Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
The answer is A because from the statement 1, you have enough information to answer the question. if you know the total is 50 gram, you have 35% of acid and 65% of water. by knowing the total, you can determine the amount in gram of acid 50*35/100 and the amount in gram of water 50*65/100, so you don't need to pursue the calculation because you know that you have enough information to answer the question.
The second statement is irrelevant because this information is already stated on the question stem, if you know the amount of acid is 35%, so the amount of water is 65%, so the ratio is 3.5:6.5 which is the same as 7:13
3.How much water (in grams) should be added to the 35%-solution of acid to obtain the 10%-solution? 1. There are 50 grams of the 35%-solution 2. In the 35%-solution the ratio of acid to water is 7:13 what is the best way to approach mixture problems ?
1) I tried to solve this problem using the mixture composition: x-original solution in grams w-water to be added in grams
0.35x + w = 0.1(x + w)
but obviously this is the wrong equation. Can some one point out why this eqn is wrong. Thanks,
First of all, the second statement tells us nothing as 35% means 7:13 or 35%:65% ratio. From the first statement we can find how much water and acid we have in 50g of 35% solution and it's sufficient. If it's not obvious, we can use following calculations:
acid = 0.35x water in the solution: 0.65x % of new solution = 0.1 = 0.35x / (w + x)
0.1*(w + x) = 0.35x - meaning: acid in new and old solutions (it should be the same number)
so, here is your mistake: 0.35x + w = 0.1(x + w)
Statement 2 gives us the information that we already have from the question Statement 1 - We can calculate the answer by using this statement... If 50 is the solution, then 17.5 is the acid Solve 17.5/(50+x)=10/100
Re: How much water (in grams) should be added to a 35%-solution [#permalink]
08 Feb 2014, 03:47
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?
(1) There are 50 grams of the 35%-solution. (2) In the 35%-solution the ratio of acid to water is 7:13.
Responding to a pm: Here is how you will use scale method here: You want to add pure water (0% acid solution) to 35% acid solution to get 10% acid solution.
w1/w2 = (35 - 10)/(10 - 0) = 25/10 = 5/2
So you will need to add 5 parts water to 2 parts 35% solution.
(1) There are 50 grams of the 35%-solution. 5/2 = Water/50 Water = 125 gms
(2) In the 35%-solution the ratio of acid to water is 7:13. This just tells us that 35% solution has 7 parts acid in every 20 parts of solution. This is same as saying that there are 35 parts acid in 100 parts of solution. No new information. Not sufficient.