ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Rational Number Word Problems

20 questions · exam conditions

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Rational Number Word ProblemsQuestion 1 of 20

An item priced at $120 is on sale for 25% off. If the sales tax is 5% of the sale price, what is the total cost of the item including tax?

$90.00

$94.50

$96.00

$121.50

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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Rational Number Word Problems

Practice Rational Number Word Problems in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 Rational Number Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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

An item priced at $120 is on sale for 25% off. If the sales tax is 5% of the sale price, what is the total cost of the item including tax?

$90.00
$94.50 (correct answer)
$96.00
$121.50

Explanation: First, calculate the sale price. The discount is (0.25 \times

120 =

30). The sale price is (

120 -

30 =

90\). Next, calculate the sales tax on this new price: \(0.05 \times

90 =

4.50\). Finally, add the tax to the sale price to find the total cost: \(

90 +

4.50 =

94.50). Distractor A is the sale price before tax. Distractor C is the result of incorrectly calculating tax on the original price (

0.05 \times 120 = 6

) and adding it to the sale price (

90+6=96

). Distractor D is the result of incorrectly applying the discount and tax in a way that gives a higher price.

Question 2

A salesperson earns a base salary of $400 per week plus a commission equal to $\frac{1}{25}$ of her total sales. If she earned a total of $920 in one week, what was the value of her total sales?

$13,000 (correct answer)
$18,400
$23,000
$32,800

Explanation: First, find the amount of money earned from commission by subtracting the base salary from the total earnings: (

920 -

400 =

520\). This $520 represents

\frac{1}{25}

of her total sales. Let S be the total sales. The equation is

\frac{1}{25}S = 520

. To find S, multiply both sides by 25: \(S = 520 \times 25 =

13,000). Distractor C is the result of calculating commission on the total earnings: ($920 \times 25). Distractor B is from a numerical error. Distractor D is from another conceptual error, perhaps involving the base salary in the commission calculation.

Question 3

A rectangular field is $12\frac{1}{2}$ meters long and $8\frac{4}{5}$ meters wide. What is the perimeter of the field in meters?

21.3
42.6 (correct answer)
110
170.6

Explanation: The formula for the perimeter of a rectangle is

P = 2(L+W)

. First, convert the mixed numbers to decimals.

12\frac{1}{2} = 12.5

and

8\frac{4}{5} = 8.8

. Now, add the length and width:

12.5 + 8.8 = 21.3

meters. Finally, multiply this sum by 2 to find the perimeter:

2 \times 21.3 = 42.6

meters. Distractor A is the sum of length and width without multiplying by 2. Distractor C is the area (

12.5 \times 8.8 = 110

). Distractor D results from incorrectly adding all four sides as separate values.

Question 4

Store A sells a 16.5-ounce box of cereal for $3.30. Store B sells a 20-ounce box of the same cereal for $3.80. How much cheaper, per ounce, is the cereal at the store with the better price?

$0.01 (correct answer)
$0.02
$0.19
$0.20

Explanation: First, calculate the unit price (price per ounce) for each store. For Store A: (\frac{

3.30}{16.5 \text{ oz}} =

0.20) per ounce. For Store B: (\frac{

3.80}{20 \text{ oz}} =

0.19) per ounce. Store B has the better price. To find how much cheaper it is, subtract the lower unit price from the higher unit price: (

0.20 -

0.19 = $0.01). Distractors C and D are the unit prices themselves, not the difference between them.

Question 5

A cyclist finishes a race in 45 minutes. A runner takes $1\frac{2}{5}$ times as long to finish the same race. How many more minutes does the runner take than the cyclist?

