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Quantitative Aptitude

VOLUME AND SURFACE AREA MCQs

Total Questions : 820 | Page 7 of 82 pages
Question 61.

  1. 10 cylindrical pillars of a building have to be painted. The diameter of each pillar is 50 cm and the height is 4 m. The cost of painting at the rate of 50 paise per square metre is

  1.    Rs 31.40
  2.    Rs 314
  3.    Rs 331.40
  4.    Rs 331
 Discuss Question
Answer: Option A. -> Rs 31.40
Question 62.

 A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:

  1.      \(12\pi\) cm3
  2.       \(15\pi\)cm3
  3.      \(16\pi\) cm3
  4.      \(20\pi\) cm3
 Discuss Question
Answer: Option A. ->   \(12\pi\) cm3

 A Right Triangle With Sides 3 Cm, 4 Cm And 5 Cm Is Rotated...


Clearly, we have r = 3 cm and h = 4 cm


So \(Volume,= \frac{1}{3}\pi r^{2}h = \left(\frac{1}{3}\times \pi\times3^{2}\times4\right)cm^{3} =12\pi cm^{3}.\)

Question 63.

In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:

  1.    75 cu. m
  2.    750 cu. m
  3.    7500 cu. m
  4.    75000 cu. m
 Discuss Question
Answer: Option B. -> 750 cu. m

1 hectare = 10,000 m2


So, Area = (1.5 x 10000) m2 = 15000 m2.


Depth = \(\frac{5}{100}m = \frac{1}{20}m\) 


So, Volume = (Area x Depth) =  \(\left(15000\times\frac{1}{20}\right)m^{3}=750m^{3.}\)

Question 64.

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

  1.    720
  2.    900
  3.    1200
  4.    1800
 Discuss Question
Answer: Option C. -> 1200

2(15 + 12) x h = 2(15 x 12)


h =  \(\frac{180}{27}m=\frac{20}{3}m.\)


So, volume =  \(\left(15\times12\times\frac{20}{3}\right)m^{3} = 1200m^{3}\)

Question 65.

66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

  1.    84
  2.    90
  3.    168
  4.    336
 Discuss Question
Answer: Option A. -> 84

Let the length of the wire be h.


Radius = \(\frac{1}{2}mm=\frac{1}{20}cm. Then,\)


\(\frac{22}{7}\times\frac{1}{20}\times\frac{1}{20}\times h= 66 \) 


h =  \(\left(\frac{66\times20\times20\times7}{22}\right)=8400cm=84 m.\)

Question 66.

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:

  1.    3.6 kg
  2.    3.696 kg
  3.    36 kg
  4.    36.9 kg
 Discuss Question
Answer: Option B. -> 3.696 kg

External radius = 4 cm,


Internal radius = 3 cm.


Volume of iron  =  \(\left(\frac{22}{7}\times\left[(4)^{2}-(3)^{2}\right]\times21
\right)cm^{3}\)


= \(\left(\frac{22}{7}\times7\times1\times21\right)cm^{3}\)


=  462 cm3.


So, Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.

Question 67.

A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:

  1.    12 kg
  2.    60 kg
  3.    72 kg
  4.    96 kg
 Discuss Question
Answer: Option B. -> 60 kg

Volume of water displaced = (3 x 2 x 0.01) m3


= (0.06 x 1000) kg


= 60 kg.

Question 68.

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m3, then the rise in the water level in the tank will be:

  1.    20 cm
  2.    25 cm
  3.    35 cm
  4.    50 cm
 Discuss Question
Answer: Option B. -> 25 cm

Total volume of water displaced = (4 x 50) m3 = 200 m3.


So, Rise in water level = \(\left(\frac{200}{40\times20}\right)m0.25m=25cm.\)

Question 69.

The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

  1.     \(30\pi\) m2
  2.        \(40\pi\) m2
  3.    \(60\pi\)  m2
  4.      \(80\pi\) m2
 Discuss Question
Answer: Option C. -> \(60\pi\)  m2

l = 10 m,


h = 8 m.


So, r = l2 - h2 = (10)2 - 82 = 6 m.


So, Curved surface area = \(\pi rl = \left(\pi\times6\times10\right)m^{2}=60\pi m^{2}\)

Question 70.

A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

  1.    49 m2
  2.    50 m2
  3.    53.5 m2
  4.    55 m2
 Discuss Question
Answer: Option A. -> 49 m2

Area of the wet surface  = [2(lb + bh + lh) - lb]


= 2(bh + lh) + lb


= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2


= 49 m2

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