Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2) = 2(15 + 2) = 34 cm
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 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 length, breadth and height of the room be l, b and h respectively As per question Cost of 2(l + b) x h = Rs. 48 ? Required cost = cost of 2 (2l + 2b) x 2h= cost of 4 [2(l + b) x h ]= 4 x Rs. 48= Rs. 192
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