Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A can contains a mixture of 2 liquids A and B in proportion [#permalink]
03 Jan 2005, 16:21

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 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.

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

Re: Tough Mixture Problem [#permalink]
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?

Re: Tough Mixture Problem [#permalink]
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......................

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

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

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..........