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# Mensuration - MCQ 3

## 20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Mensuration - MCQ 3

Description
This mock test of Mensuration - MCQ 3 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Mensuration - MCQ 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Mensuration - MCQ 3 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Mensuration - MCQ 3 exercise for a better result in the exam. You can find other Mensuration - MCQ 3 extra questions, long questions & short questions for Quant on EduRev as well by searching above.
QUESTION: 1

### A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time at the starting point, when both have started simultaneously from the same point in same direction?

Solution:

Time taken by a deer to complete one round = 9 minutes
Time taken by a rabbit to complete one round = 5 minutes
They meet together for the first time at the starting point = LCM of 9 and 5 = 45 minutes.

QUESTION: 2

### A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time, when both have started simultaneously from the same point in same direction(in min)?

Solution:

Circumference of the track = LCM of 9 and 5 = 45 m.
Ratio of time of deer and rabbit = 9 : 5
Ratio of speed of deer and rabbit = 5 : 9
Relative Speed = 4 m/min
They meet together for the first time at the starting point = 45/4 min

QUESTION: 3

### A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 14 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many centi metre will the water level rise?

Solution:

Volume of the block = 22 * 14 * 7
Radius of the cistern = 14/2 = 7
Volume of the Cylinder = 22/7 * R2 * h
22/7 * R2 * h = 22/7 * 7 * 7 * h
22/7 * 7 * 7 * h = 22 * 14 * 7 ⇒ h = 14

QUESTION: 4

A well with 28 m inside diameter is dug out 18 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embarkment. Find the height of the embarkment.

Solution:

22/7[(R2) – (r2)] * h = 22/7(7*7*18)
[(352) – (72)]h = 14 * 14 * 18
(42*28)h = 14*14*18
h = 3 m

QUESTION: 5

The radii of two cylinders are in the ratio 4:5 and their heights are in the ratio 5:7, What is the ratio of their curved surface areas?

Solution:

2πr1h1/2πr2h2= [4/5 * 5/7] = 4:7

QUESTION: 6

The ratio between the sides of a room is 3:2. The cost of white washing the ceiling of the room at 5 Rs per square metre is Rs. 2500 and the cost of papering the walls at Rs. 2 per square metre is Rs. 960. The height of the room is?

Solution:

Area of Ceiling = Total Cost / Cost of 1 sq. Unit
= 2500/5 = 500
l:b = 2x:3x
l*b = 5x2 = 500
l = 30m & b = 20 m
Area of the 4 wall = 960/2 = 480
Height = 480 / 2(30 + 20) = 4.8m

QUESTION: 7

A park is in the form of a square one of whose sides is 50 m. The area of the park excluding the circular lawn in the centre of the park is 1884 m². The radius of the circular lawn is ?

Solution:

Area of park = 50 x 50 = 2500 m²
Area of circular lawn = Area of park – area of park excluding circular lawn
= 2500 – 1884
= 616
Area of circular lawn = (22/7) x r² = 616 m²
⇒ r² = (616 x 7) / 22
= 28 x 7
= 2 x 2 x 7 x 7
∴ r = 14 m

QUESTION: 8

The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 600 cm. Find the area of the rectangle.

Solution:

Perimeter of rectangle = Perimeter of Square = 160
4a = 160 ⇒ a = 40
Area of square = 1600
1600 – lb = 600
lb = 1000 cm²

QUESTION: 9

The length of a plot is four times its breath. A playground measuring 400 square meters occupies one fourth of the total area of a plot. What is the length of the plot in meter.?

Solution:

Area of the plot = (4 x 400) m²
= 1600 m²
Length = 4y meter
Now area = 4y x y = 1600 m²
⇒ y² = 400 m²
⇒ y = 20 m
∴ Length of plot = 4y =80 m

QUESTION: 10

If the radius of the cone is doubled, keeping the height constant, what is the ratio of the volume of the smaller cone to larger cone?

