SOLUTION: I have been asked to prepare 3 liters of 30% strenth solution. My pharmacy stocks active ingredients in 8 oz. bottles of 70% strength. How many bottles of active ingredient will I
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: I have been asked to prepare 3 liters of 30% strenth solution. My pharmacy stocks active ingredients in 8 oz. bottles of 70% strength. How many bottles of active ingredient will I
Log On
Question 771153: I have been asked to prepare 3 liters of 30% strenth solution. My pharmacy stocks active ingredients in 8 oz. bottles of 70% strength. How many bottles of active ingredient will I need to open to complete this request? I tried using alligation math but it became to confusing. Is there an algebraic formula for figuring this out? Answer by mananth(16946) (Show Source):
percent ---------------- quantity
solution I 70.00% ---------------- x liters
Water 0.00% ------ 3 - x liters
Mixture 30.00% ---------------- 3
3
70.00% x + 0.00% ( 3 - x ) = 30.00% * 3
70 x + 0 ( 3 - x ) = 90
70 x + 0 - 0 x = 90
70 x - 0 x = 90 - 0
70 x = 90
/ 70
x = 1.29 liters 70.00% solution I
1.71 liters
30.00% Water
we need 1.29 liters of solution
1 liter = 33.81 fl oz.
1.29 liter = 43.62 fl. oz
1 bottle = 8 fl.oz
43.62 fl. Oz = 5.45 bottle = 6 bottles
