Drum x is 1/2 full of oil and drum y, which has twice the capacity of

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pick numbers -

drum A capacity 3 units
drum B capacity 6 units

drum A contains 1.5 units.
drum B contains4 units.

drum B gets filled 5.5 after pouring 1.5 from A to B.

Hence drum B is full 5.5/6 of its capacity. i.e. 11/12

its correct except ... you need to divide by the actual capacity of the drum to get the fraction so
(11x/6)/(2x) = 11/12

The answer is in terms of Drum Y, so I did the calc in terms of Y...

So...

X is 1/2 the size of Y and is 1/2 full... so in terms of Y, X is 1/4

Y as we already know is 2/3 full...

therefore, when you add the two drums together...

2/3 + 1/4 = (8+3)/12 = 11/12

Love the solutions here. Thx.

Heres how I solved it.

Drum x is 1/2 of y or 2x=y

Pick numbers:

x=50, y=100 Holds true to the equations above.

2/3 of y = 66.66666... 1/2 of x = 25.

Add 25+66.66666 = 91.6666666... Now just find an answer choice that has this percentage and its C.

Picking smart numbers like 6 or 100 will give the answer.

Lets say Drum x capacity is 100. Drum Y capacity will be 200.

Oil in drum x is 50
Oil in drum y is 100* 2/3 = 400/3

Fraction of oil in y after x is emptied into y is = (50 + 400/3) / 200
= 550/600
= 11/12
Answer is c

[quote="ggarr"]Drum x is 1/2 full of oil and drum y, which has twice the capacity of drum x is 2/3 full of oil. if all of the oil in drum x is poured into drum y, then drum y will be filled to what capacity?

A. 3/4
B. 5/6
C. 11/12
D. 7/6
E. 11/6

Hello All,

I took the exact approach as others to solve this problem.

Let's say drum X has a capacity of 2, which means drum Y will have a capacity of 4 (twice the capacity of drum x) . if Drum x is 1/2 full then 2.1/2=1 If drum y is 2/3 full then 4.2/3=8/3. If all the oil is poured from x to y then we can add 1+8/3 = 11/3. Now to see the what drum y's capacity will be filled to we can do 11/3 divided by four, which is equal to 11/12 (c)

Hi All,

TESTing Values will work great on this question. Based on what the question asks for and the answer choices, a great number to pick for Drum Y's capacity is 12. That number is based on the common denominator of the 5 options.

Now, we just need to follow the facts in this question and create a little table:

We're told that Drum Y is 2/3 full of oil:

Drum Y = capacity of 12; has 8 in it

We're also told that Drum Y has twice the capacity of Drum X and that Drumb X is 1/2 full of oil:

Drum X = capacity of 6; has 3 in it

Finally, we're told to pour the oil from Drum X into Drum Y….

Drum Y = 8+3 = 11
Drum Y is 11/12 full

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hello experts,
EMPOWERgmatRichC, VeritasKarishma, Bunuel, IanStewart, ArvindCrackVerbal, AaronPond, GMATinsight, JeffTargetTestPrep, ccooley
Choice D and E are improper fraction (more than 1) that means ''the position of some sorts of oil is out of drum (because the capacity is considered as 1, but here it is more than 1 in choice D and E) when we pour all of the oil of drum x into drum y.
So my question is:
Q: Will the GMAC make any correct choice where some sorts of oil's (any liquid) position will be out of the drum when one drum's oil is poured to another drum?
Thanks__
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Hi, refer the other question posted by you..
Here oil in x is put into y, which is double the capacity of x..
So if 6 is the capacity of x, 12 will be the capacity of y..
available oil is 6*1/2 + 12*2/3=11..

If we put this in y, the capacity is filled upto 11/12, so there is no spill over.
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Asad

The answer to your question is "NO"

GMAC can't give language in which there is some spillage to be considered as volume poured and available in some vassel.

Here the capacities have ensured that there is no spillage.

I solved in this way. I hope I am going in the right direction.

Let Volume of Drum X = X
Then Current oil in X = \(\frac{1}{2}\)X

Let Volume of Drum Y = Y
Then Current Oil in Y = \(\frac{2}{3}\)Y

Also Y=2X

Now the statement says oil from X is poured in Y

Hence find the total volume of oil WRT Y

\([\frac{1}{2}X+\frac{2}{3}Y)/Y\)

Replacing \(Y=2X\)

\([\frac{1}{2}X+(\frac{2}{3}*2X)]/(2X)\)

= \([\frac{11}{6}X]/(2X)\) = \(\frac{11}{12}X\)
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In my quest to use fewer algebraic expressions and more reasoning, I attempted the following approach:

Think of the drum Y as two drum X (A and B each of capacity X) combined (A sits on the top of B).

Since the capacity of drum Y is double that of drum X, 2/3 of drum Y is equivalent to 4/3 of drum X.
This volume is equal to full capacity of drum A plus 1/3 volume of the drum B. We are left with 2/3 capacity in drum B.

Now, in drum X volume of oil is 1/2 its capacity. That means we can not fill drum Y to full capacity with the oil in drum X.
Here, we can eliminate option D and E, since both are greater than 1.

Next, what would be the fraction of oil in drum X if it had the same capacity as drum Y? Simple: 1/4
So, after addition of 1/4 (2X) to 2/3 (2X), we get 1/4 + 2/3 = 11/12.

Answer C

Hello KarishmaB mam, I was struggling to solve this using the ratios approach. Would you please guide me here?

Since we are adding the amounts, it is best to assume suitable volumes (LCM of the denominators) for the two drums. Ratios are useful when one variable stays constant while 2 move with each other.

Let their volumes be 6 gallons and 12 gallons. (Since they are going to be 1/2 full and 2/3 full, 6 gallon capacity will get rid of all fractions)
Drum x has 3 gallons oil and drum y has 8 gallons oil.
When we put 3 gallons of x into y, y now has 11 gallons oil.
So it is filled to 11/12 of its capacity.

Answer (C)

Let the capacity of drum x be 2a.
Since 1/2 of x is filled of oil, the oil in the drum will have the volume = a.

Also, the capacity of drum y will be 4a.
Since it is filled to 2/3 of its capacity,
Volume of oil in drum y = 2/3*4a = 8/3a

If oil in x is poured in the drum y,
The volume of oil = 8/3a + a = 11/3a
Thus, the potion of y filled with oil = (11/3a)/4a = 11/12

Thus, the correct answer is C.

Why doesn't adding 1/2 + 2/3 -> 3/6 + 4/6 yield a correct answer?

Can someone help provide a little clarity on the logic behind this? I understand it'll give me something more than the container and I understand why 11/12 is the answer. I'm just a little unsure why my initial approach was incorrect.

Hi lmcclellan18,

The issue here is about 'scale' - since the '1/2' and '2/3' are NOT referring to the same 'whole' item (we have 1/2 a barrel and 2/3 of a barrel that is TWICE the size of the first barrel), we cannot simply add the fractions together to get the correct answer. IF both barrels were the SAME size, then we could add 1/2 and 2/3 - but that's not the case here.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

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