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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l

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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 08 Mar 2016, 18:46
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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?

A. 10
B. 20
C. 21
D. 25
E. 27
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Re: A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 08 Mar 2016, 23:17
4
3
anceer wrote:
A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?

A. 10
B. 20
C. 21
D. 25
E. 27


First of all, the wording of the question is a bit unclear. It needs to specify that 9 litres were removed and 9 litres was added back. "can is filled" is not sufficient because we do not know whether the can was full initially.

Say initial volume of mix is V1. After you remove 9 litres, the concentration of liquid A remains 7/12 and volume of the mix is (V1 - 9).
After you add back liquid B, the concentration of A becomes 7/16 and volume comes back to V1.

Ci*Vi = Cf*Vf

(7/12)*(V1 - 9) = (7/16)*V1

(V1 - 9)/V1 = 3/4

9 accounts for the difference of 1 on ratio scale so Initial volume = V1 = 4*9 = 36 litres.

7/12 of the initial mixture was liquid A so liquid A was (7/12)*36 = 21 litres.

Check this post for more on replacement: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
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Re: A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 08 Mar 2016, 19:46
2
2
anceer wrote:
A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?

A. 10
B. 20
C. 21
D. 25
E. 27


Using the values here might be the most straightforward way for this question.

As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total

If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4

Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4

Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.

Hence C is the correct answer.

FYI, the algebraic equations will become:

A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.

Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) = 7*3 = 21.

Hope this helps.
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Re: A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 08 Mar 2016, 23:37
Thanks Karishma for the link.
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Re: A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 07 May 2016, 14:04
3
let x=total liters in can
5x/12-(9)(5/12)+9=9x/16
x=36
(7/12)(36)=21 liters of A in can initially
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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 13 Dec 2018, 18:55
7x/12 - 9*7/12 = x*7/16
Solving, x=36
A = 7x/12 = 7*36/12 = 21

Method 2 ROA
5/12 - - - - 1
----9/17----
3 : 1
1 =9
4=36
A = 36*7/12 = 21

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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l  [#permalink]

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New post 21 Dec 2018, 07:06
2
anceer wrote:
A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter of liquid A was contained by the can initially?

A. 10
B. 20
C. 21
D. 25
E. 27



Suppose the can initially contain 7x and 5x litres of mixtures A and B respectively. The quantity of A in the mixture left = [7x – (7/12)×9] litres = [7x – 21/4] litres.

The quantity of B in mixture left = [5x – (5/12)×9] litres = [5x – 15/4] litres.

Therefore we can write: the ratio of the two quantities as [7x – 21/4]/ [5x – 15/4] = 7/9

In other words, we may say that 252x – 189 = 140x + 147

Hence, x = 3. Therefore the can had 7(3) = 21 litres of the quantity A. Thus the correct option is C.


Or
WAY 2 :
let A = 7x & B = 5x
total mixture = 7x+5x=12x

after 9 litre of mixture are drawn off remaining misture = 12x-9

after filling can with B:
A = (12x-9) * 7/12= 28x-21
B = {(12x-9)* 5/12}+9= 20x+21

So, (28x-21)/(20x+21)= 7/9
by solving above equation, we get
x=3

So liquid A was contained by the can initially = 7x = 7*3 = 21
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A can contains a mixture of liquids A and B is the ratio 7:5. When 9 l   [#permalink] 21 Dec 2018, 07:06
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