UPSC Prelims 2015 CSAT Paper II · Q72 of 74 Comprehension medium

Passage

Directions for the following 2 (two) items : Read the following passage and answer the 2 (two) items that follow : A, B, C, D, E and F are cousins. No two cousins are of the same age, but all have birthdays on the same day of the same month. The youngest is 17 years old and the oldest E is 22 years old. F is somewhere between B and D in age. A is older than B. C is older than D. A is one year older than C.

Which one of the following is possible ?

  1. (a) D is 20 years old
  2. (b) F is 18 years old ✓ UPSC's answer
  3. (c) F is 19 years old
  4. (d) F is 20 years old

Why the answer is (b)

• The ages of the six cousins are distinct integers ranging from 17 to 22, so the set of ages is {17, 18, 19, 20, 21, 22}.

• E is the oldest, so E = 22. The youngest is 17, so one of A, B, C, D, F is 17.

• A is one year older than C (A = C + 1), and A is older than B, and C is older than D. This implies the age order A > C > D and A > B.

• F is between B and D in age, meaning either B < F < D or D < F < B.

• Let's test the possible values for C. Since A = C + 1 and A ≤ 21 (since E=22 is max), C can be 17, 18, 19, or 20.

• If C = 17, A = 18. Then D < 17 is impossible since 17 is the minimum age. So C cannot be 17.

• If C = 18, A = 19. D < 18, so D must be 17. B < A (19) and B is not 17 (D is 17), so B can be 20 or 21. F is between B and D (17). If B=20, F is 18 or 19. But C=18, so F cannot be 18. If F=19, A=19, conflict. If B=21, F is 18, 19, or 20. C=18, A=19. F cannot be 18 or 19. So F=20. Remaining age for the last person is 21? No, ages used: E=22, A=19, C=18, D=17, F=20, B=21. Check constraints: A(19)>B(21)? No, 19 is not > 21. So this case fails because A must be older than B.

• If C = 19, A = 20. D < 19, so D is 17 or 18. B < 20. F is between B and D.

• Case 1: D = 17. B can be 18, 21, 22? No, E=22. B can be 18 or 21. If B=18, F is between 18 and 17? No integer between. If B=21, F is between 21 and 17, so F is 18, 19, or 20. A=20, C=19. F cannot be 19 or 20. So F=18. Ages: E=22, A=20, C=19, D=17, F=18, B=21. Check: A(20)>B(21)? No. Fails.

• Case 2: D = 18. B < 20. B can be 17, 21. If B=17, F between 17 and 18? No integer. If B=21, F between 21 and 18, so F is 19 or 20. A=20, C=19. F cannot be 19 or 20. Fails.

• If C = 20, A = 21. D < 20, so D is 17, 18, or 19. B < 21. F is between B and D.

• Let's check option (b) F = 18. If F=18, then B and D must be on either side of 18. So one is <18 and one is >18. The only age <18 is 17. So one of B or D is 17. The other is >18 (19, 20, 21, 22). But D < C=20, so D can be 17, 18, 19. B < A=21, so B can be 17, 18, 19, 20.

• If D=17, then B > 18. B can be 19 or 20. F=18 is between 17 and B. This works. Let's assign: E=22, A=21, C=20, D=17, F=18. Remaining age for B is 19. Check: A(21) > B(19) Yes. C(20) > D(17) Yes. F(18) between B(19) and D(17) Yes. All ages distinct: 22, 21, 20, 17, 18, 19. This is a valid arrangement.

• Therefore, F can be 18 years old.

Why the other options are wrong

(a) D is 20 years old
If D is 20, then C must be older than 20, so C is 21 or 22. If C=21, A=22, but E=22, conflict. If C=22, A=23, impossible. Thus D cannot be 20.
(c) F is 19 years old
If F is 19, testing valid age assignments shows that no combination satisfies all constraints (A=C+1, A>B, C>D, F between B and D) without age conflicts or violating the 'older than' conditions.
(d) F is 20 years old
If F is 20, similar to option (c), no valid assignment of the remaining ages to A, B, C, D, E satisfies all the given relational constraints simultaneously.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2015, held on 23 August 2015. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

Reading the answer is not the same as getting it right under a clock. Practise this question with UPSC's negative marking, and anything you miss goes into an error notebook until you get it right twice.

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