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A can contains a mixture of 2 liquids A and B in proportion

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Director
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A can contains a mixture of 2 liquids A and B in proportion [#permalink] New post 03 Jan 2005, 16:21
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 2 liquids A and B in proportion 7:5.
9 liters of of the mixture is removed and 9 liters of B is added and the proportion of A and B becomes 7:9. How many liters of liquid A was originally in the can.

OA = 63

I could not figure this out i need a break i think :)
VP
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 [#permalink] New post 03 Jan 2005, 17:28
I am getting 21 lt as the ans. If x is the total liquid, then A is 7x/12 and 5x/12 for B. Take out 9 lt we have 7(x-9)/12 and B 5(x-9)/12 (assuming ratio doesn't change). Add 9 lt of B, then B becomes 5(x-9)/12 + 9.

So...(7(x-9)/12) / (5(x-9)/12 + 9) = 7/9 => x = 36. So A in original mix was 7(36)/12 ==> 21 lt.

Not getting the OA u posted.
Director
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 [#permalink] New post 03 Jan 2005, 17:39
guys thanks for the answer My approach was

7x/12 is A and 7/12 th of 9 is removed from original value of 7x/12 similarly for B instead of your approach of 7(x-9) /12

I got the below equation

7x/12 - 7/12*9 = 7/9
____________

5x/12 - 5/12*9 + 9

solving we get x = 4.5 so A = 7*4.5 = 31.5 which is shy of double than the answer


so i am not sure which is correct
Manager
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 [#permalink] New post 03 Jan 2005, 18:31
I also got 21. I double checked to see if I got 7/9 using 21 as the answer and it checks out. Before I reduced the fraction when I was double checking I got 63/81, but that number is pretty meaningless. 21 has to be the right answer
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Re: Tough Mixture Problem [#permalink] New post 03 Jan 2005, 23:05
rxs0005 wrote:
A can contains a mixture of 2 liquids A and B in proportion 7:5.
9 liters of of the mixture is removed and 9 liters of B is added and the proportion of A and B becomes 7:9. How many liters of liquid A was originally in the can.

OA = 63

I could not figure this out i need a break i think :)


i could not get other than 21 lt. do you have explanations for 63 lt as answer?
VP
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Re: Tough Mixture Problem [#permalink] New post 03 Jan 2005, 23:05
rxs0005 wrote:
A can contains a mixture of 2 liquids A and B in proportion 7:5.
9 liters of of the mixture is removed and 9 liters of B is added and the proportion of A and B becomes 7:9. How many liters of liquid A was originally in the can.

OA = 63

I could not figure this out i need a break i think :)


i could not get other than 21 lt. do you have explanations for 63 lt as answer? if yes, pls do post......................
Senior Manager
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 [#permalink] New post 04 Jan 2005, 01:45
I got 21 as well.

Could you please post the explaination alongwith the OA?

Thanks in anticipation.
Director
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 [#permalink] New post 04 Jan 2005, 05:10
guys thanks all for your answers the OA is 63 but no official explanation was given so it must be a typo in the book it happens but thanks all for trying
Intern
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 [#permalink] New post 04 Jan 2005, 13:57
Banjeera, how did you get from:

(7(x-9)/12) / (5(x-9)/12 + 9) = 7/9

to: x=36?

Did you simply solve the the equation "(7(x-9)/12) / (5(x-9)/12 + 9) = 7/9" for x.

I could not get 36. Maybe I am making an algebraic mistake but I cannot get 36, please explain the step above. (Thanks in Advance)

I understand that if the ratio is 7 parts to 5 parts equals 12 parts in total, then if the parts total increases to 36, the the ration is 7x3=21 / 5x3=15 or a new ration of 21/15 equals 36 parts in total (21 + 15).
Thanks much
VP
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 [#permalink] New post 04 Jan 2005, 14:41
1. (7(x-9) / 12) / ((5(x-9) / 12) + 9) = 7 / 9
2. (7(x-9) / 12) / (5(x-9) + 108 / 12) = 7 / 9
3. (7x - 63 / 12) * (12 / 5x-45+108) = 7 / 9
4. 7x - 63 / 5x + 63 = 7 / 9
5. 63x-567 = 35x+441
6. 28x = 1008
7. x = 36
8. (7 / 12) * 36
9. 21
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 [#permalink] New post 04 Jan 2005, 15:05
Perrrrfect, Thanks much. I made a simple algebraic mistake, but even after doing the problem twice, I made the same mistake. I see it now. :shock:
VP
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 [#permalink] New post 04 Jan 2005, 20:18
banerjeea_98 wrote:
I am getting 21 lt as the ans. If x is the total liquid, then A is 7x/12 and 5x/12 for B. Take out 9 lt we have 7(x-9)/12 and B 5(x-9)/12 (assuming ratio doesn't change). Add 9 lt of B, then B becomes 5(x-9)/12 + 9.

So...(7(x-9)/12) / (5(x-9)/12 + 9) = 7/9 => x = 36. So A in original mix was 7(36)/12 ==> 21 lt.

Not getting the OA u posted.


banarjee,

you have asut approaches to the problems. goodluck..........

MA
  [#permalink] 04 Jan 2005, 20:18
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