Let original length = x metres and original breadth = y metres. Original area = xy sq.m Increased length = 120 100 and Increased breadth = 120 100 New area = 120 100 x * 120 100 y = 36 25 x y m 2 The difference between the Original area and New area is: 36 25 x y - x y 11 25 x y Increase % = 11 25 x y x y * 100 = 44%
Let length = L and breadth = BLet , New breadth = ZThen, New length = ( 160 / 100) L.= 8L / 5? 8L / 5 x Z = LBor Z = 5B/8Decrease in breadth = (B-5B/8)= 3B/8? Decrease in percent = (3B/8 x1/B ) x 100 %= 371/2%
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
EXAMIANS