Solve Two Part Logic - GMAT Data Insights

A pharmaceutical company conducts a two-phase clinical trial. In Phase 1, $60\%$ of participants show positive response to treatment. In Phase 2, only participants who showed positive response in Phase 1 continue, and $75\%$ of these show continued positive response. If $180$ participants started Phase 1, and the company needs at least $90$ 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?
Answer: Current success: $81$ ; additional needed: $20$ Phase 1 participants 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.
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).
Answer: 12 h, 6 h 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.
A class of 40 has boys averaging 90 and girls 70. The class average is 82. Find boys and girls.
Answer: 24 boys, 16 girls 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.
In a two-stage quality control process, products must pass both Stage 1 and Stage 2 to be approved. Let $p$ be the probability that a randomly selected product passes Stage 1, and let $q$ be the probability that a product that has passed Stage 1 also passes Stage 2. If the overall approval rate is $0.36$ and the Stage 1 pass rate is $0.6$ , what is the conditional probability $q$ , and what would be the new overall approval rate if Stage 1 pass rate increased to $0.8$ while $q$ remained constant?
Answer: $q = 0.6$ ; new overall rate = $0.48$ 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.
Positive integers $x$ and $y$ satisfy $x^2-y^2=120$ and $x+y=20$ . Find $x$ and $y$ .
Answer: 13, 7 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.
Process M produces $4y + 3$ units per hour while consuming $2y + 1$ units of raw material per hour. Process N produces $3y + 8$ units per hour while consuming $y + 4$ units of raw material per hour, where $y$ is an efficiency parameter. If the firm has $45$ units of raw material available per hour and needs to produce at least $65$ units per hour, what is the minimum value of $y$ required, and what will be the total hourly production at this minimum value?
Answer: $y = 8$ ; total production = $67$ units per hour 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).
A two-digit number's digits sum to 9; reversing yields $4/7$ of original. Find tens and units digits.
Answer: 6, 3 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.
A software company has two development teams working on separate modules of a project. Team Alpha can complete their module in $d - 3$ days, while Team Beta can complete their module in $d + 2$ days, where $d$ represents the baseline development time. The project manager wants both modules completed within $12$ days, and Team Alpha must finish at least $2$ days before Team Beta to allow for integration testing. What is the maximum value of $d$ that satisfies these constraints, and how many days will Team Beta actually take at this maximum value?
Answer: $d = 10$ ; Team Beta takes $12$ days 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.

Start this quiz