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)52(l+b)B(l+b)5\frac{(l+b)}{5}5(l+b)C(l−b)5\frac{(l-b)}{5}5(l−b)D25(l−b)\frac25 (l - b)52(l−b)Answer: (l−b)5\frac{(l-b)}{5}5(l−b) Read Explanation: (l−b)5\frac{(l-b)}{5}5(l−b) Read more in App