Multi-step problems with rational numbers

Learning goals

Translates a multi-step context into operations with rational numbers.
Plans the order of calculations before computing.
Computes with signed fractions and decimals in multi-step problems.
Interprets intermediate and final results in context.
Checks reasonableness of the result.

Difficulty: 1 – 5

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Multi-step problems with rational numbers 1

Morgan has $\$24$. They buy a museum ticket for $\$6$ and then a postcard for $\$5$. How much money is left after the ticket? How much is left after the postcard? State which amount is the intermediate result.

Hints

- Which purchase happened first? - Use the balance after the first purchase to find the balance after the second.

Solution

1. After the ticket, $24-6=18$ dollars remain. 2. After the postcard, $18-5=13$ dollars remain. 3. The $18$-dollar balance is intermediate; the final balance is $13$ dollars.

Answer

After the ticket: $\$18$. After the postcard: $\$13$. The intermediate result is $\$18$.

Multi-step problems with rational numbers 2

A ribbon was $12$ feet long. After the first cut, $9$ feet remain. Rosa then cuts off one third of what remains. How many feet are left after the second cut? Explain which length you took one third of.

Hints

- Which length does “what remains” describe? - Work out how much the second cut removes before subtracting.

Solution

1. The second cut is one third of $9$ feet, not one third of the original $12$ feet. 2. The second cut removes $\frac{1}{3}\cdot9=3$ feet. 3. The length remaining is $9-3=6$ feet.

Answer

$6$ feet. The one-third fraction applies to the $9$ feet left after the first cut.

Multi-step problems with rational numbers 3

The temperature is $-3\,\text{°C}$ in the morning. It rises by $5\,\text{°C}$ and later falls by $1.5\,\text{°C}$. What is the temperature after the rise, and what is it after the fall?

Hints

- A rise moves the temperature higher; a fall moves it lower. - Start the second change from the new temperature, not from the original one.

Solution

1. After the rise, the temperature is $-3+5=2\,\text{°C}$. 2. After the later fall, it is $2-1.5=0.5\,\text{°C}$.

Answer

After the rise: $2\,\text{°C}$. After the fall: $0.5\,\text{°C}$.

Multi-step problems with rational numbers 4

A group hikes for $1.25$ hours, takes a $0.25$-hour break, then hikes for another $0.5$ hour. How much time do they spend hiking? How much time passes altogether, including the break?

Hints

- Does the break count as hiking time? - For the total time, include every part of the trip.

Solution

1. Hiking time excludes the break: $1.25+0.5=1.75$ hours. 2. Including the break gives $1.75+0.25=2$ hours.

Answer

Hiking: $1.75$ hours. Total elapsed time: $2$ hours.

Multi-step problems with rational numbers 5

A gardener starts with a full $8$-gallon watering can. They use $\frac{3}{8}$ of the water on flowers and another $\frac{1}{4}$ of the original water on shrubs. How many gallons remain? Explain which amount each fraction refers to.

Hints

- Can you show the starting water as eight equal parts? - What fraction of the starting amount is used altogether?

Solution

1. Both fractions refer to the original $8$ gallons. 2. The gardener uses $\frac{3}{8}\cdot8=3$ gallons and $\frac{1}{4}\cdot8=2$ gallons. 3. Remaining water: $8-3-2=3$ gallons.

Answer

$3$ gallons. Both fractions are parts of the original $8$-gallon container.

Multi-step problems with rational numbers 6

A cafeteria has $2.5$ cups of rice. It uses $0.5$ cup for one meal and divides the rest equally into $4$ lunch bowls. How much rice goes into each bowl? Show the amount left before dividing.

Hints

- Could a quick sketch separate the first serving from the portions for the bowls? - Does your answer make sense when you multiply it by the number of bowls?

Solution

1. After one meal, $2.5-0.5=2$ cups remain. 2. Share the $2$ cups equally into $4$ bowls: $2\div4=0.5$ cup in each.

Answer

$2$ cups remain before dividing; each bowl gets $0.5$ cup.

Multi-step problems with rational numbers 7

Two muffins cost $\$2.25$ each, and a drink costs $\$1.75$. A cashier says the total is $\$5.00$. Write one expression for the actual total. Is the cashier right? Explain how you checked.

Hints

- How many times must you count the price of one muffin? - What additional purchase is not included in that multiplication? - Can a quick estimate show whether the cashier's amount makes sense?

Solution

1. Two muffins cost $2\cdot2.25=4.50$ dollars. 2. Include the drink: $2\cdot2.25+1.75=4.50+1.75=6.25$. 3. The cashier's total is too low. Even the muffins and a $\$1$ drink would cost more than $\$5$.

Answer

$2\cdot2.25+1.75=\$6.25$. No; $\$5.00$ is too low.

Multi-step problems with rational numbers 8

A hiker starts with $24$ ounces of water in a bottle. The hiker pours out one-third of that water, then adds $2$ ounces. Write one expression for the amount in the bottle now and find its value. What amount does one-third refer to?

Hints

- What amount was in the bottle before anything changed? - Which amount is being divided into three equal parts? - Do the pouring out and refilling change the amount in the same direction?

Solution

1. One-third of the original amount is $\frac{1}{3}\cdot24=8$ ounces. 2. Subtract that amount and add the refill: $24-\frac{1}{3}\cdot24+2=24-8+2=18$. 3. The fraction refers to the original $24$ ounces, not the later balance.

Answer

$24-\frac{1}{3}\cdot24+2=18$ ounces. One-third refers to the original $24$ ounces.

Multi-step problems with rational numbers 9

A delivery van has a maximum payload of $1.2$ tons, or $2400\,\text{lb}$. It already carries $1700\,\text{lb}$ of sand and $250\,\text{lb}$ of stone. The driver weighs $170\,\text{lb}$. A $300\,\text{lb}$ pallet of cement is to be added. Will the van exceed its payload limit? Show your calculation.

Hints

- Express all weights in the same unit. - Include the cargo and the driver. - Compare the total with the payload limit.

Solution

1. Add the existing load and the driver: $1700+250+170=2120\,\text{lb}$. 2. Include the cement pallet: $2120+300=2420\,\text{lb}$. 3. Since $2420>2400$, the payload limit would be exceeded by $20\,\text{lb}$.

Answer

Yes. The total payload would be $2420\,\text{lb}$, which is $20\,\text{lb}$ over the limit.

Multi-step problems with rational numbers 10

A book costs $\$40$ before discounts. Maple Books takes $10\%$ off the price and then charges $\$2$ for shipping. Riverbend Books takes $\$5$ off and offers free shipping. Which shop has the lower total price, and by how much? Check that your prices make sense.

Hints

- Apply each shop's discount before considering shipping. - Keep separate totals for the two offers. - Compare the final totals rather than the stated discounts alone.

Solution

1. Maple Books gives a discount of $0.10\cdot40=4$ dollars, so the final cost is $40-4+2=38$ dollars. 2. Riverbend Books charges $40-5=35$ dollars. 3. Riverbend Books costs $38-35=3$ dollars less. 4. Check: Maple's $\$4$ discount exceeds its $\$2$ shipping fee, so $\$38$ is below $\$40$. Riverbend's $\$5$ discount makes $\$35$, also below $\$40$.

Answer

Riverbend Books is $\$3$ cheaper ($\$35$ versus $\$38$). Check: Maple's discount is greater than its shipping charge, so both final prices are below $\$40$, as expected.

Multi-step problems with rational numbers 11

A lunch account has the transactions below. Find the balance after Monday and the balance after Tuesday. Explain what the plus and minus signs mean in this account. <table><tr><td>Start</td><td>$\$7.50$</td></tr><tr><td>Monday deposit</td><td>$+\$3.25$</td></tr><tr><td>Tuesday purchase</td><td>$-\$5.40$</td></tr></table>

Hints

- Does the Tuesday purchase exceed Monday's deposit? - Would a positive final balance be reasonable here? Estimate before calculating exactly.

Solution

1. A deposit adds money: $7.50+3.25=10.75$ dollars after Monday. 2. A purchase removes money: $10.75-5.40=5.35$ dollars after Tuesday. 3. The signs indicate money added and money spent, not the sign of the final balance.

Answer

After Monday: $\$10.75$. After Tuesday: $\$5.35$. Plus means a deposit; minus means a purchase.

Multi-step problems with rational numbers 12

A print shop sells posters at the same price each, then pays a supply fee. The tree shows the number of posters, the price in dollars, and the fee in dollars. Find the sales income before the fee and the money left after the fee.

