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GMAT

GMAT Quiz: Word Translations

Practice Word Translations in GMAT 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 Word Translations, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

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.

Question 1

An electronics store sets a sticker price by marking its wholesale cost up by 40%. During a sale, the store gives a 15% discount off the sticker price. If after the discount the store makes a profit of $51 on a tablet, what was the wholesale cost of the tablet?

$240
$250
$268
$280

Explanation: This is a multi-step markup and discount problem that tests your ability to work backwards from a final profit to find the original cost. When you see questions involving sequential percentage changes, always define your variable as the unknown you're solving for and track each transformation step by step. Let's call the wholesale cost

W

. The store marks up by 40%, so the sticker price becomes

W \times 1.40 = 1.4W

. During the sale, they give a 15% discount off this sticker price, making the sale price

1.4W \times 0.85 = 1.19W

. The profit is the sale price minus the wholesale cost:

1.19W - W = 0.19W = 51

. Solving for

W

W = \frac{51}{0.19} = 268.42

, which rounds to

268

. Choice A (

240

) represents a common error where students might incorrectly calculate the markup or discount percentages. Choice B (

250

) likely results from mishandling the sequential percentage calculations, perhaps treating them as simple addition rather than multiplication. Choice D (

280

) could come from confusing which direction to work the percentages or making an arithmetic error in the final division. The key insight is that after a 40% markup followed by a 15% discount, the final selling price is 119% of the original wholesale cost. Remember: when working with sequential percentage changes, multiply the decimal equivalents rather than adding the percentages, and always clearly define what each step represents before diving into calculations.

Question 2

A coffee roaster wants to make 30 pounds of a blend that is 40% Arabica beans by mixing beans that are 25% Arabica with beans that are 60% Arabica. How many pounds of the 60% Arabica beans should be used?

15.0 lb
13.5 lb
12.9 lb
18.0 lb

Explanation: This is a mixture problem where you're combining two solutions with different concentrations to achieve a target concentration. When you see questions about blending ingredients with different percentages, set up equations based on the total amount and the amount of the key ingredient (Arabica beans in this case). Let's call

x

the pounds of 60% Arabica beans needed. Since you need 30 total pounds, you'll use

(30-x)

pounds of the 25% Arabica beans. The key insight is that the total Arabica content in the final blend equals the sum of Arabica from each component:

0.60x + 0.25(30-x) = 0.40(30)

Solving:

0.60x + 7.5 - 0.25x = 12

Combining like terms:

0.35x = 4.5

Therefore:

x = \frac{4.5}{0.35} = 12.86 ≈ 12.9

pounds Answer choice (A) 15.0 lb represents exactly half the total mixture, which would only work if you were averaging two concentrations that are equidistant from 40% - but 25% and 60% aren't. Choice (B) 13.5 lb likely comes from incorrectly setting up the equation or making arithmetic errors. Choice (D) 18.0 lb is too high and would result in a blend with more than 40% Arabica. Study tip: For mixture problems, always define your variable clearly, set up equations based on the component you're tracking (here, Arabica beans), and remember that percentages must be converted to decimals in calculations. Double-check by verifying your answer produces the target percentage.

Question 3

The phrase "3 less than twice a number is at least 5 and at most 11" can be modeled by which set of all possible values of the number?

$x\ge4\text{ and }x\le5$
$4<x<7$
$x\le4\text{ or }x\ge7$
$4\le x\le7$

Explanation: When you encounter compound inequality word problems, your key task is translating the verbal description into mathematical notation, then solving step by step. Let's call the unknown number

x

. "Twice a number" means

2x

, and "3 less than twice a number" translates to

2x - 3

. The phrase "at least 5 and at most 11" creates a compound inequality:

5 \leq 2x - 3 \leq 11

. To solve this compound inequality, add 3 to all three parts:

5 + 3 \leq 2x - 3 + 3 \leq 11 + 3

, which gives us

8 \leq 2x \leq 14

. Then divide everything by 2:

4 \leq x \leq 7

. Choice A (

x \geq 4

and

x \leq 5

) represents a common error where students correctly find the lower bound but miscalculate the upper bound, possibly confusing the final division step. Choice B (

4 < x < 7

) uses strict inequalities instead of inclusive ones. This occurs when students misinterpret "at least" and "at most" as excluding the boundary values, when these phrases actually mean "greater than or equal to" and "less than or equal to." Choice C (

x \leq 4

x \geq 7

) represents the complement of the correct answer—what values are NOT solutions. This trap catches students who reverse their inequality signs or misunderstand compound inequalities. The correct answer is D:

4 \leq x \leq 7

. Remember: "At least" always means ≥, "at most" always means ≤, and compound inequalities connected by "and" create a range between two values.

Question 4

In a survey of 100 people, 55 read magazine A, 45 read magazine B, and 35 read magazine C. Exactly 20 read both A and B, 15 read both A and C, 10 read both B and C, and 5 read all three magazines. How many of the people surveyed read none of the three magazines?

