Aimathic
Login | English | Deutsch

Free Math Worksheets

Build your own math worksheets from 21,000 problems for grades 3 to 12, from fractions to calculus. Every problem includes step-by-step solutions.

Model situations with equations and inequalities

Click problems to add them to your worksheet.

5122147
In the United States, temperature is commonly measured in degrees Fahrenheit. To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius value by \(1.8\) and add \(32\). a) Write an expression for the Fahrenheit temperature. Let \(c\) be the temperature in degrees Celsius. b) Use the expression to convert \(20\,^{\circ}\text{C}\), \(35\,^{\circ}\text{C}\), and \(-10\,^{\circ}\text{C}\) to degrees Fahrenheit.

Hints

- Translate the two operations in the order they are described. - Be careful with the signs when substituting a negative temperature.

Solution

1. Following the stated operations gives \(1.8c+32\). 2. For \(20\,^{\circ}\text{C}\): \(1.8\cdot 20+32=36+32=68\,^{\circ}\text{F}\). 3. For \(35\,^{\circ}\text{C}\): \(1.8\cdot 35+32=63+32=95\,^{\circ}\text{F}\). 4. For \(-10\,^{\circ}\text{C}\): \(1.8\cdot(-10)+32=-18+32=14\,^{\circ}\text{F}\).

Answer

a) \(1.8c+32\) b) \(68\,^{\circ}\text{F}\), \(95\,^{\circ}\text{F}\), and \(14\,^{\circ}\text{F}\)
5125537
Write and solve an equation for each number riddle. a) What number must be increased by \(2.5\) to get \(1\frac{1}{4}\)? b) What number gives \(-1.5\) when \(\frac{2}{3}\) is subtracted from it? c) Three times what number is \(12.6\)?

Hints

- Translate each statement into an equation before solving. - Which operations are represented by “increased by” and “subtracted from”? - Use a variable for the unknown number. - Express fractions and decimals in compatible forms before calculating.

Solution

1. For part a, write \(x + 2.5 = 1.25\). Subtract \(2.5\): \(x = 1.25 - 2.5 = -1.25\). 2. For part b, write \(x - \frac{2}{3} = -1.5\). Since \(-1.5 = -\frac{3}{2}\), add \(\frac{2}{3}\): \(x = -\frac{3}{2} + \frac{2}{3} = -\frac{9}{6} + \frac{4}{6} = -\frac{5}{6}\). 3. For part c, write \(3x = 12.6\). Divide by \(3\): \(x = 4.2\).

Answer

a) \(x = -1.25\) b) \(x = -\frac{5}{6}\) c) \(x = 4.2\)
5126257
Three piles of counters are on a table: a left pile \(L\), a middle pile \(M\), and a right pile \(R\). The left pile has at least \(6\) counters. The middle pile starts with \(z\) counters. Follow these directions: 1. Move \(6\) counters from the left pile to the middle pile. 2. Move \(4\) counters from the middle pile to the right pile. a) Write a simplified expression for the number of counters in the middle pile after both moves. b) After both moves, the middle pile has \(15\) counters. How many counters were in the middle pile at the start?

Hints

- Decide whether each move adds counters to or removes counters from the middle pile. - Write the two changes in order. - Use the final amount to write an equation for the starting amount.

Solution

1. After the first move, the middle pile has \(z+6\) counters. 2. After the second move, it has \(z+6-4=z+2\) counters. 3. Set the final amount equal to \(15\): \(z+2=15\). 4. Subtract \(2\) from both sides to get \(z=13\).

Answer

a) \(z+2\) b) \(13\) counters
5188307
Find each number. a) A number increased by \(125\) equals \(400\). b) Four times a number equals \(320\).

Hints

- Write an equation for each statement. - Use subtraction to undo an increase. - Use division to undo multiplication.

Solution

1. For a), solve \(x + 125 = 400\): \(x = 400 - 125 = 275\). 2. For b), solve \(4x = 320\): \(x = 320 \div 4 = 80\).

Answer

a) \(275\) b) \(80\)
5188317
Luke says, “When I divide my number by \(2\), I get \(150\).” Marie says, “When I add \(150\) to my number, I get \(400\).” Who chose the greater number? Justify your answer with calculations.

Hints

- For Luke, undo division by \(2\) by multiplying. - For Marie, undo addition of \(150\) by subtracting. - Compare the two numbers.

Solution

1. Luke's number satisfies \(x \div 2 = 150\), so \(x = 150 \cdot 2 = 300\). 2. Marie's number satisfies \(y + 150 = 400\), so \(y = 400 - 150 = 250\). 3. Since \(300 > 250\), Luke chose the greater number.

Answer

Luke chose \(300\), and Marie chose \(250\). Luke chose the greater number.
5194027
Luke says, “When I subtract \(360\) from my number, I get \(240\).” Julia says, “When I add \(120\) to my number, I get \(500\).” Who chose the greater number? Justify your answer with calculations.

Hints

- Find Luke’s number first. - Then find Julia’s number. - Reverse each operation to recover the starting number. - Compare the two solutions.

Solution

1. Luke's number is \(240 + 360 = 600\). 2. Julia's number is \(500 - 120 = 380\). 3. Since \(600 > 380\), Luke chose the greater number.

Answer

Luke chose \(600\), and Julia chose \(380\). Luke chose the greater number.
5223447
A gym charges a one-time enrollment fee of \(\$s\) and a monthly fee of \(\$m\). a) Write an expression for the total cost of a membership lasting \(x\) months. b) Find the total cost for one year when \(s=35.00\) and \(m=19.90\).

Hints

- Separate the one-time fee from the fee that repeats each month. - One year has \(12\) months. - Include the currency unit in the final answer.

Solution

1. The monthly fees total \(mx\), and the one-time fee is added once, so \(G=mx+s\). 2. For one year, \(x=12\). Substitute the values: \(G=19.90\cdot 12+35.00=238.80+35.00=273.80\).

Answer

a) \(G=mx+s\) b) \(\$273.80\)
5224677
Write and solve an equation for each statement. 1) A number \(x\) is increased by \(8.4\), and the result is \(15\). 2) A number \(y\) is subtracted from \(42\), and the result is \(13\frac{1}{2}\). 3) Three times a number \(a\) is \(25.5\).

Hints

- Translate each verbal statement into an equation. - Identify the operations represented by “increased by,” “subtracted from,” and “three times.” - Use inverse operations to isolate the variable.

Solution

1. Write \(x + 8.4 = 15\). Subtract \(8.4\): \(x = 6.6\). 2. Write \(42 - y = 13.5\). Subtract \(42\): \(-y = -28.5\), so \(y = 28.5\). 3. Write \(3a = 25.5\). Divide by \(3\): \(a = 8.5\).

Answer

1) \(x = 6.6\) 2) \(y = 28.5\) 3) \(a = 8.5\)
5229107
For values of \(x\) and \(y\) that make all lengths positive and form a valid triangle, a student claims that these figures have equal perimeters: A square with side length \(s=1.5x+2y\) A triangle with side lengths \(2x+3y\), \(3x+4y\), and \(x+y\) Check the claim by writing and simplifying an expression for each perimeter.

Hints

- Multiply a square’s side length by \(4\). - Add all three side lengths of the triangle. - Compare the simplified expressions.

Solution

1. The square’s perimeter is \(4(1.5x+2y)=6x+8y\). 2. The triangle’s perimeter is \((2x+3y)+(3x+4y)+(x+y)\). 3. Combine like terms for the triangle: \(2x+3x+x=6x\) and \(3y+4y+y=8y\). 4. Both perimeters equal \(6x+8y\), so the claim is true.

Answer

The claim is true. Both perimeters are \(6x+8y\).
5229117
For values of \(x\) and \(y\) that make the lengths positive and form a valid triangle, the side lengths are \(5x+2y\), \(3x-4y\), and \(2x+3y\). Write and simplify an expression for the perimeter \(P\).

Hints

- A triangle’s perimeter is the sum of its side lengths. - Combine all \(x\)-terms and all \(y\)-terms separately. - Keep the negative sign on \(-4y\).

Solution

1. Add the three side lengths: \(P=(5x+2y)+(3x-4y)+(2x+3y)\). 2. Combine the \(x\)-terms: \(5x+3x+2x=10x\). 3. Combine the \(y\)-terms: \(2y-4y+3y=y\). 4. Therefore, \(P=10x+y\).

Answer

\(P=10x+y\)
5229157
For values of \(x\) and \(y\) that make the lengths positive and form a valid quadrilateral, its side lengths are \(s_1=3x+2\) \(s_2=2y-1\) \(s_3=x+y+4\) \(s_4=2x-y+3\). Write and simplify an expression for the perimeter \(P\).

Hints

- Perimeter is the sum of all side lengths. - Combine the \(x\)-terms, \(y\)-terms, and constants separately. - Keep each term’s sign.

Solution

1. Add all four side lengths: \(P=(3x+2)+(2y-1)+(x+y+4)+(2x-y+3)\). 2. Combine the \(x\)-terms: \(3x+x+2x=6x\). 3. Combine the \(y\)-terms: \(2y+y-y=2y\). 4. Combine the constants: \(2-1+4+3=8\). 5. Therefore, \(P=6x+2y+8\).

Answer

\(P=6x+2y+8\)
5240657
Leonie mixes \(x\) liters of fruit juice concentrate with \(y\) liters of water. The concentrate contains \(s\) grams of sugar per liter. a) Write an expression for the total amount of sugar in the pitcher. b) Write an expression for the sugar concentration, in grams per liter, of the mixture. c) Find the concentration when \(x=0.4\), \(y=1.6\), and \(s=150\).

Hints

- Multiply the concentrate volume by its sugar concentration. - Add the concentrate and water volumes. - Divide the total grams of sugar by the total volume.

Solution

1. The concentrate contains \(xs\) grams of sugar. 2. The mixture's total volume is \(x+y\) liters. 3. The sugar concentration is \(\frac{xs}{x+y}\) grams per liter. 4. For the given values, \(\frac{0.4\cdot 150}{0.4+1.6}=\frac{60}{2}=30\) grams per liter.

Answer

a) \(xs\) grams b) \(\frac{xs}{x+y}\) grams per liter c) \(30\) grams per liter
5240707
A streaming service offers two monthly plans. Basic: a monthly fee of \(\$G\) plus \(\$k\) for each rented movie. Premium: a flat monthly fee of \(\$P\) with no additional charge per movie. Let \(K_P\) be the Premium plan's monthly cost. a) Write an expression for the Basic plan's monthly cost \(K_B\) when a customer rents \(n\) movies. b) Write an expression for the difference \(K_P-K_B\). c) What does it mean when the expression from part b) equals \(0\)?

Hints

- Add the fixed fee and the per-movie charges for the Basic plan. - Subtract the entire Basic expression from the Premium cost. - A difference of \(0\) means the two quantities are equal.

Solution

1. The Basic plan costs \(K_B=G+nk\). 2. The Premium plan costs \(K_P=P\). 3. The difference is \(K_P-K_B=P-(G+nk)=P-G-nk\). 4. If the difference is \(0\), then \(P=G+nk\), so the plans cost the same for that number of movies.

Answer

a) \(K_B=G+nk\) b) \(P-(G+nk)=P-G-nk\) c) Both plans have the same monthly cost for that number of movies.
5279227
A class rents a bus for a field trip. The cost includes a fixed fee of \(\$G\) plus \(\$p\) for each of \(n\) participants. a) Write an expression for the total cost \(K\). b) Find the total cost for each scenario: 1) \(G=120.00\), \(p=14.50\), and \(n=24\) 2) \(G=85.00\), \(p=18.00\), and \(n=28\)

Hints

- Separate the fixed cost from the cost that depends on the number of people. - Multiply before adding. - Include dollars and cents in each final amount.

Solution

1. The total cost is the fixed fee plus the per-person cost: \(K=G+np\). 2. For the first scenario, \(K=120.00+24\cdot 14.50=120.00+348.00=468.00\). 3. For the second scenario, \(K=85.00+28\cdot 18.00=85.00+504.00=589.00\).

Answer

a) \(K=G+np\) b) 1) \(\$468.00\) 2) \(\$589.00\)
5368057
In a tangential quadrilateral, the sums of the lengths of opposite sides are equal. Tangential quadrilateral \(ABCD\) has side lengths \(a = 8\,\text{in.}\), \(b = 11\,\text{in.}\), and \(c = 10\,\text{in.}\). Find side length \(d\).
Figure for problem 536805

Hints

- Use the stated relationship between opposite side pairs. - Substitute the three known side lengths. - Isolate the unknown side length.

Solution

1. For a tangential quadrilateral, \(a + c = b + d\). 2. Substitute the known lengths: \(8 + 10 = 11 + d\). 3. Simplify: \(18 = 11 + d\). 4. Subtract \(11\): \(d = 7\).

Answer

\(d = 7\,\text{in.}\)
5107277
A bookshelf is \(\frac{1}{3}\) full. After \(12\) more books are added, it is \(\frac{1}{2}\) full. How many books fit on the shelf when it is completely full?

Hints

- Let a variable represent the total capacity. - Write an equation comparing the original and new fractions of the shelf. - Find the difference between \(\frac{1}{2}\) and \(\frac{1}{3}\).

