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If the length I of a room is reduced by 10% and breadth b is increased by 10%, then find the positive change in its perimeter.

A25(l+b)\frac25 (l + b)

B(l+b)5\frac{(l+b)}{5}

C(l−b)5\frac{(l-b)}{5}

D25(l−b)\frac25 (l - b)

Answer:

(l−b)5\frac{(l-b)}{5}

Read Explanation:

(l−b)5\frac{(l-b)}{5}


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