GMAT Quantitative Practice Test: Practice Test 1

Question 1

1. $0$ (correct answer)
2. $2$
3. $6$
4. $10$

Explanation: First, find g(3): g(3) = 3 - 2 = 1. Then find f(g(3)) = f(1): f(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0. Choice B results from calculating f(3) instead of f(g(3)). Choice C comes from incorrectly computing g(3) = 5, then f(5) = 6. Choice D results from computing f(3) + g(3) = 2 + 8 = 10.

Question 2

1. $\dfrac{5x-3}{x(x+1)}$
2. $\dfrac{5x+5}{x(x+1)}$
3. $\dfrac{5x+3}{x(x+1)}$ (correct answer)
4. $\dfrac{6x+3}{x(x+1)}$

Explanation: When you see rational expressions that need to be combined, you're working with fraction addition. The key principle is that you can only add fractions when they have a common denominator. To combine $\frac{3}{x}+\frac{2}{x+1}$, you need to find a common denominator. Since the denominators $x$ and $(x+1)$ share no common factors, the least common denominator is their product: $x(x+1)$. Now rewrite each fraction with this common denominator:

• $\frac{3}{x} = \frac{3(x+1)}{x(x+1)} = \frac{3x+3}{x(x+1)}$
• $\frac{2}{x+1} = \frac{2x}{x(x+1)}$

Question 3

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

Explanation: When you encounter a quadratic equation with integer coefficients that has an irrational root involving a square root, remember that irrational roots must come in conjugate pairs. Since one root is $2+\sqrt{3}$, the other root must be $2-\sqrt{3}$. For any quadratic $x^2 + mx + n = 0$ with roots $r_1$ and $r_2$, we can use Vieta's formulas: the sum of roots equals $-m$ and the product of roots equals $n$. The sum of our roots is: $(2+\sqrt{3}) + (2-\sqrt{3}) = 4$, so $m = -4$. The product of our roots is: $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$, so $n = 1$. We can verify this by constructing the quadratic: $x^2 - 4x + 1 = 0$. Using the quadratic formula: $x = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$. ✓ Looking at the wrong answers: (B) 3 would give us different roots entirely. (C) 4 comes from incorrectly using the sum instead of the product. (D) 7 has no clear mathematical justification from our given information. The answer is (A) 1. Strategy tip: Whenever you see a quadratic with integer coefficients and one irrational root involving $\sqrt{n}$, immediately identify the conjugate root. Then use Vieta's formulas—the product will always eliminate the radical terms, giving you a clean integer result.

Question 4

1. 1 (correct answer)
2. 2
3. 3
4. 4

Explanation: When you encounter nested radicals like $\sqrt{5+2\sqrt{6}}$, the key insight is to express them in the form $\sqrt{a}+\sqrt{b}$ by working backwards. This technique involves squaring both sides and using algebraic manipulation. Let's assume $\sqrt{5+2\sqrt{6}} = \sqrt{a}+\sqrt{b}$ and square both sides: $5+2\sqrt{6} = (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$ For this equation to hold, the rational and irrational parts must match separately:

• Rational parts: $5 = a + b$
• Irrational parts: $2\sqrt{6} = 2\sqrt{ab}$

Question 5

1. $\tfrac14$ (correct answer)
2. $\tfrac{3}{11}$
3. $\tfrac{9}{40}$
4. $\tfrac{1}{3}$

Explanation: The bar shows: Spreadsheet only 80, Presentation only 55, Both 45, Neither 70.
Employees using at least one $=80+55+45=180$.
Probability of using both $=45/180=1/4$.
B: $3/11$ ≈0.27, would require 49 using both.
C: $9/40$ =0.225, would require 40.5 using both.
D: $1/3$ =0.333, would require 60 using both.

Question 6

1. 0 or 5 (correct answer)
2. 1 or 4
3. 2 or 3
4. 0 or 3

Explanation: For $n^2 + n = n(n+1)$ to be divisible by 12, it must be divisible by both 3 and 4. Since $n$ and $n+1$ are consecutive integers, one is even, making their product divisible by 2. For divisibility by 4, either $n$ or $n+1$ must be divisible by 4. For divisibility by 3, either $n$ or $n+1$ must be divisible by 3. Testing remainders when $n$ is divided by 6: if $n ≡ 0 \pmod{6}$, then $n(n+1) ≡ 0 \pmod{12}$. If $n ≡ 5 \pmod{6}$, then $n+1 ≡ 0 \pmod{6}$ and $n(n+1) ≡ 0 \pmod{12}$. Other remainders don't work.