Explanation: When you encounter problems involving overlapping sets, the principle of inclusion-exclusion is your key tool. This principle helps you avoid double-counting when groups overlap. To find how many people read none of the magazines, you first need to determine how many read at least one. Using the inclusion-exclusion formula:

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|

Substituting the given values:

|A \cup B \cup C| = 55 + 45 + 35 - 20 - 15 - 10 + 5 = 95

Since 95 people read at least one magazine out of 100 surveyed, exactly

100 - 95 = 5

people read none of the magazines. Looking at the wrong answers: Choice A (3) likely results from calculation errors in the inclusion-exclusion formula. Choice C (7) might come from incorrectly subtracting the overlaps without properly adding back the triple overlap. Choice D (9) could result from forgetting to add back the 5 people who read all three magazines, leading to over-subtraction. The key insight is that when groups overlap, you must subtract the pairwise intersections (to correct for double-counting), then add back the triple intersection (since it was subtracted three times but should only be subtracted twice). Remember: Draw a Venn diagram when possible, and always double-check that your inclusion-exclusion calculation makes logical sense. The number reading at least one magazine should be less than the sum of individual magazine readers when overlaps exist.

Question 5

A certain bacterial culture doubles every 3 hours. If the culture contains $N$ bacteria after 9 hours, how many bacteria will it contain after 15 hours?

$\dfrac{N}{4}$
$\dfrac{N}{2}$
$2N$
$4N$

Explanation: This is an exponential growth problem where you need to track how a quantity changes over equal time intervals. When you see "doubles every X hours," you're dealing with exponential growth with a base of 2. First, establish the pattern. The bacteria doubles every 3 hours, so:

At 0 hours: some initial amount
At 3 hours: 2 × initial amount
At 6 hours: 4 × initial amount
At 9 hours: 8 × initial amount = $N$
At 12 hours: 16 × initial amount
At 15 hours: 32 × initial amount

Since we know the culture contains

N

bacteria at 9 hours, we need to find how many doublings occur between 9 and 15 hours. From 9 to 15 hours is 6 hours, which equals exactly 2 doubling periods (since it doubles every 3 hours). Starting with

N

bacteria at 9 hours:

At 12 hours: $2N$
At 15 hours: $2(2N) = 4N$

Choice D (

4N

) is correct. Choice A (

\frac{N}{4}

) suggests the population decreased, which contradicts growth. Choice B (

\frac{N}{2}

) also shows decrease. Choice C (

2N

) represents only one doubling period instead of two—this catches students who miscalculate the time difference or forget that 6 hours equals two 3-hour periods. Strategy tip: For exponential growth problems, always count the number of complete time periods between your starting and ending points. Don't just divide the total time—make sure you're counting intervals correctly.

Question 6

Train B leaves Central Station at 1:00 p.m. traveling 75 km/h toward Easton. Train A leaves the same station on the same track at 1:30 p.m. traveling 90 km/h toward Easton. At what time after Train A departs will Train A overtake Train B?

2 h 00 min after departure
2 h 30 min after departure
3 h 00 min after departure
3 h 30 min after departure

Explanation: This is a classic "catching up" problem where you need to find when a faster train overtakes a slower one that had a head start. The key insight is that Train B gets a 30-minute head start before Train A even begins moving. In that half hour, Train B travels:

75 \text{ km/h} \times 0.5 \text{ h} = 37.5 \text{ km}

Now both trains are moving, but Train A is faster. Train A gains on Train B at a rate of:

90 - 75 = 15 \text{ km/h}

To catch up to Train B's 37.5 km head start at a closing speed of 15 km/h takes:

\frac{37.5 \text{ km}}{15 \text{ km/h}} = 2.5 \text{ hours}

This means Train A overtakes Train B exactly 2 hours and 30 minutes after Train A's departure, making (B) correct. (A) represents a common error where students forget about Train B's head start and simply calculate when the trains would meet if they started simultaneously. (C) might result from incorrectly using 3:00 p.m. as the reference point instead of Train A's 1:30 p.m. departure time. (D) could come from adding the head start time to an incorrect calculation or mixing up the departure times. Strategy tip: In overtaking problems, always identify three key elements: the head start distance, the relative speed difference, and your reference point for measuring time. Many wrong answers exploit confusion about which moment you're measuring from.

Question 7

A swimming pool that holds 12{,}000 gallons is half full. Pipe A can fill the entire pool in 6 hours, whereas drain B can empty a full pool in 9 hours. If both pipe A and drain B are opened at the same time, how many hours will it take for the pool to become completely full?

15 hours
10 hours
12 hours
9 hours

Explanation: When you encounter work rate problems involving filling and draining, think in terms of rates per unit time. The key insight is that rates can be added when working together, but opposing forces (like filling vs. draining) subtract from each other. First, convert the given information to rates. Pipe A fills the entire 12,000-gallon pool in 6 hours, so its rate is

\frac{12,000}{6} = 2,000

gallons per hour. Drain B empties the full pool in 9 hours, so its rate is

\frac{12,000}{9} = \frac{4,000}{3}

gallons per hour. When both operate simultaneously, the net filling rate is:

2,000 - \frac{4,000}{3} = \frac{6,000 - 4,000}{3} = \frac{2,000}{3}

gallons per hour. Since the pool is half full, it needs 6,000 more gallons. At a net rate of

\frac{2,000}{3}

gallons per hour, the time required is:

\frac{6,000}{\frac{2,000}{3}} = 6,000 \times \frac{3}{2,000} = 9

hours. Choice A (15 hours) likely comes from incorrectly adding the individual times (6 + 9). Choice B (10 hours) might result from using the wrong net rate calculation. Choice C (12 hours) could stem from treating this as filling an empty pool rather than a half-full one. For work rate problems, always identify individual rates first, then combine them appropriately (add for cooperative work, subtract for opposing work), and finally apply the combined rate to the actual work remaining.

