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A can contains a mixture of two liquids A & B in

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Manager
Joined: 13 Apr 2006
Posts: 56

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A can contains a mixture of two liquids A & B in [#permalink]

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04 May 2006, 09:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A can contains a mixture of two liquids A & B in proportion 7:5. When 9 litres of mixture is drawn off and the can is filled with liquid B, the proportion of A & B becomes 7:9. How many litres of liquid A was contained by the can initially?

A. 25
B. 10
C. 20
D. 21
E. 22

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Intern
Joined: 02 Feb 2006
Posts: 2

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04 May 2006, 10:26

X/Y=7/5 - RATIO OF A to B
5X-7Y=0......(1)

9 litres would have : 7/12*9=5.25 of A and 9-5.25=3.75 of B

New ratio: (X-5.25)/(Y-3.75+9)=7/9
9X-7Y=16*5.25 ....... (2)

Solve eq 1 and 2 : answer is 21 for x

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Manager
Joined: 13 Apr 2006
Posts: 56

Kudos [?]: 13 [0], given: 0

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04 May 2006, 10:38

Kudos [?]: 13 [0], given: 0

Manager
Joined: 13 Apr 2006
Posts: 56

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04 May 2006, 10:43
anyone else have another way to solve?

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Director
Joined: 16 Aug 2005
Posts: 937

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Location: France

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04 May 2006, 10:47
A:B = 7:5

Total = 12 units

When 9 litres of mixture is removed, ((7/12) * 9) litres of A is removed, and ((5/12) * 9) litres of B is removed

Then 9 litres of B is added so that new ratio is 7:9

In new mixture, Total volume of A is 7x - ((7/12) * 9) = 7x - 21/4
And total volume of B is 5x - ((5/12) * 9) + 9 = 5x - 3 3/4 + 9 = 5x + 21/4

So, (7x - 21/4)/(5x + 21/4) = 7/9

Solving for x:
x=3
Original Volume of A = 7x = 21

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Director
Joined: 24 Oct 2005
Posts: 659

Kudos [?]: 16 [0], given: 0

Location: London

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05 May 2006, 04:06
Initial proportion of A = (7/12) T
where T is the total.

9 litres taken away, the ratio of A becomes (7/16) of the total

ie. A =(7/16)(T-9)

(7/12)T = (7/16)(T-9)
Therefore, T = 36.

Initial proprtion of A = (7/12)(36) = 21

Kudos [?]: 16 [0], given: 0

05 May 2006, 04:06
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