ExamVeda
Login
Home

Volume And Surface Area - Aptitude MCQ Questions and Solutions with Explanations

Volume And Surface Area MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Volume And Surface Area with easy and logical explanations. Volume And Surface Area MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-10 section-1

Each section contains 100 MCQs on Volume and Surface Area .
91
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Diameter of bowl = 7 cm
∴ Radius of bowl = $$\frac{2}{7}$$ cm
Height = 4 cm
∴ Volume of cylindrical bowl :
$$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \cr & = 154\,cu.cm \cr} $$
Hence, volume of soup for 250 patients :
$$\eqalign{ & = 154 \times 250 \cr & = 38500{\text{ c}}{{\text{m}}^3} \cr & = 38.5{\text{L}} \cr} $$
92
The breadth of a room is twice its height and half its length. The volume of the room is 512 cu.m. The length of the room is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Let the height of the room be x metres
Then, breadth = 2x metres and length = 4x metres
∴ Volume of the room :
= (4x × 2x × x) m3
= (8x3) m3
8x3 = 512
⇒ x3 = 64
⇒ x = 4
∴ Length of the room is :
= 4x
= (4 × 4)
= 16 m
93
A closed box made of wood of uniform thickness has length, breadth and height 12 cm, 10 cm and 8 cm respectively. If the thickness of the wood is 1 cm, the inner surface area is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Internal length = (12 - 2) cm = 10 cm
Internal breadth = (10 - 2) cm = 8 cm
Internal height = (8 - 2) cm = 6 cm
Inner surface area :
= 2 [10 × 8 + 8 × 6 + 10 × 6] cm2
= (2 × 188) cm2
= 376 cm2
94
From a cube of side 8 m, a square hole of 3 m side is hollowed from end to end. What is the volume of the remaining solid ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Volume of the remaining solid :
= Volume of the cube - Volume of the cuboid cut out from it
= [(8 × 8 × 8) - (3 × 3 × 8)] m3
= (512 - 72) m3
= 440 m3
95
The dimensions of a rectangular box are in the ratio 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rate of Rs. 8 and Rs. 9.50 per square metre is Rs. 1248. Find the dimensions of the box in metres.
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the length, breadth and height of the box be 2x, 3x and 4x respectively
Then, surface area of the box :
= 2 [2x.3x + 3x.4x + 2x.4x]
= [2(6x2 + 12x2 + 8x2)]
= 52x2
$$\eqalign{ & \therefore 52{x^2} = \frac{{1248}}{{1.50}} \cr & \Rightarrow 52{x^2} = 832 \cr & \Rightarrow {x^2} = \frac{{832}}{{52}} \cr & \Rightarrow {x^2} = 16 \cr & \Rightarrow x = 4 \cr} $$
Hence, the diameter of the box are 8 m, 12 m and 16 m

96
A cuboidal water tank contains 216 litres of water. Its depth is $$\frac{1}{3}$$ of its length and breadth is $$\frac{1}{2}$$ of $$\frac{1}{3}$$ of the difference between length and depth. The length of the tank is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the length of the tank be x dm
Then, depth of the tank = $$\frac{x}{3}$$ dm
Breadth of the tank :
$$\eqalign{ & = \left[ {\frac{1}{2}{\text{ of }}\frac{1}{3}{\text{ of }}\left( {x - \frac{x}{3}} \right)} \right]{\text{dm}} \cr & = \left( {\frac{1}{2} \times \frac{1}{3} \times \frac{{2x}}{3}} \right){\text{dm}} \cr & = \frac{x}{9}\,{\text{dm}} \cr} $$
$$\eqalign{ & \therefore x \times \frac{x}{9} \times \frac{x}{3} = 216 \cr & \Rightarrow {x^3} = 216 \times 27 \cr & \Rightarrow x = 6 \times 3 \cr & \Rightarrow x = 18\,dm \cr} $$
97
A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood :
Discuss
Answer & Solution
Answer: Option C
Solution:
The external measures of the box are (115 + 5) cm, (75 + 5) cm, and (35 + 5) cm i.e., 120 cm, 80 cm and 40 cm
Volume of the wood :
= External volume - Internal volume
= [(120 × 80 × 40) - (115 × 75 × 35)] cm3
= (384000 - 301875) cm3
= 82125 cm3
98
Find the cost of a cylinder of radius 14 m and height 3.5 m when the cost of its metal is Rs. 50 per cubic meter :
Discuss
Answer & Solution
Answer: Option B
Solution:
Volume :
$$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 14 \times 14 \times 3.5} \right){m^3} \cr & = 2156\,{m^3} \cr} $$
∴ Cost of the cylinder :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {2156 \times 50} \right) \cr & = {\text{Rs}}{\text{. 107800}} \cr} $$
99
The radii of the bases of two cylinders are in the ratio 3 : 4 and their height are in the ratio 4 : 3. The ratio of their volume is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let their radii be 3x, 4x and heights be 4y, 3y
Ratio of their volumes :
$$\eqalign{ & = \frac{{\pi \times {{\left( {3x} \right)}^2} \times 4y}}{{\pi \times {{\left( {4x} \right)}^2} \times 3y}} \cr & = \frac{{36}}{{48}} \cr & = \frac{3}{4}\,Or\,3:4 \cr} $$
100
Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Discuss
Answer & Solution
Answer: Option C
Solution:
Volume one coin :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{75}}{{100}} \times \frac{{75}}{{100}} \times \frac{2}{{10}}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{99}}{{280}}{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of larger cylinder :
$$ = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10} \right){\text{ c}}{{\text{m}}^3}$$
Number of coins :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{{280}}{{99}}} \right) \cr & = 450 \cr} $$