Hints

- Which number could count posters, and which could be a price per poster? - Can you check that the amount left is smaller than the sales income?

Solution

1. The tree gives $3$ posters at $\$7.50$ each: $3\cdot7.5=22.5$ dollars in sales. 2. After the $\$4.50$ supply fee, $22.5-4.5=18$ dollars remain.

Answer

Sales income: $\$22.50$. Money left after the supply fee: $\$18.00$.

Multi-step problems with rational numbers 13

A club has $60$ members. On Monday, one-third of them attend a workshop. On Tuesday, $25\%$ of the members who did not attend Monday come to a second workshop. How many different members attended at least one workshop? Show how many were left after Monday.

Hints

- Which group is the $25\%$ taken from? - Why should you subtract Monday's attendees before finding Tuesday's? - Do the two days count any member twice?

Solution

1. Monday has $\frac{1}{3}\cdot60=20$ attendees, leaving $60-20=40$ who did not attend. 2. Tuesday brings $0.25\cdot40=10$ different members. 3. In all, $20+10=30$ different members attended.

Answer

After Monday, $40$ members had not attended. A total of $30$ different members attended at least one workshop.

Multi-step problems with rational numbers 14

A crafter has $\frac{5}{6}$ yard of ribbon. She cuts off $\frac{1}{4}$ yard, then divides the remaining ribbon into $3$ equal pieces. Write a numerical expression and find the length of each piece in yards.

Hints

- Find the length left after cutting first. - Which entire amount is being divided into equal parts? - Keep the units attached to your answer.

Solution

1. The amount remaining is $\frac{5}{6}-\frac{1}{4}=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}$ yard. 2. Divide this remainder into $3$ equal parts: $\left(\frac{5}{6}-\frac{1}{4}\right)\div3=\frac{7}{12}\cdot\frac{1}{3}=\frac{7}{36}$.

Answer

$\left(\frac{5}{6}-\frac{1}{4}\right)\div3=\frac{7}{36}$ yard per piece.

Multi-step problems with rational numbers 15

A bank account starts at $-\$4.50$. A deposit covers one-third of the amount owed. Later, a $\$1.25$ fee is taken from the account. Find the final balance. Is the account still in debt? Explain what each change does.

Hints

- A negative account balance means money is owed. - Does a deposit move that balance toward zero or away from it? - Does a fee make the balance larger or smaller?

Solution

1. One-third of the $\$4.50$ debt is $\frac{1}{3}\cdot4.50=1.50$ dollars. 2. The deposit increases the signed balance to $-4.50+1.50=-3.00$ dollars. 3. The fee lowers it by $1.25$, leaving $-3.00-1.25=-4.25$ dollars. The negative balance means the account still owes money.

Answer

The $\$1.50$ deposit raises the balance from $-\$4.50$ to $-\$3.00$. The $\$1.25$ fee lowers it to $-\$4.25$, so the account is still in debt.

Multi-step problems with rational numbers 16

A city has a population of $36{,}000$. During the first year, the population grows by $0.5\%$. a) What is the population after the first year? b) During the second year, the new population grows by $p\%$. Write an expression for the population after the second year.

Hints

- Treat the two yearly changes separately. - The second-year increase is based on the population after the first year. - Write the percent increase using the variable $p$. - Leave the variable in the final expression.

Solution

1. For part a, the first-year increase is $36{,}000 \cdot 0.005 = 180$. The population after the first year is $36{,}000 + 180 = 36{,}180$. 2. For part b, the second-year increase is $36{,}180 \cdot \frac{p}{100} = 361.8p$. 3. Therefore, the population after the second year is $36{,}180 + 361.8p$, or equivalently $36{,}180\left(1 + \frac{p}{100}\right)$.

Answer

a) The population after the first year is $36{,}180$. b) The population after the second year is $36{,}180\left(1 + \frac{p}{100}\right)$.

Multi-step problems with rational numbers 17

A laptop battery is $\frac{5}{6}$ charged. After one hour of heavy use, the charge has dropped to $\frac{2}{3}$ of full capacity. What fraction of the full battery capacity was used during that hour? At the same constant rate, how many minutes could the laptop run from a full charge?

Hints

- Subtract the ending charge from the starting charge. - Use a common denominator. - Determine how many one-hour portions of that size make one full battery.

Solution

1. Find the fraction used in one hour: $\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{4}{6}=\frac{1}{6}$. 2. A full battery contains six portions of size $\frac{1}{6}$, so it would last $6$ hours. 3. Convert to minutes: $6\cdot60=360$ minutes.

Answer

$\frac{1}{6}$ of the full capacity was used. A full charge would last $360$ minutes.

Multi-step problems with rational numbers 18

A rainwater tank is $\frac{3}{8}$ full. During a storm, an amount of water equal to $\frac{1}{3}$ of the previously empty part is added. What fraction of the tank is now full?

Hints

- First find the fraction of the tank that was empty. - Find one third of that empty fraction. - Add the amount of water added to the original amount.

Solution

1. The empty fraction was $1-\frac{3}{8}=\frac{5}{8}$. 2. The amount added was $\frac{1}{3}\cdot\frac{5}{8}=\frac{5}{24}$ of the tank. 3. The new level is $\frac{3}{8}+\frac{5}{24}=\frac{9}{24}+\frac{5}{24}=\frac{14}{24}=\frac{7}{12}$.

Answer

The tank is $\frac{7}{12}$ full.

Multi-step problems with rational numbers 19

A carpenter has a board that is $4\,\text{ft}$ long. The carpenter cuts off pieces measuring $1\frac{1}{5}\,\text{ft}$, $\frac{3}{4}\,\text{ft}$, and $1\frac{3}{10}\,\text{ft}$. a) How long is the remaining piece? b) The carpenter then cuts off $\frac{2}{3}$ of that remaining piece. How long is the final piece left over?

Hints

- Add the lengths of the first three pieces. - Subtract their total from the original length. - If $\frac{2}{3}$ is removed, determine what fraction remains.

Solution

1. Add the first three lengths: $1\frac{1}{5}+\frac{3}{4}+1\frac{3}{10}=\frac{24}{20}+\frac{15}{20}+\frac{26}{20}=3\frac{1}{4}\,\text{ft}$. 2. Subtract from the original length: $4-3\frac{1}{4}=\frac{3}{4}\,\text{ft}$. 3. If $\frac{2}{3}$ is cut off, $\frac{1}{3}$ remains. The final length is $\frac{1}{3}\cdot\frac{3}{4}=\frac{1}{4}\,\text{ft}$.

Answer

a) $\frac{3}{4}\,\text{ft}$ b) $\frac{1}{4}\,\text{ft}$

Multi-step problems with rational numbers 20

For a sewing project, Julia needs fabric pieces measuring $1\frac{1}{2}\,\text{yd}$, $2\frac{3}{4}\,\text{yd}$, and $1\frac{1}{5}\,\text{yd}$. A fabric store offers two options: A) Buy the exact length needed for $\$3.00$ per yard. B) Buy a $6$-yard roll for $\$16.00$. Which option costs less, and by how much?

Hints

- Find the total amount of fabric needed. - Multiply that length by the price per yard for Option A. - Compare the two prices and subtract to find the difference.

Solution

1. The total length needed is $1\frac{1}{2}+2\frac{3}{4}+1\frac{1}{5}=5\frac{9}{20}=5.45\,\text{yd}$. 2. Option A costs $5.45\cdot\$3.00=\$16.35$. 3. Option B costs $\$16.00$, so it costs less. 4. The difference is $\$16.35-\$16.00=\$0.35$.

Answer

Option B costs less by $\$0.35$.

Multi-step problems with rational numbers 21

A sports club orders $5$ jump ropes at $\$4.99$ each and $3$ identical medicine balls. The total order is $\$71.45$. a) Find the price of one medicine ball. b) A $\$12.50$ discount coupon was submitted but accidentally not applied. What should the correct final order total be? c) Without calculating the exact unit price, explain why one medicine ball must cost more than $\$15.00$.

Hints

- Find the total cost of the jump ropes first. - Subtract the rope cost from the order total to find the cost of all three medicine balls. - Divide that amount by $3$ for the unit price. - For part c), use rounded values rather than an exact calculation.

Solution

1. The jump ropes cost $5\cdot\$4.99=\$24.95$. 2. The three medicine balls cost $\$71.45-\$24.95=\$46.50$. 3. For a), one medicine ball costs $\$46.50\div3=\$15.50$. 4. For b), the correct order total is $\$71.45-\$12.50=\$58.95$. 5. For c), the jump ropes cost about $\$25$, leaving a little more than $\$46$ for three medicine balls. Since $3\cdot\$15=\$45$, each ball must cost more than $\$15$.

