GMAT Quantitative Practice Test: Practice Test 4

Question 1

1. $\dfrac{x-3}{x^{2}+3x+9}$
2. $x+3$
3. $x^{2}-9$
4. $x-3$ (correct answer)

Explanation: When you encounter a rational expression that needs simplification, look for opportunities to factor and cancel common terms. This question tests your ability to recognize and apply the difference of cubes formula. The key insight is recognizing that $x^3 - 27$ is a difference of cubes, since $27 = 3^3$. The difference of cubes formula states that $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Applying this with $a = x$ and $b = 3$: $x^3 - 27 = x^3 - 3^3 = (x-3)(x^2 + 3x + 9)$ Now you can substitute this factorization into the original expression: $\frac{x^3-27}{x^2+3x+9} = \frac{(x-3)(x^2+3x+9)}{x^2+3x+9}$ Since $x \neq 3$, the factor $(x^2+3x+9)$ is never zero and can be canceled from numerator and denominator, leaving $x-3$. Choice A, $\frac{x-3}{x^2+3x+9}$, represents the expression before canceling the common factor. Choice B, $x+3$, might result from incorrectly applying the difference of squares formula instead of difference of cubes. Choice C, $x^2-9$, could come from confusing this with factoring $x^2-9 = (x-3)(x+3)$ and forgetting about the cubic terms. Remember that difference of cubes problems appear frequently on the GMAT. Memorize the formula $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ and always check whether rational expressions can be simplified by factoring and canceling common terms.

Question 2

1. $50,000 (correct answer)
2. $52,000
3. $55,000
4. $56,000

Explanation: Let S be the original salary. After the 12% increase: 1.12S. The 15% bonus is based on the new salary: 0.15 × 1.12S = 0.168S. Total compensation = 1.12S + 0.168S = 1.288S = $64,400. Therefore, S =$64,400 ÷ 1.288 = $50,000. Choice B results from calculating the bonus on the original salary instead of the increased salary. Choice C assumes the bonus is applied before the salary increase. Choice D results from arithmetic errors in the percentage calculations.

Question 3

1. 36 minutes
2. 40 minutes
3. 42 minutes (correct answer)
4. 48 minutes

Explanation: When you encounter combined work rate problems, focus on finding each machine's rate per unit time, then track their cumulative output over different time periods. First, calculate each machine's rate. Machine A prints 300 pages in 8 minutes, so its rate is $\frac{300}{8} = 37.5$ pages per minute. Machine B prints 300 pages in 12 minutes, giving a rate of $\frac{300}{12} = 25$ pages per minute. For the first 12 minutes, both machines work together. Machine A produces $37.5 \times 12 = 450$ pages, while Machine B produces $25 \times 12 = 300$ pages. Together, they complete 750 pages, leaving $1,800 - 750 = 1,050$ pages remaining. After Machine A jams, only Machine B continues. At 25 pages per minute, Machine B needs $\frac{1,050}{25} = 42$ additional minutes to finish the job. Looking at the wrong answers: (A) 36 minutes would only produce $36 \times 25 = 900$ pages, falling short by 150 pages. (B) 40 minutes yields $40 \times 25 = 1,000$ pages, still 50 pages short. (D) 48 minutes produces $48 \times 25 = 1,200$ pages, which exceeds the remaining work by 150 pages. The correct answer is (C) 42 minutes. Strategy tip: In work rate problems, always convert to "units per time" first, then multiply by time periods to find total output. Double-check by verifying your final total matches the target amount.

Question 4

1. 33 positive integers
2. 66 positive integers
3. 99 positive integers (correct answer)
4. 50 positive integers

Explanation: We need $k^3 - k = k(k^2 - 1) = k(k-1)(k+1)$ to be divisible by 6. Since this is the product of three consecutive integers, it's always divisible by 3! = 6. Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3, so their product is always divisible by 6. Therefore, $\frac{k^3 - k}{6}$ is an integer for all positive integers $k$. Since we want $k < 100$, there are 99 such values.