Question 8

Alicia is 4 years older than Benito. Five years from now, the sum of their ages will be 50. How old is Alicia now?

26 years
23 years
24 years
22 years

Explanation: This is a classic age relationship problem that tests your ability to set up and solve a system of equations. When you encounter age problems, always define your variables clearly and remember that relationships between ages remain constant over time. Let's define Alicia's current age as

A

and Benito's current age as

B

. From the problem, we know two things: Alicia is 4 years older than Benito (

A = B + 4

), and five years from now, their combined ages will be 50. In five years, Alicia will be

A + 5

and Benito will be

B + 5

. So:

(A + 5) + (B + 5) = 50

, which simplifies to

A + B + 10 = 50

, or

A + B = 40

. Now substitute the first equation into the second:

(B + 4) + B = 40

, which gives us

2B + 4 = 40

. Solving:

2B = 36

, so

B = 18

. Therefore, Alicia is currently

18 + 4 = 22

years old. Choice A (26 years) would make Benito 22, and their future sum would be 58, not 50. Choice B (23 years) would make Benito 19, giving a future sum of 52. Choice C (24 years) would make Benito 20, giving a future sum of 54. Each of these violates the "sum equals 50 in five years" condition. The correct answer is D (22 years). Strategy tip: In age problems, always check your answer by plugging it back into both original conditions. This catches arithmetic errors and confirms your solution satisfies all constraints.

Question 9

Machine X can produce 120 parts per hour. Machine Y can produce 80 parts per hour. Machine X works alone for exactly 2 hours and then breaks down; immediately, Machine Y is started and runs at its usual rate until a total of 760 parts have been produced. For how many hours does Machine Y run?

5.5 hours
6.5 hours
7.5 hours
8.5 hours

Explanation: This is a classic work rate problem where you need to track the cumulative output from two machines working sequentially. When you see machines or workers with different rates working at different times, set up the problem by calculating each phase separately. First, determine what Machine X produces during its 2-hour run:

120 \text{ parts/hour} \times 2 \text{ hours} = 240 \text{ parts}

Since the total target is 760 parts, Machine Y must produce the remaining:

760 - 240 = 520 \text{ parts}

Now find how long Machine Y needs to produce 520 parts at its rate of 80 parts per hour:

\frac{520 \text{ parts}}{80 \text{ parts/hour}} = 6.5 \text{ hours}

Looking at the wrong answers: Choice (A) 5.5 hours would only produce

5.5 \times 80 = 440

parts for Machine Y, giving a total of just 680 parts—insufficient. Choice (C) 7.5 hours would produce

7.5 \times 80 = 600

parts for Machine Y, totaling 840 parts—too many. Choice (D) 8.5 hours would produce

8.5 \times 80 = 680

parts for Machine Y, totaling 920 parts—significantly over the target. The correct answer is (B) 6.5 hours. Strategy tip: In sequential work problems, always break the problem into phases. Calculate what happens in each phase separately, then combine the results. Double-check by verifying that your final answer produces exactly the target amount when you add up all phases.

Question 10

A jar contains red and blue marbles in the ratio 3 to 4. After 10 red marbles are added and 4 blue marbles are removed, the ratio becomes 5 to 6. How many blue marbles were in the jar originally?

Explanation: When you encounter ratio problems involving changes to quantities, you need to set up equations that represent both the original and final states. This tests your ability to work with proportional relationships and algebraic manipulation. Let's define the original quantities: if red marbles are

3x

and blue marbles are

4x

, then the ratio is

3x:4x = 3:4

. After the changes, we have

(3x + 10)

red marbles and

(4x - 4)

blue marbles, with a new ratio of

5:6

. Setting up the proportion:

\frac{3x + 10}{4x - 4} = \frac{5}{6}

Cross-multiplying:

6(3x + 10) = 5(4x - 4)

18x + 60 = 20x - 20

80 = 2x

x = 40

Therefore, the original number of blue marbles was

4x = 4(40) = 160

. Looking at the wrong answers: Choice A (112) would correspond to

x = 28

, which doesn't satisfy our equation. Choice B (136) represents

x = 34

, also incorrect. Choice D (184) gives

x = 46

, which again fails to balance the proportion. Each wrong answer likely results from computational errors in cross-multiplication or solving the linear equation, or from confusing which variable represents the blue marbles. Strategy tip: In ratio change problems, always define your variables clearly (use a common factor like

x

), set up the proportion equation carefully, and verify your answer by checking that both the original and final ratios work with your solution.

Question 11

A food stand sells hot dogs for $3 each and sodas for $2 each. Yesterday the stand sold a total of 50 items and collected $128 in revenue. How many hot dogs were sold?

