Area Problems
The length of a rectangle is twice its breadth. If its length is decreased half of the 10 cm and the breadth is increased by half of the 10 cm cm, the area of the rectangle is increased by 5 sq cm, more than 70 sq cm. Find the length of the rectangle.
Given that, l = 2b [Here l = length and b = breadth]Decrease in length = Half of the 10 cm = 10/2 = 5 cmIncrease in breadth = Half of the 10 cm = 10/2 = 5 cm Increase in the area = (70 + 5) = 75 sq cm According to the question, (l - 5) (b + 5) = lb + 75 ? (2b - 5) (b + 5) = 2b2 + 75 [since l = 2b]? 5b - 25 = 75 ? 5b = 100? b = 100/ 5 = 20? l = 2b = 2 x 20 = 40 cm
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 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%
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
EXAMIANS