Question 5

1. 420
2. 840 (correct answer)
3. 1,260
4. 1,680

Explanation: When you encounter least common multiple (LCM) problems on the GMAT, the most efficient approach is prime factorization. The LCM is found by taking the highest power of each prime factor that appears in either number. First, let's find the prime factorization of each number:

• $84 = 4 \times 21 = 4 \times 3 \times 7 = 2^2 \times 3 \times 7$
• $120 = 8 \times 15 = 8 \times 3 \times 5 = 2^3 \times 3 \times 5$

• Highest power of 2: $2^3 = 8$
• Highest power of 3: $3^1 = 3$
• Highest power of 5: $5^1 = 5$
• Highest power of 7: $7^1 = 7$

Question 6

1. 12
2. 16 (correct answer)
3. 18
4. 20

Explanation: Counts: Marketing 20, Finance 16, HR 12, Operations 24, IT 8. Ordered: 8,12,16,20,24 ⇒ median is 16. A) 12 and D) 20 are adjacent values, not the middle; C) 18 is an average of 16 and 20, not an actual count.

Question 7

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 8

1. The system has exactly one solution since the coefficients are proportional
2. The system has no solution since $12 \neq 2 \times 18$ (correct answer)
3. The system has infinitely many solutions since the first equation equals twice the second
4. The system has no solution since $12 \neq 2 \times 9$

Explanation: Given $a = 2c$ and $b = 2d$, the first equation becomes $2cx + 2dy = 12$, which simplifies to $2(cx + dy) = 12$, or $cx + dy = 6$. But the second equation states $cx + dy = 18$. Since $6 ≠ 18$, we have a contradiction: the same expression ($cx + dy$) cannot equal both 6 and 18. Therefore, the system has no solution. Choice A is wrong because proportional coefficients with non-proportional constants lead to no solution, not one solution. Choice C is wrong because the equations are inconsistent. Choice D has the wrong calculation ($2 \times 9 = 18$, but we need $2 \times 18 = 36$ to match 12, which it doesn't).

Question 9

1. $30{,}240$ (correct answer)
2. $15{,}120$
3. $10{,}080$
4. $5{,}040$

Explanation: This is a constrained permutation problem where you need to treat certain elements as a single unit. When you see "consecutive" in arrangement problems, think about grouping those elements together first. To solve this, treat the three vowels (A, O, I) as one super-letter. This transforms ALGORITHM from a 9-letter word into a 7-unit arrangement: the vowel block plus the 6 consonants (L, G, R, T, H, M). First, arrange these 7 units: $7! = 5{,}040$ ways. Next, within the vowel block, the three vowels A, O, and I can be arranged among themselves in $3! = 6$ ways. Using the multiplication principle: $7! \times 3! = 5{,}040 \times 6 = 30{,}240$ total arrangements. Looking at the wrong answers: Choice B ($15{,}120$) represents $7! \times 3$, suggesting someone counted vowel arrangements incorrectly as 3 instead of $3! = 6$. Choice C ($10{,}080$) equals $7! \times 2$, perhaps from miscounting internal arrangements. Choice D ($5{,}040$) is simply $7!$, the result if you forgot to account for internal vowel arrangements entirely. The correct answer is A: $30{,}240$. Strategy tip: For "consecutive" arrangement problems, always use the block method: treat the consecutive elements as one unit, arrange the remaining elements, then multiply by the internal arrangements of the block. This two-step approach prevents common counting errors and works for any grouping constraint.