Answer

a) $\$15.50$ b) $\$58.95$ c) About $\$46$ remains for three balls, and $3\cdot\$15=\$45$, so each ball must cost more than $\$15$.

Multi-step problems with rational numbers 22

Lucas buys a notebook for $\$1.49$, four pencils for $\$0.55$ each, and a compass for $\$6.75$. He has $\$15.00$ and wants to buy three more notebooks. a) Estimate the cost of everything by rounding each price to the nearest $\$0.50$. For a price exactly halfway between two choices, round up. Can the estimate alone tell whether he has enough money? Explain. b) Find the exact total cost and the change he gets if he buys everything.

Hints

- There are four notebooks in the full purchase, not just the first one. - A rounded total can be a little higher or lower than the exact total. - How can you use the exact cost to check whether the purchase fits the budget?

Solution

1. The original items cost $1.49+4\cdot0.55+6.75=10.44$ dollars. 2. Estimate the original total as about $10.50$ dollars and the extra notebooks as $3\cdot1.50=4.50$ dollars. The estimated total is $15.00$ dollars. 3. Since the estimate equals the money available, it does not tell for sure whether the exact total is affordable. 4. The extra notebooks cost $3\cdot1.49=4.47$ dollars. The exact total is $10.44+4.47=14.91$ dollars. 5. The change is $15.00-14.91=0.09$ dollars.

Answer

a) About $\$15.00$; the estimate is too close to decide for sure. b) Total: $\$14.91$; change: $\$0.09$.

Multi-step problems with rational numbers 23

A water tank holds $42$ gallons. This is $1\frac{3}{4}$ times the amount held by a barrel. The barrel holds $1\frac{1}{3}$ times the amount held by a bucket. Use one expression to find how many gallons the bucket holds.

Hints

- Combine the two multiplicative comparisons. - Convert mixed numbers to improper fractions. - Divide the largest amount by the combined scale factor.

Solution

1. Combine the two scale factors in the expression $42\div(1\frac{3}{4}\cdot1\frac{1}{3})$. 2. Convert the mixed numbers: $1\frac{3}{4}=\frac{7}{4}$ and $1\frac{1}{3}=\frac{4}{3}$. 3. Their product is $\frac{7}{4}\cdot\frac{4}{3}=\frac{7}{3}$. 4. Divide: $42\div\frac{7}{3}=42\cdot\frac{3}{7}=18$.

Answer

The bucket holds $18$ gallons. One expression is $42\div(1\frac{3}{4}\cdot1\frac{1}{3})$.

Multi-step problems with rational numbers 24

A freight elevator can carry at most $1800\,\text{lb}$. Crate A weighs $420\,\text{lb}$. Crate B weighs $\frac{3}{4}$ as much as Crate A. Crate C weighs $90\,\text{lb}$ more than Crate B. a) Find the total weight of the three crates. b) Small packages weighing $25\,\text{lb}$ each will be added. What is the greatest number of packages that can be added without exceeding the limit?

Hints

- Find the weight of each crate in order. - Subtract the crate total from the elevator capacity. - The final answer must be a whole number of packages that does not exceed the limit.

Solution

1. Crate B weighs $\frac{3}{4}\cdot420=315\,\text{lb}$. 2. Crate C weighs $315+90=405\,\text{lb}$. 3. The crates weigh $420+315+405=1140\,\text{lb}$ altogether. 4. The remaining capacity is $1800-1140=660\,\text{lb}$. 5. Since $660\div25=26.4$, at most $26$ whole packages can be added.

Answer

a) $1140\,\text{lb}$ b) $26$ packages

Multi-step problems with rational numbers 25

Consider the expression $\frac{7}{9} - 0.25 \div \frac{3}{4}$. a) Evaluate the expression exactly by converting the decimal to a fraction. b) Explain why using only terminating decimals is not a useful way to calculate the entire expression exactly.

Hints

- Apply division before subtraction. - Convert every value to a form that supports an exact calculation. - Examine the decimal representation of $\frac{7}{9}$.

Solution

1. Convert $0.25$ to $\frac{1}{4}$. 2. Perform the division first: $\frac{1}{4} \div \frac{3}{4} = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}$. 3. Subtract: $\frac{7}{9} - \frac{1}{3} = \frac{7}{9} - \frac{3}{9} = \frac{4}{9}$. 4. Both $\frac{7}{9} = 0.\overline{7}$ and $\frac{1}{3} = 0.\overline{3}$ have repeating decimal representations. Replacing them with terminating decimals would require rounding and would not preserve the exact value.

Answer

a) $\frac{4}{9}$ b) $\frac{7}{9}$ and $\frac{1}{3}$ have repeating decimal representations, so terminating decimal approximations would not produce an exact calculation.

Multi-step problems with rational numbers 26

A city park covers $15$ acres. Forest covers $\frac{2}{5}$ of the park, and lawns cover $\frac{3}{10}$. The remaining area is a lake. Find the lake’s area in square feet. Use $1\,\text{acre}=43{,}560\,\text{ft}^2$.

Hints

- First find the fraction of the park occupied by the lake. - Apply that fraction to the total number of acres. - Convert the resulting acreage to square feet.

Solution

1. The forest and lawn fraction is $\frac{2}{5}+\frac{3}{10}=\frac{7}{10}$. 2. The lake fraction is $1-\frac{7}{10}=\frac{3}{10}$. 3. The lake covers $\frac{3}{10}\cdot15=4.5$ acres. 4. Convert to square feet: $4.5\cdot43{,}560=196{,}020\,\text{ft}^2$.

Answer

The lake covers $196{,}020\,\text{ft}^2$.

Multi-step problems with rational numbers 27

Match each story to the correct expression, and find the value. A. Leo has $\$20.00$. He buys $3$ notebooks for $\$1.50$ each and a pen for $\$2.40$. How much money remains? B. Leo has $\$20.00$. He buys $3$ notebooks for $\$1.50$ each. Then a friend repays him $\$2.40$. How much money does Leo have now? C. Three friends each have $\$20.00$. Each buys a notebook for $\$1.50$ and a pen for $\$2.40$. How much money do they have altogether afterward? (1) $3\cdot(20-1.50-2.40)$ (2) $20-3\cdot1.50-2.40$ (3) $20-3\cdot1.50+2.40$

Hints

- Decide whether money is spent or received in each story. - Notice whether the calculation applies to one person or three people. - Use the order of operations.

Solution

1. Story A matches Expression 2: $20-3\cdot1.50-2.40=20-4.50-2.40=13.10$. 2. Story B matches Expression 3: $20-3\cdot1.50+2.40=17.90$. 3. Story C matches Expression 1: $3\cdot(20-1.50-2.40)=3\cdot16.10=48.30$.

Answer

A: Expression 2, $\$13.10$ B: Expression 3, $\$17.90$ C: Expression 1, $\$48.30$

Multi-step problems with rational numbers 28

A family with $2$ adults and $4$ children visits a swimming pool. Regular admission is $\$5.50$ per adult and $\$3.50$ per child. Offer 1: Take $\$3.00$ off the total price. Offer 2: Each child pays $\$0.50$ less. Expression 1: $4\cdot3.50+2\cdot5.50-3.00$ Expression 2: $4\cdot(3.50-0.50)+2\cdot5.50$ Match each expression to its offer, and determine which offer costs less.

Hints

- Determine whether each discount is applied once or once per child. - Evaluate both expressions. - Compare the totals.

Solution

1. Expression 1 subtracts $\$3.00$ once from the total, so it represents Offer 1. 2. Expression 2 reduces the price for each of the four children, so it represents Offer 2. 3. Expression 1 equals $14.00+11.00-3.00=22.00$. 4. Expression 2 equals $4\cdot3.00+11.00=23.00$. 5. Offer 1 costs $\$1.00$ less.

Answer

Expression 1 represents Offer 1, and Expression 2 represents Offer 2. Offer 1 costs less: $\$22.00$ instead of $\$23.00$.

Multi-step problems with rational numbers 29

Harbor Nut Co. buys $15\,\text{lb}$ of cashews for $\$127.50$. The cashews are packed into bags that each hold $0.25\,\text{lb}$. The company wants a total profit of $\$67.50$ after selling all the cashews. What price should the company charge per bag?

