SOLUTION: Chas has some 12% acid solution and some 32% acid solution. Determine how much
of each he needs in order to get 40ml of a 20% acid solution.
I tried setting up the problem like
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Chas has some 12% acid solution and some 32% acid solution. Determine how much
of each he needs in order to get 40ml of a 20% acid solution.
I tried setting up the problem like
Log On
Question 27140: Chas has some 12% acid solution and some 32% acid solution. Determine how much
of each he needs in order to get 40ml of a 20% acid solution.
I tried setting up the problem like this: .12(x)+ .32(x)=.2(40) but i didn't
get the answer right Answer by AnlytcPhil(1806) (Show Source):
You can put this solution on YOUR website! Chas has some 12% acid solution and some 32% acid solution. Determine how much
of each he needs in order to get 40ml of a 20% acid solution.
I tried setting up the problem like this: .12(x)+ .32(x)=.2(40) but i didn't
get the answer right.
---------------------------------------------------------------
Your mistake is in letting the same letter x represent two different
quantities. You must let x represent only one of the quantities, and then
either y or 40-x will represent the other. Never let the same letter
represent two different quantities in the same problem. There are four ways
to solve it.
---------------------------------------------------------------
1. One unknown letting x = no of ml. of weaker solution and 40-x = no of ml. of stronger solution.
`
That equation would be .12(x) + .32(40-x) = .2(40)
---------------------------------------------------------------
2. One unknown letting x = no of ml. of stronger solution and 40-x = no of ml. of weaker solution.
`
That equation would be .12(40-x) + .32(x) = .2(40)
---------------------------------------------------------------
3. Two unknowns letting x = no of ml. of weaker solution and y = no of ml. of stronger solution.
`
This would be a system of two equations and two unknowns:
x + y = 40
.12x + .32y = .2(40)
--------------------------------------------------------------
4. Two unknowns letting x = no of ml. of stronger solution and y = no of ml.
of weaker solution.
This would be a system of two equations and two unknowns:
x + y = 40
.32x + .12y = .2(40)
--------------------------------------------------------------
For any of these four ways, the answer is 24 ml of the weaker solution
and 16 ml of the stronger solution.
Edwin
AnlytcPhil@aol.com