Question 10

1. $\dfrac25$ (correct answer)
2. $\dfrac13$
3. $\dfrac{7}{30}$
4. $\dfrac{4}{15}$

Explanation: When you encounter probability questions involving "or but not both," you're dealing with what's called the symmetric difference - elements that belong to one set or the other, but not their intersection. First, identify the relevant sets among tickets 1-30. Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 (10 tickets). Multiples of 5 are: 5, 10, 15, 20, 25, 30 (6 tickets). Multiples of both 3 and 5 (multiples of 15) are: 15, 30 (2 tickets). Since we want multiples of 3 OR multiples of 5, BUT NOT both, we need: (multiples of 3 only) + (multiples of 5 only). Multiples of 3 only: 10 - 2 = 8 tickets Multiples of 5 only: 6 - 2 = 4 tickets Total favorable outcomes: 8 + 4 = 12 tickets Probability = $\frac{12}{30} = \frac{2}{5}$ Choice B ($\frac{1}{3}$) represents $\frac{10}{30}$, which would be if you only counted multiples of 3. Choice C ($\frac{7}{30}$) doesn't correspond to any logical combination of these sets. Choice D ($\frac{4}{15}$) equals $\frac{8}{30}$, which would be just the multiples of 3 that aren't multiples of 5. The key insight is recognizing that "A or B but not both" means you must subtract the intersection twice - once from each original set. Always identify what you're excluding clearly, as "but not both" is a common source of errors on the GMAT.

Question 11

1. 0
2. 1 (correct answer)
3. 3
4. 7

Explanation: When you encounter an equation where the sum of absolute values equals zero, remember that absolute values are always non-negative. The only way for a sum of non-negative terms to equal zero is if each term individually equals zero. Given $|u-3|+|v+4|=0$, both $|u-3|=0$ and $|v+4|=0$ must be true simultaneously. From $|u-3|=0$, we get $u-3=0$, so $u=3$. From $|v+4|=0$, we get $v+4=0$, so $v=-4$. Therefore, $|u+v|=|3+(-4)|=|-1|=1$. Looking at the wrong answers: Choice A) 0 might tempt you if you incorrectly thought that since the original equation equals zero, the answer should also be zero. Choice C) 3 could trap you if you only solved for $u$ and forgot about $v$, or if you confused $|u+v|$ with just $|u|$. Choice D) 7 might result from incorrectly calculating $|u|+|v|=|3|+|-4|=3+4=7$ instead of $|u+v|$. The key insight is recognizing that when absolute values sum to zero, each component must be zero. This is a fundamental property that appears frequently on the GMAT. Always remember: if $|A|+|B|+...=0$, then each absolute value expression equals zero individually.

Question 12

1. 3 revolutions
2. 12 revolutions
3. 4 revolutions (correct answer)
4. 144 revolutions

Explanation: When you encounter gear problems on the GMAT, you're dealing with cyclical patterns that repeat at regular intervals. The key insight is finding when both gears return to their starting configuration simultaneously. To solve this, you need to determine the least common multiple (LCM) of the number of teeth on each gear. The smaller gear has 36 teeth and the larger has 48 teeth. First, find the prime factorization: 36 = 2² × 3² and 48 = 2⁴ × 3¹. The LCM is 2⁴ × 3² = 144. This means the same teeth will align again after a total of 144 teeth have passed the alignment point on either gear. Since the smaller gear has 36 teeth, it completes $\frac{144}{36} = 4$ full revolutions when this happens. Looking at the wrong answers: Choice A (3 revolutions) would mean only 108 teeth have passed (3 × 36), which isn't enough for both gears to return to their starting positions. Choice B (12 revolutions) represents 432 teeth passing (12 × 36), which is three complete cycles beyond the first realignment. Choice D (144 revolutions) confuses the total number of teeth that must pass (144) with the number of revolutions of the smaller gear. The correct answer is C: 4 revolutions. Strategy tip: In gear problems, always find the LCM of the teeth counts first, then divide by the number of teeth on the gear you're asked about. Don't confuse the LCM itself with the number of revolutions.

