# Quantitative Aptitude: Mensuration Questions Set 7

1. A solid metallic spherical ball of radius 28 cm is melted down and recast into small cones. If the diameter of the base of the cone is 28 cm and the height is 4 cm, find the number of such cones can be made ?
A) 106
B) 118
C) 112
D) 95
E) None
Option C
Solution:

Volume of sphere = (4/3)πr^3
Volume of cone = (1/3)πr^2h
Let the number of cones be ‘X’
=> (4/3) *π *28^3 = (1/3) *π *14^2*4* (X)
=> X = 112
2. The length and the breadth of a rectangular table are increased by 1 m each and due to this the area of the table increased by 27 sq. m. But if the length is increased by 1 m and breadth decreased by 1 m, area is decreased by 7 sq. m. Find the perimeter of the table.
A) 45m

B) 52m
C) 60m
D) 72m
E) None
Option B
Solution:
Let original length = l, breadth = b, so area = lb
When l and b increased by 1:
(l+1)(b+1) = lb + 27
Solve, l + b = 26
When l increased by 1, b decreased by 1:
(l+1)(b-1) = lb – 7
Solve, l – b = 6
Now solve both equations, l = 16, b = 10
Perimeter = 2(16+10)=52m
3. The water in a rectangular tank having a base 80 m by 60 m is 6.5 m deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 20 km per hour?
A) 39hrs

B) 45hrs
C) 60hrs
D) 40hrs
E) None
Option A
Solution:

Volume of water in the tank is 80*60*6.5=31200m^3
Then Volume of water flown in 1hr is 20*1000(in meter)*20/100*20/100(in meter)=800m^3
Time taken=31200/800=39hrs
4. The perimeter of a square is twice the perimeter of a rectangle. If the perimeter of a square is 140cms and the length of the rectangle is 20cm. Find the breadth of the rectangle?
A) 18

B) 20
C) 15
D) 12
E) None
Option C
Solution:

Perimeter of a Square = 4a = 140
a = 140/4 = 35cm
Perimeter of a rectangle = 140/2 = 70cm =2(l+b)
2(20+b) = 70
B = 35-20 = 15
5. A farmer wishes to grow a 100 m2 rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimension of his garden.
A) 20, 5

B) 25, 4
C) 15, 5
D) 10,10
E) None
Option A
Solution:

Area of the garden = 100 m2
⇒ l × b = 100
⇒ b= 100/l
Garden is fenced on three sides.
Length of fencing = 2l + b = 30
⇒ (200/b + b= 30
⇒ b2 – 30b + 200 = 0
⇒ (b – 20)(b – 10) = 0
⇒ b= 20 or 10
⇒ l = 100/20 = 5 or 100/10 = 10
The garden is in the shape of a rectangle.
Therefore, the length and the breadth of the garden are 5 m and 20 m respectively.
6. Inside a square plot a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 28m. What is the area of space left out in the square plot after developing the garden ?
A) 132m2
B) 140m2
C) 168m2
D) 156 m2
E) None
Option C
Solution:

area of space left = (area of square – area of circle)28*28 – (22/7*14*14)
= 784 – 616
= 168 m2
7. A room is 7.5 m long, 5.5 m broad and 5 m high. What will be the expenditure in covering the walls by paper 40 cm broad at the rate of 80 paise per metre ?
A) 255.5
B) 260
C) 282.25
D) 244
E) None
Option B
Solution:

Area of four walls = 2 × 5 (7.5 + 5.5) = 130 m^2
Area of required paper = 130 m^2
Breadth of the paper = 40 cm = 0.4 m
∴ Length of the paper =130/0.4= 325 m
∴ Cost of paper at 80 paise per meter = 325 × 0.80 = Rs.260
8. In measuring the sides of a rectangle, one side is increases by 30%, and the other side is decreased by 15%. What is the change in its area as a percentage ?
A) 7.5
B) 8
C) 10.5
D) 11
E) 12
Option C
Solution:

Let initial area of a rectangle is 100.
Then 100*130/100*85/100=110.5
The change in Diff is 110.5-100=10.5
9. The ratio between three angles of a quadrilateral is 7:11:13 respectively. the value of the fourth angle of the quadrilateral is 112°. what is the difference between the largest and smallest angles of the quadrilateral ?
A) 72°

B) 110°
C) 90°
D) 56°
E) None
Option D
Solution:

Total angles of quadrilateral is 360 °
7x+11x+13x+112=360
=>31x=360-112
=>x=248/31=8
Then 1st angle =7x=7×8=56°
2nd angle=11×8=88°
3rd angle =13×8=104
the largest angle =112°
smallest angle = 56°
difference between largest and smallest angle =112-56=56°
10. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68 m/min. The area of the field is:
A) 30 m^2

B) 40 m^2
C) 50 m^2
D) 60 m^2
E) None
Option D
Solution:

length of the diagonal= PR=52*15/60=13m
Length of its side =PQ+QR= 68*15/60=17m Then x+y=17 and From pythagoras theorem x^2+y^2=169(13^2)

Solving both x=12 and y=5
Area =12*5=60m^2

## 2 Thoughts to “Quantitative Aptitude: Mensuration Questions Set 7”

1. TR!D!P

tq az

2. Premashanthi

area =lb consider area 1/2[(2lb)] = 1/2[(l+b)² – (l²+b²)] = 1/2[17² -169]
=1/2*120 = 60 sq meter