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# A solution of acid and water is A% acid. Another solution of acid and

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Retired Moderator
Joined: 22 Aug 2013
Posts: 1443
Location: India
A solution of acid and water is A% acid. Another solution of acid and  [#permalink]

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25 Nov 2018, 22:24
1
00:00

Difficulty:

75% (hard)

Question Stats:

38% (01:29) correct 62% (01:30) wrong based on 51 sessions

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A solution of acid and water is A% acid. Another solution of acid and water is B% acid. X litres of first solution is mixed with Y litres of second solution to form a new mixture of acid and water. What is the percentage of acid in the new mixture?

(1) Ratio of A:B = 5:4.

(2) Ratio of X:Y = 2:3.
Intern
Status: Prep Mode
Joined: 16 Jan 2015
Posts: 21
Location: India
GMAT 1: 630 Q48 V28
WE: Programming (Computer Software)
Re: A solution of acid and water is A% acid. Another solution of acid and  [#permalink]

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28 Nov 2018, 12:46
Let the final percentage of acid in new mixture = k

A---------------B
-------k----------
k-B------------A-k

It implies that first acid solution and second acid solution should be mixed in a ratio of $$\frac{(k-B)}{(A-k)}$$

Given that X liters of first solution is mixed with Y liters of second solution

$$\frac{(k-B)}{(A-k)}$$= $$\frac{X}{Y}$$

$$Yk - YB = XA - Xk$$

$$k$$= $$\frac{(XA + YB)}{(Y + X)}$$

$$k$$= $$( (\frac{X}{Y})*A +B ) / (1 + \frac{X}{Y})$$

It is clear that we need $$\frac{X}{Y}$$ , A and B to solve for k.

First Statement :

$$\frac{A}{B}$$ = $$\frac{5}{4}$$ -> INSUFFICIENT

Second Statement :

$$\frac{X}{Y}$$ = $$\frac{2}{3}$$ -> Still we need A and B. So INSUFFICIENT.

First + Second Statements:

k cannot be solved unless we know one of A and B. -> INSUFFICIENT

Re: A solution of acid and water is A% acid. Another solution of acid and   [#permalink] 28 Nov 2018, 12:46
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