# Quantitative Aptitude: Mensuration Questions Set 11

1. A room of dimensions 12*8*6 in meters has one door of dimensions 0.5*0.4 in meters and one more window of dimensions 1*0.5 in meter. Find the cost of applying paint on the four walls along with the ceiling of the room but excluding the door and windows at Rs. 40/m^2.
Rs. 13298
Rs. 11267
Rs. 14259
Rs. 13284
Rs. 15250
Option D
Area of four walls along the ceiling = 2*(12+8)*6+12*8
= 336 m^2
Area of all windows = (2*0.5*0.4) + (1*0.5) = 0.9 m^2
Area of door = 2*1.5 = 3 m^2
Area to be painted = 336 â€“ 3.9 = 332.1 m^2
Required cost = 332.1*40 = Rs. 13284

2. The perimeter of a rectangular park is 178 more than the length of the rectangular park. The cost of fencing the boundary of the park at the rate of Rs. 7 per m is Rs. 1708. A path has been made inside the park of width 4 m along its boundary. Find the cost of paving the path with marbles at the rate of Rs. 9 per m^2.
Rs. 7700
Rs. 8208
Rs. 7454
Rs. 8520
Rs. 8000
Option B
Let the breadth be x meter.
Perimeter of the park = 1708/7 = 244 m
Length = 66 m
2*(66+x) = 244
=> x = 56 m
Area of path = 66*56 â€“ 58*48 = 912 m^2
Required cost = 912*9 = Rs. 8208

3. There is a square of side x cm. Another square is made by joining the mid-points of sides of that square and this process is repeated infinitely. If the perimeter of all the squares thus formed is 120(2+(2)^1/2) cm, find the value of x.
50 cm
30 cm
60 cm
40 cm
25 cm
Option B
Perimeter of square1 = 4x cm
Length of the square made by joining the mid-points of square1 = [(x/2)^2 + (x/2)^2]^1/2 = x/(2)^1/2 cm
Perimeter of the square = 4*x/(2)^1/2 cm
Now, 4x+4*x/(2)^1/2 + â€¦ to infinity = 120(2+(2)^1/2)
=> 4*x/(1 â€“ 1/(2)^1/2) = 120(2+(2)^1/2)
=> x = 30 cm

4. A cylindrical container of height 20 cm contains water which is 40% of the total volume of the container. If the longest pencil that can fit the container is of length 29 cm, find the quantity of water present in the cylindrical container. (Take pi = 3)
2646 cm^3
2885 cm^3
2256 cm^3
2770 cm^3
2600 cm^3
Option A
Diameter of the container = [(29)^2 â€“ (20)^2]^1/2
= 21 cm
Volume = 3*10.5*10.5*20 = 6615 cm^3
Amount of water = 40% of 6615 = 2646 cm^3

5. The area of a rectangle gets reduced by 9 square units,if its length is reduced by 5 units and breadth is increased by 3 units.If we increase the length by 3 units and breadth by 2 units, then the area is increased by 67 square units. Find the length and breadth of the rectangle.
15 m , 5 m
17 m , 9 m
11 m , 8 m
10 m , 9 m
17 m , 6 m
Option B
Length = l
lb â€“ (l-5)(b+3) = 9
3x â€“ 5y â€“ 6 = 0 â€”(i) (l+3)(b+2) â€“ lb = 67
2l + 3b -61 = 0 â€”(ii)
solving (i) and (ii)
l = 17 m
b = 9 m

6. The area of a rectangle gets reduced by 9 square units,if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, then the area is increased by 67 square units. Find the area of the rectangle?
150 m
144 m
162 m
149 m
153 m
Option E
L*B â€“ (L-5)(B+3) = 9
3L â€“ 5B â€“ 6 = 0 â€”(i)
(L+3)(B+2) â€“ L*B = 67
2L + 3B -61 = 0 â€”(ii)
solving (i) and (ii)
L = 17 m
B = 9 m
Area of the Rectangle = 153 m

7. The length of a plot is four times its breath. A playground measuring 900 sq. meters occupies one – fourth of the total area of a plot. What is the length of the plot in meter.?
110 m
95 m
80 m
120 m
100 m
Option D
Area of the plot = (4 x 900) mÂ² = 3600 mÂ²
Length = 4y meter
Now area = 4y x y = 3600 mÂ²
=> yÂ² = 900 mÂ²
=> y = 30 m
Length of plot = 4y = 120 m

8. If the length of a rectangle decreases by 5 m and breadth increases by 3 m, then its area reduces 9 sq m. If length and breadth of this rectangle increased by 3 m and 2 m respectively, then its area increased by 67 sq m. What is the length of rectangle?
16 m
13 m
15 m
17 m
19 m
Option D
Let length and breadth of a rectangle be x and y.
(x-5)(y+3) = xy – 9
=> 3x-5y=6 …..(i)
(x+3)(y+2) = xy + 67
=> 2x + 3y = 61 …..(ii)
On solving both the equations, we get
9x – 15y = 18 10x + 15y = 305
=> 19x = 323
=> x = 323/19
Length of rectangle = 17 m

9. The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 500 cm. If the area of the rectangle is less than that of a Square then find the area of the rectangle
1444 cmÂ²
1330 cmÂ²
1250 cmÂ²
1000 cmÂ²
1100 cmÂ²
Option E
Let â€˜aâ€™ be the side of the square.
Perimeter of rectangle = Perimeter of Square = 160
4a = 160
=> a = 40
Area of square = 1600 1600 â€“ lb = 500
Area = 1100 cmÂ²

10. The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semicircle whose diameter is equal to the side of the square ?
35.73 cm
33.15 cm
36.62 cm
38.57 cm
30.11 cm
Option D
Perimeter of square = 2 x Perimeter of rectangle
= 2 * 2 (8+7) = 60 cm.
Side of square = 60/4 = 15 cm = Diameter of semi-circle
Circumference of semi-circle = Ï€d/2 + d
= (22/7) * 2 * 15 + 15 = 38.57 cm