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| Aptitude Interview Questions And Answers |
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46.
Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.
Sol: Slant Height, l = √(r^2 + h^2) =√(21^2 + 28^2) = √1225 = 35 cm
Volume = 1/3пr^2h = (1/3 * 22/7 * 21 * 21 * 28) cm^3 = 12936 cm^3
Curved surface area = пrl = 22/7 * 21 *35 cm^3 = 2310 cm^2
Total Surface Area = (пrl + пr^2) = (2310 + 22/7 * 21 * 21) cm^2 = 3696 cm^2 |
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47.
If the radius of the sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.
Sol: Let the original radius = R. Then, new radius = 150/100 R = 3R/2
Original Volume = 4/3пR^3, New volume = 4/3п(3 R/2)^3 = 9пR^3/2
Original surface area = 4пR^2 , New surface area = 4п(3R/2)^2 = 9пR^2
Increase % in surface area = (5пR^2/4пR^2 * 100)% = 125%
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48.
If each edge of a cube is increased by 50%, find the percentage increase in its surface area.
Sol: Let the original length of each edge = a
Then, Original surface area = 6a^2
New surface area = 6 * (3a/2)^2 = 27a^2/2
Increase percent in surface area = (15/2a^2 * 1/(6a^2) * 100)% = 125%
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49.
Find the number of the bricks, each measuring 25 cm by 12.5 cm by 7.5 cm, required to build a wall 6 m long, 5 m high and 50cm thick, while the mortar occupies 5% of the volume of the wall.
Sol: Volume of the Wall = (600 * 500 * 50) cu. Cm.
Volume of the bricks = 95% of the volume of the wall.
= (95/100 * 600 * 500 * 50) cu. Cm.
Volume of 1 brick = (25 * 25/2 * 75/2) cu. Cm.
Therefore, Number of bricks = (95/100 * (600 * 500 * 50 * 2 * 10)/(25 * 25 * 75))=6080
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50.
The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Sol: Area of the field = Total cost/Rate = (333.18/24.68) hectares =13.5 hectares.
= (13.5*10000) m^2 =135000m^2.
Let altitude = x meters and base = 3x meters.
Then, ½ *3x* x= 135000 or x^2 = 9000 or x= 300.
Therefore, base =900 m & altitude = 300m.
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