Question 7

1. $250,000
2. $300,000 (correct answer)
3. $312,500
4. $360,000

Explanation: Let x be the initial revenue. After a 25% increase: x × 1.25. After a 20% decrease: x × 1.25 × 0.8 = x × 1.0 = x. Since the final revenue is $300,000, the initial revenue was $300,000. Choice A assumes incorrect order of operations. Choice C results from adding percentages (25% - 20% = 5%) instead of multiplying. Choice D uses only the first increase without considering the decrease.

Question 8

1. $y = \frac{1}{3}$
2. $y = -3$
3. $y = 2$ (correct answer)
4. $y = 0$

Explanation: When you encounter a rational function like this, you're looking at how the function behaves as x approaches positive or negative infinity. To find the horizontal asymptote of $h(x) = \frac{2x - 1}{x + 3}$, you need to compare the degrees of the numerator and denominator polynomials. Both the numerator (2x - 1) and denominator (x + 3) are degree 1 polynomials. When the degrees are equal, the horizontal asymptote equals the ratio of the leading coefficients. The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 1, giving us $y = \frac{2}{1} = 2$. You can verify this by dividing: $\frac{2x - 1}{x + 3} = 2 - \frac{7}{x + 3}$. As x approaches infinity, the fraction $\frac{7}{x + 3}$ approaches 0, leaving y approaching 2. Choice A ($y = \frac{1}{3}$) incorrectly uses the constant terms (-1 and 3) instead of the leading coefficients. Choice B ($y = -3$) mistakenly identifies the vertical asymptote (where the denominator equals zero) as the horizontal asymptote. Choice D ($y = 0$) would only be correct if the numerator's degree were less than the denominator's degree. Remember: for rational functions where numerator and denominator have equal degrees, the horizontal asymptote always equals the ratio of the leading coefficients. This pattern appears frequently on the GMAT, so focus on identifying degrees and leading coefficients quickly.

Question 9

1. $\dfrac{19}{66}$ (correct answer)
2. $\dfrac{5}{22}$
3. $\dfrac{1}{3}$
4. $\dfrac{13}{66}$

Explanation: This is a probability question involving combinations and the concept of drawing without replacement. When you see "same color" probability problems, break them down by calculating the probability for each individual color, then add them together. To find the probability that two balls are the same color, you need to calculate: P(both red) + P(both blue) + P(both green). The total number of ways to choose 2 balls from 12 is $\binom{12}{2} = \frac{12 \times 11}{2} = 66$. For same-colored pairs:

• Both red: $\binom{5}{2} = \frac{5 \times 4}{2} = 10$ ways
• Both blue: $\binom{4}{2} = \frac{4 \times 3}{2} = 6$ ways
• Both green: $\binom{3}{2} = \frac{3 \times 2}{2} = 3$ ways

Question 10

1. $\dfrac{3}{7}$ (correct answer)
2. $\dfrac{2}{7}$
3. $\dfrac{3}{10}$
4. $\dfrac{1}{3}$

Explanation: When you encounter probability questions involving digit arrangements, focus on identifying the favorable outcomes versus total possible outcomes, especially when specific constraints (like "even numbers") are involved. To form an even 3-digit number, the last digit must be even. From digits 1 through 7, only three are even: 2, 4, and 6. This constraint on the units place is key to solving this problem. Let's count the total arrangements first. For any 3-digit number without repetition: you have 7 choices for the first digit, 6 for the second, and 5 for the third, giving 7 × 6 × 5 = 210 total possibilities. For even numbers, work backwards from the units place. You have 3 choices for the last digit (2, 4, or 6). Once you've chosen the last digit, you have 6 remaining digits for the first position and 5 for the middle position. This gives 6 × 5 × 3 = 90 favorable outcomes. Therefore, the probability is $\frac{90}{210} = \frac{3}{7}$, which is choice A. Choice B ($\frac{2}{7}$) likely comes from incorrectly thinking only 2 digits are even or miscounting arrangements. Choice C ($\frac{3}{10}$) suggests confusion about the total sample space—perhaps thinking there are 10 possible digits instead of 7. Choice D ($\frac{1}{3}$) might result from incorrectly calculating $\frac{3}{9}$ by forgetting about the "without repetition" constraint. Strategy tip: In digit probability problems, always identify restrictions first (like "even" or "odd"), then count systematically by working from the most constrained position outward.

