Can A and can В are both right circular cylinders. The radiu

Question

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

Show Answer

Official Answer

E

(A) $1
(B) $2
(C) $4
(D) $8
(E) $16

dreambeliever · Answer

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

subhashghosh · Answer

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.

Onell · Answer

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$

MHIKER · Answer

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.

mohnish104 · Answer

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.

Bunuel · Answer

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

Show Answer

Official Answer

E

(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.

Answer: E.

mohnish104 · Answer

Thank you Bunuel. I guess the answer has been wrongly pointed out in the Kaplan tests.

Guilio · Answer

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?

Appreciate your help as always!

Best,
Jay

KarishmaB · Answer

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.

Guilio · Answer

Oh boy, indeed I messed up the easy calculation. Hate those mistakes so much.

Thanks for clarification Karishma!

Best,
Jay

CEdward · Answer

What I have found works for questions in which no dimensions are actually given (much like percentage word problem questions where no discrete numbers are given) is to use test numbers.

A - r = 2 and h = 1 --> π(2)^2 x 1 = 4π
B - r = 1, h = 2 --> π(1)^2 x 2 = 2π

Next, set up a ratio: 
$4 / π = $16 / 4π

E.

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