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Test Level 1: Mensuration - 2 - CAT MCQ


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20 Questions MCQ Test Level-wise Tests for CAT - Test Level 1: Mensuration - 2

Test Level 1: Mensuration - 2 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 1: Mensuration - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Mensuration - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Mensuration - 2 below.
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Test Level 1: Mensuration - 2 - Question 1

A right circular cone of radius K cm and height 24 cm rests on the base of a right circular cylinder of radius K cm (their bases lie in the same plane, as shown in the figure). The cylinder is filled with water to a height of 24 cm. If the cone is then removed, to which height will the water fall?

Detailed Solution for Test Level 1: Mensuration - 2 - Question 1

Height of water in the cylinder before the cone is taken out = 24 cm
Volume of water in the cylinder before the cone is taken out = Volume of cone + Volume of water
Let H be the height of the water level after the cone is taken out.




H = 16 cm

Test Level 1: Mensuration - 2 - Question 2

The curved surface areas of two spheres are in the ratio of 1 : 4. Find the ratio of their volumes.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 2

Let the two spheres be A and B.

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Test Level 1: Mensuration - 2 - Question 3

The ratio of the volume of a sphere to its surface area is  

Detailed Solution for Test Level 1: Mensuration - 2 - Question 3

Volume of a sphere

Surface area = 4πr2 

Ratio of the volume to the surface area = r : 3

Test Level 1: Mensuration - 2 - Question 4

A right-circular cone is carved out of a cube having the same height and radius, equal to half the side of the cube. What is the ratio of the volume of the carved cone to that of the cube left after carving?

Detailed Solution for Test Level 1: Mensuration - 2 - Question 4

Given height and diameter:

Test Level 1: Mensuration - 2 - Question 5

The sector of a circle with radius 10 cm and angle 120° is rolled up such that the two bounding radii are joined together to form a cone. Find the volume of the cone.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 5


Length of arc AXB = Circumference of the base of the cone

= 20.95 cm
Let the radius of the base be r.
So, 2πr = 20.95

44r = 146.65 
r = 146.65/44
r = 3.33 cm
Now, to find out height we can use Pythagorean theorem.
Let the height of the cone be h.
So, h2 = R2 - r2
h2 = 100 - 11.08

h = 9.42 cm
Now, volume of the cone 


= 109.3 cm3

Test Level 1: Mensuration - 2 - Question 6

A sphere is melted and recast into smaller right cones of height 7 centimeters and radius 7 centimeters. If the radius of the sphere is 7 meters, then how many cones can be made?

Detailed Solution for Test Level 1: Mensuration - 2 - Question 6

Radius of the sphere, r = 7 meters
Radius of the cone, r2 = 7 centimeters = 0.07 meters
Height of the cone, h = 7 centimeters = 0.07 meters


Solving,
Number of cones = 4 × 106

Test Level 1: Mensuration - 2 - Question 7

Water flows at a rate of 10 m per minute from a cylindrical pipe, 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter of the base is 40 cm and depth is 24 cm?

Detailed Solution for Test Level 1: Mensuration - 2 - Question 7

Volume of the conical vessel = (1/3) × π × (20)2 × 24 = 3200π cm3
Volume flown in 1 minute = (π × (2.5/10)2 × 1000) = 62.5π cm3
Therefore, time taken = 3200π/62.5π = 51 min 12 sec

Test Level 1: Mensuration - 2 - Question 8

A wall 8 m long, 6 m high and 22.5 cm thick is made up of bricks, each measuring 25 cm × 11.25 cm × 6 cm. Find the number of bricks required.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 8

Number of bricks required

= 6400

Test Level 1: Mensuration - 2 - Question 9

A piece of metal pipe is 7 cm long with inner diameter of the cross section as 4 cm. If the outer diameter is 4.5 cm and the metal weighs 8 gm/cu cm, the weight of the pipe is

Detailed Solution for Test Level 1: Mensuration - 2 - Question 9

Outer diameter = do = 4.5 cm 
Inner diameter = di = 4 cm 


Weight of the pipe  

= 187 gm

Test Level 1: Mensuration - 2 - Question 10

A building is in the form of a cuboid mounted by a semi-circular portion as shown. The inner measurements of the cuboid are 10 m × 7 m × 3 m and the radius of semi-circular portion is 3.5 m. Find the volume of the building.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 10

Radius of semi-circle = 3.5 m
Volume of the building = Volume of cubical portion + Volume of vertical cross-section

Test Level 1: Mensuration - 2 - Question 11

A toy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The base radius of the cone is 3.5 m and its volume is (2/3)rd of the volume of the hemisphere. Calculate its height.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 11

Radius of the hemisphere = Base radius of the cone = 3.5 m

Let the height of the cone be h metres.

