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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) Let \(c\) be the temperature in degrees Celsius and \(F\) be the temperature in degrees Fahrenheit. Write an equation that models the relationship. b) Use the equation 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. - Use one variable for the input temperature and one for the output temperature. - Be careful with the sign when substituting a negative Celsius temperature.

Solution

1. Following the stated operations gives \(F=1.8c+32\). 2. For \(20\,^{\circ}\text{C}\): \(F=1.8\cdot20+32=68\,^{\circ}\text{F}\). 3. For \(35\,^{\circ}\text{C}\): \(F=1.8\cdot35+32=95\,^{\circ}\text{F}\). 4. For \(-10\,^{\circ}\text{C}\): \(F=1.8\cdot(-10)+32=14\,^{\circ}\text{F}\).

Answer

a) \(F=1.8c+32\) 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) The middle pile then has \(15\) counters. Write and solve an equation for its starting amount.

Hints

- Track only how each move changes the middle pile. - Combine the two changes to simplify the expression in part a. - In part b), set the expression for the final middle pile equal to the stated final amount. - Solve the equation for the starting value.

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) Equation: \(z+2=15\); \(z=13\) counters
5223447
A gym charges a one-time enrollment fee of \(\$s\) and a monthly fee of \(\$m\). Let \(G\) be the total membership cost, in dollars, for \(x\) months. a) Write an equation that models \(G\). 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. - Use a variable for the total cost so the relationship can be written as an equation. - One year has \(12\) months.

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: \(G=19.90\cdot12+35.00=238.80+35.00=273.80\).

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

\(P=6x+2y+8\)
5237597
Anna, Beth, and Clara collect a total of \(120\,\text{lb}\) of paper for a school recycling drive. Anna collects \(10\,\text{lb}\) more than Beth and \(10\,\text{lb}\) less than Clara. Let \(x\) be Anna’s amount. Write an equation for the three amounts, simplify it, solve it, and report how many pounds each person collects.

Hints

- Use Anna’s amount as the central variable because the other two amounts are symmetric around it. - Express Beth and Clara before adding the three amounts. - Notice what happens to the constant terms when the equation is simplified. - Check the total and both difference relationships.

Solution

1. Let \(x\) be the number of pounds Anna collects. 2. Beth collects \(x - 10\), and Clara collects \(x + 10\). 3. Write the total equation \(x + (x - 10) + (x + 10) = 120\). 4. Combine like terms: \(3x = 120\). 5. Divide by \(3\): \(x = 40\). 6. Beth collects \(40 - 10 = 30\,\text{lb}\), and Clara collects \(40 + 10 = 50\,\text{lb}\). 7. Check: \(40 + 30 + 50 = 120\).

Answer

Equation: \(x+(x-10)+(x+10)=120\), which simplifies to \(3x=120\). Anna: \(40\,\text{lb}\); Beth: \(30\,\text{lb}\); Clara: \(50\,\text{lb}\)
5237717
Lucas, Mia, and Noah have \(200\) trading cards altogether. Lucas has \(20\) more cards than Mia, and Noah has twice as many cards as Mia. Let \(x\) be Mia’s number of cards. Write one equation for the total, solve it, and report each person’s number of cards.

Hints

- Use Mia’s count as the base variable for both relationships. - Translate “\(20\) more” and “twice as many” into separate expressions. - Add all three quantities before setting the total equal to \(200\). - Check both relationships after solving.

Solution

1. Let \(x\) be the number of cards Mia has. 2. Lucas has \(x + 20\), and Noah has \(2x\). 3. Write the total equation \(x + (x + 20) + 2x = 200\). 4. Combine like terms: \(4x + 20 = 200\). 5. Subtract \(20\): \(4x = 180\). Divide by \(4\): \(x = 45\). 6. Lucas has \(45 + 20 = 65\) cards, and Noah has \(2 \cdot 45 = 90\) cards.

Answer

Equation: \(x+(x+20)+2x=200\) Mia: \(45\) cards; Lucas: \(65\) cards; Noah: \(90\) cards
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.
5279229
A class rents a bus. The total cost \(K\) consists of a fixed fee \(G\) plus \(p\) dollars for each of \(n\) participants: \( K=G+np. \) a) Rearrange the formula to solve for \(n\). b) Find \(n\) when \(K=468\), \(G=120\), and \(p=14.50\). c) Rearrange the original formula to solve for \(p\).

Hints

- Remove the fixed fee before dividing by a per-person quantity. - Keep the product \(np\) intact until the additive term is removed. - Use the same starting equation for both rearrangements. - Check the numerical result in the original cost formula.

Solution

1. Subtract \(G\): \( K-G=np. \) 2. Divide by \(p\): \( n=\frac{K-G}{p}. \) 3. Substitute: \( n=\frac{468-120}{14.50}=\frac{348}{14.50}=24. \) 4. To solve for \(p\), again subtract \(G\) and then divide by \(n\): \( p=\frac{K-G}{n}. \)

Answer

a) \(n=\frac{K-G}{p}\) b) \(n=24\) c) \(p=\frac{K-G}{n}\)
5546987
A student has already earned \(26\) service points. Each community event adds \(9\) points. The student needs at least \(80\) points to qualify for an award. Write and solve an inequality for the number \(e\) of additional events needed. Interpret the solution.

Hints

- Translate “at least” into the appropriate comparison symbol. - Include both the points already earned and the points from future events. - Interpret the solution using the fact that events are counted in whole numbers.

Solution

1. “At least \(80\)” means the total must be greater than or equal to \(80\): \(26+9e\ge80\). 2. Subtract \(26\): \(9e\ge54\). 3. Divide by \(9\): \(e\ge6\). 4. Since \(e\) counts events, the student must attend at least \(6\) additional events.

Answer

\(26+9e\ge80\), so \(e\ge6\). The student needs at least \(6\) more events.
5546997
A storage shelf can hold no more than \(38\,\text{lb}\). A fixed container weighs \(14\,\text{lb}\), and each identical supply box weighs \(4\,\text{lb}\). Write and solve an inequality for the number \(b\) of supply boxes that can be placed on the shelf. What is the greatest possible number of boxes?

Hints

- Translate “no more than” into an inequality symbol. - Include the fixed container weight before the box weight. - Use the whole-number meaning of \(b\) when stating the maximum.

Solution

1. “No more than \(38\,\text{lb}\)” gives \(14+4b\le38\). 2. Subtract \(14\): \(4b\le24\). 3. Divide by \(4\): \(b\le6\). 4. Since \(b\) is a count, the greatest possible number is \(6\).

Answer

\(14+4b\le38\), so \(b\le6\). The greatest possible number is \(6\) boxes.
5107278
A bookshelf is \(\frac13\) full. After \(12\) more books are added, it is \(\frac12\) full. Let \(x\) be the number of books the shelf holds when full. a) Write an equation with \(x\) on both sides that models the situation. b) Solve your equation algebraically. Your work must show equivalent equations that remove the fractional coefficients and then isolate \(x\). c) Interpret the solution in the context.

