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Mixture

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Mixture

by dtweah » Sun May 17, 2009 5:42 am
Container A has 10 pounds of water and container B has 10 pounds of a 20% salt solution. Five pounds are transferred from A to B and mixed with B, and then y pounds are transferred from B to A and mixed with A. If the final mixture in A is a 5% salt solution then y equals

a) 3
(b) 7/2
(c) 17/6
(d) 25/9
(e) 4
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by PAB2706 » Sun May 17, 2009 6:18 am
i got answer as A ie 3 pounds

dont know if i am doing it the right way...but here goes my method

cont A =10

cont B= 10 ( 8 pound water 2 pound salt )

no after transfer cont B contains 15 pounds solution ( 86.67% water and 13.33% salt )

now when you transfer y pounds in cont A again, the water : salt proportion in every pound of solution in B is going to remain the same. ie 86.67 % and 13.33%

Thus y pounds will contain 13.33% salt

now after transfer cont A has 5% salt

thus we can form salt equation as 0.05( 5+y)=0.133y

solving this you get y=3.125.. but since 0.133 is recurring decimal the value of y will tend towards 3.

If i am wrong i am missing something.
:roll:
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by criszerriny » Sun May 17, 2009 7:16 am
A is left with 5, B now has 15, 2/15 is the salt solution in B.

Now, as you put B into A, you are going to have an equation of,

((2/15)y)/(y+5)=.05

1.25y=3.75
y=3 :)
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