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Test Level 3: Mensuration - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 3: Mensuration

Test Level 3: Mensuration for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 3: Mensuration questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Mensuration MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Mensuration below.
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Test Level 3: Mensuration - Question 1

A cylindrical pail containing water drains into a cylindrical tub, 40 cm across and 50 cm deep, while resting at an angle of 45° to the horizontal, as shown in the figure. How deep is the water in the tub when its level reaches the pail?

Detailed Solution for Test Level 3: Mensuration - Question 1

We label the points A, B and C as shown in the diagram. From symmetry, we note that ΔBAC is an isosceles right-angled triangle.
From A, we draw a line perpendicular to BC meeting the line at point D.

This construction allows us to conclude that ΔABD is also a right-angled isosceles triangle and specifically that BD = DA.
Since BD = DA; BD = DC = 20; we find DA = 20 cm.
This makes the depth of the water equal to (50 - 20) cm or 30 cm.

Test Level 3: Mensuration - Question 2

ABCD is a rectangle with AD = 1 unit. DPF and CQF are two equal arcs drawn with A and B as centres respectively. E is the midpoint of CD. Another arc with E as centre touches the two arcs DPF and CQF at P and Q, respectively. What is the area of the shaded portion?

Detailed Solution for Test Level 3: Mensuration - Question 2

As AF and BF are both radii of equal arcs, AF = BF = AD = 1 unit.
∴ AB = 2 units
Area of the rectangle = 2 sq. units

Note that BE is the diagonal of square FBCE.
Therefore, BE = √2 units
∴ EQ = BE - BQ = √2 - 1 (∵ BQ = BC = 1)
∴ Area of semicircle with centre E

Also, total area of the two segments with centres A and B 

∴ Area of the shaded portion

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Test Level 3: Mensuration - Question 3

A cylindrical container, whose diameter is 12 cm and height is 15 cm, is filled with ice-cream. The whole ice-cream is distributed among 10 children in equal cones with hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream cone.

Detailed Solution for Test Level 3: Mensuration - Question 3

Volume of the cylindrical container = 10 × volume of each cone


Diameter of each ice-cream cone = 6 cm

Test Level 3: Mensuration - Question 4

A vessel is in the form of an inverted cone, open at the top. Its depth is 8 cm and the diameter is 10 cm. It is filled with water up to the brim. When spherical chocolate bullets of radius 0.5 cm are dropped into the vessel, 1/4 of the water flows out. Find the number of bullets dropped in the vessel.

Detailed Solution for Test Level 3: Mensuration - Question 4

 Volume of the cone

Volume of all chocolate bullets = Volume of water that spilt out 

Volume of each chocolate bullet 

Number of chocolate bullets

Test Level 3: Mensuration - Question 5

What will be the mass of a 21 cm long metallic hollow cylinder with internal diameter 8 cm and thickness of metal being 4 mm, if 1 cm3 of the metal weighs 8 gm?

Detailed Solution for Test Level 3: Mensuration - Question 5

Length of cylinder = 21 cm
Internal radius = 4 cm
External radius = 4.4 cm

= 66 (19.36 – 16) = 221.76 cm3
Hence, Mass of cylinder = Volume of metal × 8 = 221.76 × 8 = 1774 gm

Test Level 3: Mensuration - Question 6

In the above figure, if the radius of the smaller circle is (√2 - 1) cm, then the area of the shaded region will be

Detailed Solution for Test Level 3: Mensuration - Question 6

Let the radius of the bigger circle be R.
OC = R √2 cm
Let r be the radius of the smaller circle.
Then, R√2 = R + r + r√2 (From the figure)


Area of the shaded region 

Test Level 3: Mensuration - Question 7

A rectangular piece of paper 16 cm long and 12 cm wide is folded along one of its diagonals. It results in some area of the paper doubling up over the portion of the paper already lying but some portion failing to double up, as it is not a square piece. Determine the area of the paper which has not doubled up.

Detailed Solution for Test Level 3: Mensuration - Question 7

Refer to the figure. △AB'E and △CDE form the area which has not doubled up.

 

Both triangles are similar since ∠AB'E = ∠CDE = 90°.
∠AEB' = ∠DEC  (vertically opposite angles)
Also, since CD and AB', the corresponding sides of the triangles are equal (= 12 cm), the triangles are congruent.
Let  and DE = B'E = x and CE = AE = 16 - x
Then, x2 + 122 = (16 - x)2

Test Level 3: Mensuration - Question 8

A cube of maximum possible volume is kept inside a sphere. A right circular cylinder of maximum possible volume is kept inside the cube. What is the ratio of the total outer surface area of the cylinder to that of the sphere?  

Detailed Solution for Test Level 3: Mensuration - Question 8

If a slice is cut parallel to the surface of the cube at the stage where the upper corners of the cube just touch the sphere.
The 3-D picture of the cube will look as shown below.

The length of a side of the cube is 2a and the radius of the sphere is R.

By Pythagoras theorem, R2 = (√2a)2 + a2 = 3a2
So, surface area of the sphere = 4 × πR2 = 4 × π × 3a2 = 12πa2
Now, the radius of the circular face of the cylinder = a.
Height of the cylinder = 2a
So, outer surface area of the cylinder = 

Thus, the desired ratio 

Test Level 3: Mensuration - Question 9

Ram cuts a semicircle of radius 8 cm from a piece of paper and folds it in such a way that it forms a right circular cone. What is the volume of this cone?

Detailed Solution for Test Level 3: Mensuration - Question 9

Let 'R' be the radius of the semicircular piece of paper.
Let the radius of the base of the cone be r.
Let the height of the cone be h.
Circumference of base of cone = 2πr = πR
So, r = R/2
Now, the slant height of the cone so formed = R
Applying Pythagoras theorem, we get

Volume of the cone,

Putting the values of r and h, we get

Putting the value of R, we get 

Test Level 3: Mensuration - Question 10

The height of a frustum made from a cone is 8 cm. If the base radii of this frustum are 6 cm and 2 cm, then find the height of the original cone.

Detailed Solution for Test Level 3: Mensuration - Question 10

Height of the frustum (h) = 8 cm

Radii of the base of the frustum are:
r1 = 6 cm and r2 = 2 cm
Height of the original cone = h1 = 8 + h2 … (i)
∵ ΔOPB and ΔOQD are similar
[∵QOD = POB (same angle) and OQD = OPB = 90°]
So, the ratio of their corresponding sides is equal.
Thus,  h1/h2 = 6/2 = 3/1  or h1 = 3h2
Putting the value h1 = 3h2 into equation (i), we get
 ⇒ 3h2 = 8 + h2 or 2h2 = 8 ⇒ h2 = 4 cm
Again putting h2 = 4 into (i), we get
⇒ h1 = 8 + 4 = 12 cm
So, height of the original cone = 12 cm

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