UPSC Prelims 2022 · CSAT
Question 55: Basic Numeracy
There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
AnswerUPSC’s answer is (c).
The PIN digits a > b > c come from 1 to 7, and neighbouring digits differ by at least 2. List by first digit: 7 with 5 then 3, 2 or 1 (753, 752, 751); 7 with 4 then 2 or 1 (742, 741); 731; 6 with 4 then 2 or 1 (642, 641); 631; and 531. That is 6 + 3 + 1 = 10 PINs. To be certain of the PIN, one may need to try all 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) 6 counts only the PINs that start with 7.
- (b) 8 misses two of the ten PINs.
- (c) Correct: ten valid PINs, so 10 attempts in the worst case.
- (d) 12 overcounts; only ten PINs meet all the conditions.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.