Expanding the walls of our classroom. This is an interactive learning ecology for students and parents in our Algebra 2 class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
Here is a rate problem: Ten liters of a 4% acid solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
So you need to set your little table: % (decimal) LITERS (of soution) LITERS(of acid) .04 10 .4 .1 x .1x .06 10+x .06(10+x)
After you have set up your table you take .4 and .1x and set them on one side of the equation and put .06(10+x) on the other. Like: .4+.1x= .06(10+x) Simplify: .4+.1x= .6 +.06x -.4 -.06 .04x=.2 x=5 So 5 liters of the 10% solution must be added. I hope this helped! Brigid
Here is a rate problem:
ReplyDeleteTen liters of a 4% acid solution must be mixed with a 10% solution to get a 6% solution. How many liters of the 10% solution are needed?
So you need to set your little table:
% (decimal) LITERS (of soution) LITERS(of acid)
.04 10 .4
.1 x .1x
.06 10+x .06(10+x)
After you have set up your table you take .4 and .1x and set them on one side of the equation and put .06(10+x) on the other.
Like:
.4+.1x= .06(10+x)
Simplify:
.4+.1x= .6 +.06x
-.4 -.06
.04x=.2
x=5
So 5 liters of the 10% solution must be added.
I hope this helped!
Brigid