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Two oil cans, X and Y, are right circular cylinders, and the height an

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Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

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New post 10 Dec 2017, 08:05
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Two oil cans, X and Y, are right circular cylinders, and the height and the radius of Y are each twice those of X. If the oil in can X, which is filled to capacity, sells for $2, then at the same rate, how much does the oil in can Y sell for if Y is filled to only half its capacity?

(A) $ 1
(B) $ 2
(C) $ 3
(D) $ 4
(E) $ 8

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Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

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New post Updated on: 10 Dec 2017, 09:37
Bunuel wrote:
Two oil cans, X and Y, are right circular cylinders, and the height and the radius of Y are each twice those of X. If the oil in can X, which is filled to capacity, sells for $2, then at the same rate, how much does the oil in can Y sell for if Y is filled to only half its capacity?

(A) $ 1
(B) $ 2
(C) $ 3
(D) $ 4
(E) $ 8


Let's suppose Cylinder X has a radius r and height h. Cylinder Y radius =2r, height 2h.
X contains πr^2h ltrs of oil which sells for 2$.

Y is half filled...so volume of oil in Y....1/2(π(2r)^2 2h) = 4πr^2h
So oil in Y will sell for 4 rate of x===> 4x2$= 8$ Hence E

Originally posted by ManishKM1 on 10 Dec 2017, 09:08.
Last edited by ManishKM1 on 10 Dec 2017, 09:37, edited 1 time in total.
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Re: Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

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New post 10 Dec 2017, 09:26
1
Hi all, IMO, answer should be E: $8.

V of cylinder = R^2*H. If height and radius of Y double height and radius of X --> V(Y) = 8V(X) ---> Price for full capacity of Y = 8 price for full capacity of X = 8*$2=$16 --> Price for half capacity of Y = $16/2=$8

Please kudo if it helps. Thanks.
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Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

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New post 10 Dec 2017, 15:38
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Bunuel wrote:
Two oil cans, X and Y, are right circular cylinders, and the height and the radius of Y are each twice those of X. If the oil in can X, which is filled to capacity, sells for $2, then at the same rate, how much does the oil in can Y sell for if Y is filled to only half its capacity?

(A) $ 1
(B) $ 2
(C) $ 3
(D) $ 4
(E) $ 8

Choose smart numbers
Calculate the amount of oil in X and Y with the formula for the volume of a right circular cylinder. Y's oil volume times the sell rate for X's oil = selling price for Y's oil

Volume of right circular cylinders: \(\pi r^2 h\)
For oil can X: Let \(r = 1\) and \(h = 1\)
For oil can Y: Let \(r = 2\) and \(h = 2\)

X is full. The amount of oil in X = X's volume
X's oil amt: \(\pi r^2 h=\pi(1)(1)= 1\pi\)

Y is half full. Y's oil amount is half of Y's volume
Y's volume: \(\pi r^2h =\pi (4)(2)=8\pi\)
Y's oil amount: \(\frac{8\pi}{2}= 4 \pi\)

X's oil sells for $2. Y has 4 times as much oil as X. Y's oil sells for ($2 * 4) = $8

OR \(\frac{$2}{1 \pi}=\frac{y}{4 \pi}\)

\(y = $8\)

Answer E

Algebraically
Volume of right circular cylinder: \(\pi r^2 h\)
Radius of Y = 2 * radius of X
Height of Y = 2 * height of X

Amount of oil in X = Volume of X = \(\pi r^2 h\)

Y's oil amount = half of Y's volume: \((\frac{1}{2})*\pi (2r)^2(2h)=(\frac{1}{2})\pi(4r^2)(2h)=4\pi r^2h\)

X's oil sells for $2. Y's oil sells for?
\(\frac{1(\pi r^2h)}{4(\pi r^2h)} = \frac{$2}{y}\)

\(y = $8\)


Answer E
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Two oil cans, X and Y, are right circular cylinders, and the height an &nbs [#permalink] 10 Dec 2017, 15:38
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