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Zigmanuir [339]
3 years ago
9

If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each sol

ution, to the nearest milliliter, is needed?
A. 167 milliliters of 7% solution and 333 milliliters of 4% solution
B. 110 milliliters of the 10% solution and 210 milliliters of 4% solution
C. 155 milliliters of the 10% solution and 300 milliliters of 4% solution
D. 170 milliliters of the 10% solution and 340 milliliters of 4% solution
Chemistry
1 answer:
GenaCL600 [577]3 years ago
4 0
The correct answer is A. <span>167 milliliters of 7% solution and 333 milliliters of 4% solution and here is how: 
</span><span>If x is the number of milliliters of the 7% saline solution and y is the number of milliliters of the 4% saline solution then add up to 500 milliliters total, so x + y = 500. 
</span>and if we do
x + y = 500 
<span>x = 500 - y </span>
<span>0.07x + 0.04y = 25 (substitute 500 - y for x) </span>
<span>0.07(500 - y) + 0.04y = 25 </span>
<span>35 - 0.07y + 0.04y = 25 </span>
<span>-0.03y + 35 = 25 </span>
<span>-0.03y = -10 </span>
<span>y = 333.333... </span>
<span>y = about 333 </span>
<span>x = 500 - y = 500 - 333 = 167 
</span>Then you know why the answer is A. 
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