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M09-18

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Bunuel
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M09-18 [#permalink] New post   16 Sep 2014, 00:39
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How many grams of water should be added to a 35% acid solution to obtain a 10% acid solution?


(1) The initial 35% acid solution weighs 50 grams.

(2) In the 35% acid solution, the ratio of acid to water is 7:13.
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Re M09-18 [#permalink] New post   16 Sep 2014, 00:40
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Official Solution:


How many grams of water should be added to a 35% acid solution to obtain a 10% acid solution?

Let \(x\) be the initial grams of the 35% acid solution and \(y\) the grams of water to be added to obtain a 10% acid solution. Since we add pure water to the solution and the amount of acid in grams remains unchanged, we can derive an equation to equate the acid content in the initial 35% solution (\(0.35x\)) and the 10% solution after water is added (\(0.1(x+y)\)). Thus, we have the equation: \(0.35x=0.1(x+y)\).

(1) The initial 35% acid solution weighs 50 grams.

Given \(x=50\), the equation becomes \(0.35*50=0.1*(50+y)\). This linear equation has one unknown, allowing us to find a single numerical value for \(y\). Sufficient.

(2) In the 35% acid solution, the ratio of acid to water is 7:13.

This information restates what we already know: a 35% acid solution implies an acid-to-water ratio of 35:65 or 7:13. Not sufficient.


Answer: A
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Re: M09-18 [#permalink] New post   19 Feb 2016, 14:42
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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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Re: M09-18 [#permalink] New post   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: M09-18 [#permalink] New post   30 Mar 2016, 09:29
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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 50gm
so 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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Re: M09-18 [#permalink] New post   28 Nov 2016, 15:41
I think this is a high-quality question and I agree with explanation. Interesting example to understand the diference between ratio and percentage
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Re: M09-18 [#permalink] New post   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 Bunuel

Please 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: M09-18 [#permalink] New post   28 Aug 2017, 03:54
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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 Bunuel

Please 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/how-much-wat ... 88479.html

18. Mixture Problems



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Re: M09-18 [#permalink] New post   14 Nov 2017, 23:29
High quality question!
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Bunuel
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Re: M09-18 [#permalink] New post   18 Apr 2023, 09:30
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M09-18 [#permalink]
18 Apr 2023, 09:30
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