# Word problem help

#### 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?

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Hello Stephen

Welcome to Math Help Forum!
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 $$\displaystyle \displaystyle x$$ cc of the first solution, and $$\displaystyle \displaystyle y$$ cc of the second. Then we can set up two simultaneous equations as follows:

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

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

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

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

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

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

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

#### Soroban

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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 $$\displaystyle x$$ cc's of solution $$\displaystyle A$$

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

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

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

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

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

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

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

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

#### Wilmer

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)

#### Stephen

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.