CAT 2020Slot 3QAQuestion & Solution
Number SystemsMedium
Question
Let m and n be natural numbers such that n is even and $0.2<\frac{m}{20},\frac{n}{m},\frac{n}{11}<0.5$. Then $m-2n$ equals
Options
3
1
2
4
Solution
$0.2<\frac{n}{11}<0.5$
=> 2.2
Since n is an even natural number, the value of n = 4
$0.2<\frac{m}{20}<0.5$ => 4< m<10. Possible values of m = 5,6,7,8,9
Since $0.2<\frac{n}{m}<0.5$, the only possible value of m is 9
Hence m-2n = 9-8 = 1