GMAT Quiz: Solve Two Part Logic
8 questions · exam conditions
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Solve Two Part LogicQuestion 1 of 8

Positive integers xx and yy satisfy x2−y2=120x^2-y^2=120 and x+y=20x+y=20. Find xx and yy.

7, 13
12, 8
14, 6
13, 7
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GMAT Quiz

GMAT Quiz: Solve Two Part Logic

Practice Solve Two Part Logic in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Solve Two Part Logic, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

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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.

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Question 1

Positive integers xx and yy satisfy x2−y2=120x^2-y^2=120 and x+y=20x+y=20. Find xx and yy.

  1. 7, 13
  2. 12, 8
  3. 14, 6
  4. 13, 7 (correct answer)
Explanation: Factor x^2 - y^2 as (x + y)(x - y). Since x + y = 20, you have 20(x - y) = 120, so x - y = 6. Add x + y = 20 and x - y = 6 to get 2x = 26, so x = 13 and y = 7. The tempting swap 7, 13 gives x - y = -6, making x^2 - y^2 = -120, not 120.

Question 2

A two-digit number's digits sum to 9; reversing yields 4/74/7 of original. Find tens and units digits.

  1. 3, 6
  2. 6, 3 (correct answer)
  3. 5, 4
  4. 4, 5
Explanation: Let the tens digit be t and the units digit be u. The original number is 10t+u, and its reverse is 10u+t. Since the reverse is 4/7 of the original, 7(10u+t)=4(10t+u), so 70u+7t=40t+4u, giving 66u=33t, or t=2u. With t+u=9, u=3 and t=6. The tempting trap is 3,6, because that makes the reverse 63, which is 7/4 of the original, not 4/7.

Question 3

Machines A and B together finish a job in 4 hours. A alone takes 6 hours longer than B. Find their solo times (A, B).

  1. 6 h, 12 h
  2. 10 h, 4 h
  3. 12 h, 6 h (correct answer)
  4. 8 h, 2 h
Explanation: Set B's solo time as b. Then A's is b + 6. Combined rate: 1/(b + 6) + 1/b = 1/4. Solving gives b^2 - 2b - 24 = 0, so b = 6 and A = 12. The trap is 6 h, 12 h: it gives the same 4-hour combined rate but reverses the condition that A takes 6 hours longer than B.

Question 4

A class of 40 has boys averaging 90 and girls 70. The class average is 82. Find boys and girls.

  1. 16 boys, 24 girls
  2. 20 boys, 20 girls
  3. 30 boys, 10 girls
  4. 24 boys, 16 girls (correct answer)
Explanation: Set boys = b, so girls = 40 - b. Total score is 90b + 70(40 - b) = 40 * 82 = 3280, which gives 20b + 2800 = 3280, so b = 24 and girls = 16. The tempting wrong answer is 16 boys, 24 girls, but that would average 78, and since 82 is closer to 90 than 70, boys must outnumber girls.

Question 5

A pharmaceutical company conducts a two-phase clinical trial. In Phase 1, 60%60\% of participants show positive response to treatment. In Phase 2, only participants who showed positive response in Phase 1 continue, and 75%75\% of these show continued positive response. If 180180 participants started Phase 1, and the company needs at least 9090 participants with positive responses in both phases for regulatory approval, how many additional Phase 1 participants would be needed if the current number falls short, and what is the current number of participants with positive responses in both phases?

  1. Current success: 8181; additional needed: 2020 Phase 1 participants (correct answer)
  2. Current success: 8181; additional needed: 1212 Phase 1 participants
  3. Current success: 108108; additional needed: 00 Phase 1 participants
  4. Current success: 135135; additional needed: 00 Phase 1 participants
Explanation: Phase 1 positive: 180 × 0.6 = 108 participants. Phase 2 positive (from those who were Phase 1 positive): 108 × 0.75 = 81 participants. Since 81 < 90, we need 9 more successful participants. If we add n Phase 1 participants, we get n × 0.6 × 0.75 = 0.45n additional successes. To get 9 more: 0.45n = 9, so n = 20. Choice B miscalculates by using only the Phase 2 success rate (9/0.75 = 12). Choice C incorrectly counts Phase 1 successes as the final number. Choice D incorrectly counts all Phase 1 successes plus Phase 2 successes.

