UPSC Prelims 2017 · CSAT
Question 51: Basic Numeracy
How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place?
AnswerUPSC’s answer is (c).
Numbers between 99 and 1000 are the three-digit numbers 100 to 999. With 8 fixed in the units place, the hundreds digit can be any of 1 to 9 (9 choices) and the tens digit any of 0 to 9 (10 choices). So there are 9 x 10 = 90 numbers, from 108 up to 998. Count by choices per digit position, taking care that the leading digit cannot be zero.
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) 64 does not fit the count of 9 x 10.
- (b) 80 would be right only if the hundreds digit had 8 choices.
- (c) Correct: 9 x 10 = 90 (108, 118, ... 998).
- (d) 104 is too many; only 90 three-digit numbers end in 8.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2017, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.