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

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

The Correct Option is C

Given \(0.2<\frac{m}{20}<0.5\) 
\(⇒ 4<m<10\)
\(0.2<\frac{n}{11}<0.5\) 
\(⇒ 2.2<n<5.5 ⇒ n=4\)
Since \(0.2<\frac{n}{m}<0.5\) and \(n=4, m=9\)
\(m-2n=9-2×4=1\)
So, the correct answer is (C): \(1\)
 

Answer 2

Given :
\(0.2<\frac{n}{11}<0.5\)
= 22 < n < 5.5
Now, n is an even natural number , therefore value of n = 4
So , \(0.2<\frac{m}{20}<0.5\)
= 4 < m < 10.
So , from the above equation , possible values of m are 5, 6, 7, 8 and 9.
\(0.2<\frac{n}{m}<0.5\)
For this, the only possible of m is 9.
So, m - 2n = (9 - 2×4)
⇒ 9 - 8
= 1
Therefore , the correct option is (C) : 1.