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GMATPREP | Mixtures | Conceptual doubt

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by [email protected] » Tue Oct 18, 2016 7:07 pm
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

C
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by [email protected] » Tue Oct 18, 2016 7:07 pm
When the solution is altered, leading to change in ratio then what is added?

1. Just water
2. Just acid
3. Both
4. Also how do we know that the amount of base remains constant
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by [email protected] » Tue Oct 18, 2016 10:01 pm
Hi nishatfarhat87,

This question was discussed here:

https://www.beatthegmat.com/ratios-and-m ... 63547.html

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by Matt@VeritasPrep » Fri Oct 28, 2016 12:27 am
Initial ratio: 4x :: 15x :: 20x

We're told that 15x = 30, so our ratio is 8 : 30 : 40. Acid : Water = 1 : 5, so need to maintain this ratio. To make Acid : Base = 3 : 5, we can keep Base the same and make Acid = (3/5)*Base = 18.

If Acid = 18 and Acid/Water = 1/5, then Water = 90. We only have 40mm of Water now, so we need to add 50 more.
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