Area Problems
The length and breadth of a square are increased by 40% and 30% respectively. The area of a resulting rectangle exceeds the area of the square by?
Let the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
Distance travelled in 1 revolution = circumference of the wheel = ?d= ( 22/7 ) x 63= 198 cmSo the distance travelled in 100 revolutions = 100 x distance travelled in 1 revolution = 100 x 198 cm= 19800 cm = 198 m
Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%
EXAMIANS