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The ratio, by volume, of acid to base to water in a certain solution
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24 Jun 2017, 03:49
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51% (02:59) correct 49% (03:17) wrong based on 168 sessions
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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
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Re: The ratio, by volume, of acid to base to water in a certain solution
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24 Jun 2017, 23:58
Shan30 wrote: 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 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




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Re: The ratio, by volume, of acid to base to water in a certain solution
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24 Jun 2017, 04:37
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.



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Re: The ratio, by volume, of acid to base to water in a certain solution
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24 Jun 2017, 04:59
Tapesh03 wrote: 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. I did everything correct.Except the last part.. Thanks a lot



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The ratio, by volume, of acid to base to water in a certain solution
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25 Jun 2017, 03:04
Phase 1Acid : 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 2We 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)
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The ratio, by volume, of acid to base to water in a certain solution
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30 Jun 2018, 02:02
Prashanth3007 wrote: 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 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 = 9040 = 50.
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