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Two solutions of acid were mixed to obtain 10 liters of a

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Re: Two solutions of acid were mixed to obtain 10 liters of a  [#permalink]

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New post 18 Apr 2017, 16:45
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bmwhype2 wrote:
Two solutions of acid were mixed to obtain 10 liters of new solution. Before they were mixed, the first solution contained 0.8 liters of acid while the second contained 0.6 liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?

A. 3 liters
B. 3.2 liters
C. 3.6 liters
D. 4 liters
E. 4.2 liters


We can let x = the number of litres of the first solution (the one with 0.8 litres of acid), so 10 - x = the number of litres of the second solution (the one with 0.6 litres of acid). Since the percentage of acid in the first solution was twice that of the second solution, we can say:

0.8/x = 2 * 0.6/(10 - x)

0.8/x = 1.2/(10 - x)

1.2x = 8 - 0.8x

2x = 8

x = 4

Answer: D
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Re: Two solutions of acid were mixed to obtain 10 liters of a  [#permalink]

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New post 06 Aug 2019, 14:06
Let the volumes of the solutions be A and B.
A+B=10 (1)
0.8/A=2(0.6/B) (2)
From (1)
B=10-A (3)

Substituting (3) in (2) we get
0.8/A=2(0.6)/(10-A)
Rearranging we get
8-0.8A=1.2A
8=2A
A=4 Litres
So the right Answer is D. The volume of the first solution was 4 litres.
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Re: Two solutions of acid were mixed to obtain 10 liters of a  [#permalink]

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New post 12 Dec 2019, 16:56
(0.8/x)= 2*(0.6/(10-x))
0.8/x = 1.2/10-x
cross multy
8 - 0.8x= 1.2x
2x = 8
x= 4
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Re: Two solutions of acid were mixed to obtain 10 liters of a   [#permalink] 12 Dec 2019, 16:56

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