Question 13

1. 140 miles
2. 160 miles
3. 168 miles (correct answer)
4. 180 miles

Explanation: When you encounter round-trip problems with different speeds, you're dealing with a classic time-distance-rate scenario where the key insight is that both legs of the journey cover the same distance but take different amounts of time. Let's call the one-way distance $d$ miles. For the trip from A to B at 60 mph, the time is $\frac{d}{60}$ hours. For the return trip at 40 mph, the time is $\frac{d}{40}$ hours. Since the total time is 7 hours: $\frac{d}{60} + \frac{d}{40} = 7$ To solve this, find a common denominator (120): $\frac{2d}{120} + \frac{3d}{120} = 7$ $\frac{5d}{120} = 7$ $\frac{d}{24} = 7$ $d = 168$ miles This confirms answer choice C is correct. Looking at the wrong answers: A) 140 miles would give a total time of $\frac{140}{60} + \frac{140}{40} = 2.33 + 3.5 = 5.83$ hours, which is too short. B) 160 miles yields $\frac{160}{60} + \frac{160}{40} = 2.67 + 4 = 6.67$ hours, still short of 7. D) 180 miles produces $\frac{180}{60} + \frac{180}{40} = 3 + 4.5 = 7.5$ hours, which exceeds the given time. Strategy tip: In round-trip problems with different speeds, always set up your equation using the fact that total time equals the sum of individual leg times. The common mistake is trying to use average speed, which doesn't work when speeds differ.

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. $x\ge4\text{ and }x\le5$
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 16

1. $(x^{2}-4)(x^{2}+1)$
2. $(x-1)(x-1)(x+2)(x+2)$
3. $(x^{2}-2)(x^{2}+2)$
4. $(x-1)(x+1)(x-2)(x+2)$ (correct answer)

Explanation: When you encounter a polynomial like $x^{4}-5x^{2}+4$, you're dealing with a quadratic in disguise. Notice that this expression only has even powers of $x$, which suggests treating $x^2$ as a single variable to simplify the factoring process. Let's substitute $u = x^2$, transforming our expression into $u^2 - 5u + 4$. This quadratic factors as $(u-1)(u-4)$. Substituting back, we get $(x^2-1)(x^2-4)$. Now we can factor further using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$:

• $x^2-1 = (x-1)(x+1)$
• $x^2-4 = (x-2)(x+2)$

Question 17

1. 5
2. 6
3. 7
4. 8 (correct answer)

Explanation: When you encounter polynomial expansion problems where you need to find an unknown coefficient, your approach should be to expand the given expression and match coefficients with the target form. Let's expand $(px+4)(x-2)+(2-x)(x+2)$ step by step. First, expand $(px+4)(x-2)$: $(px+4)(x-2) = px^2 - 2px + 4x - 8$ Next, expand $(2-x)(x+2)$: $(2-x)(x+2) = 2x + 4 - x^2 - 2x = 4 - x^2$ Combining both parts: $px^2 - 2px + 4x - 8 + 4 - x^2$ $= (p-1)x^2 + (-2p+4)x + (-4)$ Since this equals $ax^2 + bx + c$ and we're told $a = 7$, we need the coefficient of $x^2$ to equal 7: $p - 1 = 7$ $p = 8$ Choice (A) 5 would give us $a = 5 - 1 = 4$, not 7. Choice (B) 6 would give us $a = 6 - 1 = 5$. Choice (C) 7 would give us $a = 7 - 1 = 6$. Only choice (D) 8 gives us the required $a = 8 - 1 = 7$. Study tip: In coefficient-matching problems, expand systematically and collect like terms carefully. Always double-check your algebra by substituting your answer back into the original constraint—here, verify that $p = 8$ indeed produces $a = 7$.

Question 18

1. $4x+2$
2. $4x-2$
3. $2+4x$
4. $2-4x$ (correct answer)

