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m09 Q 18

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Intern
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Joined: 13 Jan 2010
Posts: 23

Kudos [?]: 6 [0], given: 10

m09 Q 18 [#permalink]

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New post 07 Aug 2010, 16:22
Please can someone help solve this problem.

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.

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Kudos [?]: 6 [0], given: 10

Manager
Manager
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Joined: 16 Apr 2010
Posts: 214

Kudos [?]: 136 [0], given: 12

Re: m09 Q 18 [#permalink]

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New post 08 Aug 2010, 23:08
Let x be the weight of the solution and w be the weight of water that will be added.
We have 0.35x acid and 0.65w

Acid Water Solution
0.35x 0.65x x
0.35x 0.65x+w x+w

Thus 0.35x/ x+w = 10...eq.1
The question is: w?

Statement 1
0.35x= 50g, thus we can solve for x and replace in in eq.1 to determine W

Statement 2
Acid / Water = 7/13 = 7x/13x
Thus the solution is 20x and solution is 100%, x =5
So Acid is 35% and water is 65%
No additional information is given.

Answer is A

Hope this helps.

Kudos [?]: 136 [0], given: 12

Manager
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Joined: 18 Jan 2011
Posts: 228

Kudos [?]: 37 [0], given: 4

Re: m09 Q 18 [#permalink]

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New post 17 Apr 2011, 12:57
jakolik wrote:

Statement 2
Acid / Water = 7/13 = 7x/13x
Thus the solution is 20x and solution is 100%, x =5
So Acid is 35% and water is 65%
No additional information is given.

Answer is A

Hope this helps.


Right, so even though we can conclude the ratio of acid to water would have been 2:18 ... but we do not know the qty
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Good Luck!!!

***Help and be helped!!!****

Kudos [?]: 37 [0], given: 4

Re: m09 Q 18   [#permalink] 17 Apr 2011, 12:57
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