If a 10 cm tall aluminum can that is a perfect right

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

by VJesus12 » Thu May 03, 2018 10:41 am

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If a 10 cm tall aluminum can that is a perfect right circular cylinder contains 250 cm^3 of soda, what is the diameter of the can in centimeters? $$\left(A\right)\ \frac{5}{\sqrt{\pi}}$$ $$\left(B\right)\ \frac{10}{\pi}$$ $$\left(C\right)\ \frac{10}{\sqrt{\pi}}$$ $$\left(D\right)\ 10\pi$$ $$\left(E\right)\ 5\sqrt{\pi}$$ The OA is the option C.

How can I find the diameter of the can? Help!!!

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Source: Beat The GMAT — Problem Solving | Thu May 03, 2018 10:41 am

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Thu May 03, 2018 12:20 pm

Hi vjesus12.

To solve this PS question we have to remember the formula of the Volume of a Cylinder: $$V=\pi\cdot r^2\cdot h$$ where "r" is the radius of the circle (the base) and "h" is the height of the cylinder.

For this question, we know that h=10cm and V=250cm^3. Hence, we can find the radius "r" and then we will calculate the diameter d=2*r.
Hence, $$V=\pi\cdot r^2\cdot h$$ $$\Rightarrow\ \ 250=\pi\cdot r^2\cdot10$$ $$\Rightarrow\ \ r^2=\frac{25}{\pi}$$ $$\Rightarrow\ \ r=\sqrt{\frac{25}{\pi}}$$ $$\Rightarrow\ \ r=\frac{5}{\sqrt{\pi}}.$$ Now, we caluclate the diameter $$d=2\cdot r$$ $$d=2\cdot\frac{5}{\sqrt{\pi}}cm$$ $$d=\frac{10}{\sqrt{\pi}}cm.$$ Therefore, the correct answer is the option C.

I really hope this explanation may help you.

Regards.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Mon May 07, 2018 10:20 am

VJesus12 wrote:If a 10 cm tall aluminum can that is a perfect right circular cylinder contains 250 cm^3 of soda, what is the diameter of the can in centimeters? $$\left(A\right)\ \frac{5}{\sqrt{\pi}}$$ $$\left(B\right)\ \frac{10}{\pi}$$ $$\left(C\right)\ \frac{10}{\sqrt{\pi}}$$ $$\left(D\right)\ 10\pi$$ $$\left(E\right)\ 5\sqrt{\pi}$$ The OA is the option C.

We can use the volume equation:

volume = Ï€r^2h

250 = Ï€r^2 x 10

25 = Ï€r^2

25/Ï€ = r^2

5/âˆšÏ€ = r

So the diameter is 10/âˆšÏ€.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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