Hints

- Express the initial and final numbers of books as fractions of the same unknown capacity. - Use the least common multiple of the denominators to clear the fractions. - After clearing fractions, collect the variable terms on one side. - Check that adding \(12\) to one-third of your answer gives one-half of it.

Solution

1. Initially the shelf holds \(\frac13x\) books. After adding \(12\) books, it holds \(\frac12x\), so \( \frac13x+12=\frac12x. \) 2. Multiply both sides by \(6\): \(2x+72=3x\). 3. Subtract \(2x\) from both sides: \(72=x\). 4. Therefore, the shelf holds \(72\) books when completely full.

Answer

a) \(\frac13x+12=\frac12x\) b) \(x=72\) c) The shelf's full capacity is \(72\) 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) Let \(x\) be the number of posters when every gap is exactly \(4\,\text{ft}\) wide. Write an equation that models the wall length, solve it, and state how many posters fit.

Hints

- Count the gaps at both ends as well as the gaps between posters. - For part a), subtract the poster widths from the wall length before sharing the remaining space among the gaps. - For part b), express both poster width and total gap width using the same variable. - The equation must account for the entire \(44\,\text{ft}\) wall.

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) Equation: \(4x+4(x+1)=44\); \(5\) 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) Write and solve an equation for the time when the tank is empty. Round to the nearest tenth of an hour.

Hints

- Model the amount remaining as the starting amount minus repeated hourly loss. - Substitute a time only after writing the expression. - For an empty tank, set the remaining amount equal to the amount an empty tank contains. - Use the requested rounding only after solving the equation.

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) Equation: \(1000-15t=0\); \(t\approx66.7\) 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 \(195\,\text{m}\).
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. Let \(x\) be the shorter base edge. Write a one-variable equation for the total length of all \(12\) edges, solve it, and report the three edge lengths.

Hints

- A rectangular prism has four edges of each of its three dimensions. - Express both longer dimensions in terms of the shorter edge. - Build one equation whose left side represents all twelve edge lengths. - Solve only after the wire-length model is complete.

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

Equation: \(4x+4(2x)+4(2x)=20\) 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. Let \(a\) be a base-edge length. Write an equation for the total rod length, solve it, and report both edge lengths.

Hints

- Count how many edges have each of the two lengths. - Express an apex edge using the base-edge variable. - The sum of all eight edge lengths must equal the rod length. - Solve the model before calculating the longer edge.

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

Equation: \(4a+4(a+5)=160\) Base edge: \(17.5\,\text{in.}\) Edge to the apex: \(22.5\,\text{in.}\)
5125628
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 \(10\,\text{in.}^2\). Let \(h\) be the original height. Write an equation with \(h\) on both sides that compares the new area with the original area, solve it algebraically, and report the original height.

Hints

- Write separate expressions for the original and new areas. - Translate “increases by \(10\,\text{in.}^2\)” as a relationship between those area expressions. - Make sure the variable appears in both area expressions before solving. - Collect the variable terms on one side after distributing.

Solution

1. The original area is \(10h\). 2. The new width is \(15\,\text{in.}\), and the new height is \(h-2\), so the new area is \(15(h-2)\). 3. The new area is \(10\,\text{in.}^2\) greater: \(15(h-2)=10h+10\). 4. Distribute: \(15h-30=10h+10\). 5. Subtract \(10h\) and add \(30\): \(5h=40\), so \(h=8\).

Answer

Equation: \(15(h-2)=10h+10\) Original height: \(8\,\text{in.}\)
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. Let \(x\) be the shorter side. Write and solve an inequality for the perimeter, then give the greatest possible length of each side.

Hints

- Express the longer side in terms of the shorter side. - Use the rectangle perimeter formula to build the left side of the inequality. - Translate “at most” as an upper-bound inequality. - After solving, use the boundary value to find the longest possible sides.

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

Inequality: \(2(x+3x)\le48\), or \(8x\le48\) Greatest possible side lengths: \(6\,\text{in.}\) 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. Let \(w\) be the width. Write and solve an inequality for the edging constraint, then state the greatest possible width.

Hints

- Express the length using the width variable. - Build a perimeter expression before applying the edging limit. - Translate “at most” into the correct inequality direction. - Interpret the inequality boundary in the physical context.

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

Inequality: \(2(w+w+4)\le40\), or \(4w+8\le40\) Greatest possible width: \(8\,\text{ft}\)
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.
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.
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.
5128037
Write an equation and solve the number riddle: “Twice a number \(x\), increased by the difference between \(15.4\) and \(8.4\), equals \(25\).” What is \(x\)?

Hints

- How can you represent twice a number? - Which operation is indicated by “increased by”? - Evaluate the fixed difference before solving the equation.

Solution

1. Translate the statement into an equation: \(2x + (15.4 - 8.4) = 25\). 2. Evaluate the difference: \(15.4 - 8.4 = 7\). 3. Simplify: \(2x + 7 = 25\). 4. Subtract \(7\): \(2x = 18\). 5. Divide by \(2\): \(x = 9\).

Answer

The equation is \(2x + (15.4 - 8.4) = 25\), and \(x = 9\).
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 composite banner is made from a parallelogram and a triangle that share an edge but do not overlap. The parallelogram has base \(8\,\text{in.}\) and height \(4.5\,\text{in.}\). The triangle has base \(12\,\text{in.}\). The entire composite banner has area \(66\,\text{in.}^2\). Find the triangle's height.

Hints

- A composite figure's area is the sum of the non-overlapping pieces. - Find the parallelogram's area first. - Subtract that area from the total to isolate the triangle's area. - Then use \(A=\frac12 bh\) for the triangle.

Solution

1. Find the parallelogram's area: \( 8\cdot4.5=36\,\text{in.}^2. \) 2. Since the two pieces do not overlap, the triangle's area is the composite area minus the parallelogram's area: \( 66-36=30\,\text{in.}^2. \) 3. Use the triangle area formula: \( 30=\frac12\cdot12\cdot h. \) 4. Thus \(30=6h\), so \(h=5\).

Answer

The triangle's height is \(5\,\text{in.}\).
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 \(30=\frac{1}{2}\cdot10\cdot h\). 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 \(\$115\).
5193836
Luke says, “When I subtract \(240\) from my number, I get \(360\).” Marie says, “When I add \(180\) to my number, I get \(800\).” Write and solve one one-step addition or subtraction equation for each person. Then determine who chose the greater number and by how much.

Hints

- Translate each statement into its own equation before calculating. - Use the inverse of subtraction for Luke’s equation and the inverse of addition for Marie’s. - Keep the two variables separate until both equations are solved. - Compare the solutions and then find their difference.

Solution

1. Luke’s statement gives \(x-240=360\). Add \(240\) to both sides: \(x=600\). 2. Marie’s statement gives \(y+180=800\). Subtract \(180\) from both sides: \(y=620\). 3. Marie’s number is greater, and \(620-600=20\).