Hints

- Find how many equal bags can be made from the total amount. - Add the desired profit to the original cost to find the needed revenue. - Divide that revenue equally across all the bags.

Solution

1. Find the number of bags: $15\div0.25=60$. 2. The required total revenue is $127.50+67.50=195.00$ dollars. 3. Divide the revenue by the number of bags: $195.00\div60=3.25$. 4. The price should be $\$3.25$ per bag.

Answer

$\$3.25$ per bag

Multi-step problems with rational numbers 30

The segment on a number line from $-1.2$ to $2.4$ is divided into three equal parts. Find the two division points.

Hints

- Find the distance between the two endpoints. - Divide the total distance into three equal lengths. - Starting at the left endpoint, add one section length at a time.

Solution

1. Find the total length: $2.4-(-1.2)=3.6$. 2. Divide by $3$ to find each part’s length: $3.6\div3=1.2$. 3. Starting at $-1.2$, the first division point is $-1.2+1.2=0$. 4. The second division point is $0+1.2=1.2$.

Answer

$0$ and $1.2$

Multi-step problems with rational numbers 31

Luke and Sarah compare their bank balances. Luke deposits $\$15.50$ and then pays $\$40.00$ for a video game. Sarah deposits $\$25.00$ and then pays $\$10.50$ for a book. After these transactions, each account has a balance of $\$5.00$. a) Find each person's starting balance. b) Who had a negative balance before the transactions? Explain briefly.

Hints

- Write a separate equation for each person. - Work backward from the same ending balance. - Keep careful track of deposits and payments. - Interpret a negative starting value in context.

Solution

1. Let $x$ be Luke's starting balance. Write $x + 15.50 - 40.00 = 5.00$. 2. Simplify: $x - 24.50 = 5.00$, so $x = 29.50$. 3. Let $y$ be Sarah's starting balance. Write $y + 25.00 - 10.50 = 5.00$. 4. Simplify: $y + 14.50 = 5.00$, so $y = -9.50$. 5. A negative account balance represents money owed, so Sarah had the negative starting balance.

Answer

a) Luke's starting balance was $\$29.50$, and Sarah's starting balance was $-\$9.50$. b) Sarah had a negative balance because her starting balance was less than $\$0$.

Multi-step problems with rational numbers 32

Fresh mushrooms are $92\%$ water. A package initially contains $500\,\text{g}$ of mushrooms. The mushrooms are dried until they are only $20\%$ water. What is their new total mass?

Hints

- Which part of the mushrooms stays unchanged during drying? - Find the initial mass of that unchanged part. - What percent of the final mass does that part represent? - Use an equation to find the final total mass.

Solution

1. Initially, the mushrooms are $100\% - 92\% = 8\%$ dry matter. The dry mass is $500\,\text{g} \cdot 0.08 = 40\,\text{g}$. 2. Drying removes water but does not change the $40\,\text{g}$ of dry matter. 3. After drying, the mushrooms are $100\% - 20\% = 80\%$ dry matter. 4. Let $m$ be the new total mass. Then $0.80m = 40$, so $m = 40 \div 0.80 = 50$.

Answer

The new total mass is $50\,\text{g}$.

Multi-step problems with rational numbers 33

A pot contains $2\,\text{L}$ of a fruit drink that is $25\%$ pure juice and $75\%$ water. The drink is heated, and only water evaporates. After heating, the mixture is $40\%$ juice. How much water evaporated?

Hints

- Find the initial amount of pure juice. - Does the amount of juice change when only water evaporates? - Use the final juice percent to find the final total volume. - Compare the initial and final total volumes.

Solution

1. The initial amount of pure juice is $2\,\text{L} \cdot 0.25 = 0.5\,\text{L}$. 2. Only water evaporates, so the amount of juice remains $0.5\,\text{L}$. 3. Let $V$ be the final total volume. Since the juice is $40\%$ of the final mixture, $0.40V = 0.5$. 4. Solving gives $V = 0.5 \div 0.40 = 1.25\,\text{L}$. 5. The amount of water that evaporated is $2\,\text{L} - 1.25\,\text{L} = 0.75\,\text{L}$.

Answer

$0.75\,\text{L}$ of water evaporated.

Multi-step problems with rational numbers 34

Lucas spends $\frac{1}{4}$ of his monthly allowance on movie tickets and $\frac{2}{5}$ on a new video game. He puts the remaining $\$14$ into savings. How much allowance did Lucas receive in all?

Hints

- Find the fraction of the allowance that remains after both purchases. - Use a common denominator to combine the fractions. - The remaining fraction of the total equals $\$14$.

Solution

1. Let $x$ be the total allowance. Write $x - \frac{1}{4}x - \frac{2}{5}x = 14$. 2. Rewrite the fractions with denominator $20$: $\frac{20}{20}x - \frac{5}{20}x - \frac{8}{20}x = 14$. 3. Combine the coefficients: $\frac{7}{20}x = 14$. 4. Multiply both sides by $\frac{20}{7}$: $x = 14 \cdot \frac{20}{7}$. 5. Calculate: $x = 40$.

Answer

Lucas received $\$40$ in allowance.

Multi-step problems with rational numbers 35

Sarah makes fruit punch using $\frac{1}{3}$ orange juice, $\frac{1}{4}$ apple juice, and $\frac{1}{6}$ grape juice. The rest is sparkling water. a) What fraction of the punch is sparkling water? b) If Sarah uses $3\,\text{cups}$ of sparkling water, how many cups of punch does she make in all? c) How many cups of orange juice are in the finished punch?

Hints

- The fractions of all ingredients must add to one whole. - Subtract the total juice fraction from $1$. - Once you know what fraction equals $3\,\text{cups}$, scale up to the whole amount.

Solution

1. Add the juice fractions: $\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4}{12} + \frac{3}{12} + \frac{2}{12} = \frac{9}{12} = \frac{3}{4}$. 2. Subtract from the whole: $1 - \frac{3}{4} = \frac{1}{4}$. Sparkling water is one-fourth of the punch. 3. Let $x$ be the total number of cups. Write $\frac{1}{4}x = 3$. 4. Multiply by $4$: $x = 12$. 5. Find the orange juice amount: $\frac{1}{3} \cdot 12 = 4\,\text{cups}$.

Answer

a) Sparkling water makes up $\frac{1}{4}$ of the punch. b) Sarah makes $12\,\text{cups}$ of punch. c) The punch contains $4\,\text{cups}$ of orange juice.

Multi-step problems with rational numbers 36

A large fruit crate contains only apples and pears. Exactly $65\%$ of the fruit are apples, and there are $42$ more apples than pears. How many pieces of fruit are in the crate altogether, and how many are pears?

Hints

- Find the percent of the fruit that are pears. - Express the difference between the apple and pear portions as a percent. - Use the fact that this percent difference represents $42$ fruit. - Find the pear count from the total.

Solution

1. Pears make up $100\% - 65\% = 35\%$ of the fruit. 2. The difference between the apple and pear portions is $65\% - 35\% = 30\%$. 3. Therefore, $30\%$ of the total number of fruit equals $42$. 4. Let $T$ be the total. Then $0.30T = 42$, so $T = 42 \div 0.30 = 140$. 5. The number of pears is $140 \cdot 0.35 = 49$.

Answer

There are $140$ pieces of fruit altogether, including $49$ pears.

Multi-step problems with rational numbers 37

A laboratory stores samples in refrigerated compartments set to the temperatures shown. <table> <tr><td>Compartment 1</td><td>$+6\,^\circ\text{C}$</td></tr> <tr><td>Compartment 2</td><td>$-2\,^\circ\text{C}$</td></tr> <tr><td>Compartment 3</td><td>$-15\,^\circ\text{C}$</td></tr> <tr><td>Compartment 4</td><td>$-28\,^\circ\text{C}$</td></tr> </table> a) Find the temperature difference between Compartments 1 and 3. b) A sample is moved from Compartment 4 to Compartment 2. By how many degrees does its temperature increase? c) Compartment 5 will be set halfway between the temperatures of Compartments 1 and 4. What should its temperature be?

Hints

- Treat a temperature difference as a distance on a number line. - For part b), subtract the starting temperature from the ending temperature. - To find the midpoint, add the two endpoint values and divide by $2$.

Solution

1. The difference between Compartments 1 and 3 is $6-(-15)=21\,^\circ\text{C}$. 2. The increase from Compartment 4 to Compartment 2 is $-2-(-28)=26\,^\circ\text{C}$. 3. The midpoint is the average of the two temperatures: $\frac{6+(-28)}{2}=\frac{-22}{2}=-11\,^\circ\text{C}$.

