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Question 1 of 5
1. Question
Water flows into a tank 200 m × 150 m through a rectangular pipe of 1.5m × 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2 metres?
Correct
Solution: B
Given:
Dimension of tank = 200 m × 150 m
Height of tank = 2 metres
Speed of flow of water = 20 kmph
Dimension of rectangular pipe = 1.5 m × 1.25 m
Formula used:
Volume of any Cuboid = Length × Breadth × Height
Calculation:
According to the given question:
Volume of the water required in the tank = (200 × 150 × 2) m^{3} = 60000 m^{3}
Length of water column flown in1 min = (20 × 1000)/60 m = 1000/3 m
Volume flown per minute = 1.5 × 1.25 × (1000/3) m^{3} = 625 m^{3}
Required time = (60000/625)min = 96 minutes
Hence, the required answer is 96 minutes
Incorrect
Solution: B
Given:
Dimension of tank = 200 m × 150 m
Height of tank = 2 metres
Speed of flow of water = 20 kmph
Dimension of rectangular pipe = 1.5 m × 1.25 m
Formula used:
Volume of any Cuboid = Length × Breadth × Height
Calculation:
According to the given question:
Volume of the water required in the tank = (200 × 150 × 2) m^{3} = 60000 m^{3}
Length of water column flown in1 min = (20 × 1000)/60 m = 1000/3 m
Volume flown per minute = 1.5 × 1.25 × (1000/3) m^{3} = 625 m^{3}
Required time = (60000/625)min = 96 minutes
Hence, the required answer is 96 minutes

Question 2 of 5
2. Question
Area of a square is given as 12800 sq miles, find the time taken by a lady to move across a diagonal with speed of 15 miles per hour.
Correct
Solution: D
We have, area of a square = 12800 sq miles.
Hence side^{2} = 12800
Side = √12800
⇒ Side = 80√2
We know that diagonal of a square = √2 × side
= √2 × 80√2
= 160 miles.
Hence we know that speed = distance/time
Time = distance/speed
Time = 160/15
= 10.66 hr.
Incorrect
Solution: D
We have, area of a square = 12800 sq miles.
Hence side^{2} = 12800
Side = √12800
⇒ Side = 80√2
We know that diagonal of a square = √2 × side
= √2 × 80√2
= 160 miles.
Hence we know that speed = distance/time
Time = distance/speed
Time = 160/15
= 10.66 hr.

Question 3 of 5
3. Question
The volume and the radius of both cone and sphere are equal, then find the ratio of height of the cone to the diameter of the sphere?
Correct
Solution: C
Volume of cone = (1/3)πr^{2}h
Where r = radius of cone
Volume of sphere = (4/3)πR^{3}
Where R = radius of sphere
According to the question
R = r and
(1/3)πr^{2}h = (4/3)πR^{3}
⇒ h = 4 × R = 2 × D (∵ 2× R = diameter of sphere = D)
⇒ h/D = 2/1
Incorrect
Solution: C
Volume of cone = (1/3)πr^{2}h
Where r = radius of cone
Volume of sphere = (4/3)πR^{3}
Where R = radius of sphere
According to the question
R = r and
(1/3)πr^{2}h = (4/3)πR^{3}
⇒ h = 4 × R = 2 × D (∵ 2× R = diameter of sphere = D)
⇒ h/D = 2/1

Question 4 of 5
4. Question
A sphere of radius 2cms is dropped into a cylinder of radius 4 cms containing water upto 2.2cms. The raise in the water level is?
Correct
Solution: A
We know that
Volume of sphere = (4/3)πr^{3}
Where r = radius of sphere
Volume of cylinder = πR^{2}H
Where R = radius of cylinder
H = height of cylinder
Let h = raise height of water in cylinder
∵ Volume of water displaced in cylinder = volume of sphere
∴ πR^{2}H = (4/3)πr^{3}
⇒42×h=43×23
⇒ h = 2/3 cms = 0.67 cms
Incorrect
Solution: A
We know that
Volume of sphere = (4/3)πr^{3}
Where r = radius of sphere
Volume of cylinder = πR^{2}H
Where R = radius of cylinder
H = height of cylinder
Let h = raise height of water in cylinder
∵ Volume of water displaced in cylinder = volume of sphere
∴ πR^{2}H = (4/3)πr^{3}
⇒42×h=43×23
⇒ h = 2/3 cms = 0.67 cms

Question 5 of 5
5. Question
The diameter of a wheel is 1.4m. If this wheel rotates 500 rotations, how long it can travel?
Correct
Solution: D
Diameter = 1.4 m.
Circumference of the wheel = πd = 22/7 × 1.4
= 4.4
We have distance traveled in 1 rotation = 4.4 meter
Distance covered in 500 rotations = 4.4 × 500 = 2200 meters
Incorrect
Solution: D
Diameter = 1.4 m.
Circumference of the wheel = πd = 22/7 × 1.4
= 4.4
We have distance traveled in 1 rotation = 4.4 meter
Distance covered in 500 rotations = 4.4 × 500 = 2200 meters