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# Word problems :: M24-04

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Director
Joined: 03 Sep 2006
Posts: 875
Followers: 7

Kudos [?]: 891 [0], given: 33

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02 Jan 2013, 07:12
X grams of water were added to the 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution became $$\frac{1}{Y}$$ times of the initial concentration. What was the concentration of acid in the original solution?

(1) X=80

(2) Y=2

Please correct what's wrong in my approach:

Initial: 80gms
Acid: x gms

$$H_{2}O$$: y gms

concentration of acid is : x/80

Later: 80+X
Acid: x gms

$$H_{2}O$$: y+X gms

concentration of acid is : x/(80+X)

Given: $$\frac{1}{Y}*(\frac{x}{80})=\frac{x}{(80+X)}$$

Clearly even if we combine 1 and 2 x can't be found, which is required.
Manager
Joined: 28 Dec 2012
Posts: 112
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
Followers: 3

Kudos [?]: 70 [0], given: 90

Re: Word problems :: M24-04 [#permalink]

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02 Jan 2013, 12:30
Given x= 80 and Y = 2... Since acid was also 80, and on adding 80 gm water to acid concentration is halved => acid was pure acid. 100% concnetration.
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Re: Word problems :: M24-04   [#permalink] 02 Jan 2013, 12:30
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# Word problems :: M24-04

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