Answer

Luke: \(x-240=360\), so \(x=600\). Marie: \(y+180=800\), so \(y=620\). Marie’s number is greater by \(20\).
5222476
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

- Separate each plan’s fixed cost from its per-hour cost. - Translate “per hour” as a multiplication involving the number of hours. - Substitute the requested values only after writing each expression. - Use the context to interpret both the constant and the variable.

Solution

1. Plan A is \(12+4t\), and Plan B is \(7t\). 2. For \(t=2\), Plan A costs \(12+4\times2=20\), and Plan B costs \(7\times2=14\). Plan B is less expensive. 3. For \(t=5\), Plan A costs \(12+4\times5=32\), and Plan B costs \(7\times5=35\). 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.
5223226
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

- Identify the starting quantity and the amount removed each day. - Express repeated daily use as a product before subtracting it. - Interpret a value of zero using the physical situation. - Replace the known daily rate with the new variable only after the original expression is clear.

Solution

1. Subtract the total amount used from the starting amount: \(V-12n\). 2. For \(V=500\) and \(n=10\), \(500-12\times10=500-120=380\) 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. The total cost is \(\$1900\). Let \(x\) be the material cost. Write a one-variable equation that models all three cost components, solve it, and report the material cost.

Hints

- Express the labor cost as a multiple of the material cost. - Include both variable cost components and the fixed machine fee in one total-cost equation. - Combine like terms only after the model is written. - Check that the three cost components sum to the stated total.

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

Equation: \(x+1.25x+100=1900\) Material cost: \(\$800\)
5224967
A bus for a class field trip costs a flat \(\$450\). Zoo admission costs \(\$12\) per person. The total trip cost is \(\$774\). Let \(x\) be the number of people. Write and solve an equation for the total trip cost, then report how many people are attending.

Hints

- Separate the fixed bus cost from the per-person admission cost. - Use one variable for the number of people and build the complete total-cost equation. - Remove the fixed cost before isolating the variable. - Check the total cost with the resulting attendance.

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

Equation: \(450+12x=774\) \(27\) people
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.
5225449
A water tank initially contains \(B\) gallons. Pipe A adds \(a\) gallons per minute, Pipe B adds \(b\) gallons per minute, and a drain removes \(c\) gallons per minute. After \(m\) minutes, the tank contains \(W\) gallons: \( W=B+m(a+b-c). \) a) Rearrange the formula to solve for \(m\). b) The tank's capacity is \(V\). Write \(V-W\) in terms of \(B,m,a,b,c\), and explain what that expression represents. c) State the physical conditions on \(m\) under which the model is meaningful if the tank cannot contain less than \(0\) or more than \(V\) gallons.

Hints

- Isolate the term containing \(m\) before dividing. - Treat \(a+b-c\) as one grouped net-rate factor. - Capacity minus current contents is unused capacity. - Separate the algebraic rearrangement from the physical-domain restrictions.

Solution

1. Subtract \(B\): \( W-B=m(a+b-c). \) 2. Divide by the net rate: \( m=\frac{W-B}{a+b-c}, \) provided \(a+b-c\ne0\). 3. Substitute the formula for \(W\): \( V-W=V-[B+m(a+b-c)]. \) This is the unused tank capacity after \(m\) minutes. 4. Physically, \(m\ge0\) and \( 0\le B+m(a+b-c)\le V. \)

Answer

a) \(m=\frac{W-B}{a+b-c}\), for \(a+b-c\ne0\) b) \(V-[B+m(a+b-c)]\); it is the unused capacity. c) \(m\ge0\) and \(0\le B+m(a+b-c)\le V\).
5226026
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 fixed monthly charge from the per-rental charge. - Use the number of rentals as the variable in the expression. - Substitute a rental count only after the expression has been written. - Calculate both plans independently before comparing them.

Solution

1. The monthly cost is \(8.50+2.50n\). 2. For \(n=6\), \(8.50+2.50\times6=8.50+15.00=23.50\). 3. For \(5\) rentals, the first service costs \(8.50+2.50\times5=21.00\). The competing service costs \(4.00\times5=20.00\). 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 \(\$21.00\).
5228217
Two adjacent angles form a straight angle, so their measures add to \(180^\circ\). One angle is \(44^\circ\) greater than the other. Find both angle measures by writing and solving an equation.

Hints

- Recall the sum of two angles that form a straight angle. - Represent the smaller angle with a variable. - Express the larger angle using the given difference. - Add the two expressions and solve.

Solution

1. Let \(x\) degrees be the smaller angle. The larger angle is \(x + 44\) degrees. 2. Their sum is \(180^\circ\), so \(x + (x + 44) = 180\). 3. Combine like terms: \(2x + 44 = 180\). 4. Subtract \(44\): \(2x = 136\). 5. Divide by \(2\): \(x = 68\). 6. The larger angle is \(68^\circ + 44^\circ = 112^\circ\).

Answer

The angle measures are \(68^\circ\) and \(112^\circ\).
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 \(90^\circ\).
5228317
The sum of four consecutive integers is \(66\). Use an equation to find the four integers.

Hints

- Express each integer after the first in terms of \(x\). - Add the four expressions and set their sum equal to \(66\). - Combine like terms before solving.

Solution

1. Let \(x\) be the first integer. 2. The next three integers are \(x + 1\), \(x + 2\), and \(x + 3\). 3. Write the equation \(x + (x + 1) + (x + 2) + (x + 3) = 66\). 4. Combine like terms: \(4x + 6 = 66\). 5. Subtract \(6\): \(4x = 60\). 6. Divide by \(4\): \(x = 15\). 7. The integers are \(15\), \(16\), \(17\), and \(18\).

Answer

The four integers are \(15\), \(16\), \(17\), and \(18\).
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 \(128\,\text{in.}^2\).
5228767
A \(100\,\text{in.}\) rope is cut into three pieces. The first piece is twice as long as the second piece. The third piece is exactly \(5\,\text{in.}\) shorter than the first piece. Let \(x\) be the second piece’s length. Write one equation that models all three pieces and the total length, solve it, and report every piece length.

Hints

- Choose the second piece as the base quantity because the other two are described from it. - Express all three lengths with the same variable before adding them. - Set their sum equal to the original rope length. - Use the solved variable to recover all three piece lengths.

Solution

1. Let \(x\) be the length of the second piece in inches. 2. The first piece is \(2x\), and the third piece is \(2x - 5\). 3. Write the total-length equation \(2x + x + (2x - 5) = 100\). 4. Combine like terms: \(5x - 5 = 100\). 5. Add \(5\): \(5x = 105\). Divide by \(5\): \(x = 21\). 6. The first piece is \(2 \cdot 21 = 42\,\text{in.}\), the second is \(21\,\text{in.}\), and the third is \(42 - 5 = 37\,\text{in.}\).

