Test: Mensuration (March 18) - CAT MCQ

From statement I, we know that the height of the initial cone is 20cm. However, nothing is said about the small cone. Hence, we cannot answer the question using statement A. So, we can eliminate choices (A) and (D).

We are down to choices (A), (B) or (D).

From Statement II, we know that the ratio of the volume of the small cone to that of the large cone is 1 : 15.

(r is the base radius of the smaller cone and h is the height of the smaller cone)
or r² * h : R² * H is 1 : 15

From this information, we will not be able to find the answer to h. Hence, we can eliminate choice (A). 

Combining the information in the two statements: 

When a section is made the two cones are similar triangles. so 



 Substituting H = 20, we can get the value for h.
Choice (B) is therefore, the correct answer.

Correct Answer: If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together.

Area = 4πr²

This question requires a good deal of visualization. Since, both the box and cans are hard solids, simply dividing the volume won’t work because the shape can’t be deformed. 

Each cylindrical can has a diameter of 14 cm and while they are kept erect in the box will occupy height of 30 cm 

Number of such cans that can be placed in a row

Number of such rows that can be placed

Thus 5 x 3 = 15 cans can be placed in an erect position.

However, height of box = 45cm and only 30 cm has been utilized so far 

Remaining height = 15 cm > 14 cm (Diameter of the can)

So, some cans can be placed horizontally on the base.

Number of cans in horizontal row
Number of such rows
∴ 2 x 3 = 6 cans can be placed horizontally

∴ Maximum number of cans = 15+6 = 21 

Choice (D) is therefore, the correct answer.

Although the figure looks like a 3D figure but on reading through the question, it is clear that the diagram is on a single plane.

Explanation : As, tangent from an exterior point makes right angle with the radius.

=> ADXB and ODX’E are squares,

∴ AO = BO = 40 cm

OD = DE = 30 cm

Length of string wound around circle x => 270°/360° ∗ 2 ∗ π ∗ 80

= 3/4 ∗ 80π

=60π

Similarly, Length of string wound around circle x’ => 270°/360° ∗ 2 ∗ π ∗ 30

= 3/4 ∗ 30π

= 45π/2

∴ Length of the string = 40 + 30 + 60π + 45π/2 ∗ π

= 70 + 165π/2

Total surface area of the box = 2(4x6 + 1x6 + 1x4)

= 2(24 + 6 + 4)

= 68 cm²

As the problem says the paper can’t be torn/cut a portion of paper will need to be fold, so, the area of paper required would be greater than 68 cm². Only option b) gives the area greater 68 cm²

Choice (B) is therefore, the correct answer.

Correct Answer: 12 cm by 6 cm

Given,

=) a² + z² – 2az = 2az (Please note how the solution is being managed here. You must always be aware of what you are looking for. Here, as equation -? we are looking for (a-z)2 in terms of az) 

The question demands visualization. The shortest distance required to be travelled by the fly would be diagonally and be given by:-