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

User Ed Rushton
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1 Answer

4 votes

Answer:

10%-saline solution = 60mL

15%-saline solution = 40mL

Step-by-step explanation:

Let

10%-saline solution = x

15%-saline solution = y

x + y = 100

0.1x + 0.15y = 100*0.12

x + y = 100 (1)

0.1x + 0.15y = 12 (2)

from (1)

x = 100 - y

Substitute x = 100 - y into

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 = 2 / 0.05

y = 40mL

Substitute y = 40 into

x + y = 100

x + 40 = 100

x = 100 - 40

x = 60mL

10%-saline solution = 60mL

15%-saline solution = 40mL

User Rohit Soni
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