Question

A chemist has one solution containing 30% insecticide and another solution containing 50% insecticide. How much of each solution should the chemist mix to get 200 L of a 42% insecticide?

My teacher explained how to set up the problem:
3x+5y=84
x+y=200
what I don't get is why the first equation is equal to 84, could someone could explain please?

Answer 1

This really enters the boundaries of Chemistry, but I will explain it in simpler terms.

Since solution one contains 30% insecticide, this means 70% is not insecticide but of some other matter.

This applies to solution 2, 50% insecticide and 50% of some other matter.

It tells you to find 42% insecticide therefore when you combine the solution, in which your teacher kindly tells you that x+y=200, you would only want 42% insecticide.

(.42)(200)= 84

Answer 2

amount of 30% solution needed ---- x L
amount of 50% solution needed ---- 200-x L

.3x + .5(200-x) = .42(200)
.3x + 100 - .5x = 84
-.2x = -16
x = -16/-.2 = 80

The chemist will need 80 L of the 30%, and 20 L of the 50% solution.
check:
.3(80) + .5(120) = 84

or , using the 2 equations in 2 unknowns ...

3x+5y=84
x+y=200 ---> y = 200-x , note that is how I started
sub that into the 1st
3x + 5(200-x) = 84
3x + 1000 - 5x = 84
-2x = -916
x = 458

this answer makes no sense,
because your first equation makes no sense.
it should have been
.3x + .5y = 84

either your teacher gave you the wrong first equation, or
you copied it incorrectly.

Answer 3

Yes, I was wondering the same thing too, the first equation indeed must be .3x + .5y = 84. This way it will make much more sense.

Answer 4

Just noticed a typo in my conclusion statement,
should have read:
The chemist will need 80 L of the 30%, and <b>120 L</b> of the 50% solution.

notice I did use 120L in my check