Explanation: When you encounter algebraic expressions with multiple binomial products, your goal is to expand each product using FOIL (First, Outer, Inner, Last), then combine like terms carefully. Let's expand each product systematically. For $(x+2)(x-5)$: multiply $x \cdot x = x^2$, then $x \cdot (-5) = -5x$, then $2 \cdot x = 2x$, and finally $2 \cdot (-5) = -10$. This gives us $x^2 - 5x + 2x - 10 = x^2 - 3x - 10$. For $(x-3)(x+4)$: multiply $x \cdot x = x^2$, then $x \cdot 4 = 4x$, then $(-3) \cdot x = -3x$, and finally $(-3) \cdot 4 = -12$. This gives us $x^2 + 4x - 3x - 12 = x^2 + x - 12$. Now substitute back into the original expression: $(x^2 - 3x - 10) - (x^2 + x - 12)$. Distribute the negative sign: $x^2 - 3x - 10 - x^2 - x + 12$. Combining like terms: the $x^2$ terms cancel, $-3x - x = -4x$, and $-10 + 12 = 2$. The result is $2 - 4x$. Choice A gives $4x + 2$ (wrong sign on the $x$ term), choice B gives $4x - 2$ (wrong signs on both terms), and choice C gives $2 + 4x$ (correct constant but wrong sign on $x$). These errors typically occur from sign mistakes during distribution or combining like terms. Remember: when subtracting polynomials, distribute the negative sign to every term in the second polynomial, then combine like terms methodically to avoid sign errors.

Question 19

1. $2^4 \times 3^3 \times 5^3 \times 7^2$
2. $2 \times 3^3 \times 5^3 \times 7^2$ (correct answer)
3. $2^3 \times 3^2 \times 5^3 \times 7^2$
4. $2^5 \times 3^4 \times 5^3 \times 7^2$

Explanation: GCF(m,n) = 2³ × 3² (minimum powers of common factors). LCM(m,n) = 2⁴ × 3⁴ × 5³ × 7² (maximum powers of all factors). Therefore, LCM/GCF = (2⁴ × 3⁴ × 5³ × 7²)/(2³ × 3²) = 2¹ × 3² × 5³ × 7² = 2 × 3³ × 5³ × 7². Choice A uses maximum powers throughout. Choice C incorrectly uses minimum powers for some factors. Choice D adds exponents instead of finding the ratio.

Question 20

1. $\dfrac{7}{11}$ (correct answer)
2. $\dfrac{1}{2}$
3. $\dfrac{4}{11}$
4. $\dfrac{7}{20}$

Explanation: When you encounter a problem asking for the probability that something came from a specific source given that a certain outcome occurred, you're dealing with conditional probability and Bayes' theorem. The key insight is that you need to consider both the likelihood of selecting each bag and the probability of drawing gold from each bag. Let's work through this systematically. First, calculate the probability of drawing a gold coin through each possible path:

• Path 1 (Bag X): P(Bag X) × P(Gold|Bag X) = $\frac{1}{2} \times \frac{4}{10} = \frac{1}{5}$
• Path 2 (Bag Y): P(Bag Y) × P(Gold|Bag Y) = $\frac{1}{2} \times \frac{7}{10} = \frac{7}{20}$

Question 21

1. $18x^{2}y^{2}$
2. $6x^{3}y^{2}$
3. $12x^{2}y^{3}$
4. $6x^{2}y^{2}$ (correct answer)

Explanation: When you see a question asking for the greatest common factor (GCF) of algebraic expressions, you need to find the largest expression that divides evenly into both terms. This means taking the GCF of the numerical coefficients and the lowest power of each variable that appears in both expressions. To find the GCF of $18x^{3}y^{2}$ and $30x^{2}y^{4}$, work with each part separately: For the coefficients: The GCF of 18 and 30 is 6 (since $18 = 2 \times 3^2$ and $30 = 2 \times 3 \times 5$, sharing factors of 2 and 3). For the x terms: You have $x^3$ and $x^2$. The GCF uses the lowest power, which is $x^2$. For the y terms: You have $y^2$ and $y^4$. The GCF uses the lowest power, which is $y^2$. Therefore, the GCF is $6x^{2}y^{2}$, which is choice D. Choice A ($18x^{2}y^{2}$) incorrectly uses 18 instead of 6 as the coefficient. Choice B ($6x^{3}y^{2}$) mistakenly uses $x^3$ (the higher power) instead of $x^2$. Choice C ($12x^{2}y^{3}$) has two errors: using 12 instead of 6 for the coefficient, and $y^3$ instead of $y^2$. Strategy tip: Always take the GCF of coefficients and the minimum power for each variable. A quick check is to verify that your answer divides evenly into both original expressions.

