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%
speed = 12 km/h = 12 × 5 18 = 10 3 m / s distance covered = 20 × 2 × 22 7 × 50 = 44000 7 m time taken = distance /speed = 44000 7 × 3 10 s e c = 4400 × 3 7 × 1 60 m i n = 220 7 m i n
Speed = 12 x (5/18) m/sec =10/3 m/sec there4; perimeter = (10/3) x 15 x 60 m=3000 m? 2( a + 4a) = 3000 m? a = 300 mSo, length = 1200 m and breadth = 300 m ? Area = (1200 x 300 ) m2 = 360000m2
Original area = (22/7) x 9 x 9 cm2New area = (22/7) x 7 x 7 cm2? Decrease = 22/7 x [(9)2 -(7)2] cm2=(22/7) x 16 x 2 cm2Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %= 39.5 %
Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%
EXAMIANS