Explanation: When you encounter a problem involving two unknowns and total quantities, you're looking at a classic system of equations scenario. Set up variables for each item type and create equations based on the given constraints. Let

h

= number of hot dogs and

s

= number of sodas. From the problem, you can write two equations:

Total items: $h + s = 50$
Total revenue: $3h + 2s = 128$

From the first equation,

s = 50 - h

. Substitute this into the revenue equation:

3h + 2(50 - h) = 128

3h + 100 - 2h = 128

h = 28

You can verify: 28 hot dogs and 22 sodas gives us

28 × 3 + 22 × 2 = 84 + 44 = 128

✓ Choice A (22) represents the number of sodas, not hot dogs—a common trap where students might solve correctly but report the wrong variable. Choice B (24) likely comes from calculation errors in the substitution step. Choice D (32) could result from setting up the revenue equation incorrectly, perhaps switching the prices or making algebraic mistakes. Strategy tip: Always define your variables clearly and double-check which quantity the question asks for. GMAT word problems frequently include answer choices that represent other calculated values from the same problem. After solving, substitute your answer back into both original equations to confirm it works—this catches most arithmetic errors and ensures you've answered the right question.

Question 12

The sum of three consecutive integers is 75. What is the least of the three integers?

Explanation: When you encounter consecutive integer problems, set up an algebraic equation using the middle integer as your variable—this approach minimizes the arithmetic and reduces errors. Let's call the three consecutive integers

n-1

n

, and

n+1

, where

n

is the middle integer. Since their sum equals 75:

(n-1) + n + (n+1) = 75

Simplifying:

3n = 75

, so

n = 25

Therefore, the three consecutive integers are 24, 25, and 26, making the least integer 24. Let's examine why each answer choice appears: (A) 24 is correct—it's the smallest of our three integers (24, 25, 26). (B) 25 represents the middle integer in our sequence. This is a common trap answer because 25 is the most prominent number in your calculation, but the question asks for the least integer, not the middle one. (C) 26 is the largest integer in our sequence. Students might select this if they made an error in their algebraic setup or misread the question as asking for the greatest integer. (D) 27 likely results from incorrectly setting up the problem as

n + (n+1) + (n+2) = 75

and then confusing which integer the question requests, or from arithmetic errors in the solving process. Strategy tip: Always define your variable as the middle term in consecutive integer problems—it eliminates negative coefficients and makes the algebra cleaner. After solving, double-check by verifying that your three integers actually sum to the given total.

Question 13

A food stand sells hot dogs for $3 each and sodas for $2 each. Yesterday the stand sold a total of 50 items and collected $128 in revenue. How many hot dogs were sold?

Explanation: This is a classic system of equations problem that tests your ability to translate word problems into mathematical relationships. When you see questions involving two unknowns with two constraints (total quantity and total revenue), you'll need to set up two equations. Let

h

= number of hot dogs and

s

= number of sodas. From the given information:

Total items: $h + s = 50$
Total revenue: $3h + 2s = 128$

From the first equation,

s = 50 - h

. Substituting into the second equation:

3h + 2(50 - h) = 128

3h + 100 - 2h = 128

h = 28

Let's verify: If 28 hot dogs were sold, then

50 - 28 = 22

sodas were sold. Revenue check:

28 \times 3 + 22 \times 2 = 84 + 44 = 128

✓ Choice A (22) likely comes from solving incorrectly and finding the number of sodas instead of hot dogs. Choice B (24) might result from arithmetic errors in the substitution step or incorrectly setting up the revenue equation. Choice D (32) could come from reversing the prices (treating hot dogs as

2 and sodas as

3) or other computational mistakes. Strategy tip: Always define your variables clearly and double-check your final answer by substituting back into both original conditions. On the GMAT, word problems often include answer choices that represent common algebraic mistakes, so verification is crucial for catching errors before you submit.

Question 14

If $y$ is 3 more than the square of $x$ and $z$ is twice the difference between $y$ and 4, which of the following expresses $z$ solely in terms of $x$ ?

$z=2x^2-6$
$z=2x^2+2$
$z=2x^2+6$
$z=2x^2-2$

Explanation: When you encounter word problems that ask you to express one variable in terms of another, you need to systematically translate each piece of information into mathematical expressions, then substitute to eliminate intermediate variables. Let's work through this step by step. First, "

y

is 3 more than the square of

x

" translates to

y = x^2 + 3

. Next, "

z

is twice the difference between

y

and 4" means

z = 2(y - 4)

. To express

z

solely in terms of

x

, substitute the expression for

y

into the equation for

z

z = 2(y - 4) = 2((x^2 + 3) - 4) = 2(x^2 + 3 - 4) = 2(x^2 - 1) = 2x^2 - 2

This matches answer choice D. Looking at the wrong answers: Choice A (

z = 2x^2 - 6

) likely results from incorrectly calculating

2(x^2 + 3 - 4)

2x^2 + 2(3) - 2(4) = 2x^2 + 6 - 8

, making an error in the arithmetic. Choice B (

z = 2x^2 + 2

) comes from writing the difference as

(4 - y)

instead of

(y - 4)

, which flips the sign. Choice C (

z = 2x^2 + 6

) results from misinterpreting "the difference between

y

and 4" as addition rather than subtraction. Study tip: In multi-step algebraic translations, write out each relationship separately before substituting. This prevents you from trying to do too much mental math at once and helps you catch sign errors early.

