1. ## Word problem help

could you please tell me how I set this up to solve? thanks

a veterinarian has two solutions that contain different concentrations of a certain medicine. One is 15% concentration and the other is 5 % concentration. How many cubic centimeters of each should the veterinarian mix to get 20 cc of a 6 % solution?

2. Hello Stephen

Welcome to Math Help Forum!
Originally Posted by Stephen
could you please tell me how I set this up to solve? thanks

a veterinarian has two solutions that contain different concentrations of a certain medicine. One is 15% concentration and the other is 5 % concentration. How many cubic centimeters of each should the veterinarian mix to get 20 cc of a 6 % solution?
Suppose he uses $\displaystyle x$ cc of the first solution, and $\displaystyle y$ cc of the second. Then we can set up two simultaneous equations as follows:

The total volume must be 20 cc. Therefore
$\displaystyle x+y = ...$ ?
This is equation (1).

The quantity of the medicine in $\displaystyle x$ cc of the first solution is $\displaystyle \frac{15x}{100}$.

The quantity of the medicine in $\displaystyle y$ cc of the second solution is $\displaystyle \frac{5y}{100}$.

The total quantity of medicine in the mixture is therefore ... ?

This must be 6% of the total volume, $\displaystyle 20$ cc, which is $\displaystyle \frac{6\times 20}{100} = ...$ ?

The second equation is therefore
$\displaystyle \frac{15x}{100}+\frac{5y}{100}=...$ ?
This is equation (2).

Can you complete what I have started, and then solve the simultaneous equations to find $\displaystyle x$ and $\displaystyle y$?

3. Hello, Stephen!

Here's a one-variable approach . . .

A vet has two solutions that contain different concentrations of a certain medicine.
One is 15% concentration and the other is 5% concentration.
How many cc's of each should the vet mix to get 20 cc of a 6% solution?

We will use $x$ cc's of solution $A$

. . It will contain: . $15\%\times x \:=\:0.15x$ cc's of the med.

We will use $(20-x)$ cc's of solution $B.$

. . It will contain: . $5\%\times(20-x) \:=\:0.05(20-x)$ cc's of the med.

The mixture will contain: . $0.15x + 0.05(20-x)$ cc's of the med.

But we know that the mixture will be 20 cc's with 6% med.

. . It contains: . $6\% \times 20 \:=\:0.06(20) \:=\:1.2$ cc's of the med.

We just described the final amount of med in two ways.

There is our equation! . . . $0.15x + 0.05(20-x) \;=\;1.2$

4. ## Well...

To each his own; I find this way removes confusion:
Code:
   x  :  15
20-x  :   5
===========
20  :   6
15x + 5(20-x) = 6(20) ; solve: x = 2

In general:
Code:
 x  :  a
y  :  b
========
x+y :  c
ax + by = c(x+y)

5. Thanks so much for the help! Ive been taking a algebra course over the summer and Ive been doing well. But im studying for a cumulative end of course exam and i have a hard time retaining some of the stuff i learned earlier on. thanks again for the help.