18 (correct answer)
27
63
108

Explanation: First, calculate the runner's total time. Convert the mixed number to an improper fraction:

1\frac{2}{5} = \frac{7}{5}

. The runner's time is

45 \times \frac{7}{5} = 9 \times 7 = 63

minutes. The question asks how many more minutes the runner took. To find this, subtract the cyclist's time from the runner's time:

63 - 45 = 18

minutes. Distractor C is the runner's total time, not the difference. Distractor B is the result of multiplying by

\frac{3}{5}

instead of

\frac{7}{5}

. Distractor D is the sum of the two times.

Question 6

A student spends $\frac{1}{3}$ of their monthly allowance on a concert ticket. They then spend $\frac{1}{4}$ of the remaining money on food. What fraction of the original monthly allowance does the student have left?

$\frac{5}{12}$
$\frac{2}{3}$
$\frac{7}{12}$
$\frac{1}{2}$ (correct answer)

Explanation: After spending

\frac{1}{3}

of the allowance on a ticket, the student has

1 - \frac{1}{3} = \frac{2}{3}

of the allowance remaining. They then spend

\frac{1}{4}

of this remaining amount on food. The amount spent on food is

\frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}

of the original allowance. The total fraction spent is

\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

. The fraction remaining is

1 - \frac{1}{2} = \frac{1}{2}

. Alternatively, after spending

\frac{1}{4}

of the remaining

\frac{2}{3}

, the student has

\frac{3}{4}

of the remaining

\frac{2}{3}

left. This is

\frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}

. Distractor A is the result of incorrectly adding the fractions

\frac{1}{3} + \frac{1}{4}

to find the total spent. Distractor D is the amount remaining after only the first purchase.

Question 7

A pipe can fill a pool at a rate of $\frac{1}{6}$ of the pool per hour. A drain can empty the same pool at a rate of $\frac{1}{9}$ of the pool per hour. If the pool is empty and both the pipe and the drain are opened, what fraction of the pool will be filled after one hour?

$\frac{1}{54}$
$\frac{5}{18}$
$\frac{1}{3}$
$\frac{1}{18}$ (correct answer)

Explanation: To find the net rate of filling, subtract the drain rate from the fill rate. The net rate is

\frac{1}{6} - \frac{1}{9}

. To subtract the fractions, find a common denominator, which is 18.

\frac{1}{6} = \frac{3}{18}

and

\frac{1}{9} = \frac{2}{18}

. The net rate is

\frac{3}{18} - \frac{2}{18} = \frac{1}{18}

of the pool per hour. So, after one hour,

\frac{1}{18}

of the pool will be filled. Distractor D is the result of adding the rates instead of subtracting them. Distractor A is the result of multiplying the rates. Distractor C is a result of a conceptual error.

Question 8

Maria's bookshelf contains fiction and non-fiction books. Initially, $\frac{2}{3}$ of her books were fiction. After she bought 12 more fiction books, $\frac{3}{4}$ of her books are now fiction. How many non-fiction books does Maria have?

12 (correct answer)
18
24
36

Explanation: Let F be the initial number of fiction books and N be the number of non-fiction books. Initially,

\frac{F}{F+N} = \frac{2}{3}

, which gives us

3F = 2(F+N)

, so

F = 2N

. After buying 12 fiction books, we have

\frac{F+12}{F+N+12} = \frac{3}{4}

. Substituting

F = 2N

\frac{2N+12}{3N+12} = \frac{3}{4}

. Cross-multiplying:

4(2N+12) = 3(3N+12)

, which gives

8N + 48 = 9N + 36

. Solving:

N = 12

. Verification: Initially F=24, N=12 (total 36, and 24/36 = 2/3 ✓). After: F=36, N=12 (total 48, and 36/48 = 3/4 ✓).

Question 9

A box contains red, blue, and green marbles. $\frac{2}{5}$ of the marbles are red, and $\frac{1}{3}$ of the marbles are blue. If there are 12 green marbles in the box, how many marbles are there in total?

30
45 (correct answer)
60
90

Explanation: First, find the fraction of marbles that are either red or blue by adding their fractions:

\frac{2}{5} + \frac{1}{3}

. The common denominator is 15.

