# A plant food is to be made from 3 chemicals. The mix must include 60% of the first and second chemicals. The second and third chemicals must be in the ratio of 4:3 by weight. How much of each chemical is needed to make 750kg of the plant food?

Feb 22, 2016

The weights of the first, second and third chemicals are 50kg, 300kg, and 400kg respectively.

#### Explanation:

Let's let the weight of each of the first, second and third chemicals be $a$, $b$, and $c$, and the total weight of the plant food be $d$. The total of each individual component in the mixture must add up to the total weight, so we have our first equation:

$a + b + c = d$

We have also been told that the weight of the first and second chemical represents 60%, or $0.60$ of the total weight, which gives us a second equation:

$a + b = 0.60 d$

Finally, we are given the ratio of the second and third chemicals as $4 : 3$ leading us to a third equation:

$\frac{b}{c} = \frac{4}{3}$

which we can solve for $c$:

$c = \frac{3}{4} b$

We now have 3 equations and 3 unknowns. Looking at the equations, we notice that the first and second equations differ only by the presence of $c$. Lets take the final equaion for $c$ and put it into the first equation to eliminate $c$:

$a + 1 \frac{3}{4} b = d$

if we subtract the second equation from this equation we get:

$\frac{3}{4} b = 0.4 d$

Now we can substitute the total weight of plant food for $d$ and solve for $b$:

$b = \frac{4 \cdot 0.4}{3} \cdot 750 k g = 400 k g$

Now we can solve for $c$ using the rearranged third equation above:

$c = \frac{3}{4} b = \frac{3}{4} \cdot 200 k g = 300 k g$

Finally, lets use the first equation for the total weight rearranged to solve for $a$:

$a = d - b - c = 750 k g - 400 k g - 300 k g = 50 k g$

Its always a good idea to check the answers, so lets make sure that we got the percentage of first and second ingredients and ratio of the second and third ingredients correct:

$a + b = 50 k g + 400 k g = 450 k g$

which is 60% of $750 k g$ and

$\frac{b}{c} = \frac{400 k g}{300 k g} = \frac{4}{3}$ which is the ratio we wanted!