Solution

1. Let \(x\) be the total number of book spaces. 2. The equation is \(\frac{1}{3}x+12=\frac{1}{2}x\). 3. Subtract \(\frac{1}{3}x\): \(12=\left(\frac{1}{2}-\frac{1}{3}\right)x=\frac{1}{6}x\). 4. Multiply by \(6\): \(x=72\).

Answer

\(72\) books
5119207
A cookbook gives this rule for roasting meat: “Allow \(45\) minutes per pound, plus \(20\) additional minutes for preheating and resting.” a) Write an equation for the total time \(T\), in minutes, for a roast weighing \(m\) pounds. b) Find the total time for a \(2.5\)-pound roast. c) A roast takes \(155\) minutes in all. Use your equation to find its weight.

Hints

- The phrase “per pound” indicates multiplication, and “plus” indicates a fixed addition. - Identify what the variable \(m\) represents. - To find the weight from the total time, reverse the operations in the equation.

Solution

1. The cooking time is \(45\) minutes for each pound plus \(20\) fixed minutes, so \(T=45m+20\). 2. For \(m=2.5\), \(T=45\cdot 2.5+20=112.5+20=132.5\) minutes. 3. Set \(T=155\): \(155=45m+20\). Subtract \(20\) to get \(135=45m\), then divide by \(45\): \(m=3\).

Answer

a) \(T=45m+20\) b) \(132.5\) minutes c) \(3\) pounds
5120387
Posters will be hung along a \(44\,\text{ft}\) wall. Each poster is \(4\,\text{ft}\) wide. All gaps—including the gaps from the end posters to the ends of the wall—must have the same width. a) What is the width of each gap when \(4\) posters are hung? b) How many posters fit when every gap is exactly \(4\,\text{ft}\) wide?

Hints

- Find the total width occupied by the posters. - With \(n\) posters, count the gaps at both ends as well as those between posters. - For part b), write an equation using poster width and gap width.

Solution

1. Four posters use \(4 \cdot 4\,\text{ft} = 16\,\text{ft}\). 2. The remaining length is \(44\,\text{ft} - 16\,\text{ft} = 28\,\text{ft}\). 3. Four posters create \(5\) equal gaps: one at each end and three between posters. 4. Each gap is \(28\,\text{ft} \div 5 = 5.6\,\text{ft}\). 5. For b), let \(x\) be the number of posters. There are \(x + 1\) gaps, so \(4x + 4(x + 1) = 44\). 6. Solve: \(4x + 4x + 4 = 44\), so \(8x = 40\) and \(x = 5\).

Answer

a) \(5.6\,\text{ft}\) b) \(5\) posters
5120747
Square posters with side length \(2.5\,\text{ft}\) will be hung in one row along a \(43\,\text{ft}\) school hallway wall. Equal gaps of width \(x\) will be left between adjacent posters and at both ends of the row. a) Explain why the total-length model for \(n\) posters is \(L=2.5n+x(n+1)\). b) Exactly \(12\) posters will be used. Find \(x\) when the entire wall is filled. c) How would the model change if the posters touched each other but the row had a fixed \(4\,\text{ft}\) margin at each end?

Hints

- Visualize a short row with two or three posters and count the gaps. - Substitute the known values into the length model. - Subtract the total poster width before solving for the gap width.

Solution

1. a) The posters contribute \(2.5n\) feet. There is one more gap than posters, so the gaps contribute \(x(n+1)\) feet. 2. b) Substitute \(n=12\) and \(L=43\): \(2.5\cdot12+13x=43\). 3. Simplify: \(30+13x=43\), so \(13x=13\) and \(x=1\,\text{ft}\). 4. c) The posters contribute \(2.5n\), and the two fixed margins contribute \(8\). The model is \(L=2.5n+8\).

Answer

a) There are \(n\) posters and \(n+1\) equal gaps. b) \(x=1\,\text{ft}\) c) \(L=2.5n+8\)
5121747
A water tank starts with \(1000\) gallons. Because of a leak, it loses water at a constant rate of \(15\) gallons per hour. a) Write an expression for the amount of water remaining after \(t\) hours. b) Use the expression to find how much water remains after \(12\) hours. c) After how many hours will the tank be empty? Round to the nearest tenth of an hour.

Hints

- Identify the starting amount and the amount lost each hour. - After \(t\) hours, the hourly loss has occurred \(t\) times. - An empty tank contains \(0\) gallons.

Solution

1. Start with \(1000\) and subtract \(15\) gallons for each hour: \(1000-15t\). 2. For \(t=12\), \(1000-15\cdot 12=1000-180=820\). 3. Set the remaining amount equal to \(0\): \(1000-15t=0\). Then \(15t=1000\), so \(t=\frac{1000}{15}=66.666\ldots\approx 66.7\).

Answer

a) \(1000-15t\) b) \(820\) gallons c) About \(66.7\) hours
5121877
A bank statement shows a \(\$450.00\) rent payment and a \(\$1250.60\) paycheck deposit. After both transactions, the account balance is \(\$840.25\). Write and solve an equation to find the account balance immediately before the two transactions.

Hints

- Decide whether each transaction increases or decreases the account balance. - Write an equation that begins with the unknown starting balance and ends with the given final balance. - Combine the two known changes before isolating the variable.

Solution

1. Let \(x\) be the starting balance. Write the equation \(x - 450.00 + 1250.60 = 840.25\). 2. Combine the known changes: \(-450.00 + 1250.60 = 800.60\), so \(x + 800.60 = 840.25\). 3. Subtract \(800.60\) from both sides: \(x = 840.25 - 800.60\). 4. Calculate the starting balance: \(x = 39.65\).

Answer

The account balance was \(\$39.65\) before the two transactions.
5121887
A hot-air balloon rises \(245\,\text{m}\) and then descends \(312\,\text{m}\). Its final altitude is \(128\,\text{m}\). Write and solve an equation to find the balloon’s altitude before these two changes.

Hints

- Decide which operation represents rising and which represents descending. - Find the net change in altitude after both movements. - Use an inverse operation to undo the net change and isolate the starting altitude.

Solution

1. Let \(h\) be the starting altitude. Write the equation \(h + 245 - 312 = 128\). 2. Combine the altitude changes: \(245 - 312 = -67\), so \(h - 67 = 128\). 3. Add \(67\) to both sides: \(h = 128 + 67\). 4. Calculate the starting altitude: \(h = 195\).

Answer

The balloon’s starting altitude was \(195\,\text{m}\).
5124137
A quadrilateral has a perimeter of \(24\,\text{cm}\). Find its four side lengths from each description. a) The quadrilateral has \(180^\circ\) rotational symmetry, and all four sides are congruent. b) The quadrilateral has exactly one line of symmetry, and that line is one of its diagonals. Two sides are each \(5\,\text{cm}\) long. c) The quadrilateral is an isosceles trapezoid with parallel sides of lengths \(10\,\text{cm}\) and \(4\,\text{cm}\).

Hints

- Use the symmetry description to determine which side lengths must be equal. - Write a perimeter equation for each quadrilateral. - In part b, remember that a rhombus would have more than one line of symmetry. - In part c, the two legs of an isosceles trapezoid are congruent.

Solution

1. In part a, all four sides are congruent, so each side is \(24\div4=6\,\text{cm}\). 2. In part b, the quadrilateral is a kite with two pairs of adjacent congruent sides. Let each unknown side be \(x\). Then \(2\cdot5+2x=24\). Solving gives \(2x=14\), so \(x=7\). The side lengths are \(5\,\text{cm}\), \(5\,\text{cm}\), \(7\,\text{cm}\), and \(7\,\text{cm}\). 3. In part c, the two legs are congruent. Let each leg have length \(x\). Then \(10+4+2x=24\). Solving gives \(2x=10\), so \(x=5\). The side lengths are \(10\,\text{cm}\), \(5\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\).

Answer

a) \(6\,\text{cm}\), \(6\,\text{cm}\), \(6\,\text{cm}\), \(6\,\text{cm}\). b) \(5\,\text{cm}\), \(5\,\text{cm}\), \(7\,\text{cm}\), \(7\,\text{cm}\). c) \(10\,\text{cm}\), \(5\,\text{cm}\), \(4\,\text{cm}\), \(5\,\text{cm}\).
5124907
A rectangular garden bed has width \(x\) meters. Its length is \(3\,\text{m}\) more than twice its width. a) Write an expression for the perimeter \(P\). b) Simplify the expression. c) Find the total fence length needed when the width is \(4.5\,\text{m}\).

Hints

- Express the length in terms of \(x\). - Use the perimeter formula for a rectangle. - Distribute before combining like terms.

Solution

1. The width is \(x\), and the length is \(2x+3\). 2. The perimeter is \(P=2x+2(2x+3)\). 3. Distribute and combine like terms: \(P=2x+4x+6=6x+6\). 4. For \(x=4.5\), \(P=6\cdot4.5+6=27+6=33\,\text{m}\).

Answer

a) \(P=2x+2(2x+3)\) b) \(P=6x+6\) c) \(33\,\text{m}\)
5125137
A row of adjacent squares is built with matchsticks. The first square uses \(4\) matchsticks. Each additional square shares one side with the square before it, so it requires only \(3\) more matchsticks. a) Write an equation for the number of matchsticks \(S\) needed for \(n\) squares. b) How many matchsticks are needed for \(15\) squares? c) A row uses \(76\) matchsticks. How many squares are in the row?

Hints

- Identify how many matchsticks each new square adds. - Account for the first square separately before simplifying. - Test your equation with \(2\) or \(3\) squares. - For part c), undo the added \(1\) before dividing by \(3\).

Solution

1. The first square uses \(4\) matchsticks, and the remaining \(n-1\) squares each add \(3\), so \(S=4+3(n-1)\). 2. Simplify: \(S=4+3n-3=3n+1\). 3. For \(n=15\), \(S=3\cdot15+1=46\). 4. For \(S=76\), solve \(76=3n+1\): \(75=3n\), so \(n=25\).

Answer

a) \(S=3n+1\) b) \(46\) matchsticks c) \(25\) squares
5125597
A wireframe model of a rectangular prism is made from \(20\,\text{ft}\) of wire. One base edge is twice the length of the other base edge, and the height equals the longer base edge. Find the three edge lengths.

Hints

- A rectangular prism has four edges of each dimension. - Express all three edge lengths using one variable. - Set the sum of all twelve edge lengths equal to the total wire length.

Solution

1. Let \(x\) feet be the shorter base edge. Then the longer base edge and the height are each \(2x\) feet. 2. A rectangular prism has four edges of each of the three lengths. Therefore, the total wire length is \(4x + 4(2x) + 4(2x)\). 3. Write and simplify the equation: \(4x + 8x + 8x = 20\), so \(20x = 20\). 4. Divide by \(20\): \(x = 1\). 5. The shorter base edge is \(1\,\text{ft}\), and the longer base edge and height are each \(2\,\text{ft}\).

Answer

The three dimensions are \(1\,\text{ft}\), \(2\,\text{ft}\), and \(2\,\text{ft}\).
5125617
A wireframe square pyramid is made from a \(160\,\text{in.}\) aluminum rod. Its four base edges have equal length. Each of the four edges from the base to the apex is exactly \(5\,\text{in.}\) longer than a base edge. Find the length of a base edge and the length of an edge to the apex.

Hints

- Count the base edges and the edges that meet at the apex. - Represent each longer edge in terms of a base edge. - Set the sum of all eight edge lengths equal to the rod length.

Solution

1. Let \(a\) inches be the length of a base edge. Each edge to the apex has length \(a + 5\). 2. The pyramid has \(4\) base edges and \(4\) edges to the apex, so \(4a + 4(a + 5) = 160\). 3. Distribute and combine like terms: \(4a + 4a + 20 = 160\), so \(8a = 140\). 4. Divide by \(8\): \(a = 17.5\). 5. Each edge to the apex is \(17.5\,\text{in.} + 5\,\text{in.} = 22.5\,\text{in.}\).

Answer

A base edge is \(17.5\,\text{in.}\), and an edge to the apex is \(22.5\,\text{in.}\).
5125627
A rectangular poster is \(10\,\text{in.}\) wide. If the width is increased by \(5\,\text{in.}\) and the height is decreased by \(2\,\text{in.}\), the area increases by \(10\,\text{in.}^2\). Find the original height.

Hints

- Use length times width for the area of a rectangle. - Express the new dimensions in terms of the original height. - Relate the new area to the original area with an equation. - Include the stated increase in area.

Solution

1. Let \(x\) inches be the original height. The original area is \(10x\). 2. The new width is \(15\,\text{in.}\), and the new height is \(x - 2\) inches. The new area is \(15(x - 2)\). 3. Because the new area is \(10\,\text{in.}^2\) greater, write \(15(x - 2) = 10x + 10\). 4. Distribute: \(15x - 30 = 10x + 10\). 5. Subtract \(10x\) and add \(30\): \(5x = 40\). Divide by \(5\): \(x = 8\).

Answer

The original height was \(8\,\text{in.}\).
5125657
Paul is selling \(40\) used video games at a flea market. His total cost for the booth and the games was \(\$120.00\). He has already sold \(15\) games for \(\$2.50\) each. At what price must he sell each of the remaining games to recover exactly his total cost? Write and solve an equation.