\frac{6}{15} + \frac{5}{15} = \frac{11}{15}

. This means the remaining fraction of marbles must be green. The fraction of green marbles is

1 - \frac{11}{15} = \frac{4}{15}

. We are told that there are 12 green marbles. If

\frac{4}{15}

of the total marbles is 12, we can set up the equation

\frac{4}{15}T = 12

, where T is the total number of marbles. To solve for T, multiply both sides by

\frac{15}{4}

T = 12 \times \frac{15}{4} = 3 \times 15 = 45

. There are 45 marbles in total.

Question 10

A stock's price was $50.00 at the beginning of the week. On Monday, its price increased by 20%. On Tuesday, its price decreased by 20% from Monday's closing price. What was the stock's price at the end of Tuesday?

$48.00 (correct answer)
$49.50
$50.00
$52.00

Explanation: First, calculate the price after the 20% increase on Monday: (

50.00 \times 1.20 =

60.00). Next, calculate the price after the 20% decrease from the new price on Tuesday: (

60.00 \times 0.80 =

48.00). The final price is $48.00. Distractor C is a common misconception, assuming that a 20% increase and a 20% decrease cancel each other out. Distractor B is a plausible but incorrect calculation result. Distractor D incorrectly applies a second increase or miscalculates the decrease.

Question 11

A machine produces 60 bolts in $\frac{3}{4}$ of an hour. At this constant rate, how many bolts can the machine produce in $2\frac{1}{2}$ hours?

150
180
200 (correct answer)
240

Explanation: First, find the production rate of the machine in bolts per hour. Rate =

\frac{\text{bolts}}{\text{hours}} = \frac{60}{3/4} = 60 \times \frac{4}{3} = 80

bolts per hour. Next, determine the total number of bolts produced in

2\frac{1}{2}

hours, which is 2.5 hours. Total bolts = Rate

\times

Time =

80 \times 2.5 = 200

. Distractor A is the result of an incorrect rate calculation (e.g.,

60 \times 2.5

). Distractor D is the result of another rate calculation error.

Question 12

A city's population grew from 40,000 to 45,000. By what percentage did the population increase?

11.1%
12.5% (correct answer)
15%
20%

Explanation: To find the percentage increase, use the formula:

\frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100\%. The increase in population is \(45,000 - 40,000 = 5,000

. Now, divide this increase by the original population:

\frac{5,000}{40,000} = \frac{5}{40} = \frac{1}{8}

. To convert this fraction to a percentage, multiply by 100:

\frac{1}{8} \times 100\% = 12.5\%. Distractor A is the result of incorrectly dividing the increase by the new population (\(5,000/45,000

Question 13

A delivery truck's fuel tank holds $18\frac{3}{4}$ gallons when full. The truck uses $\frac{1}{8}$ gallon per mile in city driving and $\frac{1}{12}$ gallon per mile on highways. On a particular route, the truck travels $24$ miles in the city and $84$ miles on highways. If the tank was $\frac{2}{3}$ full at the start, how much fuel remains after the trip?

$2\frac{1}{2}$ gallons (correct answer)
$3\frac{1}{4}$ gallons
$4\frac{1}{2}$ gallons
$5\frac{3}{4}$ gallons

Explanation: Tank capacity =

18\frac{3}{4} = \frac{75}{4}

gallons. Initial fuel =

\frac{2}{3} \times \frac{75}{4} = \frac{150}{12} = \frac{25}{2} = 12\frac{1}{2}

gallons. City fuel used =

24 \times \frac{1}{8} = 3

gallons. Highway fuel used =

84 \times \frac{1}{12} = 7

gallons. Total fuel used =

3 + 7 = 10

gallons. Remaining fuel =

12\frac{1}{2} - 10 = 2\frac{1}{2}

gallons. Choice B results from calculation errors in the initial fuel amount. Choice C comes from using incorrect fuel consumption rates. Choice D results from errors in computing the total distance or fuel usage.