Answer

Equation: \(2x+x+(2x-5)=100\) Piece lengths: \(42\,\text{in.}\), \(21\,\text{in.}\), and \(37\,\text{in.}\)
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. Let \(x\) be Class A’s amount. Write one equation that models the three class amounts, solve it, and report how much each class raised.

Hints

- Use Class A as the base quantity for the fractional relationship. - Express Class C from the Class B expression, not directly from Class A. - Add all three expressions and set the sum equal to the fundraiser total. - Keep the fractional coefficient exact while solving.

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

Equation: \(x+\frac{2}{3}x+\left(\frac{2}{3}x-15\right)=510\) Class A: \(\$225\); Class B: \(\$150\); Class C: \(\$135\)
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 \(\$1.50\). Total sales were \(\$110.00\), and exactly twice as many sandwiches as juice bottles were sold. Let \(x\) be the number of juice bottles. Write a one-variable revenue equation, solve it, and report both quantities.

Hints

- Represent the smaller item count with a variable. - Use the “twice as many” condition to express the other count. - Multiply each item count by its price before adding the revenues. - Check both the count relationship and the total sales.

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

Equation: \(2.00(2x)+1.50x=110\) \(40\) sandwiches and \(20\) bottles of apple juice
5229387
A hiking group travels a total of \(46\,\text{miles}\) over three days. On the second day, the group hikes \(1.5\) times the distance from the first day. On the third day, the group hikes \(2\,\text{miles}\) less than on the second day. Let \(x\) be the first-day distance. Write one equation for the three-day total, solve it, and report the distance hiked each day.

Hints

- Use the first-day distance as the base variable. - Build the second- and third-day expressions from that same variable. - Add the three daily expressions and equate them to the total distance. - Check that the solved distances satisfy both stated relationships.

Solution

1. Let \(x\) be the distance hiked on the first day. 2. The second-day distance is \(1.5x\), and the third-day distance is \(1.5x - 2\). 3. Write the total equation \(x + 1.5x + (1.5x - 2) = 46\). 4. Combine like terms: \(4x - 2 = 46\). 5. Add \(2\): \(4x = 48\). Divide by \(4\): \(x = 12\). 6. The second-day distance is \(1.5 \cdot 12 = 18\,\text{miles}\), and the third-day distance is \(18 - 2 = 16\,\text{miles}\).

Answer

Equation: \(x+1.5x+(1.5x-2)=46\) Day 1: \(12\,\text{miles}\); Day 2: \(18\,\text{miles}\); Day 3: \(16\,\text{miles}\)
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) Let \(x\) be Class A’s amount. Write and solve one equation that models all three contributions, then report each amount. 2) What fraction or percent of the total is raised by Class C?

Hints

- Express Class B from Class A before building Class C’s expression. - Class C equals the sum of the first two class expressions. - Add all three contributions to create the total-fundraising equation. - For part 2, compare Class C’s solved amount with the full total.

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) Equation: \(x+(x+40)+(2x+40)=720\). Class A: \(\$160\); Class B: \(\$200\); Class C: \(\$360\). 2) \(\frac{1}{2}\), or \(50\%\)
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 \(27\).
5238279
A beverage plant uses \(4\) Type A machines and \(5\) Type B machines at the same time. Each Type A machine fills \(x\) bottles per minute, each Type B machine fills \(y\) bottles per minute, the order contains \(F\) bottles, and the filling time is \(t\) minutes. The quantities satisfy \( F=t(4x+5y). \) a) Rearrange the formula to solve explicitly for \(t\). b) Find \(t\) when \(x=25\), \(y=20\), and \(F=4000\). c) Explain why the denominator in your rearranged formula represents a rate.

Hints

- Treat \(4x+5y\) as one complete factor multiplying \(t\). - Use the same division on both sides to isolate \(t\). - After rearranging, substitute values only into the solved formula. - Track the units of \(F\) and \(4x+5y\).

Solution

1. Start with \(F=t(4x+5y)\). 2. Divide both sides by the entire factor \(4x+5y\): \( t=\frac{F}{4x+5y}. \) 3. Substitute the values: \( t=\frac{4000}{4\cdot25+5\cdot20} =\frac{4000}{200}=20. \) 4. The quantity \(4x+5y\) is the combined number of bottles filled per minute by all nine machines, so it is a rate.

Answer

a) \(t=\frac{F}{4x+5y}\) b) \(20\) minutes c) \(4x+5y\) is the combined filling rate in bottles per minute.
5238289
A landscaping crew has \(2\) experienced workers and \(3\) assistants. Each experienced worker mows \(f\) square feet per hour, each assistant mows \(h\) square feet per hour, the lawn area is \(A\) square feet, and the work takes \(d\) hours. For the original crew, \( A=d(2f+3h). \) a) Rearrange the formula to solve for \(d\). b) One experienced worker is replaced by two assistants. Write the corresponding area equation using a new time \(d_{\text{new}}\), then rearrange it to solve for \(d_{\text{new}}\). c) Use the two solved formulas to determine when the new crew is faster than the original crew. d) Find the original time when \(f=1200\), \(h=800\), and \(A=19{,}200\).

Hints

- Isolate a time variable by dividing by the entire crew-rate factor. - Recount each worker type after the replacement before writing the new equation. - For the same positive area, a larger rate gives a smaller time. - Compare the two denominators before substituting numerical values.

Solution

1. Divide \(A=d(2f+3h)\) by \(2f+3h\): \( d=\frac{A}{2f+3h}. \) 2. The new crew has \(1\) experienced worker and \(5\) assistants, so \( A=d_{\text{new}}(f+5h). \) Dividing by \(f+5h\) gives \( d_{\text{new}}=\frac{A}{f+5h}. \) 3. For the same positive area \(A\), the new crew is faster when its denominator is larger: \( f+5h>2f+3h. \) Thus \(2h>f\). 4. For the given original crew, \( d=\frac{19{,}200}{2\cdot1200+3\cdot800} =\frac{19{,}200}{4800}=4. \)

Answer

a) \(d=\frac{A}{2f+3h}\) b) \(A=d_{\text{new}}(f+5h)\), so \(d_{\text{new}}=\frac{A}{f+5h}\) c) The new crew is faster when \(2h>f\). d) \(4\) hours
5238329
An empty water tank holds \(V\) gallons. Pipe A adds \(x\) gallons per minute, and Pipe B adds \(5\) gallons per minute more than Pipe A. If both pipes run for \(t\) minutes, then \( V=t(2x+5). \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(V=750\) and \(x=35\). c) Rearrange the same formula to solve for \(x\) in terms of \(V\) and \(t\).

Hints

- Treat \(2x+5\) as one factor when isolating \(t\). - For part c), undo multiplication by \(t\) first. - Then undo the added \(5\) before dividing by \(2\). - Keep the symbolic quantities intact until the requested variable is isolated.