Answer

a) $21\,^\circ\text{C}$ b) $26\,^\circ\text{C}$ c) $-11\,^\circ\text{C}$

Multi-step problems with rational numbers 38

A bus leaves its starting stop. At the first stop, $8$ passengers get on. At the second stop, the number of passengers doubles. At the third stop, $5$ passengers get off. There are then $15$ passengers on the bus. How many passengers were on the bus at the starting stop?

Hints

- Work backward from the final number of passengers. - Use the inverse of “get off” and the inverse of “double.” - Identify what happened immediately before the final stop.

Solution

1. Work backward from $15$ passengers. 2. Undo the $5$ passengers getting off: $15 + 5 = 20$. 3. Undo the doubling: $20 \div 2 = 10$. 4. Undo the $8$ passengers getting on: $10 - 8 = 2$.

Answer

$2$ passengers.

Multi-step problems with rational numbers 39

At a charity walk, Class A raises $\$x$. Class B raises $20\%$ more than Class A. Class C raises $\$50$ less than Class B. 1. Write an expression for the amount raised by Class B. 2. Write an expression for the amount raised by Class C. 3. Write and simplify an expression for the total amount raised by all three classes.

Hints

- Write the percent increase as a decimal. - Express each class's amount separately before adding. - Combine like terms when simplifying the total. - Distinguish variable terms from fixed amounts.

Solution

1. A $20\%$ increase gives $x+0.20x=1.20x$. 2. Class C raises $1.20x-50$. 3. Add the three amounts: $x+1.20x+(1.20x-50)=3.40x-50$.

Answer

1. $1.20x$, or $x+0.20x$ 2. $1.20x-50$ 3. $3.40x-50$

Multi-step problems with rational numbers 40

A farm dries lavender for scented sachets. Fresh lavender loses $65\%$ of its weight when dried. a) How many pounds of dried lavender are produced from $14\,\text{lb}$ of fresh lavender? b) A large order requires $7\,\text{lb}$ of dried lavender. The farm has $18\,\text{lb}$ of fresh lavender available. Is that enough? Justify your answer.

Hints

- Find the percent of the weight that remains after drying. - Part a asks for the remaining part; part b asks for the original whole. - You can also find how much dried lavender $18\,\text{lb}$ of fresh lavender would produce.

Solution

1. After drying, $100\% - 65\% = 35\%$ of the original weight remains. 2. For part a, the dried weight is $14\,\text{lb} \cdot 0.35 = 4.9\,\text{lb}$. 3. For part b, let $F$ be the required fresh weight. Then $0.35F = 7$, so $F = 7 \div 0.35 = 20\,\text{lb}$. 4. Since $20\,\text{lb} > 18\,\text{lb}$, the available lavender is not enough.

Answer

a) The farm produces $4.9\,\text{lb}$ of dried lavender. b) No. Producing $7\,\text{lb}$ of dried lavender requires $20\,\text{lb}$ of fresh lavender.

Multi-step problems with rational numbers 41

A $750\,\text{ft}^2$ school garden is divided into a vegetable garden, a flower garden, and a lawn. The lawn has the same area as the vegetable and flower gardens combined. The vegetable garden has two-thirds the area of the flower garden. Find the area of each section.

Hints

- If the lawn equals the other two sections combined, what fraction of the total is the lawn? - Express the vegetable-garden area in terms of the flower-garden area. - Write an equation for the combined area of the vegetable and flower gardens.

Solution

1. Let $V$, $F$, and $L$ be the areas of the vegetable garden, flower garden, and lawn. 2. Since $L = V + F$ and $V + F + L = 750$, it follows that $2(V + F) = 750$. Thus $V + F = 375$ and $L = 375$. 3. The vegetable garden is two-thirds the flower garden, so $V = \frac{2}{3}F$. 4. Substitute into $V + F = 375$: $\frac{2}{3}F + F = 375$. 5. Combine like terms: $\frac{5}{3}F = 375$. Multiply by $\frac{3}{5}$: $F = 225$. 6. Find the vegetable-garden area: $V = \frac{2}{3} \cdot 225 = 150$.

Answer

The vegetable garden is $150\,\text{ft}^2$, the flower garden is $225\,\text{ft}^2$, and the lawn is $375\,\text{ft}^2$.

Multi-step problems with rational numbers 42

A hiker plans to complete a trail. In the morning, the hiker walks one-third of the total distance plus $2\,\text{miles}$. After lunch, the hiker walks $50\%$ of the remaining distance. Then $6\,\text{miles}$ remain. What is the total length of the trail?

Hints

- Use a variable for the total trail length. - Write an expression for the distance remaining after the morning. - After half of a remaining distance is walked, what fraction of that distance is still left? - You can also work backward from the final $6\,\text{miles}$.

Solution

1. Let $x$ be the total trail length in miles. 2. The morning distance is $\frac{1}{3}x + 2$. 3. The distance remaining after the morning is $x - (\frac{1}{3}x + 2) = \frac{2}{3}x - 2$. 4. After the hiker walks half of that remaining distance, the other half is still left. Write $0.5(\frac{2}{3}x - 2) = 6$. 5. Simplify: $\frac{1}{3}x - 1 = 6$. 6. Add $1$: $\frac{1}{3}x = 7$. Multiply by $3$: $x = 21$.

Answer

The trail is $21\,\text{miles}$ long.

Multi-step problems with rational numbers 43

A water tank contains an unknown amount of water. First, one-fourth of the water and an additional $20\,\text{gal}$ are removed for irrigation. Then $60\%$ of the remaining water is used to wash a car. At the end, $28\,\text{gal}$ remain. How much water was originally in the tank?

Hints

- If $60\%$ is removed, what percent remains? - Write an expression for the amount left after the first removal. - You can work backward to find the amount present before the car was washed. - Each percent is applied to the amount remaining at that stage.

Solution

1. Let $V$ be the original volume in gallons. 2. After the first removal, the volume is $V - (\frac{1}{4}V + 20) = \frac{3}{4}V - 20$. 3. Removing $60\%$ leaves $40\%$, so write $0.40(\frac{3}{4}V - 20) = 28$. 4. Divide by $0.40$: $\frac{3}{4}V - 20 = 70$. 5. Add $20$: $\frac{3}{4}V = 90$. 6. Multiply by $\frac{4}{3}$: $V = 120$.

Answer

The tank originally contained $120\,\text{gal}$ of water.

Multi-step problems with rational numbers 44

A science lab uses a blue-dye solution that is $25\%$ dye by volume. It needs a diluted solution that is $5\%$ dye. Assume the volumes add. a) How many milliliters of water must be added to $100\,\text{mL}$ of the concentrate? b) To prepare exactly $1\,\text{L}$ of the diluted solution, how many milliliters of concentrate and water are needed?

Hints

- Find how much dye is in the undiluted solution. - Use that unchanged dye amount to find the larger mixture volume. - In part b), start from the desired total volume and its dye percentage. - Subtract the concentrate volume to obtain the water volume.

Solution

1. The original $100\,\text{mL}$ contains $100\cdot0.25=25\,\text{mL}$ of dye. 2. If that is $5\%$ of the mixture, the final volume is $25\div0.05=500\,\text{mL}$. Add $500-100=400\,\text{mL}$ water. 3. One liter is $1000\,\text{mL}$, requiring $1000\cdot0.05=50\,\text{mL}$ dye. 4. The concentrate needed is $50\div0.25=200\,\text{mL}$; the remaining $800\,\text{mL}$ is water.

Answer

a) Add $400\,\text{mL}$ of water. b) Use $200\,\text{mL}$ concentrate and $800\,\text{mL}$ water.

Multi-step problems with rational numbers 45

For a school celebration, one-third of the budget is spent on renting the space, two-fifths is spent on food, and one-sixth is spent on drinks. Exactly $\$30$ remains for decorations. What is the total budget?

Hints

- Use a variable for the total budget. - Add the three fractional parts of the budget. - Determine what fraction of the whole remains for decorations. - Use a common denominator when adding the fractions.

Solution

1. Let $x$ be the total budget in dollars. 2. Write $x = \frac{1}{3}x + \frac{2}{5}x + \frac{1}{6}x + 30$. 3. Combine the fractional parts: $\frac{1}{3} + \frac{2}{5} + \frac{1}{6} = \frac{10}{30} + \frac{12}{30} + \frac{5}{30} = \frac{9}{10}$. 4. Then $x = \frac{9}{10}x + 30$, so $\frac{1}{10}x = 30$. 5. Multiply by $10$: $x = 300$.

