UPSC Prelims 2024 · CSAT
Question 49: Basic Numeracy
Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?
AnswerUPSC’s answer is (d).
x > 2y > 3z with x, y, z from 1 to 7. If z = 2, then 2y > 6 means y is at least 4, so x > 8, which is impossible; larger z is worse. So z = 1 and y is at least 2. Also x > 2y with x at most 7 means y is at most 3. For y = 2: x = 5, 6 or 7 (three triplets). For y = 3: x = 7 (one). The triplets are (5,2,1), (6,2,1), (7,2,1) and (7,3,1): four.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) Wrong. There are four triplets, not one.
- (b) Wrong. Two misses other triplets: (5,2,1), (6,2,1), (7,2,1) and (7,3,1) all work.
- (c) Wrong. Three misses (7,3,1).
- (d) Right. Four triplets, all with z = 1.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2024, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.