# Quantitative Aptitude: Mensuration Set 1

Mensuration Questions for Banking, Insurance and SSC Exams

1. What will be the area of trapezium whose parallel sides are 22 cm and 16 cm long, and the distance between them is 11 cm?
A) 209 cm2
B) 282 cm2
C) 265 cm2
D) 179 cm2
E) 302 cm2
Option A
Solution:

Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (22 + 16) * (11) = 209 cm2
2. The perimeter of a rectangle is 42 m. If the area of the square formed on the diagonal of the rectangle as its side is 1 1/12 % more than the area of the rectangle, find the longer side of the rectangle.
A) 19 m
B) 16 m
C) 9 m
D) 5 m
E) 12 m
Option E
Solution:

Let the sides of the rectangle be l and b respectively.
From the given data,
âˆš(l2 + b2) = (1 + 1 1/12) lb
=> l2 + b2 = (1 + 13/12) lb = 25/12 * lb
12(l2 + b2 ) = 25 lb
Adding 24 lb on both sides
12 l2 + 12b2 + 24lb = 25 lb
12(l2 + b2 + 2lb) = 49 lb
12(l + b)2 = 49lb
but 2(l + b) = 42 => l + b = 21
So 12(21)2 = 49lb
Solve, we get lb = 108
Since l + b = 21, longer side = 12 m
3. At the rate of Rs. 2 per sq m, cost of painting a rectangular floor is Rs 5760. If the length of the floor is 80% more than its breadth, then what is the length of the floor?
A) 25 m
B) 72 m
C) 67 m
D) 56 m
E) 46 m
Option B
Solution:

Let the length and the breadth of the floor be l m and b m respectively.
l = b + 80% of b = l + 0.8 b = 1.8b
Area of the floor = 5760/2 = 2880 sq m
l*b = 2880 i.e., l * l/1.8 = 2880
l = 72
4. A 7 m wide path is to be made around a circular garden having a diameter of 7 m. What will be the area of the path in square metre?
A) 298
B) 256
C) 308
D) 365
E) 387
Option C
Solution:

Area of the path = Area of the outer circle – Area of the inner circle = Ï€{7/2 + 7}2 – Ï€ [7/2]2
= 308 sq m
5. The perimeter of a rectangle of length 62 cm and breadth 50 cm is four times perimeter of a square. What will be the circumference of a semicircle whose diameter is equal to the side of the given square?
A) 36 cm
B) 25 cm
C) 29 cm
D) 17 cm
E) 16 cm
Option B
Solution:

Let the side of the square be a cm.
Parameter of the rectangle = 2(62 + 50) = 224 cm Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(Ï€)(r) + d
= 1/2(22/7)(7) + 14 = 25 cm
6. What is the volume of a cylinder whose curved surface area is 1408 cm2 and height is 16 cm?
A) 7715 cm3
B) 9340 cm3
C) 8722 cm3
D) 7346 cm3
E) 9856 cm3
Option E
Solution:

2Ï€rh = 1408, h = 16
Solve both, so r = 14
Volume = Ï€ r2h = (22/7) * 14 * 14 * 16 = 9856
7. A cone with diameter of its base as 30 cm is formed by melting a spherical ball of diameter 10 cm. What is the approximate height of the cone?
A) 6 cm
B) 3 cm
C) 2 m
D) 5 cm
E) None of these
Option C
Solution:

Radius of cone = 30/2 = 15, radius of ball = 10/2 = 5
Volumes will be equal, so
(1/3) Ï€ r2h = (4/3) Ï€ R3
152h = 4* 53
So h = 2.2
8. A cylinder whose base of circumference is 6 m can roll at a rate of 3 rounds per second. How much distance will the cylinder cover in 9 seconds?
A) 125 m
B) 162 m
C) 149 m
D) 173 m
E) 157 m
Option B
Solution:

Distance covered in one round = 2 x Ï€ x r = 6 m
Distance covered in 1 second = 3 x 6 = 18 m
So distance covered in 9 seconds = 18Ã—9= 162m
9. A container is formed by surmounting a hemisphere on a right circular cylinder of same radius as that of hemisphere. If the volume of the container is 576Ï€ m3 and radius of cylinder is 6 m, then find the height of the container.
A) 14 m
B) 12 m
C) 20 m
D) 18 m
E) 22 m
Option D
Solution:

Volume of the container = Volume of the cylinder + Volume of the hemisphere
Volume of the container = Ï€ 62h + (2/3) Ï€ 63 = 576Ï€
= Ï€ 36 (h + 4) = 576Ï€
Solving we get h = 12
So the height of the container = 12 + 6 = 18 m
10. The radii of two cylinders are in the ratio 3 : 2 and their curved surface areas are in the ratio 3 : 5. What is the ratio of their volumes?
A) 8 : 11
B) 5 : 9
C) 7 : 4
D) 9 : 10
E) 13 : 7
Option D
Solution:

r1/r2 = 3/2 or r1 = 3/2 * r2
CSA1/CSA2 = 2Ï€r1h1/2Ï€r2h2 = 3/5
So h1/h2 = 2/5
Volume1/ Volume2 = Ï€r12h1/ Ï€r22h2 = 9/10

## 15 Thoughts to “Quantitative Aptitude: Mensuration Set 1”

1. Preeti

mam, ek doubt h

2. Preeti

A,B and C can complete a work in 40% lesser time than C while B and C can complete the same work in 60% lesser time than A . if A, B and C together can complete the work in 20 days . then in how many days they alone do the work ?

1. A, B and C in 40% lesser time than C. So C in 40/(100-40) * 100 = 200/3% more time than A, B and C together.
So C in (1 + 200/3)% of 20 = 100/3 days.

1/A + 1/B + 3/100 = 1/20
So 1/A + 1/B = 1/50

Now
B and C can complete the same work in 60% lesser time than A:
60% lesser than A means 40/100 of A or 2A/5
1/B + 3/100 = 5/2A.
Now solve these 2 equations in A and B.
A 70 days. B 175 days. C 100/3 days

3. Smarty

Distance covered in one round = 2 x Ï€ x r = 6 m
Distance covered in 1 second = 3 x 6 = 18 m
So distance covered in 9 seconds = 18/9 = 2m

if in 1 sec 18 m….then it should be more in 9 sec

somthing is wrong

1. Smarty

qstn no 7

2. Ohh yes. 18Ã—9

3. po aspirant

ohhhk

1. Smarty

arey ye Purani ID hai BA pe block hai….isko b delete krta hoon haha …..distract kr deti ho yar ðŸ˜€

1. po aspirant

haha,,ok padho srry distraction k liy

1. Smarty

hmm aap bhi pdho acche see, no sorry ðŸ™‚
bye tc:)

2. po aspirant

ans wat i asked ..pls if u kno…

3. Mr X

muje nhi pta tha

4. po aspirant

no prob…himanshu has replied

4. Nishu

thanks

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