Word problem help

Jul 2010
7
0
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?
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
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 \(\displaystyle \displaystyle y\)?

Grandad
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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! . . . \(\displaystyle 0.15x + 0.05(20-x) \;=\;1.2\)

 
Dec 2007
3,184
558
Ottawa, Canada
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)
 
Jul 2010
7
0
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.