CAT 2020 Exam Shift 3 Solved Paper

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Question : 76 of 76

Marks: +1, -0

Let

{m}

and

{n}

be natural numbers such that

{n}

is even and

0.2<\;{m}/{20}, \;{n}/{m}, \;{n}/{11}<0.5

. Then

m-2 n

equals

3
4
1
2

Solution:

Given

0.2 < \;{m}/{20} < 0.5 ⇒ 4 < m < 10

0.2<\;{n}/{11} < 0.5 ⇒ 2.2 < n < 5.5 ⇒ n=4

since

0.2 < \;{n}/{m}<0.5

and

n=4, m=9

m-2 n=9-2 × 4=1

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