Solution

1. Divide both sides of \(V=t(2x+5)\) by \(2x+5\): \( t=\frac{V}{2x+5}. \) 2. Substitute \(V=750\) and \(x=35\): \( t=\frac{750}{70+5}=\frac{750}{75}=10. \) 3. To solve for \(x\), divide by \(t\): \(\frac{V}{t}=2x+5\). 4. Subtract \(5\): \(\frac{V}{t}-5=2x\). 5. Divide by \(2\): \( x=\frac{\frac{V}{t}-5}{2}. \)

Answer

a) \(t=\frac{V}{2x+5}\) b) \(10\) minutes c) \(x=\frac{\frac{V}{t}-5}{2}\)
5238449
Boxes are stacked on a pallet in \(r\) rows with \(k\) boxes per row. Each box weighs \(m\) pounds, and the empty pallet weighs \(50\) pounds. If the loaded pallet weighs \(M\) tons, then \( M=\frac{rkm+50}{2000}. \) a) Rearrange the formula to solve for the box weight \(m\). b) Find \(m\) when \(M=0.905\), \(r=5\), and \(k=8\). c) Explain the purpose of the factor \(2000\) in the original formula.

Hints

- Clear the denominator before moving the pallet's fixed weight. - Subtract the \(50\)-pound pallet weight before dividing by the number of boxes. - The product \(rk\) is the total number of boxes. - Check the units after rearranging.

Solution

1. Multiply both sides by \(2000\): \( 2000M=rkm+50. \) 2. Subtract \(50\): \( 2000M-50=rkm. \) 3. Divide by \(rk\): \( m=\frac{2000M-50}{rk}. \) 4. Substitute: \( m=\frac{2000(0.905)-50}{5\cdot8} =\frac{1810-50}{40} =\frac{1760}{40}=44. \) 5. The denominator \(2000\) converts the total weight from pounds to tons because \(1\) ton is \(2000\) pounds.

Answer

a) \(m=\frac{2000M-50}{rk}\) b) \(m=44\,\text{lb}\) c) The factor \(2000\) converts pounds to tons.
5239349
A sports club buys equipment costing \(\$K\). A city grant contributes \(\$Z\), and the remaining cost is divided equally among \(s\) active members. If each member pays \(P\) dollars, then \( P=\frac{K-Z}{s}. \) a) Rearrange the formula to solve for \(Z\). b) Find the grant \(Z\) when \(K=540\), \(s=15\), and \(P=28\). c) Explain why increasing \(Z\) lowers \(P\) when \(K\) and \(s\) stay fixed.

Hints

- Clear the denominator first. - Keep track of the subtraction sign attached to \(Z\). - Isolate \(Z\) only after multiplying by \(s\). - Interpret the formula after rearranging, not only numerically.

Solution

1. Multiply both sides by \(s\): \(Ps=K-Z\). 2. Add \(Z\) to both sides and subtract \(Ps\): \(Z=K-Ps\). 3. Substitute the values: \(Z=540-28\cdot15=540-420=120\). 4. A larger grant removes more of the fixed total cost before the remainder is divided among members, so each member pays less.

Answer

a) \(Z=K-Ps\) b) \(Z=\$120\) c) A larger grant reduces the amount left for the members to share.
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 \(\$6\).
5239807
A cycling group plans a three-day, \(180\,\text{mile}\) trip. The second-day distance will be twice the first-day distance. The third-day distance will be \(15\,\text{miles}\) shorter than the second-day distance. a) Write an equation to find the first-day distance \(x\). b) Find the distance for each day. c) Without solving again, explain how the equation from part a would change if the third-day distance were instead \(10\,\text{miles}\) longer than the first-day distance.

Hints

- The third-day distance in parts a and b is compared with the second-day distance. - Add the three daily expressions to equal the total distance. - In part c, change only the expression for the third day.

Solution

1. For part a, the three daily distances are \(x\), \(2x\), and \(2x - 15\). Write \(x + 2x + (2x - 15) = 180\). 2. Combine like terms: \(5x - 15 = 180\). 3. Add \(15\): \(5x = 195\). Divide by \(5\): \(x = 39\). 4. The second-day distance is \(2 \cdot 39 = 78\,\text{miles}\), and the third-day distance is \(78 - 15 = 63\,\text{miles}\). 5. For part c, replace the third-day expression with \(x + 10\). The new equation is \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).

Answer

a) \(x + 2x + (2x - 15) = 180\) b) The first-day distance is \(39\,\text{miles}\), the second-day distance is \(78\,\text{miles}\), and the third-day distance is \(63\,\text{miles}\). c) The equation becomes \(x + 2x + (x + 10) = 180\), or \(4x + 10 = 180\).
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 \(375\,\text{ft}^2\).
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) Let \(s\) be the power draw of one floodlight. Write and solve an equation for the original system, then report 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 a heater’s power draw using the floodlight variable. - Model the original total before solving for either device’s power. - A \(20\%\) reduction means the upgraded heater uses \(80\%\) of its original power. - Recalculate the system total with the solved device powers.

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) Equation: \(8s+2(5s)=3600\). One floodlight draws \(200\,\text{W}\), and one heater draws \(1000\,\text{W}\). b) \(3200\,\text{W}\)
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.
5240659
A fruit-juice mixture contains \(x\) liters of concentrate and \(y\) liters of water. The concentrate contains \(s\) grams of sugar per liter. If the final mixture has sugar concentration \(q\) grams per liter, then \( q=\frac{xs}{x+y}. \) a) Rearrange the formula to solve for \(s\). b) Find \(s\) when \(q=30\), \(x=0.4\), and \(y=1.6\). c) Explain why the factor \(x+y\) appears in your rearranged formula.

Hints

- Clear the denominator before trying to isolate \(s\). - Treat \(x+y\) as one grouped quantity. - Divide by the coefficient multiplying \(s\) only after clearing the fraction. - Use the units to interpret \(x+y\).

Solution

1. Multiply both sides by \(x+y\): \( q(x+y)=xs. \) 2. Divide by \(x\): \( s=\frac{q(x+y)}{x}. \) 3. Substitute the values: \( s=\frac{30(0.4+1.6)}{0.4}=\frac{60}{0.4}=150. \) 4. The factor \(x+y\) is the total volume of the final mixture.

Answer

a) \(s=\frac{q(x+y)}{x}\) b) \(150\) grams per liter c) \(x+y\) is the total mixture volume.
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. - Find the fraction of the book that remains unread. - Relate that remaining fraction directly to \(125\) pages.

Solution

1. Let \(x\) be the total number of pages. 2. The fraction left is \(1-\frac{1}{3}-\frac{1}{4}=\frac{5}{12}\). 3. Model the remaining pages directly: \(\frac{5}{12}x=125\). 4. Multiply both sides by \(\frac{12}{5}\): \(x=300\).

