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How much water (in grams) should be added to a 35%solution of acid to obtain a 10%solution of acid? (1) There are 50 grams of the 35%solution. (2) In the 35%solution of acid the ratio of acid to water is 7:13.
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Official Solution: Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient. Answer: A
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Re: M0918
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27 Sep 2014, 07:27
Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.
Answer: A Are there anymore Questions on similar lines which are Not part of Gmat Club Tests but are of the forum? I tend to get lost in x%solution of y kind of questions. If possible please provide the links. Thanks in advance.



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Re: M0918
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27 Sep 2014, 13:04
earnit wrote: Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.
Answer: A Are there anymore Questions on similar lines which are Not part of Gmat Club Tests but are of the forum? I tend to get lost in x%solution of y kind of questions. If possible please provide the links. Thanks in advance. DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114Hope it helps.
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Re: M0918
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27 Sep 2014, 13:12
(1) Given x=50
0.35*50=0.1*(50+y).
linear equation, one unknown,Sufficient.
(2) acid to water ratio, 7:13. same as information given that it is a 35%solution. Not sufficient.



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Re: M0918
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13 Dec 2014, 12:45
Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.
Answer: A It seems that there is a small typing error in your explanation for (2) the ratio should be 35:65=7:13.



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Re: M0918
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15 Dec 2014, 08:36
haihai89 wrote: Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=5:13. Not sufficient.
Answer: A It seems that there is a small typing error in your explanation for (2) the ratio should be 35:65=7:13. Thank you. Edited.
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Re: M0918
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19 Feb 2016, 14:42
With an allegation chart it might look like this:
35%0% 25(10)10
So the ratio of Solution to Pure Water is 2:5
A) 2X=50 x= 25 5*25= 125 grams of water need to be added B) 2x=7x/13x InSuf
Ans. A



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30 Mar 2016, 08:42
I'm super confused here. Look there are 35% of acid and 65% of water and as per the statement 1, there is 50gms of acid.
Now 35% gives 50gms of acid and 65% gives to 92gms of water. Approx 142.86 is the total amt of mixture of acid and water.
35/100( total mixture of acid and water)=50gm of acid
Hence total mixture= 142.86
Now I add 125L of water (as per the soln) to 142.86 which gives 267. Now if i calculate 50/267= 18% of acid and it is not giving me 10% of acid. Where am I going wrong?



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Re: M0918
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30 Mar 2016, 09:29
harish1986 wrote: I'm super confused here. Look there are 35% of acid and 65% of water and as per the statement 1, there is 50gms of acid.
Now 35% gives 50gms of acid and 65% gives to 92gms of water. Approx 142.86 is the total amt of mixture of acid and water.
35/100( total mixture of acid and water)=50gm of acid
Hence total mixture= 142.86
Now I add 125L of water (as per the soln) to 142.86 which gives 267. Now if i calculate 50/267= 18% of acid and it is not giving me 10% of acid. Where am I going wrong? Hi, few points.. 1)when we say 50 gms of 35%, it means the entire solution is 50gmso acid= 35*0.5=17.5gm and water=65*0.5=32.5 2) when we add x quantity of water, acid becomes 10%so \(\frac{17.5}{(x+50)} = \frac{10}{100}\).. so 175=x+50.. x=125.. this means 125 gm of water is to be added so A is correct
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28 Nov 2016, 15:41
I think this is a highquality question and I agree with explanation. Interesting example to understand the diference between ratio and percentage



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Re: M0918
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28 Aug 2017, 03:43
Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient.
Answer: A Hi BunuelPlease help me with some another approach,which i can use with every questions like this.Example Acid+x/Total+x=Desired Qty/Total



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Re: M0918
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28 Aug 2017, 03:54
himanshukamra2711 wrote: Bunuel wrote: Official Solution:
Say there are \(x\) grams of 35%solution of acid and \(y\) grams of water should be added to obtain a 10%solution of acid. Equate amount of acid: \(0.35*x=0.1*(x+y)\). Explanation: since after we add pure water to the solution the amount of acid in grams is not changed, then we are simply equating the amount of acid in grams in initial 35% solution (0.35x) and the amount of acid in grams in 10% solution after water is added (0.1(x+y)). (1) There are 50 grams of the 35%solution of acid. Given \(x=50\), so \(0.35*50=0.1*(50+y)\). We have a linear equation with only one unknown, hence we can get the single numerical value of \(y\). Sufficient. (2) In the 35%solution of acid the ratio of acid to water is 7:13. The same info as we already have: 35%solution of acid means that the ratio of acid to water is 35:65=7:13. Not sufficient.
Answer: A Hi BunuelPlease help me with some another approach,which i can use with every questions like this.Example Acid+x/Total+x=Desired Qty/Total Check here: https://gmatclub.com/forum/howmuchwat ... 88479.html18. Mixture Problems For more check Ultimate GMAT Quantitative Megathread Hope it helps.
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Re: M0918
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14 Nov 2017, 23:29
High quality question!



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Re: M0918
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01 Jan 2018, 21:30
WRONG SENTENCE STRUCTURE! If you add the water to the strong acid, you will get a BIG explosion! That is a CHEM 101 lab test question! Moron test writers...
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