Middle School Math Quiz: Converting Forms Strategically

9 questions · exam conditions

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Converting Forms StrategicallyQuestion 1 of 9

A car's value depreciates by $\frac{1}{8}$ each year. After two years, the car is worth $19,250. What was the car's original value?

$25,000

$22,000

$24,640

$28,000

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Middle School Math Quiz

Middle School Math Quiz: Converting Forms Strategically

Practice Converting Forms Strategically in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Converting Forms Strategically, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A car's value depreciates by $\frac{1}{8}$ each year. After two years, the car is worth $19,250. What was the car's original value?

$25,000
$22,000
$24,640 (correct answer)
$28,000

Explanation: If the car depreciates by

\frac{1}{8}

each year, it retains

\frac{7}{8}

of its value each year. After two years, it retains

(\frac{7}{8})^2 = \frac{49}{64}

of its original value. If

\frac{49}{64} \times \text{original value} = 19,250

, then original value =

19,250 \times \frac{64}{49} = \$24,640

. Choice A results from incorrectly treating depreciation as simple rather than compound. Choice B uses wrong fraction calculations. Choice D results from using

\frac{1}{8}

appreciation instead of depreciation.

Question 2

A store marks up items by $37.5\%$ above cost. If an item costs the store $24, what fraction of the selling price represents the store's profit?

$\frac{3}{8}$
$\frac{3}{11}$ (correct answer)
$\frac{8}{11}$
$\frac{5}{13}$

Explanation: First convert the markup percentage:

37.5\% = \frac{3}{8} = 0.375

. The selling price is

\$24 + \$24 \times 0.375 = \$24 + \$9 = \$33

. The profit is $9, so the fraction of selling price that is profit is $

\frac{9}{33} = \frac{3}{11}

. Choice A represents the markup rate as a fraction of cost, not selling price. Choice C represents cost as a fraction of selling price. Choice D results from incorrectly calculating

37.5\%

as

\frac{5}{13}

Question 3

Maria's test scores decreased by $12\%$ from her first test to her second test, then increased by $25\%$ from her second test to her third test. If her third test score was $77$ , what was her first test score?

$69.44$
$70$ (correct answer)
$68.32$
$72$

Explanation: Working backwards: if the third test score is

77

and this represents a

25\%

increase from the second test, then the second test score was

77 ÷ 1.25 = 61.6

. If the second test score of

61.6

represents a

12\%

decrease from the first test, then the first test score was

61.6 ÷ 0.88 = 70

. Choice A results from incorrectly applying the percentages in forward direction. Choice C comes from calculation errors in the decimal conversions. Choice D assumes simple arithmetic relationships rather than percentage calculations.

Question 4

A recipe calls for ingredients in the ratio $2:3:5$ by weight. If the total weight of these three ingredients is $4.8$ kg, and one ingredient makes up $30\%$ of the mixture, what is the weight of the heaviest ingredient?

$2.4$ kg (correct answer)
$2.0$ kg
$1.8$ kg
$1.44$ kg

Explanation: The ratio

2:3:5

means the parts are

\frac{2}{10} = 20\%

\frac{3}{10} = 30\%

, and

\frac{5}{10} = 50\%

of the total. The ingredient that makes up

30\%

corresponds to the middle ratio part (3). The heaviest ingredient corresponds to ratio part 5, which is

50\%

4.8

kg =

2.4

kg. Choice B assumes the

30\%

ingredient is the heaviest. Choice C miscalculates

30\%

4.8

. Choice D calculates

30\%

4.8

, identifying the wrong ingredient as heaviest.

Question 5

A bakery uses a mixture that is $60\%$ flour and $40\%$ other ingredients by weight. If they want to create a new mixture that is $75\%$ flour by adding pure flour, how much pure flour must they add to $20$ pounds of the original mixture?

$12$ pounds (correct answer)
$15$ pounds
$8$ pounds
$10$ pounds

Explanation: Original mixture has

20 \times 0.6 = 12

pounds of flour. Let

x

= pounds of pure flour added. New mixture will have

(12 + x)

pounds of flour out of

(20 + x)

total pounds. Setting up equation:

\frac{12 + x}{20 + x} = 0.75

. Solving:

12 + x = 0.75(20 + x) = 15 + 0.75x

, so

0.25x = 3

, thus

x = 12

. Choice B calculates the final amount of flour needed, not the amount to add. Choice C uses incorrect percentage calculations. Choice D results from setting up the proportion incorrectly.

Question 6

A rectangular garden has dimensions in the ratio $3:4$ . If the area is increased by $44\%$ while keeping the same ratio, and the new perimeter is $42$ meters, what were the original dimensions?