Question 15

A farmer has $180$ feet of fencing to enclose a rectangular garden. One side of the garden will be against an existing wall, so fencing is needed for only three sides. If the farmer wants to maximize the area of the garden, what should be the length of the side parallel to the wall?

$45$ feet
$60$ feet
$90$ feet
$120$ feet

Explanation: Let

x

be the length of the side parallel to the wall, and

y

be the length of each side perpendicular to the wall. Since fencing is needed for only three sides, we have

x + 2y = 180

, so

y = \frac{180 - x}{2} = 90 - \frac{x}{2}

. The area is

A = xy = x(90 - \frac{x}{2}) = 90x - \frac{x^2}{2}

. To maximize area, we take the derivative and set it equal to zero:

\frac{dA}{dx} = 90 - x = 0

, which gives

x = 90

. We can verify this is a maximum by checking the second derivative:

\frac{d^2A}{dx^2} = -1 < 0

, confirming a maximum. Choice A gives

A = 45(67.5) = 3037.5

. Choice B gives

A = 60(60) = 3600

. Choice C gives

A = 90(45) = 4050

. Choice D gives

A = 120(30) = 3600

Question 16

A consulting firm charges clients based on a complex fee structure: a base rate of $200 per hour for the first $$10$$ hours of work, then$ 150 per hour for the next $20$ hours, and $100 per hour for any hours beyond $30$ . If the firm worked $h$ hours on a project where $h > 30$ , which expression represents the total fee charged?

$200h + 150h + 100h$
$200(10) + 150(20) + 100(h - 30)$
$200(10) + 150(h - 10) + 100(h - 30)$
$200h + 150(h - 10) + 100(h - 30)$

Explanation: Since

h > 30

, we need to calculate the fee for each tier separately. The first 10 hours are charged at

200 per hour, giving $$200(10) = \

2000

. The next 20 hours (hours 11-30) are charged at $150 per hour, giving

150(20) = $3000

. The remaining

(h - 30)

hours are charged at $100 per hour, giving

100(h - 30)

. Choice A incorrectly applies all three rates to all

hours. Choice C incorrectly calculates the middle tier as

(h - 10)

hours instead of exactly 20 hours. Choice D incorrectly applies the $200 rate to all

h$$ hours instead of just the first 10.

Question 17

A car rental company offers two pricing plans. Plan A charges $40 per day plus$ 0.25 per mile. Plan B charges $25 per day plus$ 0.40 per mile. For what daily mileage would both plans cost the same amount?

$100$ miles
$150$ miles
$200$ miles
$250$ miles

Explanation: Let

m

be the number of miles driven per day. Plan A costs

40 + 0.25m

dollars. Plan B costs

25 + 0.40m

dollars. Setting them equal:

40 + 0.25m = 25 + 0.40m

. Subtracting

0.25m

from both sides:

40 = 25 + 0.15m

. Subtracting

25

from both sides:

15 = 0.15m

. Dividing by

0.15

m = \frac{15}{0.15} = 100

miles. We can verify: Plan A at 100 miles costs

40 + 0.25(100) = 40 + 25 = \$65

. Plan B at 100 miles costs

25 + 0.40(100) = 25 + 40 = \$65

. Choice B (150 miles): Plan A costs

77.50, Plan B costs

85. Choice C (200 miles): Plan A costs

90, Plan B costs

105. Choice D (250 miles): Plan A costs

102.50, Plan B costs

125.

Question 18

A delivery service charges a base fee of $5 plus$ 0.75 per mile for the first $20$ miles, and $0.50 per mile for each additional mile beyond $$20$$ miles. If a customer's total bill was$ 23, how many miles was the delivery?

$24$ miles
$26$ miles
$30$ miles
$36$ miles

Explanation: Let

d

be the total distance. The base fee is

5. For the first 20 miles, the charge is $$0.75 \times 20 = \

. Since the total bill is $23, which exceeds

5 + 15 = $20

, the delivery must be more than 20 miles. For miles beyond 20, the charge is $0.50 per mile. So we have:

5 + 15 + 0.50(d - 20) = 23

. Simplifying:

20 + 0.50(d - 20) = 23

, so

0.50(d - 20) = 3

, which gives

d - 20 = 6

, therefore

d = 26

miles. Choice A (24 miles) would cost

5 + 15 + 0.50(4) = $22

. Choice C (30 miles) would cost

5 + 15 + 0.50(10) = $25

. Choice D (36 miles) would cost

5 + 15 + 0.50(16) = $28$$.

