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CAT Exam  >  CAT Test  >  Quantitative Aptitude (Quant)  >  CAT Practice: Mensuration - CAT MCQ

CAT Practice: Mensuration - Free MCQ Test with solutions for Quant Aptitude


MCQ Practice Test & Solutions: CAT Practice: Mensuration (15 Questions)

You can prepare effectively for CAT Quantitative Aptitude (Quant) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "CAT Practice: Mensuration". These 15 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 15

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CAT Practice: Mensuration - Question 1
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There is a right circular cone with base radius 3 units and height 4 units. The surface of this right circular cone is painted. It is then cut into two parts by a plane parallel to the base so that the volume of the top part (the small cone) divided by the volume of the frustum equals the painted area of the top part divided by the painted area of the bottom part. The height of the small cone is

Detailed Solution: Question 1

Take the total surface area of the initial cone into consideration and then proceed.

CAT Practice: Mensuration - Question 2
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A large solid cube of steel of side 1 metre is molten and recast into a number of smaller cubes of side 5cm or 10cm. If it is known that the number of 5cm cubes was at least double the number of 10cm cubes, what is the minimum percentage increase in the total surface area in this process?

Detailed Solution: Question 2

We know that if we have to minimise the total surface area, we need to keep the number of 5cm cubes minimum.

Let the number of 10cm cubes be b and the number of 5 cm cubes be a

Therefore,

100 x 100 x 100 = a x 5 x 5 x 5 + b x 10 x 10 x 10

8000 = a + 8b

Now, a > = 2b, since a needs to be minimum, a = 2b

Therefore, 10b = 8000

b = 800

a = 1600

Initial surface area = 6 x 100 x 100 = 60000

Final surface area = 6 (800 x 10 x 10 + 1600 x 5 x 5) = 720000

Hence, percentage increase = 720000 - 60000/60000 x 100 = 1100%

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CAT Practice: Mensuration - Question 3
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What is the ratio of the shaded region to non-shaded region in the following diagram

ABC is an equilateral triangle and D, E and F are the midpoint of the sides.

Detailed Solution: Question 3

Since ABC is an equilateral triangle and DEF is the midpoint. Thus they will divide the ABC in 4 equilateral triangle of side half of that of ABC
Let side of ABC = 2a units. Then the side of the smaller triangle is a unit.
Shaded region = Area of 3 small circles + Area DEF - Area of 1 small circle
Shaded region = Area of 2 small circles + Area DEF 
Circles are an incircle of equilateral triangle of side "a"
Radius of this incircle 
Area of 1 small circle = 
Area of 2 small circle 
Area of triangle DEF = 
Shaded area = 
Unshaded area = Total area - Shaded area
Unshaded area =  - Shaded area
Unshaded area = 
Unshaded area = 
Ratio = 

CAT Practice: Mensuration - Question 4
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In the diagram, square ABCD has a side length of 6 units. Circular arcs of radius 6 units are drawn with centres B and D. What is the area of the shaded region?

Detailed Solution: Question 4

Shaded Area = 2 [Area of Sector ACB - Area of Right ΔABC]

= 2 [Area of Sector ACD - Area of Right ΔADC]

CAT Practice: Mensuration - Question 5
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The faces of a cube of n cm is first painted red and then the cube is cut into smaller cubes of 1cm. If the difference between the number of cubes with 1 face painted and the number with 2 faces painted is 90, what is the number of cubes with no face painted?


Detailed Solution: Question 5

For n cm sized cube the number of cubes with 3 faces painted = 8

Number of cubes with 2 faces painted = 12x(n - 2)

Number of cubes with 1 face painted = 6(n - 2)2

Number of cubes with 0 face painted = (n - 2)3

Given 6(n - 2)2 - 12(n - 2) = 90

So (n - 2)(n - 4) = 15

Or n = 7

Therefore number of cubes with 0 face painted = (7 - 2)3 = 125

CAT Practice: Mensuration - Question 6
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Water flows at a speed of 30 km/h through a cylindrical pipe of inner radius 1.5 m into a tank of 100 m  150 m. In what time (in minutes rounded off to the nearest integer) will the water rise by 4 m?

Detailed Solution: Question 6

Volume required = (100  150  4) m3
Speed of water in cylindrical pipe = m/min
Volumetric flow of water in cylindrical pipe =  m3/min = 1125 m3/min
Required time =  min
= 17 minutes (approx.)

