Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample from **Mensuration** that are important for all the competitive exams. We have included some questions that are repeatedly asked in exams !!

**A rectangular box has dimensions 1.6 m × 1 m × 0.6 m. How many cubical boxes each of side 20 cm can be fit inside the rectangular box?**

A) 220

B) 205

C) 120

D) 165

E) 124**C) 120**

Explanation:

20 cm = 0.2 m

So number of boxes that can fit = 1.6 × 1 × 0.6/0.2 × 0.2 × 0.2**What is the volume of a cylinder whose curved surface area is 704 cm**^{2}and height is 14 cm?

A) 2878 cm^{3}

B) 2914 cm^{3}

C) 2396 cm^{3}

D) 2816 cm^{3}

E) 2916 cm^{3}**D) 2816 cm**^{3}

Explanation:

2ᴨrh = 704, h = 14

Solve both, so r = 8

Volume = ᴨr^{2}h**If the volume of wire remains the same but its radius decreases to 1/3rd of previous, then the new length of the wire will be how many times of the previous length?**

A) 3 times

B) 4 times

C) 6 times

D) 10 times

E) 9 times**E) 9 times**

Explanation:

Wire will be taken as cylinder.

R = r/3

ᴨr^{2}h = ᴨR^{2}H

r^{2}h = (r/3)^{2}H

so H = 9h**A cone with diameter of its base as 12 cm is formed by melting a spherical ball of diameter 6 cm. What is the height of the cone?**

A) 6 cm

B) 3 cm

C) 32 m

D) 15 cm

E) None of these**B) 3 cm**

Explanation:

Radius of cone = 12/2 = 6, radius of ball = 6/2 = 3

Volumes will be equal, so

(1/3)ᴨ r^{2}h = (4/3)ᴨ R^{3}

6^{2}h = 4* 3^{3}

So h = 3**The total cost of painting the walls of a room is Rs 475. Find the cost of painting the walls of another room whose length, breadth and height each are double than the dimensions of the previous room.**

A) Rs 1780

B) Rs 1900

C) Rs 1846

D) Rs 1960

E) Rs 1600**B) Rs 1900**

Explanation:

Area of first room = 2(l+b)*h

After all dimensions doubled, new area = 2(2l+2b)*2h = 4[2(l+b)*h ] = 4 times previous area, so cost of painting = 4*475**A circular wire of diameter 84 cm is bent into a rectangle with sides ratio 6 : 5. What are the respective sides of the rectangle?**

A) 60 cm, 72 cm

B) 78 cm, 65 cm

C) 72 cm, 60 cm

D) 90 cm, 75 cm

E) None of these**C) 72 cm, 60 cm**

Explanation:

Length of wire = 2ᴨr = 2(22/7)*42 = 264 cm

Perimeter of rectangle = 2(6x+5x) = 264

Solve, x = 12

So dimensions – 12*6, 12*5**The ratio of the outer and the inner perimeters of a circular path is 9 : 8. The path is 3 metres wide. What is the diameter of the outer circle?**

A) 20 m

B) 27 m

C) 35 m

D) 39 m

E) 24 m**B) 27 m cm**^{3}

Explanation:

2ᴨr/2ᴨR = 9/8

So r/R = 9/8, so r = (9/8)R

r-R = 3, so (9/8)R – R = 3

R = 24, so r = (9/8)*24 = 27**Four circular cardboard pieces, each of diameter 28 cm are placed in such a way that each piece touches two other pieces. Find the area of the space enclosed by the four pieces.**

A) 195 cm^{2}

B) 134 cm^{2}

C) 168 cm^{2}

D) 162 cm^{2}

E) None of these**C) 168 cm**^{2}

Explanation:

Required area = area of square shown in figure – 4*areas of 1/4th parts of each circle, so

Required area = 28*28 – 4*(1/4)*(22/7)*14*14**A solid sphere of radius 20 cm is melted to form 12 spherical balls. Find the radius of each of the ball.**

A) 12 cm

B) 8 cm

C) 20 cm

D) 10 cm

E) 15 cm**D) 10 cm**

Explanation:

Volume of each of the 8 ball = [(4/3)*(22/7)*20^{3}]/8

So (4/3)*(22/7)*r^{3}= [(4/3)*(22/7)*20^{3}]/8

Solve, r = 10**The perimeter of a rectangular plot is 210 m. Find the cost of gardening 2 m broad boundary around rectangular plot whose perimeter The cost of gardening is Rs 14 per m**^{2}.

A) Rs 6450

B) Rs 6400

C) Rs 6480

D) Rs 6104

E) Rs 6880**D) Rs 6104**

Explanation:

Given 2(l+b) = 210

2 m broad boundary means increase in l and b by 4 m

So area of the boundary will be [(l+4)(b+4) – lb] = 4(l+b) + 16 = 2*[2(l+b)] + 16 = 2*210 + 16 = 436 m^{2}

So cost of gardening = 436*14

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