Question 19

$240
$250
$268
$280

Explanation: This problem tests your ability to work backwards through sequential percentage changes—a common GMAT trap area where students often make calculation errors. Let's call the wholesale cost

W

. The store marks up by 40%, so the sticker price becomes

W \times 1.40 = 1.4W

. During the sale, they give a 15% discount, so the final selling price is

1.4W \times 0.85 = 1.19W

. Since profit equals selling price minus cost, we have:

1.19W - W = 0.19W = 51

. Solving:

W = \frac{51}{0.19} = 268.42

, which rounds to

268

. Let's verify: wholesale cost

268

, sticker price

268 \times 1.40 = 375.20

, sale price

375.20 \times 0.85 = 318.92

, profit

318.92 - 268 = 50.92 \approx 51

. This confirms answer C. Choice A (

240

) gives a profit of only

45.60

—this likely comes from incorrectly calculating the percentage changes. Choice B (

250

) yields a profit of

47.50

, probably from rounding errors in the markup/discount calculations. Choice D (

280

) produces a profit of

53.20

, which might result from confusing the order of operations or miscalculating one of the percentage changes. The key strategy here is to set up your equation carefully with the correct multipliers (1.40 for a 40% markup, 0.85 for a 15% discount) and always verify your answer by working forward through the problem. Sequential percentage problems require precision—double-check each step.

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GMAT Quiz: Word Translations

Practice Word Translations in GMAT 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 Word Translations, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

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.

Question 1

$240
$250
$268
$280

W

. The store marks up by 40%, so the sticker price becomes

W \times 1.40 = 1.4W

. During the sale, they give a 15% discount off this sticker price, making the sale price

1.4W \times 0.85 = 1.19W

. The profit is the sale price minus the wholesale cost:

1.19W - W = 0.19W = 51

. Solving for

W

W = \frac{51}{0.19} = 268.42

, which rounds to

268

. Choice A (

240

) represents a common error where students might incorrectly calculate the markup or discount percentages. Choice B (

250

) likely results from mishandling the sequential percentage calculations, perhaps treating them as simple addition rather than multiplication. Choice D (

280

Question 2

15.0 lb
13.5 lb
12.9 lb
18.0 lb

x

the pounds of 60% Arabica beans needed. Since you need 30 total pounds, you'll use

(30-x)

pounds of the 25% Arabica beans. The key insight is that the total Arabica content in the final blend equals the sum of Arabica from each component:

0.60x + 0.25(30-x) = 0.40(30)

Solving:

0.60x + 7.5 - 0.25x = 12

Combining like terms:

0.35x = 4.5

Therefore:

x = \frac{4.5}{0.35} = 12.86 ≈ 12.9

Question 3

The phrase "3 less than twice a number is at least 5 and at most 11" can be modeled by which set of all possible values of the number?

$x\ge4\text{ and }x\le5$
$4<x<7$
$x\le4\text{ or }x\ge7$
$4\le x\le7$

x

. "Twice a number" means

2x

, and "3 less than twice a number" translates to

2x - 3

. The phrase "at least 5 and at most 11" creates a compound inequality:

5 \leq 2x - 3 \leq 11

. To solve this compound inequality, add 3 to all three parts:

5 + 3 \leq 2x - 3 + 3 \leq 11 + 3

, which gives us

8 \leq 2x \leq 14

. Then divide everything by 2:

4 \leq x \leq 7

. Choice A (

x \geq 4

and

x \leq 5

) represents a common error where students correctly find the lower bound but miscalculate the upper bound, possibly confusing the final division step. Choice B (

4 < x < 7

x \leq 4

x \geq 7

4 \leq x \leq 7

. Remember: "At least" always means ≥, "at most" always means ≤, and compound inequalities connected by "and" create a range between two values.

Question 4

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|

Substituting the given values:

|A \cup B \cup C| = 55 + 45 + 35 - 20 - 15 - 10 + 5 = 95

Since 95 people read at least one magazine out of 100 surveyed, exactly

100 - 95 = 5

Question 5

A certain bacterial culture doubles every 3 hours. If the culture contains $N$ bacteria after 9 hours, how many bacteria will it contain after 15 hours?

$\dfrac{N}{4}$
$\dfrac{N}{2}$
$2N$
$4N$

At 0 hours: some initial amount
At 3 hours: 2 × initial amount
At 6 hours: 4 × initial amount
At 9 hours: 8 × initial amount = $N$
At 12 hours: 16 × initial amount
At 15 hours: 32 × initial amount

Since we know the culture contains

N

N

bacteria at 9 hours:

At 12 hours: $2N$
At 15 hours: $2(2N) = 4N$

Choice D (

4N

) is correct. Choice A (

\frac{N}{4}

) suggests the population decreased, which contradicts growth. Choice B (

\frac{N}{2}

) also shows decrease. Choice C (

2N

Question 6

2 h 00 min after departure
2 h 30 min after departure
3 h 00 min after departure
3 h 30 min after departure

75 \text{ km/h} \times 0.5 \text{ h} = 37.5 \text{ km}

Now both trains are moving, but Train A is faster. Train A gains on Train B at a rate of:

90 - 75 = 15 \text{ km/h}

To catch up to Train B's 37.5 km head start at a closing speed of 15 km/h takes:

\frac{37.5 \text{ km}}{15 \text{ km/h}} = 2.5 \text{ hours}

Question 7

15 hours
10 hours
12 hours
9 hours

\frac{12,000}{6} = 2,000

gallons per hour. Drain B empties the full pool in 9 hours, so its rate is

\frac{12,000}{9} = \frac{4,000}{3}

gallons per hour. When both operate simultaneously, the net filling rate is:

2,000 - \frac{4,000}{3} = \frac{6,000 - 4,000}{3} = \frac{2,000}{3}

gallons per hour. Since the pool is half full, it needs 6,000 more gallons. At a net rate of

\frac{2,000}{3}

gallons per hour, the time required is:

\frac{6,000}{\frac{2,000}{3}} = 6,000 \times \frac{3}{2,000} = 9

Question 8

Alicia is 4 years older than Benito. Five years from now, the sum of their ages will be 50. How old is Alicia now?