$9$ m by $12$ m
$10.5$ m by $14$ m
$6$ m by $8$ m
$7.5$ m by $10$ m (correct answer)

Explanation: When you encounter problems involving ratios and area changes, you need to work systematically through the scaling relationships. Since the garden maintains a

3:4

ratio, you can express the dimensions as

3x

and

4x

for some scale factor

x

. The original area is

3x \times 4x = 12x^2

. After a

44\%

increase, the new area becomes

12x^2 \times 1.44 = 17.28x^2

. Since the ratio stays

3:4

, the new dimensions are

3y

and

4y

where

12y^2 = 17.28x^2

. Solving this:

y^2 = 1.44x^2

, so

y = 1.2x

. The new dimensions are

3.6x

and

4.8x

. With a perimeter of

42

meters:

2(3.6x + 4.8x) = 42

, which gives

16.8x = 42

, so

x = 2.5

. Therefore, the original dimensions were

7.5

m by

10

m. Choice A (

9

m by

12

m) would give

x = 3

, resulting in a new perimeter of

50.4

m, not

42

m. Choice B (

10.5

m by

14

m) corresponds to

x = 3.5

, yielding a new perimeter of

58.8

m. Choice C (

6

m by

8

m) has

x = 2

, producing a new perimeter of

33.6

m. Strategy tip: In ratio problems with area scaling, always establish your variable relationships first, then use the constraint (here, the perimeter) to solve for the scale factor. Double-check by verifying both the ratio and the given measurements work out correctly.

Question 7

A solution is $15\%$ salt by volume. If $2.4$ liters of water is added to $3.6$ liters of this solution, what percent of the new mixture is salt?

$10\%$
$7.5\%$
$9\%$ (correct answer)
$12\%$

Explanation: When you encounter mixture problems involving percentages, focus on tracking the actual amount of the substance (salt) and the total volume separately, then find the new percentage. Start with what you know:

3.6

liters of

15\%

salt solution contains

3.6 \times 0.15 = 0.54

liters of pure salt. When you add

2.4

liters of water, you're adding zero salt but increasing the total volume. The new mixture has

0.54

liters of salt in a total volume of

3.6 + 2.4 = 6.0

liters. The new percentage is

\frac{0.54}{6.0} = 0.09 = 9\%

, making C correct. Looking at the wrong answers: A)

10\%

likely comes from incorrectly assuming the percentage drops proportionally to the volume increase (

15\% \times \frac{3.6}{6.0} = 9\%

would be correct, but students might miscalculate). B)

7.5\%

results from the common error of thinking the percentage is simply halved because you're roughly doubling the volume (

15\% \div 2 = 7.5\%

). D)

12\%

might come from subtracting the percentage of water added relative to original solution (

15\% - 3\% = 12\%

), which is mathematically meaningless. For mixture problems, always calculate the actual amount of the key substance first, then find the new total, and finally compute the new percentage. Don't try to work directly with percentages—track the concrete quantities.

Question 8

A store offers a discount such that customers pay $\frac{4}{5}$ of the marked price. During a special sale, an additional $15\%$ is taken off the already discounted price. If a customer pays $68 for an item during the special sale, what was the original marked price?

$95
$120
$85
$100 (correct answer)

Explanation: When you encounter multi-step discount problems, work backwards from the final price through each discount in reverse order. This systematic approach prevents calculation errors and helps you track the original value. Let's trace backwards from the $68 final price. First, the customer received an additional $

15\%

off the already discounted price, meaning they paid

85\%

(or

\frac{85}{100} = \frac{17}{20}

) of the discounted price. So the discounted price before the special sale was:

\

68 \div \frac{17}{20} = $68 \times \frac{20}{17} = $80

. Next, this $80 represents the price after the original discount where customers pay $

\frac{4}{5}$$ of the marked price. Therefore: $$\

80 \div \frac{4}{5} = $80 \times \frac{5}{4} = $100$$. The original marked price was $100, confirming answer D. Answer A (

95) likely comes from incorrectly applying the $$15\%$$ additional discount to the original price instead of the already discounted price. Answer B (

120) results from miscalculating one of the fractions, possibly treating

\frac{4}{5}

\frac{3}{5}

. Answer C ($85) appears to be the result of working forward incorrectly or confusing which discount applies first. Remember: when solving backwards through multiple discounts, divide by each discount rate (expressed as a decimal or fraction). When working forward, you multiply. Always identify which direction you're working before starting your calculations.

Question 9

The ratio of boys to girls in a class is $7:5$ . If $25\%$ of the boys and $40\%$ of the girls wear glasses, what fraction of the entire class wears glasses?

$\frac{5}{16}$ (correct answer)
$\frac{1}{3}$
$\frac{5}{12}$
$\frac{7}{20}$

Explanation: In the ratio

7:5

, there are

7

boys and

5

girls out of

12

total students. Boys with glasses:

7 \times 0.25 = 1.75

. Girls with glasses:

5 \times 0.40 = 2

. Total with glasses:

1.75 + 2 = 3.75

out of

12

, so

\frac{3.75}{12} = \frac{15}{48} = \frac{5}{16}

. Choice B averages the two percentages incorrectly. Choice C results from calculation errors. Choice D uses the wrong total.