Question 6

In a two-stage quality control process, products must pass both Stage 1 and Stage 2 to be approved. Let pp be the probability that a randomly selected product passes Stage 1, and let qq be the probability that a product that has passed Stage 1 also passes Stage 2. If the overall approval rate is 0.360.36 and the Stage 1 pass rate is 0.60.6, what is the conditional probability qq, and what would be the new overall approval rate if Stage 1 pass rate increased to 0.80.8 while qq remained constant?

  1. q=0.6q = 0.6; new overall rate = 0.480.48 (correct answer)
  2. q=0.6q = 0.6; new overall rate = 0.520.52
  3. q=0.45q = 0.45; new overall rate = 0.480.48
  4. q=0.45q = 0.45; new overall rate = 0.360.36
Explanation: The overall approval rate equals P(Stage 1) × P(Stage 2|Stage 1) = p × q. Given: 0.6 × q = 0.36, so q = 0.36/0.6 = 0.6. With the new Stage 1 rate of 0.8 and q remaining 0.6, the new overall rate = 0.8 × 0.6 = 0.48. Choice B incorrectly adds the improvements rather than multiplying. Choice C miscalculates q as 0.36 - 0.6 = -0.24, then uses |−0.24| = 0.24, but realizes this is wrong and guesses 0.45. Choice D assumes the overall rate cannot change if only one stage changes.

Question 7

A software company has two development teams working on separate modules of a project. Team Alpha can complete their module in d−3d - 3 days, while Team Beta can complete their module in d+2d + 2 days, where dd represents the baseline development time. The project manager wants both modules completed within 1212 days, and Team Alpha must finish at least 22 days before Team Beta to allow for integration testing. What is the maximum value of dd that satisfies these constraints, and how many days will Team Beta actually take at this maximum value?

  1. d=15d = 15; Team Beta takes 1717 days
  2. d=13d = 13; Team Beta takes 1515 days
  3. d=12d = 12; Team Beta takes 1414 days
  4. d=10d = 10; Team Beta takes 1212 days (correct answer)
Explanation: We have three constraints: (1) Team Alpha finishes within 12 days: d - 3 ≤ 12, so d ≤ 15. (2) Team Beta finishes within 12 days: d + 2 ≤ 12, so d ≤ 10. (3) Alpha finishes at least 2 days before Beta: (d + 2) - (d - 3) ≥ 2, which gives 5 ≥ 2 (always satisfied). The binding constraint is d ≤ 10, so maximum d = 10. At d = 10: Team Alpha takes 7 days, Team Beta takes 12 days. Choice A violates constraint (2). Choice B violates constraint (2). Choice C violates constraint (2). Only choice D satisfies all constraints.

Question 8

A consulting firm analyzes the efficiency of two manufacturing processes, Process M and Process N, across different operational scenarios.

Process M produces 4y+34y + 3 units per hour while consuming 2y+12y + 1 units of raw material per hour. Process N produces 3y+83y + 8 units per hour while consuming y+4y + 4 units of raw material per hour, where yy is an efficiency parameter. If the firm has 4545 units of raw material available per hour and needs to produce at least 6565 units per hour, what is the minimum value of yy required, and what will be the total hourly production at this minimum value?

  1. y=8y = 8; total production = 6767 units per hour (correct answer)
  2. y=6y = 6; total production = 5959 units per hour
  3. y=7y = 7; total production = 6262 units per hour
  4. y=9y = 9; total production = 7171 units per hour
Explanation: Both processes run simultaneously. Total production = (4y + 3) + (3y + 8) = 7y + 11. Total material consumption = (2y + 1) + (y + 4) = 3y + 5. Constraints: (1) Material limit: 3y + 5 ≤ 45, so 3y ≤ 40, thus y ≤ 13.33. (2) Production requirement: 7y + 11 ≥ 65, so 7y ≥ 54, thus y ≥ 7.71. Since y must be an integer, minimum y = 8. At y = 8: production = 7(8) + 11 = 67, material use = 3(8) + 5 = 29 ≤ 45 ✓. Choice B has y = 6 < 7.71 (insufficient production). Choice C has y = 7 < 7.71 (insufficient production). Choice D has y = 9 > 8 (not minimum).