UPSC Prelims 2019 · CSAT
Question 73: Basic Numeracy
How many triplets (x, y, z) satisfy the equation x + y + z = 6, where x, y and z are natural numbers ?
AnswerUPSC’s answer is (d).
Natural numbers start at 1. Put x = 1 + a, y = 1 + b and z = 1 + c with a, b, c at least 0, so a + b + c = 3. The number of such solutions is C(5, 2) = 10. Listing confirms it: 1, 1, 4 in 3 orders, 1, 2, 3 in 6 orders and 2, 2, 2 in 1 order, giving 3 + 6 + 1 = 10.
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: undercounts the 10 ordered triplets.
- (b) Wrong: 5 ignores the different orders.
- (c) Wrong: 9 misses one triplet.
- (d) Right: 3 + 6 + 1 = 10.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.