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Mixture Problem

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Director
Joined: 16 May 2007
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04 Jul 2007, 20:02
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Can someone please explain how to solved there kinds of problems. Im having a tough time with mixture problems.

Some part of the 50% soultion of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced ?

A. 1/5
B. 1/4
C. 1/2
D. 3/4
E. 4/5
VP
Joined: 10 Jun 2007
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04 Jul 2007, 21:49
trahul4 wrote:
Can someone please explain how to solved there kinds of problems. Im having a tough time with mixture problems.

Some part of the 50% soultion of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced ?

A. 1/5
B. 1/4
C. 1/2
D. 3/4
E. 4/5

C.

Solution A: acid = 50%, Volume = A-B
Solution B: acid = 30%, Volume = B
Solution C: acid = 40%, Volume = A
50*(A-B) + 30*B = 40*A
50A - 40A = 50B - 30B
10A = 20B
B/A = 1/2
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Location: Singapore
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04 Jul 2007, 22:15
assume acid is 1 litre, and assume part removed is x litres

then, 50(1-x) + 30 = 40
x = 0.8 -> 8/10 = 4/5
VP
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04 Jul 2007, 22:30
ywilfred wrote:
assume acid is 1 litre, and assume part removed is x litres

then, 50(1-x) + 30 = 40
x = 0.8 -> 8/10 = 4/5

Shouldn't it be 50(1-x) + 30x = 40?
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Posts: 5045
Location: Singapore
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04 Jul 2007, 22:48
bkk145 wrote:
ywilfred wrote:
assume acid is 1 litre, and assume part removed is x litres

then, 50(1-x) + 30 = 40
x = 0.8 -> 8/10 = 4/5

Shouldn't it be 50(1-x) + 30x = 40?

my bad. yes, it should be 50(1-x) + 30x = 40 to get x = 1/2
04 Jul 2007, 22:48
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