}\).

Answer

a) \(8\,\text{minutes}\) b) \(27\,\text{minutes}\) c) \(35\,\text{minutes}\) d) \(90\,\text{minutes}\), or \(1\,\text{hour}\ 30\,\text{minutes}\)
Rate problems with constant rates 17
A garden hose delivers water at a constant rate of \(12\,\text{gal/min}\). How long will it take to fill a small pool that holds \(1500\,\text{gal}\)? Give your answer in hours and minutes.

Hints

- What operation gives the time when you know the total amount and the amount delivered each minute? - Divide the pool capacity by the hose's rate. - How many minutes are in one hour?

Solution

1. Divide the pool capacity by the flow rate: \(1500 \div 12=125\) minutes. 2. Convert \(125\) minutes to hours and minutes: \(125=2 \times 60+5\). Therefore, the time is \(2\) hours \(5\) minutes.

Answer

It will take \(2\) hours \(5\) minutes.
Rate problems with constant rates 18
A delivery driver travels at a constant rate of \(50\) miles per hour. Complete the table. <table> <tr> <td>Time</td> <td>\(1\,\text{h}\)</td> <td>\(2\,\text{h}\)</td> <td>\(4\,\text{h}\)</td> <td>\(10\,\text{h}\)</td> <td>\(\frac{1}{2}\,\text{h}\)</td> </tr> <tr> <td>Distance</td> <td>\(50\,\text{mi}\)</td> <td>?</td> <td>?</td> <td>?</td> <td>?</td> </tr> </table>

Hints

- Use the rate of \(50\) miles for each hour. - When the time doubles, the distance doubles. - For half an hour, find half of the one-hour distance. - For \(10\) hours, multiply the one-hour distance by \(10\).

Solution

1. For \(2\) hours: \(50 \times 2 = 100\) miles. 2. For \(4\) hours: \(50 \times 4 = 200\) miles. 3. For \(10\) hours: \(50 \times 10 = 500\) miles. 4. For \(\frac{1}{2}\) hour: \(50 \div 2 = 25\) miles. 5. Complete the table. <table> <tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(10\,\text{h}\)</td><td>\(\frac{1}{2}\,\text{h}\)</td></tr> <tr><td>Distance</td><td>\(50\,\text{mi}\)</td><td>\(100\,\text{mi}\)</td><td>\(200\,\text{mi}\)</td><td>\(500\,\text{mi}\)</td><td>\(25\,\text{mi}\)</td></tr> </table>

Answer

<table> <tr><td>Time</td><td>\(1\,\text{h}\)</td><td>\(2\,\text{h}\)</td><td>\(4\,\text{h}\)</td><td>\(10\,\text{h}\)</td><td>\(\frac{1}{2}\,\text{h}\)</td></tr> <tr><td>Distance</td><td>\(50\,\text{mi}\)</td><td>\(100\,\text{mi}\)</td><td>\(200\,\text{mi}\)</td><td>\(500\,\text{mi}\)</td><td>\(25\,\text{mi}\)</td></tr> </table>
Rate problems with constant rates 19
A hiker walks at a constant rate of \(4\) miles per hour. a) How far does the hiker walk in \(3\) hours? b) How far does the hiker walk in half an hour? c) How many hours does it take the hiker to walk \(24\) miles?

Hints

- Use the rate of \(4\) miles for each hour. - Half an hour gives half the one-hour distance. - For part c), determine how many groups of \(4\) miles are in \(24\) miles.

Solution

1. In \(3\) hours, the hiker walks \(4 \times 3 = 12\) miles. 2. Half an hour is half of one hour, so the hiker walks \(4 \div 2 = 2\) miles. 3. For \(24\) miles, divide distance by the unit rate: \(24 \div 4 = 6\) hours.

Answer

a) \(12\,\text{mi}\) b) \(2\,\text{mi}\) c) \(6\,\text{h}\)
Rate problems with constant rates 20
A parking garage charges by the minute at a rate of \(\$4\) per hour. Complete the table. <table> <tr> <td>Parking time</td> <td>\(1\,\text{hr}\)</td> <td>\(2\,\text{hr}\)</td> <td>\(30\,\text{min}\)</td> <td>\(1\,\text{hr}\ 30\,\text{min}\)</td> <td>\(3\,\text{hr}\)</td> </tr> <tr> <td>Cost</td> <td>\(\$4\)</td> <td></td> <td></td> <td></td> <td></td> </tr> </table>

Hints

- What is the cost for half of an hour? - Combine the costs for one hour and half an hour when needed. - Determine how many hourly rates are included in each parking time.

Solution

1. For \(2\) hours, \(2 \times \$4 = \$8\). 2. Thirty minutes is half an hour, so \(\$4 \div 2 = \$2\). 3. For \(1\) hour \(30\) minutes, \(\$4 + \$2 = \$6\). 4. For \(3\) hours, \(3 \times \$4 = \$12\).

Answer

<table> <tr> <td>Parking time</td> <td>\(1\,\text{hr}\)</td> <td>\(2\,\text{hr}\)</td> <td>\(30\,\text{min}\)</td> <td>\(1\,\text{hr}\ 30\,\text{min}\)</td> <td>\(3\,\text{hr}\)</td> </tr> <tr> <td>Cost</td> <td>\(\$4\)</td> <td>\(\$8\)</td> <td>\(\$2\)</td> <td>\(\$6\)</td> <td>\(\$12\)</td> </tr> </table>
Rate problems with constant rates 21
Renting an electric bicycle costs \(\$12\) per hour. The shop also charges for parts of an hour at the same rate. Find each rental price. <table><tr><th>Rental time</th><td>\(30\,\text{min}\)</td><td>\(15\,\text{min}\)</td><td>\(45\,\text{min}\)</td><td>\(2\,\text{h}\ 15\,\text{min}\)</td><td>\(3\,\text{h}\ 30\,\text{min}\)</td></tr><tr><th>Price</th><td>\(\square\)</td><td>\(\square\)</td><td>\(\square\)</td><td>\(\square\)</td><td>\(\square\)</td></tr></table>

Hints

- Think about how much of an hour each shorter rental lasts. - What part of an hour is fifteen minutes? - Combine full hours and extra minutes for the longer rentals.

Solution

1. Thirty minutes costs \(\$12\div2=\$6\). 2. Fifteen minutes costs \(\$12\div4=\$3\). 3. Forty-five minutes costs \(\$6+\$3=\$9\). 4. Two hours and fifteen minutes costs \(2\times\$12+\$3=\$27\). 5. Three hours and thirty minutes costs \(3\times\$12+\$6=\$42\).

Answer

<table><tr><th>Rental time</th><td>\(30\,\text{min}\)</td><td>\(15\,\text{min}\)</td><td>\(45\,\text{min}\)</td><td>\(2\,\text{h}\ 15\,\text{min}\)</td><td>\(3\,\text{h}\ 30\,\text{min}\)</td></tr><tr><th>Price</th><td>\(\$6\)</td><td>\(\$3\)</td><td>\(\$9\)</td><td>\(\$27\)</td><td>\(\$42\)</td></tr></table>
Rate problems with constant rates 22
A gardener works in a park from \(8{:}30\) a.m. to \(11{:}30\) a.m. The gardener charges \(\$42\) per hour. What is the total charge?

Hints

- How many hours pass between the start and end times? - Multiply the number of hours by the hourly rate.

Solution

1. The time from \(8{:}30\) a.m. to \(11{:}30\) a.m. is \(3\) hours. 2. Multiply the hours by the hourly rate: \(3 \times \$42 = \$126\).

Answer

The total charge is \(\$126\).
Rate problems with constant rates 23
An IT technician works in an office from \(2{:}15\) p.m. to \(4{:}45\) p.m. The rate is \(\$64\) per hour. How much must the office pay?

Hints

- Find the elapsed time in hours and minutes. - What is half of the hourly charge? - Add the charges for the full hours and the remaining half hour.

Solution

1. The work time is \(2\) hours \(30\) minutes, or \(2.5\) hours. 2. The charge for \(2\) full hours is \(2 \times \$64 = \$128\). 3. The charge for half an hour is \(\$64 \div 2 = \$32\). 4. The total is \(\$128 + \$32 = \$160\).

Answer

The office must pay \(\$160\).
Rate problems with constant rates 24
A campground charges \(\$12\) per night for a small tent site. a) Complete the price table. <table> <tr><th>Nights</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr> <tr><th>Price</th><td>\(\$12\)</td><td></td><td></td><td></td><td></td><td></td></tr> </table> b) What is the total price for \(14\) nights?

Hints

- How much does the price increase for each additional night? - Combine the prices for \(10\) nights and \(4\) nights to find the price for \(14\) nights. - Multiply the number of nights by the nightly rate.

Solution

1. Multiply each number of nights by \(\$12\) to complete the table. <table> <tr><th>Nights</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr> <tr><th>Price</th><td>\(\$12\)</td><td>\(\$24\)</td><td>\(\$36\)</td><td>\(\$60\)</td><td>\(\$96\)</td><td>\(\$120\)</td></tr> </table> 2. For \(14\) nights, \(14 \times \$12 = \$168\).

Answer

a) <table><tr><th>Nights</th><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(5\)</td><td>\(8\)</td><td>\(10\)</td></tr><tr><th>Price</th><td>\(\$12\)</td><td>\(\$24\)</td><td>\(\$36\)</td><td>\(\$60\)</td><td>\(\$96\)</td><td>\(\$120\)</td></tr></table> b) \(14\) nights cost \(\$168\).
Rate problems with constant rates 25
At an ice cream shop, each scoop costs the same amount, with no discounts for larger orders. Four scoops cost \(\$6.00\). a) How much do \(8\) scoops cost? b) How much do \(2\) scoops cost?

Hints

- Compare each requested number of scoops with the original four. - Since the price per scoop stays the same, what happens when the amount doubles or halves?

Solution

1. Eight scoops are twice four scoops, so they cost \(\$6.00\times2=\$12.00\). 2. Two scoops are half of four scoops, so they cost \(\$6.00\div2=\$3.00\).

Answer

a) \(\$12.00\) b) \(\$3.00\)
Rate problems with constant rates 26
An amusement park has two kinds of boats for a canal ride. Wave: total weight limit \(560\,\text{kg}\), with \(8\) seats. Anchor: total weight limit \(450\,\text{kg}\), with \(6\) seats. What is the average share of the total weight limit per seat for each boat? Which boat has the larger share? This is not a weight limit for an individual rider.

Hints

- What two quantities are being compared for each boat? - How can you make a fair comparison between boats with different seat counts?

Solution

1. Wave has \(560\div8=70\,\text{kg}\) per seat if the capacity is divided equally. 2. Anchor has \(450\div6=75\,\text{kg}\) per seat on the same basis. 3. Anchor's equal per-seat share is larger, since \(75>70\).

Answer

Wave: \(70\,\text{kg}\) per seat; Anchor: \(75\,\text{kg}\) per seat. Anchor's share is larger.
Rate problems with constant rates 27
An orchard produces \(48{,}000\,\text{L}\) of apple juice in a year. On average, one mature apple tree provides enough apples for about \(60\,\text{L}\) of juice. About how many apple trees are needed to produce that amount?

Hints

- Divide the total juice volume by the average volume from one tree. - Simplify the division by removing a factor of \(10\). - Use approximation language in the answer.

Solution

1. Divide the total juice volume of \(48{,}000\,\text{L}\) by the average volume of \(60\,\text{L}\) per tree. The quotient is about \(800\). 2. The quotient is approximate because the yield per tree is an average.

Answer

About \(800\) apple trees are needed.
Rate problems with constant rates 28
A region uses about \(900{,}000\,\text{L}\) of milk each month. A tanker truck carries \(15{,}000\,\text{L}\) per full load. About how many full truckloads equal the monthly amount?

Hints

- Divide the monthly volume by the volume in one full truck. - Simplify by dividing both numbers by \(1000\). - Use approximation language in the answer.

Solution

1. Divide the approximate monthly volume of \(900{,}000\,\text{L}\) by the truck capacity of \(15{,}000\,\text{L}\). The quotient is about \(60\). 2. The result is approximate because the monthly volume is approximate.

Answer

About \(60\) full truckloads equal the monthly amount.
Rate problems with constant rates 29
A large aquarium holds \(200\) gallons of water. Water costs \(\$5.00\) per \(1000\) gallons. a) What does it cost to fill the aquarium completely? b) Once a month, half of the water is replaced for cleaning. What does one water change cost?

Hints

- How many \(200\)-gallon amounts fit in \(1000\) gallons? - Use that relationship to find the cost of \(200\) gallons. - Half the water has half the cost.

Solution

1. Since \(1000 \div 200 = 5\), \(200\) gallons is one fifth of \(1000\) gallons. A full tank costs \(\$5.00 \div 5 = \$1.00\). 2. Half of \(200\) gallons is \(100\) gallons, so a half-tank water change costs \(\$1.00 \div 2 = \$0.50\).

Answer

a) A complete fill costs \(\$1.00\). b) One half-tank water change costs \(\$0.50\).
Rate problems with constant rates 30
A packing machine fills \(12\) bags of apples in \(3\) minutes. At the same constant rate, how many bags does it fill in \(10\) minutes?

Hints

- First find how many bags the machine fills in one minute. - Use the unit rate to find the amount for \(10\) minutes.

Solution

1. Find the unit rate: \(12 \div 3 = 4\) bags per minute. 2. Multiply by \(10\) minutes: \(4 \times 10 = 40\) bags.

Answer

The machine fills \(40\) bags in \(10\) minutes.
Rate problems with constant rates 31
A baker uses \(35\,\text{lb}\) of flour in \(5\) days. He bakes the same amount of bread each day. a) How many pounds of flour does he use each day? b) How many days will a \(56\,\text{lb}\) supply last if he continues using flour at the same daily rate?

Hints

- First find how much flour is used in one day. - Which operation divides an amount equally among several days? - Once you know the daily amount, determine how many times it fits into the new supply.

Solution

1. Find the daily flour use: \(35 \div 5 = 7\,\text{lb}\). 2. Find how many days the new supply will last: \(56 \div 7 = 8\) days.

Answer

a) He uses \(7\,\text{lb}\) of flour each day. b) The \(56\,\text{lb}\) supply will last \(8\) days.
Rate problems with constant rates 32
A bakery machine packs \(40\) bags of dinner rolls in \(5\) minutes. At the same constant rate, how many bags does it pack in \(15\) minutes?

Hints

- How many times does \(5\) minutes fit into \(15\) minutes? - At a constant rate, how does the output change when the time is multiplied by that factor?

Solution

1. Compare the time intervals: \(15 \div 5 = 3\), so \(15\) minutes is \(3\) times as long as \(5\) minutes. 2. Multiply the number of bags by the same factor: \(40 \times 3 = 120\) bags.

Answer

The machine packs \(120\) bags in \(15\) minutes.
Rate problems with constant rates 33
A warehouse divides \(72\,\text{L}\) of apple juice equally among \(6\) large containers. A delivery requires \(156\,\text{L}\) of juice. How many containers of the same size are needed?

Hints

- Find how many liters one container holds. - Divide the delivery volume by the volume in one container. - A ratio table can organize the two situations.

Solution

1. Find the volume in one container: \(72\,\text{L} \div 6 = 12\,\text{L}\). 2. Divide the delivery volume by the volume per container: \(156\,\text{L} \div 12\,\text{L} = 13\).

Answer

\(13\) containers are needed.
Rate problems with constant rates 34
A fruit salad recipe uses \(1.5\,\text{lb}\) of apples for \(4\) people. A class wants to make enough fruit salad for \(20\) people. How many pounds of apples are needed?

Hints

- Determine how many times as many people will be served. - Multiply the amount of apples by the same factor. - Check that your answer is larger than the original amount.

Solution

1. Find the scale factor: \(20 \div 4 = 5\). 2. Multiply the apple amount by the scale factor: \(1.5\,\text{lb} \times 5 = 7.5\,\text{lb}\).

Answer

The class needs \(7.5\,\text{lb}\) of apples.
Rate problems with constant rates 35
A truck delivers \(48\) tons of sand in \(6\) full loads. At the same rate, how many loads are needed to deliver \(72\) tons?

Hints

- First compare the total sand delivered with the number of full loads. - What number of loads would deliver the new amount at the same size per load?

Solution

1. The truck delivers \(48\div6=8\) tons per load. 2. The number of loads needed is \(72\div8=9\).

Answer

\(9\) loads
Rate problems with constant rates 36
Four apples have a total weight of \(1.25\,\text{lb}\). How many of these apples are in a bag weighing \(5\,\text{lb}\)?

Hints

- Determine how many times the smaller group's weight fits into the bag's weight. - Each equal-weight group contains \(4\) apples. - Multiply the number of groups by \(4\).

Solution

1. Find how many groups of \(1.25\,\text{lb}\) are in \(5\,\text{lb}\): \(5\,\text{lb} \div 1.25\,\text{lb} = 4\). 2. Each group contains \(4\) apples, so the bag contains \(4 \times 4 = 16\) apples.

Answer

The bag contains \(16\) apples.
Rate problems with constant rates 37
A hot-air balloon takes off when the ground temperature is \(7\,^{\circ}\text{C}\). In a simplified model, the temperature drops by \(1\,^{\circ}\text{C}\) for every \(100\,\text{m}\) the balloon rises. The balloon rises \(1800\,\text{m}\). What temperature does the model predict?

Hints

- Count the \(100\)-meter rises in the trip. - Find the total temperature drop before comparing it with the starting temperature. - If the temperature reaches zero, how much farther will it fall?

Solution

1. The balloon rises through \(1800\div100=18\) intervals of \(100\,\text{m}\). 2. The temperature drops by \(18\times1=18\,^{\circ}\text{C}\). 3. A drop of \(7\,^{\circ}\text{C}\) takes the temperature to zero. The remaining drop is \(18-7=11\,^{\circ}\text{C}\) below zero, giving \(-11\,^{\circ}\text{C}\).

Answer

\(-11\,^{\circ}\text{C}\)
Rate problems with constant rates 38
A package of \(5\) markers costs \(\$4.50\). A value pack of \(12\) markers costs \(\$9.60\). How many cents less does each marker cost in the value pack?

Hints

- Find the price of one marker in each package. - Divide each total price by its number of markers. - Subtract the two unit prices. - Express the final difference in cents.

Solution

1. Find the unit price in the small package: \(\$4.50 \div 5 = \$0.90\) per marker. 2. Find the unit price in the value pack: \(\$9.60 \div 12 = \$0.80\) per marker. 3. Find the difference: \(\$0.90 - \$0.80 = \$0.10\), which is \(10\) cents.

Answer

Each marker costs \(10\) cents less in the value pack.
Rate problems with constant rates 39
A recycling process turns \(12\,\text{kg}\) of used paper into \(9\,\text{kg}\) of new paper. A school collects \(480\,\text{kg}\) of used paper. How many kilograms of new paper can be produced at the same rate?

Hints

- Find how many groups of \(12\,\text{kg}\) are in \(480\,\text{kg}\). - Scale the amount of new paper by the same factor. - A ratio table can organize the quantities.

Solution

1. Find the scale factor: \(480\,\text{kg} \div 12\,\text{kg} = 40\). 2. Multiply the output by the same factor: \(40 \times 9\,\text{kg} = 360\,\text{kg}\).

Answer

The collected paper can produce \(360\,\text{kg}\) of new paper.
Rate problems with constant rates 40
A movie theater concession stand sells popcorn at a constant rate of \(5\) bags per minute. How many bags are sold in half an hour?

Hints

- How many minutes are in one hour? - How many minutes are in half an hour? - Multiply the number of bags sold per minute by the number of minutes.

Solution

1. Convert the time: half an hour is \(30\) minutes. 2. Multiply the rate by the time: \(5 \times 30 = 150\) bags.

Answer

The concession stand sells \(150\) bags of popcorn in half an hour.
Rate problems with constant rates 41
When a freezer is turned on, its internal temperature decreases from \(18\,^{\circ}\text{C}\) to \(-12\,^{\circ}\text{C}\). The temperature decreases at a constant rate of \(5\,^{\circ}\text{C}\) per hour. How many hours does the cooling process take?

Hints

- Find the total distance between the starting and ending temperatures. - Include the change on both sides of zero. - Divide the total temperature change by the hourly change.

Solution

1. Find the total temperature decrease. From \(18\) to \(0\) is \(18\,^{\circ}\text{C}\), and from \(0\) to \(-12\) is another \(12\,^{\circ}\text{C}\). The total decrease is \(18 + 12 = 30\,^{\circ}\text{C}\). 2. Divide the total decrease by the hourly rate: \(30 \div 5 = 6\).

Answer

The cooling process takes \(6\) hours.
Rate problems with constant rates 42
A sports shop sells tennis balls individually at the same price per ball. Four balls cost \(\$6.40\). a) How much do \(8\) balls cost? b) How much does \(1\) ball cost? c) How much do \(15\) balls cost?

Hints

- Find a simple multiple of the original group for part a). - What operation gives the price for one ball? - Use the one-ball price when the new number is not a multiple of four.

Solution

1. Eight balls cost twice as much as four: \(\$6.40\times2=\$12.80\). 2. One ball costs \(\$6.40\div4=\$1.60\). 3. Fifteen balls cost \(15\times\$1.60=\$24.00\).

Answer

a) \(\$12.80\) b) \(\$1.60\) c) \(\$24.00\)
Rate problems with constant rates 43
A recipe for \(4\) servings of rice pudding uses exactly \(1\,\text{qt}\) of milk. How much milk is needed for \(20\) servings?

Hints

- Determine how many times as many servings are needed. - More servings require proportionally more milk. - Multiply the original milk amount by the scale factor.

Solution

1. Find the scale factor: \(20 \div 4 = 5\). 2. Multiply the milk amount by the same factor: \(1\,\text{qt} \times 5 = 5\,\text{qt}\).

Answer

The recipe needs \(5\,\text{qt}\) of milk.
Rate problems with constant rates 44
A package of \(500\) sheets of copier paper weighs \(5\,\text{lb}\). How much do \(100\) sheets of the same paper weigh?

Hints

- Compare the new number of sheets with the original number. - Determine the factor that changes \(500\) to \(100\). - Apply the same operation to the weight.

Solution

1. One hundred sheets is one-fifth of \(500\) sheets because \(500 \div 5 = 100\). 2. Divide the weight by the same factor: \(5\,\text{lb} \div 5 = 1\,\text{lb}\).

Answer

\(100\) sheets weigh \(1\,\text{lb}\).
Rate problems with constant rates 45
A school copier prints \(30\) pages in \(2\) minutes at a constant rate. a) How many pages does it print in \(7\) minutes? b) How long does it take to print \(225\) pages?

Hints

- Find the number of pages printed in one minute. - A table of time and pages may help. - Use the unit rate to scale up or work backward.

Solution

1. Find the unit rate: \(30 \div 2 = 15\) pages per minute. 2. In \(7\) minutes, the copier prints \(7 \times 15 = 105\) pages. 3. To print \(225\) pages, it takes \(225 \div 15 = 15\) minutes.

Answer

a) \(105\) pages b) \(15\) minutes
Rate problems with constant rates 46
At a farmers market, \(4\,\text{lb}\) of apples cost \(\$6.00\). a) How much do \(7\,\text{lb}\) of apples cost? b) How many pounds of apples can a customer buy for \(\$12.00\)?

Hints

- Find the price of one pound first. - Use the unit price to calculate either a cost or a weight.

Solution

1. Find the unit price: \(\$6.00 \div 4 = \$1.50\) per pound. 2. Seven pounds cost \(7 \times \$1.50 = \$10.50\). 3. For \(\$12.00\), the customer can buy \(\$12.00 \div \$1.50 = 8\,\text{lb}\).

Answer

a) \(\$10.50\) b) \(8\,\text{lb}\)
Rate problems with constant rates 47
Eighteen students order pizza for a class party. Their order costs \(\$108\). Four more students join, and the class orders extra pizza at the same cost per student. How much does the extra pizza cost?

Hints

- What stays the same when more students join the order? - How could you use the original order to find a price for one student? - Check that your answer is only for the added order, not the total party.

Solution

1. The first order costs \(\$108\div18=\$6\) per student. 2. The extra pizza costs \(4\times\$6=\$24\).

Answer

\(\$24\)
Rate problems with constant rates 48
A recycling process uses \(20\,\text{kg}\) of used paper to make \(16\,\text{kg}\) of new paper. How many kilograms of used paper are needed to make \(80\,\text{kg}\) of new paper at the same rate?

Hints

- Find how many times \(16\,\text{kg}\) fits into \(80\,\text{kg}\). - Multiply both quantities in the ratio by the same factor. - A ratio table may help organize the amounts.

Solution

1. Find the scale factor from \(16\,\text{kg}\) to \(80\,\text{kg}\): \(80\,\text{kg} \div 16\,\text{kg} = 5\). 2. Multiply the used-paper amount by the same factor: \(20\,\text{kg} \times 5 = 100\,\text{kg}\).

Answer

\(100\,\text{kg}\) of used paper are needed.
Rate problems with constant rates 49
A cyclist travels \(9\,\text{km}\) in one-half hour. On average, how many meters does the cyclist travel per minute?

Hints

- Convert kilometers to meters. - Express one-half hour in minutes. - Divide the total distance by the total time.

Solution

1. Convert the distance: \(9\,\text{km} = 9000\,\text{m}\). 2. One-half hour is \(30\) minutes. 3. Find the unit rate: \(9000\,\text{m} \div 30\,\text{min} = 300\,\text{m/min}\).

Answer

The cyclist travels an average of \(300\,\text{m}\) per minute.
Rate problems with constant rates 50
Three apples cost \(90\) cents. How much do \(6\) apples cost?

Hints

- Compare \(6\) apples with \(3\) apples. - When the number of apples doubles, determine what happens to the cost. - You can also find the cost of one apple first. - A ratio table can help match numbers of apples with costs.

Solution

1. Six apples are twice as many as \(3\) apples because \(6 \div 3 = 2\). 2. Double the cost: \(2 \times 90 = 180\) cents. 3. Convert: \(180\) cents is \(\$1.80\).

Answer

Six apples cost \(\$1.80\).
Rate problems with constant rates 51
A cyclist travels \(s\) miles in \(t\) hours. a) Write an expression for the cyclist's average speed \(v\), in miles per hour. b) Find \(v\) for each set of values: 1) \(s=52.5\) miles and \(t=2.5\) hours 2) \(s=18.9\) miles and \(t=1.5\) hours

Hints

- Average speed compares distance with time. - Divide the distance by the number of hours. - Check that your answer has units of miles per hour.

Solution

1. Average speed is distance divided by time, so \(v=\frac{s}{t}\). 2. For the first trip, \(v=\frac{52.5}{2.5}=21\) miles per hour. 3. For the second trip, \(v=\frac{18.9}{1.5}=12.6\) miles per hour.

Answer

a) \(v=\frac{s}{t}\) b) 1) \(21\) miles per hour 2) \(12.6\) miles per hour
Rate problems with constant rates 52
A customer buys \(6\) apples for a total of \(\$4.50\). 1) Find the price of one apple. 2) Write a formula for the total cost \(G\) of \(n\) apples at \(p\) dollars per apple. 3) Use the formula to find each total cost: a) \(15\) apples at \(\$0.60\) each b) \(8\) apples at \(\$1.10\) each

Hints

- Divide the total cost by the number of apples to find the unit price. - Multiply the number of items by the price per item. - Substitute each pair of values into the formula.

Solution

1. The unit price is \(\$4.50\div 6=\$0.75\) per apple. 2. The total cost is \(G=np\). 3. For part a), \(G=15\times 0.60=9.00\), so the cost is \(\$9.00\). 4. For part b), \(G=8\times 1.10=8.80\), so the cost is \(\$8.80\).

Answer

1) \(\$0.75\) 2) \(G=np\) 3) a) \(\$9.00\) b) \(\$8.80\)
Rate problems with constant rates 53
A mobile phone plan charges \(\$0.05\) per megabyte \((\text{MB})\) of data used while traveling abroad. Find the cost for each amount of data: a) \(10\,\text{MB}\) b) \(100\,\text{MB}\) c) \(250\,\text{MB}\) d) \(1.5\,\text{GB}\), using \(1\,\text{GB}=1000\,\text{MB}\) e) Write an expression for the cost of using \(m\,\text{MB}\).

Hints

- Multiply the number of megabytes by the unit price. - Convert gigabytes to megabytes before calculating. - Use the same multiplication rule with the variable \(m\).

Solution

1. Multiply the data amount by \(\$0.05\) per megabyte. 2. For \(10\,\text{MB}\), the cost is \(10\times 0.05=0.50\), or \(\$0.50\). 3. For \(100\,\text{MB}\), the cost is \(100\times 0.05=5.00\), or \(\$5.00\). 4. For \(250\,\text{MB}\), the cost is \(250\times 0.05=12.50\), or \(\$12.50\). 5. Since \(1.5\,\text{GB}=1500\,\text{MB}\), the cost is \(1500\times 0.05=75.00\), or \(\$75.00\). 6. For \(m\,\text{MB}\), the cost is \(0.05m\) dollars.

Answer

a) \(\$0.50\) b) \(\$5.00\) c) \(\$12.50\) d) \(\$75.00\) e) \(0.05m\) dollars
Rate problems with constant rates 54
The two aligned number lines show a bus traveling at a constant rate. Panel a) shows time in minutes, and panel b) shows distance in miles. How far has the bus traveled after \(4\) minutes?
Figure for problem 554204

Hints

- Read the two number lines at matching horizontal positions. - Find the tick for \(4\) minutes on the time line. - Move to the corresponding position on the distance line and read its value.

Solution

1. The aligned lines pair \(2\) minutes with \(6\) miles, so each additional \(2\) minutes corresponds to another \(6\) miles. 2. The point aligned with \(4\) minutes on panel a) is \(12\) miles on panel b).

Answer

\(12\) miles
Rate problems with constant rates 55
The aligned number lines show a cyclist’s constant rate. Panel a) shows time in minutes, and panel b) shows distance in kilometers. The cyclist’s route is \(4500\,\text{m}\) long. Use the number-line relationship to determine how many minutes the route takes. Include the needed unit conversion.
Figure for problem 554208

Hints

- The route length and the distance number line must use the same unit before you compare them. - Locate the converted distance on panel b). - Read the aligned time from panel a), working backward from distance to time.

Solution

1. Convert the route length: \(4500\,\text{m}=4.5\,\text{km}\). 2. On panel b), \(4.5\) kilometers aligns with \(6\) minutes on panel a). 3. Therefore the cyclist takes \(6\) minutes to travel the route at the shown constant rate.

Answer

\(6\) minutes
Rate problems with constant rates 56
A bakery packs \(24\) muffins equally into \(4\) trays. Ari divides \(4\div24\) and calls the result muffins per tray. What rate does Ari's calculation actually find? Give the correct rate in muffins per tray.

Hints

- How do the units change when you reverse the order of a division? - Which quantity should a rate in muffins per tray count for each tray?

Solution

1. Ari divided trays by muffins: \(4\div24=\frac{1}{6}\) tray per muffin. 2. To find muffins per tray, divide \(24\) muffins by \(4\) trays: \(24\div4=6\) muffins per tray.

Answer

Ari found \(\frac{1}{6}\) tray per muffin. The requested rate is \(6\) muffins per tray.
Rate problems with constant rates 57
Three movie tickets cost \(\$21\). Tickets all have the same price. Complete the table. <table><tr><th>Tickets</th><td>\(3\)</td><td>\(1\)</td><td>\(5\)</td></tr><tr><th>Cost</th><td>\(\$21\)</td><td>\(\square\)</td><td>\(\square\)</td></tr></table>

Hints

- How could you use the known cost of three tickets to find other prices? - The cost and ticket count must change together at a fixed rate.

Solution

1. One ticket costs \(\$21\div3=\$7\). 2. Five tickets cost \(5\times\$7=\$35\).

Answer

One ticket: \(\$7\); five tickets: \(\$35\)
Rate problems with constant rates 58
The graph shows the distance a jogger travels at a steady pace. Use a plotted point after more than one hour to find the speed in miles per hour. Show your calculation.
Figure for problem 559302

Hints

- Read the labels on the axes first. - What do the coordinates of a plotted point say about the time and distance? - How do you find a one-hour rate from a trip lasting several hours?

Solution

1. The point \((2,8)\) means the jogger covers \(8\) miles in \(2\) hours. 2. The speed is \(8\div2=4\,\text{mi/h}\). The point \((4,16)\) agrees.

Answer

\(4\,\text{mi/h}\), because \(8\div2=4\)
Rate problems with constant rates 59
A hose fills \(20\) gallons in \(4\) minutes at a steady rate. Find both its rate in gallons per minute and the time needed for one gallon in minutes per gallon. Explain why the two numbers have different units.

Hints

- In each requested unit, what quantity must equal one? - What would you divide by what for the first rate, and how does the order change for the second?

Solution

1. The volume rate is \(20\div4=5\,\text{gal/min}\). 2. The time per gallon is \(4\div20=\frac{1}{5}\,\text{min/gal}\). 3. The numbers describe the same flow, but their quantities and units are reversed.

Answer

\(5\,\text{gal/min}\) and \(\frac{1}{5}\,\text{min/gal}\); the two rates exchange the amount and time being measured per one unit.
Rate problems with constant rates 60
A ferry travels at \(8\,\text{mi/h}\). How many hours does it take to travel \(32\) miles at that speed?

Hints

- The distance in one hour is known. - How many groups of eight miles make the whole trip?

Solution

1. Each hour covers \(8\) miles. 2. The time is \(32\div8=4\) hours.

Answer

\(4\) hours
Rate problems with constant rates 61
Panel a) measures hours, and panel b) measures miles on the same trip. Each colored band shows one total time or distance. What is the travel speed in miles per hour?
Figure for problem 559306

Hints

- Read the length of each colored interval using its own number-line scale. - One interval is a time and the other is a distance. - The rate compares distance with each single hour.

Solution

1. The first interval spans \(2\) hours. 2. The second interval spans \(12\) miles. 3. The speed is \(12\div2=6\,\text{mi/h}\).

Answer

\(6\,\text{mi/h}\)
Rate problems with constant rates 62
A bookstore sells \(4\) notebooks for \(\$10\). Another store sells \(6\) of the same notebooks for \(\$13.50\). Which store has the lower price per notebook?

Hints

- The pack with the lower total price does not automatically offer the better deal. - Compare the price for one notebook in each store.

Solution

1. The first store charges \(\$10\div4=\$2.50\) per notebook. 2. The second store charges \(\$13.50\div6=\$2.25\) per notebook. 3. The second store has the lower unit price.

Answer

The second store, at \(\$2.25\) per notebook
Rate problems with constant rates 63
Juice bottles are sold only in packs of \(6\). Each pack costs \(\$12\). A team needs at least \(9\) bottles. How many whole packs must it buy, and how much will they cost?

Hints

- You cannot purchase only part of a pack. - Find the smallest whole number of packs that gives at least nine bottles. - Use the per-pack price, not a per-bottle fraction of a pack.

Solution

1. One pack has too few bottles. 2. Two packs have \(2\times6=12\) bottles, enough for the team. 3. The two packs cost \(2\times\$12=\$24\).

Answer

\(2\) packs, costing \(\$24\)
Rate problems with constant rates 64
A bike shop charges \(\$3\) to start a rental and \(\$2\) for each full hour. a) What does a one-hour rental cost? What does a two-hour rental cost? b) Ken says two hours must cost twice as much as one hour. Explain whether that works here.

Hints

- Separate the price charged once from the price charged each hour. - What changes when the rental grows from one hour to two? - Is every part of the bill doubled?

Solution

1. One hour costs \(3+2=\$5\). 2. Two hours cost \(3+2\times2=\$7\). 3. Twice the one-hour total is \(\$10\), not \(\$7\), because the start fee is paid only once.

Answer

a) One hour: \(\$5\); two hours: \(\$7\) b) No. The \(\$3\) start fee is paid once, not once per hour.
Rate problems with constant rates 65
A team mixes apple juice and sparkling water in the same ratio in two containers. A pitcher holds \(1.5\,\text{L}\), and a dispenser holds \(6\,\text{L}\). The pitcher contains \(600\,\text{mL}\) of apple juice. How much apple juice is in the dispenser? Justify your answer using equivalent ratios.

Hints

- Express all volumes in the same unit. - Find the scale factor from the pitcher to the dispenser. - Determine the fraction of the pitcher that is apple juice. - Apply the same ratio to the larger container.

Solution

1. Convert the total volumes to milliliters: \(1.5\,\text{L}=1500\,\text{mL}\) and \(6\,\text{L}=6000\,\text{mL}\). 2. The apple-juice fraction in the pitcher is \(\frac{600}{1500}=\frac{2}{5}\). 3. The dispenser has the same ratio, so its apple-juice amount is \(6000\times\frac{2}{5}=2400\,\text{mL}\). Equivalently, the dispenser is \(4\) times as large, so \(600\times4=2400\).

Answer

The dispenser contains \(2400\,\text{mL}\), or \(2.4\,\text{L}\), of apple juice.
Rate problems with constant rates 66
A bag of coffee beans is worth \(\$144\). After \(3\,\text{lb}\) of beans are removed, the remaining beans are worth \(\$108\). Assume the price per pound is constant. a) What was the original weight of the full bag? b) What would the full bag cost if the price were reduced by \(\$2\) per pound?

Hints

- Find the value of the beans that were removed. - Use the removed value and weight to find the unit rate. - In b), the full bag's weight stays the same while the price per pound changes.

Solution

1. The removed beans were worth \(\$144-\$108=\$36\). 2. The unit price was \(\$36\div 3\,\text{lb}=\$12\) per pound. 3. The full bag weighed \(\$144\div(\$12\text{ per pound})=12\,\text{lb}\). 4. The reduced price is \(\$12-\$2=\$10\) per pound. 5. At the reduced price, the full bag would cost \(12\,\text{lb}\times\$10\text{ per pound}=\$120\).

Answer

a) \(12\,\text{lb}\) b) \(\$120\)
Rate problems with constant rates 67
A car averages \(25\) miles per gallon. a) How many gallons of gas does it use to travel \(15\) miles? b) How far can it travel with \(21.25\) gallons of gas, assuming the same average fuel economy?

Hints

- Use the meaning of miles per gallon to connect distance and fuel. - For part a), ask how many groups of \(25\) miles fit into \(15\) miles. - For part b), use the distance traveled for each gallon.

Solution

1. For a), divide distance by miles per gallon: \(15\div25=0.6\) gallon. 2. For b), multiply the amount of gas by the distance per gallon: \(21.25\times25=531.25\) miles.

Answer

a) \(0.6\) gallon b) \(531.25\) miles
Rate problems with constant rates 68
Luke rides his bike \(5.4\) miles in \(15\) minutes. His friend Sarah rides \(4.5\) miles in \(12\) minutes. a) Who has the greater average speed? Compare their speeds in miles per hour. b) At Sarah's speed, how many feet does she travel in one second? Use \(1\,\text{mile}=5280\,\text{ft}\) and \(1\,\text{hour}=3600\,\text{s}\).

Hints

- First compare the distance covered in one minute by each cyclist. - How many minutes are in an hour? - Use the two given unit conversions to express Sarah's speed in feet per second.

Solution

1. Luke's speed is \(5.4\div15=0.36\) mile per minute, or \(0.36\times60=21.6\) miles per hour. 2. Sarah's speed is \(4.5\div12=0.375\) mile per minute, or \(0.375\times60=22.5\) miles per hour. 3. Sarah is faster since \(22.5>21.6\). 4. Her distance in one hour is \(22.5\times5280=118{,}800\,\text{ft}\). Per second, that is \(118{,}800\div3600=33\,\text{ft/s}\).

Answer

a) Sarah, at \(22.5\) miles per hour compared with Luke's \(21.6\) miles per hour. b) \(33\,\text{ft/s}\)
Rate problems with constant rates 69
In a drone race, Drone Alpha travels \(45.5\,\text{m}\) in \(3.5\,\text{s}\). Drone Beta travels \(61.2\,\text{m}\) in \(4.8\,\text{s}\). a) Find each drone's speed in meters per second. Which drone is faster? b) How many centimeters does the faster drone travel in \(0.01\) second?

Hints

- Divide distance by time to find each speed. - Use the meaning of meters per second to find a distance over a shorter time. - Convert meters to centimeters at the end.

Solution

1. Drone Alpha: \(45.5\div3.5=13\,\text{m/s}\). 2. Drone Beta: \(61.2\div4.8=12.75\,\text{m/s}\). 3. Since \(13>12.75\), Drone Alpha is faster. 4. In \(0.01\) second, Alpha travels \(13\times0.01=0.13\,\text{m}=13\,\text{cm}\).

Answer

a) Drone Alpha is faster at \(13\,\text{m/s}\); Drone Beta travels \(12.75\,\text{m/s}\). b) \(13\,\text{cm}\)
Rate problems with constant rates 70
Two garden hoses are tested while filling a small pool. Hose A delivers \(25.5\) gallons in \(1.5\) minutes. Hose B delivers \(32.4\) gallons in \(1.8\) minutes. a) Which hose has the greater flow rate in gallons per minute? b) How many gallons does the faster-flowing hose deliver in one second?

Hints

- Find how much water each hose delivers in one minute. - Compare the two unit rates. - Divide the faster rate by the number of seconds in one minute for part b).

Solution

1. Hose A: \(25.5\div1.5=17\) gal/min. 2. Hose B: \(32.4\div1.8=18\) gal/min. 3. Hose B has the greater flow rate. 4. Since one minute has \(60\) seconds, \(18\div60=0.3\) gallon per second.

Answer

a) Hose B, at \(18\) gal/min; Hose A has \(17\) gal/min. b) \(0.3\) gallon per second
Rate problems with constant rates 71
A large water tank can hold \(450\,\text{m}^3\) of water. Use \(1\,\text{m}^3=1000\,\text{L}\). a) Convert this volume to liters. b) A pump removes water at a rate of \(1500\,\text{L}\) per minute. How many hours will it take to empty the full tank?

Hints

- Express the tank's volume in the unit used for the pump's rate. - What does the rate tell you about the amount removed in each minute? - Convert the resulting number of minutes to hours.

Solution

1. The tank holds \(450\times1000=450{,}000\,\text{L}\). 2. The pumping time is \(450{,}000\div1500=300\,\text{min}\). 3. Convert to hours: \(300\div60=5\,\text{h}\).

Answer

a) \(450{,}000\,\text{L}\) b) \(5\,\text{h}\)
Rate problems with constant rates 72
An empty swimming pool is \(10\,\text{m}\) long and \(5\,\text{m}\) wide. A hose adds water at a constant rate of \(40\,\text{L}\) per minute. Use \(1\,\text{m}^3=1000\,\text{L}\). a) When the water reaches a depth of \(1.6\,\text{m}\), how many liters will the pool contain? b) How many hours and minutes will it take to reach that depth?

Hints

- What solid shape models the water in the pool? - How can you find the water volume from its dimensions? - Make sure the water volume and the hose rate use compatible units.

Solution

1. The water volume is \(10\times5\times1.6=80\,\text{m}^3\). 2. Convert to liters: \(80\times1000=80{,}000\,\text{L}\). 3. At \(40\,\text{L}\) per minute, the time is \(80{,}000\div40=2000\,\text{min}\). 4. Since \(2000=33\times60+20\), the time is \(33\) hours \(20\) minutes.

Answer

a) \(80{,}000\,\text{L}\) b) \(33\) hours \(20\) minutes
Rate problems with constant rates 73
Two weather models predict how air temperature changes with altitude. Model A predicts a decrease of \(3.5^\circ\text{F}\) for every \(1000\,\text{ft}\) of altitude. Model B predicts a decrease of \(5.5^\circ\text{F}\) for every \(1000\,\text{ft}\). At sea level, the temperature is \(59^\circ\text{F}\). At an altitude of \(10{,}000\,\text{ft}\), what is the difference between the temperatures predicted by the two models? Before calculating, explain which model predicts the lower temperature.

Hints

- A larger decrease per \(1000\,\text{ft}\) leads to a lower final temperature. - How many \(1000\,\text{ft}\) intervals are in \(10{,}000\,\text{ft}\)? - Find each predicted temperature separately, then compare them.

Solution

1. Model B predicts the lower temperature because it has the greater temperature decrease for each \(1000\,\text{ft}\). 2. An altitude of \(10{,}000\,\text{ft}\) contains \(10\) intervals of \(1000\,\text{ft}\). Model A predicts \(59-10\times3.5=59-35=24^\circ\text{F}\). 3. Model B predicts \(59-10\times5.5=59-55=4^\circ\text{F}\). 4. The difference is \(24-4=20^\circ\text{F}\).

Answer

Model B predicts the lower temperature. The predictions differ by \(20^\circ\text{F}\): Model A gives \(24^\circ\text{F}\) and Model B gives \(4^\circ\text{F}\).
Rate problems with constant rates 74
Aoife wrote the following expression to model a temperature change during a mountain hike: \(50-\frac{9000-3000}{1000}\times3.5\) a) Evaluate the expression step by step. b) Describe a situation that this expression could represent. Explain what \(50\), \(3000\), \(9000\), and \(3.5\) mean in your situation.

Hints

- Follow the order of operations. - Which number could represent the starting temperature? - Which numbers could represent elevations? - Since the hiker climbs to a higher elevation, should the temperature change be added to or subtracted from the starting temperature?

Solution

1. Find the change in altitude: \(9000-3000=6000\,\text{ft}\). 2. Find the number of \(1000\,\text{ft}\) intervals: \(6000\div1000=6\). 3. Find the total temperature decrease: \(6\times3.5=21^\circ\text{F}\). 4. Subtract from the starting temperature: \(50-21=29\). The expression equals \(29\). 5. One possible situation is a hike that starts at \(3000\,\text{ft}\) with a temperature of \(50^\circ\text{F}\) and ends at \(9000\,\text{ft}\), with temperature decreasing \(3.5^\circ\text{F}\) for every \(1000\,\text{ft}\) gained.

Answer

a) \(29\) b) Example: A hike starts at \(3000\,\text{ft}\) when the temperature is \(50^\circ\text{F}\) and ends at \(9000\,\text{ft}\). The temperature decreases \(3.5^\circ\text{F}\) for every \(1000\,\text{ft}\) of elevation gain.
Rate problems with constant rates 75
A hiking trail is \(2.5\) miles long. Use \(1\,\text{mile}=5280\,\text{ft}\). a) Leon's average step length is \(2.2\,\text{ft}\). About how many steps does he take to cover the whole trail? b) His younger sister takes \(2000\) more steps on the same trail. What is her average step length in feet?

Hints

- Work with the trail distance in the same unit used for step length. - A distance per step can tell you how many steps fit into a total distance. - For part b), the distance stays fixed while the number of steps changes.

Solution

1. The trail is \(2.5\times5280=13{,}200\,\text{ft}\) long. 2. Leon takes approximately \(13{,}200\div2.2=6000\) steps. 3. His sister takes about \(6000+2000=8000\) steps. 4. Her average step length is approximately \(13{,}200\div8000=1.65\,\text{ft}\).

Answer

a) About \(6000\) steps b) About \(1.65\,\text{ft}\)
Rate problems with constant rates 76
A town of \(80{,}000\) people collects \(2400\) metric tons of food scraps for compost each year. Use \(1\) metric ton \(=1000\,\text{kg}\). a) How many kilograms of food scraps does the town collect per person each year? b) Another town has \(120{,}000\) people. How many metric tons would it collect each year at the same per-person rate?

Hints

- Which two quantities should you compare to find kilograms for one resident? - Keep the weight unit the same when working out the rate. - Use the rate from the first town for the second town, then convert back to metric tons.

Solution

1. Convert the yearly amount: \(2400\times1000=2{,}400{,}000\,\text{kg}\). 2. Divide by residents: \(2{,}400{,}000\div80{,}000=30\,\text{kg}\) per resident per year. 3. Scale the same rate to the second town: \(120{,}000\times30=3{,}600{,}000\,\text{kg}\) each year. 4. Convert back: \(3{,}600{,}000\div1000=3600\) metric tons.

Answer

a) \(30\,\text{kg}\) per resident per year b) \(3600\) metric tons per year
Rate problems with constant rates 77
A snail travels an average of \(0.14\,\text{m}\) each minute. a) How far does it travel in \(7.5\) minutes? b) A second snail moves \(1.3\) times as fast as the first. How far does the second snail travel in \(7.5\) minutes?

Hints

- Use the rate and time to find distance. - For part b), you can scale the first distance by the speed factor. - Keep track of decimal place value in each product.

Solution

1. For a), use distance equals rate times time: \(0.14\,\text{m/min}\times7.5\,\text{min}=1.05\,\text{m}\). 2. The second snail travels \(1.3\) times as far in the same time, so \(1.05\times1.3=1.365\,\text{m}\).

Answer

a) \(1.05\,\text{m}\) b) \(1.365\,\text{m}\)
Rate problems with constant rates 78
A leaky faucet wastes \(0.045\) gallons of water per minute. How many gallons are wasted if the faucet leaks continuously for \(4.5\) hours?

Hints

- Make sure the time and rate use compatible units. - Recall how many minutes are in one hour. - Multiply the rate per minute by the total number of minutes.

Solution

1. Convert the time to minutes: \(4.5\times60=270\) minutes. 2. Multiply the time by the leak rate: \(270\,\text{min}\times0.045\,\text{gal/min}=12.15\,\text{gal}\). 3. The faucet wastes \(12.15\) gallons.

Answer

\(12.15\) gallons of water are wasted.
Rate problems with constant rates 79
A swimmer completes \(400\,\text{m}\) in \(4\) minutes \(24.8\) seconds. The pool is \(50\,\text{m}\) long. What was the swimmer's average time, in seconds, for one length of the pool?

Hints

- Convert the entire time to seconds first. - Determine how many pool lengths make up \(400\,\text{m}\). - Divide the total time by the number of equal lengths.

Solution

1. Convert the total time to seconds: \(4\times60+24.8=264.8\) seconds. 2. The swimmer completes \(400\div50=8\) pool lengths. 3. Divide the total time by the number of lengths: \(264.8\div8=33.1\) seconds per length.

Answer

\(33.1\) seconds per pool length
Rate problems with constant rates 80
A bottling machine fills \(112.5\) gallons of juice in \(15\) minutes. a) How many gallons does it fill per minute on average? b) How many seconds does it take, on average, to fill one gallon?

Hints

- For part a), find the amount filled in one minute. - For part b), convert the total time to seconds first. - Divide total time by total gallons to find time per gallon.

Solution

1. For a), \(112.5\div15=7.5\) gal/min. 2. Convert \(15\) minutes to \(900\) seconds. 3. For b), \(900\div112.5=8\) seconds per gallon.

Answer

a) \(7.5\) gal/min b) \(8\) seconds per gallon
Rate problems with constant rates 81
A peregrine falcon dives at \(324\,\text{km/h}\). A swift travels \(336\,\text{m}\) in \(12\) seconds. Use \(1\,\text{m/s}=3.6\,\text{km/h}\). a) Find the swift's speed in meters per second. b) Which bird has the greater speed? Convert the swift's speed to kilometers per hour before comparing.

Hints

- Compare the measurements only after they use the same units. - First find the swift's distance per second. - Which of the given conversions changes meters per second to kilometers per hour?

Solution

1. The swift's speed is \(336\div12=28\,\text{m/s}\). 2. Convert: \(28\times3.6=100.8\,\text{km/h}\). 3. Since \(324>100.8\), the peregrine falcon's stated speed is faster.

Answer

a) \(28\,\text{m/s}\) b) The peregrine falcon; the swift travels at \(100.8\,\text{km/h}\).
Rate problems with constant rates 82
A remote-controlled car travels \(188\,\text{m}\) in \(40\) seconds. A second model has a constant speed of \(16.2\,\text{km/h}\). Use \(1\,\text{m/s}=3.6\,\text{km/h}\). a) Find the first car's speed in meters per second. b) Which car is faster? Convert the first car's speed to kilometers per hour before comparing.

Hints

- How far does the first model travel in one second? - Why do the speed units need to match before comparing? - Use the given conversion and compare the two decimals carefully.

Solution

1. The first car's speed is \(188\div40=4.7\,\text{m/s}\). 2. Convert: \(4.7\times3.6=16.92\,\text{km/h}\). 3. Since \(16.92>16.2\), the first car is faster.

Answer

a) \(4.7\,\text{m/s}\) b) The first car, at \(16.92\,\text{km/h}\) versus \(16.2\,\text{km/h}\).
Rate problems with constant rates 83
A car uses \(12\) gallons of gas to travel \(300\) miles. a) Find the average fuel used per mile, in gallons per mile. b) How many miles can the car travel on \(6\) gallons if the rate stays the same?

Hints

- Divide total fuel by total distance to find fuel used per mile. - For part b), determine how many \(0.04\)-gallon portions fit into \(6\) gallons. - A rate table can help keep the quantities organized.

Solution

1. For a), \(12\div300=0.04\) gallon per mile. 2. For b), \(6\div0.04=150\) miles.

Answer

a) \(0.04\) gal/mi b) \(150\) miles
Rate problems with constant rates 84
A school-supply store sells notebooks in two package sizes: Package A: \(8\) notebooks for \(\$6.40\). Package B: \(12\) notebooks for \(\$9.00\). a) Which package has the lower price per notebook? b) Suppose the store also sells notebooks individually at that lower unit price. How much would \(20\) notebooks cost?

Hints

- Which calculation gives the price of one notebook in each package? - You may compare unit prices without purchasing either package. - In part b), use the individual-sale price stated in the question.

Solution

1. Package A costs \(6.40\div8=\$0.80\) per notebook. 2. Package B costs \(9.00\div12=\$0.75\) per notebook. 3. Package B has the lower unit price. 4. At \(\$0.75\) per notebook, \(20\) notebooks cost \(20\times0.75=\$15.00\).

Answer

a) Package B, at \(\$0.75\) per notebook instead of \(\$0.80\). b) \(\$15.00\)
Rate problems with constant rates 85
A car uses \(6\) gallons of gasoline to travel \(150\) miles. Its fuel tank holds \(15\) gallons and is currently \(40\%\) full. Is there enough gasoline for a planned \(140\)-mile trip? Justify your answer with calculations.

Hints

- Find the amount of gasoline used per mile. - Find the current amount of gasoline as a percent of tank capacity. - Find the gasoline needed for the planned distance. - Compare the amount available with the amount needed.

Solution

1. The car uses \(6 \div 150 = 0.04\) gallon per mile. 2. The tank currently contains \(15 \times 0.40 = 6\) gallons. 3. The planned trip requires \(140 \times 0.04 = 5.6\) gallons. 4. Since \(5.6 \le 6\), there is enough gasoline.

Answer

Yes. The tank contains \(6\) gallons, and the trip requires \(5.6\) gallons.
Rate problems with constant rates 86
In a school cafeteria, every serving of pasta costs the same amount. Three servings cost \(\$13.50\). a) Find the cost of \(5\) servings. b) How many servings can be purchased for \(\$31.50\)? c) Is the relationship between number of servings and total price proportional or not proportional? Justify your answer.

Hints

- First find the price of one serving. - Does buying more servings make the total price increase or decrease? - What happens to the price when the number of servings doubles?

Solution

1. One serving costs \(13.50\div 3=\$4.50\). 2. Five servings cost \(5\times 4.50=\$22.50\). 3. The number of servings for \(\$31.50\) is \(31.50\div 4.50=7\). 4. The relationship is proportional because the unit price is constant and doubling the number of servings doubles the total price.

Answer

a) \(\$22.50\). b) \(7\) servings. c) Proportional, because the price per serving is constant.
Rate problems with constant rates 87
Consider the hour hand of an analog clock. a) What angle does the hour hand sweep through in exactly one hour? b) What angle does it sweep through in \(15\) minutes? c) Through how many degrees does the hour hand turn from \(8{:}00\) a.m. to \(11{:}30\) a.m.?

Hints

- How many hours does the hour hand take to complete one revolution? - Find its angle per hour. - Convert \(15\) minutes to a fraction of an hour. - Find the total elapsed time in part c.

Solution

1. The hour hand completes a \(360^\circ\) revolution in \(12\) hours, so it moves \(360^\circ \div 12 = 30^\circ\) per hour. 2. Fifteen minutes is one-fourth of an hour, so the hand moves \(30^\circ \div 4 = 7.5^\circ\). 3. The elapsed time from \(8{:}00\) a.m. to \(11{:}30\) a.m. is \(3.5\) hours. 4. The hour hand turns \(3.5 \times 30^\circ = 105^\circ\).

Answer

a) \(30^\circ\) b) \(7.5^\circ\) c) \(105^\circ\)
Rate problems with constant rates 88
A scooter uses \(1.5\) gallons of gasoline for every \(60\) miles traveled. Gas costs \(\$3.60\) per gallon. Tim plans to ride to a concert \(27\) miles away and then ride home. a) How many gallons of gasoline will the round trip use? b) What will the gasoline cost for the trip?

Hints

- Find the total round-trip distance. - Find the gasoline used per mile. - Multiply the number of gallons by the price per gallon.

Solution

1. The round-trip distance is \(2\times 27=54\) miles. 2. The gasoline rate is \(1.5\div 60=0.025\) gallon per mile. 3. The trip uses \(54\times 0.025=1.35\) gallons. 4. The cost is \(1.35\times 3.60=\$4.86\).

Answer

a) \(1.35\) gallons. b) \(\$4.86\).
Rate problems with constant rates 89
A scooter has a \(2\)-gallon fuel tank. At current prices, a full tank costs \(\$7.20\). The scooter uses an average of \(1.2\) gallons for every \(60\) miles. a) Find the price of one gallon of gasoline. b) How far can the scooter travel on a full tank, in theory? c) A day trip is \(90\) miles long. Is \(\$6.50\) enough to pay for the gasoline needed? Justify your answer with calculations.

Hints

- Divide the full-tank price by the tank capacity. - Find how many miles the scooter travels per gallon. - First find the gasoline needed for \(90\) miles, then its cost. - Compare the calculated cost with the budget.

Solution

1. The price per gallon is \(7.20\div 2=\$3.60\). 2. The scooter travels \(60\div 1.2=50\) miles per gallon. With \(2\) gallons, it can travel \(2\times 50=100\) miles. 3. The trip uses \(90\times \frac{1.2}{60}=1.8\) gallons. 4. The cost is \(1.8\times 3.60=\$6.48\). Since \(6.48<6.50\), the money is enough.

Answer

a) \(\$3.60\) per gallon. b) \(100\) miles. c) Yes. The trip costs \(\$6.48\).
Rate problems with constant rates 90
A family fills a small backyard pool with a garden hose. The hose delivers \(0.75\) gallon in \(15\) seconds. Water costs \(\$6.00\) per \(1000\) gallons. a) How many gallons per minute flow from the hose? b) The pool holds \(180\) gallons. How long does it take to fill? c) What does one complete filling cost? d) During a backyard party, the same hose supplies all \(\$2.70\) worth of water used, at its usual flow rate. How many hours did the hose run in total?

Hints

- How many short intervals make up one minute? - Use the same water-flow rate in every part. - The billing rate describes a cost for each group of \(1000\) gallons. - How are the party's water cost, the amount used, and the time connected?

Solution

1. There are four \(15\)-second intervals in one minute, so the rate is \(4\times0.75=3\) gallons per minute. 2. Filling the pool takes \(180\div3=60\) minutes, or one hour. 3. The filling costs \(\frac{180}{1000}\times6.00=\$1.08\). 4. The party uses \(\frac{2.70}{6.00}\times1000=450\) gallons. At three gallons per minute, the time is \(450\div3=150\) minutes, or \(2.5\) hours.

Answer

a) \(3\) gallons per minute b) \(60\) minutes, or \(1\) hour c) \(\$1.08\) d) \(2.5\) hours, or \(2\) hours \(30\) minutes
Rate problems with constant rates 91
For this water-use model, suppose one showerhead uses \(3\) gallons of water per minute and a water-saving showerhead uses \(1.75\) gallons per minute. Water costs \(\$6.00\) per \(1000\) gallons. a) How much water does the water-saving showerhead save during an \(8\)-minute shower? b) Find the cost savings for the \(8\)-minute shower. c) Jan uses the same amount of water with the water-saving showerhead as a \(7\)-minute shower with the other showerhead. How long does Jan shower?

Hints

- First find the difference in gallons used per minute. - Use the cost per \(1000\) gallons to find the cost of the saved water. - For part c, find the total water used during the \(7\)-minute shower first.

Solution

1. The savings per minute is \(3 - 1.75 = 1.25\) gallons. 2. In \(8\) minutes, the water saved is \(1.25 \times 8 = 10\) gallons. 3. The cost savings is \(\frac{10}{1000} \times 6.00 = \$0.06\). 4. A \(7\)-minute shower at \(3\) gallons per minute uses \(7 \times 3 = 21\) gallons. With the water-saving showerhead, the time is \(21 \div 1.75 = 12\) minutes.

Answer

a) \(10\) gallons b) \(\$0.06\) c) \(12\) minutes
Rate problems with constant rates 92
A large water tank is filled at a constant rate. It initially contains \(10\) gallons. After \(10\) minutes, it contains \(35\) gallons. The owner claims, “The \(140\)-gallon tank will be full in another half hour.” Check the claim. Is the stated filling time realistic?

Hints

- Find the volume added during the first \(10\) minutes. - Find how much space remains in the tank. - Divide the remaining volume by the filling rate and compare with \(30\) minutes.

Solution

1. In \(10\) minutes, the volume increases by \(35-10=25\) gallons, so the filling rate is \(25\div10=2.5\) gallons per minute. 2. The tank still needs \(140-35=105\) gallons. 3. At \(2.5\) gallons per minute, the remaining time is \(105\div2.5=42\) minutes. 4. The owner's estimate of \(30\) more minutes is too short. The tank needs \(42\) more minutes.

Answer

The claim is not realistic. The filling rate is \(2.5\) gallons per minute, and the tank needs another \(42\) minutes to fill.
Rate problems with constant rates 93
Lucas is charging a tablet. At \(2{:}00\) p.m., the battery level is \(15\%\). At \(2{:}30\) p.m., it is \(75\%\). Assume the tablet charges at a constant rate from \(0\%\) to \(100\%\). a) How many percentage points does the battery level increase per minute? b) How many minutes does a complete charge from \(0\%\) to \(100\%\) take?

Hints

- Find the total percentage-point increase during the \(30\)-minute interval. - Divide the increase by the number of minutes to find the unit rate. - Use the constant rate to find the time needed for \(100\) percentage points.

Solution

1. The battery level increases by \(75\% - 15\% = 60\) percentage points. 2. The elapsed time is \(30\) minutes. 3. The charging rate is \(60 \div 30 = 2\) percentage points per minute. 4. A full charge takes \(100 \div 2 = 50\) minutes.

Answer

a) \(2\) percentage points per minute b) \(50\) minutes
Rate problems with constant rates 94
Two copy shops charge the following prices: Shop A: \(80\) copies for \(\$6.40\). Shop B: \(150\) copies for \(\$9.00\). a) Find the price per copy at each shop. b) Find the cost of \(400\) copies at each shop. c) How many copies can be purchased for \(\$12.00\) at each shop?

Hints

- Find the cost of one copy at each shop. - Use the unit price for a larger number of copies. - When money is given, divide by the price per copy.

Solution

1. Shop A costs \(6.40\div 80=\$0.08\) per copy. Shop B costs \(9.00\div 150=\$0.06\) per copy. 2. For \(400\) copies, Shop A costs \(400\times 0.08=\$32.00\), and Shop B costs \(400\times 0.06=\$24.00\). 3. For \(\$12.00\), Shop A provides \(12.00\div 0.08=150\) copies, and Shop B provides \(12.00\div 0.06=200\) copies.

Answer

a) Shop A: \(\$0.08\) per copy. Shop B: \(\$0.06\) per copy. b) Shop A: \(\$32.00\). Shop B: \(\$24.00\). c) Shop A: \(150\) copies. Shop B: \(200\) copies.
Rate problems with constant rates 95
Two cars are tested for gasoline use. Car A uses \(6\) gallons over \(120\) miles. Car B uses \(8\) gallons over \(128\) miles. a) Find each car's gasoline use per \(100\) miles. Which car uses less gasoline? b) How many gallons would each car use on a \(1400\)-mile trip?

Hints

- Convert each rate to gasoline used for the same distance. - Use a per-\(100\)-mile rate for comparison. - Determine how many groups of \(100\) miles are in the trip.

Solution

1. Car A uses \((6\div 120)\times 100=5\) gallons per \(100\) miles. 2. Car B uses \((8\div 128)\times 100=6.25\) gallons per \(100\) miles. 3. Car A uses less gasoline. 4. For \(1400\) miles, Car A uses \(14\times 5=70\) gallons, and Car B uses \(14\times 6.25=87.5\) gallons.

Answer

a) Car A uses \(5\) gallons per \(100\) miles; Car B uses \(6.25\) gallons per \(100\) miles. Car A uses less. b) Car A uses \(70\) gallons; Car B uses \(87.5\) gallons.
Rate problems with constant rates 96
Two factory machines produce parts. Machine A makes \(480\) parts in \(4\) hours. Machine B makes \(525\) parts in \(5\) hours. a) Which machine works faster? Justify your answer by finding parts per hour. b) How long does each machine take to complete an order of \(1050\) parts? c) An urgent order must be completed within \(9\) hours. Which machine can complete the order by itself?

Hints

- Interpret “faster” as more parts produced per hour. - Divide the target number of parts by each hourly rate. - Compare each time with the deadline.

Solution

1. Machine A makes \(480\div 4=120\) parts per hour. Machine B makes \(525\div 5=105\) parts per hour. Machine A is faster. 2. Machine A takes \(1050\div 120=8.75\) hours, or \(8\) hours \(45\) minutes. 3. Machine B takes \(1050\div 105=10\) hours. 4. Only Machine A meets the \(9\)-hour deadline.

Answer

a) Machine A, at \(120\) parts per hour; Machine B makes \(105\) parts per hour. b) Machine A: \(8.75\) hours. Machine B: \(10\) hours. c) Only Machine A.
Rate problems with constant rates 97
A copier in a school library prints \(150\) pages in \(4\) minutes. Assume printing time and page count are proportional. a) How many pages can the copier print in \(10\) minutes? b) How long does it take to print a \(525\)-page document? c) Give one reason why the proportional model might not remain exact for a very large job, such as \(50{,}000\) pages.

Hints

- First find the number of pages printed in one minute. - Use the unit rate to scale up or work backward. - Think about technical or human interruptions during a very large job.

Solution

1. The copier's rate is \(150\div 4=37.5\) pages per minute. 2. In \(10\) minutes, it prints \(37.5\times 10=375\) pages. 3. Printing \(525\) pages takes \(525\div 37.5=14\) minutes. 4. A very large job may require paper refills, toner replacement, clearing jams, or cooling pauses, so the rate may not remain constant.

Answer

a) \(375\) pages. b) \(14\) minutes. c) Possible reasons include paper refills, toner replacement, paper jams, or overheating.
Rate problems with constant rates 98
During steady rain, a family observes a rain barrel. At \(2{:}00\) p.m., it contains \(12\) gallons. At \(2{:}12\) p.m., it contains \(27\) gallons. Find the average inflow rate during this period in gallons per hour.

Hints

- Find the change in volume. - Divide by the elapsed number of minutes. - Convert a per-minute rate to a per-hour rate.

Solution

1. The volume increases by \(27-12=15\) gallons. 2. The elapsed time is \(12\) minutes, so the rate is \(15\div12=1.25\) gallons per minute. 3. Multiply by \(60\) minutes per hour: \(1.25\times60=75\) gallons per hour.

Answer

\(75\) gallons per hour
Rate problems with constant rates 99
A high-capacity pump fills a swimming pool. The pump delivers \(125\,\text{gal}\) every \(30\) seconds. a) Find the pump's rate in gallons per hour. b) The pool holds \(60{,}000\,\text{gal}\). How many hours will it take to fill the pool? c) Filling begins at \(7{:}45\) a.m. At what time will the pool be full?

Hints

- First determine how much water the pump delivers in one minute. - How many minutes are in one hour? - Divide the pool capacity by the hourly rate.

Solution

1. In one minute, the pump delivers \(125 \times 2=250\,\text{gal}\). 2. In one hour, it delivers \(250 \times 60=15{,}000\,\text{gal}\). 3. The filling time is \(60{,}000 \div 15{,}000=4\) hours. 4. Four hours after \(7{:}45\) a.m. is \(11{:}45\) a.m.

Answer

a) \(15{,}000\,\text{gal/h}\) b) \(4\) hours c) \(11{:}45\) a.m.
Rate problems with constant rates 100
Three trail signs show the remaining distance to Pine Ridge. <table> <tr><th>Sign</th><th>Distance to Pine Ridge</th></tr> <tr><td>1</td><td>\(21\,\text{mi}\)</td></tr> <tr><td>2</td><td>\(17\,\text{mi}\)</td></tr> <tr><td>3</td><td>\(12\,\text{mi}\)</td></tr> </table> The hiking group takes \(1\) hour to travel from Sign 1 to Sign 2 and \(1.25\) hours to travel from Sign 2 to Sign 3. a) Find the group's average speed on each interval. b) Did the group have the same average speed on both intervals?

Hints

- A decrease in the distance remaining tells how far the group traveled between signs. - Compare each interval's traveled distance with its own elapsed time. - Put both interval rates in miles per hour before comparing them.

Solution

1. From Sign 1 to Sign 2, the remaining distance decreases by \(21 - 17 = 4\) miles, so the group travels \(4\) miles in \(1\) hour. The average speed is \(4\,\text{mi/h}\). 2. From Sign 2 to Sign 3, the remaining distance decreases by \(17 - 12 = 5\) miles. The average speed is \(5 \div 1.25 = 4\,\text{mi/h}\). 3. The two average speeds are equal.

Answer

a) Sign 1 to Sign 2: \(4\,\text{mi/h}\); Sign 2 to Sign 3: \(4\,\text{mi/h}\). b) Yes. The average speed is the same on both intervals.
Rate problems with constant rates 101
A damaged pipe leaks about \(15\,\text{L}\) of water each minute. How many liters leak from the pipe in one full \(24\)-hour day?

Hints

- How many minutes are in one hour? - How many hours are in one day? - First find the hourly amount, then the daily amount.

Solution

1. Find the approximate amount leaked in one hour: \(15\,\text{L} \times 60 \approx 900\,\text{L}\). 2. Find the approximate amount leaked in \(24\) hours: \(900\,\text{L} \times 24 \approx 21{,}600\,\text{L}\).

Answer

The pipe leaks about \(21{,}600\,\text{L}\) in one day.
Rate problems with constant rates 102
A freight elevator can carry up to \(1200\,\text{kg}\) in total and at most \(15\) people. a) If the weight limit is shared equally among \(15\) riders, what is the average share per rider? This is not an individual weight limit. b) A group of \(12\) workers has a combined mass of \(980\,\text{kg}\). May they all ride together? Explain by checking both limits.

Hints

- Dividing total capacity evenly gives a comparison rate, not a rule about one rider's weight. - For part b), check people and total kilograms separately. - Do both conditions hold at the same time?

Solution

1. With \(15\) riders, the equal share of the total weight limit is \(1200\div15=80\,\text{kg}\) per person. 2. The group has \(12\) people, fewer than the \(15\)-person limit. 3. Their total mass \(980\,\text{kg}\) is also below \(1200\,\text{kg}\). 4. Both limits are met, so all \(12\) may ride together.

Answer

a) \(80\,\text{kg}\) per person as an equal share of total capacity b) Yes. The group is below both the person limit and the total weight limit.
Rate problems with constant rates 103
A truck uses \(8\) gallons of diesel for every \(100\) miles traveled. Its fuel tank holds \(80\) gallons. How far can the truck travel on a full tank at this rate?

Hints

- How many equal groups of the given fuel amount fit into the tank? - Every group of fuel corresponds to the same distance.

Solution

1. There are \(80\div8=10\) groups of \(8\) gallons in the tank. 2. Each group allows \(100\) miles, so the range is \(10\times100=1000\) miles.

Answer

\(1000\) miles
Rate problems with constant rates 104
A farmer has \(20\) hazelnut bushes. Each bush produces about \(1500\) nuts. A sample shows that \(100\) nuts have a mass of \(240\,\text{g}\). About how many kilograms of hazelnuts do all the bushes produce?

Hints

- Find the total number of nuts first. - Determine how many groups of \(100\) nuts are in the total. - Use the sample rate, then convert grams to kilograms.

Solution

1. Estimate the total number of nuts: \(20 \times 1500 \approx 30{,}000\). 2. Estimate the number of groups of \(100\) nuts: \(30{,}000 \div 100 \approx 300\). 3. Estimate the total mass: \(300 \times 240\,\text{g} \approx 72{,}000\,\text{g}\). 4. Convert to kilograms: \(72{,}000\,\text{g} = 72\,\text{kg}\), so the estimated mass is about \(72\,\text{kg}\).

Answer

The bushes produce about \(72\,\text{kg}\) of hazelnuts.
Rate problems with constant rates 105
At a camp, one water container holds \(24\,\text{L}\). Each child drinks about \(3\,\text{L}\) of water per day. a) About how many children can one container supply for one day? b) The camp has \(120\) children. About how many containers are needed each day?

Hints

- Divide the container's volume by the daily amount per child. - Then divide the total number of children by the number supplied by one container. - Use approximation language because the daily amount is an average.

Solution

1. Estimate the number of children supplied by one container: \(24\,\text{L} \div 3\,\text{L} \approx 8\) children. 2. Estimate the number of containers for \(120\) children: \(120 \div 8 \approx 15\) containers.

Answer

a) One container supplies about \(8\) children for one day. b) The camp needs about \(15\) containers per day.
Rate problems with constant rates 106
One pear tree produces about \(150\,\text{kg}\) of fruit in a year. The pears are packed into crates that each hold \(15\,\text{kg}\). a) About how many crates can be filled with the pears from one tree? b) An orchard has \(200\) such trees. About how many crates can be filled with the entire harvest?

Hints

- Determine how many \(15\)-kilogram groups are in \(150\,\text{kg}\). - Use the crates-per-tree rate for all \(200\) trees. - Check the multiplication by working backward.

Solution

1. Estimate the number of crates per tree: \(150\,\text{kg} \div 15\,\text{kg} \approx 10\). 2. Estimate the total for \(200\) trees: \(10 \times 200 \approx 2000\) crates.

Answer

a) One tree fills about \(10\) crates. b) The orchard fills about \(2000\) crates.
Rate problems with constant rates 107
In a simplified zoo-planning model, use these rounded daily quantities: an adult elephant drinks about \(80\,\text{L}\) of water, one trunkful is modeled as \(16\,\text{L}\), and a newborn elephant drinks about \(10\,\text{L}\) of milk. a) About how many modeled trunkfuls of water does an adult elephant drink each day? b) About how many liters of water do \(5\) adult elephants drink in one day? c) About how many liters of milk does a newborn elephant drink in \(7\) days?

Hints

- Divide the modeled daily water amount by the amount in one modeled trunkful. - Multiply one adult's daily amount by \(5\). - Multiply the newborn's daily amount by \(7\). - Use approximation language throughout.

Solution

1. Estimate the number of modeled trunkfuls: \(80\,\text{L} \div 16\,\text{L} \approx 5\). 2. Estimate the herd's daily water: \(80\,\text{L} \times 5 \approx 400\,\text{L}\). 3. Estimate the newborn's weekly milk: \(10\,\text{L} \times 7 \approx 70\,\text{L}\).

Answer

a) About \(5\) modeled trunkfuls b) About \(400\,\text{L}\) c) About \(70\,\text{L}\)
Rate problems with constant rates 108
For this walking-rate model, suppose an elephant walks at \(5\) miles per hour and a person walks at \(4\) miles per hour. a) The elephant travels \(20\) miles in one day. How many hours does it walk? b) How far does the person walk in \(4\) hours? c) After \(4\) hours, how much farther has the elephant walked than the person?

Hints

- Use the elephant’s modeled hourly rate to find the time for \(20\) miles. - For part c), first find each distance traveled in \(4\) hours. - Subtract the two distances to find how much farther one traveled.

Solution

1. For a), \(20\,\text{mi} \div 5\,\text{mi/h} = 4\,\text{h}\). 2. For b), \(4\,\text{h} \times 4\,\text{mi/h} = 16\,\text{mi}\). 3. In \(4\) hours, the elephant travels \(4\,\text{h} \times 5\,\text{mi/h} = 20\,\text{mi}\). 4. The difference is \(20\,\text{mi} - 16\,\text{mi} = 4\,\text{mi}\).

Answer

a) \(4\) hours b) \(16\) miles c) \(4\) miles farther
Rate problems with constant rates 109
In a simplified daily-activity model, suppose an elephant walks \(18\) miles each day at \(6\) miles per hour and spends \(18\) hours per day eating. a) How many hours per day does it spend walking? b) How many hours remain in a \(24\)-hour day for sleeping, bathing, and other activities?

Hints

- Divide the modeled daily distance by the walking rate. - Add the time spent walking and eating. - Subtract that total from the \(24\) hours in a day.

Solution

1. Find the walking time: \(18\,\text{mi} \div 6\,\text{mi/h} = 3\,\text{h}\). 2. Walking and eating take \(3\,\text{h} + 18\,\text{h} = 21\,\text{h}\). 3. The remaining time is \(24\,\text{h} - 21\,\text{h} = 3\,\text{h}\).

Answer

a) The elephant walks for \(3\) hours. b) It has \(3\) hours left for other activities.
Rate problems with constant rates 110
A water-saving showerhead uses \(1\) gallon less water per minute than an older model. A family showers for a total of \(30\) minutes each day. a) How many gallons does the family save each day? b) How many gallons does it save in \(7\) days? c) If \(1000\) gallons cost \(\$4.00\), how much money does the family save in \(10\) weeks?

Hints

- Multiply the savings per minute by the daily shower time. - Use \(7\) days for one week. - Compare the total gallons saved with groups of \(1000\) gallons.

Solution

1. The daily savings are \(1\,\text{gal/min} \times 30\,\text{min} = 30\,\text{gal}\). 2. The weekly savings are \(30\,\text{gal} \times 7 = 210\,\text{gal}\). 3. In \(10\) weeks, the family saves \(210\,\text{gal} \times 10 = 2100\,\text{gal}\). 4. At \(\$4.00\) per \(1000\) gallons, \(2100\) gallons cost \(2.1 \times \$4.00 = \$8.40\).

Answer

a) \(30\) gallons b) \(210\) gallons c) \(\$8.40\)
Rate problems with constant rates 111
An older toilet uses \(12\,\text{L}\) of water per flush, while a water-saving toilet uses \(5\,\text{L}\). A person flushes about \(6\) times per day. a) About how many liters does one person save in one day by using the water-saving toilet? b) About how many liters does a four-person family save in \(7\) days? c) About how many liters does the family save in \(365\) days?

Hints

- Find the water saved with one flush. - Scale the per-flush savings to one person per day. - Then scale by the number of people and days. - Use approximation language because the daily number of flushes is an average.

Solution

1. Find the savings per flush: \(12\,\text{L} - 5\,\text{L} = 7\,\text{L}\). 2. Estimate one person's daily savings: \(6 \times 7\,\text{L} \approx 42\,\text{L}\). 3. Estimate the family's daily savings: \(4 \times 42\,\text{L} \approx 168\,\text{L}\). 4. Estimate the weekly savings: \(7 \times 168\,\text{L} \approx 1176\,\text{L}\). 5. Estimate the yearly savings: \(365 \times 168\,\text{L} \approx 61{,}320\,\text{L}\).

Answer

a) About \(42\,\text{L}\) b) About \(1176\,\text{L}\) c) About \(61{,}320\,\text{L}\)
Rate problems with constant rates 112
A city charges \(\$2.00\) for \(1000\) gallons of water. a) How many cents do \(100\) gallons cost? b) A family uses \(80\) gallons of water per day for washing dishes, laundry, and showers. How many gallons does it use in \(30\) days? c) What does that month's water cost? Give the amount in dollars and cents.

Hints

- Compare the amount in part a) with the amount named in the price. - For the monthly use, what quantity repeats each day? - The same charge applies to each group of \(1000\) gallons.

Solution

1. Since \(\$2.00\) is \(200\) cents, \(100\) gallons cost \(200\div10=20\) cents. 2. The monthly use is \(80\times30=2400\) gallons. 3. At \(\$2.00\) per \(1000\) gallons, the cost is \(\frac{2400}{1000}\times\$2.00=\$4.80\).

Answer

a) \(20\) cents b) \(2400\) gallons c) \(\$4.80\)
Rate problems with constant rates 113
Tim compares two ways to use water when brushing his teeth. <table><tr><th>Method</th><th>Water per brushing</th></tr><tr><td>Leave the tap running</td><td>\(3\) gallons</td></tr><tr><td>Rinse using a cup</td><td>\(0.25\) gallon</td></tr></table> a) Tim brushes twice a day. How many gallons can he save each day by using a cup both times? b) How many gallons can he save in \(100\) days? c) If \(1000\) gallons cost about \(\$3.00\), how much money can he save in \(100\) days?

Hints

- Compare the water used by the two methods for one brushing. - How many brushings happen each day and over \(100\) days? - Compare the amount saved with the amount named in the water price.

Solution

1. Each brushing saves \(3-0.25=2.75\) gallons. 2. Twice a day saves \(2\times2.75=5.5\) gallons per day. 3. In \(100\) days the savings are \(5.5\times100=550\) gallons. 4. At about \(\$3.00\) per \(1000\) gallons, the cost savings are \(\frac{550}{1000}\times\$3.00\approx\$1.65\).

Answer

a) \(5.5\) gallons b) \(550\) gallons c) About \(\$1.65\)
Rate problems with constant rates 114
A family pays \(\$0.20\) for \(40\) gallons of water. A bath uses about \(60\) gallons, while a shower uses \(20\) gallons. a) Find the water cost of one bath. b) How much is saved by showering instead of taking a bath? c) A person usually takes two baths each week. If the person showers instead every time for \(52\) weeks, how much money is saved?

Hints

- Find the cost of \(20\) gallons from the cost of \(40\) gallons. - Compare the costs of a bath and a shower. - How many baths are replaced during \(52\) weeks?

Solution

1. Twenty gallons costs half of \(\$0.20\), or \(\$0.10\). Therefore, about \(60\) gallons costs \(\$0.20 + \$0.10 \approx \$0.30\). 2. The savings per shower are approximately \(\$0.30 - \$0.10 \approx \$0.20\). 3. The weekly savings are approximately \(2 \times \$0.20 \approx \$0.40\). 4. The yearly savings are approximately \(52 \times \$0.40 \approx \$20.80\).

Answer

a) About \(\$0.30\) b) About \(\$0.20\) c) About \(\$20.80\)
Rate problems with constant rates 115
A nursery pays \(\$2.00\) for \(1000\) gallons of water. Flower beds can be watered in two ways: - With watering cans: \(12\) cans holding \(2\) gallons each - With a sprinkler: \(3\) gallons per minute for \(10\) minutes a) How many gallons does each method use? b) What is the water cost of using the sprinkler once? c) The flowers are watered every day for \(30\) days. How much money is saved by always using the watering cans instead of the sprinkler?

Hints

- Find the total water used by each method. - Use the cost per \(1000\) gallons to find the sprinkler’s cost. - Find the daily difference, then scale it to \(30\) days.

Solution

1. Watering cans use \(12 \times 2\,\text{gal} = 24\,\text{gal}\). The sprinkler uses \(3\,\text{gal/min} \times 10\,\text{min} = 30\,\text{gal}\). 2. At \(\$2.00\) per \(1000\) gallons, \(30\) gallons costs \(\frac{30}{1000} \times \$2.00 = \$0.06\). 3. The watering cans save \(30\,\text{gal} - 24\,\text{gal} = 6\,\text{gal}\) each day. 4. In \(30\) days, the savings are \(6\,\text{gal} \times 30 = 180\,\text{gal}\). 5. The monetary savings are \(\frac{180}{1000} \times \$2.00 = \$0.36\).

Answer

a) Watering cans: \(24\) gallons; sprinkler: \(30\) gallons b) \(\$0.06\) c) \(\$0.36\)
Rate problems with constant rates 116
A garden owner pays \(\$4.00\) for \(1000\) gallons of tap water. A large watering cart holds \(10\) gallons, and filling a small garden pond requires \(200\) gallons. a) How many cents do \(10\) gallons cost? b) How many cents does the pond filling cost? c) Would using \(25\) full watering carts cost more or less than filling the pond? Explain.

Hints

- Convert \(\$4.00\) to cents. - Determine how many \(10\)-gallon groups fit in \(1000\) gallons and in \(200\) gallons. - Find the total volume and cost of \(25\) carts. - Compare either the water amounts or their costs.

Solution

1. Since \(\$4.00 = 400\) cents, \(10\) gallons costs \(400 \div 100 = 4\) cents. 2. The pond uses \(200 \div 10 = 20\) groups of \(10\) gallons, so it costs \(20 \times 4 = 80\) cents. 3. Twenty-five carts hold \(25 \times 10\,\text{gal} = 250\,\text{gal}\) and cost \(25 \times 4 = 100\) cents. This is more than the \(80\)-cent pond filling.

Answer

a) \(4\) cents b) \(80\) cents c) Twenty-five carts would cost more because they use \(250\) gallons and cost \(100\) cents.
Rate problems with constant rates 117
A school pays \(30\) cents for every \(100\) gallons of water. Its restrooms have two types of toilets: - Older toilet: \(3\) gallons per flush - Newer toilet: \(1.5\) gallons per flush a) What does the water for \(100\) flushes of the older toilet cost? b) How much money is saved over \(100\) flushes by using the newer toilet?

Hints

- Find the total water used in \(100\) flushes. - Compare that amount with groups of \(100\) gallons. - You can first find the water saved by one flush. - Convert cents to dollars at the end.

Solution

1. The older toilet uses \(100 \times 3\,\text{gal} = 300\,\text{gal}\). Since \(300\) gallons is three groups of \(100\) gallons, the cost is \(3 \times 30 = 90\) cents, or \(\$0.90\). 2. The newer toilet saves \(3\,\text{gal} - 1.5\,\text{gal} = 1.5\,\text{gal}\) per flush. 3. Over \(100\) flushes, it saves \(150\) gallons. At \(30\) cents per \(100\) gallons, the savings are \(1.5 \times 30 = 45\) cents, or \(\$0.45\).

Answer

a) \(\$0.90\) b) \(\$0.45\)
Rate problems with constant rates 118
A car wash pays \(\$2.00\) for every \(1000\) gallons of water. - A pressure-washer cleaning uses \(60\) gallons. - An automatic wash uses \(150\) gallons. a) How many cents does the water for one pressure-washer cleaning cost? b) How many cents does the water for one automatic wash cost? c) A taxi company has \(10\) cars washed automatically. What is the total water cost?

Hints

- Find the cost of a convenient amount such as \(10\) or \(100\) gallons. - Use that smaller unit rate for \(60\) and \(150\) gallons. - For part c), first find the total gallons used.

Solution

1. Since \(\$2.00 = 200\) cents, \(10\) gallons costs \(2\) cents. 2. A \(60\)-gallon cleaning costs \(6 \times 2 = 12\) cents. 3. A \(150\)-gallon wash costs \(15 \times 2 = 30\) cents. 4. Ten automatic washes use \(10 \times 150\,\text{gal} = 1500\,\text{gal}\), which costs \(\$3.00\).

Answer

a) \(12\) cents b) \(30\) cents c) \(\$3.00\)
Rate problems with constant rates 119
At a class fitness event, Maya and Noah each run a \(1\)-mile route. Maya takes \(12\) minutes. Noah takes \(10\) minutes. a) How many miles would Maya cover in one hour at the same pace? b) How many miles would Noah cover in one hour at the same pace? c) How many hours would Maya need to cover \(10\) miles at that pace?

Hints

- Use the time for one mile to decide how many miles fit in one hour. - How does Maya's time for one mile relate to the time for a longer run at the same pace? - Would her time for ten miles be greater or less than one hour?

Solution

1. Maya covers \(60\div12=5\) miles in one hour. 2. Noah covers \(60\div10=6\) miles in one hour. 3. Maya needs \(10\times12=120\) minutes, or \(2\) hours.

Answer

a) \(5\) miles b) \(6\) miles c) \(2\) hours
Rate problems with constant rates 120
A regional train and an express train travel along the same route. - The regional train takes \(2\) minutes to travel \(1\) mile. - The express train takes \(30\) seconds to travel \(1\) mile. a) How many miles does the regional train travel in one hour? b) How many miles does the express train travel in one minute? c) How many miles does the express train travel in one hour?

Hints

- How many seconds are in one minute? - If the express train travels \(1\) mile in \(30\) seconds, how far does it go in \(60\) seconds? - Use the result from part b) for \(60\) minutes.

Solution

1. The regional train travels \(60 \div 2 = 30\) miles in one hour. 2. Since one minute is \(60\) seconds, the express train travels \(60 \div 30 = 2\) miles in one minute. 3. In one hour, the express train travels \(60 \times 2 = 120\) miles.

Answer

a) \(30\) miles b) \(2\) miles c) \(120\) miles
Rate problems with constant rates 121
For a paper-production model, use \(40\) gallons of water per pound for one paper process and \(10\) gallons per pound for a recycled-paper process. A set of \(100\) invitation cards weighs \(4\,\text{lb}\). How much water is saved by making \(25\) of the cards with the recycled-paper process instead of the first process?

Hints

- What fraction of \(100\) cards is \(25\) cards? - Use that fraction to find the weight of \(25\) cards. - Find the difference between the two modeled water-use rates per pound.

Solution

1. Since \(25\) cards is one fourth of \(100\) cards, \(25\) cards weigh \(4\,\text{lb} \div 4 = 1\,\text{lb}\). 2. The modeled water savings per pound are \(40\,\text{gal} - 10\,\text{gal} = 30\,\text{gal}\). 3. Therefore, making \(25\) cards with the recycled-paper process saves \(30\) gallons in this model.

Answer

\(30\) gallons of water
Rate problems with constant rates 122
An audiobook narrator reads a very long book containing exactly \(1{,}000{,}000\) words. The narrator reads an average of \(125\) words per minute. How many days, hours, and minutes of reading time are needed to finish the book, not including breaks?

Hints

- First find the total number of minutes from the word total and the words-per-minute rate. - How many minutes are in one hour? - How many hours are in one day? - Use remainders to keep track of leftover minutes and hours.

Solution

1. Find the total number of minutes: \(1{,}000{,}000 \div 125 = 8000\,\text{min}\). 2. Convert minutes to hours: \(8000 \div 60 = 133\) remainder \(20\), so the time is \(133\,\text{hr}\) and \(20\,\text{min}\). 3. Convert hours to days: \(133 \div 24 = 5\) remainder \(13\), so the time is \(5\) days, \(13\) hours, and \(20\) minutes.

Answer

The narrator needs \(5\) days, \(13\) hours, and \(20\) minutes of reading time.
Rate problems with constant rates 123
A painter paints \(35\,\text{ft}\) of fence in \(5\) hours on Monday. On Tuesday, the painter works for \(3\) hours at the same rate. How many feet of fence does the painter complete during both days altogether?

Hints

- Find how many feet the painter completes in one hour. - Use that rate to find Tuesday's distance. - Add the amounts from Monday and Tuesday.

Solution

1. Find the unit rate: \(35 \div 5 = 7\) feet per hour. 2. On Tuesday, the painter completes \(3 \times 7 = 21\) feet. 3. Altogether, the painter completes \(35 + 21 = 56\) feet.

Answer

The painter completes \(56\,\text{ft}\) of fence during both days altogether.
Rate problems with constant rates 124
A gardener uses a hose to fill \(4\) watering cans in \(5\) minutes. Each can holds \(2\) gallons. The same hose fills a large barrel in \(15\) minutes. How many gallons does the barrel hold?

Hints

- Find the total water in the four cans. - Use the \(5\)-minute amount to find the hose’s flow rate. - Apply the same rate for \(15\) minutes.

Solution

1. In \(5\) minutes, the hose delivers \(4 \times 2\,\text{gal} = 8\,\text{gal}\). 2. The flow rate is \(8\,\text{gal} \div 5\,\text{min} = 1.6\,\text{gal/min}\). 3. In \(15\) minutes, the hose delivers \(1.6\,\text{gal/min} \times 15\,\text{min} = 24\,\text{gal}\).

Answer

The barrel holds \(24\) gallons.
Rate problems with constant rates 125
Two families buy a \(14\,\text{lb}\) box of apples for \(\$28\). The Miller family takes \(6\,\text{lb}\), and the Schmidt family takes the rest. If they divide the cost fairly by weight, how much should each family pay?

Hints

- Find the price per pound first. - Determine how many pounds the second family receives. - Multiply each family’s weight by the unit price.

Solution

1. Find the price per pound: \(\$28 \div 14\,\text{lb} = \$2\) per pound. 2. The Miller family pays \(6\,\text{lb} \times \$2/\text{lb} = \$12\). 3. The Schmidt family receives \(14\,\text{lb} - 6\,\text{lb} = 8\,\text{lb}\). 4. The Schmidt family pays \(8\,\text{lb} \times \$2/\text{lb} = \$16\).

Answer

The Miller family pays \(\$12\), and the Schmidt family pays \(\$16\).
Rate problems with constant rates 126
Leon shops at a farmers market. He pays \(\$6.00\) for \(3\,\text{lb}\) of oranges. He also buys \(2\,\text{lb}\) of bananas, and his total is \(\$11.00\). At the same prices, Sarah buys \(1\,\text{lb}\) of oranges and \(3\,\text{lb}\) of bananas. How much does Sarah pay?

Hints

- First find the price of \(1\,\text{lb}\) of oranges. - How much of Leon’s total was for bananas? - Use that amount to find the price of \(1\,\text{lb}\) of bananas. - Then find and add Sarah’s two costs.

Solution

1. Find the price per pound of oranges: \(\$6.00 \div 3 = \$2.00\) per pound. 2. Find what Leon paid for the bananas: \(\$11.00 - \$6.00 = \$5.00\). 3. Find the price per pound of bananas: \(\$5.00 \div 2 = \$2.50\) per pound. 4. Find Sarah’s cost for oranges: \(1 \times \$2.00 = \$2.00\). 5. Find Sarah’s cost for bananas: \(3 \times \$2.50 = \$7.50\). 6. Add Sarah’s costs: \(\$2.00 + \$7.50 = \$9.50\).

Answer

Sarah pays \(\$9.50\).
Rate problems with constant rates 127
An apple grower harvests \(450\,\text{lb}\) of apples in \(5\) days. He works exactly \(6\) hours each day. On average, how many pounds of apples does he harvest per hour? Show two different ways to calculate the answer.

Hints

- One method can begin by finding the total number of hours worked. - Another method can begin by finding the number of pounds harvested each day. - For each method, use the information step by step to find an hourly rate.

Solution

1. Method 1: Find the total work time: \(5 \times 6 = 30\) hours. Then find the hourly rate: \(450 \div 30 = 15\,\text{lb}\) per hour. 2. Method 2: Find the amount harvested each day: \(450 \div 5 = 90\,\text{lb}\) per day. Then find the hourly rate: \(90 \div 6 = 15\,\text{lb}\) per hour.

Answer

The grower harvests an average of \(15\,\text{lb}\) of apples per hour.
Rate problems with constant rates 128
An older printing machine works \(5\) days for \(8\) hours each day and prints \(4000\) posters. A newer machine prints the same \(4000\) posters in \(4\) days while also running \(8\) hours each day. Find each machine’s average number of posters per hour. Which machine is faster, and by how many posters per hour?

Hints

- Find the total operating hours for each machine. - Divide the total number of posters by the total number of hours. - Subtract the smaller hourly rate from the larger hourly rate.

Solution

1. The older machine runs for \(5 \times 8 = 40\) hours. Its rate is \(4000 \div 40 = 100\) posters per hour. 2. The newer machine runs for \(4 \times 8 = 32\) hours. Its rate is \(4000 \div 32 = 125\) posters per hour. 3. Compare the rates: \(125 - 100 = 25\) posters per hour. The newer machine is faster.

Answer

The older machine prints \(100\) posters per hour, and the newer machine prints \(125\) posters per hour. The newer machine is faster by \(25\) posters per hour.
Rate problems with constant rates 129
A machine fills \(240\) juice bottles in \(4\) minutes. a) How many bottles does the machine fill in one minute? b) A newer machine fills \(20\) more bottles per minute. How many bottles does the newer machine fill in \(8\) minutes?

Hints

- First find the rate for one minute. - Increase that rate by \(20\) bottles per minute. - Multiply the new rate by \(8\) minutes.

Solution

1. Find the first machine’s unit rate: \(240 \div 4 = 60\) bottles per minute. 2. Find the newer machine’s rate: \(60 + 20 = 80\) bottles per minute. 3. Find the number filled in \(8\) minutes: \(80 \times 8 = 640\) bottles.

Answer

a) The first machine fills \(60\) bottles per minute. b) The newer machine fills \(640\) bottles in \(8\) minutes.
Rate problems with constant rates 130
Jonas and Sarah run on a track. Jonas runs \(800\,\text{m}\) in \(4\) minutes. Sarah runs \(500\,\text{m}\) in \(2\) minutes. Who runs faster? How far would the faster runner travel in \(10\) minutes at the same rate?

Hints

- Find how far each runner travels in \(1\) minute. - Compare the two unit rates. - Use the faster runner’s rate to find the distance in \(10\) minutes.

Solution

1. Find Jonas’s rate: \(800 \div 4 = 200\,\text{m}\) per minute. 2. Find Sarah’s rate: \(500 \div 2 = 250\,\text{m}\) per minute. 3. Since \(250 > 200\), Sarah runs faster. 4. Find Sarah’s distance in \(10\) minutes: \(250 \times 10 = 2500\,\text{m}\).

Answer

Sarah runs faster. At the same rate, she would run \(2500\,\text{m}\) in \(10\) minutes.
Rate problems with constant rates 131
A juice press fills \(12\) bottles in \(20\) minutes. a) How many bottles does it fill in \(60\) minutes? b) How long does it take to fill \(6\) bottles? c) How many bottles does it fill in \(30\) minutes?

Hints

- Compare each new amount or time with the original pair. - Use the same scale factor for time and number of bottles. - Half an hour is \(30\) minutes.

Solution

1. For \(60\) minutes, the scale factor is \(60 \div 20 = 3\). Therefore, \(12 \times 3 = 36\) bottles. 2. Six bottles is half of \(12\), so the time is half of \(20\) minutes: \(20 \div 2 = 10\) minutes. 3. Thirty minutes is \(1.5\) times \(20\) minutes, so \(12 \times 1.5 = 18\) bottles.

Answer

a) \(36\) bottles b) \(10\) minutes c) \(18\) bottles
Rate problems with constant rates 132
A cider mill presses \(3\,\text{qt}\) of cider from \(5\,\text{lb}\) of apples. a) How many quarts of cider can be made from \(20\,\text{lb}\) of apples? b) How many pounds of apples are needed to make \(15\,\text{qt}\) of cider? c) How many pounds of apples are needed to make \(9\,\text{qt}\) of cider?

Hints

- Compare each new apple amount with \(5\,\text{lb}\). - If the amount of apples is multiplied by a number, the amount of cider is multiplied by the same number. - A ratio table with apples in one column and cider in the other may help. - For a target cider amount, determine the scale factor from \(3\,\text{qt}\).

Solution

1. For \(20\,\text{lb}\) of apples, the scale factor is \(20 \div 5 = 4\). The cider amount is \(4 \times 3 = 12\,\text{qt}\). 2. For \(15\,\text{qt}\) of cider, the scale factor is \(15 \div 3 = 5\). The apple amount is \(5 \times 5 = 25\,\text{lb}\). 3. For \(9\,\text{qt}\) of cider, the scale factor is \(9 \div 3 = 3\). The apple amount is \(3 \times 5 = 15\,\text{lb}\).

Answer

a) \(12\,\text{qt}\) of cider b) \(25\,\text{lb}\) of apples c) \(15\,\text{lb}\) of apples
Rate problems with constant rates 133
A printer uses \(120\,\text{mL}\) of ink to print \(40\) posters. a) How many milliliters of ink are needed for \(200\) posters at the same rate? b) How many posters can be printed with \(300\,\text{mL}\) of ink? c) How many posters can be printed with \(30\,\text{mL}\) of ink?

Hints

- Find the amount of ink used for one poster. - Multiply the unit rate to find the ink for \(200\) posters. - Divide each available ink amount by the unit rate.

Solution

1. Find the ink used per poster: \(120\,\text{mL} \div 40 = 3\,\text{mL}\) per poster. 2. For part a, multiply: \(200 \times 3\,\text{mL} = 600\,\text{mL}\). 3. For part b, divide: \(300\,\text{mL} \div 3\,\text{mL} = 100\) posters. 4. For part c, divide: \(30\,\text{mL} \div 3\,\text{mL} = 10\) posters.

Answer

a) \(600\,\text{mL}\) b) \(100\) posters c) \(10\) posters
Rate problems with constant rates 134
Paul rides his bicycle a total of \(72\,\text{mi}\) in \(6\) hours at a constant speed. He rides \(24\,\text{mi}\) in the morning and the rest in the afternoon. How many hours does he ride in the morning, and how many hours does he ride in the afternoon?

Hints

- First find how many miles Paul rides in one hour. - How many times does that hourly distance fit into the morning distance? - How many miles remain for the afternoon? - Check that the two riding times add to the total time.

Solution

1. Find Paul’s speed: \(72 \div 6 = 12\,\text{mi/h}\). 2. Find the morning riding time: \(24 \div 12 = 2\) hours. 3. Find the afternoon distance: \(72 - 24 = 48\,\text{mi}\). 4. Find the afternoon riding time: \(48 \div 12 = 4\) hours.

Answer

Paul rides for \(2\) hours in the morning and \(4\) hours in the afternoon.
Rate problems with constant rates 135
For a school festival, Mr. Weber buys fruit. A \(2\,\text{lb}\) bag of apples costs \(\$4\), and a \(3\,\text{lb}\) bag of pears costs \(\$9\). He wants to buy exactly \(12\,\text{lb}\) of each fruit. How much more will the pears cost than the apples?

Hints

- Find the price of \(1\,\text{lb}\) of each fruit. - Determine how many bags of each fruit make \(12\,\text{lb}\). - Find the two total costs, and then compare them.

Solution

1. Find the price per pound of apples: \(\$4 \div 2 = \$2\) per pound. 2. Find the cost of \(12\,\text{lb}\) of apples: \(12 \times \$2 = \$24\). 3. Find the price per pound of pears: \(\$9 \div 3 = \$3\) per pound. 4. Find the cost of \(12\,\text{lb}\) of pears: \(12 \times \$3 = \$36\). 5. Find the difference: \(\$36 - \$24 = \$12\).

Answer

The pears will cost \(\$12\) more than the apples.
Rate problems with constant rates 136
At the base of a mountain, the elevation is \(650\,\text{m}\) and the temperature is \(2\,^{\circ}\text{C}\). A station higher up is at \(2250\,\text{m}\). In a simplified model, the temperature drops by \(1\,^{\circ}\text{C}\) for every \(200\,\text{m}\) of elevation gain. What temperature does the model predict at the upper station?

Hints

- Use the difference in elevation, not the height of the upper station alone. - Think about how many \(200\)-meter rises there are. - Track the temperature down to zero and then below zero.

Solution

1. The station is \(2250-650=1600\,\text{m}\) higher. 2. There are \(1600\div200=8\) intervals, so the temperature drops by \(8\,^{\circ}\text{C}\). 3. A drop of \(2\,^{\circ}\text{C}\) reaches zero. Another \(8-2=6\,^{\circ}\text{C}\) of drop gives \(-6\,^{\circ}\text{C}\).

Answer

\(-6\,^{\circ}\text{C}\)
Rate problems with constant rates 137
A taxi company charges a base fare of \(\$4.50\) for every ride. It also charges: - daytime rides: \(\$2\) per mile - nighttime rides: \(\$3\) per mile Ms. Maier takes a \(12\,\text{mi}\) daytime ride. Mr. Schmidt takes a \(10\,\text{mi}\) nighttime ride. Who pays more? Calculate both fares.

Hints

- Find each rider’s mileage charge separately. - Add the base fare to each mileage charge. - Compare the two total fares.

Solution

1. Find Ms. Maier’s mileage charge: \(12 \times \$2 = \$24\). 2. Add the base fare: \(\$24 + \$4.50 = \$28.50\). 3. Find Mr. Schmidt’s mileage charge: \(10 \times \$3 = \$30\). 4. Add the base fare: \(\$30 + \$4.50 = \$34.50\). 5. Since \(\$34.50 > \$28.50\), Mr. Schmidt pays more.

Answer

Ms. Maier pays \(\$28.50\), and Mr. Schmidt pays \(\$34.50\). Mr. Schmidt pays more.
Rate problems with constant rates 138
A hiker walks \(8\) miles in \(2\) hours at a constant rate. The full route is \(20\) miles long. After the first \(2\) hours, how many more hours must the hiker walk?

Hints

- Find the number of miles the hiker walks in one hour. - Use the unit rate to find the total time for \(20\) miles. - Subtract the \(2\) hours already completed.

Solution

1. Find the unit rate: \(8 \div 2 = 4\) miles per hour. 2. Find the total time for \(20\) miles: \(20 \div 4 = 5\) hours. 3. Subtract the time already walked: \(5 - 2 = 3\) hours.

Answer

The hiker must walk for \(3\) more hours.
Rate problems with constant rates 139
An excursion boat travels for \(8\) hours each day for \(6\) consecutive days. During that time, it covers \(864\,\text{mi}\). What is the boat’s average distance traveled per hour?

Hints

- First find the total number of hours the boat travels. - Use the total distance and total time to find the distance traveled in one hour. - Break the problem into two calculations.

Solution

1. Find the total travel time: \(6 \times 8 = 48\) hours. 2. Find the average distance per hour: \(864 \div 48 = 18\,\text{mi/h}\).

Answer

The boat travels an average of \(18\,\text{mi}\) per hour.
Rate problems with constant rates 140
A delivery driver compares two workweeks. During the first week, he drives \(8\) hours per day for \(4\) days and travels \(960\,\text{mi}\). During the second week, he drives \(6\) hours per day for \(5\) days and travels \(990\,\text{mi}\). In which week is his average speed greater?

Hints

- Find the driver’s total number of hours for each week. - Find the average miles per hour for each week. - Compare the two unit rates.

Solution

1. For the first week, find the total driving time: \(4 \times 8 = 32\) hours. Then find the average speed: \(960 \div 32 = 30\,\text{mi/h}\). 2. For the second week, find the total driving time: \(5 \times 6 = 30\) hours. Then find the average speed: \(990 \div 30 = 33\,\text{mi/h}\). 3. Since \(33 > 30\), his average speed is greater during the second week.

Answer

His average speed is greater during the second week: \(33\,\text{mi/h}\) compared with \(30\,\text{mi/h}\).
Rate problems with constant rates 141
A grower delivers \(5\) crates of tomatoes with a combined weight of \(60\,\text{lb}\). The grower also delivers \(8\) crates of peppers. The pepper crates weigh \(12\,\text{lb}\) more altogether than the tomato crates. How many pounds heavier is one tomato crate than one pepper crate?

Hints

- Find the unit weight of one tomato crate. - Use the total-weight comparison to find the pepper crates' combined weight. - Find one pepper crate's weight and compare the two unit weights.

Solution

1. Find the weight of one tomato crate: \(60\,\text{lb} \div 5 = 12\,\text{lb}\). 2. Find the combined weight of the pepper crates: \(60\,\text{lb} + 12\,\text{lb} = 72\,\text{lb}\). 3. Find the weight of one pepper crate: \(72\,\text{lb} \div 8 = 9\,\text{lb}\). 4. Find the difference: \(12\,\text{lb} - 9\,\text{lb} = 3\,\text{lb}\).

Answer

One tomato crate is \(3\,\text{lb}\) heavier than one pepper crate.
Rate problems with constant rates 142
A ferry travels at a constant rate of \(12\) miles per hour. a) How far does the ferry travel in \(4\) hours? b) After \(3\) hours, the ferry has \(24\) miles left to reach its destination. How long is the entire route?

Hints

- Multiply the rate by the number of hours. - For part b), find the distance already traveled. - Add the distance traveled and the distance remaining.

Solution

1. In \(4\) hours, the ferry travels \(12 \times 4 = 48\) miles. 2. In \(3\) hours, it travels \(12 \times 3 = 36\) miles. 3. The entire route is \(36 + 24 = 60\) miles.

Answer

a) \(48\,\text{mi}\) b) \(60\,\text{mi}\)
Rate problems with constant rates 143
Mia and Jonas take a bicycle trip. Mia rides for \(3\) hours at \(12\,\text{mi/h}\). Jonas rides for \(4\) hours at \(8\,\text{mi/h}\). Who travels farther, and by how many miles?

Hints

- Find the distance each rider travels. - Use the relationship between rate, time, and distance. - Subtract the two distances to find the difference.

Solution

1. Find Mia’s distance: \(3 \times 12 = 36\,\text{mi}\). 2. Find Jonas’s distance: \(4 \times 8 = 32\,\text{mi}\). 3. Since \(36 > 32\), Mia travels farther. 4. Find the difference: \(36 - 32 = 4\,\text{mi}\).

Answer

Mia travels farther. She rides \(4\,\text{mi}\) farther than Jonas.
Rate problems with constant rates 144
Mia's step length is \(24\,\text{in}\), and her father's step length is \(36\,\text{in}\). a) How far does each person travel in \(10\) steps? b) Mia's father takes \(2\) steps. How many steps must Mia take to travel the same distance?

Hints

- Restate what each part asks you to compare. - Multiply each step length by the number of steps. - A table can help match steps with distances. - For part b), determine how many of Mia's steps fit the father's distance.

Solution

1. Mia travels \(10 \times 24\,\text{in} = 240\,\text{in}\). 2. Her father travels \(10 \times 36\,\text{in} = 360\,\text{in}\). 3. Her father travels \(2 \times 36\,\text{in} = 72\,\text{in}\) in \(2\) steps. 4. Mia needs \(72 \div 24 = 3\) steps to travel the same distance.

Answer

a) Mia travels \(240\,\text{in}\), and her father travels \(360\,\text{in}\). b) Mia must take \(3\) steps.
Rate problems with constant rates 145
Lukas rides his bike at a constant speed of \(14\,\text{mi/h}\). He rides for \(4\) hours on Saturday and \(5\) hours on Sunday. How many miles does he ride altogether?

Hints

- Does the riding speed change between the two days? - Think about combining the riding times before calculating distance. - You can also find each day's distance and add them.

Solution

1. Add the riding times: \(4+5=9\) hours. 2. Find the total distance: \(14\times9=126\,\text{mi}\). 3. Check another way: Saturday's distance is \(14\times4=56\,\text{mi}\), and Sunday's is \(14\times5=70\,\text{mi}\). The sum is \(56+70=126\,\text{mi}\).

Answer

\(126\,\text{mi}\)
Rate problems with constant rates 146
During an apple harvest, workers pick \(1350\,\text{lb}\) of apples from Field A in \(6\) days. During the same \(6\) days, they pick \(2088\,\text{lb}\) from Field B. What is the average number of pounds picked per day from each field? What is the difference between the two daily averages?

Hints

- Find the amount picked in one day for each field. - Once you have both daily averages, subtract to find how much greater one is. - Check each division by \(6\) carefully.

Solution

1. Find the daily average for Field A: \(1350 \div 6 = 225\,\text{lb}\) per day. 2. Find the daily average for Field B: \(2088 \div 6 = 348\,\text{lb}\) per day. 3. Find the difference: \(348 - 225 = 123\,\text{lb}\) per day.

Answer

Field A averages \(225\,\text{lb}\) per day, and Field B averages \(348\,\text{lb}\) per day. The difference is \(123\,\text{lb}\) per day.
Rate problems with constant rates 147
A leaking faucet loses about \(12\) drops of water each minute. a) About how many drops are lost in one hour? b) About how many drops are lost in a full \(24\)-hour day? c) About \(20\) drops equal \(1\,\text{mL}\). About how many milliliters of water are lost in one day?

Hints

- Use the constant drops-per-minute rate to scale to an hour and then a day. - For part c, determine how many groups of \(20\) drops are in the daily total. - Keep track of the units at each step.

Solution

1. a) In one hour, the faucet loses about \(12\times60=720\) drops. 2. b) In one day, it loses about \(720\times24=17{,}280\) drops. 3. c) Convert drops to milliliters: \(17{,}280\div20=864\). The faucet loses about \(864\,\text{mL}\) per day.

Answer

a) About \(720\) drops b) About \(17{,}280\) drops c) About \(864\,\text{mL}\)
Rate problems with constant rates 148
Lucas reads for \(20\) minutes every evening at a constant rate of \(2\) pages per minute. a) How many pages does he read in one week of \(7\) days? b) Each book in his favorite adventure series has \(160\) pages. At the same rate, how many complete books does he read in \(52\) weeks?

Hints

- First find the number of pages read each day. - Scale the daily amount to one week and then to \(52\) weeks. - Divide the \(52\)-week page total by the number of pages per book.

Solution

1. Lucas reads \(20\times2=40\) pages per day. 2. a) In one week, he reads \(40\times7=280\) pages. 3. In \(52\) weeks, he reads \(280\times52=14{,}560\) pages. 4. b) The number of complete books is \(14{,}560\div160=91\).

Answer

a) \(280\) pages b) \(91\) complete books
Rate problems with constant rates 149
A hiking group covers \(15\,\text{mi}\) per day for \(6\) days. They plan to cover the same total distance on the return trip in only \(5\) days. What average distance must they hike each day on the return trip?

Hints

- Find the total distance of the first trip. - If the same distance is covered in fewer days, should the daily distance be greater or less? - Divide the total distance by the number of return-trip days.

Solution

1. Find the total distance: \(6 \times 15 = 90\,\text{mi}\). 2. Divide the same distance among \(5\) days: \(90 \div 5 = 18\,\text{mi}\) per day.

Answer

The group must hike an average of \(18\,\text{mi}\) per day on the return trip.
Rate problems with constant rates 150
A pump fills a water tank at \(120\,\text{gal}\) per hour and takes exactly \(6\) hours to fill the tank. A stronger pump fills the same tank at \(180\,\text{gal}\) per hour. How long will the stronger pump take, and how many hours shorter is the wait?

Hints

- First find the tank’s total capacity. - Use the stronger pump’s rate to find its filling time. - Subtract the two times to find how much shorter the wait is.

Solution

1. Find the tank’s capacity: \(6 \times 120 = 720\,\text{gal}\). 2. Find the time needed by the stronger pump: \(720 \div 180 = 4\) hours. 3. Find the time saved: \(6 - 4 = 2\) hours.

Answer

The stronger pump takes \(4\) hours and reduces the wait by \(2\) hours.
Rate problems with constant rates 151
A large pump moves \(2240\,\text{gal}\) of water in \(7\) minutes. A small garden pump moves \(48\,\text{gal}\) in \(6\) minutes. How many times as much water per minute does the large pump move as the small pump?

Hints

- Find how many gallons each pump moves in one minute. - Compare the two per-minute rates. - Which operation tells how many times one amount is another?

Solution

1. Find the large pump’s rate: \(2240 \div 7 = 320\,\text{gal}\) per minute. 2. Find the small pump’s rate: \(48 \div 6 = 8\,\text{gal}\) per minute. 3. Compare the rates: \(320 \div 8 = 40\).

Answer

The large pump moves \(40\) times as much water per minute as the small pump.
Rate problems with constant rates 152
A bakery has two machines for packing dinner rolls. Machine A packs \(960\) rolls in \(4\) hours. Machine B packs the same number of rolls in \(6\) hours. How many more rolls per hour does Machine A pack than Machine B?

Hints

- Find how many rolls each machine packs in one hour. - Divide the same total amount by each machine’s time. - Subtract the smaller hourly rate from the larger one.

Solution

1. Find Machine A’s rate: \(960 \div 4 = 240\) rolls per hour. 2. Find Machine B’s rate: \(960 \div 6 = 160\) rolls per hour. 3. Find the difference: \(240 - 160 = 80\) rolls per hour.

Answer

Machine A packs \(80\) more rolls per hour than Machine B.
Rate problems with constant rates 153
Two large haul trucks carry rock at a mine. The blue truck carries \(435\) tons in \(5\) trips. The red truck carries \(656\) tons in \(8\) trips. Which carries more rock per trip, and by how many tons?

Hints

- Find each truck's load per trip, not its total carried. - Both trucks make a different number of trips. - Compare the two per-trip amounts.

Solution

1. The blue truck carries \(435\div5=87\) tons per trip. 2. The red truck carries \(656\div8=82\) tons per trip. 3. The blue truck carries \(87-82=5\) more tons per trip.

Answer

The blue truck carries \(5\) tons more per trip.
Rate problems with constant rates 154
Three crates of apples have the same total weight as \(5\) baskets of strawberries. A farmer delivers \(345\) apple crates. How many strawberry baskets would have the same total weight?

Hints

- Determine how many groups of \(3\) crates are in \(345\) crates. - Each group corresponds to \(5\) baskets. - Multiply the number of groups by \(5\).

Solution

1. Find the number of groups of \(3\) apple crates: \(345 \div 3 = 115\). 2. Each group is equivalent to \(5\) strawberry baskets, so \(115 \times 5 = 575\) baskets.

Answer

\(575\) strawberry baskets would have the same total weight.
Rate problems with constant rates 155
At a school field day, the PTA sells pizza. Volunteers prepare \(8\) large pizzas and cut each pizza into \(8\) equal slices. The dough costs \(\$14.50\), the toppings cost \(\$26.80\), and napkins cost \(\$3.50\). The sale must earn a profit of \(\$51.20\) for new sports equipment. What price should be charged for each slice if every slice is sold?

Hints

- Determine the total number of slices. - Add all expenses. - Add the desired profit to the expenses. - Divide the required revenue by the number of slices.

Solution

1. Find the number of slices: \(8 \times 8 = 64\). 2. Find the total expenses: \(\$14.50 + \$26.80 + \$3.50 = \$44.80\). 3. Add the desired profit to find the required revenue: \(\$44.80 + \$51.20 = \$96.00\). 4. Divide the required revenue by the number of slices: \(\$96.00 \div 64 = \$1.50\).

Answer

Each pizza slice should cost \(\$1.50\).
Rate problems with constant rates 156
Two printing machines produce \(1560\) flyers altogether. The first machine runs for \(14\) minutes, and the second runs for \(12\) minutes. Both machines print the same number of flyers per minute. How many flyers does each machine print?

Hints

- Add the two operating times. - Use the total flyers and combined time to find the shared unit rate. - Multiply that rate by each machine’s operating time.

Solution

1. Add the operating times: \(14 + 12 = 26\) minutes. 2. Find the shared printing rate: \(1560 \div 26 = 60\) flyers per minute. 3. Find the first machine’s output: \(14 \times 60 = 840\) flyers. 4. Find the second machine’s output: \(12 \times 60 = 720\) flyers.

Answer

The first machine prints \(840\) flyers, and the second machine prints \(720\) flyers.
Rate problems with constant rates 157
At a farmers market, \(12\,\text{lb}\) of apples cost \(\$24\). A note shows these calculations: 1) \(24 \div 12 = 2\) 2) \(2 \times 7 = 14\) a) What quantity is found in the first calculation? b) What does the result \(14\) mean in the second calculation? c) How much would \(10\,\text{lb}\) of the apples cost?

Hints

- Think about the units that belong with each number. - What does dividing a total cost by the number of pounds find? - Once you know the price per pound, how can you find the cost of another amount?

Solution

1. The first calculation divides the total cost by the total number of pounds, so it finds the unit price: \(\$2\) per pound. 2. The second calculation multiplies the unit price by \(7\), so it shows that \(7\,\text{lb}\) of apples cost \(\$14\). 3. Find the cost of \(10\,\text{lb}\): \(10 \times \$2 = \$20\).

Answer

a) It finds the price per pound, which is \(\$2\). b) It means that \(7\,\text{lb}\) of apples cost \(\$14\). c) \(10\,\text{lb}\) of apples cost \(\$20\).
Rate problems with constant rates 158
At a print shop, the blue press prints \(120\) pages in \(4\) minutes. The green press prints \(150\) pages in \(6\) minutes. Both work at steady rates. a) Which press prints more pages per minute? b) How many pages would the faster press print in \(15\) minutes? c) How long would the blue press take to print \(300\) pages?

Hints

- How could you compare the presses fairly when they run for different lengths of time? - Which press would make more pages in one minute? - For part c), which quantity is unknown: pages or minutes?

Solution

1. The blue press prints \(120\div4=30\) pages per minute. 2. The green press prints \(150\div6=25\) pages per minute, so the blue press is faster. 3. In \(15\) minutes, the blue press prints \(30\times15=450\) pages. 4. For \(300\) pages, the blue press takes \(300\div30=10\) minutes.

Answer

a) The blue press b) \(450\) pages c) \(10\) minutes
Rate problems with constant rates 159
An organic farm grows strawberries. Last year, the field produced an average of \(3\,\text{lb}\) per square yard. This year, the yield increased to \(5\,\text{lb}\) per square yard. Before the farm switched to organic growing methods, the yield was \(1\,\text{lb}\) per square yard. a) How many more pounds does a \(100\,\text{yd}^2\) area produce this year than last year? b) How many more pounds does the same area produce this year than before the switch?

Hints

- First find each difference for one square yard. - How does the difference change for an area \(100\) times as large? - Be sure to compare the correct two yields in each part.

Solution

1. Find the difference per square yard between this year and last year: \(5 - 3 = 2\,\text{lb}\). 2. Find the difference for \(100\,\text{yd}^2\): \(2 \times 100 = 200\,\text{lb}\). 3. Find the difference per square yard between this year and before the switch: \(5 - 1 = 4\,\text{lb}\). 4. Find the difference for \(100\,\text{yd}^2\): \(4 \times 100 = 400\,\text{lb}\).

Answer

a) The \(100\,\text{yd}^2\) area produces \(200\,\text{lb}\) more than last year. b) It produces \(400\,\text{lb}\) more than before the switch.
Rate problems with constant rates 160
A remote-controlled car and a small robot race for exactly \(1\) minute. The car travels \(25\,\text{m}\) every \(10\) seconds. The robot travels \(2\,\text{m}\) every second. a) How far does the car travel in \(1\) minute? b) How far does the robot travel in \(1\) minute? c) Which one wins, and by how many meters?

Hints

- How many seconds are in one minute? - How many \(10\)-second intervals are in one minute? - Find both distances for the same amount of time. - Subtract the shorter distance from the longer distance.

Solution

1. Convert the time: \(1\) minute is \(60\) seconds. 2. There are \(60 \div 10 = 6\) ten-second intervals, so the car travels \(6 \times 25 = 150\,\text{m}\). 3. The robot travels \(60 \times 2 = 120\,\text{m}\). 4. Since \(150 > 120\), the car wins. 5. Find the winning margin: \(150 - 120 = 30\,\text{m}\).

Answer

a) The car travels \(150\,\text{m}\). b) The robot travels \(120\,\text{m}\). c) The car wins by \(30\,\text{m}\).
Rate problems with constant rates 161
A conveyor belt moves packages at \(2\,\text{m}\) per second. A package must travel \(300\,\text{m}\). Is \(2\) minutes enough time for the package to reach the end? Justify your answer with a calculation.

Hints

- How many seconds are in \(2\) minutes? - Find how far the package travels in that time. - Compare the result with \(300\,\text{m}\).

Solution

1. Convert the time to seconds: \(2\) minutes is \(120\) seconds. 2. Find the distance traveled in that time: \(120 \times 2 = 240\,\text{m}\). 3. Since \(240 < 300\), the package does not reach the end. 4. It is \(300 - 240 = 60\,\text{m}\) short.

Answer

No. In \(2\) minutes, the package travels \(240\,\text{m}\), so it is \(60\,\text{m}\) short of the end.
Rate problems with constant rates 162
At \(5{:}00\) p.m., a weather station records \(-4\,^{\circ}\text{C}\). Until \(10{:}00\) p.m., the temperature falls by \(3\,^{\circ}\text{C}\) each hour. a) What is the temperature at \(10{:}00\) p.m.? b) At what time is it exactly \(-10\,^{\circ}\text{C}\)?

Hints

- Count the number of hours that pass. - Starting below zero, a drop makes the distance below zero larger. - For part b), compare the distances of both temperatures below zero.

Solution

1. Five hours pass, so the temperature falls by \(5\times3=15\,^{\circ}\text{C}\). 2. It starts \(4\) degrees below zero and falls \(15\) more degrees, reaching \(4+15=19\) degrees below zero: \(-19\,^{\circ}\text{C}\). 3. From \(-4\) to \(-10\) is a further drop of \(10-4=6\,^{\circ}\text{C}\). 4. That takes \(6\div3=2\) hours after \(5{:}00\) p.m., so it happens at \(7{:}00\) p.m.

Answer

a) \(-19\,^{\circ}\text{C}\) b) \(7{:}00\) p.m.
Rate problems with constant rates 163
A laboratory sample is cooled steadily from \(5\,^{\circ}\text{C}\) at \(8{:}00\) a.m. to \(-19\,^{\circ}\text{C}\) at \(2{:}00\) p.m. By how many degrees Celsius does its temperature fall each hour?

Hints

- First find how many hours the sample cools. - Count the drop from the starting temperature to zero, then the drop below zero. - Divide the whole drop evenly among the hours.

Solution

1. The sample cools for \(6\) hours. 2. It cools \(5\) degrees to reach zero and then \(19\) degrees more below zero: \(5+19=24\,^{\circ}\text{C}\) in all. 3. The hourly drop is \(24\div6=4\,^{\circ}\text{C}\).

Answer

\(4\,^{\circ}\text{C}\) per hour
Rate problems with constant rates 164
A solar water-heating system warms \(12\,\text{gal}\) of water per hour on a cloudy day and \(28\,\text{gal}\) per hour on a sunny day. How much more water does it warm in \(6\) hours on a sunny day? What is the difference after \(9\) hours?

Hints

- How much more water is warmed in one hour on a sunny day? - Use the one-hour difference to find the difference after several hours. - The same hourly difference applies to both time periods.

Solution

1. Find the difference per hour: \(28 - 12 = 16\,\text{gal}\). 2. Find the difference after \(6\) hours: \(16 \times 6 = 96\,\text{gal}\). 3. Find the difference after \(9\) hours: \(16 \times 9 = 144\,\text{gal}\).

Answer

After \(6\) hours, the difference is \(96\,\text{gal}\). After \(9\) hours, the difference is \(144\,\text{gal}\).
Rate problems with constant rates 165
Gardeners planned to plant \(180\) flowers per day and finish a park project in \(12\) days. With help from volunteers, they finished \(4\) days early. On average, how many more flowers per day did they plant than planned?

Hints

- Find the total number of flowers in the project. - Determine how many days the gardeners actually worked. - Divide the total flowers by the actual days. - Compare the actual daily rate with the planned rate.

Solution

1. Find the total number of flowers: \(180 \times 12 = 2160\) flowers. 2. Find the actual number of workdays: \(12 - 4 = 8\) days. 3. Find the actual daily rate: \(2160 \div 8 = 270\) flowers per day. 4. Find the increase: \(270 - 180 = 90\) flowers per day.

Answer

The gardeners planted an average of \(90\) more flowers per day than planned.
Rate problems with constant rates 166
At a cookout stand, hot dogs cost \(\$6\) for \(8\). They can also be bought individually at the same price per hot dog. a) How much do \(24\) hot dogs cost? b) How much do \(12\) hot dogs cost? c) Explain how you found the price for \(12\) hot dogs without buying only whole packs of eight.

Hints

- Compare the first quantity with the stated group of eight. - What does the same price per hot dog let you do when the order size changes? - Can you find the price for twelve by scaling or splitting an order of eight?

Solution

1. Twenty-four hot dogs are \(3\) groups of \(8\), so they cost \(3\times\$6=\$18\). 2. Four hot dogs cost half as much as eight, or \(\$3\), since the per-hot-dog price is unchanged. 3. Twelve hot dogs are eight plus four, so the cost is \(\$6+\$3=\$9\). 4. Dividing the order into eight and four hot dogs works because the per-hot-dog price stays the same.

Answer

a) \(\$18\) b) \(\$9\) c) One way is to use eight hot dogs at \(\$6\) and four at \(\$3\), for \(\$9\) in all.
Rate problems with constant rates 167
At a game booth, \(20\) marbles cost \(\$4\). a) How many marbles can you buy for \(\$12\)? b) Tim wants \(100\) marbles and has \(\$25\). Does he have enough money? Support your answer with a calculation.

Hints

- Compare \(\$12\) with the price of one group of \(20\) marbles. - Determine how many groups of \(20\) make \(100\). - Compare the cost of \(100\) marbles with Tim’s budget.

Solution

1. Since \(\$12\) is \(3\) times \(\$4\), the number of marbles is \(3 \times 20 = 60\). 2. One hundred marbles are \(5\) groups of \(20\), so they cost \(5 \times \$4 = \$20\). 3. Since \(\$20 < \$25\), Tim has enough money.

Answer

a) You can buy \(60\) marbles for \(\$12\). b) Yes. The \(100\) marbles cost \(\$20\), so Tim has enough money.
Rate problems with constant rates 168
A car uses \(6\,\text{gal}\) of gasoline to travel \(300\,\text{mi}\). a) How much gasoline does it use to travel \(900\,\text{mi}\)? b) How far can it travel with \(30\,\text{gal}\) of gasoline? c) How much gasoline does it use to travel \(150\,\text{mi}\)?

Hints

- Compare \(900\,\text{mi}\) with the given \(300\,\text{mi}\). - For part b, compare \(30\,\text{gal}\) with \(6\,\text{gal}\). - How is \(150\,\text{mi}\) related to \(300\,\text{mi}\)?

Solution

1. The distance \(900\,\text{mi}\) is \(3\) times \(300\,\text{mi}\), so the gasoline use is \(3 \times 6 = 18\,\text{gal}\). 2. The amount \(30\,\text{gal}\) is \(5\) times \(6\,\text{gal}\), so the distance is \(5 \times 300 = 1500\,\text{mi}\). 3. The distance \(150\,\text{mi}\) is half of \(300\,\text{mi}\), so the gasoline use is \(6 \div 2 = 3\,\text{gal}\).

Answer

a) \(18\,\text{gal}\) b) \(1500\,\text{mi}\) c) \(3\,\text{gal}\)
Rate problems with constant rates 169
A candle burns at a constant rate. After \(15\) minutes, it is exactly \(12\,\text{mm}\) shorter. The candle was originally \(144\,\text{mm}\) tall. How many hours will the candle burn before it is completely used up?

Hints

- Find how many equal burned sections make up the full candle. - Multiply that number by \(15\) minutes. - Convert the final time from minutes to hours.

Solution

1. Find how many \(12\)-millimeter sections are in the candle: \(144\,\text{mm} \div 12\,\text{mm} = 12\). 2. Each section takes \(15\) minutes to burn, so the total time is \(12 \times 15 = 180\) minutes. 3. Convert to hours: \(180 \div 60 = 3\) hours.

Answer

The candle will burn for \(3\) hours.
Rate problems with constant rates 170
A copier prints \(250\) flyers in exactly \(6\) minutes \(15\) seconds. At the same rate, how long will it take to print \(1000\) flyers? Give the answer in minutes.

Hints

- Determine the scale factor from \(250\) flyers to \(1000\) flyers. - Multiply the minutes and seconds by that factor. - Convert any complete group of \(60\) seconds to minutes.

Solution

1. The number of flyers is multiplied by \(1000 \div 250 = 4\). 2. Multiply the time by \(4\): \(4 \times 6\) minutes is \(24\) minutes, and \(4 \times 15\) seconds is \(60\) seconds. 3. Since \(60\) seconds is \(1\) minute, the total time is \(24 + 1 = 25\) minutes.

Answer

The copier will take \(25\) minutes.
Rate problems with constant rates 171
At a farmers market, \(16\,\text{oz}\) of strawberries cost \(\$3.20\). Leo has exactly \(\$1.20\). How many ounces of strawberries can he buy if the price is proportional to the weight?

Hints

- Find the cost of one ounce. - Divide the money available by the unit price. - Keep the relationship between price and weight proportional.

Solution

1. Find the unit price: \(\$3.20 \div 16 = \$0.20\) per ounce. 2. Divide Leo's money by the price per ounce: \(\$1.20 \div \$0.20 = 6\).

Answer

Leo can buy \(6\,\text{oz}\) of strawberries.
Rate problems with constant rates 172
Liam, Mia, and Tom buy glass craft beads together. All the beads have the same price per ounce. Liam takes \(12\,\text{oz}\), Mia takes \(20\,\text{oz}\), and Tom takes \(18\,\text{oz}\). Liam pays the full bill of \(\$12.50\). How much should Mia and Tom each repay Liam?

Hints

- Find the total weight of all the beads. - Divide the total cost by the total weight to find the unit price. - Multiply the unit price by each person's weight.

Solution

1. Find the total weight: \(12 + 20 + 18 = 50\,\text{oz}\). 2. Find the unit price: \(\$12.50 \div 50 = \$0.25\) per ounce. 3. Mia's share is \(20 \times \$0.25 = \$5.00\). 4. Tom's share is \(18 \times \$0.25 = \$4.50\).

Answer

Mia should repay \(\$5.00\), and Tom should repay \(\$4.50\).
Rate problems with constant rates 173
Liam wants to buy \(12\) boxes of pasta and \(4\) jars of tomato sauce. He compares prices at three stores: <table> <tr><th>Store</th><th>Pasta quantity</th><th>Pasta price</th><th>Sauce quantity</th><th>Sauce price</th></tr> <tr><td>Fresh Market</td><td>\(3\) boxes</td><td>\(\$2.40\)</td><td>\(2\) jars</td><td>\(\$2.50\)</td></tr> <tr><td>City Shop</td><td>\(4\) boxes</td><td>\(\$3.12\)</td><td>\(4\) jars</td><td>\(\$4.80\)</td></tr> <tr><td>Super Saver</td><td>\(6\) boxes</td><td>\(\$4.68\)</td><td>\(1\) jar</td><td>\(\$1.29\)</td></tr> </table> Find the total cost at each store. Which store is least expensive?

Hints

- Determine how many of each offered package are needed. - Find the pasta and sauce costs separately for each store. - Add each store's two costs and compare the totals.

Solution

1. Fresh Market: \(12 \div 3 = 4\) pasta offers, costing \(4 \times \$2.40 = \$9.60\). The sauce requires \(4 \div 2 = 2\) offers, costing \(2 \times \$2.50 = \$5.00\). The total is \(\$14.60\). 2. City Shop: \(12 \div 4 = 3\) pasta offers, costing \(3 \times \$3.12 = \$9.36\). The four jars of sauce cost \(\$4.80\). The total is \(\$14.16\). 3. Super Saver: \(12 \div 6 = 2\) pasta offers, costing \(2 \times \$4.68 = \$9.36\). Four jars of sauce cost \(4 \times \$1.29 = \$5.16\). The total is \(\$14.52\). 4. Since \(\$14.16 < \$14.52 < \$14.60\), City Shop is least expensive.

Answer

Fresh Market: \(\$14.60\) City Shop: \(\$14.16\) Super Saver: \(\$14.52\) City Shop is least expensive.
Rate problems with constant rates 174
For a class snack, Ms. Wilson needs \(15\) apples and \(5\) bags of oranges. She compares two offers: Offer 1: \(3\) apples cost \(\$1.20\), and one bag of oranges costs \(\$2.45\). Offer 2: \(5\) apples cost \(\$1.90\), and a value pack of \(5\) bags of oranges costs \(\$12.00\). Which offer is less expensive for the full purchase? Find the difference in total price.

Hints

- Calculate the full cost under each offer separately. - Determine how many apple groups are needed to make \(15\) apples. - Compare the two total costs.

Solution

1. Offer 1: Fifteen apples require \(15 \div 3 = 5\) groups, so they cost \(5 \times \$1.20 = \$6.00\). The oranges cost \(5 \times \$2.45 = \$12.25\). The total is \(\$18.25\). 2. Offer 2: Fifteen apples require \(15 \div 5 = 3\) groups, so they cost \(3 \times \$1.90 = \$5.70\). The oranges cost \(\$12.00\). The total is \(\$17.70\). 3. The difference is \(\$18.25 - \$17.70 = \$0.55\).

Answer

Offer 2 is less expensive. It costs \(\$17.70\), which is \(\$0.55\) less than Offer 1.
Rate problems with constant rates 175
Ms. Smith and Mr. Webb clean an office building. Ms. Smith works \(20\) hours per week, and Mr. Webb works \(15\) hours per week. They earn the same hourly wage. Together, they earn \(\$2240\) over \(4\) weeks. a) How much do they earn together in one week? b) What is their hourly wage? c) Mr. Webb claims that he earns more than \(\$1000\) in \(4\) weeks. Is he correct? Show a calculation.

Hints

- Divide the four-week total by \(4\). - Add the two weekly work times. - Divide weekly earnings by weekly hours to find the hourly wage. - Calculate Mr. Webb's four-week earnings separately.

Solution

1. Their combined weekly earnings are \(\$2240 \div 4 = \$560\). 2. Together, they work \(20 + 15 = 35\) hours per week. 3. Their hourly wage is \(\$560 \div 35 = \$16\) per hour. 4. Mr. Webb earns \(15 \times 4 \times \$16 = \$960\) over four weeks. Since \(\$960 < \$1000\), his claim is incorrect.

Answer

a) \(\$560\) b) \(\$16\) per hour c) No. Mr. Webb earns \(\$960\) in \(4\) weeks.
Rate problems with constant rates 176
A beekeeper uses \(15\,\text{kg}\) of honey to fill \(30\) jars. a) How many jars can be filled with \(45\,\text{kg}\) of honey at the same rate? b) How many kilograms of honey are needed to fill \(100\) jars?

Hints

- For part a, compare \(45\,\text{kg}\) with \(15\,\text{kg}\). - For part b, find the amount needed for a convenient number of jars, such as \(10\). - Keep the honey-to-jars ratio equivalent.

Solution

1. For part a, find the scale factor: \(45\,\text{kg} \div 15\,\text{kg} = 3\). 2. Scale the number of jars: \(30 \times 3 = 90\) jars. 3. For part b, find the honey needed for \(10\) jars: \(15\,\text{kg} \div 3 = 5\,\text{kg}\). 4. Scale from \(10\) jars to \(100\) jars: \(5\,\text{kg} \times 10 = 50\,\text{kg}\).

Answer

a) The beekeeper can fill \(90\) jars. b) The beekeeper needs \(50\,\text{kg}\) of honey.
Rate problems with constant rates 177
For trip planning, a ferry company estimates fuel use at \(6\,\text{L}\) per passenger for every \(100\,\text{km}\) traveled. A ferry carries \(850\) passengers on a \(120\,\text{km}\) trip. According to this estimate, how many liters of fuel will the trip use?

Hints

- Find the fuel estimate for one passenger over \(120\,\text{km}\). - Compare \(120\,\text{km}\) with the reference distance of \(100\,\text{km}\). - Multiply the per-passenger result by the number of passengers.

Solution

1. A \(120\)-kilometer trip is \(120 \div 100 = 1.2\) times the reference distance. 2. The fuel estimate per passenger is \(6\,\text{L} \times 1.2 = 7.2\,\text{L}\). 3. For \(850\) passengers, the total is \(850 \times 7.2\,\text{L} = 6120\,\text{L}\).

Answer

The estimated fuel use is \(6120\,\text{L}\).
Rate problems with constant rates 178
Ms. Miller walks \(1.2\,\text{km}\) in \(15\) minutes. Her dog runs \(450\,\text{m}\) across a field in \(5\) minutes. Which one travels more meters per minute? Show the two unit rates.

Hints

- Use the same distance unit for both rates. - Convert kilometers to meters. - Divide each distance by its time.

Solution

1. Convert Ms. Miller's distance: \(1.2\,\text{km} = 1200\,\text{m}\). Her unit rate is \(1200\,\text{m} \div 15\,\text{min} = 80\,\text{m/min}\). 2. The dog's unit rate is \(450\,\text{m} \div 5\,\text{min} = 90\,\text{m/min}\). 3. Since \(90 > 80\), the dog travels more meters per minute.

Answer

The dog travels farther per minute: \(90\,\text{m/min}\), compared with Ms. Miller's \(80\,\text{m/min}\).
Rate problems with constant rates 179
In a simplified observation, a hummingbird beats its wings \(80\) times per second. A mosquito makes \(15{,}000\) wingbeats in \(30\) seconds. a) Which animal makes more wingbeats per second? b) How many wingbeats would each make during a \(2\)-minute flight at these constant rates?

Hints

- Find the mosquito's wingbeats in one second. - Convert \(2\) minutes to seconds. - Multiply each per-second rate by the total number of seconds.

Solution

1. The mosquito's unit rate is \(15{,}000 \div 30 = 500\) wingbeats per second. Since \(500 > 80\), the mosquito has the greater rate. 2. Convert the flight time: \(2\) minutes is \(120\) seconds. 3. The hummingbird makes \(80 \times 120 = 9600\) wingbeats. 4. The mosquito makes \(500 \times 120 = 60{,}000\) wingbeats.

Answer

a) The mosquito, at \(500\) wingbeats per second compared with \(80\) for the hummingbird. b) Hummingbird: \(9600\) wingbeats; mosquito: \(60{,}000\) wingbeats.
Rate problems with constant rates 180
Two fans are tested. Model A makes \(1200\) revolutions in \(2\) minutes. Model B makes \(15\) revolutions per second. a) Which model rotates faster? Show a calculation using the same time unit. b) How many revolutions does each model make in one-half hour?

Hints

- Express both rates using minutes or both using seconds. - One minute has \(60\) seconds. - One-half hour has \(30\) minutes. - Use each per-minute rate to find the half-hour total.

Solution

1. Model A makes \(1200 \div 2 = 600\) revolutions per minute. Model B makes \(15 \times 60 = 900\) revolutions per minute. Therefore, Model B rotates faster. 2. One-half hour is \(30\) minutes. Model A makes \(600 \times 30 = 18{,}000\) revolutions. 3. Model B makes \(900 \times 30 = 27{,}000\) revolutions.

Answer

a) Model B rotates faster. b) Model A: \(18{,}000\) revolutions; Model B: \(27{,}000\) revolutions.
Rate problems with constant rates 181
The train and road routes from Lakeview to Riverton are both \(120\,\text{km}\) long. The train takes \(1\) hour \(15\) minutes. Ms. Miller drives there in \(1\) hour \(30\) minutes. a) How much time does the train save? b) What is Ms. Miller's average driving speed in kilometers per hour?

Hints

- Express both travel times in minutes before comparing them. - To find driving speed, first express the driving time in hours. - Divide the distance by the driving time.

Solution

1. The train saves \(15\) minutes: \(90-75=15\). 2. The drive takes \(1.5\) hours, so the average speed is \(120\div1.5=80\,\text{km/h}\).

Answer

a) \(15\) minutes b) \(80\,\text{km/h}\)
Rate problems with constant rates 182
A small fountain fills a collection tank. In \(15\) minutes, \(120\,\text{L}\) of water flows into the tank. a) What is the flow rate in liters per minute? b) The empty tank holds \(800\,\text{L}\). How long will it take to fill completely? Give the answer in hours and minutes.

Hints

- Divide the volume by the time to find the per-minute rate. - Divide the tank capacity by that rate. - Regroup every \(60\) minutes as one hour.

Solution

1. The flow rate is \(120\,\text{L} \div 15\,\text{min} = 8\,\text{L/min}\). 2. The filling time is \(800\,\text{L} \div 8\,\text{L/min} = 100\) minutes. 3. Since \(100\) minutes is \(1\) hour \(40\) minutes, that is the total filling time.

Answer

a) \(8\,\text{L/min}\) b) \(1\) hour \(40\) minutes
Rate problems with constant rates 183
Mr. Webb drives to work and back \(4\) days each week. His one-way commute is \(50\,\text{mi}\). His car averages \(25\,\text{mi}\) per gallon, and gasoline costs \(\$4.10\) per gallon. How much does he spend on gasoline for one week of commuting?

Hints

- Count both the trip to work and the trip home. - Find the total weekly mileage. - Divide miles by miles per gallon to find gallons used. - Multiply gallons by the price per gallon.

Solution

1. The daily round trip is \(50 \times 2 = 100\,\text{mi}\). 2. The weekly distance is \(100 \times 4 = 400\,\text{mi}\). 3. The car uses \(400 \div 25 = 16\) gallons. 4. The weekly cost is \(16 \times \$4.10 = \$65.60\).

Answer

Mr. Webb spends \(\$65.60\) on gasoline each week.
Rate problems with constant rates 184
A clock's second hand moves at a constant rate. It takes exactly \(5\,\text{s}\) to move from one number to the next, such as from \(12\) to \(1\). a) How many full revolutions does the second hand make from \(2{:}45\) p.m. to \(3{:}15\) p.m.? b) During that half hour, how much total time does the hand spend moving between \(12\) and \(1\)? Give the result in minutes and seconds.

Hints

- How much time passes between the two clock times? - How long does one full revolution of a second hand take? - What happens between \(12\) and \(1\) during each revolution? - How many seconds are in one minute?

Solution

1. The elapsed time from \(2{:}45\) p.m. to \(3{:}15\) p.m. is \(30\) minutes. 2. A second hand makes one full revolution each minute, so it makes \(30\) revolutions. 3. During each revolution, the hand spends \(5\) seconds between \(12\) and \(1\). The total is \(30 \times 5 = 150\) seconds. 4. Convert \(150\) seconds: \(150\,\text{s} = 2\,\text{min}\,30\,\text{s}\).

Answer

a) \(30\) revolutions b) \(2\,\text{min}\,30\,\text{s}\)
Rate problems with constant rates 185
At a grocery store, \(20\,\text{lb}\) of potatoes cost \(\$30\). Leo writes the following work to find the price of \(8\,\text{lb}\): \(20\,\text{lb}\) costs \(\$30\) \(2\,\text{lb}\) costs \(\$3\) \(8\,\text{lb}\) costs \(\$9\) Check his work. Where is the error, and what is the correct price for \(8\,\text{lb}\)?

Hints

- Determine the factor that changes \(2\) pounds to \(8\) pounds. - Apply the same factor to both quantities. - Recheck the multiplication on the price side.

Solution

1. The first scale-down step is correct: dividing both quantities by \(10\) gives \(2\,\text{lb}\) for \(\$3\). 2. To change \(2\,\text{lb}\) to \(8\,\text{lb}\), multiply by \(4\). The price must also be multiplied by \(4\). 3. Since \(\$3 \times 4 = \$12\), not \(\$9\), the correct price is \(\$12\).

Answer

The error is in the final step. Leo should calculate \(\$3 \times 4 = \$12\). Eight pounds of potatoes cost \(\$12\).
Rate problems with constant rates 186
A remote-controlled car travels \(12\,\text{m}\) in \(4\) minutes at a constant rate. Priya writes: \(12 \div 4 = 3\,\text{m/min}\) \(30\,\text{s} = 0.5\,\text{min}\) \(3 \div 0.5 = 6\,\text{m}\) Which line is the first incorrect step? Explain the rate error and give the correct distance traveled in \(30\) seconds.

Hints

- Check whether the unit rate itself is correct. - Compare \(30\) seconds with one full minute. - Ask whether traveling for less than one minute should produce more or less than the one-minute distance.

Solution

1. The unit rate \(12 \div 4 = 3\,\text{m/min}\) is correct. 2. The conversion \(30\,\text{s} = 0.5\,\text{min}\) is also correct. 3. The first incorrect step is \(3 \div 0.5\). A distance equals rate multiplied by time, so the distance is \(3 \times 0.5 = 1.5\,\text{m}\). 4. Dividing by \(0.5\) would make the distance larger even though the car travels for less than one minute, which is inconsistent with the rate.

Answer

The first incorrect step is \(3 \div 0.5 = 6\,\text{m}\). The correct distance is \(1.5\,\text{m}\).
Rate problems with constant rates 187
For a class party, \(12\,\text{qt}\) of juice cost \(\$18\). Mateo tries to find the cost of \(5\,\text{qt}\): \(12\,\text{qt}\) cost \(\$18\) \(1\,\text{qt}\) costs \(\$1.80\) \(5\,\text{qt}\) cost \(\$9\) Explain the error in the first calculation and find the correct cost of \(5\,\text{qt}\).

Hints

- Recalculate \(18 \div 12\). - Multiply the proposed unit price by \(12\) to check it. - Use the correct unit price to find the cost of \(5\) quarts.

Solution

1. The unit price must be found by dividing \(\$18\) by \(12\), not by \(10\). 2. The correct unit price is \(\$18 \div 12 = \$1.50\) per quart. 3. Five quarts cost \(5 \times \$1.50 = \$7.50\).

Answer

Mateo divided incorrectly. The unit price is \(\$1.50\) per quart, so \(5\,\text{qt}\) cost \(\$7.50\).
Rate problems with constant rates 188
At a farm stand, \(5\) pounds of potatoes cost \(\$10\). A customer wants to buy \(8\) pounds and says, “I will have to pay \(\$18\).” Is the customer correct? Explain with a calculation.

Hints

- Find the price of one pound first. - Use the unit price to find the cost of \(8\) pounds. - Compare your result with the customer's claim.

Solution

1. Find the unit price: \(\$10 \div 5 = \$2\) per pound. 2. Find the price of \(8\) pounds: \(8 \times \$2 = \$16\). 3. Since \(\$16 \ne \$18\), the customer is not correct.

Answer

No. The customer is not correct because \(8\) pounds of potatoes cost \(\$16\).
Rate problems with constant rates 189
A class takes a three-day hiking trip. Each day, the class hikes for \(4\) hours in the morning and \(2\) hours in the afternoon at an average rate of \(3\,\text{mi/h}\). How many miles does the class hike in all?

Hints

- How many hours does the class hike each day? - Use the hourly rate to find one day’s distance. - Then find the distance for all \(3\) days.

Solution

1. Find the hiking time each day: \(4 + 2 = 6\) hours. 2. Find the distance each day: \(6 \times 3 = 18\,\text{mi}\). 3. Find the total distance for \(3\) days: \(3 \times 18 = 54\,\text{mi}\).

Answer

The class hikes \(54\,\text{mi}\) in all.
Rate problems with constant rates 190
Lukas delivers mail by electric bicycle for \(5\) hours per day from Monday through Friday. He rides \(12\,\text{mi}\) each hour. Sarah delivers mail for \(7\) hours per day on \(4\) days and rides \(10\,\text{mi}\) each hour. Who travels farther during the week?

Hints

- Find Lukas’s total weekly distance. - Find Sarah’s total weekly distance. - Compare the two distances. - Be careful to use the correct number of workdays for each person.

Solution

1. Find Lukas’s total riding time: \(5 \times 5 = 25\) hours. 2. Find Lukas’s distance: \(25 \times 12 = 300\,\text{mi}\). 3. Find Sarah’s total riding time: \(4 \times 7 = 28\) hours. 4. Find Sarah’s distance: \(28 \times 10 = 280\,\text{mi}\). 5. Since \(300 > 280\), Lukas travels farther.

Answer

Lukas travels farther during the week.
Rate problems with constant rates 191
Ms. Miller buys two pieces of fabric from the same bolt at the same price per yard. One piece is \(2\,\text{yd}\) long, and the other is \(6\,\text{yd}\) long. She pays \(\$72\) in all. What is the cost of each piece?

Hints

- Find the total number of yards purchased. - Use the total cost and total length to find the price per yard. - Use the unit price to find the cost of each piece.

Solution

1. Find the total length: \(2 + 6 = 8\,\text{yd}\). 2. Find the price per yard: \(\$72 \div 8 = \$9\) per yard. 3. Find the cost of the \(2\,\text{yd}\) piece: \(2 \times \$9 = \$18\). 4. Find the cost of the \(6\,\text{yd}\) piece: \(6 \times \$9 = \$54\).

Answer

The \(2\,\text{yd}\) piece costs \(\$18\), and the \(6\,\text{yd}\) piece costs \(\$54\).
Rate problems with constant rates 192
A remote-controlled car moves at a constant rate. Its wheels turn \(10\) times while the car travels \(4\,\text{m}\). A full battery powers \(150\) wheel rotations. How far can the car travel on one full battery?

Hints

- Determine how many groups of \(10\) rotations fit into \(150\) rotations. - Each group of rotations corresponds to the same distance. - Multiply the number of groups by \(4\,\text{m}\).

Solution

1. Find how many groups of \(10\) rotations are in \(150\) rotations: \(150 \div 10 = 15\). 2. Each group corresponds to \(4\,\text{m}\), so the distance is \(15 \times 4\,\text{m} = 60\,\text{m}\).

Answer

The car can travel \(60\,\text{m}\).
Rate problems with constant rates 193
Lucas and Emilia walk together along a \(120\,\text{m}\) path. Lucas takes \(15\) steps for every \(10\,\text{m}\), while Emilia takes \(18\) steps for every \(10\,\text{m}\). How many more steps does Emilia take over the full path?

Hints

- Find how many \(10\)-meter sections are in the full path. - Use each student's constant number of steps per section. - Compare the two totals.

Solution

1. Find the number of \(10\)-meter sections: \(120\,\text{m} \div 10\,\text{m} = 12\). 2. Lucas takes \(12 \times 15 = 180\) steps. 3. Emilia takes \(12 \times 18 = 216\) steps. 4. Find the difference: \(216 - 180 = 36\) steps.

Answer

Emilia takes \(36\) more steps than Lucas.
Rate problems with constant rates 194
A school copier prints \(50\) worksheets in \(3\) minutes at a constant rate. a) How many worksheets does it print in \(15\) minutes? b) How many minutes does it take to print \(300\) worksheets?

Hints

- For part a, determine how many \(3\)-minute intervals fit into \(15\) minutes. - For part b, determine how many groups of \(50\) worksheets make \(300\). - Use the constant rate for each equal interval.

Solution

1. For part a, find how many \(3\)-minute intervals are in \(15\) minutes: \(15 \div 3 = 5\). Then \(5 \times 50 = 250\), so the copier prints \(250\) worksheets. 2. For part b, find how many groups of \(50\) worksheets are in \(300\): \(300 \div 50 = 6\). Then \(6 \times 3 = 18\), so printing \(300\) worksheets takes \(18\) minutes.

Answer

a) The copier prints \(250\) worksheets. b) It takes \(18\) minutes.
Rate problems with constant rates 195
A class plans a bicycle trip to a lake \(60\,\text{mi}\) away. The group rides \(12\,\text{mi}\) each hour. a) How far has the group traveled after \(2\) hours? b) The group takes a long break after \(3\) hours. How far is it from the lake then? c) How many hours of riding time are needed for the entire trip?

Hints

- Use the distance traveled in one hour. - Multiply the hourly distance by the number of hours. - Subtract the distance already traveled from the total distance. - Determine how many hourly distances fit into the full trip.

Solution

1. Find the distance after \(2\) hours: \(12 \times 2 = 24\,\text{mi}\). 2. Find the distance after \(3\) hours: \(12 \times 3 = 36\,\text{mi}\). 3. Find the remaining distance: \(60 - 36 = 24\,\text{mi}\). 4. Find the total riding time: \(60 \div 12 = 5\) hours.

Answer

a) The group has traveled \(24\,\text{mi}\). b) The group is \(24\,\text{mi}\) from the lake. c) The entire trip requires \(5\) hours of riding.
Rate problems with constant rates 196
A truck is delivering a load to a city \(260\,\text{mi}\) away. It travels at \(65\,\text{mi/h}\). a) How far has the truck traveled after \(3\) hours? b) After how many hours has the truck traveled exactly half the total distance? c) A faster delivery van travels the same route at \(80\,\text{mi/h}\). How far is the van from the city after \(3\) hours?

Hints

- Find half of the total distance. - Work with each vehicle separately. - Use multiplication to find a distance after a given time. - Use division to find how many hourly distances fit into half the trip.

Solution

1. Find the truck’s distance after \(3\) hours: \(65 \times 3 = 195\,\text{mi}\). 2. Find half the total distance: \(260 \div 2 = 130\,\text{mi}\). 3. Find the time to travel half the distance: \(130 \div 65 = 2\) hours. 4. Find the van’s distance after \(3\) hours: \(80 \times 3 = 240\,\text{mi}\). 5. Find the van’s remaining distance: \(260 - 240 = 20\,\text{mi}\).

Answer

a) The truck has traveled \(195\,\text{mi}\). b) It reaches halfway after \(2\) hours. c) The van is \(20\,\text{mi}\) from the city.
Rate problems with constant rates 197
Two farms compare strawberry yields. Sunny Acres harvests \(480\,\text{lb}\) from \(12{,}000\,\text{ft}^2\). Fresh Fields harvests \(900\,\text{lb}\) from \(20{,}000\,\text{ft}^2\). Which farm has the greater yield per \(1000\,\text{ft}^2\)? Show both rates.

Hints

- Express both yields using the same area unit. - Determine how many groups of \(1000\,\text{ft}^2\) each farm uses. - Divide each harvest by its number of area groups.

Solution

1. Sunny Acres produces \(480 \div 12 = 40\,\text{lb}\) per \(1000\,\text{ft}^2\). 2. Fresh Fields produces \(900 \div 20 = 45\,\text{lb}\) per \(1000\,\text{ft}^2\). 3. Since \(45 > 40\), Fresh Fields has the greater yield.

Answer

Fresh Fields has the greater yield: \(45\,\text{lb}\) per \(1000\,\text{ft}^2\), compared with \(40\,\text{lb}\) per \(1000\,\text{ft}^2\) at Sunny Acres.
Rate problems with constant rates 198
A seller offers three rectangular garden plots: 1) \(40\,\text{ft} \times 50\,\text{ft}\) for \(\$12{,}000\) 2) \(60\,\text{ft} \times 30\,\text{ft}\) for \(\$9000\) 3) \(25\,\text{ft} \times 80\,\text{ft}\) for \(\$11{,}000\) Find the price per square foot for each plot. Order the plots from least expensive to most expensive by unit price.

Hints

- Find each rectangular area first. - Divide each total price by its area. - Compare the three unit prices.

Solution

1. Plot 1 has area \(40 \times 50 = 2000\,\text{ft}^2\), so its unit price is \(\$12{,}000 \div 2000 = \$6.00\) per square foot. 2. Plot 2 has area \(60 \times 30 = 1800\,\text{ft}^2\), so its unit price is \(\$9000 \div 1800 = \$5.00\) per square foot. 3. Plot 3 has area \(25 \times 80 = 2000\,\text{ft}^2\), so its unit price is \(\$11{,}000 \div 2000 = \$5.50\) per square foot. 4. Since \(\$5.00 < \$5.50 < \$6.00\), the order is Plot 2, Plot 3, Plot 1.

Answer

Plot 1: \(\$6.00/\text{ft}^2\) Plot 2: \(\$5.00/\text{ft}^2\) Plot 3: \(\$5.50/\text{ft}^2\) Order: Plot 2, Plot 3, Plot 1
Rate problems with constant rates 199
A print shop's copier prints \(p\) pages in \(t\) minutes at a steady rate. Both numbers are positive. a) Write its printing rate in pages per minute. b) Write an expression for the number of minutes it takes to print \(x\) pages at the same rate. c) The copier prints \(180\) pages in \(4.5\) minutes. Find its printing rate. d) At that rate, how long will it take to print \(540\) pages?

Hints

- What do pages per minute and minutes per page tell you about the same printing job? - If the copier prints twice as many pages at the same rate, what happens to the time? - Check that the units in each expression match what is asked.

Solution

1. Divide pages by minutes to find the rate: \(\frac{p}{t}\) pages per minute. 2. Each page takes \(\frac{t}{p}\) minutes. For \(x\) pages, the time is \(\frac{xt}{p}\) minutes. 3. The given numerical rate is \(180\div 4.5=40\) pages per minute. 4. Printing \(540\) pages takes \(540\div 40=13.5\) minutes.

Answer

a) \(\frac{p}{t}\) pages per minute b) \(\frac{xt}{p}\) minutes c) \(40\) pages per minute d) \(13.5\) minutes
Rate problems with constant rates 200
The number line begins at \(0\) and is divided into \(20\) equal intervals. The endpoint is labeled “?”. Point \(A\) is marked on the number line. Find the value at “?” in each case. a) Point \(A\) represents \(0.8\). b) Point \(A\) represents \(0.08\). c) Point \(A\) represents \(\frac{2}{5}\).
Figure for problem 531830

Hints

- Use the image to compare the number of intervals from \(0\) to \(A\) with the total number of intervals. - Determine how many times the distance from \(0\) to \(A\) fits in the whole scale. - Multiply the value at \(A\) by that scale factor.

Solution

1. From the image, point \(A\) is \(4\) of the \(20\) equal intervals from \(0\) to the endpoint. Thus, the whole scale is \(20 \div 4 = 5\) times the value at \(A\). 2. For a), the endpoint is \(5 \times 0.8 = 4\). 3. For b), the endpoint is \(5 \times 0.08 = 0.4\). 4. For c), the endpoint is \(5 \times \frac{2}{5} = 2\).

Answer

a) \(4\) b) \(0.4\) c) \(2\)
Rate problems with constant rates 201
The diagram shows the dimensions of a wall that needs wallpaper. You can choose between two types of rolls: <table><tr><th>Wallpaper</th><th>Coverage per roll</th><th>Price per roll</th></tr><tr><td>Type A</td><td>\(50\,\text{ft}^2\)</td><td>\(\$12.50\)</td></tr><tr><td>Type B</td><td>\(75\,\text{ft}^2\)</td><td>\(\$18.00\)</td></tr></table> Find the total cost for each wallpaper type. Which type costs less? Assume each roll must be purchased whole.
Figure for problem 535701

Hints

- Read the wall dimensions from the diagram. - Find the wall’s area first. - Divide the wall area by each roll’s coverage. - Multiply the number of rolls by the price per roll.

Solution

1. Read the wall dimensions from the diagram: \(15\,\text{ft}\) by \(10\,\text{ft}\). 2. The wall area is \(15\,\text{ft} \times 10\,\text{ft} = 150\,\text{ft}^2\). 3. For Type A, the number of rolls is \(150 \div 50 = 3\). The cost is \(3 \times \$12.50 = \$37.50\). 4. For Type B, the number of rolls is \(150 \div 75 = 2\). The cost is \(2 \times \$18.00 = \$36.00\). 5. Since \(\$36.00 < \$37.50\), Type B costs less.

Answer

Type A costs \(\$37.50\). Type B costs \(\$36.00\). Type B costs less.
Rate problems with constant rates 202
The graph shows the number of pages a printer produces at a constant rate. a) What is the unit rate in pages per minute? b) At the same rate, how many pages will the printer produce in \(6\) minutes? c) How many minutes will it take to produce \(150\) pages?
Figure for problem 551204

Hints

- Read one convenient time-and-pages point from the graph. - Use that point to determine how many pages correspond to \(1\) minute. - Use the same constant rate both forward and backward.

Solution

1. From the graph, the printer produces \(120\) pages in \(4\) minutes, so its unit rate is \(120 \div 4 = 30\) pages per minute. 2. In \(6\) minutes, it produces \(30 \times 6 = 180\) pages. 3. To produce \(150\) pages, the time is \(150 \div 30 = 5\) minutes.

Answer

a) \(30\) pages per minute b) \(180\) pages c) \(5\) minutes
Rate problems with constant rates 203
Lines a and b show the distances traveled by two hikers moving at constant speeds. Time is measured in hours and distance in miles. a) Find each hiker’s unit rate. b) Which hiker is faster, and by how many miles per hour?
Figure for problem 554205

Hints

- Choose a clear point on each line whose coordinates you can read exactly. - For each line, compare distance with time in miles per hour. - After finding both unit rates, compare them and then find their difference.

Solution

1. Line a passes through \((4,20)\), so its unit rate is \(20 \div 4=5\,\text{mi/h}\). 2. Line b passes through \((4,24)\), so its unit rate is \(24 \div 4=6\,\text{mi/h}\). 3. Since \(6>5\), hiker b is faster. 4. The speed difference is \(6-5=1\,\text{mi/h}\).

Answer

a) Hiker a: \(5\,\text{mi/h}\); hiker b: \(6\,\text{mi/h}\) b) Hiker b is faster by \(1\,\text{mi/h}\).
Rate problems with constant rates 204
A pump is described by the table. A second pump is described by the graph. <table> <tr><th>Time (minutes)</th><th>Water added (gallons)</th></tr> <tr><td>\(2\)</td><td>\(50\)</td></tr> <tr><td>\(5\)</td><td>\(125\)</td></tr> </table> a) Find the unit rate represented by the table. b) The graph has a marked point P. Read the coordinates of P and use them to find the graph's unit rate. c) Do the table and graph represent the same constant rate? Explain.
Figure for problem 554206

Hints

- Find gallons per one minute from a nonzero table row. - Read point P using the graph's x-axis for time and y-axis for water. - Compare the two unit rates, not just the sizes of the displayed numbers.

Solution

1. The table gives \(50 \div 2 = 25\,\text{gal/min}\), which agrees with \(125 \div 5 = 25\,\text{gal/min}\). 2. The graph's marked point is \(P=(4,120)\), so its rate is \(120 \div 4 = 30\,\text{gal/min}\). 3. The rates are different, \(25\,\text{gal/min}\) and \(30\,\text{gal/min}\), so the table and graph do not represent the same constant-rate relationship.

Answer

a) \(25\,\text{gal/min}\) b) \(P=(4,120)\), so \(30\,\text{gal/min}\) c) No. The table shows \(25\,\text{gal/min}\), while the graph shows \(30\,\text{gal/min}\).
Rate problems with constant rates 205
A machine fills bottles at a constant rate. <table> <tr><th>Time (minutes)</th><th>Bottles filled</th></tr> <tr><td>0</td><td>0</td></tr> <tr><td>2</td><td>10</td></tr> <tr><td>4</td><td>20</td></tr> <tr><td>6</td><td>30</td></tr> </table> Talia predicts that the machine will fill \(32\) bottles in \(8\) minutes because both the time and bottle count should go up by \(2\). Explain why this reasoning is incorrect and find the correct number of bottles.

Hints

- Look for what stays the same across the nonzero rows of the table. - Compare bottles per minute rather than adding the same number to quantities with different units. - Use the constant multiplicative relationship to extend the table to \(8\) minutes.

Solution

1. The table shows a multiplicative constant rate: \(10 \div 2=5\) bottles per minute. 2. Talia used the same additive change for quantities with different units. The table does not say that adding \(2\) minutes means adding \(2\) bottles. 3. At the constant rate, \(8\) minutes gives \(5 \times 8=40\) bottles.

Answer

Talia’s additive reasoning is incorrect. The constant rate is \(5\) bottles per minute, so the machine fills \(40\) bottles in \(8\) minutes.
Rate problems with constant rates 206
A team needs at least \(10\) water bottles. A store sells only \(4\)-packs for \(\$6\) and \(7\)-packs for \(\$11\). Which whole packs should the team buy to pay the least? How much will they cost?

Hints

- All packs must be bought whole; extra bottles are allowed. - Try a few ways to reach at least the needed number of bottles. - A pack with a lower price per bottle is not always the cheapest choice for a small order.

Solution

1. Three \(4\)-packs provide \(12\) bottles for \(3\times\$6=\$18\). 2. Two \(7\)-packs provide \(14\) bottles for \(2\times\$11=\$22\). 3. One of each provides \(11\) bottles for \(\$6+\$11=\$17\). Using more packs cannot beat this because each pack costs at least \(\$6\). 4. The cheapest valid purchase is one \(4\)-pack and one \(7\)-pack.

Answer

One \(4\)-pack and one \(7\)-pack, costing \(\$17\)
Rate problems with constant rates 207
On a number line, point \(A=\frac{5}{6}\) and point \(B=1\frac{1}{4}\) are exactly \(5\,\text{cm}\) apart. a) How long is the interval from \(0\) to \(1\) on this number line? b) What number is located exactly \(2\,\text{cm}\) to the right of point \(B\)? Give the result as a simplified fraction or mixed number.

Hints

- Find the numerical distance between points \(A\) and \(B\). - Use the relationship between that numerical distance and \(5\,\text{cm}\) to find the length of one whole unit. - Convert \(2\,\text{cm}\) back to a numerical amount before adding it to \(B\).

Solution

1. Find the numerical distance: \(1\frac{1}{4}-\frac{5}{6}=\frac{5}{4}-\frac{5}{6}=\frac{15}{12}-\frac{10}{12}=\frac{5}{12}\). 2. A numerical distance of \(\frac{5}{12}\) corresponds to \(5\,\text{cm}\), so \(\frac{1}{12}\) corresponds to \(1\,\text{cm}\). Therefore, one whole unit corresponds to \(12\,\text{cm}\). 3. A distance of \(2\,\text{cm}\) corresponds to \(\frac{2}{12}=\frac{1}{6}\). 4. Add this value to point \(B\): \(\frac{5}{4}+\frac{1}{6}=\frac{15}{12}+\frac{2}{12}=\frac{17}{12}=1\frac{5}{12}\).

Answer

a) \(12\,\text{cm}\) b) \(\frac{17}{12}=1\frac{5}{12}\)
Rate problems with constant rates 208
A cyclist rides at a constant speed of \(23.4\,\text{km/h}\). A runner covers \(400\,\text{m}\) in \(1\) minute \(15\) seconds. Use \(1\,\text{m/s}=3.6\,\text{km/h}\). a) How many meters does the cyclist travel per second? b) Who travels farther in \(10\) minutes? Find each distance in meters.

Hints

- Express both travel times in the same units. - The given conversion lets you compare the cyclist's speed with distances measured in meters. - For the runner, how many complete \(75\)-second intervals fit in ten minutes?

Solution

1. The cyclist's speed is \(23.4\div3.6=6.5\,\text{m/s}\). 2. Ten minutes is \(600\) seconds, so the cyclist travels \(6.5\times600=3900\,\text{m}\). 3. The runner needs \(75\) seconds per \(400\,\text{m}\). In \(600\) seconds, the runner covers \(600\div75=8\) such intervals, or \(8\times400=3200\,\text{m}\). 4. The cyclist travels farther because \(3900>3200\).

Answer

a) \(6.5\,\text{m/s}\) b) The cyclist: \(3900\,\text{m}\) compared with \(3200\,\text{m}\).
Rate problems with constant rates 209
A school break begins at \(9{:}35\) a.m. and ends at \(9{:}55\) a.m. a) Through what angle does the minute hand move during the break? b) Through what angle does the hour hand move during the same time? c) Find the smaller angle between the two hands at exactly \(9{:}55\) a.m.

Hints

- How many minutes does the break last? - Use the number of degrees each hand moves per minute. - For part c, find each hand’s position measured clockwise from 12.

Solution

1. The break lasts \(55 - 35 = 20\) minutes. 2. The minute hand moves \(6^\circ\) per minute, so it sweeps through \(20 \times 6^\circ = 120^\circ\). 3. The hour hand moves \(0.5^\circ\) per minute, so it sweeps through \(20 \times 0.5^\circ = 10^\circ\). 4. At \(9{:}55\), the minute hand is \(55 \times 6^\circ = 330^\circ\) clockwise from 12. 5. The hour hand is \(9 \times 30^\circ + 55 \times 0.5^\circ = 297.5^\circ\) clockwise from 12. 6. The smaller angle between the hands is \(330^\circ - 297.5^\circ = 32.5^\circ\).

Answer

a) \(120^\circ\) b) \(10^\circ\) c) \(32.5^\circ\)
Rate problems with constant rates 210
A faulty toilet leaks about \(0.5\) gallon of water per minute. Water costs \(\$6.00\) per \(1000\) gallons. a) How many gallons are wasted in one full day? b) What does the leak cost per day? c) After how many complete days will the accumulated cost first exceed the \(\$18.50\) price of a replacement part?

Hints

- Find the number of minutes in one day. - Use the leak rate to find the daily volume. - Compare multiples of the daily cost with the replacement-part price.

Solution

1. One day has \(24\times 60=1440\) minutes. 2. The daily waste is \(0.5\times 1440=720\) gallons. 3. The daily cost is \(\frac{720}{1000}\times 6.00=\$4.32\). 4. After \(4\) days, the cost is \(4\times 4.32=\$17.28\), which is below \(\$18.50\). After \(5\) days, it is \(5\times 4.32=\$21.60\), which exceeds \(\$18.50\).

Answer

a) \(720\) gallons per day. b) \(\$4.32\) per day. c) After \(5\) complete days.
Rate problems with constant rates 211
A passenger train travels at a constant speed of \(60\,\text{mi/h}\). Use \(1\,\text{mi}=5280\,\text{ft}\) and \(1\,\text{h}=3600\,\text{s}\). a) Find the train's speed in feet per second. b) The train is \(440\,\text{ft}\) long. How many seconds does it take to completely cross a bridge that is \(1320\,\text{ft}\) long? The crossing is complete when the back of the train leaves the bridge. c) How many miles does a passenger travel in \(12\) minutes?

Hints

- Convert the given speed to the length and time units requested in part a). - Picture the front and back of the train when it has just cleared the bridge. - Express the trip in part c) as a fraction of an hour.

Solution

1. Convert the speed: \(60\times\frac{5280}{3600}=88\,\text{ft/s}\). 2. The front of the train must cover \(1320+440=1760\,\text{ft}\) before the back clears the bridge. 3. The crossing time is \(1760\div88=20\,\text{s}\). 4. Twelve minutes is \(12\div60=0.2\) hour, so the passenger travels \(60\times0.2=12\) miles.

Answer

a) \(88\,\text{ft/s}\) b) \(20\) seconds c) \(12\) miles
Rate problems with constant rates 212
A pipe leaks about \(0.2\,\text{gal}\) of water every \(5\) minutes. a) How many gallons of water leak out in one week? b) An empty \(3\)-gallon bucket is placed under the leak at the beginning of a day. The bucket must be empty again at the end of the day. How many times must the bucket be emptied during that day so that it never overflows? Justify your answer.

Hints

- First find the number of gallons lost each minute. - How long does it take the leak to fill a \(3\)-gallon bucket? - Remember that the bucket must also be empty at the end of the day.

Solution

1. The leak rate is \(0.2 \div 5=0.04\,\text{gal/min}\). 2. One week has \(7 \times 24 \times 60=10{,}080\) minutes. The weekly loss is \(10{,}080 \times 0.04=403.2\) gallons. 3. In one day, the leak produces \(24 \times 60 \times 0.04=57.6\) gallons. 4. The bucket fills every \(3 \div 0.04=75\) minutes. It becomes completely full \(19\) times during the day, accounting for \(19 \times 3=57\) gallons. 5. At the end of the day, \(57.6-57=0.6\) gallon remains. Including the required final emptying, the bucket must be emptied \(20\) times.

Answer

a) \(403.2\) gallons b) The bucket must be emptied \(20\) times: \(19\) times when full and once more at the end of the day.
Rate problems with constant rates 213
Harvesting a strawberry field by hand used to take \(165\) minutes, and sorting the berries took another \(75\) minutes. A new machine completes both jobs in \(40\) minutes. a) How long did both jobs take before the machine was used? b) How many minutes does the machine save? c) How many times as great as the manual work rate is the machine’s work rate?

Hints

- Add the two original work times first. - To find time saved, subtract the new time from the original time. - For the same job, divide the longer time by the shorter time to compare the work rates.

Solution

1. Find the original total time: \(165 + 75 = 240\) minutes. 2. Find the time saved: \(240 - 40 = 200\) minutes. 3. For the same amount of work, compare the times: \(240 \div 40 = 6\). The machine works at \(6\) times the manual rate.

Answer

a) The work originally took \(240\) minutes. b) The machine saves \(200\) minutes. c) The machine works at \(6\) times the manual rate.
Rate problems with constant rates 214
A climber is at a summit \(4100\,\text{m}\) above sea level, where the temperature is \(-14\,^{\circ}\text{C}\). The climber descends to a shelter where it is \(-6\,^{\circ}\text{C}\). In a simplified model, the temperature rises by \(1\,^{\circ}\text{C}\) for every \(250\,\text{m}\) of descent. What is the shelter's elevation?

Hints

- Compare how far each temperature is below zero. - How much vertical distance corresponds to that temperature change? - Descending lowers the elevation.

Solution

1. Moving from \(14\) degrees below zero to \(6\) degrees below zero is a rise of \(14-6=8\,^{\circ}\text{C}\). 2. The climber descends \(8\times250=2000\,\text{m}\). 3. The shelter is at \(4100-2000=2100\,\text{m}\) above sea level.

Answer

\(2100\,\text{m}\) above sea level
Rate problems with constant rates 215
A school trip includes \(88\) students and \(4\) teachers. Two bus sizes are available: - Bus A: \(20\) seats for \(\$130\) - Bus B: \(30\) seats for \(\$180\) a) Sari claims, “The larger bus costs less per seat.” Check the claim. b) Find the least expensive combination of buses that can carry all \(92\) people. Give the total cost.

Hints

- Find the total number of passengers. - Divide each bus price by its number of seats to compare unit rates. - Test combinations with enough total seats. - Compare the costs of all valid combinations.

Solution

1. There are \(88+4=92\) people. 2. For a), Bus A costs \(\$130\div20=\$6.50\) per seat. Bus B costs \(\$180\div30=\$6.00\) per seat. Sari's claim is correct. 3. For b), for each possible number of Bus B buses from \(0\) through \(4\), use the fewest Bus A buses that provide at least \(92\) seats: five Bus A buses cost \(\$650\); four Bus A buses and one Bus B bus cost \(\$700\); two buses of each type cost \(\$620\); one Bus A bus and three Bus B buses cost \(\$670\); and four Bus B buses cost \(\$720\). 4. The least expensive valid combination is two buses of each type for \(\$620\).

Answer

a) Yes. Bus B costs \(\$6.00\) per seat, compared with \(\$6.50\) for Bus A. b) Use \(2\) buses of each type. The total cost is \(\$620\).
Rate problems with constant rates 216
A school festival needs at least \(115\) apples. A store offers two packages: - Bag A: \(15\) apples for \(\$3.00\) - Bag B: \(25\) apples for \(\$4.00\) a) Which bag has the lower cost per apple? Explain. b) What combination of bags gives at least \(115\) apples for the least total cost? Find the cost.

Hints

- Divide each package price by its number of apples. - A lower unit rate does not automatically mean using only that package gives the least total cost. - Test combinations that reach at least \(115\) apples and compare their costs.

Solution

1. a) Bag A costs \(\$3.00\div15=\$0.20\) per apple. Bag B costs \(\$4.00\div25=\$0.16\) per apple, so Bag B has the lower unit rate. 2. b) For \(0,1,2,3,4\), or \(5\) Bag B packages, the least-cost valid combinations have total costs of \(\$24.00, \$22.00, \$23.00, \$21.00, \$19.00\), and \(\$20.00\), respectively. 3. The \(\$19.00\) combination uses \(4\) Bag B packages and \(1\) Bag A package. It gives \(4\times25+15=115\) apples. 4. Therefore, buy \(4\) Bag B packages and \(1\) Bag A package.

Answer

a) Bag B, at \(\$0.16\) per apple b) Buy \(4\) Bag B packages and \(1\) Bag A package for \(\$19.00\).
Rate problems with constant rates 217
A scout group of \(54\) people is planning an overnight trip. Two types of tents are available: - Small tent: sleeps \(4\) people and costs \(\$15\) per night. - Large tent: sleeps \(6\) people and costs \(\$20\) per night. a) Show which tent type has the lower cost per sleeping space. b) Find the least expensive combination of tents that provides space for all \(54\) people. What is the total cost?

Hints

- Compare costs for the same number of sleeping spaces. - Check whether one tent type can provide exactly \(54\) spaces. - Compare any mixed arrangement with an arrangement using the lower-rate tent.

Solution

1. a) Three small tents provide \(12\) spaces and cost \(3\times\$15=\$45\). Two large tents also provide \(12\) spaces and cost \(2\times\$20=\$40\). Therefore, large tents have the lower cost per space. 2. b) Nine large tents provide exactly \(9\times6=54\) spaces and cost \(9\times\$20=\$180\). 3. Replacing two large tents with three small tents keeps the same number of spaces but increases the cost by \(\$5\). Other mixtures either cost more or require extra unused space. Therefore, nine large tents give the least cost.

Answer

a) Large tents have the lower cost per sleeping space. b) Rent \(9\) large tents for \(\$180\).
Rate problems with constant rates 218
A gardener buys two bags of the same potting soil at the same price per pound. The first bag costs \(\$10.50\), and the second bag costs \(\$12.00\). Together, the bags weigh \(75\,\text{lb}\). a) How many pounds does each bag weigh? b) A third bag of the same soil weighs \(50\,\text{lb}\). How much does it cost?

Hints

- First find the combined price for the known combined weight. - What is the price of \(1\,\text{lb}\) of soil? - Use the unit price to find each of the first two weights. - Use the same unit price for the third bag.

Solution

1. Find the combined cost of the first two bags: \(\$10.50 + \$12.00 = \$22.50\). 2. Express the total in cents: \(\$22.50 = 2250\) cents. 3. Find the price per pound: \(2250 \div 75 = 30\) cents per pound. 4. Find the first bag’s weight: \(1050 \div 30 = 35\,\text{lb}\). 5. Find the second bag’s weight: \(1200 \div 30 = 40\,\text{lb}\). 6. Find the third bag’s cost: \(50 \times 30 = 1500\) cents, or \(\$15.00\).

Answer

a) The first bag weighs \(35\,\text{lb}\), and the second bag weighs \(40\,\text{lb}\). b) The third bag costs \(\$15.00\).
Rate problems with constant rates 219
An organic farm sells \(10\,\text{lb}\) bags of potatoes at the same price per bag. The farm earns \(\$180\) from morning sales and \(\$264\) from afternoon sales. It sells exactly \(7\) more bags in the afternoon than in the morning. How many pounds of potatoes are sold that day?

Hints

- Use the difference in revenue and the difference in the number of bags to find one bag’s price. - Find the number of bags sold in each part of the day. - Add the numbers of bags. - Convert the total number of bags to a total weight.

Solution

1. Find the difference in sales revenue: \(\$264 - \$180 = \$84\). 2. Since the difference represents \(7\) bags, find the price per bag: \(\$84 \div 7 = \$12\). 3. Find the number of bags sold in the morning: \(\$180 \div \$12 = 15\) bags. 4. Find the number sold in the afternoon: \(\$264 \div \$12 = 22\) bags. 5. Find the total number of bags: \(15 + 22 = 37\). 6. Find the total weight: \(37 \times 10 = 370\,\text{lb}\).

Answer

The farm sells \(370\,\text{lb}\) of potatoes that day.
Rate problems with constant rates 220
Two hiking groups walk a combined distance of \(46\,\text{mi}\). Group B walks \(4\,\text{mi}\) farther than Group A. Group A walks at \(3\,\text{mi/h}\), and Group B walks at \(5\,\text{mi/h}\). How many hours does each group walk?

Hints

- First determine the distance each group walks. - Subtract Group B’s extra distance from the combined distance; the remainder can be split equally. - Use distance and speed to find each group’s time.

Solution

1. Find Group A’s distance by removing the \(4\,\text{mi}\) difference and dividing the remainder equally: \((46 - 4) \div 2 = 21\,\text{mi}\). 2. Find Group B’s distance: \(21 + 4 = 25\,\text{mi}\). 3. Find Group A’s time: \(21 \div 3 = 7\) hours. 4. Find Group B’s time: \(25 \div 5 = 5\) hours.

Answer

Group A walks for \(7\) hours, and Group B walks for \(5\) hours.
Rate problems with constant rates 221
Two trucks carry a combined total of \(1260\) tons of sand. Yellow carries \(60\) tons more than Blue. Blue carries \(15\) tons per trip, and Yellow carries \(22\) tons per trip. How many trips does each truck make?

Hints

- First split the combined amount using the \(60\)-ton difference. - Then divide each truck's total by its tons-per-trip rate. - Check that the two totals add to \(1260\) tons.

Solution

1. Find Blue's total amount: \((1260\,\text{tons} - 60\,\text{tons}) \div 2 = 600\,\text{tons}\). 2. Find Yellow's total amount: \(600\,\text{tons} + 60\,\text{tons} = 660\,\text{tons}\). 3. Find Blue's trips: \(600\,\text{tons} \div 15\,\text{tons} = 40\). 4. Find Yellow's trips: \(660\,\text{tons} \div 22\,\text{tons} = 30\).

Answer

Blue makes \(40\) trips, and Yellow makes \(30\) trips.
Rate problems with constant rates 222
A hardware store sells a full roll of underground cable for \(\$630\). The price per foot is constant throughout the roll. A gardener buys \(12\,\text{ft}\) of cable, leaving cable worth \(\$450\) on the roll. How many feet of cable were on the roll at first?

Hints

- First find how much the gardener paid for the \(12\,\text{ft}\) piece. - Use that amount to find the price per foot. - Determine how many one-foot prices fit into the full roll’s value.

Solution

1. Find the value of the \(12\,\text{ft}\) piece sold: \(\$630 - \$450 = \$180\). 2. Find the price per foot: \(\$180 \div 12 = \$15\) per foot. 3. Find the original length: \(\$630 \div \$15 = 42\,\text{ft}\).

Answer

The roll originally held \(42\,\text{ft}\) of cable.
Rate problems with constant rates 223
A runner covers \(200\,\text{m}\) in \(25\) seconds. a) What is the average speed in meters per second? b) A student says, “If the runner could hold that speed for \(10\,\text{km}\), the run would take less than \(25\) minutes.” Check this calculation. Use \(1\,\text{km}=1000\,\text{m}\) and \(1\) minute \(=60\) seconds. c) Why might the predicted time not be realistic?

Hints

- Find the distance covered in one second. - Express the longer distance in the same unit. - Separate what the constant-speed calculation predicts from what a runner could sustain.

Solution

1. The speed is \(200\div25=8\,\text{m/s}\). 2. Convert \(10\,\text{km}\) to \(10{,}000\,\text{m}\). At that speed, the time is \(10{,}000\div8=1250\) seconds. 3. This is \(20\) minutes \(50\) seconds, less than \(25\) minutes. The student's calculation is correct for the stated constant-speed model. 4. A fast speed held for only \(200\,\text{m}\) is unlikely to be maintained over \(10\,\text{km}\).

Answer

a) \(8\,\text{m/s}\) b) Yes, under the assumption: \(20\) minutes \(50\) seconds c) The runner may slow down over the much longer distance.
Rate problems with constant rates 224
Ms. Miller is seeding a \(15{,}000\,\text{ft}^2\) lawn. The seed label recommends \(2\) to \(3\,\text{lb}\) of seed per \(1000\,\text{ft}^2\). Seed is sold in \(2\)-pound bags, and she will spread the entire contents of every bag she buys. What is the minimum number and maximum number of bags she can buy while staying within the recommended rate?

Hints

- Determine how many groups of \(1000\,\text{ft}^2\) are in the lawn. - Calculate the minimum and maximum total seed amounts separately. - Remember that only whole bags can be purchased and every bag is fully used.

Solution

1. The lawn contains \(15\) groups of \(1000\,\text{ft}^2\). 2. The minimum seed amount is \(15 \times 2 = 30\,\text{lb}\), which requires \(30 \div 2 = 15\) bags. 3. The maximum recommended amount is \(15 \times 3 = 45\,\text{lb}\). 4. Twenty-two bags contain \(44\,\text{lb}\), which is within the recommendation. Twenty-three bags contain \(46\,\text{lb}\), which exceeds it. Therefore, the maximum is \(22\) bags.

Answer

She can buy at least \(15\) bags and at most \(22\) bags.
Rate problems with constant rates 225
A rectangular sports field is \(315\,\text{ft}\) long and \(200\,\text{ft}\) wide. Fertilizer is applied at a rate of \(4\,\text{lb}\) for every \(1000\,\text{ft}^2\). Each bag contains \(25\,\text{lb}\) and costs \(\$32\). What is the total cost of buying enough whole bags for the field?

Hints

- Find the area of the rectangular field. - Determine how many \(1000\)-square-foot groups are in the field. - Calculate the required fertilizer weight. - Round the number of bags up because only whole bags can be purchased.

Solution

1. Find the field area: \(315 \times 200 = 63{,}000\,\text{ft}^2\). 2. There are \(63{,}000 \div 1000 = 63\) groups of \(1000\,\text{ft}^2\). 3. The required fertilizer amount is \(63 \times 4 = 252\,\text{lb}\). 4. Since \(252 \div 25 = 10\) remainder \(2\), the buyer needs \(11\) whole bags. 5. The total cost is \(11 \times \$32 = \$352\).

Answer

The fertilizer costs \(\$352\).
Rate problems with constant rates 226
A park rents lawn space for a festival at \(\$4356\) per acre. Smaller spaces cost the same amount per square foot. A rectangular space \(20\,\text{ft}\) wide costs \(\$140\) to rent. How long is it? Use \(1\) acre \(=43{,}560\,\text{ft}^2\).

Hints

- Which unit of area matches the dimensions of the rectangular space? - How can you compare the rent for an acre with the rent for a smaller area? - How are a rectangle's area and side lengths related?

Solution

1. The rent per square foot is \(\$4356\div43{,}560=\$0.10\). 2. The rented area is \(\$140\div\$0.10=1400\,\text{ft}^2\). 3. The length is \(1400\div20=70\,\text{ft}\).

Answer

\(70\,\text{ft}\)

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