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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,
Re: mixture problem [#permalink]
25 Feb 2011, 14:39
This post received KUDOS
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, 02:47
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