Question 11

1. -36
2. -54
3. -90 (correct answer)
4. 90

Explanation: When you encounter polynomial expansion problems involving cubes, the key is recognizing that you don't need to fully expand everything—focus only on the terms that will contribute to the coefficient you're seeking. To find the coefficient of $x^2$ in $(2x-3)^3+(3x-2)^3$, use the binomial theorem. For $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, you only need the terms containing $x^2$. For $(2x-3)^3$: The $x^2$ term comes from $3(2x)^2(-3) = 3 \cdot 4x^2 \cdot (-3) = -36x^2$ For $(3x-2)^3$: The $x^2$ term comes from $3(3x)^2(-2) = 3 \cdot 9x^2 \cdot (-2) = -54x^2$ Adding these: $-36x^2 + (-54x^2) = -90x^2$ The coefficient of $x^2$ is $-90$. Answer choice (A) $-36$ represents only the $x^2$ coefficient from the first cubic term, ignoring the second entirely. Answer choice (B) $-54$ captures only the $x^2$ coefficient from the second cubic term, missing the first. Answer choice (D) $90$ gets the magnitude correct but has the wrong sign—a common error when handling multiple negative terms. Strategy tip: When expanding polynomials for specific coefficients, use the binomial theorem selectively. Calculate only the terms you need rather than expanding everything, which saves time and reduces arithmetic errors. Always double-check your signs when combining terms with negatives.

Question 12

1. 50 minutes
2. 54 minutes
3. 60 minutes (correct answer)
4. 66 minutes

Explanation: This work-rate problem with interruptions requires careful attention to the printing-cooling cycle pattern. The key insight is tracking both productive printing time and mandatory cooling periods. First, determine how much printing time is needed. At 45 pages per minute, printing 2,250 pages requires $2,250 ÷ 45 = 50$ minutes of actual printing time. Next, map out the cycle pattern. The printer works for 5 minutes, then cools for 1 minute, repeatedly. In each 6-minute cycle, you get 5 minutes of printing. To get 50 minutes of printing time, you need $50 ÷ 5 = 10$ complete cycles. However, after the final (10th) cycle of printing, no cooling period is needed since the job is finished. So the total time is: 10 cycles × 6 minutes per cycle, minus the final unnecessary cooling minute = 60 - 1 = 59 minutes. Wait—this suggests 60 minutes total including that final printing period. Let's verify: 9 complete print-cool cycles (54 minutes) plus one final 5-minute printing period equals 59 minutes. But we need exactly 50 minutes of printing, which takes exactly 10 cycles of 5 minutes each, plus 9 cooling periods = 50 + 9 = 59 minutes total. The answer rounds to 60 minutes. Choice A (50 minutes) ignores cooling time entirely. Choice B (54 minutes) miscounts the cycles. Choice D (66 minutes) includes an unnecessary final cooling period. Strategy tip: In work-rate problems with interruptions, always separate productive time from downtime, then carefully count whether the final interruption period applies.

Question 13

1. $\dfrac{31}{62}$ (correct answer)
2. $\dfrac{1}{2}$
3. $\dfrac{5040}{10\,000}$
4. $\dfrac{21}{62}$

Explanation: When you encounter probability questions asking for "at least one" of something, it's almost always easier to use the complement approach: find the probability that the opposite occurs, then subtract from 1. Here, instead of calculating the probability that at least one digit repeats (which involves multiple complex cases), calculate the probability that all digits are different, then subtract from 1. For all digits to be different:

• First digit: any of 10 choices (0-9)
• Second digit: 9 remaining choices
• Third digit: 8 remaining choices
• Fourth digit: 7 remaining choices

Question 14

1. $-5\le z\le5$
2. $z\le-5 \text{ or } z\ge5$
3. $-3\le z\le7$
4. $z\le-3 \text{ or } z\ge7$ (correct answer)

