Let $m$ and $n$ be natural numbers such that $n$ is even and $0.2<\frac{\mathrm{m}}{20}, \frac{\mathrm{n}}{\mathrm{m}}, \frac{\mathrm{n}}{11}<0.5$. Then $m-2n$ equals

We have three inequalities to work with:

• $0.2 < \frac{m}{20} < 0.5$
• $0.2 < \frac{n}{m} < 0.5$
• $0.2 < \frac{n}{11} < 0.5$

Plus the constraints: $m, n$ are natural numbers and $n$ is even.

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From $0.2 < \frac{n}{11} < 0.5$

Multiply all parts by $11$:

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

$2.2 < n < 5.5$

Since $n$ is a natural number: $n \in \{3, 4, 5\}$

Since $n$ must be even: $n = 4$

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From $0.2 < \frac{m}{20} < 0.5$

Multiply all parts by $20$:

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

$4 < m < 10$

So $m \in \{5, 6, 7, 8, 9\}$

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We need $0.2 < \frac{n}{m} < 0.5$ where $n = 4$

So: $0.2 < \frac{4}{m} < 0.5$

For the right inequality:

$\frac{4}{m} < 0.5$

Cross multiply: $4 < 0.5m$

Therefore: $m > 8$

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Combining constraints: $4 < m < 10$ and $m > 8$

This gives us: $8 < m < 10$

Since $m$ is a natural number: $m = 9$

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

Answer: $1$