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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

- 4
- 2
- 1
- 3

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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" **

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