Hints

- Find the revenue from the games already sold. - Determine how many games remain. - Set total revenue equal to the total cost. - Divide the remaining amount by the number of remaining games.

Solution

1. There are \(40 - 15 = 25\) games remaining. Let \(x\) dollars be the price of each remaining game. 2. The total revenue must equal \(\$120.00\): \(15 \cdot 2.50 + 25x = 120\). 3. Simplify: \(37.50 + 25x = 120\). 4. Subtract \(37.50\): \(25x = 82.50\). 5. Divide by \(25\): \(x = 3.30\).

Answer

Paul must sell each remaining game for \(\$3.30\).
5125667
A student newspaper club prints \(200\) copies of a new issue. Printing costs \(\$180.00\), and the club wants to earn an additional \(\$60.00\) for the class fund. The first \(120\) copies were sold for \(\$1.00\) each. At what price must the remaining copies be sold? Write and solve an equation.

Hints

- Add the printing cost and desired profit to find the target revenue. - Find the revenue already earned. - Determine how many copies remain. - Use an equation to distribute the remaining revenue across those copies.

Solution

1. The target revenue is \(180 + 60 = 240\) dollars. There are \(200 - 120 = 80\) copies remaining. 2. Let \(x\) dollars be the price of each remaining copy. Write \(120 \cdot 1.00 + 80x = 240\). 3. Simplify: \(120 + 80x = 240\). 4. Subtract \(120\): \(80x = 120\). 5. Divide by \(80\): \(x = 1.50\).

Answer

The remaining copies must be sold for \(\$1.50\) each.
5125677
A raffle has \(500\) tickets. The original plan was to sell every ticket for \(\$1.50\). During a promotion, \(300\) tickets were sold for an average of \(\$1.10\) each. What price must be charged for each of the remaining \(200\) tickets to reach exactly the originally planned total revenue? Write and solve an equation.

Hints

- Find the originally planned total revenue. - Calculate the revenue already earned during the promotion. - Subtract to find the revenue still needed. - Divide that amount among the remaining tickets.

Solution

1. The originally planned revenue was \(500 \cdot 1.50 = 750\) dollars. 2. Let \(x\) dollars be the price of each remaining ticket. Write \(300 \cdot 1.10 + 200x = 750\). 3. Simplify: \(330 + 200x = 750\). 4. Subtract \(330\): \(200x = 420\). 5. Divide by \(200\): \(x = 2.10\).

Answer

Each remaining ticket must cost \(\$2.10\).
5125897
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 \(50\,\text{g}\).
5125907
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.
5125957
A rectangle has a perimeter of at most \(48\,\text{in.}\). One side is exactly three times as long as the other side. Find the greatest possible length of each side.

Hints

- Write the longer side in terms of the shorter side. - Use the perimeter formula for a rectangle. - Translate “at most” into an inequality symbol. - Solve the inequality for the shorter side.

Solution

1. Let \(x\) inches be the shorter side. The longer side is \(3x\) inches. 2. The perimeter is \(2(x + 3x) = 8x\). 3. Because the perimeter is at most \(48\,\text{in.}\), write \(8x \leq 48\). 4. Divide by \(8\): \(x \leq 6\). 5. The greatest possible shorter side is \(6\,\text{in.}\), and the greatest possible longer side is \(3 \cdot 6\,\text{in.} = 18\,\text{in.}\).

Answer

The greatest possible side lengths are \(6\,\text{in.}\) and \(18\,\text{in.}\).
5125967
A rectangular flower bed can use at most \(40\,\text{ft}\) of edging. Its length will be exactly \(4\,\text{ft}\) greater than its width. What is the greatest possible width of the flower bed?

Hints

- Express the length in terms of the width. - Write a perimeter expression using one variable. - Use the amount of available edging to form an inequality. - Solve the inequality step by step.

Solution

1. Let \(w\) feet be the width. The length is \(w + 4\) feet. 2. The perimeter is \(2(w + w + 4) = 4w + 8\). 3. Because no more than \(40\,\text{ft}\) of edging is available, write \(4w + 8 \leq 40\). 4. Subtract \(8\): \(4w \leq 32\). 5. Divide by \(4\): \(w \leq 8\).

Answer

The flower bed can be at most \(8\,\text{ft}\) wide.
5126087
A small birdbath starts with \(15\,\text{L}\) of water. During a rain shower, \(0.5\,\text{L}\) flows in each minute. At the same time, a leak drains a constant amount \(y\), in liters per minute. After 20 minutes, the birdbath contains exactly \(21\,\text{L}\). A student proposes the equation \(15 + 0.5 \cdot 20 - y \cdot 20 = 21\). a) Does the equation correctly model the situation? Briefly explain. b) Find the water loss \(y\) in liters per minute.

Hints

- Identify what the \(15\) and \(21\) represent. - Match each term in the equation to a change during the 20 minutes. - Simplify the numerical terms before isolating \(y\).

Solution

1. The equation is correct: \(15\) is the starting amount, \(0.5 \cdot 20\) is the rain added in 20 minutes, \(y \cdot 20\) is the amount lost, and \(21\) is the final amount. 2. Simplify: \(15 + 10 - 20y = 21\). 3. Combine constants: \(25 - 20y = 21\). 4. Subtract \(21\) from \(25\): \(4 = 20y\). 5. Divide by \(20\): \(y = 0.2\).

Answer

a) Yes. The equation combines the starting amount, the rain added, and the leak loss to produce the final amount. b) The birdbath loses \(0.2\,\text{L}\) per minute.
5126137
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.
5126237
A mental math performer at a carnival says, “Choose a secret number, double it, add \(10\), and divide the result by \(2\). Tell me your final result.” a) Let the starting number be \(x\). Write and simplify an expression for the final result. b) One participant says the final result is \(42\). What was the starting number? c) Describe the quick mental rule the performer can use to recover the starting number.

Hints

- Write an expression that follows the operations in order. - Can you divide each term in the numerator by \(2\)? - Once you know how the final result compares with the starting number, reverse that change.

Solution

1. The final result is \(\frac{2x+10}{2}\). 2. Dividing both terms in the numerator by \(2\) gives \(x+5\). 3. For a final result of \(42\), solve \(x+5=42\). Subtracting \(5\) gives \(x=37\). 4. Since the final result is always \(5\) more than the starting number, the performer subtracts \(5\) from the reported result.

Answer

a) \(\frac{2x+10}{2}=x+5\) b) \(37\) c) Subtract \(5\) from the final result.
5126267
There are three piles of counters. The left and right piles each start with exactly \(12\) counters. The middle pile starts with \(z\) counters. Follow these steps: 1. Move \(5\) counters from the left pile to the middle pile. 2. Move \(3\) counters from the right pile to the middle pile. 3. Count the counters remaining in the left pile. Remove that same number of counters from the middle pile and set them aside. a) How many counters remain in the left pile after step \(1\)? b) Write and simplify an expression for the number of counters in the middle pile after step \(3\). c) How many counters are in the middle pile at the end if \(z=10\)?

Hints

- Track the number of counters in each pile after each step. - First find the fixed number left in the left pile. - In step \(3\), subtract that fixed number from the middle-pile expression.

Solution

1. After step \(1\), the left pile has \(12-5=7\) counters. 2. After steps \(1\) and \(2\), the middle pile has \(z+5+3=z+8\) counters. 3. Step \(3\) removes \(7\) counters from the middle pile, so the final amount is \((z+8)-7=z+1\). 4. When \(z=10\), the final amount is \(10+1=11\).

Answer

a) \(7\) counters b) \(z+1\) c) \(11\) counters
5127717
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.
5135917
A bag contains red, yellow, and white gummy bears. There are exactly \(12\) red gummy bears, and the probability of selecting a red gummy bear is \(0.25\). a) Find the total number of gummy bears. b) How many yellow and white gummy bears are there altogether? c) There are twice as many yellow gummy bears as white gummy bears. Find the number of yellow gummy bears.

Hints

- Write probability as red count divided by total count. - Subtract the red count from the total. - Represent the white count with a variable and the yellow count as twice that variable.

Solution

1. a) Let \(T\) be the total. Since \(\frac{12}{T} = 0.25\), \(T = \frac{12}{0.25} = 48\). 2. b) There are \(48 - 12 = 36\) yellow and white gummy bears altogether. 3. c) Let \(x\) be the number of white gummy bears. Then there are \(2x\) yellow gummy bears. The equation \(x + 2x = 36\) gives \(x = 12\), so there are \(24\) yellow gummy bears.

Answer

a) \(48\) gummy bears b) \(36\) yellow and white gummy bears altogether c) \(24\) yellow gummy bears
5135927
Two bags contain colored counters. Bag A contains \(5\) blue and \(15\) red counters. Bag B contains only green and yellow counters. The probability of drawing a red counter from Bag A equals the probability of drawing a green counter from Bag B. a) Bag B contains \(12\) green counters. Find the number of yellow counters in Bag B. b) All counters from both bags are combined. Find the probability of drawing a red counter from the combined container.

Hints

- First find the red probability in Bag A. - Use an equation to find the total in Bag B. - When the bags are combined, update the total but not the red count.

Solution

1. The probability of red from Bag A is \(\frac{15}{20} = \frac{3}{4}\). 2. a) Let \(T\) be the total number of counters in Bag B. Then \(\frac{12}{T} = \frac{3}{4}\), so \(T = 16\). Therefore, Bag B has \(16 - 12 = 4\) yellow counters. 3. b) The combined container has \(20 + 16 = 36\) counters, including \(15\) red counters. Thus, \(P(\text{red}) = \frac{15}{36} = \frac{5}{12}\).

Answer

a) \(4\) yellow counters b) \(\frac{5}{12} \approx 41.67\%\)
5135937
A school raffle has grand-prize tickets, small-prize tickets, and losing tickets. The probability of a grand prize is \(\frac{1}{50}\), and the probability of a small prize is \(\frac{1}{5}\). The raffle box contains exactly \(390\) losing tickets. a) How many tickets were prepared altogether? b) How many grand-prize tickets are there?

Hints

- Find the losing-ticket probability, then set losing tickets divided by total tickets equal to that probability. - Multiply the total number of tickets by the grand-prize probability.

Solution

1. The losing-ticket probability is \(1 - \frac{1}{50} - \frac{1}{5} = 1 - \frac{11}{50} = \frac{39}{50}\). 2. a) Let \(T\) be the total number of tickets. Since \(\frac{390}{T} = \frac{39}{50}\), \(T = 500\). 3. b) The number of grand-prize tickets is \(500 \cdot \frac{1}{50} = 10\).

Answer

a) \(500\) tickets b) \(10\) grand-prize tickets
5136297
A container holds \(40\) red, blue, and green counters. It contains \(15\) red counters and \(10\) blue counters. a) Find the probability of drawing a green counter. b) Find the probability of not drawing a blue counter. c) Some red counters are removed. Afterward, the probability of drawing a blue counter is exactly \(\frac{1}{3}\). How many red counters were removed?

Hints

- Find the green count from the total and the known colors. - Not blue includes red and green. - Removing counters changes both the selected color count and the total count. - For part c, write an equation using the unchanged blue count and the new total.

Solution

1. There are \(40 - 15 - 10 = 15\) green counters. 2. a) \(P(\text{green}) = \frac{15}{40} = \frac{3}{8}\). 3. b) There are \(30\) counters that are not blue, so \(P(\text{not blue}) = \frac{30}{40} = \frac{3}{4}\). 4. c) Let \(x\) be the number of red counters removed. Then \(\frac{10}{40 - x} = \frac{1}{3}\). Solving gives \(30 = 40 - x\), so \(x = 10\).

Answer

a) \(\frac{3}{8} = 37.5\%\) b) \(\frac{3}{4} = 75\%\) c) \(10\) red counters
5137627
Max and Julia start \(18\) miles apart and walk toward each other. Max walks at \(4\,\text{mph}\). Julia walks \(50\%\) faster than Max. They start at the same time. Use a linear equation to determine who travels farther before they meet and by how many miles.

Hints

- Find Julia's speed first. - Write each distance as rate times the same meeting time. - The two distances add to the original separation. - Subtract the smaller distance from the larger distance.

Solution

1. Julia's speed is \(4 \cdot 1.5 = 6\,\text{mph}\). 2. Let \(t\) be the meeting time in hours. Their distances add to \(18\) miles, so \(4t + 6t = 18\). 3. Combine like terms: \(10t = 18\), so \(t = 1.8\) hours. 4. Max travels \(4 \cdot 1.8 = 7.2\) miles, and Julia travels \(6 \cdot 1.8 = 10.8\) miles. 5. Julia travels farther by \(10.8 - 7.2 = 3.6\) miles.

Answer

Julia travels farther, by \(3.6\) miles.
5141417
A parallelogram and a triangle have the same area. The parallelogram has a base of \(8\,\text{in.}\) and a height of \(4.5\,\text{in.}\). The triangle has a base of \(12\,\text{in.}\). Find the triangle''s height.

Hints

- First find the area of the parallelogram. - Use the fact that the two areas are equal. - Substitute the triangle's base into its area formula. - Solve the resulting equation for the height.

