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How much water (in grams) should be added to a 35%-solution

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How much water (in grams) should be added to a 35%-solution [#permalink] New post 27 Dec 2009, 14:59
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64% (02:07) correct 36% (00:59) wrong based on 138 sessions
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.
[Reveal] Spoiler: OA
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Re: m09#18 [#permalink] New post 27 Dec 2009, 17:37
Expert's post
tania wrote:
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

There are 50 grams of the 35%-solution.
In the 35%-solution the ratio of acid to water is 7:13.

I know the above one is easy, can someone please explain to me in detail, how the above problem should be solved?


Set the equation 0.35S=0.1(S+W), W=2.5S --> W=? You can see that only thing we need to know to calculate the value of W is S (initial amount of solution in grams).

The equation basically is saying that acid in grams in initial 35% solution (0.35S), equals to the acid in grams in 10% solution after water is added (0.1(S+W)), as amount of acid is not changed.

(1) Directly gives the value of S=50 --> W=2.5*50=125gr. Sufficient.

(2) A/W=7/13, this information is the same as given in the stem: there is 35% of acid in the solution. Out of 20 shares (7+13=20) 7 is acid or 7/20=35%. Not sufficient.

Answer: A.

Hope it's clear.
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Re: m09#18 [#permalink] New post 28 Dec 2009, 09:12
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

Therefore the answer is A
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mixture problem [#permalink] New post 25 Feb 2011, 12:50
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,

-Mike
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Re: mixture problem [#permalink] New post 25 Feb 2011, 14:39
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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)
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Re: mixture problem [#permalink] New post 25 Feb 2011, 15:17
Thanks a lot, now I get it.
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Re: m09#18 [#permalink] New post 26 Feb 2011, 20:53
statement 2) is useless. Its Truism analogous to "sun rises in the east". Hence A.
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Re: m09#18 [#permalink] New post 23 Apr 2011, 08:34
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

So answer - option A
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Re: How much water (in grams) should be added to a 35%-solution [#permalink] New post 08 Feb 2014, 02:47
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Re: How much water (in grams) should be added to a 35%-solution   [#permalink] 08 Feb 2014, 02:47
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