GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Aug 2018, 04:14

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47961
Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

### Show Tags

10 Dec 2017, 08:05
00:00

Difficulty:

35% (medium)

Question Stats:

69% (01:08) correct 31% (00:21) wrong based on 38 sessions

### HideShow timer Statistics

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

_________________
Manager
Joined: 22 Apr 2017
Posts: 112
Location: India
GMAT 1: 620 Q46 V30
GMAT 2: 620 Q47 V29
GMAT 3: 630 Q49 V26
GMAT 4: 690 Q48 V35
GPA: 3.7
Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

### Show Tags

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. Intern Joined: 25 Jul 2017 Posts: 38 Re: Two oil cans, X and Y, are right circular cylinders, and the height an [#permalink] ### Show Tags 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.
SC Moderator
Joined: 22 May 2016
Posts: 1903
Two oil cans, X and Y, are right circular cylinders, and the height an  [#permalink]

### Show Tags

10 Dec 2017, 15:38
1
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?
$$\frac{1(\pi r^2h)}{4(\pi r^2h)} = \frac{2}{y}$$

$$y = 8$$

_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

Two oil cans, X and Y, are right circular cylinders, and the height an &nbs [#permalink] 10 Dec 2017, 15:38
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.