UPSC Prelims 2024 · CSAT
Question 6: Basic Numeracy
In the expression 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1, ∗ is chosen from +, –, × each at most two times. What is the smallest non-negative value of the expression?
AnswerUPSC’s answer is (d).
Four operators sit between five numbers, and each of +, −, × may be used at most twice. Because the value cannot go below zero by requirement, check whether 0 can be reached: 5 − 4 − 3 + 2 × 1 = 5 − 4 − 3 + 2 = 0, which uses − twice, + once and × once. All limits are met, so 0 is attainable and is the smallest non-negative value. Remember that multiplication is done first.
Why each option is right or wrong
- (a) Wrong. 3 is larger than 0, which is attainable.
- (b) Wrong. 2 is reachable (for example 5 + 4 − 3 × 2 − 1) but 0 is smaller.
- (c) Wrong. 1 is larger than 0, which is attainable.
- (d) Right. 5 − 4 − 3 + 2 × 1 = 0, within the usage limits.
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.