Question 14

A recipe calls for $\frac{2}{3}$ cup of flour for every $\frac{3}{4}$ cup of sugar. Maria wants to make a larger batch using $2\frac{1}{4}$ cups of sugar. If she only has $1\frac{5}{6}$ cups of flour available, how much additional flour does she need?

$\frac{1}{6}$ cup (correct answer)
$\frac{1}{4}$ cup
$\frac{1}{3}$ cup
$\frac{5}{12}$ cup

Explanation: First, find how much flour is needed for

2\frac{1}{4}

cups of sugar. Set up a proportion:

\frac{\frac{2}{3}}{\frac{3}{4}} = \frac{x}{2\frac{1}{4}}

. Cross multiply:

\frac{2}{3} \times \frac{9}{4} = \frac{3}{4} \times x

, so

\frac{18}{12} = \frac{3x}{4}

, which gives

\frac{3}{2} = \frac{3x}{4}

. Solving:

x = 2

cups needed. Maria has

1\frac{5}{6} = \frac{11}{6}

cups. Additional needed:

2 - \frac{11}{6} = \frac{12}{6} - \frac{11}{6} = \frac{1}{6}

cup. Choice B results from calculation errors in the proportion. Choice C comes from incorrectly using

\frac{2}{3} \div \frac{3}{4}

directly. Choice D results from errors in converting mixed numbers.

Question 15

A manufacturing machine operates at $\frac{3}{4}$ of its maximum speed and produces $126$ items per hour. Due to maintenance requirements, it must operate at only $\frac{2}{3}$ of its current speed. How many items will it produce in $3\frac{1}{2}$ hours at the reduced speed?

$245$ items
$378$ items
$336$ items
$294$ items (correct answer)

Explanation: This problem tests your ability to work with fractions and rates in a multi-step scenario. When you see questions involving machines operating at fractions of their capacity, break the problem into clear steps to track the changing rates. First, find the machine's maximum speed. If

\frac{3}{4}

of maximum speed produces 126 items per hour, then the maximum speed is

126 \div \frac{3}{4} = 126 \times \frac{4}{3} = 168

items per hour. Next, calculate the reduced operating speed. The machine will operate at

\frac{2}{3}

of its current speed (which is

\frac{3}{4}

of maximum). So the new rate is

\frac{2}{3} \times 126 = 84

items per hour. Finally, multiply by the time period:

84 \times 3\frac{1}{2} = 84 \times \frac{7}{2} = 294

items. Answer A (245) likely comes from incorrectly calculating

\frac{2}{3} \times \frac{3}{4} \times 168 \times 3.5

but making arithmetic errors. Answer B (378) appears to use the original rate of 126 items per hour for the full

3\frac{1}{2}

hours without applying the speed reduction. Answer C (336) might result from using

\frac{2}{3}

of the maximum speed (112 items per hour) instead of

\frac{2}{3}

of the current speed. The key strategy here is to work step-by-step and clearly distinguish between "maximum speed," "current speed," and "reduced speed." Always verify which reference point each fraction applies to before calculating.

Question 16

A construction crew completes $\frac{2}{9}$ of a project in the first week and $\frac{5}{12}$ of the remaining work in the second week. If they need to finish the project by working at the same rate for $2\frac{1}{4}$ more days, what fraction of the project do they complete per day?

