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

Question

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

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

Bunuel · Accepted Answer

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

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

The volume of can X is  \pi{r^2}h , which when filled to capacity sells for $2;

The volume of can Y is  \pi{(2r)^2}(2h)=8\pi{r^2}h . Can Y is only half fill so volume of oil in it is  4\pi{r^2}h . Since  \pi{r^2}h  sells for $2 then  4\pi{r^2}h  sells for 4*$2=8$.

Answer: E.

Hope its clear.

jvujuc · Answer

Volume of x(Vx) is r^2*PI*h (r-radius, h height)

Volume of y (Vy) is (2r)^2*PI*2h = 4*r^2*PI*2h=8r^2*PI*h = 8*Vx

If Vx is 2$
Vy/2=4*Vx=8$

generis · Answer

Choose values. Volume of a right circular cylinder is  \pi{r^2}h .

Let radius of X = 1, and height of X = 2.

Radius and height of Y are each twice that of X. Radius of Y = 2, height = 4.

So volume of X is  (\pi)(1^2)(2)  =  2\pi

Volume of Y =  (\pi)(2^2)(4)  =  16\pi . 
 But Y is only half full , so the volume of oil that Y contains is  8\pi

Let Z = the sell price of the oil in can Y.

\frac{Price.Of.X.Oil}{Volume.X}=\frac{Z}{Volume.Y}

\frac{$2}{(2\pi)} = \frac{Z}{(8\pi)}

Z = $8, Answer E

ScottTargetTestPrep · Answer

We can let the radius of X = r and the height of X = h; thus, the radius of Y = 2r and the height of Y = 2h.
 
Thus, the volume of cylinder X is πr^2h and the volume of cylinder Y is:
 
π (2h) = 8πr^2h
 
Since cylinder X is filled to its capacity, the volume of oil in X is πr^2h. However, since cylinder Y is only filled to half of its capacity, the volume of oil in Y is 4πr^2h. Since the oil in X sells for $2, at the same rate, the oil in Y should sell for 4 times as much (since it has 4 times as much oil), or $8. 
 
Answer: E

generis · Answer

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

BrentGMATPrepNow · Answer

Volume of cylinder equals = π(radius²)(height)

Let Can X have radius 1 and height 1. 
Volume of Can X = π(1²)(1) =   π

This means Can Y has radius 2 and height 2. 
Volume of Can X = π(2²)(2) =   8π

8π  /  π   = 8, which means the volume of Can Y is 8 times the volume of Can X
Since Can Y is only HALF FULL, then Can Y contains   4 TIMES   the volume that Can X contains

So, if Can X sells for $2, Can Y must sell for $8 (since $2   TIMES 4   is $8)

Answer: E

Cheers,
Brent

CrackverbalGMAT · Answer

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

Oil Can X: radius =  r  , height =  h , Volume =  πr^2h

Oil Can Y : radius =  2r , height =  2h , Volume =  π(2r)^2*2h  =  8*πr^2h

Oil in Can X , which is filled to capacity( πr^2h)  is sold for 2 $.

Oil in Can Y, which is filled to half the capapcity (  4*πr^2h  ) will be sold for 2 * 4 = 8 $.

Option E is the answer.

Thanks,
  Clifin J Francis,
GMAT SME

mysterymanrog · Answer

Fast approach that took me 40 seconds: Smart Numbers

Radius_x=Height_x=1

Volume_x=\pi

\pi=$2

Radius_y=2*Radius_x=2

Height_y=2*Height_x=2

Y is filled to half, so height of Y is effectively 1

Volume_y=4\pi

4*\pi=$2*4=$8

ArnauG · Answer

The volume of a right circular cylinder is given by V = πr^2h, where "r" represents the radius and "h" represents the height of the cylinder.

Let's assume the radius and height of can X are "r" and "h", respectively.

The volume of can X is V(X) = πr^2h.

Now, let's consider can Y. The height and radius of can Y are each twice those of can X. So, the radius and height of can Y are 2r and 2h, respectively.

The volume of can Y is V(Y) = π(2r)^2(2h) = 4πr^2(2h) = 8πr^2h.

We are given that can X is filled to capacity and sells for $2.

Now, let's consider can Y. It is filled to only half its capacity. Therefore, the volume of oil in can Y is (1/2) * V(Y) = (1/2) * 8πr^2h = 4πr^2h.

Since the price is directly proportional to the volume, we can conclude that the oil in can Y will sell for 4 times the price of can X.

Therefore, if can X sells for $2, the oil in can Y will sell for 4 * $2 = $8.

Hence, the answer is (E) $8.

Vegita · Answer

Hi  BrentGMATPrepNow

I used the same method as you to solve this question but  generis  method is new to me. Could you please help me understand why this ratio would be equal?

(price of oil X/ volume of x) = (price of y/ volume of y)

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