A clown blows up a spherical balloon such that its...

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

A clown blows up a spherical balloon such that its...

by AAPL » Wed Nov 22, 2017 2:58 pm

A clown blows up a spherical balloon such that its volume increases at a constant rate. I takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches. How many seconds does it take for the radius of the balloon to increase from 3 inches to 5 inches?

NOTE: The volume of a single sphere is 4/3*Ï€r^3.

A. 6
B. 9
C. 24
D. 30
E. 42

The OA is E.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

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Source: Beat The GMAT — Problem Solving | Wed Nov 22, 2017 2:58 pm

EconomistGMATTutor: GMAT Instructor; Posts: 555; Joined: Wed Oct 04, 2017 4:18 pm; Thanked: 180 times; Followed by:12 members

Answer

by EconomistGMATTutor » Thu Nov 23, 2017 9:51 am

Hello AAPL.

This is an interesting question. I will try to explain how to solve it.

It takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches, so the volume increases from $$V_1=\frac{4\pi\left(1\right)^3}{3}=\frac{4\pi}{3}\ \ \ to\ \ V_2=\frac{4\pi\left(2\right)^3}{3}=\frac{32\pi}{3}$$ in 3 seconds.

We know that the volume increases at a constant rate. So it increase at $$\frac{28\pi}{3}\ each\ 3\ \sec onds,\ or\ \frac{28\pi}{9}\ per\ \sec ond.$$ Now, $$V_3=\frac{4\pi\left(3\right)^3}{3}=36\pi\ \ and\ V_5=\frac{4\pi\left(5\right)^3}{3}=\frac{500\pi}{3}.$$ So $$V_5-V_4=\frac{500\pi}{3}-36\pi=\frac{392\pi}{3}.$$ Finally, $$\frac{\frac{392\pi}{3}}{\frac{28\pi}{9}}=\frac{3528\pi}{84\pi}=42\ .$$ It implies that it takes 42 seconds to increase the radius from 3 inches to 5 inches. The correct answer is E .

I hope it can help you.

Feel free to ask me again if you have a doubt.

Regards.

GMAT Prep From The Economist
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

PS

by Scott@TargetTestPrep » Mon Dec 09, 2019 5:18 pm

AAPL wrote:A clown blows up a spherical balloon such that its volume increases at a constant rate. I takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches. How many seconds does it take for the radius of the balloon to increase from 3 inches to 5 inches?

NOTE: The volume of a single sphere is 4/3*Ï€r^3.

A. 6
B. 9
C. 24
D. 30
E. 42

The OA is E.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

If the radius of the balloon is 1 inch, the volume of the balloon is 4/3*π(1)^3 = 4π/3. If the radius is 2 inches, the volume is 4/3*π(2)^3 = 32π/3. Since it takes 3 seconds for the radius to increase from 1 inch to 2 inches, the rate at which the volume is increasing is (32π/3 - 4π/3)/3 = 28π/9 cubic inches per second.

Now, if the radius of the balloon is 3 inches, the volume of the balloon is 4/3*π(3)^3 = 108π/3. If the radius is 5 inches, the volume is 4/3*π(5)^3 = 500π/3. Since the rate at which the volume is increasing is 28π/9 cubic inches per second, then it takes (500π/3 - 108π/3)/(28π/9) = 392π/3 * 9/(28π) = 14 * 3 = 42 seconds to increase the radius of the balloon from 3 inches to 5 inches.

Answer: E

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