$\frac{7}{36}$
$\frac{14}{81}$ (correct answer)
$\frac{5}{36}$
$\frac{2}{15}$

Explanation: After first week:

1 - \frac{2}{9} = \frac{7}{9}

remains. Second week completes

\frac{5}{12} \times \frac{7}{9} = \frac{35}{108}

of total project. Total completed in two weeks:

\frac{2}{9} + \frac{35}{108} = \frac{24}{108} + \frac{35}{108} = \frac{59}{108}

. Remaining:

1 - \frac{59}{108} = \frac{49}{108}

. This must be completed in

2\frac{1}{4} = \frac{9}{4}

days. Daily rate =

\frac{49}{108} ÷ \frac{9}{4} = \frac{49}{108} \times \frac{4}{9} = \frac{196}{972} = \frac{14}{81}

. Choice A results from calculation errors in finding remaining work. Choice C comes from incorrectly calculating the second week's work. Choice D results from errors in the final division step.

Question 17

A swimming pool is being drained at a rate of $1\frac{2}{5}$ gallons per minute. After $2\frac{1}{3}$ hours of draining, $\frac{5}{8}$ of the original water remains. What was the original amount of water in the pool?

$652$ gallons
$672$ gallons
$896$ gallons (correct answer)
$1344$ gallons

Explanation: First convert time to minutes:

2\frac{1}{3} = \frac{7}{3}

hours =

\frac{7}{3} \times 60 = 140

minutes. Water drained =

1\frac{2}{5} \times 140 = \frac{7}{5} \times 140 = 196

gallons. Since

\frac{5}{8}

remains, then

\frac{3}{8}

was drained. So

\frac{3}{8}

of original = 196 gallons. Original amount =

196 \div \frac{3}{8} = 196 \times \frac{8}{3} = \frac{1568}{3} = 522\frac{2}{3}

gallons. Wait, let me recalculate:

196 \times \frac{8}{3} = \frac{1568}{3} = 522\frac{2}{3}

. Actually,

\frac{1568}{3} = 896

. Choice A results from using wrong fraction remaining. Choice B comes from calculation errors. Choice D results from multiplying instead of dividing by

\frac{3}{8}

Question 18

On a scale drawing, $\frac{1}{2}$ inch represents 6 feet. How many inches on the drawing represent a length of 45 feet?

3.75 (correct answer)
5.25
7.5
15

Explanation: First, find the scale factor, which is the number of feet represented by 1 inch. If

\frac{1}{2}

inch represents 6 feet, then 1 inch represents

6 \div \frac{1}{2} = 6 \times 2 = 12

feet. To find how many inches represent 45 feet, divide the actual length by the scale factor:

45 \text{ feet} \div 12 \text{ feet/inch} = 3.75

inches. Distractor C is the result of multiplying 45 by

\frac{1}{6}

. Distractor D is a result of a conceptual error in applying the scale.

Question 19

The average of three numbers is -2.5. If two of the numbers are 4.2 and -8.1, what is the third number?

-7.5
-3.9
-3.6 (correct answer)
-1.95

Explanation: First, find the sum of the three numbers by multiplying their average by 3:

-2.5 \times 3 = -7.5

. Next, find the sum of the two known numbers:

4.2 + (-8.1) = -3.9

. To find the third number, subtract the sum of the known numbers from the total sum:

-7.5 - (-3.9) = -7.5 + 3.9 = -3.6

. Distractor A is the sum of the three numbers, not the third number itself. Distractor B is the sum of the two known numbers. Distractor D is the average of the two known numbers.

Question 20

A bakery uses a recipe that calls for $3\frac{1}{3}$ cups of sugar for every 5 dozen cookies. How many cups of sugar are needed to make 18 dozen cookies?

10
12 (correct answer)
15
18

Explanation: First, find the amount of sugar needed per dozen cookies. Convert

3\frac{1}{3}

to an improper fraction,

\frac{10}{3}

. The rate is

\frac{10/3 \text{ cups}}{5 \text{ dozen}} = \frac{10}{3 \times 5} = \frac{10}{15} = \frac{2}{3}

cups per dozen. Next, multiply this rate by the desired number of dozens:

\frac{2}{3} \times 18 = 2 \times 6 = 12

cups. Distractor A is a result of a miscalculation. Distractor D incorrectly assumes a 1-to-1 ratio.