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%
Length = (40 x 10 ) dm = 400 dm.Breadth = (15 x 10 ) dm = 150 dm.Area of veranda = (400 x 150 ) dm2Area of one stone = (6 x 5 ) dm2? Required number of stones = (400 x 150) /(6 x 5) = 2000
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3
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
Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2) = 2(15 + 2) = 34 cm
EXAMIANS