Answer

A suitable equation is \(\frac{5}{12}x=125\). 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.
5279249
A water tank initially contains \(W\) gallons. A leak drains water at a constant rate of \(k\) gallons per hour. If \(A\) is the amount of water remaining after \(h\) hours, then \( A=W-kh. \) a) Rearrange the formula to solve for \(h\). b) Find \(A\) when \(W=500\), \(k=12\), and \(h=8\). c) Use your rearranged formula to give the empty-tank time in terms of \(W\) and \(k\). d) State the physically meaningful interval of \(h\) while the tank is draining.

Hints

- Treat \(A=W-kh\) as a literal equation and isolate the term containing \(h\). - After moving the constant term, account carefully for the negative sign on \(kh\). - For an empty tank, substitute \(A=0\) into the formula you already solved for \(h\). - The model stops being physical when the remaining amount would become negative.

Solution

1. Start with \(A=W-kh\). Subtract \(W\): \(A-W=-kh\). 2. Multiply by \(-1\): \(W-A=kh\). Divide by \(k\): \( h=\frac{W-A}{k}. \) 3. For the given values, \(A=500-12\cdot8=404\) gallons. 4. The tank is empty when \(A=0\), so the rearranged formula gives \(h=\frac{W}{k}\). 5. During the physical draining process, \(0\le h\le\frac{W}{k}\).

Answer

a) \(h=\frac{W-A}{k}\) b) \(404\) gallons c) \(h=\frac{W}{k}\) d) \(0\le h\le\frac{W}{k}\)
5279474
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. Let \(v\) be the faster train’s speed. Write one equation using distance equals rate times time, solve it, and report both speeds.

Hints

- Express the slower speed as a fraction of the faster speed. - Write each train’s distance using the common travel time. - Because the trains move in opposite directions, their distances add to the separation. - Check that the two speeds have the required ratio.

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

Equation: \(1.5v+1.5\left(\frac{2}{3}v\right)=150\) Faster train: \(60\,\text{mph}\); slower train: \(40\,\text{mph}\)
5280329
Two pumps fill a tank. Pump A moves \(x\) gallons per minute and Pump B moves \(y\) gallons per minute. After both pumps have run for \(t\) minutes, the tank still needs \(R\) gallons to reach its capacity \(V\). The quantities satisfy \( V=(x+y)t+R. \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(V=630\), \(x=15\), \(y=20\), and \(R=210\). c) Rearrange the original formula to solve for \(R\).

Hints

- Remove the additive term \(R\) before dividing. - Treat \(x+y\) as a single factor multiplying \(t\). - For part c), isolate \(R\) with one subtraction. - Check that the units of your solved formula for \(t\) are minutes.

Solution

1. Subtract \(R\): \( V-R=(x+y)t. \) 2. Divide by \(x+y\): \( t=\frac{V-R}{x+y}. \) 3. Substitute: \( t=\frac{630-210}{15+20}=\frac{420}{35}=12. \) 4. To solve for \(R\), subtract \((x+y)t\) from both sides: \( R=V-(x+y)t. \)

Answer

a) \(t=\frac{V-R}{x+y}\) b) \(12\) minutes c) \(R=V-(x+y)t\)
5320647
The diagram shows the composition of an urn. 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 gray balls must be added so that the probability of selecting a blue ball is exactly \(25\%\)?
Figure for problem 532064

Hints

- Count the blue balls and the total number of balls in the urn. - Probability is favorable outcomes divided by total outcomes. - Adding gray balls changes the total number of balls but does not change the number of blue balls. - For part b), set the new blue-ball probability equal to \(\frac14\).

Solution

1. The urn contains \(6\) blue balls and \(9\) gray balls, for a total of \(15\) balls. 2. Therefore, \(P(\text{blue})=\frac{6}{15}=\frac{2}{5}=40\%\). 3. For part b), the number of blue balls remains \(6\). Let \(x\) be the number of gray balls added. Then the new total is \(15+x\), so \(\frac{6}{15+x}=\frac14\). 4. Cross-multiplying gives \(24=15+x\), so \(x=9\). 5. Therefore, \(9\) gray balls must be added.

Answer

a) \(\frac25=40\%\) b) \(9\) gray balls
5352767
The number line shows a temperature change from the previous evening to the morning low. Find the previous evening's temperature.
Figure for problem 535276

Hints

- Read the endpoint and signed jump from the number line. - Work backward from the morning low by reversing the displayed change. - Check that applying the displayed change to your answer reaches the shown endpoint.

Solution

1. Let the starting temperature be \(x\). The number line shows a fall of \(15\,^{\circ}\text{F}\) to \(-6\,^{\circ}\text{F}\), so \(x-15=-6\). 2. Add \(15\) to both sides: \(x=-6+15=9\). 3. The previous evening's temperature was \(9\,^{\circ}\text{F}\).

Answer

\(9\,^{\circ}\text{F}\)
53680510
The diagram shows quadrilateral \(ABCD\) with a circle tangent to all four sides. a) State the tangent-segment fact about two tangent segments drawn from the same external point. b) Use that fact to derive a relationship among the four side lengths of this quadrilateral. c) Use the side lengths shown in the diagram to find \(d\).
Figure for problem 536805

Hints

- Focus on the two tangent segments that meet at each vertex. - Give the equal tangent lengths temporary variables and express each side as a sum of two of them. - Compare the sums of opposite sides only after deriving the relationship. - Then substitute the three lengths shown in the diagram.

Solution

1. Tangent segments from the same external point to a circle have equal lengths. 2. Let the tangent lengths from vertices \(A,B,C,D\) be \(p,q,r,s\), respectively. Then \( AB=p+q,\quad BC=q+r,\quad CD=r+s,\quad DA=s+p. \) 3. Therefore, \( AB+CD=(p+q)+(r+s)=(q+r)+(s+p)=BC+DA. \) 4. Read the side lengths from the diagram: \( 8+10=11+d. \) 5. Thus \(18=11+d\), so \(d=7\).

Answer

a) Tangent segments from the same external point are equal. b) \(AB+CD=BC+DA\) c) \(d=7\)
5547007
A recreation center charges a \(\$25\) room fee plus \(\$7\) per participant. Situation A: One group is charged exactly \(\$74\). Situation B: Another group has a budget of no more than \(\$74\). For each situation, decide whether an equation or an inequality is appropriate, write the model, solve it, and explain the difference in the meanings of the solutions.

Hints

- Compare the meanings of “exactly” and “no more than.” - Use the same cost expression for both situations. - After solving, decide whether the result should be one value or a set of values.

Solution

1. For Situation A, “exactly” requires an equation: \(25+7p=74\). 2. Solve: \(7p=49\), so \(p=7\). Exactly \(7\) participants produce a \(\$74\) charge. 3. For Situation B, “no more than” requires an inequality: \(25+7p\le74\). 4. Solve: \(7p\le49\), so \(p\le7\). With a nonnegative whole-number count, any \(p\) from \(0\) through \(7\) stays within the budget. 5. The equation identifies one exact count; the inequality describes a set of allowable counts.

