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Question 12 : 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

Video Explanation

Explanatory Answer

0.2 < $\frac{m}{20}$, $\frac{n}{m}$, $\frac{n}{11}$ < 0.5
$\frac{1}{5}$ < $\frac{m}{20}$, $\frac{n}{m}$, $\frac{n}{11}$ < $\frac{1}{2}$

$\frac{1}{5}$ < $\frac{m}{20}$ < $\frac{1}{2}$
Observe that if m = 4; $\frac{m}{20}$ = $\frac{4}{20}$ = $\frac{1}{5}$
In order for $\frac{m}{20}$to be greater than $\frac{1}{5}$, m > 4
Similarly, observe that if m = 10; $\frac{m}{20}$ = $\frac{10}{20}$ = $\frac{1}{2}$
In order for $\frac{m}{20}$to be lesser than $\frac{1}{2}$, m < 10
So, the possible values that m can take = {5, 6, 7, 8, 9}

$\frac{1}{5}$ < $\frac{n}{11}$ < $\frac{1}{2}$
Observe that if n = 2.2; $\frac{n}{11}$ = $\frac{2.2}{11}$ = $\frac{1}{5}$
In order for $\frac{n}{11}$to be greater than $\frac{1}{5}$, n > 2.2
Similarly, observe that if n = 5.5; $\frac{n}{11}$ = $\frac{5.5}{11}$ = $\frac{1}{2}$
In order for $\frac{n}{11}$to be lesser than $\frac{1}{2}$, n < 5.5
So, the possible values that n can take = {3, 4, 5}
We are told that n is even; n = 4.

If n = 4; $\frac{n}{m}$ will be greater than $\frac{1}{5}$ if m is less than 20.
Similarly at n = 4; $\frac{n}{m}$ will be lesser than $\frac{1}{2}$ if m is greater than 8.
So, for n = 4; 8 ≤ m ≤ 20.
m can only take one of the values in {5, 6, 7, 8, 9}
Hence m = 9.

m - 2n = 9 - 2(4) = 9 - 8 = 1.

The question is "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"

Hence, the answer is, "1"

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