Solution:

v1/v2 = (r1²) * h1 / (r2²) * h2 (h1=h2)
(r)²/(2r)²
v1/v2 = 1:4

QUESTION: 11

A room has floor size of 15*6sq cm. What is the height of the room , if the sum of the areas of the base and roof is equal to the sum of the areas of the four walls ?

Solution:

lb+lb = lh+hb+lh+hb
2lb = 2h(l+b)
h = lb/l+b
h=15*6/15+6 = 4.29 cm

QUESTION: 12

What is the volume of a right cone whose cross section is isosceles triangle with a base 10cm and slant height 13cm ?

Solution:

Base of the cone(r) =10/2 = 5 cm
Height of the cone (h)= √169-25 =√144 = 12
Volume = 1/3πr2 h
= 22*5*5*12/7*3 = 314.2 Sq cm

QUESTION: 13

Smallest side of a right angled triangle is 6 cm less than the side of a square of perimeter 60 cm. Second largest side of the right angled triangle is 4 cm less than the length of rectangle of area 80 sq. cm and breadth 5 cm. What is the largest side of the right angled triangle?

Solution:

Side of 1st square = 60/4 = 15 cm.
Smallest side of right angled triangle= 15 −6 = 9 cm.
Length of 2nd rectangle = 80/5 = 16 cm.
Second largest side of the 1strectangle = 16−4 = 12 cm.
Largest side = hypotenuse=√92+122=15cm

QUESTION: 14

Circumference of a circle-A is 5/4 times perimeter of a square. Area of the square is 961 sq. cm. What is the area of another circle-B whose diameter is half the radius of the circle-A?

Solution:

The side of the square = a=√961=31 cm.
Perimeter = 4 X 31 = 124 cm The circumference of the circle = 5/4 x 124 = 22/7 x 2 x r
Radius = 155*7/44 = 24.6 cm
Half the radius of Circle-A = 12.3 cm.
Radius of Circle-B = 14/2 = 6.15 cm
The area = 22/7 x 6.15 x 6.15 = 54 cm2

QUESTION: 15

A horse is tethered to a peg with a 16 cm long rope at the corner of a 31 cm long and 27cm wide rectangular grass field. What area of the field will the horse graze?

Solution:

Area = [22*16*16/7] / 4
= 5632/7*4 = 201.1 cm2

QUESTION: 16

The biggest possible circle is inscribed in rectangle of length 10 m and breadth 7 m. Then the area of circle is?

Solution:

Area = 22*3.5*3.5/7 = 269.5/7 = 38.5 m2

QUESTION: 17

In a swimming pool measuring 80 cm x 30 cm, 120 men take a dip. If the average displacement of water by a man is 5 cm cube, What will be the rise in water level ?

Solution:

Total volume displaced by 8 men= 120×5 cm cube
However volume=lxbxh=80x30xh
80x30xh=120×5
h=120*5/80*30 = 600/2400 = 0.25.
so the water level rises by 0.25 cm = ¼ cm

QUESTION: 18

A cylindrical tank of diameter 14 cm is full of water. If 9 litres of water is drawn off, the water level in the tank will drop by

Solution:

1 litres = 1000 cm
9 litres = 9000 cm
Volume of cylinder = πr2h
22/7*7*7*h = 11000
h = 9000*7*4/22*7*7 = 233.8 cm

QUESTION: 19

A circular wire of diameter 42cm is folded in the shape of a rectangle whose sides are in the ratio 6:5. Find the ratio of the enclosed rectangle ?

Solution:

Circumference = 2*22*21/7 = 132cm
L:b = 6x:5x
Perimeter of the rectangle = 2(6x+5x) = 22x
22x = 132
X = 6
Area = 6x * 5x = 30x*x
=30*6*6 = 1080cm2

QUESTION: 20

A cylindrical tub of radius 10 cm contains water up to a depth of 25cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 5cm. What is the radius of the iron ball(approximate) ?

Solution:

Volume of the ball = volume of the raised water
4*22*r3 /3*7 = 22*100*5
r3 = 11000*3*7/88 = 2625
r = 13.79 or 14