Solution

1. Find the parallelogram's area: \(8 \cdot 4.5 = 36\,\text{in.}^2\). 2. The triangle also has area \(36\,\text{in.}^2\). 3. Use the triangle area formula: \(36 = \frac{1}{2} \cdot 12 \cdot h\). 4. Simplify: \(36 = 6h\). 5. Divide by \(6\): \(h = 6\).

Answer

The triangle''s height is \(6\,\text{in.}\).
5141427
A rectangular garden is \(12\,\text{ft}\) long and has an area of \(120\,\text{ft}^2\). A triangular flower bed is attached along one of the garden''s shorter sides, so the triangle''s base equals the garden''s width. The combined area is \(150\,\text{ft}^2\). Find the height of the triangular flower bed.

Hints

- Use the rectangle's area to find its width. - Subtract the rectangle's area from the combined area. - Use the garden's width as the triangle's base. - Solve the triangle area equation for its height.

Solution

1. Let \(w\) feet be the garden's width. From \(12w = 120\), obtain \(w = 10\). 2. The triangular flower bed has area \(150 - 120 = 30\,\text{ft}^2\). 3. Its base is the garden's width, \(10\,\text{ft}\). Write \(30 = \frac{1}{2} \cdot 10 \cdot h\). 4. Simplify: \(30 = 5h\). 5. Divide by \(5\): \(h = 6\).

Answer

The triangular flower bed is \(6\,\text{ft}\) high.
5155107
Two classes sell raffle tickets for a fundraiser. Class A has \(120\) tickets, including \(30\) winning tickets. Class B has \(150\) tickets, including \(40\) winning tickets. a) Which class has the greater probability of drawing a winning ticket? Compare the probabilities. b) How many losing tickets must be added to Class B so that both classes have exactly the same winning probability?

Hints

- Compare winning tickets divided by total tickets for each class. - In part b, the number of winning tickets stays \(40\). - Write an equation for the new total that gives probability \(0.25\).

Solution

1. Class A has winning probability \(\frac{30}{120} = \frac{1}{4} = 0.25\). 2. Class B has winning probability \(\frac{40}{150} = \frac{4}{15} \approx 0.2667\), so Class B is initially greater. 3. b) Let \(x\) be the new total number of Class B tickets. To match Class A, \(\frac{40}{x} = 0.25\), so \(x = 160\). 4. Class B currently has \(150\) tickets, so \(160 - 150 = 10\) losing tickets must be added.

Answer

a) Class B: \(\frac{4}{15} \approx 26.67\%\), compared with \(25\%\) for Class A b) \(10\) losing tickets
5183107
Complete the account table. In each row, an amount is withdrawn from the previous balance to produce the new balance. <table> <tr><th>Previous balance</th><th>Withdrawal</th><th>New balance</th></tr> <tr><td>\(\$12\)</td><td>\(\$20\)</td><td>(1)</td></tr> <tr><td>\(-\$8\)</td><td>\(\$15\)</td><td>(2)</td></tr> <tr><td>(3)</td><td>\(\$25\)</td><td>\(-\$10\)</td></tr> </table>

Hints

- A withdrawal decreases the balance. - Use an inverse operation to find a missing previous balance. - Pay close attention to the sign when the account already has a negative balance.

Solution

1. For (1), \(12-20=-8\), so the new balance is \(-\$8\). 2. For (2), \(-8-15=-23\), so the new balance is \(-\$23\). 3. For (3), let the previous balance be \(x\). Solve \(x-25=-10\), which gives \(x=15\).

Answer

(1) \(-\$8\) (2) \(-\$23\) (3) \(\$15\)
5183797
Ms. Weber’s bank account balance is \(-\$120\). After her garage rent is withdrawn, the new balance is \(-\$185\). Write an equation using \(x\) for the garage rent, and solve for \(x\).

Hints

- The rent decreases the account balance. - Write the relationship as previous balance minus rent equals new balance. - Use inverse operations to isolate \(x\). - Check that subtracting your answer produces the stated new balance.

Solution

1. The previous balance minus the rent equals the new balance, so \(-120-x=-185\). 2. Add \(120\) to both sides: \(-x=-65\). 3. Multiply both sides by \(-1\): \(x=65\). 4. The garage rent is \(\$65\).

Answer

Equation: \(-120-x=-185\) Solution: \(x=65\) The garage rent is \(\$65\).
5183807
Write and solve an equation for each number puzzle. a) Subtracting \(55\) from a number \(y\) gives \(-25\). b) Adding a number \(z\) to \(-140\) gives \(60\).

Hints

- Translate each sentence into an equation one phrase at a time. - Use inverse operations to isolate the variable. - Pay close attention to the signs when the value crosses zero.

Solution

1. In a), the equation is \(y-55=-25\). Add \(55\) to both sides to get \(y=30\). 2. In b), the equation is \(-140+z=60\). Add \(140\) to both sides to get \(z=200\).

Answer

a) \(y-55=-25\), so \(y=30\) b) \(-140+z=60\), so \(z=200\)
5184017
Mr. Schmidt pays a \(\$545\) bill from his checking account. Soon afterward, a credit of \(\$210\) is deposited. His balance is then exactly \(-\$185\). What was his account balance before these two transactions?

Hints

- Determine the combined effect of the two transactions. - You can also undo the transactions in reverse order, starting from the final balance. - A bill decreases the balance, while a credit increases it.

Solution

1. The net change is \(-545+210=-335\). 2. Let \(x\) be the initial balance. Then \(x-335=-185\). 3. Add \(335\) to both sides: \(x=-185+335=150\). 4. Check: \(150-545+210=-185\).

Answer

The account balance before the transactions was \(\$150\).
5184027
An elevator in a parking garage first travels up \(5\) levels and then down \(8\) levels. It ends on Level \(-2\). On which level did the elevator start?

Hints

- Picture the garage levels as a vertical number line. - Combine the upward and downward movements into one net change. - Starting from the ending level, reverse the net change.

Solution

1. The total change is \(5-8=-3\), so the elevator ends \(3\) levels below where it started. 2. Let \(x\) be the starting level. Then \(x-3=-2\). 3. Add \(3\) to both sides: \(x=1\).

Answer

The elevator started on Level \(1\).
5184607
During one level of a video game, Tim's score changes several times. He first earns \(120\) bonus points, then loses \(250\) points, and finally earns another \(40\) points. His score at the end of the level is \(-30\). What was Tim's score at the beginning of the level?

Hints

- Calculate the total change in the score. - You can work backward from the final score one change at a time. - Decide whether the beginning score must have been greater or less than the ending score.

Solution

1. Find the total change in the score: \(120-250+40=-90\). 2. Let \(x\) be the beginning score. Then \(x-90=-30\). 3. Add \(90\) to both sides: \(x=-30+90=60\).

Answer

Tim began the level with \(60\) points.
5187237
A withdrawal of \(\$150\) changes an account balance. After the withdrawal, the balance is \(-\$35\). Write and solve an equation to find the balance before the withdrawal.

Hints

- Define a variable for the unknown starting balance. - How does a withdrawal change an account balance? - Which inverse operation will undo the withdrawal?

Solution

1. Let \(x\) represent the balance before the withdrawal. 2. The withdrawal gives the equation \(x-150=-35\). 3. Add \(150\) to both sides: \(x=-35+150=115\).

Answer

The balance before the withdrawal was \(\$115\).
5193837
Luke says, “When I subtract \(240\) from my number, I get \(360\).” Marie says, “When I add \(180\) to my number, I get \(800\).” Who chose the greater number, and by how much?

Hints

- Find Luke’s number first. - Then find Marie’s number. - Compare the two numbers. - Subtract to find how much greater one number is.

Solution

1. Luke's number is \(360 + 240 = 600\). 2. Marie's number is \(800 - 180 = 620\). 3. Since \(620 > 600\), Marie chose the greater number. 4. The difference is \(620 - 600 = 20\).

Answer

Marie chose \(620\), which is \(20\) greater than Luke's number, \(600\).
5222477
A kayak rental company offers two plans. Plan A: a \(\$12\) base fee plus \(\$4\) per hour. Plan B: no base fee and \(\$7\) per hour. a) Write an expression for the total cost of each plan for \(t\) hours. b) Find the cost of each plan for \(2\) hours and for \(5\) hours. Which plan is less expensive each time? c) Explain what the number \(12\) and the variable \(t\) mean in Plan A.

Hints

- Translate “per hour” as multiplication by the number of hours. - Separate each plan's fixed cost from its hourly cost. - Substitute the given values of \(t\). - Compare the two totals for each rental time.

Solution

1. Plan A is \(A(t)=12+4t\), and Plan B is \(B(t)=7t\). 2. For \(t=2\), \(A(2)=12+4\cdot 2=20\), and \(B(2)=7\cdot 2=14\). Plan B is less expensive. 3. For \(t=5\), \(A(5)=12+4\cdot 5=32\), and \(B(5)=7\cdot 5=35\). Plan A is less expensive. 4. In Plan A, \(12\) is the one-time base fee in dollars, and \(t\) is the rental time in hours.

Answer

a) Plan A: \(12+4t\); Plan B: \(7t\) b) At \(2\) hours: Plan A costs \(\$20\), Plan B costs \(\$14\); Plan B is less expensive. At \(5\) hours: Plan A costs \(\$32\), Plan B costs \(\$35\); Plan A is less expensive. c) The \(12\) is the one-time base fee, and \(t\) is the number of rental hours.
5223227
A storage tank contains \(V\) gallons of oil. A machine uses \(12\) gallons each day. a) Write an expression for the amount of oil remaining after \(n\) days. b) Evaluate the expression for \(V=500\) and \(n=10\). c) What does it mean in this context if the expression has a value of \(0\)? d) How does the expression change if the machine uses an unknown amount of \(x\) gallons per day?

Hints

- Look for a pattern in the amount remaining after one, two, and three days. - Removing oil from the tank corresponds to subtraction. - Interpret what an amount of \(0\) gallons means physically.

Solution

1. Subtract the total amount used from the starting amount: \(V-12n\). 2. For \(V=500\) and \(n=10\), \(500-12\cdot 10=500-120=380\) gallons. 3. A value of \(0\) means the oil supply has been completely used after \(n\) days. 4. Replacing the daily use of \(12\) gallons with \(x\) gallons gives \(V-xn\).

Answer

a) \(V-12n\) b) \(380\) gallons c) The oil supply is completely used up. d) \(V-xn\)
5224907
The total cost of manufacturing a part includes material cost, labor cost, and a fixed \(\$100\) machine fee. Labor costs \(25\%\) more than materials. Find the material cost when the total cost is \(\$1900\).

Hints

- Express the labor cost as a multiple of the material cost. - Add material, labor, and the fixed fee. - Subtract the fixed fee before dividing.

Solution

1. Let \(x\) dollars be the material cost. 2. Labor costs \(25\%\) more, so the labor cost is \(1.25x\). 3. Add all cost components: \(x + 1.25x + 100 = 1900\). 4. Combine like terms: \(2.25x + 100 = 1900\). 5. Subtract \(100\): \(2.25x = 1800\). 6. Divide by \(2.25\): \(x = 800\).

Answer

The material cost is \(\$800\).
5224967
A bus for a class field trip costs a flat \(\$450\). Zoo admission costs \(\$12\) per person. The total trip cost is \(\$774\). How many people are attending?

Hints

- Identify the fixed cost and the per-person cost. - Write an equation for the total cost. - Subtract the fixed cost before dividing by the admission price.

Solution

1. Let \(x\) be the number of people. 2. Add the fixed bus cost and the admission cost: \(450 + 12x = 774\). 3. Subtract \(450\): \(12x = 324\). 4. Divide by \(12\): \(x = 27\).

Answer

\(27\) people are attending the field trip.
5225067
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.
5225447
A rectangular water tank has a maximum capacity of \(V\) gallons and initially contains \(B\) gallons. Pipe A adds \(a\) gallons per minute, Pipe B adds \(b\) gallons per minute, and an open drain removes \(c\) gallons per minute. a) Write an expression for the amount of water \(W\) after \(m\) minutes. b) Explain what \(V-[B+m(a+b-c)]\) represents in this context. Consider only values of \(m\ge 0\) for which \(0\le B+m(a+b-c)\le V\).

Hints

- Combine the two inflow rates and subtract the outflow rate. - Multiply the net rate by the elapsed time. - Capacity minus current contents gives the space still available.

Solution

1. The net change per minute is \(a+b-c\) gallons. 2. After \(m\) minutes, the total change is \(m(a+b-c)\). 3. Adding the initial amount gives \(W=B+m(a+b-c)\). 4. In part b), \(B+m(a+b-c)\) is the current amount of water. Subtracting it from the capacity \(V\) gives the unused capacity, or the number of additional gallons the tank can hold. 5. The stated domain keeps the modeled amount between empty and full.

Answer

a) \(W=B+m(a+b-c)\) b) It is the unused tank capacity after \(m\) minutes, in gallons. The model applies for \(m\ge 0\) and \(0\le B+m(a+b-c)\le V\).
5226027
A streaming service charges a monthly base fee of \(\$8.50\) plus \(\$2.50\) for each rented movie. a) Write an expression for the monthly cost when \(n\) movies are rented. b) Find the cost for a month with \(6\) rentals. c) A competing service has no base fee but charges \(\$4.00\) per rental. Which service costs less for \(5\) rentals? Show the calculation.

