Area Problems
The dimensions of a room are 10m x 7m x 5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m x 3m. One window is of size 2m x 1.5m and the other 2 windows are of size 1m x 1.5m. The cost of painting the walls at Rs. 3 per sq m is
Area of 4 walls = 2(l+b)h=2(10+7) x 5 = 170 sq mArea of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq marea to be planted = 170 -12 = 158 sq m Cost of painting = Rs. 158 x 3 = Rs. 474
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 = 5y meters and breadth = 3y meters.Then, perimeter = 2 x (5y + 3y ) m = 16y meters ...(i)But perimeter = Total Cost / Rate = 3000 / 7.50 m = 400 m ...(ii)from eqs. (i) and (ii)16y = 400? y = 25? Length - Breadth = (5 x 25 - 3 x 25 ) m = 2 x 25 m = 50 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%
let the side of the square be x meterslength of two sides = 2x metersdiagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0 . 59 x 2 x * 100 % = 30% (approx)
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