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A cylinder is being filled with sand weighing 200 kg per cub [#permalink]

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17 Feb 2013, 10:01

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A cylinder is being filled with sand weighing 200 kg per cubic foot. The cylinder has a diameter of 1 foot and is 5 feet tall. How much sand is being used??

A 250pi kg B 500pi kg C 500pi^2 kg D 1000pi kg E 1000pi^2 kg

Lots of geometry questions! The area of a cylinder is Pi*Radius^2*Height. A cylinder is like a bunch of circles stacked on top of each other and that is where the height comes in. So in this case (to find the volume) you need to find the area of the circle: (.5^2)*Pi and multiply that by the height of 5. So the volume is 5/4PI cubic feet. That's how much sand you need to fill the cylinder. Now, you know that the sand weighs 200Kg per cubic foot. So, multiply 200 by 5/4Pi and you get 250Pi Kg.

Let me know if this makes sense to you.

HG.
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"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

Re: A cylinder is being filled with sand weighing 200 kg per cub [#permalink]

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22 Nov 2014, 15:11

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A cylinder is being filled with sand weighing 200 kg per cub [#permalink]

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08 Jan 2015, 12:50

What is quite interesting is that, while I love geometry problems I didn't even think that the diameter is double the radius.

Anyway, what I did was to find r using the circumference of the circle, 2πr. So, I basically said that the circumference of the circle the buttom of the cylinder) is 1, so 2πr=1 --> r= 1/2π.

Then I substituted this into πr^2*h, which is the capacity of the cylinder. This ended in a capacity of 5/4.

Finally, I did 5/4*200 = 2500. At this point I realised that 2πr=1 is not correct (I confused the word diameter with perimeter), but it still has the elements I needed and went with answer A, which is the correct one.

Even though I understand the correct solution, could you tell me what made my result this similar? I mean, in terms of "number properties". Because I guess there is a relationship I don't see...

The reason why you still got to the correct answer is more about co-incidence and "off-setting errors" than anything else.

The co-incidence is that whoever wrote the question chose to make the diameter 1 ft, so the radius = 1/2 ft. When you tried to use the circumference....2r(pi) = 1....you ended up with 1/(2pi).

The offsetting errors take a bit more explanation.... 1/(2pi) is close to 1/6, so you SHOULD have come up with V = (5)(pi)(1/2pi)^2 = 5/(4pi) = about 5/12.... which would have given you a final answer that should have been MUCH smaller (a little over 100 ft^3). But you ignored the pi in the denominator when you did the calculation, so you ended up with 5/4....which is what you get when you do the math correctly.

This is certainly a funny set of circumstances, but you have to 'weed out' these little mistakes from your 'process' - on Test Day, they can seriously hurt your performance.

Re: A cylinder is being filled with sand weighing 200 kg per cub [#permalink]

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17 Jul 2016, 08:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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