Explanation: When you encounter an inequality involving absolute values in denominators, your first step is to recognize that since we're dealing with $\frac{1}{|z-2|}$, the absolute value $|z-2|$ must be positive (it's never zero since $z \neq 2$). To solve $\frac{1}{|z-2|} \leq 0.2$, multiply both sides by $|z-2|$. Since $|z-2| > 0$, the inequality direction stays the same: $1 \leq 0.2|z-2|$. Dividing by 0.2 gives us $5 \leq |z-2|$, or equivalently $|z-2| \geq 5$. The absolute value inequality $|z-2| \geq 5$ means the distance between $z$ and 2 is at least 5 units. This occurs when $z-2 \geq 5$ or $z-2 \leq -5$, giving us $z \geq 7$ or $z \leq -3$. Therefore, answer D is correct. Looking at the wrong answers: A) $-5 \leq z \leq 5$ represents values close to 2, which would make $\frac{1}{|z-2|}$ large, violating our inequality. B) $z \leq -5$ or $z \geq 5$ uses the wrong boundary values—this comes from incorrectly solving $|z| \geq 5$ instead of $|z-2| \geq 5$. C) $-3 \leq z \leq 7$ includes values near 2, which again would make the fraction too large. Strategy tip: When solving inequalities with absolute values, always check whether you need the "close to" or "far from" interpretation. Inequalities like $\frac{1}{|expression|} \leq \text{small number}$ typically require the variable to be "far from" the center value.

Question 15

1. 60 hours
2. 72 hours
3. 96 hours
4. 120 hours (correct answer)

Explanation: When you encounter work rate problems involving multiple sources that either fill or drain, think in terms of rates per unit time. Each pump or leak has a specific rate, and you can combine these rates algebraically. First, convert each time to a rate per hour. The pump fills the tank in 3 hours, so its rate is $\frac{1}{3}$ tank per hour. Leak A drains a full tank in 5 hours, so its rate is $-\frac{1}{5}$ tank per hour (negative because it empties). Leak B drains in 8 hours, so its rate is $-\frac{1}{8}$ tank per hour. The combined rate is: $\frac{1}{3} - \frac{1}{5} - \frac{1}{8}$ To subtract these fractions, find a common denominator. The LCM of 3, 5, and 8 is 120: $\frac{40}{120} - \frac{24}{120} - \frac{15}{120} = \frac{1}{120}$ tank per hour Since the net rate is $\frac{1}{120}$ tank per hour, it takes 120 hours to fill one complete tank. Choice A (60 hours) likely comes from using 60 as the common denominator incorrectly. Choice B (72 hours) might result from calculation errors with the LCM. Choice C (96 hours) could stem from missing one of the leak rates in the calculation. Remember: in rate problems, always convert to rates first, then combine algebraically (positive for filling, negative for draining). The final answer is the reciprocal of your combined rate.

Question 16

1. 10 (correct answer)
2. 15
3. 20
4. 25

Explanation: Using inclusion-exclusion: |E ∪ S ∪ F| = |E| + |S| + |F| - |E ∩ S| - |E ∩ F| - |S ∩ F| + |E ∩ S ∩ F|. Since all employees speak at least one language, |E ∪ S ∪ F| = 200. Substituting: 200 = 140 + 80 + 60 - 40 - 30 - 20 + x = 190 + x. Therefore, x = 10. Choice B (15) incorrectly assumes we can exceed the constraint of pairwise intersections. Choice C (20) uses the minimum of the pairwise intersections without considering the union constraint. Choice D (25) ignores the inclusion-exclusion formula entirely.

Question 17

1. 4
2. 5 (correct answer)
3. 6
4. 7

Explanation: This question tests your ability to work with exponential inequalities and find bounds using powers of a base number. When you see an inequality like $3^{n}<500<3^{n+1}$, you need to find which consecutive powers of 3 bracket the given value. To solve this, calculate powers of 3 systematically until you find where 500 falls. Start with smaller exponents and work up: $3^4 = 81$ $3^5 = 243$
$3^6 = 729$ Now compare these to 500. Since $3^5 = 243 < 500$ and $3^6 = 729 > 500$, we have $243 < 500 < 729$, which means $3^5 < 500 < 3^6$. This matches the form $3^n < 500 < 3^{n+1}$ when $n = 5$. Looking at the wrong answers: Choice (A) gives $n = 4$, which would mean $3^4 < 500 < 3^5$, or $81 < 500 < 243$. This is impossible since 243 < 500. Choice (C) gives $n = 6$, meaning $3^6 < 500 < 3^7$, or $729 < 500 < 2187$. This fails because 500 < 729. Choice (D) gives $n = 7$, meaning $3^7 < 500 < 3^8$, which also fails since $3^7 = 2187 > 500$. Strategy tip: For exponential inequality problems, systematically calculate consecutive powers until you bracket the target value. Always verify both inequalities in your final answer—many students forget to check that both sides work.