26 years
23 years
24 years
22 years

A

and Benito's current age as

B

. From the problem, we know two things: Alicia is 4 years older than Benito (

A = B + 4

), and five years from now, their combined ages will be 50. In five years, Alicia will be

A + 5

and Benito will be

B + 5

. So:

(A + 5) + (B + 5) = 50

, which simplifies to

A + B + 10 = 50

, or

A + B = 40

. Now substitute the first equation into the second:

(B + 4) + B = 40

, which gives us

2B + 4 = 40

. Solving:

2B = 36

, so

B = 18

. Therefore, Alicia is currently

18 + 4 = 22

Question 9

5.5 hours
6.5 hours
7.5 hours
8.5 hours

120 \text{ parts/hour} \times 2 \text{ hours} = 240 \text{ parts}

Since the total target is 760 parts, Machine Y must produce the remaining:

760 - 240 = 520 \text{ parts}

Now find how long Machine Y needs to produce 520 parts at its rate of 80 parts per hour:

\frac{520 \text{ parts}}{80 \text{ parts/hour}} = 6.5 \text{ hours}

Looking at the wrong answers: Choice (A) 5.5 hours would only produce

5.5 \times 80 = 440

parts for Machine Y, giving a total of just 680 parts—insufficient. Choice (C) 7.5 hours would produce

7.5 \times 80 = 600

parts for Machine Y, totaling 840 parts—too many. Choice (D) 8.5 hours would produce

8.5 \times 80 = 680

Question 10

A jar contains red and blue marbles in the ratio 3 to 4. After 10 red marbles are added and 4 blue marbles are removed, the ratio becomes 5 to 6. How many blue marbles were in the jar originally?

3x

and blue marbles are

4x

, then the ratio is

3x:4x = 3:4

. After the changes, we have

(3x + 10)

red marbles and

(4x - 4)

blue marbles, with a new ratio of

5:6

. Setting up the proportion:

\frac{3x + 10}{4x - 4} = \frac{5}{6}

Cross-multiplying:

6(3x + 10) = 5(4x - 4)

18x + 60 = 20x - 20

80 = 2x

x = 40

Therefore, the original number of blue marbles was

4x = 4(40) = 160

. Looking at the wrong answers: Choice A (112) would correspond to

x = 28

, which doesn't satisfy our equation. Choice B (136) represents

x = 34

, also incorrect. Choice D (184) gives

x = 46

x

), set up the proportion equation carefully, and verify your answer by checking that both the original and final ratios work with your solution.

Question 11

A food stand sells hot dogs for $3 each and sodas for $2 each. Yesterday the stand sold a total of 50 items and collected $128 in revenue. How many hot dogs were sold?

h

= number of hot dogs and

s

= number of sodas. From the problem, you can write two equations:

Total items: $h + s = 50$
Total revenue: $3h + 2s = 128$

From the first equation,

s = 50 - h

. Substitute this into the revenue equation:

3h + 2(50 - h) = 128

3h + 100 - 2h = 128

h = 28

You can verify: 28 hot dogs and 22 sodas gives us

28 × 3 + 22 × 2 = 84 + 44 = 128

Question 12

The sum of three consecutive integers is 75. What is the least of the three integers?

n-1

n

, and

n+1

, where

n

is the middle integer. Since their sum equals 75:

(n-1) + n + (n+1) = 75

Simplifying:

3n = 75

, so

n = 25

n + (n+1) + (n+2) = 75

Question 13

A food stand sells hot dogs for $3 each and sodas for $2 each. Yesterday the stand sold a total of 50 items and collected $128 in revenue. How many hot dogs were sold?

h

= number of hot dogs and

s

= number of sodas. From the given information:

Total items: $h + s = 50$
Total revenue: $3h + 2s = 128$

From the first equation,

s = 50 - h

. Substituting into the second equation:

3h + 2(50 - h) = 128

3h + 100 - 2h = 128

h = 28

Let's verify: If 28 hot dogs were sold, then

50 - 28 = 22

sodas were sold. Revenue check:

28 \times 3 + 22 \times 2 = 84 + 44 = 128

2 and sodas as

Question 14

If $y$ is 3 more than the square of $x$ and $z$ is twice the difference between $y$ and 4, which of the following expresses $z$ solely in terms of $x$ ?

