GMAT Quiz: Word Translations

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

1. $240
2. $250
3. $268 (correct answer)
4. $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

1. 15.0 lb
2. 13.5 lb
3. 12.9 lb (correct answer)
4. 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

1. 3
2. 5 (correct answer)
3. 7
4. 9

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 4

1. 2 h 00 min after departure
2. 2 h 30 min after departure (correct answer)
3. 3 h 00 min after departure
4. 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 5

1. 15 hours
2. 10 hours
3. 12 hours
4. 9 hours (correct answer)

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 6

1. 5.5 hours
2. 6.5 hours (correct answer)
3. 7.5 hours
4. 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 7

1. $45$ feet
2. $60$ feet
3. $90$ feet (correct answer)
4. $120$ 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 8

1. $200h + 150h + 100h$
2. $200(10) + 150(20) + 100(h - 30)$ (correct answer)
3. $200(10) + 150(h - 10) + 100(h - 30)$
4. $200h + 150(h - 10) + 100(h - 30)$

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 $h$ 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 9

1. $100$ miles (correct answer)
2. $150$ miles
3. $200$ miles
4. $250$ 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 10

1. $24$ miles
2. $26$ miles (correct answer)
3. $30$ miles
4. $36$ 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 = \15$$. Since the total bill is $23, which exceeds $5 + 15 = \20, the delivery must be more than 20 miles. For miles beyond 20, the charge is $0.50 per mile. So we have: $$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 11

1. $240
2. $250
3. $268 (correct answer)
4. $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.

Question 12

1. $x\ge4\text{ and }x\le5$
2. $4<x<7$
3. $x\le4\text{ or }x\ge7$
4. $4\le x\le7$ (correct answer)

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 ($x \leq 4$ or $x \geq 7$) 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 13

1. $\dfrac{N}{4}$
2. $\dfrac{N}{2}$
3. $2N$
4. $4N$ (correct answer)

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

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

Question 14

1. 26 years
2. 23 years
3. 24 years
4. 22 years (correct answer)

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 15

1. 112
2. 136
3. 160 (correct answer)
4. 184

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 16

1. 22
2. 24
3. 28 (correct answer)
4. 32

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: $3h + 2s = 128$

Question 17

1. 24 (correct answer)
2. 25
3. 26
4. 27

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 18

1. 22
2. 24
3. 28 (correct answer)
4. 32

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: $3h + 2s = 128$

Question 19

1. $z=2x^2-6$
2. $z=2x^2+2$
3. $z=2x^2+6$
4. $z=2x^2-2$ (correct answer)

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 $2(x^2 + 3 - 4)$ as $2x^2 + 2(3) - 2(4) = 2x^2 + 6 - 8$, 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.