Answer

Situation A: \(25+7p=74\), so \(p=7\). Situation B: \(25+7p\le74\), so \(p\le7\); feasible counts are nonnegative whole numbers through \(7\).
5547017
A fundraiser already has \(\$32\) and earns \(\$6\) for each batch of cards sold. The goal is to raise at least \(\$80\). A student writes \(32+6b\le80\). a) Explain the modeling error. b) Write the correct inequality. c) Solve it and interpret the answer.

Hints

- Translate “at least” before examining the student's symbol. - Ask whether a total greater than \(80\) should satisfy the model. - Solve the corrected inequality and interpret \(b\) as a count.

Solution

1. The symbol \(\le\) represents totals no greater than \(80\), but the fundraiser needs a total of \(80\) or more. 2. The correct model is \(32+6b\ge80\). 3. Subtract \(32\): \(6b\ge48\). 4. Divide by \(6\): \(b\ge8\). 5. The fundraiser must sell at least \(8\) batches.

Answer

a) The student's inequality points in the wrong direction for “at least.” b) \(32+6b\ge80\) c) \(b\ge8\), so at least \(8\) batches must be sold.
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.
5120398
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. Let \(x\) be the number of posts in the first arrangement. Write an equation with \(x\) on both sides by equating the two expressions for the same fence length. Solve that equation algebraically, then find the fence length.

Hints

- Count gaps, including both end gaps, for each arrangement. - Write one length expression for each spacing choice using the same variable. - Because the physical fence length is unchanged, set those two variable expressions equal. - Solve the resulting equation by collecting the variable terms on one side.

Solution

1. With \(x\) posts, the first arrangement has \(x+1\) equal gaps, so its length is \(12(x+1)\). 2. The second arrangement has \(x+3\) posts and \(x+4\) gaps, so its length is \(7.5(x+4)\). 3. Equate the two length expressions: \(12(x+1)=7.5(x+4)\). 4. Distribute: \(12x+12=7.5x+30\). Subtract \(7.5x\) and \(12\): \(4.5x=18\), so \(x=4\). 5. The fence length is \(12(4+1)=60\,\text{ft}\).

Answer

Equation: \(12(x+1)=7.5(x+4)\) Solution: \(x=4\) Fence length: \(60\,\text{ft}\)
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. a) Let \(s\) be Noah's cube edge length. Write and solve an equation for \(s\). b) Let \(a\) be Maya's base-edge length. Write and solve an equation for \(a\) using the wire lengths of all \(12\) prism edges. c) Use your two equations to find the difference between the model heights.

Hints

- Count all edges that use wire before writing each equation. - For Maya's model, express the vertical edges in terms of the same variable used for a base edge. - Your response must include an equation for each model, not only arithmetic calculations. - Interpret the solved variable before comparing the heights.

Solution

1. A cube has \(12\) equal edges, so Noah's equation is \(12s=48\). Thus \(s=4\), so Noah's model is \(4\,\text{ft}\) high. 2. Maya's prism has \(8\) horizontal base/top edges of length \(a\) and \(4\) vertical edges of length \(2a\). Her equation is \( 8a+4(2a)=48. \) 3. Simplify: \(16a=48\), so \(a=3\). Maya's height is \(2a=6\,\text{ft}\). 4. The height difference is \(6-4=2\,\text{ft}\).

Answer

a) \(12s=48\), so \(s=4\) b) \(8a+4(2a)=48\), so \(a=3\) c) Maya's model is \(2\,\text{ft}\) taller.
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 \(90\,\text{lb}\).
5125947
A class sells waffles at a school carnival. Fixed costs for equipment, batter, and supplies are \(\$42.00\). Each waffle also costs \(\$0.35\) in toppings and is sold for \(\$1.50\). a) First find the net amount from each waffle that is available to cover fixed costs and profit. Use that net amount to write and solve an inequality for avoiding a loss. Then give the least whole number of waffles the class must sell. b) Write and solve an inequality for earning a profit of at least \(\$120.00\). Then give the least whole number of waffles the class must sell.

Hints

- Find how much of each sale remains after paying the per-waffle topping cost. - Represent the number of waffles with one variable and use the net amount as its coefficient. - Translate “at least” with a lower-bound inequality. - Interpret a decimal boundary using the fact that the number of waffles must be whole.

Solution

1. Let \(w\) be the number of waffles sold. The net amount from each waffle after toppings is \(1.50-0.35=1.15\) dollars. 2. To avoid a loss, the net waffle revenue must cover the fixed costs: \(1.15w\ge42\). 3. Solve: \(w\ge36.521\ldots\). Because \(w\) is a whole number, at least \(37\) waffles are needed. 4. A profit of at least \(\$120.00\) means net waffle revenue minus fixed costs is at least \(120\): \(1.15w-42\ge120\). 5. Add \(42\): \(1.15w\ge162\), so \(w\ge140.869\ldots\). 6. Therefore, at least \(141\) waffles are needed.

Answer

a) Net amount per waffle: \(\$1.15\). Inequality: \(1.15w\ge42\); at least \(37\) waffles. b) Inequality: \(1.15w-42\ge120\); at least \(141\) waffles.
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 of \(40\) students, \(60\%\) regularly play on organized sports teams. Of those team athletes, \(25\%\) play on a soccer team. a) How many students play on a soccer team? b) Some students who do not play organized sports hike regularly. Let \(h\) be the number of these hikers. The team athletes and hikers together must make up more than \(80\%\) of the class. Write and solve an inequality for \(h\), then give the least whole number of hikers that satisfies the condition. c) If the number of hikers is the least value from part b), how many students are neither team athletes nor hikers?

Hints

- Find the number of team athletes before finding the soccer subset. - Convert \(80\%\) of the class into a student-count threshold. - Add the unknown hiker count to the known number of team athletes, then translate “more than” with a strict inequality. - Interpret the inequality using the fact that a student count must be a whole number.

Solution

1. There are \(0.60\cdot40=24\) team athletes. Of these, \(0.25\cdot24=6\) play soccer. 2. More than \(80\%\) of \(40\) students means more than \(0.80\cdot40=32\) students altogether. 3. With \(h\) non-athlete hikers, write \(24+h>32\). 4. Subtract \(24\): \(h>8\). Since \(h\) is a whole number, the least possible value is \(9\). 5. With \(9\) hikers, \(40-(24+9)=7\) students are neither team athletes nor hikers.

Answer

a) \(6\) students b) Inequality: \(24+h>32\), so \(h>8\); least value: \(9\) hikers c) \(7\) students
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 \(\$228.00\).
5238359
A wood-pellet bin holds \(M\) pounds when full. Filling it completely costs \(\$S\), and the pellets have a constant price per pound. A heating system uses pellets worth \(\$k\) each day. After \(t\) days, \(R\) pounds remain. These quantities satisfy \( R=M-\frac{kMt}{S}. \) a) Rearrange the formula to solve for \(t\). b) Find \(t\) when \(M=3000\), \(S=1200\), \(k=10\), and \(R=2250\). c) Explain why \(S\) appears in the numerator of your solved formula for \(t\).