Answer

The total budget is $\$300$.

Multi-step problems with rational numbers 46

In a school library, two-fifths of the items are novels and one-fourth are nonfiction books. Exactly half of the remaining items are children’s books. The final $105$ items are magazines. How many items are in the library altogether?

Hints

- First find the fraction represented by the novels and nonfiction books together. - Subtract that fraction from $1$ to find the remaining fraction. - If half of the remainder is children’s books, what fraction of the remainder is magazines? - Set the magazine fraction of the total equal to $105$.

Solution

1. Let $x$ be the total number of library items. 2. The novels and nonfiction books make up $\frac{2}{5}x + \frac{1}{4}x = \frac{13}{20}x$. 3. The remaining portion is $x - \frac{13}{20}x = \frac{7}{20}x$. 4. Half of that remainder consists of magazines, so the magazines make up $\frac{1}{2} \cdot \frac{7}{20}x = \frac{7}{40}x$. 5. Write $\frac{7}{40}x = 105$. Multiply by $\frac{40}{7}$: $x = 600$.

Answer

The library has $600$ items altogether.

Multi-step problems with rational numbers 47

A school project budget assigns $\frac{3}{8}$ to materials, $\frac{1}{5}$ to transportation, $\frac{1}{10}$ to food, and $\frac{1}{4}$ to equipment rental. The remaining $\$15$ is set aside as an emergency fund. Find the total budget and the amount budgeted for materials.

Hints

- Add the four budget fractions first. - Subtract their sum from 1 to find the emergency-fund fraction. - Set that fraction of the total equal to $\$15$. - After finding the total, calculate $\frac{3}{8}$ of it.

Solution

1. Let $x$ be the total budget in dollars. 2. The assigned fractions add to $\frac{3}{8} + \frac{1}{5} + \frac{1}{10} + \frac{1}{4} = \frac{15}{40} + \frac{8}{40} + \frac{4}{40} + \frac{10}{40} = \frac{37}{40}$. 3. The emergency fund is the remaining $1 - \frac{37}{40} = \frac{3}{40}$ of the budget. 4. Write $\frac{3}{40}x = 15$. Multiply by $\frac{40}{3}$: $x = 200$. 5. The materials amount is $\frac{3}{8} \cdot 200 = 75$.

Answer

The total budget is $\$200$, and $\$75$ is budgeted for materials.

Multi-step problems with rational numbers 48

Jordan saved money for a project. First, Jordan spends $\frac{1}{4}$ of the money on reference books. Then Jordan spends $\frac{2}{5}$ of the remaining money on supplies. After both purchases, $\$9.00$ remains. Find the original amount and the amounts spent on books and supplies.

Hints

- Find the fraction remaining after the first purchase. - Express the supplies cost as a fraction of the original amount. - Determine what fraction of the original amount remains after both purchases. - Set the final remaining amount equal to $\$9.00$.

Solution

1. Let $x$ be the original amount in dollars. 2. After the books are purchased, $x - \frac{1}{4}x = \frac{3}{4}x$ remains. 3. The supplies cost $\frac{2}{5} \cdot \frac{3}{4}x = \frac{3}{10}x$. 4. The amount left after both purchases is $\frac{3}{4}x - \frac{3}{10}x = \frac{9}{20}x$. 5. Write $\frac{9}{20}x = 9$. Multiply by $\frac{20}{9}$: $x = 20$. 6. The books cost $\frac{1}{4} \cdot 20 = 5$, and the supplies cost $\frac{3}{10} \cdot 20 = 6$.

Answer

Jordan originally had $\$20.00$, spent $\$5.00$ on books, and spent $\$6.00$ on supplies.

Multi-step problems with rational numbers 49

A composting facility analyzes the organic waste delivered in one day. The pie chart shows the composition of the waste. The facility found exactly $12$ tons of contaminants, such as plastic or metal, in the deliveries. How many tons of food scraps and yard waste were delivered altogether?

Hints

- Use the pie chart to identify the contaminant percent and the combined percent for food scraps and yard waste. - Use the $12$-ton contaminant amount to find the total mass. - Then find $85\%$ of the total. - You can also compare the two percentages directly as a ratio.

Solution

1. The pie chart shows that contaminants are $5\%$ of the total, while food scraps and yard waste together are $55\% + 30\% = 85\%$. 2. Let $T$ be the total mass of waste. Since $5\%$ of the total is $12$ tons, $0.05T = 12$, so $T = 12 \div 0.05 = 240$ tons. 3. The mass of food scraps and yard waste is $240 \cdot 0.85 = 204$ tons. 4. Equivalently, $85\%$ is $17$ times $5\%$, so the desired mass is $17 \cdot 12 = 204$ tons.

Answer

A total of $204$ tons of food scraps and yard waste were delivered.

Multi-step problems with rational numbers 50

A climbing gym offers a group discount on its regular ticket price and charges one equipment-rental fee for the whole group. Which calculation tree gives the correct total cost, A or B? Use the numbers in the trees to find the cost. Explain what goes wrong in the other tree.

Hints

- Which amount is charged once to the group? - Which part of each tree repeats for every person? - Translate the order of operations in each tree into words before calculating.

Solution

1. Tree A subtracts the $\$3$ per-person discount from the $\$18$ regular ticket: $18-3=15$ dollars per person. 2. For $15$ people, admission costs $15\cdot15=225$ dollars. Adding the $\$10$ group fee gives $235$ dollars. 3. Tree B adds the one-time $\$10$ fee before multiplying by the number of people, incorrectly charging it to each person.

Answer

Tree A is correct. The total cost is $\$235$. Tree B incorrectly multiplies the one-time fee by the group size.

Multi-step problems with rational numbers 51

A garden tank first loses $\frac{1}{5}$ of its water. Then $\frac{1}{4}$ of the water that remains is used. There are $12$ gallons left. How many gallons were in the tank at the start? Work backward and explain each step.

Hints

- Start from the amount after the second use. - What fraction remains after each removal? - Reverse the stages in the opposite order.

Solution

1. After the second use, $\frac{3}{4}$ of the water from the first stage remains. Before that use there were $12\div\frac{3}{4}=16$ gallons. 2. The first use left $\frac{4}{5}$ of the starting amount. Starting amount: $16\div\frac{4}{5}=20$ gallons. 3. Check: after losing $4$ gallons there are $16$; after losing another $4$, there are $12$.

Answer

$20$ gallons. Before the second use there were $12\div\frac{3}{4}=16$ gallons; before the first use there were $16\div\frac{4}{5}=20$ gallons.

Multi-step problems with rational numbers 52

A roll of copper wire is $30\,\text{ft}$ long. An electrician cuts off pieces measuring $4\frac{1}{2}\,\text{ft}$, $7\frac{3}{4}\,\text{ft}$, $2\frac{1}{5}\,\text{ft}$, and $5\frac{2}{5}\,\text{ft}$. After one final piece is cut, exactly $25\%$ of the original roll should remain. How long must the final piece be?

Hints

- Find the total length already removed. - Find $25\%$ of the original length. - Determine the total amount that must be removed, then subtract the amount already removed.

Solution

1. The pieces already cut total $4.5+7.75+2.2+5.4=19.85\,\text{ft}$. 2. The target remaining length is $25\%$ of $30\,\text{ft}$: $0.25\cdot30=7.5\,\text{ft}$. 3. Therefore, a total of $30-7.5=22.5\,\text{ft}$ must be removed. 4. The final piece must be $22.5-19.85=2.65\,\text{ft}$.

Answer

The final piece must be $2.65\,\text{ft}$ long.

Multi-step problems with rational numbers 53

A warehouse lifting system can support a maximum combined weight of $1200\,\text{kg}$, including the platform itself and its cargo. During one job, the cargo changes as follows: 1. Load $350.5\,\text{kg}$. 2. Unload $120\frac{1}{4}\,\text{kg}$. 3. Load $480.75\,\text{kg}$. 4. Unload $200\,\text{kg}$. 5. Load $150.5\,\text{kg}$. a) How much cargo is on the platform after step 5? b) What is the greatest possible platform weight that keeps the combined weight at or below $1200\,\text{kg}$ at every step?

Hints

- Track the cargo weight after every step. - Identify when the cargo is heaviest. - Write an inequality using the maximum cargo weight and the $1200\,\text{kg}$ limit.