CAT Practice: Mensuration - Question 7
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The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

Detailed Solution: Question 7

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.
In this question, we are going to assume that the L is greater than B.
Hence, the one side painted at Rs 200/ft should be B to minimise costs.
Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L
Now, L x B = 60000 B = 60000/L
Hence, total cost = 300B + 200L = 18000000/L + 200L
To minimise this cost, we can use AM>=GM,

Hence, minimum cost = Rs 120000.

*Answer can only contain numeric values
CAT Practice: Mensuration - Question 8
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If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is


Detailed Solution: Question 8

Let the diagonals of the rhombus be d1 and d2
We know that the

For a rhombus with side a=36, the diagonals intersect at right angle. Giving a right-angle triangle with the side as hypotenuse.

CAT Practice: Mensuration - Question 9
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ABCD is a square of side 10 cm. What is the area of the least-sized square that may be inscribed in ABCD with its vertices on the sides of ABCD?

Detailed Solution: Question 9

  1. The side of square ABCD is 10 cm.

  2. If a square is inscribed such that its diagonal is equal to the side of ABCD, the diagonal of the inscribed square will be 10 cm.

  3. For a square, the relationship between its side length (s) and its diagonal (d) is given by the formula:
    d = s√2

  4. Here, the diagonal is 10 cm, so:
    10 = s√2

  5. Solving for s:
    s = 10 / √2 = 10 × √2 / 2 = 5√2

  6. The area of the inscribed square is s² = (5√2)² = 25 × 2 = 50 cm².

CAT Practice: Mensuration - Question 10
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A cube is inscribed in a hemisphere of radius R, such that four of its vertices lie on the base of the hemisphere and the other four touch the hemispherical surface of the half-sphere. What is the volume of the cube?

Detailed Solution: Question 10

Let ABCDEFGH be the cube of side a and O be the centre of the hemisphere.

AC = √2 a

OD = OC = R
Let P be the mid-point of AC

OP = a

Now in  Δ AOC

CAT Practice: Mensuration - Question 11
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All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. The total area of all its surfaces (in cm2) is

Detailed Solution: Question 11

Equate the area of the square ABCD and triangle PDC and find a relation between the slant height and the length of the base of the pyramid

CAT Practice: Mensuration - Question 12
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A cuboid of length 20 m, breadth 15 m and height 12 m is lying on a table. The cuboid is cut into two equal halves by a plane which is perpendicular to the base and passes through a pair of diagonally opposite points of that surface. Then, a second cut is made by a plane which is parallel to the surface of the table again dividing the cuboid into two equal halves. Now this cuboid is divided into four pieces. Out of these four pieces, one piece is now removed from its place. What is the total surface area of the remaining portion of the cuboid?

Detailed Solution: Question 12

Since we don’t know that the cut is made parallel to which face, we cannot determine the surface area.

CAT Practice: Mensuration - Question 13
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Four spheres each of radius 10 cm lie on a horizontal table so that the centres of the spheres form a square of side 20 cm. A fifth sphere also of radius 10 cm is placed on them so that it touches each of these spheres without disturbing them. How many cm above the table is the centre of the fifth sphere?

Detailed Solution: Question 13

OA= 10

OA= 10
To find the value of PA, go through the options now.

CAT Practice: Mensuration - Question 14
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The square of side 1 cm are cut from four comers of a sheet of tin (having length = 1 and breadth = b) in order to form an open box. If the whole sheet of tin was rolled along its length to form a cylinder, then the volume of the cylinder is equal to (343/4) cm3. Find the volume of the box. (1 and b are integers)

Detailed Solution: Question 14

The length of the rectangle will be equal to the circumference of the base of the cylinder.

CAT Practice: Mensuration - Question 15
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John Nash, an avid mathematician, had his room constructed such that the floor of the room was an equilateral triangle in shape instead of the usual rectangular shape. One day he brought home a bird and tied it to one end of a string and then tied the other end of the string to one of the corners of his room. The next day, he untied the other end of the string from the corner of the room and tied it to a point exactly at the center of the floor of the room. Assuming that the dimensions of the room are relatively large compared to the length of the string, find the number of times, by which the maximum possible space in which the bird can fly, increase.

Detailed Solution: Question 15

Consider the length of the string less than or equal to the inradius of the floor. At the comer of the floor you will find a sixth part of a hemisphere and at the centre it will be a hemisphere.

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About this CAT Practice Test

This test contains 15 MCQs on CAT Practice: Mensuration with detailed solutions for each question. It is part of the Quantitative Aptitude (Quant) course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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