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

Let \( m \) and \( n \) be natural numbers. We know: \( n \) is even, \( 0.2 < \frac{m}{20} \), \( \frac{n}{m} \), and \( \frac{n}{11} < 0.5 \). We need to find \( m - 2n \). 

First, consider \( 0.2 < \frac{m}{20} \). Multiplying both sides by 20 gives \( 4 < m \). Since \( m \) is a natural number, \( m \geq 5 \).

Next, analyze \( \frac{n}{11} < 0.5 \). Multiplying by 11, we get \( n < 5.5 \). Since \( n \) is a natural even number, \( n = 2, 4 \).

For \( n = 2 \), check \( \frac{n}{m} < 1 \) implies \( 2 < m \), already true since \( m \geq 5 \).

For \( n = 4 \), check \( \frac{n}{m} < 1 \) implies \( 4 < m \), also within our bounds.

Test pairs \((m,n)\):

\((m,n)\)	\(m-2n\)
\((5,2)\)	1
\((6,2)\)	2
\((7,2)\)	3
\((8,4)\)	0
\((9,4)\)	1

Among the options, only \( m-2n = 1 \) is given, which occurs for both pairs \((5,2)\) and \((9,4)\) consistent with all conditions. Thus, \( m-2n = 1 \).

Answer 2

Given: 

$0.2 < \dfrac{n}{11} < 0.5$

Multiplying all parts by 11:

$0.2 \times 11 < n < 0.5 \times 11$

$2.2 < n < 5.5$

Since $n$ is an even natural number, the only possible value of $n$ in this range is:

$n = 4$

Now consider:

$0.2 < \dfrac{m}{20} < 0.5$

Multiplying all parts by 20:

$0.2 \times 20 < m < 0.5 \times 20$

$4 < m < 10$

So, the possible integer values of $m$ are: $5, 6, 7, 8, 9$

Next, we are given:

$0.2 < \dfrac{n}{m} < 0.5$

Substitute $n = 4$ and test for each value of $m$:

• $\dfrac{4}{5} = 0.8$ ❌
• $\dfrac{4}{6} \approx 0.667$ ❌
• $\dfrac{4}{7} \approx 0.571$ ❌
• $\dfrac{4}{8} = 0.5$ ❌ (not strictly less than 0.5)
• $\dfrac{4}{9} \approx 0.444$ ✅ (satisfies the condition)

So, only valid value of $m$ is $9$.

Now compute: $m - 2n = 9 - 2 \times 4 = 9 - 8 = \boxed{1}$

Correct option: (C): $\boxed{1}$