Hints

- First isolate the term containing \(t\). - Remove the denominator \(S\) before dividing by the other factors. - Keep \(kM\) together as the factor multiplying \(t\). - Check the direction of the relationship between \(S\) and \(t\) from the solved formula.

Solution

1. Subtract \(M\): \( R-M=-\frac{kMt}{S}. \) Equivalently, \( M-R=\frac{kMt}{S}. \) 2. Multiply by \(S\): \( S(M-R)=kMt. \) 3. Divide by \(kM\): \( t=\frac{S(M-R)}{kM}. \) 4. Substitute: \( t=\frac{1200(3000-2250)}{10\cdot3000} =\frac{1200\cdot750}{30000}=30. \) 5. A larger full-bin cost \(S\), with the same dollar use \(k\) per day, means fewer pounds are used per day, so reaching the same remaining amount takes more time.

Answer

a) \(t=\frac{S(M-R)}{kM}\) b) \(30\) days c) A larger \(S\) means each daily dollar amount buys fewer pounds, so the same mass decrease takes more days.
5239269
A youth group with \(n\) members rents a bus costing \(\$B\). The cost is shared equally. Then \(m\) more people join, the bus cost rises by \(\$s\), and each person's savings relative to the original per-person cost is \(C\) dollars. These quantities satisfy \( C=\frac{B}{n}-\frac{B+s}{n+m}. \) a) Rearrange the formula to solve for \(s\). b) Use your rearranged formula to find \(s\) when \(B=400\), \(n=20\), \(m=5\), and \(C=2\). c) Explain what your value of \(s\) means in the situation.

Hints

- First isolate the fraction that contains \(s\). - Clear its denominator only after the fraction is isolated. - Subtract \(B\) last. - Check the result by comparing the old and new per-person costs.

Solution

1. Start with \( C=\frac{B}{n}-\frac{B+s}{n+m}. \) 2. Move the second fraction to the other side: \( \frac{B+s}{n+m}=\frac{B}{n}-C. \) 3. Multiply by \(n+m\): \( B+s=(n+m)\left(\frac{B}{n}-C\right). \) 4. Subtract \(B\): \( s=(n+m)\left(\frac{B}{n}-C\right)-B. \) 5. Substitute the values: \( s=25(20-2)-400=25\cdot18-400=50. \) 6. The larger bus can cost \(\$50\) more while still giving each person a \(\$2\) savings.

Answer

a) \(s=(n+m)\left(\frac{B}{n}-C\right)-B\) b) \(s=50\) c) The new bus may cost \(\$50\) more and each person will still save \(\$2\).
5239468
A chemist has \(200\,\text{g}\) of a \(10\%\) salt solution. The chemist adds \(x\) grams of a \(40\%\) salt solution to make a mixture that is exactly \(28\%\) salt. a) Write an equation with \(x\) on both sides that equates the amount of salt before and after mixing. b) Solve the equation algebraically. Show at least two equivalent-equation steps after writing the model. c) Interpret the solution.

Hints

- Compute the grams of salt already present. - Express both the added salt and the final total mass using \(x\). - The final salt amount equals \(28\%\) of the final mixture mass. - After distributing, collect the variable terms on one side.

Solution

1. The original solution contains \(0.10\cdot200=20\) grams of salt. The added solution contributes \(0.40x\) grams of salt. 2. The final mixture has mass \(200+x\) grams and is \(28\%\) salt, so \( 20+0.40x=0.28(200+x). \) 3. Distribute: \(20+0.40x=56+0.28x\). 4. Subtract \(0.28x\): \(20+0.12x=56\). 5. Subtract \(20\): \(0.12x=36\). 6. Divide by \(0.12\): \(x=300\). 7. The chemist must add \(300\,\text{g}\) of the \(40\%\) solution.

Answer

a) \(20+0.40x=0.28(200+x)\) b) \(x=300\) c) Add \(300\,\text{g}\) of the \(40\%\) solution.
5240689
A cyclist travels for \(t_1\) hours at \(v_1\) miles per hour and then for \(t_2\) hours at \(v_2\) miles per hour. The average speed over the whole trip is \( v_{\text{avg}}=\frac{v_1t_1+v_2t_2}{t_1+t_2}. \) a) Rearrange the formula to solve for \(v_2\). b) Find \(v_2\) when \(v_{\text{avg}}=14\), \(v_1=10\), \(t_1=2\), and \(t_2=3\). c) Check your result in the original average-speed formula.

Hints

- Clear the fraction by multiplying by the entire denominator. - Move the known distance term \(v_1t_1\) before dividing. - The last factor attached to \(v_2\) is \(t_2\). - Check using total distance divided by total time.

Solution

1. Multiply both sides by \(t_1+t_2\): \( v_{\text{avg}}(t_1+t_2)=v_1t_1+v_2t_2. \) 2. Subtract \(v_1t_1\): \( v_{\text{avg}}(t_1+t_2)-v_1t_1=v_2t_2. \) 3. Divide by \(t_2\): \( v_2=\frac{v_{\text{avg}}(t_1+t_2)-v_1t_1}{t_2}. \) 4. Substitute: \( v_2=\frac{14(2+3)-10\cdot2}{3} =\frac{70-20}{3} =\frac{50}{3}=16\frac23. \) 5. Check: total distance \(=10\cdot2+\frac{50}{3}\cdot3=70\) miles and total time \(=5\) hours, so the average is \(70/5=14\) mph.

Answer

a) \(v_2=\frac{v_{\text{avg}}(t_1+t_2)-v_1t_1}{t_2}\) b) \(v_2=16\frac23\) mph c) Substitution gives an average speed of \(14\) mph.
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.
5547027
A theater group buys \(6\) identical tickets. Each ticket costs \(x\) dollars plus a \(\$2.75\) booking fee. The total charge is \(\$76.50\). a) Write an equation that models the purchase. b) Find the ticket price \(x\) before the booking fee. c) If the ticket price before the fee were \(\$1.50\) higher, how much would the total charge increase? Explain using the structure of your model.

Hints

- First express the complete cost of one ticket, including the fee. - The same complete cost is charged six times. - Solve by undoing the multiplication before undoing the added fee. - For part c), consider how a change in one ticket's base price is repeated across all six tickets.

Solution

1. Each complete ticket charge is \(x+2.75\), and there are \(6\) identical charges, so \(6(x+2.75)=76.50\). 2. Divide by \(6\): \(x+2.75=12.75\). 3. Subtract \(2.75\): \(x=10.00\). 4. Increasing \(x\) by \(\$1.50\) increases each of the \(6\) ticket charges by \(\$1.50\), so the total increases by \(6\cdot1.50=\$9.00\).

Answer

a) \(6(x+2.75)=76.50\) b) \(x=\$10.00\) c) The total charge would increase by \(\$9.00\).

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