Solution

1. Track the cargo after each step: $350.5$, $350.5-120.25=230.25$, $230.25+480.75=711$, $711-200=511$, and $511+150.5=661.5$, all in kilograms. 2. The greatest cargo weight is $711\,\text{kg}$, after step 3. 3. Let $P$ be the platform weight. The limiting inequality is $P+711\le1200$. 4. Therefore, $P\le489$.

Answer

a) $661.5\,\text{kg}$ b) The greatest possible platform weight is $489\,\text{kg}$.

Multi-step problems with rational numbers 54

Ms. Meyer buys office supplies: - $4$ notepads at $\$2.35$ each - $2$ packs of highlighters at $\$4.89$ each - $1$ calendar for $\$12.50$ She uses store credits of $\$5.50$, $\$3.75$, and $\$8.20$. a) Find the exact amount she still owes after all credits are applied. b) A cashier incorrectly says she owes $\$16.58$. Explain a likely scanning error that would produce exactly that amount.

Hints

- Find the purchase total before applying the store credits. - Add the three store-credit values. - Compare the correct amount owed with $\$16.58$. - Check whether the difference matches an item price.

Solution

1. The notepads cost $4\cdot\$2.35=\$9.40$. 2. The highlighters cost $2\cdot\$4.89=\$9.78$. 3. The purchase total is $\$9.40+\$9.78+\$12.50=\$31.68$. 4. The store credits total $\$5.50+\$3.75+\$8.20=\$17.45$. 5. For a), she owes $\$31.68-\$17.45=\$14.23$. 6. For b), the incorrect amount is $\$16.58-\$14.23=\$2.35$ too high. That is exactly the price of one notepad, so one notepad was likely scanned twice or a fifth notepad was entered.

Answer

a) $\$14.23$ b) One extra $\$2.35$ notepad was likely charged.

Multi-step problems with rational numbers 55

A juice bar has $20$ gallons of juice to package. It can use small jugs that hold $0.33$ gallon or large jugs that hold $0.75$ gallon. a) How many large jugs can be filled completely, and how much juice remains? b) A manager claims that using only small jugs leaves less juice unused than using only large jugs. Check the claim. c) If $20$ large jugs are filled first, how many small jugs can then be filled completely while using as much of the $20$ gallons as possible?

Hints

- “Filled completely” means only whole containers count. - Find the amount used by the full containers, then subtract from the total. - For part c), first find how much juice the $20$ large jugs use.

Solution

1. For a), $26\cdot0.75=19.5$, so $26$ large jugs can be filled and $20-19.5=0.5$ gallon remains. 2. With small jugs, $60\cdot0.33=19.8$, so $0.2$ gallon remains. Since $0.2<0.5$, the manager's claim is true. 3. For c), $20$ large jugs hold $20\cdot0.75=15$ gallons, leaving $5$ gallons. 4. Fifteen small jugs use $15\cdot0.33=4.95$ gallons, leaving $0.05$ gallon. A sixteenth small jug cannot be filled completely.

Answer

a) $26$ large jugs; $0.5$ gallon remains b) The claim is true. Small jugs leave $0.2$ gallon, while large jugs leave $0.5$ gallon. c) $15$ small jugs

Multi-step problems with rational numbers 56

Consider the expression $(12.50+3\cdot4.50)\div2$. a) Write a realistic word problem that can be represented by this expression. Include appropriate units. b) Evaluate the expression.

Hints

- The parentheses can represent a combined total. - Dividing by $2$ can represent sharing equally between two people. - Use a context in which all quantities and units make sense.

Solution

1. One possible story is: Two friends share the cost of a $\$12.50$ pizza and three drinks that cost $\$4.50$ each. 2. Evaluate the multiplication: $3\cdot4.50=13.50$. 3. Add inside the parentheses: $12.50+13.50=26.00$. 4. Divide equally: $26.00\div2=13.00$.

Answer

a) Answers will vary. One example is two friends equally sharing a $\$12.50$ pizza and three $\$4.50$ drinks. b) $13$, or $\$13.00$ in the example

Multi-step problems with rational numbers 57

A strawberry farm harvests $120$ lb of strawberries. Production costs are $\$1.80$ per pound, and the farm wants a total profit of $\$240$. In the morning, $70$ lb are sold for $\$4.50$ per pound. The remaining strawberries will be sold for a lower price in the afternoon. How much lower, in dollars per pound, must the afternoon price be than the morning price to meet the profit goal exactly?

Hints

- Find the total production cost first. - Add the target profit to find the total revenue needed. - Determine the remaining revenue and remaining pounds after the morning sales. - Compare the required afternoon price with the morning price.

Solution

1. Total production cost: $120\cdot1.80=216$ dollars. 2. Required total revenue: $216+240=456$ dollars. 3. Morning revenue: $70\cdot4.50=315$ dollars. 4. Remaining revenue needed: $456-315=141$ dollars. 5. Remaining amount: $120-70=50$ lb. 6. Afternoon price: $141\div50=2.82$ dollars per pound. 7. Price difference: $4.50-2.82=1.68$ dollars per pound.

Answer

The afternoon price must be $\$1.68$ per pound lower.

Multi-step problems with rational numbers 58

A stack of freshly cut wood weighs $120\,\text{lb}$ and is $85\%$ water. After drying, the wood is $80\%$ water. a) Find the new total weight of the wood. b) Explain why the total weight decreases substantially even though the water percentage decreases by only $5$ percentage points.

Hints

- Find the initial weight of the dry wood. - What percent of the final weight is dry wood? - The amount of dry wood stays constant while the total changes.

Solution

1. Initially, the dry wood is $100\% - 85\% = 15\%$ of the total. Its weight is $120\,\text{lb} \cdot 0.15 = 18\,\text{lb}$. 2. The dry wood remains $18\,\text{lb}$. After drying, it is $100\% - 80\% = 20\%$ of the total weight. 3. Let $w$ be the new total weight. Then $0.20w = 18$, so $w = 18 \div 0.20 = 90$. 4. The dry portion rises from $15\%$ to $20\%$ of the total. Because its weight stays fixed, the total must fall enough for $18\,\text{lb}$ to become one-fifth of the whole.

Answer

a) The new total weight is $90\,\text{lb}$. b) The dry wood stays at $18\,\text{lb}$, but its share rises from $15\%$ to $20\%$, so the total weight must decrease to $90\,\text{lb}$.

Multi-step problems with rational numbers 59

Class A has $25$ students, including $12$ boys. Class B has $20$ students. If $4$ new boys joined Class B, boys would make up exactly $50\%$ of that class. Find the original percent of students in Class B who were boys. Which class originally had the greater percent of boys?

Hints

- First find the percent for Class A. - For Class B, find the total number of students after the new students join. - Use the future $50\%$ condition to work backward to the original number of boys. - Compare the two original percentages.

Solution

1. In Class A, the percent of boys is $\frac{12}{25} \cdot 100\% = 48\%$. 2. After the change, Class B would have $20 + 4 = 24$ students. 3. If $50\%$ of the $24$ students were boys, there would be $24 \cdot 0.50 = 12$ boys. 4. Before the $4$ boys joined, Class B had $12 - 4 = 8$ boys. 5. The original percent of boys in Class B was $\frac{8}{20} \cdot 100\% = 40\%$. 6. Since $48\% > 40\%$, Class A originally had the greater percent of boys.

Answer

The original percent in Class B was $40\%$. Class A had the greater original percent of boys, at $48\%$.

Multi-step problems with rational numbers 60

At an amusement park, $500$ visitors were surveyed. Of them, $42\%$ said they love roller coasters. Of those roller-coaster fans, about $15\%$, rounded to the nearest whole percent, said wooden coasters are their favorite. a) Using the reported percentages, what percent of all surveyed visitors prefer wooden coasters? b) What numerical value does this give for the number of people out of $500$? Explain why it is not an exact head count. c) An employee says exactly $31$ people preferred wooden coasters. Is this consistent with the reported $15\%$, rounded to the nearest whole percent?

Hints

- Consider what “rounded to the nearest whole percent” says about precision. - First calculate the reported portion of all $500$ visitors. - A head count must be a whole number. - Check $31$ as a percent of the roller-coaster fans and round it.

Solution

1. For part a, the calculated percent of all visitors is $0.42 \cdot 0.15 = 0.063 = 6.3\%$. 2. For part b, $500 \cdot 0.063 = 31.5$. This is not an exact count because people are counted in whole numbers and the $15\%$ value was rounded. 3. There were $500 \cdot 0.42 = 210$ roller-coaster fans. 4. If $31$ of them preferred wooden coasters, the actual subgroup percent was $\frac{31}{210} \cdot 100\% \approx 14.76\%$. 5. Since $14.76\%$ rounds to $15\%$, the employee’s statement is consistent with the survey report.