Question 22

1. $(14,42)$
2. $(21,28)$ (correct answer)
3. $(12,49)$
4. $(28,35)$

Explanation: When you encounter problems involving both least common multiple (LCM) and greatest common factor (GCF), remember the fundamental relationship: $\text{LCM}(m,n) \times \text{GCF}(m,n) = m \times n$. This gives you a powerful constraint to check potential answers. Given that $\text{LCM}(m,n) = 84$ and $\text{GCF}(m,n) = 7$, we know that $m \times n = 84 \times 7 = 588$. Additionally, since the GCF is 7, both $m$ and $n$ must be multiples of 7. Let's verify each option by checking if $m \times n = 588$ and confirming the LCM and GCF: For choice B: $(21,28)$. First, $21 \times 28 = 588$ ✓. To find the LCM, factor: $21 = 3 \times 7$ and $28 = 4 \times 7 = 2^2 \times 7$. The LCM takes the highest power of each prime: $2^2 \times 3 \times 7 = 84$ ✓. The GCF is the common factor: $7$ ✓. Choice A: $(14,42)$ gives $14 \times 42 = 588$, but $\text{LCM}(14,42) = 42$ and $\text{GCF}(14,42) = 14$. Choice C: $(12,49)$ gives $12 \times 49 = 588$, but $\text{GCF}(12,49) = 1$ since 12 and 49 share no common factors. Choice D: $(28,35)$ gives $28 \times 35 = 980 \neq 588$. Study tip: Always use the product relationship $\text{LCM} \times \text{GCF} = m \times n$ as your first filter—it quickly eliminates impossible answers before you calculate LCM and GCF directly.

Question 23

1. 0 (correct answer)
2. 2
3. 6
4. 12

Explanation: First solve for x: $3(2x - 5) + 4 = 2(x + 1) - 7$ becomes $6x - 15 + 4 = 2x + 2 - 7$, which simplifies to $6x - 11 = 2x - 5$. Solving: $4x = 6$, so $x = \frac{3}{2}$. Now substitute into $x^2 - 3x + 2$: $\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9}{4} - \frac{18}{4} + \frac{8}{4} = \frac{-1}{4} + \frac{8}{4} = 0$. Choice B results from incorrectly solving the linear equation as $x = 2$. Choice C comes from finding $x = \frac{3}{2}$ correctly but then calculating $x^2 + 3x + 2$ instead. Choice D results from solving incorrectly to get $x = 3$.

Question 24

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

Explanation: First, express everything in terms of base 3: $3^{2x} \cdot (3^2)^{x-1} = (3^3)^{x+1}$. This becomes $3^{2x} \cdot 3^{2(x-1)} = 3^{3(x+1)}$, which simplifies to $3^{2x} \cdot 3^{2x-2} = 3^{3x+3}$. Using the product rule: $3^{2x+2x-2} = 3^{3x+3}$, so $3^{4x-2} = 3^{3x+3}$. Since the bases are equal, the exponents must be equal: $4x-2 = 3x+3$. Solving: $x = 5$. Choice A results from solving $2x = x+3$ (forgetting the exponent rules). Choice C comes from $4x-2 = 2x+3$ (misreading $27$ as $3^2$). Choice D results from $4x-2 = x+11$ (computational error).

Question 25

1. 5 years
2. 7 years
3. 6 years (correct answer)
4. 8 years

Explanation: This is a compound growth problem where you need to find when a population reaches a specific threshold. When you see "grows at X% per year," think exponential growth using the formula $P = P_0(1 + r)^t$, where $P_0$ is the initial population, $r$ is the growth rate, and $t$ is time in years. You want to find when the population first exceeds 150% of its original size, meaning $P > 1.5P_0$. Setting up the inequality: $P_0(1.08)^t > 1.5P_0$. Dividing both sides by $P_0$ gives us $(1.08)^t > 1.5$. Let's test each year systematically:

• Year 5: $(1.08)^5 = 1.469$
• Year 6: $(1.08)^6 = 1.586$