Dukegraz464idn
Answered

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A pharmacist mixed some 10%-saline solution with some 15%-saline solution to obtain 100 mL of a 12%-saline solution. How much of the 10%-saline solution did the pharmacist use in the mixture?

Sagot :

Shinalcantaraidn
Let 'x' be the amount @ 10% saline solution
      'y' be the amount @ 15% saline solution
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-since the total mixture amounts to 100mL then
x + y = 100   ----equation 1
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-the resulting solution is 12% concentrated
0.1x + 0.15y = 100(0.12)
 0.1x + 0.15y = 12             ------equation 2
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From equation 1
x + y = 100
x = 100 - y    -------equation 1'
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Substitute equation 1' to equation 2
0.1x + 0.15y = 12
0.1(100-y) + 0.15y = 12
10 - 0.1y + 0.15y = 12
-0.1y + 0.15y = 12 - 10
0.05y = 2
y = 40mL
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Substitute y=40 to equation 1'
x = 100 - y
x = 100 - 40
x = 60mL
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Therefore he used 40mL of 15% saline solution and 60mL of 10% saline solution.
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