Quantitative Aptitude
VOLUME AND SURFACE AREA MCQs
Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
So, Volume of the box = (32 x 20 x 8) m3 = 5120 m3
\(\frac{\pi r^{2}h}{2\pi rh} = \frac{924}{264}\Rightarrow r= \left(\frac{924}{264}\times2\right) = 7m\)
And, \(2\pi rh\Rightarrow h= \left(264\times\frac{7}{22}\times\frac{1}{2}\times\frac{1}{7}\right) = 6m.\)
So, Required ratio = \(\frac{2r}{h}=\frac{14}{6}= 7:3.\)
Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) = \(\frac{8000\times1000}{320\times250}=100\)
x = 10 cm = 1 dm.
h = 14 cm, r = 7 cm.
So, l = (7)2 + (14)2 = 245 = 75 cm.
So, Total surface area = \(\pi rl+\pi r^{2}\)
= \(\left(\frac{22}{7}\times7\times75+\frac{22}{7}\times7\times7\right)cm^{2}\)
= [154(5 + 1)] cm2
= (154 x 3.236) cm2
= 498.35 cm2.
Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
Let the edge of the large cube be a.
So, a3 = 216 \(\Rightarrow\) a = 6 cm.
So, Required ratio = \(\left(\frac{6\times(3^{2}\times4^{2}\times5^{2})}{6\times6^{2}}\right)=\frac{50}{36}= 25:18.\)
Number of bricks = \(\frac{Volume of the wall}{Volume of 1brick}=\left(\frac{800\times600\times22.5}{25\times11.25\times6}\right)= 6400.\)
 - Volume of rectangular tank = 650 litres = 650 × 1000 cm 3 = 650000 cm 3
We know that Volume of a cuboid = l × b × h
650000 = 130 × 250 × h
⇒ h = (650000)/( 130 × 250 )= 20 cm
Therefore, height of the tank = 20 cm
 - ∴Volume of the hall = length × breadth × height
= 16 × 14 × 5
= 1120 m3
Volume occupied my 1 man = 3.5 m3
3.5 m3 of air occupied my 1 man
1120 m3 of air occupied my ( 1/3.5)x 1120= 320 man
Therefore, number of men that can be accommodated = 320men
- Diameter of cylinder = 48 cm
⇒ Radius of cylinder (r) = 24 cm
Now, when the Cuboid is completely submerged
Then, Volume of cylindrical portion formed by original water level and after increase in water level =
volume of cuboid
⇒ πr 2 h = 33 cm × 18 cm × 12 cm
22/7 x 24 x 24 x h = 33 cm × 18 cm × 12 cm
h = (33 cm × 18 cm × 12 cm x7)/ 22 x 24 x 24=3.94cm
Hence, the required increase in water level is 3.94 cm
We are given a cylinder container with diameter 48cm which contains sufficient water to submerge a rectangular solid of iron with dimensions 33cm x 18cm x 12cm. We need to find the rise in the level of water when the solid is completely submerged.
Let's first find the volume of the rectangular solid of iron:
Volume of the rectangular solid = Length x Breadth x Height
= 33cm x 18cm x 12cm
= 7128 cubic cm
Now, let's find the volume of water displaced when the solid is completely submerged in the cylinder container. Since the volume of water displaced will be equal to the volume of the solid, we have:
Volume of water displaced = 7128 cubic cm
Let's find the height by which the level of water will rise in the cylinder container:
Volume of water displaced = πr^2h
where r is the radius of the cylinder and h is the height by which the level of water will rise.
Since the diameter of the cylinder is given as 48cm, the radius will be half of it, i.e., r = 24cm.
Substituting the values in the formula above, we get:
7128 = π x 24^2 x h
h = 7128 / (π x 24^2)
h ≈ 3.94 cm
Therefore, the rise in the level of water when the solid is completely submerged is 3.94 cm.
Hence, the correct option is B. 3.94 cm.
If you think the solution is wrong then please provide your own solution below in the comments section .
 - circumference - radius = 74
pie d- d/2 =74
22d/7 - d/2 =74
(44d-7d)/14 =74
37d = 74x14
d= (74x14)/37
d = 28cm