Answer

a) The calculated overall percent is $6.3\%$. b) The calculation gives $31.5$ people, but this is not an exact count because the subgroup percent was rounded and people are counted in whole numbers. c) Yes. $31$ out of $210$ is approximately $14.76\%$, which rounds to $15\%$.

Multi-step problems with rational numbers 61

A bakery has fresh pretzels in a basket. The first customer buys half of them. The second customer then buys $15$ pretzels. The third customer buys half of the pretzels that remain. Afterward, $10$ pretzels are left. How many pretzels were in the basket at first?

Hints

- Start with the final amount and reverse each step. - If half remained after a purchase, how many were there before it? - Add back any fixed number that was removed.

Solution

1. Work backward from the final $10$ pretzels. 2. Before the third customer bought half, there were $10 \cdot 2 = 20$ pretzels. 3. Before the second customer bought $15$, there were $20 + 15 = 35$ pretzels. 4. Before the first customer bought half, there were $35 \cdot 2 = 70$ pretzels. 5. Checking forward leaves $10$ pretzels, so the result is correct.

Answer

$70$ pretzels.

Multi-step problems with rational numbers 62

A school garden is divided into different areas. Flower beds cover $\frac{2}{7}$ of the garden, and vegetables cover exactly one-half of the garden. The rest is a paved path. The path has an area that is exactly $30\,\text{ft}^2$ less than the flower-bed area. a) What fraction of the garden is the path? b) What is the total area of the garden?

Hints

- Subtract the other two fractions from the whole to find the path’s fraction. - Use a variable for the total garden area. - Translate “$30\,\text{ft}^2$ less than” into an equation. - Compare the path area with the flower-bed area.

Solution

1. The fraction used for the path is $1 - \frac{2}{7} - \frac{1}{2} = \frac{14}{14} - \frac{4}{14} - \frac{7}{14} = \frac{3}{14}$. 2. Let $x$ be the total garden area in square feet. 3. The flower-bed area is $\frac{2}{7}x$, and the path area is $\frac{3}{14}x$. 4. Because the path is $30\,\text{ft}^2$ smaller than the flower beds, write $\frac{3}{14}x = \frac{2}{7}x - 30$. 5. Rewrite $\frac{2}{7}x$ as $\frac{4}{14}x$. Then $\frac{3}{14}x = \frac{4}{14}x - 30$. 6. Subtract $\frac{3}{14}x$: $\frac{1}{14}x = 30$, so $x = 420$.

Answer

a) The path covers $\frac{3}{14}$ of the garden. b) The total garden area is $420\,\text{ft}^2$.

Multi-step problems with rational numbers 63

A hiking group plans a three-day route. On the first day, the group hikes $\frac{2}{5}$ of the total distance. On the second day, the group hikes $\frac{1}{3}$ of the distance that remained after the first day. On the third day, the group hikes the final $12\,\text{miles}$. Find the total route length and identify the days on which the group hiked the same distance.

Hints

- Find the fraction of the route remaining after the first day. - Express the second-day distance as a fraction of the total route. - Determine what fraction remains for the third day. - Find each day’s distance before comparing them.

Solution

1. Let $d$ be the total route length in miles. The first-day distance is $\frac{2}{5}d$. 2. After the first day, $d - \frac{2}{5}d = \frac{3}{5}d$ remains. 3. The second-day distance is $\frac{1}{3} \cdot \frac{3}{5}d = \frac{1}{5}d$. 4. The third-day distance is $\frac{3}{5}d - \frac{1}{5}d = \frac{2}{5}d$. 5. Write $\frac{2}{5}d = 12$. Multiply by $\frac{5}{2}$: $d = 30$. 6. The daily distances are $12\,\text{miles}$, $6\,\text{miles}$, and $12\,\text{miles}$. Therefore, the first and third days have equal distances.

Answer

The route is $30\,\text{miles}$ long. The group hikes $12\,\text{miles}$ on both the first and third days.

Multi-step problems with rational numbers 64

At an apple-processing facility, $5\%$ of a delivery is removed because the apples are damaged. The remaining apples are pressed, and the juice produced weighs $75\%$ as much as the apples that are pressed. How many pounds of apples must be delivered to produce $570\,\text{lb}$ of juice?

Hints

- Work backward in two stages. - First find how many pounds of apples must be pressed to produce the desired juice mass. - Then find the original delivery before the damaged apples were removed. - You can also use one combined multiplier.

Solution

1. The mass of apples that must be pressed is $570\,\text{lb} \div 0.75 = 760\,\text{lb}$. 2. Those $760\,\text{lb}$ are $95\%$ of the original delivery because $5\%$ was removed. 3. Let $x$ be the original delivery. Then $0.95x = 760$, so $x = 760 \div 0.95 = 800$. 4. Equivalently, $x \cdot 0.95 \cdot 0.75 = 570$, which also gives $x = 800$.

Answer

The facility must receive $800\,\text{lb}$ of apples.

Multi-step problems with rational numbers 65

A hiker carries water for a three-day trip. Each day, the hiker uses exactly half of the amount available that morning plus $0.25\,\text{qt}$. At the end of the third day, $0.5\,\text{qt}$ remains. How much water did the hiker have at the beginning? Write an equation or work backward.

Hints

- Begin with the amount left at the end of the third day. - If $x$ is the morning amount, the evening amount is $\frac{x}{2} - 0.25$. - Reverse “use half plus one-fourth quart” one day at a time. - A three-row table may help organize the backward steps.

Solution

1. Work backward from the end of the third day. If $x_3$ is the morning amount, then $\frac{x_3}{2} - 0.25 = 0.5$. Thus $\frac{x_3}{2} = 0.75$, so $x_3 = 1.5\,\text{qt}$. 2. Before the second day, solve $\frac{x_2}{2} - 0.25 = 1.5$. Then $\frac{x_2}{2} = 1.75$, so $x_2 = 3.5\,\text{qt}$. 3. Before the first day, solve $\frac{x_1}{2} - 0.25 = 3.5$. Then $\frac{x_1}{2} = 3.75$, so $x_1 = 7.5\,\text{qt}$.

Answer

The hiker started with $7.5\,\text{qt}$ of water.

Multi-step problems with rational numbers 66

A class manages a field-trip fund. In the first month, the class spends one-third of the original fund on tickets and another $\$50$ on a bus reservation. In the second month, the class spends one-fourth of the remaining money on food and another $\$30$ on a registration fee. After these expenses, $\$300$ remains. How much money was originally in the fund?

Hints

- Work backward from the final $\$300$. - After one-fourth of an amount is spent, three-fourths remains before the fixed fee is considered. - Find the amount present before the second month, then find the original amount. - Keep the percentage-based expenses and fixed expenses separate.

Solution

1. Let $R$ be the amount remaining after the first month. After the second-month expenses, $R - (\frac{1}{4}R + 30) = 300$. 2. Simplify: $\frac{3}{4}R - 30 = 300$, so $\frac{3}{4}R = 330$ and $R = 440$. 3. Let $x$ be the original fund. After the first month, $x - (\frac{1}{3}x + 50) = 440$. 4. Simplify: $\frac{2}{3}x - 50 = 440$, so $\frac{2}{3}x = 490$. 5. Multiply by $\frac{3}{2}$: $x = 735$.

Answer

The fund originally contained $\$735$.

Multi-step problems with rational numbers 67

**Scoring in an Adventure Game** In a computer game, a player’s raw score is converted to a level score by the calculation tree. a) Find the level score for each raw score. 1) $100$ points 2) $250$ points 3) $75$ points b) A player earns exactly $120$ level-score points. Work backward to find the raw score. c) Write the process as one expression using $x$ for the raw score.

Hints

- Follow the tree from the raw score toward the final result. - To work backward, apply inverse operations in reverse order. - Use parentheses so the subtraction occurs before division.

Solution

1. The calculation is $(4x-200)\div5$. For $x=100$: $(400-200)\div5=40$. For $x=250$: $(1000-200)\div5=160$. For $x=75$: $(300-200)\div5=20$. 2. Work backward from $120$: $120\cdot5=600$, $600+200=800$, and $800\div4=200$. The raw score was $200$. 3. The expression is $(4x-200)\div5$.

Answer

a) 1) $40$ 2) $160$ 3) $20$ b) $200$ raw-score points c) $(4x-200)\div5$

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

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Multi-step problems with rational numbers

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