SOLUTION: A pharmacist wants to prepare 300 ml of a 20 percent alcohol solution. How much of a 30 percent solution and a 15 percent solution can be used to form the desired mixture?

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Question 891887: A pharmacist wants to prepare 300 ml of a 20 percent alcohol solution. How much of a 30 percent solution and a 15 percent solution can be used to form the desired mixture?
Found 3 solutions by josgarithmetic, josmiceli, MathTherapy:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website!
Your are looking for this lesson which also contains the general solution - http://www.algebra.com/tutors/Mixtures%3A-All-in-Symbols.lesson?content_action=show_dev

Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website!
Let = ml of 30% solution needed
Let = ml of 15% solution needed
-------------
(1)
(2)
-------------------------------
(2)
(2)
(2)
--------------------------
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
--------------------------

and, since
(1)
(1)
100 ml of 30% solution is needed
200 ml of 15% solution is needed
check:
(2)
(2)
(2)
(2)
OK

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

A pharmacist wants to prepare 300 ml of a 20 percent alcohol solution. How much of a 30 percent solution and a 15 percent solution can be used to form the desired mixture?

Let amount of 30% solution to be mixed be T
Then amount of 15% solution to be mixed = 300 � T
Therefore, we have: .3(T) + .15(300 � T) = .2(300)
.3T + 45 - .15T = 60
.3T - .15T = 60 � 45
.15T = 15
T, or amount of 30% solution to be mixed = , or mL
Amount of 15% solution to be mixed = 300 � 100, or mL