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

The correct option is 1.

To find the value of $m\&\#x2212;2n$, we need to determine the natural numbers $m$ and $n$ that satisfy the given conditions.
We are given that $n$ is an even natural number and that the following three inequalities hold simultaneously:
$0.2\&lt;\frac{m}{20}\&lt;0.5$
$0.2\&lt;\frac{n}{11}\&lt;0.5$
$0.2\&lt;\frac{n}{m}\&lt;0.5$

Step 1: Solve for the range of $m$ from the first inequality
Multiplying the inequality $0.2\&lt;\frac{m}{20}\&lt;0.5$ by $20$, we get:
$20\&\#x00D7;0.2\&lt;m\&lt;20\&\#x00D7;0.5$
$4\&lt;m\&lt;10$
Since $m$ is a natural number, $m$ must belong to the set:
$m\&\#x2208;\{5,6,7,8,9\}$

Step 2: Solve for $n$ from the third inequality and given conditions
Multiplying the inequality $0.2\&lt;\frac{n}{11}\&lt;0.5$ by $11$, we get:
$11\&\#x00D7;0.2\&lt;n\&lt;11\&\#x00D7;0.5$
$2.2\&lt;n\&lt;5.5$
Since $n$ is given to be an even natural number, the only even integer in this range is $4$.
Therefore, we have:
$n=4$

Step 3: Solve for $m$ using the middle inequality
Substituting $n=4$ into $0.2\&lt;\frac{n}{m}\&lt;0.5$, we get:
$0.2\&lt;\frac{4}{m}\&lt;0.5$
Taking the reciprocal of all terms and reversing the inequalities yields:
$\frac{1}{0.5}\&lt;\frac{m}{4}\&lt;\frac{1}{0.2}$
Simplifying the fractions:
$2\&lt;\frac{m}{4}\&lt;5$
Multiplying by $4$:
$8\&lt;m\&lt;20$

Step 4: Find the unique value for $m$ and calculate the expression
Combining the constraints from Step 1 ($4\&lt;m\&lt;10$) and Step 3 ($8\&lt;m\&lt;20$), we get the intersection:
$8\&lt;m\&lt;10$
Since $m$ is a natural number, the only integer satisfying this condition is:
$m=9$
Finally, we compute $m\&\#x2212;2n$:
$m\&\#x2212;2n=9\&\#x2212;2\left(4\right)=9\&\#x2212;8=1$