The ratio, by volume, of acid to base to water in a certain solution

Question

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

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

sashiim20 · Accepted Answer

Acid : Base : Water = 4 : 15 : 20 Acid : Base = 4 : 15 = \frac{4}{15} Acid : Water = 4: 20 = \frac{4}{20} = \frac{1}{5} Given initially Base was 30mm . Acid : Base : Water = 2 * (4 : 15 : 20) = 8 : 30 : 40 Ratio of Acid : Base changed = 3 : 5 \frac{Acid}{Base} = \frac{3}{5} \frac{Acid}{30} = \frac{3}{5} Acid = \frac{3}{5} * 30 => Acid = 18mm \frac{Acid}{Water} = \frac{1}{5} \frac{18}{Water} = \frac{1}{5} Water = 18 * 5 = 90mm . Therefore minimum amount of water that could be added in the second phase = 90 - 40 = 50mm . Answer (C)... _________________ Please, consider giving kudos if you find my answer helpful in any way

Tapesh03 · Answer

Solution - Acid:Base:Water - 4:15:20
Base is 30 mm, so as per 4:15:20, the other components are 8mm: 30mm: 40mm
Acid:Base changed to 3:5 & assuming there is not change in the base, Acid needs to be increased to 18mm to keep 3:5 ratio
if Acid increases to 18, water would also increase to 90mm to keep the same ratio i.e 4:20 or 1:5
Initially, water was 40mm, so the increase is 50 mm & thus C is the correct answer.

Prashanth3007 · Answer

I did everything correct.Except the last part.. Thanks a lot

pushpitkc · Answer

Phase 1

Acid : Base : Water is 4:15:20
Since the initial concentration of base is 30mm,
the inital concentration of water is 40mm and acid is 8mm.

Phase 2

We have to alter the solution such that the Acid:Base ratio is 3:5
we must add 10mm of acid , to the existing 8mm to make the ratio 3:5
It has also been given that the ratio of Acid : Water must remain the same, which is 4:20(1:5)
So, for 18mm of acid, we must have 90mm of acid.

Old solution : 8mm(Acid) : 30mm(Base) : 40mm(Water)
New solution : 18mm(Acid) : 30mm(Base) : 90mm(Water)
 
Since we have been asked the minimum water that needs to be added, it must be  50mm(Option C)

GMATGuruNY · Answer

Original A:B:W = 4:15:20. 
B = 30, so the actual values of A, B and W are all doubled: 
 A = 8 
B = 30 
W = 40 .

New ratio of  A:B  = 3:5 
Unchanged ratio of  A:W  = 4:20 = 1:5 = 3:15 (Since A is common to the two red ratios, we want A to be represented by the same value in each ratio so that the two ratios can be combined.) 
Combining the two red ratios, we get:
A:B:W = 3:5:15.

We must minimize the amount of water added to the values in blue so that A:B:W = 3:5:15.
A:B:W = 3:5:15 =  18:30:90 .
The ratio in green indicates the following:
New A=18.
B=30. 
New W=90. 
Added water = new W - old W = 90-40 = 50.

C

GmatPoint · Answer

Initial ratio = 4:15:20
Since initial content of base is 30, the initial content of water = (20/15)*30 = 40

Initial ratio of acid and water = 4:20 = 1:5

The ratio in the second phase = 3:5:15(since 3:5 is the ratio of acid and base, and 1:5 is that of acid and water)

If the base content is the same, the minimum amount of water will be added.

Thus, if base is 30mm in second phase, water = (15/5)*30 = 90

The quantity of water added = 90 - 40 = 50

Thus, the correct option is C.

ThatDudeKnows · Answer

Initial
A : B : W
4: 15: 20
8: 30: 40

We want to keep values as low as possible.
Add acid to make A : B = 3 : 5
18 :30 : W
Add water to make A : W = 1 : 5
18 : 30 : 90

Water at end is 90. We started with 40. That means we added 50.

Answer choice C.

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The ratio, by volume, of acid to base to water in a certain solution

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