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 rectangle garden having length and breadth as 110 m and 65 m respectively has 2.5 m wide path around the sides inside the garden. Find the cost of gravelling the path at 50 paise per sq. metre.**

A) Rs 420

B) Rs 325

C) Rs 425

D) Rs 405

E) Rs 375**C) Rs 425**

Explanation:

Area of garden = 110*65 = 7150 m^{2}

Area of path = (110-5)*(65-5) = 6300 m^{2}

So area of path = 7150 – 6300 = 850 m^{2}

So cost = 850*(50/100) = Rs 425**The cost of painting a triangular board at Rs 24.68 per m**^{2}is Rs 33318. If the base of this board is thrice its height, find the height of the board.

A) 66 m

B) 75 m

C) 30 m

D) 90 m

E) 45 m**D) 90 m**

Explanation:

Area of triangle = 33318/24.68 = 1350 m^{2}

If height is x, then base = 3x

So (1/2)*3x*x = 1350

x^{2}= 900, x = 30

so base = 3*30 = 90**The radius of a wheel of car is 70 cm. How many revolutions per minute the wheel will make in order to keep a speed of 66 km/hr?**

A) 234

B) 272

C) 225

D) 300

E) 250**E) 250**

Explanation:

Distance to be covered in 1 min = 66*(1000/60) = 1100 m

70 cm = 0.70 m

Circumference of wheel = 2*(22/7)*0.70 = 4.4 m

Number of revolutions = (1100/4.4) = 250**The length and breadth of a rectangle are in the ratio 3 : 2. If the length is increased by 5 m keeping the breadth same, the new area of rectangle is 2600 m**^{2}. What is the breadth of the rectangle?

A) 30 m

B) 20 m

C) 32 m

D) Cannot be determined

E) None of these**B) 20 m**

Explanation:

3x, 2x

So (3x+5)*2x = 2600

6x^{2}+ 10x = 2600

3x^{2}+ 5x – 1300 = 0

3x^{2}– 60x + 65x – 1300 = 0

3x(x-20) + 65(x-20) = 0

Solve, x = 20

So breadth = 20 m**What will be the percentage increase in the surface area of the cube whose side is increased by 50%**

A) 75%

B) 125%

C) 150%

D) 100%

E) 92%**B) 125%**

Explanation:

Surface area of cube = 6a^{2}

A increases by 50%

So area increases by 50 + 50 + (50)(50)/100 = 125%**The radius of base and height of a cylinder are in the ratio 2 : 3. Find the total surface area of the cylinder if its volume is 12936 cm**^{3}.

A) 3080 cm^{2}

B) 2680 cm^{2}

C) 4940 cm^{2}

D) 3280 cm^{2}

E) None of these**A) 3080 cm**^{2}

Explanation:

(22/7)*(2x)^{2}*3x = 12936

Solve, x = 7

So radius = 7 cm, height = 21 cm

Total surface area = 2*(22/7)*14*(21+14) = 3080**The diameter and the height of a right circular cylinder are 11.2 cm and 21 cm respectively. The metal around its outer body is 0.4 cm thick. What is the volume of the metal?**

A) 186.34 cm^{3}

B) 306.24 cm^{3}

C) 156.56 cm^{3}

D) 336.24 cm^{3}

E) None of these**B) 306.24 cm**^{3}

Explanation:

Internal radius = (11.2)/2 = 5.6, so outer radius = 5.6+0.4 = 6cm

So volume of metal = (22/7)*6^{2}*21 – (22/7)*(5.6)^{2}*21 = (22/7)*[6^{2}– (5.6)^{2}]*21**What is the volume of a right circular cone whose radius of base is 70 cm and curved surface area is 40040 cm**^{2}?

A) 839500 cm^{3}

B) 713400 cm^{3}

C) 862400 cm^{3}

D) 462320 cm^{3}

E) None of these**C) 862400 cm**^{3}

Explanation:

r = 70

(22/7)*70*l = 40040

So l = 182

h = √(182)^{2}– 70^{2}=168 cm

so vol = (1/3)*(22/7)*70^{2}*168**Find the radius of each of the 8 spherical balls which are made from a solid sphere of radius 10 cm by melting it.**

A) 12 cm

B) 8 cm

C) 20 cm

D) 5 cm

E) 15 cm**D) 5 cm**

Explanation:

Volume of each of the 8 ball = [(4/3)*(22/7)*10^{3}]/8

So (4/3)*(22/7)*r^{3}= [(4/3)*(22/7)*10^{3}]/8

Solve, r = 5**The metal used in the cylinder having external radius 6 cm, height 15 cm and thickness 0.25 cm is to be cast from a cylinder of radius 1 cm. What is the approximate height of the cylinder from which casting is to be done?**

A) 8 1/4 days

B) 8 3/4 days

C) 9 days

D) 10 3/4 days

E) 11 1/3 days**B) 8 3/4 days**

Explanation:

Volume of metal = (22/7)*[6^{2}– (5.75)^{2}]*15

So (22/7)*1^{2}*h = (22/7)*[6^{2}– (5.75)^{2}]*15

Solve, h ≈ 44

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