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?

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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?

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Answer 1 · 2016-02-22T15:02:09

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

$a + b + c = d$ $a+b+c=d$

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

$a + b = 0.60 d$ $a+b=0.60d$

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

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

which we can solve for $c$ $c$ :

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

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 $c$ $c$ . Lets take the final equaion for $c$ $c$ and put it into the first equation to eliminate $c$ $c$ :

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

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

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

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

$b = \frac{4 \cdot 0.4}{3} \cdot 750 k g = 400 k g$ $b=(4*0.4)/3*750kg=400kg$

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

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

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

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

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:

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

which is $60 %$ $60%$ of $750 k g$ $750kg$ and

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

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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?

1 Answer

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