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

61. The length, breadth and height of a cuboid are in the ratio 1 : 2 : 3. The length, breadth and height of the cuboid are increased by 100%, 200% and 200% respectively. Then the increase in the volume of the cuboid is :

A. 5 Times

B. 6 Times

C. 12 Times

D. 17 Times

Discuss

Solution:

Let the original length, breadth and height of the cuboid be x, 2x and 3x units respectively
Then, original volume = (x × 2x × 3x) cu.units = 6x³ cu.units
New length = 200% of x = 2x
New breadth = 300% of 2x = 6x
New height = 300% of 3x = 9x
∴ New volume :
= (2x × 6x × 9x) cu.units
= 108x³ cu.units
Increase in volume :
= (108x³ - 6x³) cu.units
= (102x³) cu.units
∴ Required ratio :

\begin{aligned} = \frac{102 x^{3}}{6 x^{3}} \\ = 17 (Times) \end{aligned}

62. A water tank is 30 m long, 20 m wide and 12 m deep. It is made of iron sheet which is 3 m wide. The tank is open at the top. If the cost of the iron sheet is Rs. 10 per metre, then the total cost of the iron sheet required to build the tank is :

A. Rs. 6000

B. Rs. 8000

C. Rs. 9000

D. Rs. 10000

Discuss

Solution:

Since the tank is open at the top, we have :
Area of sheet required = Surface area of the tank

\begin{aligned} = l b + 2 (b h + l h) \\ = [30 \times 20 + 2 (20 \times 12 + 30 \times 12)] m^{2} \\ = (600 + 1200) m^{2} \\ = 1800 m^{2} \end{aligned}

Length of sheet required :

\begin{aligned} = (\frac{Area}{Width}) \\ = \frac{1800}{3} m \\ = 600 m \end{aligned}

∴ Cost of the sheet
= Rs. (600 × 10)
= Rs. 6000

63. If the total length of diagonals of a cube is 12 cm, then what is the total length of the edges of the cube ?

A. 6 $\sqrt{3}$ cm

B. 12 cm

C. 15 cm

D. 12 $\sqrt{3}$ cm

Discuss

Solution:

Since a cube has 4 diagonals, we have :
Length of a diagonal

\begin{aligned} = (\frac{12}{4}) c m \\ = 3 c m \end{aligned}

Let the length of each edge of the cube be a cm
Then,

\begin{aligned} \sqrt{3} a = 3 \\ o r, a = \sqrt{3} \end{aligned}

∴ Total length of the edges of the cube =

12 \sqrt{3}

64. By what percent the volume of a cube increases if the length of each edge was increased by 50%

A. 50%

B. 125%

C. 237.5%

D. 273.5%

Discuss

Solution:

Let original edge = a
Then, original volume = a³
New edge :

\begin{aligned} = \frac{150}{100} a \\ = \frac{3 a}{2} \end{aligned}

New volume :

\begin{aligned} = {(\frac{3 a}{2})}^{3} \\ = \frac{27 a^{3}}{8} \end{aligned}

Increase in volume :

\begin{aligned} = (\frac{27 a^{3}}{8} - a^{3}) \\ = \frac{19 a^{3}}{8} \end{aligned}

∴ Increase % :

\begin{aligned} = (\frac{19 a^{3}}{8} \times \frac{1}{a^{3}} \times 100) % \\ = 237.5 % \end{aligned}

65. The height of a closed cylinder of given volume and the minimum surface area is :

A. Equal to its diameter

B. Half of its diameter

C. Double of its diameter

D. None of these

Discuss

Solution:

\begin{aligned} V = π r^{2} h and \\ S = 2 π r h + 2 π r^{2} \\ = 2 π r (h + r) \\ Where, h = \frac{V}{π r^{2}} \\ \Rightarrow S = 2 π r (\frac{V}{π r^{2}} + r) \\ \Rightarrow S = \frac{2 V}{r} + 2 π r^{2} \\ \Rightarrow \frac{d S}{d r} = \frac{- 2 V}{r^{2}} + 4 π r and \\ \frac{d^{2} S}{d r^{2}} = (\frac{4 V}{r^{3}} + 4 π) > 0 \end{aligned}

