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Aptitude Volume and Surface Area Test Yourself

43955.A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
49 $m^2$
50 $m^2$
53.5$ m^2$
55 $m^2$
Explanation:

Area of the wet surface = [2(lb + bh + lh) - lb]

= 2(bh + lh) + lb

= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] $m^2$

= 49 $m^2. $

43960.The curved surface area of a cylindrical pillar is 264 $m^2$ and its volume is 924 $m^3$. Find the ratio of its diameter to its height.
3 : 7
7 : 3
6 : 7
7 : 6
Explanation:
$\dfrac{\pi r^2 h}{2 \pi rh} = \dfrac{924}{264} => r = (\dfrac{924}{264} \times 2) = 7 m.$
And, $2 \pi rh = 264 => h = (264 \times \dfrac{7}{22} \times \dfrac{1}{2} \times \dfrac{1}{7}) = 6 m.$
ஃ Required ratio = $ \dfrac{2r}{h} = \dfrac{14}{6} = 7 : 3.$
43962.A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/$cm^3$, then the weight of the pipe is:
3.6 kg
3.696 kg
36 kg
36.9 kg
Explanation:

External radius = 4 cm,

Internal radius = 3 cm.

Volume of iron = $(\dfrac{22}{7} \times [(4)^2 - (3)^2] \times 21) cm^3$

= $(\dfrac{22}{7} \times 7 \times 1 \times 21) cm^3 $

= 462 cm^3

ஃ Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
43963.A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:
720
900
1200
1800
Explanation:

2(15 + 12) x h = 2(15 x 12)

=> h = $\dfrac{180}{27}m = \dfrac{20}{3}m$

ஃVolume = $(15 \times 12 \times \dfrac{20}{3})m^3 = 1200 m^3$

43966.50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is $4 m^3$, then the rise in the water level in the tank will be:
20 cm
25 cm
35 cm
50 cm
Explanation:

Total volume of water displaced = (4 x 50) $m^3 = 200 m^3.$

ஃ Rise in water level =$( \dfrac{200}{40 \times 20})$m 0.25 m = 25cm.

44167.Calculate the volume of a cone with radius 5cm and height 12cm
214.21 $cm^3$
314.16 $cm^3$
358.58 $cm^3$
287.36 $cm^3$
Explanation:

V = $\dfrac {1}{3} \pi r2 h$

= $\dfrac {1}{3} \times \pi \times 52 \times 12$

= 314.159.

= 314.16 $cm^3$
44172.The cost of the paint is Rs. 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint outside of a cube having 8 feet each side.
Rs. 850
Rs. 860
Rs. 876
Rs. 886
Explanation:

We will first calculate the Surface area of cube, then we will calculate the quantity of paint required to get answer.
Here we go,

Surface area =6$a^2$=6∗$8^2$=384sq feet

Quantity required =384/16=24kg

Cost of painting =36.50∗24

=Rs.876
44174.Calculate the volume of a sphere with radius 6cm.
904.78 $cm^3$
845.47 $cm^3$
642.87 $cm^3$
918.57 $cm^3$
Explanation:

V = $\dfrac {4}{3} \pi r3$

= $\dfrac {4}{3} x \pi x 63 $

= 904.7786

= 904.78 $cm^3$
44179.How many cubes of 10 cm edge can be put in a cubical box of 1 m edge.
10000 cubes
1000 cubes
100 cubes
50 cubes
Explanation:

Number of cubes= (100x100x100) / (10x10x10)= 1000

44180.Savitri had to make a model of a cylindrical kaleidoscope for her science project.
She wanted to use chart paper to make the curved surface of the kaleidoscope.
What would be the area of chart paper required by her, if she wanted
to make a kaleidoscope of length 25 cm with a 3.5 cm radius?
348 $cm^2$
468 $cm^2$
352 $cm^2$
550 $cm^2$
Explanation:

Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.

Height (length) of kaleidoscope (h) = 25 cm.

Area of chart paper required = curved surface area of the kaleidoscope

= 2πrh = $2 \times \dfrac{22}{7} \times 3.5 \times 25$

= 550 $cm^2$
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