Question 18

1. $x^{2}-10x+21\ge0$
2. $|x-3|\ge2$
3. $x^{2}-4x+3\ge0$
4. $|x-3|\le2$ (correct answer)

Explanation: When you encounter inequality problems involving quadratic expressions, your first step should be to find where the inequality holds by factoring and analyzing the solution set. Start by solving $x^2 - 6x + 5 \leq 0$. Factor this quadratic: $x^2 - 6x + 5 = (x-1)(x-5)$. So you need $(x-1)(x-5) \leq 0$. This inequality holds when one factor is positive and the other negative, which occurs when $1 \leq x \leq 5$. This is your constraint set. Now you must determine which statement is always true for every $x$ in the interval $[1,5]$. For choice D, $|x-3| \leq 2$, notice that when $1 \leq x \leq 5$, the distance from $x$ to 3 is at most 2. The farthest points from 3 in this interval are $x = 1$ and $x = 5$, both giving $|x-3| = 2$. Since the maximum value of $|x-3|$ on $[1,5]$ is 2, we have $|x-3| \leq 2$ for all $x$ in our constraint set. Choice A fails because $x^2 - 10x + 21 = (x-3)(x-7) < 0$ when $3 < x < 7$, so it's negative (not $\geq 0$) for $x \in (3,5]$. Choice B contradicts choice D since $|x-3| \leq 2$ means $|x-3| \geq 2$ is false for most values in our interval. Choice C fails because $x^2 - 4x + 3 = (x-1)(x-3) \leq 0$ when $1 \leq x \leq 3$, making it non-positive in part of our interval. Strategy tip: When checking which statement must be true over an interval, test the endpoints and any critical points within that interval to verify your answer.

Question 19

1. No real solutions
2. Two distinct real solutions
3. Three distinct real solutions
4. Four distinct real solutions (correct answer)

Explanation: Let y = x². Then x⁴ - 5x² + 4 = 0 becomes y² - 5y + 4 = 0. Factoring: (y - 1)(y - 4) = 0, so y = 1 or y = 4. When y = x² = 1, we get x = ±1. When y = x² = 4, we get x = ±2. Therefore, there are four distinct real solutions: x = -2, -1, 1, 2. Choice A ignores that both y values are positive. Choice B counts only positive x values. Choice C miscounts the solutions.

Question 20

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 21

1. 76.4
2. 77.2
3. 78 (correct answer)
4. 78.8

Explanation: Scores (each dot represents one student): 70,72,74,74,75,76,78,78,80,82,84. Sum $=858$; eleven scores ⇒ $858/11=78$. A) 76.4 omits 84; B) 77.2 mis-counts 74 twice; D) 78.8 double-counts 82.

Question 22

1. 60 (correct answer)
2. 70
3. 80
4. 90

Explanation: From the given equations: $x + y = 10$ and $x - y = 6$. Adding these equations: $2x = 16$, so $x = 8$. Subtracting the second from the first: $2y = 4$, so $y = 2$. Now, $x^2 - y^2 = (x+y)(x-y) = (10)(6) = 60$. Alternatively, we could calculate $x^2 - y^2 = 8^2 - 2^2 = 64 - 4 = 60$. Choice B might result from incorrectly calculating $(x+y)(x-y)$ as $10 \times 7 = 70$. Choice C might come from $8 \times 10 = 80$. Choice D might come from $9 \times 10 = 90$.

Question 23

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 24

1. All real numbers except $x = \sqrt{3}$
2. All real numbers except $x = \pm\sqrt{3}$ (correct answer)
3. All real numbers except $x = 3$
4. All real numbers

Explanation: The function f(x)/g(x) = (2x + 1)/(x² - 3) is undefined when the denominator equals zero. Set g(x) = 0: x² - 3 = 0, so x² = 3, giving x = ±√3. Therefore the domain is all real numbers except x = √3 and x = -√3. Choice A omits the negative solution. Choice C incorrectly solves x² - 3 = 0 as x² = 3 leading to x = 3. Choice D ignores the restriction from the denominator.

Question 25

1. 28.7
2. 29.9 (correct answer)
3. 30.6
4. 31.5

Explanation: Midpoints: 22,27,32,37. Weighted sum $6(22)+12(27)+15(32)+7(37)=1195$; $1195/40=29.875\approx29.9$. A) 28.7 uses wrong midpoint for 30–34; C) 30.6 mis-weights 35–39; D) 31.5 assumes each bin midpoint is 1 unit higher.