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one of mixtures


Malegria

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Container A and Container B both contain a mixture of alcohol and water. The ratio of alcohol to water in container A is 1 to 3. The ratio of alcohol to water in container B is 1 to 9. 8 ounces of the mixture in container A is taken and mixed into container B. Then, 16 ounces of the mixture in container B is taken and mixed into container A. If both containers originally contained 40 ounces of mixture, what is the final ratio of alcohol to water in container A?

 

A. 1:7

 

B. 2:7

 

C. 5:19

 

D. 1:3

 

E. 10:27

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C.

Container A has Alcohol:Water in 1:3 ratio. i.e. 1/4th alcohol out of 40oz

Thus orig alcohol in A, Aaorig = 10oz, orig water in A, Aworig = 30oz

Similarly, orig alcohol in B, Baorig = 4oz and Bworig = 36oz

 

Now 8oz of A is transferred to B, this contains 2oz of Alcohol and 6oz of water

Hence the new amount of alcohol in A is Aanew = 10-2 = 8oz, Awnew = 24oz

the new amounts in B are Banew = 4+2 = 6, and Bwnew = 42oz

The new concentration of alcohol in B is then 6oz in 48oz of solution

 

Now 16oz of B is transferred, this contains 6oz*16/48 = 2oz alcohol and 14oz water

The new value of Alcohol in A = Aanew + 2 = 10oz,new value of water in A = Awnew+14 = 38

Thus the ratio of alcohol to water in A is 10oz : 38oz or, 5:19

Hence C

 

Whew!

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I got the same answer, C.

 

It took me about five minutes to solve though... I used the same approach as Greycellz. Is there a shorter way to solve this problem?

Put it in the form of a table and just add and substract on each side

1-3 1-9

10-30 4-36

8- 24 6-42

etc.

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:mad:

Hear me through here for a second - In A we have a ratio of 1 to 3 or 25% alchohol. We take 8 ounces of it, meaning 2 ounces of alcohol and put it in B. B was 1 to 9 or 10% alcohol so it's original 40 contained 4 ounces of alcohol. after the addition of A we have 48 total ounces with 4+2 = 6 ounces of alcohol. Take 16 ounces of this , which will contain 2 ounces of alchol and add it back to A, which was 32 with 8 and now will be 10 ounes of alcohol out of 48 total ounces. So in fractional terms we're at 5/19. Is this the same as the ratio though? it seems like it would be 5 parts alcohol to 14 parts water???? Where am i screwing this up?

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:mad:

Hear me through here for a second - In A we have a ratio of 1 to 3 or 25% alchohol. We take 8 ounces of it, meaning 2 ounces of alcohol and put it in B. B was 1 to 9 or 10% alcohol so it's original 40 contained 4 ounces of alcohol. after the addition of A we have 48 total ounces with 4+2 = 6 ounces of alcohol. Take 16 ounces of this , which will contain 2 ounces of alchol and add it back to A, which was 32 with 8 and now will be 10 ounes of alcohol out of 48 total ounces. So in fractional terms we're at 5/19. Is this the same as the ratio though? it seems like it would be 5 parts alcohol to 14 parts water???? Where am i screwing this up?

 

10 ounes of alcohol out of 48 total ounces

Hence water = 48-10 = 38 ounces

Hence Ratio of A:W = 10:38 = 5:19

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