⇒  h = 4.67 m

Test Level 1: Mensuration - 2 - Question 12

A sphere of diameter 18 cm made up of copper is melted and converted into a wire of diameter 4 mm. Find the length of the wire (in metres).

Detailed Solution for Test Level 1: Mensuration - 2 - Question 12

Volume of the sphere = Volume of the wire (cylinder)

⇒ h = 24,300 cm = 243 metres

Test Level 1: Mensuration - 2 - Question 13

A solid metal sphere is melted and smaller spheres, all with the same radius, are formed. 20% of the material is lost in this process. The radius of each smaller sphere is 1/8 the radius of the original sphere. If 20 litres of paint was needed to paint the original sphere, then how many litres of paint would be required to paint all the smaller spheres?

Detailed Solution for Test Level 1: Mensuration - 2 - Question 13

Let the radius of the original sphere be R and that of each smaller one be r.
⇒ r = R/8
Then, volumes of the larger and smaller spheres are 


Number of smaller spheres formed

Surface area of original sphere = 4πR2 
Surface area of each smaller sphere 

Surface area of all smaller spheres = Number of small spheres × Surface area of each smaller sphere 83 × 0.8 × 4πR2/64 = 6.4 × (4πR2)
Therefore, ratio of the total surface area of all smaller spheres to the surface area of original sphere 

Volume of paint required = 6.4 × Volume of paint used on original sphere = 6.4 × 20 = 128 litres

Test Level 1: Mensuration - 2 - Question 14

In a right angled triangle, find the hypotenuse if base and perpendicular are respectively 36015 cm and 48020 cm.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 14


Let hypotenuse = x cm
Then, by Pythagoras theorem:
x2 = (48020)2 + (36015)2
x ⇒ 60025 cm

Test Level 1: Mensuration - 2 - Question 15

The inner circumference of a circular track is 440 cm. The track is 14 cm wide. Find the diameter of the outer circle of the track.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 15

Let inner radius = r; then 2π r = 440
∴ r = 70
Radius of outer circle = 70 + 14 = 84 cm ∴ Outer diameter = 2 × Radius = 2 × 84 = 168

Test Level 1: Mensuration - 2 - Question 16

The outer circumference of a circular track is 220 meter. The track is 7 meter wide everywhere. Calculate the cost of levelling the track at the rate of 50 paise per square meter.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 16

Let outer radius = R; then inner radius = r = R – 7
2πR = 220 ⇒ R = 35 m;
r = 35 – 7 = 28 m
Area of track = π R2 – πr2 ⇒ p(R2 – r2)
= 1386 m2

Test Level 1: Mensuration - 2 - Question 17

A pit 7.5 meter long, 6 meter wide and 1.5 meter deep is dug in a field. Find the volume of soil removed in cubic meters.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 17

Volume of soil removed = l × b × h = 7.5 × 6 × 1.5 = 67.5 m3

Test Level 1: Mensuration - 2 - Question 18

The length, breadth and height of a room are in the ratio of 3 : 2 : 1. If its volume be 1296 m3, find its breadth 

Detailed Solution for Test Level 1: Mensuration - 2 - Question 18

Let the common ratio be = x
Then; length = 3x, breadth = 2x and height = x
Then; as per question 3x · 2x · x = 1296 ⇒ 6x3 = 1296
⇒ x = 6 m
Breadth = 2x = 12 m

Test Level 1: Mensuration - 2 - Question 19

The whole surface of a rectangular block is 8788 square cm. If length, breadth and height are in the ratio of 4 : 3 : 2, find length.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 19

Let the common ratio be = x
Then, length = 4x, breadth = 3x and height = 2x
As per question;
2(4x · 3x + 3x · 2x + 2x · 4x) = 8788
2(12x2 + 6x2 + 8x2) = 8788 ⇒ 52x2 = 8788
⇒ x = 13 Length = 4x = 52 cm

Test Level 1: Mensuration - 2 - Question 20

Find curved and total surface area of a conical flask of radius 6 cm and height 8 cm.

Detailed Solution for Test Level 1: Mensuration - 2 - Question 20

Slant length 
Then curved surface area = πrl = π × 6 × 10⇒ 60p
And total surface area = πrl + πr2 ⇒ π ((6 × 10) + 62)
= 96π cm2

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