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21. The diameter of the base of a cylindrical drum is 35 dm and the height is 24 dm. It is full of kerosene. How many tins each of size 25 cm × 22 cm × 35 cm can be filled with kerosene from the drum ?

A. 120

B. 600

C. 1020

D. 1200

Discuss

Solution:

\begin{aligned} Number of tins \\ = \frac{Voulme of the drum}{Volume of each tin} \\ = \frac{(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 24)}{(\frac{25}{10} \times \frac{22}{10} \times \frac{35}{10})} \\ = 1200 \end{aligned}

22. The radius of a cylindrical cistern is 10 metres and its height is 15 metres. Initially the cistern is empty. We start filling the cistern with water through a pipe whose diameter is 50 cm. Water is coming out of the pipe with a velocity of 5 m/sec. How many minutes will it take in filling the cistern with water ?

A. 20 min

B. 40 min

C. 60 min

D. 80 min

Discuss

Solution:

Volume of cistern :

\begin{aligned} = (π \times 10^{2} \times 15) m^{3} \\ = 1500 π m^{3} \end{aligned}

Volume of water flowing through the pipe in 1 sec :

\begin{aligned} = (π \times 0.25 \times 0.25 \times 5) m^{3} \\ = 0.3125 π m^{3} \end{aligned}

∴ Time taken to fill the cistern :

\begin{aligned} = (\frac{1500 π}{0.3125 π}) \\ = (\frac{1500 \times 10000}{3125}) \\ = 4800 \sec \\ = (\frac{4800}{60}) min \\ = 80 min \end{aligned}

23. What length of solid cylinder 2 cm in diameter must be taken to cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick and 15 cm long ?

A. 42.3215 cm

B. 44.0123 cm

C. 44.0625 cm

D. 44.6023 cm

Discuss

Solution:

External radius = 6 cm
Internal radius = (6 - 0.25) = 5.75 cm
Volume of material in hollow cylinder :

\begin{aligned} = [\frac{22}{7} \times {{(6)}^{2} - {(5.75)}^{2}} \times 15] c m^{3} \\ = (\frac{22}{7} \times 11.75 \times 0.25 \times 15) c m^{3} \\ = (\frac{22}{7} \times \frac{1175}{100} \times \frac{25}{100} \times 15) c m^{3} \\ = (\frac{11 \times 705}{56}) c m^{3} \end{aligned}

Let the length of solid cylinder be h
Then,

\begin{aligned} \frac{22}{7} \times 1 \times 1 \times h = (\frac{11 \times 705}{56}) \\ \Rightarrow h = (\frac{11 \times 705}{56} \times \frac{7}{22}) \\ \Rightarrow h = 44.0625 cm \end{aligned}

24. If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :

A. remains unaltered

B. decreases by 25%

C. increases by 25%

D. increases by 50%

Discuss

Solution:

Let the original radius and height of the cone be r and h respectively
Then, Original volume =

\frac{1}{3} π r^{2} h

New radius =

\frac{r}{2}

and new hight = 2h
New volume :

\begin{aligned} = \frac{1}{3} \times π \times {(\frac{r}{2})}^{2} \times 3 h \\ = \frac{3}{4} \times \frac{1}{3} π r^{2} h \end{aligned}

∴ Decrease % :

\begin{aligned} = (\frac{\frac{1}{4} \times \frac{1}{3} π r^{2} h}{\frac{1}{3} π r^{2} h} \times 100) % \\ = 25 % \end{aligned}

25. A bucket is in the from of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

A. 15 cm

B. 20 cm

C. 25 cm

D. 30 cm

Discuss

Solution:

Volume of bucket :
= 28.490 litres
= (28.490 ×1000) cm³
= 28490 cm³
Let the height of the bucket be h cm
We have : r = 21 cm. and R = 28 cm

∴ \frac{π}{3} h [{(28)}^{2} + {(21)}^{2} + 28 \times 21]

= 28490

\Rightarrow h (784 + 441 + 588) =

\frac{28490 \times 21}{22}

\begin{aligned} \Rightarrow 1813 h = 27195 \\ \Rightarrow h = \frac{27195}{1813} \\ \Rightarrow h = 15 c m \end{aligned}

26. If the volume and surface area of a sphere are numerically the same, then its radius is :

A. 1 unit

B. 2 units

C. 3 units

D. 4 units

Discuss

Solution:

\begin{aligned} \frac{4}{3} π r^{3} = 4 π r^{2} \\ \Rightarrow r = 3 units \end{aligned}

27. A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is :

A. 2.5 cm

B. 3 cm

C. 3.5 cm

D. 4 cm

Discuss

Solution:

Let the radius of the sphere be r cm
Then,

\begin{aligned} \frac{4}{3} π r^{3} = π \times {(0.1)}^{2} \times 3600 \\ \Leftrightarrow r^{3} = 36 \times \frac{3}{4} \\ \Leftrightarrow r^{3} = 27 \\ \Leftrightarrow r = 3 c m \end{aligned}

28. The capacities of two hemispherical vessels are 6.4 litres and 21.6 litres. The areas of inner curved surfaces of the vessels will be in the ratio of :

A. $\sqrt{2}$ : $\sqrt{3}$

B. 2 : 3

C. 4 : 9

D. 16 : 81

Discuss

Solution:

Let their radii be R and r
Then,

\begin{aligned} \frac{\frac{2}{3} π R^{3}}{\frac{2}{3} π r^{3}} = \frac{6.4}{21.6} \\ \Rightarrow {(\frac{R}{r})}^{3} = \frac{8}{27} \\ \Rightarrow {(\frac{R}{r})}^{3} = {(\frac{2}{3})}^{3} \\ \Rightarrow \frac{R}{r} = \frac{2}{3} \end{aligned}

∴ Ratio of curved surface area :

= \frac{2 π R^{2}}{2 π r^{2}} = {(\frac{R}{r})}^{2} = \frac{4}{9} o r 4 : 9

29. The base of a pyramid is an equilateral triangle of side 1 m. If the height of the pyramid is 4 metres, then the volume is :

A. 0.550 m³

B. 0.577 m³

C. 0.678 m³

D. 0.750 m³

Discuss

Solution:

Area of the base :

\begin{aligned} = (\frac{\sqrt{3}}{4} \times 1^{2}) m^{2} \\ = \frac{\sqrt{3}}{4} m^{2} \end{aligned}

∴ Volume of pyramid :

\begin{aligned} = (\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 4) m^{3} \\ = (\frac{\sqrt{3}}{3}) m^{3} \\ = (\frac{1.732}{3}) m^{3} \\ = 0.577 m^{3} \end{aligned}

30. Each side of a cube is decreased by 25%. Find the ratio of the volume of the original cube and the resulting cube = ?

A. 64 : 1

B. 27 : 64

C. 64 : 27

D. 8 : 1

Discuss

Solution:

Let the side of cube = 10 cm
∴ Original volume = 10 × 10 × 10 = 1000 cm³
Now, side of new cube := 10 - 25% of 10 = 7.5
∴ New volume = 7.5 × 7.5 × 7.5 = 421.875 cm³
∴ Required ratio :

\begin{aligned} = \frac{1000}{421.875} \\ = \frac{1000000}{421875} \\ = \frac{64}{27} O r 64 : 27 \end{aligned}

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