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Test: Mensuration (March 19) - CAT MCQ


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10 Questions MCQ Test - Test: Mensuration (March 19)

Test: Mensuration (March 19) for CAT 2024 is part of CAT preparation. The Test: Mensuration (March 19) questions and answers have been prepared according to the CAT exam syllabus.The Test: Mensuration (March 19) MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Mensuration (March 19) below.
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Test: Mensuration (March 19) - Question 1

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 for Test: Mensuration (March 19) - Question 1

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

Test: Mensuration (March 19) - Question 2

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?

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Test: Mensuration (March 19) - Question 3

The cost of white washing one m2 is Rs 50. What will be the maximum amount saved in painting the room in the most economical way, if the sum of the length, breadth and height is 21 m and all the sides are integers (floor is not to be white washed)?

Detailed Solution for Test: Mensuration (March 19) - Question 3

The amount saved would be maximum only when the difference of maximum area and minimum area to be painted is maximum.

Test: Mensuration (March 19) - Question 4

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 for Test: Mensuration (March 19) - Question 4

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

Test: Mensuration (March 19) - Question 5

The maximum distance between two points of the unit cube is

Detailed Solution for Test: Mensuration (March 19) - Question 5

The distance from any vertex at the base of the cube to the vertex that is perpendicular along height to the diametrically opposite vertex is required.
We have to calculate DF.

Test: Mensuration (March 19) - Question 6

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 for Test: Mensuration (March 19) - Question 6

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

Test: Mensuration (March 19) - Question 7

Under the Indian Posts and Telegraph Act 1885, any package in the form of a right circular cylinder will not be accepted if the sum of its height and the diameter of its base exceeds 10 inches. The height (in inches) of a package of maximum volume that would be accepted is

Detailed Solution for Test: Mensuration (March 19) - Question 7

Volume is maximum when radius is equal to height.

Test: Mensuration (March 19) - Question 8

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 for Test: Mensuration (March 19) - Question 8

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

Test: Mensuration (March 19) - Question 9

The area of a rectangular field is 52000 m². This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is : 

Detailed Solution for Test: Mensuration (March 19) - Question 9

Test: Mensuration (March 19) - Question 10

 What is the area of an equilateral triangle of side 16 cm?

Detailed Solution for Test: Mensuration (March 19) - Question 10

Area of an equilateral triangle = √3/4 S2 

If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm2

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