Mensuration Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 2 EXERCISES

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The following question based on Mensuration topic of quantitative aptitude

Questions : If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius R, then

(a) $R_1 + R_2$ < R

(b) $R_1 + R_2$ = R

(c) $R_1 + R_2 > R_3$

(d) Nothing definite can be said about the relation among $R_1.R_2$ and R

The correct answers to the above question in:

Answer: (b)

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Read more model questions set 2 Based Quantitative Aptitude Questions and Answers

Question : 1

Assertion (A)
Triangles on the same base and between the same parallel lines are equal in area.
Reason (R)
The distance between two parallel lines is same everywhere.

a) A is true but R is false

b) Both A and R individually true and R is the correct explanation of A

c) Both A and R are individually true but R is not the correct explanation of A

d) A is false but R is true

Answer: (b)

mensuration-area-and-volume-aptitude-mcq

A. By the properties of triangle, it is true.

R. It is also true that the distance between two parallel lines is same everywhere.

Hence, A and R are true and R is the correct explanation of A.

Question : 2

The cross section of a canal is a trapezium in shape. If the canal is 7 metres wide at the top and 9 metres at the bottom and the area of cross-section is 1280 square metres, find the height of the canal.

a) 154 metres

b) 160 metres

c) 172 metres

d) None of these

Answer: (b)

mensuration-aptitude-mcq

Let the height of canal = h.

Then, area of canal = $1/2$ × h(9 + 7)

or = 1280 = $1/2$ h(16)

∴ h = ${1280 × 2}/{16}$ = 160m

Question : 3

mensuration-area-and-volume-aptitude-mcq
In the figure given above, what is ∠CBA?

a) 50°

b) 30°

c) 45°

d) 60°

Answer: (d)

The sum of opposite angles in cyclic quadrilateral is always 180°.

∴ ∠ACQ + ∠APQ = 180°

75° + ∠APQ = 180°

∴ ∠APQ = 180° – 75° = 105°

∠ACQ + ∠QCR = 180° (∵ Straight line)

75° + ∠QCR = 180°

∠QCR = 180° – 75° = 105°

∠CQR = 180° – 105° – 30° = 45°

Since, ∠APQ + ∠BPQ = 180° (Straight line)

∴ 105° + ∠BPQ = 180°

∠BPQ = 75°

In ΔBPQ ∠B + ∠P + ∠Q = 180°

∠B + 75° + 45° = 180°

⇒ ∠B = 60° ∴ ∠CBA = 60°

Question : 4

mensuration-area-and-volume-aptitude-mcq
In the figure given above, AB is a diameter of the circle with centre O and EC = ED. What is ∠EFO?

a) 25°

b) 15°

c) 20°

d) 30°

Answer: (c)

The Given,

mensuration-area-and-volume-aptitude-mcq

⇒ ∠EDC = ∠ECD = 35°

Since, ∠OCD = 55°

Then, ∠OCE = 20°

By using then theorem that triangle on the same segment of a circle makes as equal angles.

Here, OE is a segment, which makes a ΔOFE and ΔOCE.

Therefore, ∠OCE = ∠EFO = 20°

Question : 5

mensuration-area-and-volume-aptitude-mcq
In the given triangle, AB is parallel to PQ. AP = c, PC = b, PQ = a, AB = x. What is the value of x?

a) $b + {ca}/b$

b) $a + {ab}/c$

c) $a + {bc}/a$

d) $a + {ac}/b$

Answer: (d)

In ΔABC and ΔPQC,

mensuration-area-and-volume-aptitude-mcq

∴ ${PC}/{AC} = {PQ}/{AB}$

⇒ $b/{c + a} = a/x$

∴ x = ${a(c + b)}/b = {ac}/b + a$

Question : 6

Using the data in the above question, what is the breadth of the rectangle ?

a) 8

b) 6

c) 12

d) 10

Answer: (a)

Breadth of the rectangle is 8 units, as solved above

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