Hints

- Separate the one-time monthly fee from the per-rental charge. - Multiply the per-rental price by the number of rentals. - Calculate both plans separately before comparing them.

Solution

1. The monthly cost is \(8.50+2.50n\). 2. For \(n=6\), \(8.50+2.50\cdot 6=8.50+15.00=23.50\). 3. For \(5\) rentals, the first service costs \(8.50+2.50\cdot 5=21.00\). The competing service costs \(4.00\cdot 5=20.00\). The competing service costs less.

Answer

a) \(8.50+2.50n\) dollars b) \(\$23.50\) c) The competing service is less expensive: \(\$20.00\) instead of \(\$21.00\).
5228227
Three adjacent angles lie along a straight line. The second angle is three times the first angle. The third angle equals the sum of the first two angles. Find all three angle measures by writing and solving an equation.

Hints

- A straight angle measures \(180^\circ\). - Express all three angles using the first angle. - Use the relationship between the third angle and the first two. - Check that the three results add to \(180^\circ\).

Solution

1. Let \(x\) degrees be the first angle. 2. The second angle is \(3x\), and the third angle is \(x + 3x = 4x\). 3. The three angles form a straight angle, so \(x + 3x + 4x = 180\). 4. Combine like terms: \(8x = 180\). 5. Divide by \(8\): \(x = 22.5\). 6. The angles are \(22.5^\circ\), \(3 \cdot 22.5^\circ = 67.5^\circ\), and \(4 \cdot 22.5^\circ = 90^\circ\).

Answer

The three angle measures are \(22.5^\circ\), \(67.5^\circ\), and \(90^\circ\).
5228437
Two rectangles have the same width. The first rectangle is \(8\,\text{in.}\) high, and the second is twice as high. Their combined area is \(192\,\text{in.}^2\). Find their common width and the area of each rectangle.

Hints

- Write the two heights first. - Use area equals width times height for each rectangle. - Add the two area expressions and set their sum equal to the combined area. - Use the width to calculate each individual area.

Solution

1. Let \(w\) inches be the common width. 2. The heights are \(8\,\text{in.}\) and \(16\,\text{in.}\). 3. Add the two area expressions: \(8w + 16w = 192\). 4. Combine like terms: \(24w = 192\). 5. Divide by \(24\): \(w = 8\). 6. The first area is \(8\,\text{in.} \times 8\,\text{in.} = 64\,\text{in.}^2\), and the second area is \(16\,\text{in.} \times 8\,\text{in.} = 128\,\text{in.}^2\).

Answer

The common width is \(8\,\text{in.}\). The areas are \(64\,\text{in.}^2\) and \(128\,\text{in.}^2\).
5228867
Three seventh-grade classes raise a total of \(\$510\) for an environmental project. Class B raises exactly two-thirds as much as Class A. Class C raises \(\$15\) less than Class B. Find the amount raised by each class.

Hints

- Choose the Class A amount as the base value. - Express two-thirds of that amount algebraically. - Add the three class amounts and set the total equal to \(\$510\). - Use a common denominator when combining the variable terms.

Solution

1. Let \(x\) be the amount raised by Class A. 2. Class B raises \(\frac{2}{3}x\), and Class C raises \(\frac{2}{3}x - 15\). 3. Write the total equation \(x + \frac{2}{3}x + \left(\frac{2}{3}x - 15\right) = 510\). 4. Combine like terms: \(\frac{7}{3}x - 15 = 510\). 5. Add \(15\): \(\frac{7}{3}x = 525\). 6. Multiply by \(\frac{3}{7}\): \(x = 225\). 7. Class B raises \(\$150\), and Class C raises \(\$135\).

Answer

Class A raised \(\$225\), Class B raised \(\$150\), and Class C raised \(\$135\).
5229097
Three sections of temporary fencing are installed at a construction site. The first section is \(4a+3b\) feet long. The second section is \(a-b+2\) feet longer than the first section. The third section is \(2a+4b-5\) feet long. For values of \(a\) and \(b\) that make all three lengths positive, write and simplify an expression for the total length of the fencing.

Hints

- Write an expression for the second section first. - “Longer than” indicates addition. - Combine the \(a\)-terms, \(b\)-terms, and constants separately.

Solution

1. The second section has length \((4a+3b)+(a-b+2)=5a+2b+2\). 2. Add all three section lengths: \((4a+3b)+(5a+2b+2)+(2a+4b-5)\). 3. Combine like terms: \(4a+5a+2a=11a\), \(3b+2b+4b=9b\), and \(2-5=-3\). 4. The total length is \(11a+9b-3\) feet.

Answer

The total length is \(11a+9b-3\) feet.
5229127
For values of \(a\) and \(b\) that make all original and new dimensions positive, a rectangle has length \(L=4a+3b\) and width \(W=2a+b\). A second rectangle is built with a length that is \(a-b\) greater than \(L\) and a width that is \(b\) less than \(W\). Write and simplify an expression for the perimeter of the second rectangle.

Hints

- Find the new length and width separately. - “Greater than” indicates addition, and “less than” indicates subtraction. - Use \(P=2(L+W)\) for a rectangle.

Solution

1. The new length is \(L_2=(4a+3b)+(a-b)=5a+2b\). 2. The new width is \(W_2=(2a+b)-b=2a\). 3. Use the perimeter formula: \(P=2(L_2+W_2)\). 4. Substitute and simplify: \(P=2[(5a+2b)+2a]=2(7a+2b)=14a+4b\).

Answer

\(P=14a+4b\)
5229167
For values of \(a\) and \(b\) that make all side lengths positive and form a valid triangle, the first two sides are Side 1: \(a+4b\) Side 2: \(3a-2b+5\). The third side is \(2a+b+3\) less than the sum of the first two sides. Write and simplify an expression for the perimeter \(P\).

Hints

- Add the first two side lengths first. - “Less than” means subtract the entire given expression. - Add all three side lengths to find the perimeter.

Solution

1. Add the first two sides: \((a+4b)+(3a-2b+5)=4a+2b+5\). 2. Find the third side: \((4a+2b+5)-(2a+b+3)=2a+b+2\). 3. Add all three sides: \(P=(4a+2b+5)+(2a+b+2)=6a+3b+7\).

Answer

\(P=6a+3b+7\)
5229337
A school cafeteria sells sandwiches for \(\$2.00\) and bottles of apple juice for \(\$1.50\). At the end of the day, total sales were \(\$110.00\). The cafeteria sold exactly twice as many sandwiches as juice bottles. How many of each item were sold?

Hints

- Represent the number of one item with a variable. - Express the other quantity using the given relationship. - Multiply each price by its quantity and add the revenues. - Check that the final revenue is \(\$110.00\).

Solution

1. Let \(x\) be the number of juice bottles sold. Then \(2x\) sandwiches were sold. 2. Write the revenue equation: \(2.00(2x) + 1.50x = 110\). 3. Combine like terms: \(4x + 1.5x = 110\), so \(5.5x = 110\). 4. Divide by \(5.5\): \(x = 20\). 5. The number of sandwiches is \(2 \cdot 20 = 40\).

Answer

The cafeteria sold \(40\) sandwiches and \(20\) bottles of apple juice.
5229407
Three classes raise a total of \(\$720\) for a community project. Class B raises \(\$40\) more than Class A. Class C raises the same amount as Classes A and B combined. 1. Find the amount raised by each class. 2. What fraction or percent of the total is raised by Class C?

Hints

- Let the Class A amount be \(x\). - Express the Class C amount using the expressions for Classes A and B. - Set the sum of all three amounts equal to \(\$720\). - Divide Class C’s amount by the total to find its share.

Solution

1. Let \(x\) be the amount raised by Class A. 2. Class B raises \(x + 40\), and Class C raises \(x + (x + 40) = 2x + 40\). 3. Write \(x + (x + 40) + (2x + 40) = 720\). 4. Combine like terms: \(4x + 80 = 720\). 5. Subtract \(80\): \(4x = 640\). Divide by \(4\): \(x = 160\). 6. Class B raises \(\$200\), and Class C raises \(\$360\). 7. Class C’s share is \(\frac{360}{720} = \frac{1}{2} = 50\%\).

Answer

1) Class A raised \(\$160\), Class B raised \(\$200\), and Class C raised \(\$360\). 2) Class C raised \(\frac{1}{2}\), or \(50\%\), of the total.
5229417
A triangle has a perimeter of \(27\,\text{in.}\). The second side is \(3\,\text{in.}\) longer than the first side, and the third side is twice as long as the first side. a) Write an equation that can be used to find the first side length \(x\). b) Find all three side lengths. c) How would the equation in part a) change if the third side were \(3\,\text{in.}\) shorter than the first side instead of twice as long?

Hints

- Add the three side lengths to represent the perimeter. - Translate “twice as long” and “shorter than” into algebraic expressions. - Express every side in terms of \(x\). - For part c), replace only the expression for the third side.

Solution

1. Add the three side expressions: \(x + (x + 3) + 2x = 27\). 2. Combine like terms: \(4x + 3 = 27\). 3. Subtract \(3\): \(4x = 24\). Divide by \(4\): \(x = 6\). 4. The three side lengths are \(6\,\text{in.}\), \(9\,\text{in.}\), and \(12\,\text{in.}\). 5. Under the changed condition, the third side would be \(x - 3\), so the equation would be \(x + (x + 3) + (x - 3) = 27\).

Answer

a) \(x + (x + 3) + 2x = 27\), or \(4x + 3 = 27\) b) The side lengths are \(6\,\text{in.}\), \(9\,\text{in.}\), and \(12\,\text{in.}\). c) The equation would be \(x + (x + 3) + (x - 3) = 27\).
5229427
Consider the equation \(x+(x-2)+3x=43\). a) Create a number puzzle or real-world situation represented by this equation. Briefly explain what \(x\), \(x-2\), and \(3x\) mean in your situation. b) Solve the equation. Give the value of \(x\) and the values of the other two expressions.

Hints

- Think of three related quantities whose total is \(43\). - In a context, what could “\(2\) fewer” and “three times as many” describe? - Combine all the \(x\)-terms before solving. - Check that the three values add to \(43\).

Solution

1. One possible situation is: Three boxes contain \(43\) apples altogether. The first box contains \(x\) apples, the second contains \(2\) fewer than the first, and the third contains three times as many as the first. 2. Combine like terms: \(x+x-2+3x=43\), so \(5x-2=43\). 3. Add \(2\) to both sides to get \(5x=45\), then divide by \(5\) to get \(x=9\). 4. The three values are \(x=9\), \(x-2=7\), and \(3x=27\). The check is \(9+7+27=43\).

Answer

a) Answers will vary. One example is three boxes containing \(x\), \(x-2\), and \(3x\) apples, for a total of \(43\). b) \(x=9\). The three values are \(9\), \(7\), and \(27\).
5230157
A post office sells stamps for \(\$0.85\) and \(\$1.10\). A customer spends \(\$33.60\) and buys exactly twice as many \(\$0.85\) stamps as \(\$1.10\) stamps. How many stamps of each type does the customer buy?

Hints

- Represent the number of one stamp type with a variable. - Use the “twice as many” relationship for the other type. - Add the values of both groups of stamps. - Report both quantities.

Solution

1. Let \(x\) be the number of \(\$1.10\) stamps. Then the number of \(\$0.85\) stamps is \(2x\). 2. Write the total-value equation: \(1.10x + 0.85(2x) = 33.60\). 3. Combine like terms: \(1.10x + 1.70x = 33.60\), so \(2.80x = 33.60\). 4. Divide by \(2.80\): \(x = 12\). 5. The number of \(\$0.85\) stamps is \(2 \cdot 12 = 24\).

Answer

The customer buys \(12\) stamps at \(\$1.10\) and \(24\) stamps at \(\$0.85\).
5238277
A beverage plant uses two types of bottling machines. Each Type A machine fills \(x\) bottles per minute, and each Type B machine fills \(y\) bottles per minute. The plant runs \(4\) Type A machines and \(5\) Type B machines at the same time to fill an order of \(F\) bottles. a) Write an expression for the time \(t\), in minutes, needed to fill the order. b) Find \(t\) when \(x=25\), \(y=20\), and \(F=4000\).

Hints

- Find the total number of bottles all the machines fill in one minute. - Divide the order size by the combined rate. - Evaluate the denominator before dividing.

Solution

1. The Type A machines fill \(4x\) bottles per minute, and the Type B machines fill \(5y\) bottles per minute. 2. The combined rate is \(4x+5y\) bottles per minute. 3. The time is \(t=\frac{F}{4x+5y}\). 4. Substituting the given values gives \(t=\frac{4000}{4\cdot 25+5\cdot 20}=\frac{4000}{200}=20\) minutes.

