Question 1158396
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(1) A standard algebraic solution....<br>
x mL of the 50% dextrose, plus (100-x) mL of water (0% dextrose), equals 15% of 100 mL:<br>
{{{.50(x)+0(100-x) = .15(100)}}}
{{{.5x = 15}}}
{{{x = 15/.5 = 30}}}
ANSWER: 30 mL of 50%; 70 mL of water<br>
(2) Basically the same solution, informally....<br>
The water is 0% dextrose, so all of the dextrose comes from the 50% dextrose solution.
In the final mixture, the amount of dextrose is 15% of 100 mL, or 15 mL.
The 50% dextrose solution is one-half dextrose; to get 15mL of dextrose, you need 30 mL of solution.<br>
(3) A completely different method of solving....<br>
You are mixing 0% and 50% dextrose to get 15% dextrose.
15% is 15/50 = 3/10 of the way from 0% to 50%.
So 3/10 of the mixture is the 50% dextrose.
3/10 of 100 mL is 30 mL.<br>