∴ S is minimum when :

\begin{aligned} \frac{d S}{d r} = 0 \\ \Rightarrow \frac{- 2 V}{r^{2}} + 4 π r = 0 \\ \Rightarrow V = 2 π r^{3} \\ \Rightarrow π r^{2} h = 2 π r^{3} \\ \Rightarrow h = 2 r \end{aligned}

66. Water is poured into an empty cylindrical tank at a constant rate for 5 minutes. After the water has been poured into the tank. the depth of the water is 7 feet. The radius of the tank is 100 feet. Which of the following is the best approximation for the rate at which the water was poured into the tank ?

A. 140 cubic feet/sec

B. 440 cubic feet/sec

C. 700 cubic feet/sec

D. 2200 cubic feet/sec

Discuss

Solution:

Volume of water flown into the tank in 5 min :

\begin{aligned} = (\frac{22}{7} \times 100 \times 100 \times 7) cu .feet \\ = 220000 cu .feet \end{aligned}

∴ Rate of flow of water :

\begin{aligned} = (\frac{220000}{5 \times 60}) cu .feet/sec \\ = 733.3 \approx 700 cu .feet/sec \end{aligned}

67. The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is :

A. 60π cm²

B. 68π cm²

C. 120π cm²

D. 136π cm²

Discuss

Solution:

h = 15 cm, r = 8 cm
So,

\begin{aligned} l = \sqrt{r^{2} + h^{2}} \\ = \sqrt{8^{2} + {(15)}^{2}} \\ = 17 c m \end{aligned}

∴ Curved surface area :

\begin{aligned} = π r l \\ = (π \times 8 \times 17) c m^{2} \\ = 136 π c m^{2} \end{aligned}

68. A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and the height are in the ratio 5 : 12, then the ratio of the total surface area of the cylinder to that of the cone is :

A. 3 : 1

B. 13 : 9

C. 17 : 9

D. 34 : 9

Discuss

Solution:

Let their radius and height be 5x and 12x respectively
Slant height of the cone,

l = \sqrt{{(5 x)}^{2} + {(12 x)}^{2}} = 13 x

\begin{aligned} \frac{Total surface area of cylinder}{Total surface area of cone} \\ = \frac{2 π r (h + r)}{π r (l + r)} \\ = \frac{2 (h + r)}{(l + r)} \\ = \frac{2 \times (12 x + 5 x)}{(13 x + 5 x)} \\ = \frac{34 x}{18 x} \\ = \frac{17}{9} O r 17 : 9 \end{aligned}

69. The curved surface area of a sphere is 5544 sq.cm. Its volume is :

A. 22176 cm³

B. 33951 cm³

C. 38808 cm³

D. 42304 cm³

Discuss

Solution:

\begin{aligned} 4 π r^{2} = 5544 \\ \Rightarrow r^{2} = (5544 \times \frac{1}{4} \times \frac{7}{22}) \\ \Rightarrow r^{2} = 441 \\ \Rightarrow r = 21 \end{aligned}

∴ Volume :

\begin{aligned} = (\frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21) c m^{3} \\ = 38808 c m^{3} \end{aligned}

70. If a hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cylinder of base diameter 8 cm, then the height of the cylinder is :

A. 4 cm

B. $\frac{13}{3}$ cm

C. $\frac{14}{3}$ cm

D. 5 cm

Discuss

Solution:

Let the height of the cylinder be h cm
Then,

\begin{aligned} \frac{4}{3} π [{(4)}^{3} - {(2)}^{3}] = π \times 4^{2} \times h \\ \Rightarrow \frac{4}{3} \times π \times 56 = π \times 16 h \\ \Rightarrow h = \frac{4 \times 56}{3 \times 16} \\ \Rightarrow h = \frac{14}{3} c m \end{aligned}

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