Answer

a) \(t=\frac{F}{4x+5y}\) minutes b) \(20\) minutes
5238287
A landscaping crew mows lawns. Each experienced worker mows \(f\) square feet per hour, and each assistant mows \(h\) square feet per hour. The current crew has \(2\) experienced workers and \(3\) assistants. They need to mow \(A\) square feet. a) Write an expression for the time \(d\), in hours, needed by the current crew. b) One experienced worker is replaced by two assistants. How does the crew's hourly rate change? Under what condition will the new crew work faster than the original crew? c) Find \(d\) for the original crew when \(f=1200\), \(h=800\), and \(A=19{,}200\).

Hints

- Add the hourly rates of all workers on each crew. - A faster crew mows more area per hour. - Compare the new rate with the original rate. - In part c), evaluate the denominator before dividing.

Solution

1. The original crew's rate is \(2f+3h\) square feet per hour. 2. Its mowing time is \(d=\frac{A}{2f+3h}\). 3. The new crew has \(1\) experienced worker and \(5\) assistants, so its rate is \(f+5h\). 4. The change in rate is \((f+5h)-(2f+3h)=2h-f\). The new crew is faster when \(2h>f\). 5. For the given values, \(d=\frac{19{,}200}{2\cdot 1200+3\cdot 800}=\frac{19{,}200}{4800}=4\) hours.

Answer

a) \(d=\frac{A}{2f+3h}\) hours b) The rate changes by \(2h-f\) square feet per hour. The new crew is faster when \(2h>f\). c) \(4\) hours
5238327
An empty water tank holds \(V\) gallons and is filled by two pipes. Pipe A adds \(x\) gallons per minute. Pipe B adds \(5\) gallons per minute more than Pipe A. a) Write an expression for the amount of water in the tank after both pipes have been open for \(10\) minutes. b) Write an expression for the time \(t\), in minutes, needed to fill the tank. c) Find \(t\) when \(V=750\) and \(x=35\).

Hints

- Add the two pipe rates to find the amount added each minute. - Multiply the combined rate by \(10\) for part a). - Divide the tank capacity by the combined rate for part b).

Solution

1. Pipe A's rate is \(x\) gallons per minute, and Pipe B's rate is \(x+5\) gallons per minute. 2. Their combined rate is \(x+(x+5)=2x+5\) gallons per minute. 3. After \(10\) minutes, the tank contains \(10(2x+5)=20x+50\) gallons. 4. The filling time is \(t=\frac{V}{2x+5}\). 5. For \(V=750\) and \(x=35\), \(t=\frac{750}{2\cdot 35+5}=\frac{750}{75}=10\) minutes.

Answer

a) \(10(2x+5)=20x+50\) gallons b) \(t=\frac{V}{2x+5}\) minutes c) \(10\) minutes
5238357
A wood-pellet storage bin has a capacity of \(M\) pounds. Filling the bin completely costs \(\$S\). The heating system uses pellets worth \(\$k\) each day. a) Write an expression for the number of pounds of pellets remaining after \(t\) days. b) Find the remaining amount when \(M=3000\), \(S=1200\), \(k=10\), and \(t=30\).

Hints

- First find the cost of one pound of pellets. - Divide the daily cost by the cost per pound to find the daily amount used. - Multiply the daily amount by \(t\), then subtract it from the full amount.

Solution

1. The price per pound is \(\frac{S}{M}\) dollars. 2. The system uses \(\frac{k}{S/M}=\frac{kM}{S}\) pounds per day. 3. In \(t\) days, it uses \(\frac{kMt}{S}\) pounds. 4. The amount remaining is \(M-\frac{kMt}{S}\). 5. For the given values, \(M-\frac{kMt}{S}=3000-\frac{10\cdot 3000\cdot 30}{1200}=3000-750=2250\) pounds.

Answer

a) \(M-\frac{kMt}{S}\) pounds b) \(2250\) pounds
5238447
A shipping company stacks boxes on pallets. Each pallet has \(r\) rows with \(k\) boxes in each row. Each box weighs \(m\) pounds. a) Write an expression for the weight \(G\), in pounds, of all the boxes. b) An empty pallet weighs \(50\) pounds. Write an expression for the total weight \(M\), in tons, of a loaded pallet. Use \(1\) ton \(=2000\) pounds. c) A forklift can lift at most \(1\) ton. Determine whether it can lift a pallet when \(r=5\), \(k=8\), and \(m=44\).

Hints

- Multiply to find the total weight of the boxes. - Add the empty pallet's weight. - Divide pounds by \(2000\) to convert to tons. - Compare the result with the forklift's limit.

Solution

1. The boxes weigh \(G=rkm\) pounds. 2. Including the pallet, the total weight in pounds is \(rkm+50\). 3. Converting to tons gives \(M=\frac{rkm+50}{2000}\). 4. For the given values, \(M=\frac{5\cdot 8\cdot 44+50}{2000}=\frac{1810}{2000}=0.905\) tons. 5. Since \(0.905\le 1\), the forklift can lift the pallet.

Answer

a) \(G=rkm\) pounds b) \(M=\frac{rkm+50}{2000}\) tons c) Yes. The loaded pallet weighs \(0.905\) tons, which is below the \(1\)-ton limit.
5239347
A sports club plans to buy equipment costing \(\$K\). The city provides a grant of \(\$Z\). The remaining cost is divided equally among \(s\) active members. a) Write an expression for the amount each member must pay. b) Find the amount per member when \(K=540\), \(Z=120\), and \(s=15\). c) How does the amount per member change if \(s\) decreases while \(K\) and \(Z\) stay the same? Explain.

Hints

- Subtract the grant from the total cost. - Divide the remaining cost equally among the members. - Think about what happens when a fixed amount is divided among fewer people.

Solution

1. After the grant, the remaining cost is \(K-Z\) dollars. 2. Each member pays \(\frac{K-Z}{s}\) dollars. 3. For the given values, \(\frac{540-120}{15}=\frac{420}{15}=28\), so each member pays \(\$28\). 4. If \(s\) decreases, the same remaining cost is divided among fewer people, so the amount per member increases.

Answer

a) \(\frac{K-Z}{s}\) dollars b) \(\$28\) c) The amount increases because the same cost is divided among fewer members.
5239377
Consider the formula \(x=\frac{S-b}{k}\). a) Create a real-world problem that can be modeled by this formula. Explain what \(S\), \(b\), and \(k\) represent in your problem. b) Find \(x\) when \(S=85\), \(b=13\), and \(k=12\).

Hints

- Think of a situation in which part of a total is removed first. - What quantity could be divided into \(k\) equal groups? - The fraction bar groups the entire numerator, so subtract before dividing.

Solution

1. One possible context is dividing a remaining balance into equal payments. Let \(S\) be the total price, \(b\) the amount paid at the start, \(k\) the number of equal payments, and \(x\) the amount of each payment. 2. Substitute the given values: \(x=\frac{85-13}{12}\). 3. Simplify the numerator: \(85-13=72\). 4. Divide: \(x=\frac{72}{12}=6\).

Answer

a) Answers will vary. Example: A used bicycle costs \(\$85\). A buyer pays \(\$13\) at the start and divides the remaining balance into \(12\) equal monthly payments. Here, \(S\) is the total price, \(b\) is the initial payment, \(k\) is the number of monthly payments, and \(x\) is one payment. b) \(x=6\), so each payment is \(\$6\).
5239867
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 \(375\,\text{ft}^2\).
5240077
Two parking lots, A and B, contain cars in the ratio \(2:3\). If \(12\) cars move from Lot A to Lot B, the new ratio is \(1:2\). How many cars were originally in each lot?

Hints

- Represent the original ratio with equal-sized parts. - Update both counts after \(12\) cars move. - Translate the new \(1:2\) ratio into an equation. - Cross multiply to remove the fractions.

Solution

1. Let Lot A originally contain \(2x\) cars and Lot B contain \(3x\) cars. 2. After the move, the counts are \(2x - 12\) and \(3x + 12\). 3. Write the new ratio equation \(\frac{2x - 12}{3x + 12} = \frac{1}{2}\). 4. Cross multiply: \(2(2x - 12) = 3x + 12\). 5. Distribute and solve: \(4x - 24 = 3x + 12\), so \(x = 36\). 6. Lot A originally had \(2 \cdot 36 = 72\) cars, and Lot B had \(3 \cdot 36 = 108\) cars.

Answer

Lot A originally had \(72\) cars, and Lot B had \(108\) cars.
5240097
A teacher buys \(6\) packs of highlighters and \(12\) pencils for \(\$54.00\). One pack of highlighters costs four times as much as one pencil. a) Use an equation to find the price of one pencil and one pack of highlighters. b) A student claims, “If a pack of highlighters cost only three times as much as a pencil, the savings would buy exactly \(6\) more pencils.” Determine whether the claim is correct.

Hints

- Use the smaller price as the variable. - Express the larger price as a multiple of that variable. - Write an equation for the original total cost. - For part b), compare the total savings with the cost of six pencils.

Solution

1. Let \(x\) dollars be the price of one pencil. A pack of highlighters costs \(4x\) dollars. 2. Write \(12x + 6(4x) = 54\). 3. Combine like terms: \(12x + 24x = 54\), so \(36x = 54\) and \(x = 1.50\). 4. A pencil costs \(\$1.50\), and a pack of highlighters costs \(4 \cdot 1.50 = 6.00\) dollars. 5. Under the changed ratio, a pack would cost \(3 \cdot 1.50 = 4.50\) dollars. The savings per pack would be \(6.00 - 4.50 = 1.50\) dollars. 6. Across \(6\) packs, the savings would be \(6 \cdot 1.50 = 9.00\) dollars. 7. Six pencils cost \(6 \cdot 1.50 = 9.00\) dollars, so the claim is correct.

Answer

a) One pencil costs \(\$1.50\), and one pack of highlighters costs \(\$6.00\). b) The claim is correct because the \(\$9.00\) savings equals the cost of \(6\) pencils.
5240107
A workshop uses \(8\) identical LED floodlights and \(2\) identical heaters. Together, they draw \(3600\,\text{W}\). Each heater draws five times as much power as one floodlight. a) Find the power draw of one floodlight and one heater. b) The heaters are upgraded so each uses \(20\%\) less power. What is the new total power draw of all \(8\) floodlights and \(2\) heaters?

Hints

- Express the total power as the sum of all device powers. - Use the five-to-one relationship to write the equation with one variable. - A \(20\%\) reduction leaves \(80\%\) of the original heater power. - Recalculate the total using both types of devices.

Solution

1. Let \(s\) be the power draw of one floodlight. Each heater draws \(5s\). 2. Write \(8s + 2(5s) = 3600\). 3. Combine like terms: \(18s = 3600\), so \(s = 200\). 4. One heater draws \(5 \cdot 200 = 1000\,\text{W}\). 5. A \(20\%\) reduction is \(0.20 \cdot 1000 = 200\,\text{W}\), so each upgraded heater draws \(800\,\text{W}\). 6. The new total is \(8 \cdot 200 + 2 \cdot 800 = 1600 + 1600 = 3200\,\text{W}\).

Answer

a) One floodlight draws \(200\,\text{W}\), and one heater draws \(1000\,\text{W}\). b) The upgraded system draws \(3200\,\text{W}\) in all.
5240507
A concentrated vinegar solution is \(25\%\) acetic acid by volume. A recipe requires vinegar that is \(5\%\) acetic acid. Assume the volumes add. a) How many milliliters of water must be added to \(100\,\text{mL}\) of the concentrate to make a \(5\%\) solution? b) A cook wants to make exactly \(1\,\text{L}\) of the \(5\%\) solution. How many milliliters of concentrate and how many milliliters of water are needed?

Hints

- Find the acetic-acid volume in the given concentrate. - Determine the total volume for which that acid volume is \(5\%\). - For part b, first find the acid volume required in \(1\,\text{L}\) of the target solution. - Use the concentrate’s percentage to find how much concentrate supplies that acid volume.

Solution

1. For part a, \(100\,\text{mL}\) of concentrate contains \(100\,\text{mL} \cdot 0.25 = 25\,\text{mL}\) of acetic acid. 2. If \(25\,\text{mL}\) is \(5\%\) of the final mixture, the final volume is \(25 \div 0.05 = 500\,\text{mL}\). 3. The water added is \(500\,\text{mL} - 100\,\text{mL} = 400\,\text{mL}\). 4. For part b, \(1\,\text{L} = 1000\,\text{mL}\), and a \(5\%\) solution contains \(1000\,\text{mL} \cdot 0.05 = 50\,\text{mL}\) of acetic acid. 5. The concentrate volume needed is \(50 \div 0.25 = 200\,\text{mL}\). The water volume is \(1000\,\text{mL} - 200\,\text{mL} = 800\,\text{mL}\).

Answer

a) Add \(400\,\text{mL}\) of water. b) Mix \(200\,\text{mL}\) of concentrate with \(800\,\text{mL}\) of water.
5240687
A cyclist takes a two-day trip. On the first day, the cyclist rides for \(t_1\) hours at \(v_1\) miles per hour. On the second day, the cyclist rides for \(t_2\) hours at \(v_2\) miles per hour. Write an expression for the average speed \(v_{\text{avg}}\) over the entire trip.

Hints

- Use distance equals rate times time for each day. - Add the two distances and the two travel times. - Average speed is total distance divided by total time, not usually the mean of the two speeds.

