speed = 12 km/h =
distance covered =
time taken = distance /speed =
Let breadth = x metres, length = 3x metres, height = H metres.
Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m
So, breadth = 6 m and length = = 9 m.
Now, papered area = (1720/10) = 172 sq.m
Area of 1 door and 2 windows = 8 sq.m
Total area of 4 walls = (172 + 8) sq.m = 180 sq.m
From statement (A),
20b = (1/2) × b × h
h = 40 cm.
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) =
- 5s - 3s - 15 =
2s = 15
Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2)
= 2(15 + 2) = 34 cm
Area of square = 40 x 40
= 1600 sq.cm
Given that the areas of Square and Rectangle are equal
=> Area of rectangle = 1600 Sq.cm
We know that, Area of rectangle = L x B
Given L = 64 cm
Breadth of rectangle = 1600/64 = 25 cm
Perimeter of the rectangle = 2(L + B) = 2(64+25) = 178 cm.
Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m
Cost of painting = Rs. 158 x 3 = Rs. 474
area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m
cost of graveling = 344 x (75/100) = Rs. 258
perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800
Area = 5.5 × 3.75 sq. metre.
Cost for 1 sq. metre. = Rs. 800
Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m
Let inner radius be r metres. Then, 2 r = 440 ; r = = 70 m.
Radius of outer circle = (70 + 14) m = 84 m.
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