Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%
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%
let original radius = r and new radius = (50/100) r = r/2 original area = ?r 2 and new area = ? r / 2 2 decrease in area = 3 ?r 2 / 4 * 1 ?r 2 *100 = 75%
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
EXAMIANS