M09-18

Question

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.

Bunuel · Accepted Answer

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

DJ1986 · Answer

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

harish1986 · Answer

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?

chetan2u · Answer

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

lndncalling · Answer

I think this is a  high-quality  question and I agree with explanation. Interesting example to understand the diference between ratio and percentage

himanshukamra2711 · Answer

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

Bunuel · Answer

Check here: https://gmatclub.com/forum/how-much-wat ... 88479.html 18. Mixture Problems Theory Weighted Average and Mixture Problems on the GMAT Tips and Tricks: Mixtures Questions DS Questions PS Questions For more check Ultimate GMAT Quantitative Megathread Hope it helps.

BobsterGMAT · Answer

High quality question!

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

ISHAAN96 · Answer

I think this is a  high-quality  question.

Dbrunik · Answer

I think this is a  high-quality  question and I agree with explanation.

IIIJOHNNYSIII · Answer

I like the solution - it’s helpful. Thank you!

Mayank_1206 · Answer

I like the solution - it’s helpful.

M09-18

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards