<PRE><B>Question 476652</B>
Ten liters of 30% solution needs to be mixed with 80% solution to produce a 50 % solution.
 How much of the 80% solution is needed.
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Let x = amt of 80% solution
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you can solve this with a single equation using the decimal equiv of per cent;
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.30(10) + .80x = .50(x + 10)
3 + .80x = .50x + 5
.80x - .50x = 5 - 3
.30x = 2
x = {{{2/.3}}}
x = 6{{{2/3}}} liters of 80% solution required
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:
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Check this decimals
.3(10) + .8(6.67) = .50(6.67+10)
3 + 5.333 = .50(16.67)
8.333 ~ 8.335
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