UPSC NDA 2 2019 Mathematics Solved Paper

Concept:Two-digit numbers divisible by 4 form an arithmetic progression (AP).The $n$th term of an AP is $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.Explanation:The smallest two-digit number divisible by 4 is 12.The largest such number is 96.So the sequence is 12, 16, 20, ..., 96.Here $a = 12$, $d = 4$, and $a_n = 96$.Using the formula: $12 + (n-1) \times 4 = 96$.Subtract 12: $(n-1) \times 4 = 84$.Divide by 4: $n-1 = 21$.Thus $n = 22$.Therefore, there are 22 two-digit numbers divisible by 4.Answer:22 (Option B).