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# M25-37

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Math Expert
Joined: 02 Sep 2009
Posts: 49208

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16 Sep 2014, 01:24
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Difficulty:

5% (low)

Question Stats:

93% (00:48) correct 7% (01:15) wrong based on 297 sessions

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A cylindrical tank has a base radius of 1 meters and a height equal to 2 meters. The tank is half full of water. If all the water from that tank is poured into another cylindrical tank with the base radius equal to 2 meters and a height of 1 meters, what percent of this new tank will be filled with water?

A. 20%
B. 25%
C. 32%
D. 35%
E. 40%

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Math Expert
Joined: 02 Sep 2009
Posts: 49208

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16 Sep 2014, 01:24
1
Official Solution:

A cylindrical tank has a base radius of 1 meters and a height equal to 2 meters. The tank is half full of water. If all the water from that tank is poured into another cylindrical tank with the base radius equal to 2 meters and a height of 1 meters, what percent of this new tank will be filled with water?

A. 20%
B. 25%
C. 32%
D. 35%
E. 40%

The volume of a cylinder is $$\pi*r^2*h$$.

The volume of tank #1 is $$\pi*1^2*2=2 \pi$$. Since, the tank is half full, then the volume of water in it is $$\frac{2 \pi}{2}=\pi$$.

The volume of tank #2 is $$\pi*2^2*1=4 \pi$$.

Percent of water in tank #2 is $$\frac{\pi}{4 \pi}*100=25%$$.

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18 Jul 2018, 20:19
Bunuel wrote:
A cylindrical tank has a base radius of 1 meters and a height equal to 2 meters. The tank is half full of water. If all the water from that tank is poured into another cylindrical tank with the base radius equal to 2 meters and a height of 1 meters, what percent of this new tank will be filled with water?

A. 20%
B. 25%
C. 32%
D. 35%
E. 40%

This can be solved by simple application of the formula of the Volume of the Cylinder = πr^2h

Volume of water in Tank 1= π*1^2*2/2 (since it is half full)= π
Volume of Tank 2= π*2^2*1=4 π
Now Percent of 2nd tank filled = (π/4π)*100 = 25%
Re: M25-37 &nbs [#permalink] 18 Jul 2018, 20:19
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# M25-37

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