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Home > Practice > Arithmetic Aptitude > Volume and Surface Area > Miscellaneous

91. 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 :

A. 264 cm²

B. 376 cm²

C. 456 cm²

D. 696 cm²

Discuss

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

92. 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 ?

A. 440 m³

B. 480 m³

C. 508 m³

D. 520 m³

Discuss

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

93. 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.

A. 2 m, 12 m, 8 m

B. 4 m, 9 m, 16 m

C. 8 m, 12 m, 16 m

D. None of these

Discuss

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

\begin{aligned} ∴ 52 x^{2} = \frac{1248}{1.50} \\ \Rightarrow 52 x^{2} = 832 \\ \Rightarrow x^{2} = \frac{832}{52} \\ \Rightarrow x^{2} = 16 \\ \Rightarrow x = 4 \end{aligned}

Hence, the diameter of the box are 8 m, 12 m and 16 m

94. 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 :

A. 2 dm

B. 6 dm

C. 18 dm

D. 72 dm

Discuss

Solution:

Let the length of the tank be x dm
Then, depth of the tank =

\frac{x}{3}

dm
Breadth of the tank :

\begin{aligned} = [\frac{1}{2} of \frac{1}{3} of (x - \frac{x}{3})] dm \\ = (\frac{1}{2} \times \frac{1}{3} \times \frac{2 x}{3}) dm \\ = \frac{x}{9} dm \end{aligned}

\begin{aligned} ∴ x \times \frac{x}{9} \times \frac{x}{3} = 216 \\ \Rightarrow x^{3} = 216 \times 27 \\ \Rightarrow x = 6 \times 3 \\ \Rightarrow x = 18 d m \end{aligned}

95. 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 :

A. 81000 cu.cm

B. 81775 cu.cm

C. 82125 cu.cm

D. None of these

Discuss

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

96. 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 :

A. Rs. 100208

B. Rs. 107800

C. Rs. 10800

D. Rs. 109800

Discuss

Solution:

Volume :

\begin{aligned} = π r^{2} h \\ = (\frac{22}{7} \times 14 \times 14 \times 3.5) m^{3} \\ = 2156 m^{3} \end{aligned}

∴ Cost of the cylinder :

\begin{aligned} = Rs . (2156 \times 50) \\ = Rs . 107800 \end{aligned}

97. 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 :

A. 2 : 3

B. 3 : 2

C. 3 : 4

D. 4 : 3

Discuss

Solution:

Let their radii be 3x, 4x and heights be 4y, 3y
Ratio of their volumes :

\begin{aligned} = \frac{π \times {(3 x)}^{2} \times 4 y}{π \times {(4 x)}^{2} \times 3 y} \\ = \frac{36}{48} \\ = \frac{3}{4} O r 3 : 4 \end{aligned}

98. 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.

A. 430

B. 440

C. 450

D. 460

Discuss

Solution:

Volume one coin :

\begin{aligned} = (\frac{22}{7} \times \frac{75}{100} \times \frac{75}{100} \times \frac{2}{10}) c m^{3} \\ = \frac{99}{280} c m^{3} \end{aligned}

Volume of larger cylinder :

= (\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10) c m^{3}

Number of coins :

\begin{aligned} = (\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{280}{99}) \\ = 450 \end{aligned}

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