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Vial V is 2/3 full of a certain solution and vial W, which has 50%

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Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 08 Mar 2013, 11:16
2
4
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

80% (02:31) correct 20% (02:39) wrong based on 178 sessions

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Vial V is \(\frac{2}{3}\) full of a certain solution and vial W, which has 50% more capacity than vial V, is \(\frac{3}{4}\) full of the same solution. If half of the solution in vial V is poured into vial W, vial W will be filled to what fraction of its capacity?

(A) \(\frac{27}{36}\)

(B) \(\frac{10}{11}\)

(C) \(\frac{11}{12}\)

(D) \(\frac{23}{24}\)

(E) \(\frac{35}{36}\)

I was able to solve it in under 2 mins, but curious to know if there is a 10-sec approach?

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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 08 Mar 2013, 13:32
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Hello,

Let me try helping you with this one.

It is mentioned in the question that the volume of W is 50% larger than the volume of V. Hence, W=\(\frac{3}{2}\)*V
This implies that V=\(\frac{2}{3}\)*W
Now, V is filled with 2/3 with the solution and W is filled 3/4 with the solution.
Half of the solution in V is filled into W. We need to find the fraction of W that is filled. So, we have to express everything in terms of W.

Half of the solution in V in terms of W is 1/3. When we express that in terms of W, it becomes

half the volume of solution in V=\(\frac{1}{3}\)*\(\frac{2}{3}\)*W
This is added to 3/4 of the volume W.
The total volume is (\(\frac{1}{3}\)*\(\frac{2}{3}\)*W)+(\(\frac{3}{4}\)*W)=\(\frac{35}{36}\)*W

Hence, the answer is E.

Hope this helps! Let me know if I could help you any further.

megafan wrote:
Vial V is \(\frac{2}{3}\) full of a certain solution and vial W, which has 50% more capacity than vial V, is \(\frac{3}{4}\) full of the same solution. If half of the solution in vial V is poured into vial W, vial W will be filled to what fraction of its capacity?

(A)\(\frac{27}{36}\)

(B)\(\frac{10}{11}\)

(C)\(\frac{11}{12}\)

(D)\(\frac{23}{24}\)

(E)\(\frac{35}{36}\)

I was able to solve it in under 2 mins, but curious to know if there is a 10-sec approach?
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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 09 Mar 2013, 02:24
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let the capacity of V = 6, then the capacity of W = 6+0.5*6=9

the amount of solution in V = (2/3)*6=4 and the half of it is 2 (4/2 :) )
the amount of solution in W = (3/4)*9=27/4

(half of solution in V + the amount of solution in W )/ capacity of W = (2+27/4)/9 =35/36
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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 09 Mar 2013, 04:21
megafan wrote:
Vial V is \(\frac{2}{3}\) full of a certain solution and vial W, which has 50% more capacity than vial V, is \(\frac{3}{4}\) full of the same solution. If half of the solution in vial V is poured into vial W, vial W will be filled to what fraction of its capacity?

(A)\(\frac{27}{36}\)

(B)\(\frac{10}{11}\)

(C)\(\frac{11}{12}\)

(D)\(\frac{23}{24}\)

(E)\(\frac{35}{36}\)

I was able to solve it in under 2 mins, but curious to know if there is a 10-sec approach?


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Hope it helps.
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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 12 Aug 2013, 06:47
Let capacity be denoted by "C"

V = C ===> currently filled = 2/3 C
Given : Capacity of W is 50% more than V

W = C + 0.5 C = 1.5 C ===> currently filled = 3/4(1.5C)

half of contents of V ==> 2/3C * 1/2 = 2/6C are then poured in W
Required to calculate the fraction to which W will be filled after half of contents of V are filled in W
2/6C +3/4(1.5C)/1.5C
Solving, we get
35/36

Ans:E
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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 12 Aug 2013, 10:20
megafan wrote:
Vial V is \(\frac{2}{3}\) full of a certain solution and vial W, which has 50% more capacity than vial V, is \(\frac{3}{4}\) full of the same solution. If half of the solution in vial V is poured into vial W, vial W will be filled to what fraction of its capacity?

(A)\(\frac{27}{36}\)

(B)\(\frac{10}{11}\)

(C)\(\frac{11}{12}\)

(D)\(\frac{23}{24}\)

(E)\(\frac{35}{36}\)

I was able to solve it in under 2 mins, but curious to know if there is a 10-sec approach?


...........
just 30 seconds to solve with this techniques:
Attachments

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volume and ratio own.png [ 33.83 KiB | Viewed 2350 times ]


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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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New post 26 Sep 2016, 06:57
i worked with 24 and 36, as it seemed easier to work with...
24*2/3 = 16.
36*3/4 = 27

16/2 = 8
8+27 = 35.
we have 35/36
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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%  [#permalink]

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Re: Vial V is 2/3 full of a certain solution and vial W, which has 50%   [#permalink] 25 Dec 2017, 21:04
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