Water enters a cylindrical barrel at a constant speed

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Mon Jul 02, 2018 12:08 am

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[GMAT math practice question]

Water enters a cylindrical barrel at a constant speed of 500 cm^3/min, and the height of the barrel increases at a constant speed of 10 cm per minute. What is the approximate radius of the barrel, in centimeters?

A. 1
B. 2
C. 3
D. 4
E. 5

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Shahrukh_mbabreakspace: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Mon Jul 02, 2018 1:56 amFollowed by: 2 members

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by Shahrukh_mbabreakspace » Mon Jul 02, 2018 4:52 am

V/T= A*H/T, where V/T is volumetric flow rate, A is the cross-section area and H/T is change in height with time.
So, 500=A*10
=> A= 50
So pi*r^2= 50
And r~4

Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

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by Max@Math Revolution » Wed Jul 04, 2018 12:38 am

=>

Let r be the radius of the barrel.
The area of the water surface 3.14*r2.
The volume of water poured in 1 minute is 10*3.14*r^2.
Then, 10*3.14*r^2 = 500 or 31.4*r^2 = 500.
r^2=500/31.4 ~16.

Thus, the radius is approximately 4 cm.

Therefore, the answer is D.
Answer: D

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Jul 04, 2018 6:15 pm

Max@Math Revolution wrote:[GMAT math practice question]

Water enters a cylindrical barrel at a constant speed of 500 cm^3/min, and the height of the barrel increases at a constant speed of 10 cm per minute. What is the approximate radius of the barrel, in centimeters?

A. 1
B. 2
C. 3
D. 4
E. 5

The rate the water enters a cylinder is the area of the circular base of the cylinder times the rate the height increases. Since the area of the circular base of the cylinder is πr^2, we have:

500 = πr^2 x 10

50/π = r^2

√(50/π) = r

Since π is approximately 3, then 50/π is between 16 and 17. Let’s round it down to 16, so we have:

r ≈ √16 = 4

Answer: D

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