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 the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30%
Cross section area = 1/2 x ( a + b ) x d where a and b are the parallel sides, d is the perpendicular distance between them.? 1/2 x ( a + b ) x d = 640? d = (640 x 2) / 16 = 80m
perimeter = total cost / cost per m = 10080 /20 = 504mside of the square = 504/4 = 126mbreadth of the pavement = 3mside of inner square = 126 - 6 = 120marea of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq mcost of pavement = 246*6*50 = Rs. 73800
Let breadth = b meters. then, length = 3b/2 meters ? b x 3b/2 = 2/3 X 10000? b2 = (4 x 10000)/9? b = ( 2 X 100)/3 m ? Length = (3/2) x (2/3) x 100 m= 100 m
EXAMIANS