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Question 59
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
Solution
$$0.2<\frac{n}{11}<0.5$$
=> 2.2<n<5.5
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
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