Ratio problem : Petersons

[Khalid]

The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm of base, what is the minimum amount of water that could be added in the second phase?

A) 18
B) 36
C) 50
D) 60
E) 90

OA C

I could go as far as,

Base = 15x = 30

x = 2

Initial quantities:

A:B:W = 8:30:40

Also, since the ratio of Acid to water remains same"

Previous Ratio ( A1 :W1) = New Ratio ( A2:W2)
W2 = 15

Not sure what to do next. Thanks for your help

[lilpetunia]

this is how I did it:
A:B:W= 4:15:20
knowing that B=30 I found out how much acid and water original solution contained=> A:B=x:30 => x=12 ( so we know original solution contained 12mm of acid). Similarly B:W=30:y => y=40 ( now we know original solution contained 40mm of water)

After solution was altered the ratios are A:B=3:5 and A:W=4:20 ( unchanged from original). Now if you simplify A:W you can say that ratio of A:W=1:5
3:5
A:B:W
1: : 5 (multiply by 3 since you know you need three parts of acid for correct ratio to base) and you get:
3:5:15 is the new ratio

then similarly like at the beginning 3:5=z:30=>z=18 ( new volume of acid in solution)
similarly 5:15=30:q=>q=90 ( new volume of water is 90ml)

lastly if original volume of water in solution was 40 and now it's 90, you must have added 50ml, correct answer is C. I am not sure if this is the quickest solution but that's how I got to it.

[Khalid]

this is how I did it:
A:B:W= 4:15:20
knowing that B=30 I found out how much acid and water original solution contained=> A:B=x:30 => x=12 ( so we know original solution contained 12mm of acid). Similarly B:W=30:y => y=40 ( now we know original solution contained 40mm of water)

After solution was altered the ratios are A:B=3:5 and A:W=4:20 ( unchanged from original). Now if you simplify A:W you can say that ratio of A:W=1:5
3:5
A:B:W
1: : 5 (multiply by 3 since you know you need three parts of acid for correct ratio to base) and you get:
3:5:15 is the new ratio

then similarly like at the beginning 3:5=z:30=>z=18 ( new volume of acid in solution)
similarly 5:15=30:q=>q=90 ( new volume of water is 90ml)

lastly if original volume of water in solution was 40 and now it's 90, you must have added 50ml, correct answer is C. I am not sure if this is the quickest solution but that's how I got to it.

Thanks much!

[RonPurewal]

first, i'm moving this thread to the general math folder, since it's not a gmatprep problem.

second, the problem statement displays an appalling lack of understanding of chemistry. acid and base can't coexist in a solution; if they're thrown into a solution together, they'll neutralize each other to produce water. therefore, an acid:base:water ratio of 4:15:20 will quickly become an acid:base:water ratio of 0:11:24.
but i digress.

The ratio, by volume, of acid to base to water in a certain solution is 4:15:20. The solution is altered so that the ratio of acid to base is 3:5 and the ratio of acid to water remains same. If the solution initially contained 30mm of base, what is the minimum amount of water that could be added in the second phase?

A) 18
B) 36
C) 50
D) 60
E) 90

OA C

I could go as far as,

Base = 15x = 30

x = 2

Initial quantities:

A:B:W = 8:30:40

Also, since the ratio of Acid to water remains same"

Previous Ratio ( A1 :W1) = New Ratio ( A2:W2)
W2 = 15

Not sure what to do next. Thanks for your help

the solution above works wonderfully.

here's another, more abstract solution. it's more conceptually difficult, but it's faster.

1) realize that you're going to have to add both water and acid (this is the only way that you can keep that ratio constant). you could also attain the requisite ratios by removing base, but that option isn't permitted.

2) the current ratio of acid:base is 4:15 (grrr...). you want it to be 9:15. this means that you need to add 5/4 as much acid as is already in there, since you need to turn the ratio coefficient of "4" into a "9". note that it's not necessary to calculate the actual amount of acid.

3) you have to increase the water in exactly the same proportion.
there are 40 ml of water, as calculated above (you do have to figure this out, still).
therefore, you need to add (5/4)(40), or 50 ml of water.

if you have any trouble following this explanation, you should immediately change over to the explanation(s) above, in which you calculate the specific numbers involved.

[netcaesar]

I do not understand the solution.

Can anyone explain it, indicating all the quantities?

Regards.

[JonathanSchneider]

Check lilpetunia's post above.

[700+]

then similarly like at the beginning 3:5=z:30=>z=18 ( new volume of acid in solution)

I have a doubt in the above statement

3:5 the ratio of acid is to base when the solution was altered and the solution initially contained 30 mm of base. So how could 3:5 be equal to z:30?

[tim]

The base hasn’t changed; it stays 30, and we need to know how much water there must be (z) in order to attain a 3:5 ratio..

[700+]

The base hasn’t changed; it stays 30, and we need to know how much water there must be (z) in order to attain a 3:5 ratio..

Could you please let me know how did we reach the conclusion that the base hasn't changed?

[tim]

the question is asking for the minimum amount of water that can be added. if ANY base at all is added, that will require a corresponding addition of water to compensate. so the minimum amount of additional water will also involve the minimum amount of additional base, which is zero..

[555]

I got B when i did it.

[tim]

Okay...have you figured out what you did wrong?