Asked by Anonymous
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?
A 60 mL
B 45 mL
C 40 mL
D 25 mL CAN SOME GIVE AN EQUATION TO HELP MY SOLVE THIS PROBLEM THIS IS THE ONLY WAY TO SOLVE THis problem
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?
A 60 mL
B 45 mL
C 40 mL
D 25 mL CAN SOME GIVE AN EQUATION TO HELP MY SOLVE THIS PROBLEM THIS IS THE ONLY WAY TO SOLVE THis problem
Answers
Answered by
Damon
x ml of 10% --->0.1 x salt
(100-x ml) of 15% ----> 0.15(100-x) salt
total salt = .1x + 15 - .15x = 15-.05x salt
so
15-.05x = .12(100) = 12
3 = .05x
x = 60 ml
(100-x ml) of 15% ----> 0.15(100-x) salt
total salt = .1x + 15 - .15x = 15-.05x salt
so
15-.05x = .12(100) = 12
3 = .05x
x = 60 ml
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