Solution

1. The first-day distance is \(v_1t_1\) miles. 2. The second-day distance is \(v_2t_2\) miles. 3. The total distance is \(v_1t_1+v_2t_2\), and the total time is \(t_1+t_2\). 4. Therefore, \(v_{\text{avg}}=\frac{v_1t_1+v_2t_2}{t_1+t_2}\).

Answer

\(v_{\text{avg}}=\frac{v_1t_1+v_2t_2}{t_1+t_2}\)
5240767
A smartphone battery is at \(80\%\) at the start of an observation. During intensive use, its charge decreases by \(12\) percentage points per hour. Write an inequality and find the times \(t\) when the charge is below \(20\%\). Consider only times up to complete discharge.

Hints

- Write a linear expression for the battery charge after \(t\) hours. - Translate “below” with a strict inequality. - Reverse the inequality sign when dividing by a negative number. - Find when the battery reaches \(0\%\) to set the physical endpoint.

Solution

1. The charge after \(t\) hours is \(80 - 12t\). 2. Write the inequality \(80 - 12t < 20\). 3. Subtract \(80\): \(-12t < -60\). 4. Divide by \(-12\) and reverse the inequality sign: \(t > 5\). 5. Complete discharge occurs when \(80 - 12t = 0\), so \(t = \frac{20}{3} \approx 6.67\). 6. Therefore, within the physical time interval, the charge is below \(20\%\) when \(5 < t \le \frac{20}{3}\).

Answer

The inequality is \(80 - 12t < 20\), and the relevant solution is \(5 < t \le \frac{20}{3}\) hours.
5241197
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 \(\$300\).
5241207
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.
5241217
Maya is reading a new book. On the first day, she reads exactly one-third of the pages. On the second day, she reads one-fourth of the total number of pages. She then has exactly \(125\) pages left. Write an equation and find the total number of pages in the book.

Hints

- Use a variable for the total number of pages. - Express each amount read as a fraction of the total. - The pages read and the pages left must add to the total. - Combine the fractional terms before solving.

Solution

1. Let \(x\) be the total number of pages. 2. The pages read and the pages remaining add to the total, so write \(\frac{1}{3}x + \frac{1}{4}x + 125 = x\). 3. Combine the fractions: \(\frac{4}{12}x + \frac{3}{12}x + 125 = x\), so \(\frac{7}{12}x + 125 = x\). 4. Subtract \(\frac{7}{12}x\): \(125 = \frac{5}{12}x\). 5. Multiply by \(\frac{12}{5}\): \(x = 300\).

Answer

A suitable equation is \(\frac{1}{3}x + \frac{1}{4}x + 125 = x\). The book has \(300\) pages.
5241287
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.
5241297
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.
5279247
A water tank initially contains \(W\) gallons. A leak drains water at a constant rate of \(k\) gallons per hour. a) Write an expression for the amount of water remaining after \(h\) hours. b) Find the remaining amount when \(W=500\), \(k=12\), and \(h=8\). c) Write a formula, using \(W\) and \(k\), for the time when the tank becomes empty. d) A student says that any value can be substituted for \(h\). Explain why that is not meaningful in this situation.

Hints

- Multiply the leak rate by the number of hours. - Subtract the leaked amount from the starting amount. - Set the remaining amount equal to \(0\) to find when the tank is empty. - Consider the physical meaning of a negative result.

Solution

1. In \(h\) hours, the leak drains \(kh\) gallons, so \(W-kh\) gallons remain. 2. For the given values, \(500-12\cdot 8=500-96=404\) gallons remain. 3. The tank is empty when \(kh=W\), so \(h=\frac{W}{k}\). 4. The model is meaningful only for \(0\le h\le\frac{W}{k}\). After the tank is empty, the amount cannot continue into negative values.

Answer

a) \(W-kh\) gallons b) \(404\) gallons c) \(h=\frac{W}{k}\) hours d) The model is limited to \(0\le h\le\frac{W}{k}\) because a tank cannot contain a negative amount of water.
5279477
Two rectangular beds in a school garden have the same width. The first bed is \(45\,\text{ft}\) long, and the second is \(60\,\text{ft}\) long. The second bed has \(375\,\text{ft}^2\) more area than the first. Find their common width.

Hints

- Write an area expression for each garden bed. - Subtract the smaller area from the larger area. - Use the shared width as the variable. - Solve the resulting equation.

Solution

1. Let \(x\) feet be the common width. 2. The difference in area is \(60x - 45x = 375\). 3. Combine like terms: \(15x = 375\). 4. Divide by \(15\): \(x = 25\).

Answer

The common width is \(25\,\text{ft}\).
5280107
Two trains leave the same station at the same time and travel in opposite directions. After \(1.5\) hours, the trains are \(150\) miles apart. The slower train travels at exactly \(\frac{2}{3}\) the speed of the faster train. Find both speeds.

Hints

- Write the slower train's speed as a fraction of the faster train's speed. - Each distance equals speed times \(1.5\) hours. - Because the trains move in opposite directions, their distances add.

Solution

1. Let \(v\,\text{mph}\) be the faster train's speed. The slower train's speed is \(\frac{2}{3}v\). 2. In \(1.5\) hours, their distances add to \(150\) miles: \(1.5v + 1.5(\frac{2}{3}v) = 150\). 3. Simplify: \(1.5v + v = 150\), so \(2.5v = 150\). 4. Divide by \(2.5\): \(v = 60\). 5. The slower train's speed is \(\frac{2}{3} \cdot 60 = 40\,\text{mph}\).

Answer

The faster train travels at \(60\,\text{mph}\), and the slower train travels at \(40\,\text{mph}\).
5280327
Two pumps fill a water tank with total capacity \(V\). Pump A moves \(x\) gallons per minute, and Pump B moves \(y\) gallons per minute. a) After both pumps have run for \(t\) minutes, the tank still needs \(R\) gallons. Write a formula for \(V\). b) Find \(V\) when \(x=15\), \(y=20\), \(t=12\), and \(R=80\). c) Suppose the tank's actual capacity is \(630\) gallons. After the first \(12\) minutes, how many additional minutes must both pumps run to fill it?

Hints

- Add the two pump rates. - The total capacity equals the amount already pumped plus the amount still needed. - For part c), find the remaining volume and divide by the combined rate.

Solution

1. The combined rate is \(x+y\) gallons per minute. 2. In \(t\) minutes, the pumps add \((x+y)t\) gallons. Including the amount still needed gives \(V=(x+y)t+R\). 3. For the given values, \(V=(15+20)\cdot 12+80=35\cdot 12+80=500\) gallons. 4. After \(12\) minutes, the pumps have added \(35\cdot 12=420\) gallons. 5. For a \(630\)-gallon tank, \(630-420=210\) gallons remain. At \(35\) gallons per minute, this takes \(210\div 35=6\) additional minutes.

Answer

a) \(V=(x+y)t+R\) b) \(500\) gallons c) \(6\) additional minutes
5320647
An urn contains \(15\) balls: \(6\) blue and \(9\) white. The diagram shows the composition. One ball is selected at random. a) Find the probability of selecting a blue ball. Write the probability as a fraction in simplest form and as a percent. b) How many white balls must be added so that the probability of selecting a blue ball is exactly \(25\%\)?
Figure for problem 532064

Hints

- Use the stated blue count and total number of balls. - Write the initial probability as blue balls divided by total balls. - For part b, the number of blue balls does not change. - Write an equation in which \(6\) divided by the new total equals \(25\%\).

Solution

1. There are \(6\) blue balls out of \(15\) total balls, so \(P(\text{blue}) = \frac{6}{15} = \frac{2}{5} = 40\%\). 2. For part b, the number of blue balls stays \(6\). Let \(x\) be the new total number of balls. The equation \(\frac{6}{x} = \frac{1}{4}\) gives \(x = 24\). 3. Since the urn currently contains \(15\) balls, \(24 - 15 = 9\) white balls must be added.

Answer

a) \(\frac{2}{5} = 40\%\) b) \(9\) white balls
5352767
During a cold night, the temperature falls by \(15\,^\circ\text{F}\) and reaches a low of \(-6\,^\circ\text{F}\) in the morning. Use the number line to find the temperature the previous evening.
Figure for problem 535276

Hints

- A decrease is represented by a leftward jump. - Work backward from \(-6\) by moving \(15\) units right. - Check that subtracting \(15\) from your answer gives \(-6\).

Solution

1. Let the starting temperature be \(x\). The situation is modeled by \(x-15=-6\). 2. Add \(15\) to both sides: \(x=-6+15=9\). 3. The previous evening’s temperature was \(9\,^\circ\text{F}\).

Answer

\(9\,^\circ\text{F}\)
5106487
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 \(489\,\text{kg}\).
5119257
Hikers use this rule to estimate travel time: “Allow one hour for every \(3\) miles of horizontal distance. Add one more hour for every \(2000\) feet of elevation gain.” a) Write an equation for the total time \(T\), in hours. Let \(s\) be the horizontal distance in miles and \(h\) be the elevation gain in feet. b) Estimate the time for a hike of \(7.5\) miles with \(3000\) feet of elevation gain. c) A group claims, “If we double the distance but keep the elevation gain the same, the estimated time also doubles.” Use your equation to evaluate the claim.

Hints

- Express “one hour for every \(3\) miles” as a quotient. - Keep track of the units: \(s\) is in miles and \(h\) is in feet. - Compare the new expression with twice the original expression.

Solution

1. The horizontal-distance time is \(\frac{s}{3}\), and the elevation-gain time is \(\frac{h}{2000}\). Therefore, \(T=\frac{s}{3}+\frac{h}{2000}\). 2. Substitute \(s=7.5\) and \(h=3000\): \(T=\frac{7.5}{3}+\frac{3000}{2000}=2.5+1.5=4\) hours. 3. The original time is \(T_1=\frac{s}{3}+\frac{h}{2000}\). Doubling only the distance gives \(T_2=\frac{2s}{3}+\frac{h}{2000}\). But twice the original time is \(2T_1=\frac{2s}{3}+\frac{h}{1000}\). These are not equal unless \(h=0\), so the claim is generally false.

Answer

a) \(T=\frac{s}{3}+\frac{h}{2000}\) b) \(4\) hours c) The claim is false in general. Only the distance portion doubles; the elevation portion stays the same.
5120397
Fence posts are installed with equal spacing. The gap from the first post to the beginning of the fence line and the gap from the last post to the end are the same as the gaps between posts. With \(12\,\text{ft}\) gaps, a certain number of posts is needed. With \(7.5\,\text{ft}\) gaps, \(3\) more posts are needed. Find the total length of the fence line.

Hints

- Express the number of gaps in terms of the number of posts. - Write one length expression for each spacing choice. - The total length is the same in both arrangements, so set the expressions equal.

Solution

1. Let \(x\) be the original number of posts. Then there are \(x + 1\) equal gaps. 2. With \(12\,\text{ft}\) gaps, the length is \(L = 12(x + 1)\). 3. The second arrangement has \(x + 3\) posts and therefore \(x + 4\) gaps. Its length is \(L = 7.5(x + 4)\). 4. Set the expressions equal: \(12(x + 1) = 7.5(x + 4)\). 5. Distribute: \(12x + 12 = 7.5x + 30\). Subtract \(7.5x\) and \(12\): \(4.5x = 18\), so \(x = 4\). 6. The length is \(L = 12(4 + 1) = 60\,\text{ft}\).

Answer

\(60\,\text{ft}\)
5120757
A temporary construction fence uses \(12\,\text{ft}\) panels and connecting posts that are \(4\,\text{in.}\) wide. A fence section begins and ends with a post, with one panel between each pair of posts. a) Write a formula for the total length \(L\), in feet, of a fence with \(n\) panels. b) How many panels are needed to cover at least \(160\,\text{ft}\)? c) How many posts are needed for that number of panels? Explain the relationship.

Hints

- Convert the post width to feet. - Count the posts for a small number of panels to identify the pattern. - “At least” requires an inequality and a whole-number result.

Solution

1. Convert the post width: \(4\,\text{in.}=\frac{1}{3}\,\text{ft}\). A fence with \(n\) panels has \(n+1\) posts, so \(L=12n+\frac{1}{3}(n+1)\). 2. To cover at least \(160\,\text{ft}\), solve \(12n+\frac{1}{3}(n+1)\ge160\). 3. Multiply by \(3\): \(36n+n+1\ge480\), so \(37n\ge479\) and \(n\ge\frac{479}{37}\approx12.95\). 4. Since \(n\) must be a whole number, at least \(13\) panels are needed. 5. The number of posts is \(n+1=14\).

Answer

a) \(L=12n+\frac{1}{3}(n+1)\) b) \(13\) panels c) \(14\) posts; there is one more post than panel.
5125607
Noah and Maya each have exactly \(48\,\text{ft}\) of wire for a wireframe model. Noah builds a cube. Maya builds a rectangular prism with a square base whose height is twice a base edge. Find the difference between the heights of the two models.

Hints

- Find the height of each model separately. - Count how many edges of each length appear in each wireframe. - Express all edge lengths of Maya's prism using one variable.