$z=2x^2-6$
$z=2x^2+2$
$z=2x^2+6$
$z=2x^2-2$

y

is 3 more than the square of

x

" translates to

y = x^2 + 3

. Next, "

z

is twice the difference between

y

and 4" means

z = 2(y - 4)

. To express

z

solely in terms of

x

, substitute the expression for

y

into the equation for

z

z = 2(y - 4) = 2((x^2 + 3) - 4) = 2(x^2 + 3 - 4) = 2(x^2 - 1) = 2x^2 - 2

This matches answer choice D. Looking at the wrong answers: Choice A (

z = 2x^2 - 6

) likely results from incorrectly calculating

2(x^2 + 3 - 4)

2x^2 + 2(3) - 2(4) = 2x^2 + 6 - 8

, making an error in the arithmetic. Choice B (

z = 2x^2 + 2

) comes from writing the difference as

(4 - y)

instead of

(y - 4)

, which flips the sign. Choice C (

z = 2x^2 + 6

) results from misinterpreting "the difference between

y

Question 15

$45$ feet
$60$ feet
$90$ feet
$120$ feet

Explanation: Let

x

be the length of the side parallel to the wall, and

y

be the length of each side perpendicular to the wall. Since fencing is needed for only three sides, we have

x + 2y = 180

, so

y = \frac{180 - x}{2} = 90 - \frac{x}{2}

. The area is

A = xy = x(90 - \frac{x}{2}) = 90x - \frac{x^2}{2}

. To maximize area, we take the derivative and set it equal to zero:

\frac{dA}{dx} = 90 - x = 0

, which gives

x = 90

. We can verify this is a maximum by checking the second derivative:

\frac{d^2A}{dx^2} = -1 < 0

, confirming a maximum. Choice A gives

A = 45(67.5) = 3037.5

. Choice B gives

A = 60(60) = 3600

. Choice C gives

A = 90(45) = 4050

. Choice D gives

A = 120(30) = 3600

Question 16

$200h + 150h + 100h$
$200(10) + 150(20) + 100(h - 30)$
$200(10) + 150(h - 10) + 100(h - 30)$
$200h + 150(h - 10) + 100(h - 30)$

Explanation: Since

h > 30

, we need to calculate the fee for each tier separately. The first 10 hours are charged at

200 per hour, giving $$200(10) = \

2000

. The next 20 hours (hours 11-30) are charged at $150 per hour, giving

150(20) = $3000

. The remaining

(h - 30)

hours are charged at $100 per hour, giving

100(h - 30)

. Choice A incorrectly applies all three rates to all

hours. Choice C incorrectly calculates the middle tier as

(h - 10)

hours instead of exactly 20 hours. Choice D incorrectly applies the $200 rate to all

h$$ hours instead of just the first 10.

Question 17

$100$ miles
$150$ miles
$200$ miles
$250$ miles

Explanation: Let

m

be the number of miles driven per day. Plan A costs

40 + 0.25m

dollars. Plan B costs

25 + 0.40m

dollars. Setting them equal:

40 + 0.25m = 25 + 0.40m

. Subtracting

0.25m

from both sides:

40 = 25 + 0.15m

. Subtracting

25

from both sides:

15 = 0.15m

. Dividing by

0.15

m = \frac{15}{0.15} = 100

miles. We can verify: Plan A at 100 miles costs

40 + 0.25(100) = 40 + 25 = \$65

. Plan B at 100 miles costs

25 + 0.40(100) = 25 + 40 = \$65

. Choice B (150 miles): Plan A costs

77.50, Plan B costs

85. Choice C (200 miles): Plan A costs

90, Plan B costs

105. Choice D (250 miles): Plan A costs

102.50, Plan B costs

125.

Question 18

$24$ miles
$26$ miles
$30$ miles
$36$ miles

Explanation: Let

d

be the total distance. The base fee is

5. For the first 20 miles, the charge is $$0.75 \times 20 = \

. Since the total bill is $23, which exceeds

5 + 15 = $20

, the delivery must be more than 20 miles. For miles beyond 20, the charge is $0.50 per mile. So we have:

5 + 15 + 0.50(d - 20) = 23

. Simplifying:

20 + 0.50(d - 20) = 23

, so

0.50(d - 20) = 3

, which gives

d - 20 = 6

, therefore

d = 26

miles. Choice A (24 miles) would cost

5 + 15 + 0.50(4) = $22

. Choice C (30 miles) would cost

5 + 15 + 0.50(10) = $25

. Choice D (36 miles) would cost

5 + 15 + 0.50(16) = $28$$.

Question 19

$240
$250
$268
$280

W

. The store marks up by 40%, so the sticker price becomes

W \times 1.40 = 1.4W

. During the sale, they give a 15% discount, so the final selling price is

1.4W \times 0.85 = 1.19W

. Since profit equals selling price minus cost, we have:

1.19W - W = 0.19W = 51

. Solving:

W = \frac{51}{0.19} = 268.42

, which rounds to

268

. Let's verify: wholesale cost

268

, sticker price

268 \times 1.40 = 375.20

, sale price

375.20 \times 0.85 = 318.92

, profit

318.92 - 268 = 50.92 \approx 51

. This confirms answer C. Choice A (

240

) gives a profit of only

45.60

—this likely comes from incorrectly calculating the percentage changes. Choice B (

250

) yields a profit of

47.50

, probably from rounding errors in the markup/discount calculations. Choice D (

280

) produces a profit of

53.20