SOLUTION: A 5% salt solution is mixed with a 10% salt solution. How many milliliters of each solution are needed to obtain 400 milliliters of an 8% solution.
s= number of milliliters of
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s= number of milliliters of
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Question 696115: A 5% salt solution is mixed with a 10% salt solution. How many milliliters of each solution are needed to obtain 400 milliliters of an 8% solution.
s= number of milliliters of 5% salt solution
t= number of milliliters of 10% s alt solution
I wrote an equation of
(s+t=400
(.05s+.1t=40
i tried solving but got stuck. am i on the right track? Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
A 5% salt solution is mixed with a 10% salt solution. How many milliliters of each solution are needed to obtain 400 milliliters of an 8% solution.
s= number of milliliters of 5% salt solution
t= number of milliliters of 10% s alt solution
I wrote an equation of
(s+t=400
(.05s+.1t=40
i tried solving but got stuck. am i on the right track?
Let amount of 5% salt solution to be mixed be F
Then amount of 10% salt solution to be mixed = 400 – F
Therefore, .05F + .1(400 – F) = .08(F + 400 – F)
.05F + 40 - .1F = 32
.05F - .1F = 32 – 40
- .05F = - 8
F, or amount of 5% salt solution to be mixed = , or mL
Amount of 10% salt solution to be mixed = 400 – 160, or mL
OR
Doing it your way:
s + t = 400
.05s + .1t = .08(400) ---- .05s + .1t = 32
You can now solve for s and then t, or vice-versa
If you notice, you'll see that you're on the right track, with the exception that the 2nd equation should be .05s + .1t = 32, instead of .05s + .1t = 40
You can do the check!!
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