Can A and can В are both right circular cylinders. The radiu : GMAT Problem Solving (PS)
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# Can A and can В are both right circular cylinders. The radiu

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Can A and can В are both right circular cylinders. The radiu [#permalink]

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17 Mar 2011, 13:46
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Can A and can В are both right circular cylinders. The radius of can A is twice the radius of can B, while the height of can A is half the height of can B. If it costs $4.00 to fill half of can B with a certain brand of gasoline, how much would it cost to completely fill can A with the same brand of gasoline? (A)$1
(B) $2 (C)$4
(D) $8 (E)$16
[Reveal] Spoiler: OA
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Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink]

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17 Mar 2011, 14:09
banksy wrote:
2. (KP) Can A and can В are both right circular cylinders. The radius of can A is twice the radius of can B, while the height of can A is half the height of can B. If it costs $4.00 to fill half of can B with a certain brand of gasoline, how much would it cost to completely fill can A with the same brand of gasoline? (A)$1
(B) $2 (C)$4
(D) $8 (E)$16

Let x be the radius of b and 2h be the height of B. Therefore, radius of A = 2x and height = h

Vol of b = 3.14*x^2*2h
Vol of a = 3.14*4x^2*h

cost to fill half of B = $4 --> cost to fill full B =$8
--> 3.14*x^2*2h = 8 --> 3.14*x^2*h = 4 --> 4*(3.14*x^2*h) = $16 Ans E SVP Joined: 16 Nov 2010 Posts: 1672 Location: United States (IN) Concentration: Strategy, Technology Followers: 33 Kudos [?]: 504 [0], given: 36 Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink] ### Show Tags 19 Mar 2011, 00:57 B -> pi * r^2*h A -> pi * 4r^2 * h/2 = 2 * pi * r^2 * h So, Vol of 1/2 of can B = vol of 1/4 can A So can A will take 4 * 4 =$16 for gasoline.
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Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink]

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19 Mar 2011, 01:52
banksy wrote:
2. (KP) Can A and can В are both right circular cylinders. The radius of can A is twice the radius of can B, while the height of can A is half the height of can B. If it costs $4.00 to fill half of can B with a certain brand of gasoline, how much would it cost to completely fill can A with the same brand of gasoline? (A)$1
(B) $2 (C)$4
(D) $8 (E)$16

Let radius of B be r and A be 2r
vol of B=πr^2h
vol of a=4πr^2h (radius= (2r)) == twice the volume of B

To fill 1/2 of B it cost 4$to fill B it cost 8$
to fill 2b ie (a) it cost 16$Director Status: No dream is too large, no dreamer is too small Joined: 14 Jul 2010 Posts: 650 Followers: 42 Kudos [?]: 833 [0], given: 39 Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink] ### Show Tags 19 Mar 2011, 02:05 banksy wrote: 2. (KP) Can A and can В are both right circular cylinders. The radius of can A is twice the radius of can B, while the height of can A is half the height of can B. If it costs$4.00 to fill half of can B with a certain brand of gasoline, how much would it cost to completely fill can A with the same brand of gasoline?
(A) $1 (B)$2
(C) $4 (D)$8
(E) $16 Volume of right circular cylinder = 2πr^2h Let Radius of Can A= 4 Height of can A = 2 Volume of Can A = 2*π*r^2*h = 2*π*4^2*2 = 64π Radius of can B = 2 Height of can B = 4 Volume of Can B= 2*π*r^2*h = 2*π*2^2*4 = 32π 1/2 of B= 32π/2=16π thus cost required to fill can A = (4/16π)*64π=$16
Ans E.
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Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink]

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22 Sep 2013, 19:34
sorry guys the answer is $8. lets consider the radius of can A to be 2x. radius of can B to be x. let the height of can A be y. Let the height of can B be 2y. Volume of a can A (pie 4x^2y) Volume of can B (pie 2 x^2 y). Since volume of can B is exactly half of can A. It costs$8 to fill can A completely.
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Re: 2. (KP) Can A and can В are both right circular cylinders. T [#permalink]

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22 Sep 2013, 23:26
mohnish104 wrote:
sorry guys the answer is $8. lets consider the radius of can A to be 2x. radius of can B to be x. let the height of can A be y. Let the height of can B be 2y. Volume of a can A (pie 4x^2y) Volume of can B (pie 2 x^2 y). Since volume of can B is exactly half of can A. It costs$8 to fill can A completely.

Notice that we a re told that it costs $4.00 to fill half of can B. so the correct answer is E, not D. Can A and can В are both right circular cylinders. The radius of can A is twice the radius of can B, while the height of can A is half the height of can B. If it costs$4.00 to fill half of can B with a certain brand of gasoline, how much would it cost to completely fill can A with the same brand of gasoline?

(A) $1 (B)$2
(C) $4 (D)$8
(E) $16 Volume of B = $$\pi{r^2}h$$ Volume of A = $$\pi{(2r)^2}(\frac{h}{2})=2\pi{r^2}h$$. Thus the volume of can A is twice the volume of cam B. If it costs$4.00 to fill half of can B, then it costs 4*$4=$16 to completely fill can A.

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Re: Can A and can В are both right circular cylinders. The radiu [#permalink]

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23 Sep 2013, 04:25
Thank you Bunuel. I guess the answer has been wrongly pointed out in the Kaplan tests.
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27 Jan 2016, 23:07
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Re: Can A and can В are both right circular cylinders. The radiu [#permalink]

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02 Feb 2016, 14:48
Hey guys,

one question. I first tried to pick some smart numbers. Apparently they weren't smart enough as I did not get the volume of cylinder A to be twice the volume of cylinder B. I picked 4 and 2 for radii and 3 and 6 for heights. Of course, if we pick 4 and 2 for radii and 2 and 4 for heights, we are fine. So my question then is, if I for whatever reason at the actual test decide to go with smart numbers strategy for such a problem, how can I avoid this mistake? What am I missing here?

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Re: Can A and can В are both right circular cylinders. The radiu [#permalink]

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02 Feb 2016, 22:31
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MrSobe17 wrote:
Hey guys,

one question. I first tried to pick some smart numbers. Apparently they weren't smart enough as I did not get the volume of cylinder A to be twice the volume of cylinder B. I picked 4 and 2 for radii and 3 and 6 for heights. Of course, if we pick 4 and 2 for radii and 2 and 4 for heights, we are fine. So my question then is, if I for whatever reason at the actual test decide to go with smart numbers strategy for such a problem, how can I avoid this mistake? What am I missing here?

Best,
Jay

Nothing! These numbers are fine too. You probably messed up the calculations somewhere.

Volume of a right circular cylinder $$= \pi*r^2*h$$

Volume of can A $$= \pi * 4^2* 3 = 48* \pi$$
Volume of can B $$= \pi * 2^2 * 6 = 24* \pi$$

Volume of can A is twice the volume of can B.

OF course, you can stick with radii as 2r and r and heights as h and 2h since you just need the comparative volumes of the two cans.
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Re: Can A and can В are both right circular cylinders. The radiu [#permalink]

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03 Feb 2016, 00:50
Oh boy, indeed I messed up the easy calculation. Hate those mistakes so much.

Thanks for clarification Karishma!

Best,
Jay
Re: Can A and can В are both right circular cylinders. The radiu   [#permalink] 03 Feb 2016, 00:50
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