Solution

1. A cube has \(12\) equal edges. Let \(s\) be the edge length. Then \(12s = 48\), so \(s = 4\). Noah's model is \(4\,\text{ft}\) high. 2. Let \(a\) be a base edge of Maya's prism. Its height is \(2a\). 3. The prism has \(8\) base edges of length \(a\) and \(4\) vertical edges of length \(2a\). Therefore, \(8a + 4(2a) = 48\). 4. Simplify: \(16a = 48\), so \(a = 3\). Maya's model is \(2a = 6\,\text{ft}\) high. 5. The difference in height is \(6\,\text{ft} - 4\,\text{ft} = 2\,\text{ft}\).

Answer

Maya''s model is \(2\,\text{ft}\) taller than Noah''s model.
5125917
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 \(90\,\text{lb}\).
5126157
A school library receives a shipment of new books. One-half are novels, one-fifth are nonfiction books, and one-eighth are graphic novels. The remaining \(63\) books are children’s books. Write and solve an equation to find the total number of new books \(x\). Briefly explain what \(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\) represents in this situation.

Hints

- Subtract the known category fractions from one whole to find the children’s-book fraction. - Express the fractions with a common denominator. - Set the remaining fraction of the total equal to \(63\).

Solution

1. Write the equation \(x\left(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\right) = 63\). 2. Use denominator \(40\): \(\frac{1}{2} + \frac{1}{5} + \frac{1}{8} = \frac{20}{40} + \frac{8}{40} + \frac{5}{40} = \frac{33}{40}\). 3. Find the remaining fraction: \(1 - \frac{33}{40} = \frac{7}{40}\). 4. Solve \(\frac{7}{40}x = 63\): \(x = 63 \cdot \frac{40}{7}\). 5. Calculate: \(x = 360\). 6. The expression represents the fraction of all new books that are children’s books.

Answer

The equation can be written as \(\frac{7}{40}x = 63\), and the total is \(x = 360\) books. The expression \(1 - \left(\frac{1}{2} + \frac{1}{5} + \frac{1}{8}\right)\) is the fraction of the shipment made up of children’s books.
5126277
Isabelle has \(40\) counters divided among three piles: left \(L\), middle \(M\), and right \(R\). She does not know the amount in each pile. The left pile has at least \(7\) counters, and the right pile has at least \(5\). She moves \(7\) counters from \(L\) to \(M\), then moves \(5\) counters from \(R\) to \(M\). After the moves, a student reports that the left and right piles contain \(15\) counters altogether. Isabelle immediately says, “Then the middle pile started with exactly \(13\) counters.” Use equations with variables to show that Isabelle is correct.

Hints

- Write an equation for the total number of counters before any moves. - Express the reported outside-pile total using the original amounts \(L\) and \(R\). - How are \(L+R\), \(M\), and the total of \(40\) related?

Solution

1. At the start, \(L+M+R=40\). 2. After the moves, the two outside piles contain \((L-7)+(R-5)\) counters. The student says this total is \(15\), so \((L-7)+(R-5)=15\). 3. Simplifying gives \(L+R-12=15\), so \(L+R=27\). 4. Substitute this into the total-count equation: \(27+M=40\). 5. Solving gives \(M=13\). Therefore, Isabelle is correct.

Answer

From \((L-7)+(R-5)=15\), we get \(L+R=27\). Since \(L+M+R=40\), it follows that \(27+M=40\), so \(M=13\).
5127237
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 \(48\%\).
5127467
In a class, \(60\%\) of the students regularly play on organized sports teams. Of those students, \(25\%\) play on a soccer team. a) What percent of the entire class plays on a soccer team? b) In addition, \(5\) students who do not play organized sports hike regularly. The team athletes and these hikers together make up more than \(80\%\) of the class. Find the total number of students in the class, given that the number of soccer players must be a whole number. c) How many students are neither team athletes nor hikers?

Hints

- Find a percent of a percent for part a. - Write an inequality for the athletes and hikers together. - Use the whole-number condition on the number of soccer players to narrow the possible class sizes.

Solution

1. For part a, the percent of the class on a soccer team is \(0.60 \cdot 0.25 = 0.15 = 15\%\). 2. Let \(T\) be the class size. The condition in part b gives \(0.60T + 5 > 0.80T\). 3. Solving the inequality gives \(5 > 0.20T\), so \(T < 25\). 4. The number of soccer players is \(0.15T = \frac{3}{20}T\). For this to be a whole number, \(T\) must be a multiple of \(20\). 5. The positive multiple of \(20\) less than \(25\) that can contain the stated groups is \(T = 20\). 6. There are \(0.60 \cdot 20 + 5 = 17\) team athletes and hikers. Therefore, \(20 - 17 = 3\) students are in neither group.

Answer

a) \(15\%\) of the class plays on a soccer team. b) The class has \(20\) students. c) \(3\) students are neither team athletes nor hikers.
5135697
A box contains \(40\) marbles. Of the marbles, \(25\%\) are blue and the rest are white. How many blue marbles must be added so that the probability of drawing a blue marble is exactly \(40\%\)? Do not change the number of white marbles. Justify your reasoning.

Hints

- Find the initial blue and white counts. - Adding blue marbles changes both the favorable count and the total count. - The number of white marbles stays unchanged while blue marbles are added. - Let a variable represent the number added and write a probability equation.

Solution

1. Initially, there are \(0.25 \cdot 40 = 10\) blue marbles and \(30\) white marbles. 2. Let \(x\) be the number of blue marbles added. Then the new blue count is \(10 + x\), and the new total is \(40 + x\). 3. Set up the equation \(\frac{10 + x}{40 + x} = 0.4\). 4. Solve: \(10 + x = 0.4(40 + x)\), so \(10 + x = 16 + 0.4x\). Then \(0.6x = 6\), giving \(x = 10\). 5. Check: after adding \(10\) blue marbles, \(\frac{20}{50} = 0.4\).

Answer

\(10\) blue marbles must be added.
5142477
A new candle is \(10\,\text{in.}\) tall. Once lit, it burns down at a constant rate of \(0.4\,\text{in.}\) per hour. For what burning times \(h\) is the candle shorter than \(4\,\text{in.}\)? Consider only times up to the moment when the candle burns out completely.

Hints

- Write a linear expression for the candle's height after \(h\) hours. - Use a strict inequality because the candle must be shorter than \(4\,\text{in.}\). - Reverse the inequality sign when dividing by a negative number. - Find the time when the candle reaches a height of \(0\).

Solution

1. The candle's height after \(h\) hours is \(10 - 0.4h\). 2. Write the inequality for a height below \(4\,\text{in.}\): \(10 - 0.4h < 4\). 3. Subtract \(10\): \(-0.4h < -6\). 4. Divide by \(-0.4\) and reverse the inequality sign: \(h > 15\). 5. The candle burns out when \(10 - 0.4h = 0\), which gives \(h = 25\). 6. Therefore, within the physical time interval, the candle is shorter than \(4\,\text{in.}\) when \(15 < h \le 25\).

Answer

\(15 < h \le 25\) hours
5224867
A rancher is asked how many sheep are in the flock and replies, “If I tripled my flock, sold one-fourth of the original number of sheep, and then received \(2\) more sheep, I would have exactly \(68\) sheep.” a) Write and solve an equation to find the original number of sheep. b) An apprentice claims the statement must be impossible because calculations involving fourths can produce fractional sheep. Determine whether that concern applies here.

Hints

- Translate “triple” and “one-fourth of the original number” into algebraic terms. - Every fractional part refers to the original flock size. - A solution involving animals must be checked for whole-number reasonableness. - Verify that one-fourth of your result is a whole number.

Solution

1. Let \(x\) be the original number of sheep. 2. Write the equation \(3x - \frac{1}{4}x + 2 = 68\). 3. Combine like terms: \(\frac{11}{4}x + 2 = 68\). 4. Subtract \(2\): \(\frac{11}{4}x = 66\). 5. Multiply by \(\frac{4}{11}\): \(x = 24\). 6. The result is meaningful because \(24\) is a whole number and one-fourth of \(24\) is \(6\), also a whole number.

Answer

a) The original flock had \(24\) sheep. b) The apprentice is not correct in this case. Both the original number, \(24\), and one-fourth of it, \(6\), are whole numbers.
5228447
Two rectangular flower beds have the same length. Bed A is \(3\,\text{ft}\) wide, and Bed B is \(5\,\text{ft}\) wide. Bed B has \(14\,\text{ft}^2\) more area than Bed A. a) Use an equation to find the common length. b) Bed C has the same length and a width equal to the sum of the other two widths, \(8\,\text{ft}\). Find the area of Bed C. c) Explain why the area of Bed C equals the sum of the areas of Beds A and B.

Hints

- Compare the two area expressions using their common length. - Find how much more area is added for each foot of length. - Notice that \(8 = 3 + 5\). - Use the distributive property to explain the area relationship.

Solution

1. Let \(L\) feet be the common length. The difference in area is \(5L - 3L = 14\). 2. Simplify: \(2L = 14\), so \(L = 7\). 3. The areas of Beds A and B are \(7 \cdot 3 = 21\,\text{ft}^2\) and \(7 \cdot 5 = 35\,\text{ft}^2\). 4. Bed C has area \(7 \cdot 8 = 56\,\text{ft}^2\). 5. Because all three beds have length \(L\), the distributive property gives \(L(3 + 5) = 3L + 5L\). Therefore, Bed C's area is the sum of the other two areas.

Answer

a) The common length is \(7\,\text{ft}\). b) The area of Bed C is \(56\,\text{ft}^2\). c) Since the common length is a shared factor, \(L(3 + 5) = 3L + 5L\), so the combined width produces the sum of the two areas.
5229347
A store sells large wooden crates for \(\$12.00\) and small crates for \(\$4.50\). A customer spends \(\$153.00\) and buys three times as many small crates as large crates. a) Use an equation to find how many crates of each size the customer bought. b) The customer bought \(24\) crates in all. What would the total cost have been if the customer had instead bought twice as many large crates as small crates?

Hints

- Express the number of small crates in terms of the number of large crates. - Write an equation for the original total cost. - For part b), keep the total number of crates unchanged. - Translate the new ratio into two quantities that add to \(24\).

Solution

1. Let \(x\) be the number of large crates. Then the number of small crates is \(3x\). 2. Write \(12x + 4.50(3x) = 153\). 3. Combine like terms: \(12x + 13.5x = 153\), so \(25.5x = 153\). 4. Divide by \(25.5\): \(x = 6\). The customer bought \(6\) large crates and \(18\) small crates. 5. For part b), let \(y\) be the number of small crates. Then there are \(2y\) large crates, and \(y + 2y = 24\). 6. Solve: \(3y = 24\), so \(y = 8\). The new quantities are \(8\) small and \(16\) large crates. 7. The new cost is \(16 \cdot 12 + 8 \cdot 4.50 = 192 + 36 = 228\) dollars.

Answer

a) The customer bought \(6\) large crates and \(18\) small crates. b) The new total cost would be \(\$228.00\).
5239267
A youth group with \(n\) members plans a trip. The bus costs \(\$B\), divided equally among the group. Then \(m\) more people join. A larger bus is needed, increasing the total cost by \(\$s\). a) Write an expression for the decrease in cost per person after the additional people join. b) Find the savings per person when \(n=20\), \(B=400\), \(m=5\), and \(s=50\). c) Under what condition would the cost per person stay the same even though more people join and the bus costs more? Explain without calculating specific values.

Hints

- Find the original and new costs per person. - Savings is the original amount minus the new amount. - For part c), compare the added cost per new participant with the original cost per person.

Solution

1. The original cost per person is \(\frac{B}{n}\). 2. The new cost per person is \(\frac{B+s}{n+m}\). 3. The decrease in cost per person is \(\frac{B}{n}-\frac{B+s}{n+m}\). 4. For the given values, \(\frac{400}{20}-\frac{400+50}{20+5}=20-18=2\), so each person saves \(\$2\). 5. The price stays unchanged when the added cost per new participant equals the original cost per person: \(\frac{s}{m}=\frac{B}{n}\).

Answer

a) \(\frac{B}{n}-\frac{B+s}{n+m}\) dollars b) \(\$2\) per person c) The cost stays the same when \(\frac{s}{m}=\frac{B}{n}\), so each new participant's share exactly covers the added bus cost.
5239467
A chemist has \(200\,\text{g}\) of a \(10\%\) salt solution. How many grams of a \(40\%\) salt solution must be added to make a mixture that is exactly \(25\%\) salt?

Hints

- Find the mass of salt already present. - Use a variable for the mass of the second solution. - Express both the added salt and the final total mass in terms of the variable. - Set total salt equal to \(25\%\) of the final mixture.

Solution

1. The original solution contains \(200\,\text{g} \cdot 0.10 = 20\,\text{g}\) of salt. 2. Let \(x\) be the mass, in grams, of the \(40\%\) solution added. It contributes \(0.40x\) grams of salt, and the total mixture mass is \(200 + x\) grams. 3. Set up the equation \(20 + 0.40x = 0.25(200 + x)\). 4. Solving gives \(20 + 0.40x = 50 + 0.25x\), so \(0.15x = 30\) and \(x = 200\).

Answer

The chemist must add \(200\,\text{g}\) of the \(40\%\) salt solution.
5241227
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 \(420\,\text{ft}^2\).
5279427
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.
5318877
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?
Figure for problem 531887

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.

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.