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3 Answers

Here is a quick way to get the answer.

37-20 = 17, and 40-37 = 3

That means (3/20)(200) = 30 kg of a 20% fat-content chocolate, and 200-30 = 170 kg of a 40% fat-content chocolate.

Ideas: Inverse proportion between mass and difference from the equilibrium position.


Let's state what we know:  both fat-content chocolates added together will be 200 kg

Let's let x = the 20% fat-content chocolate, and y = the 40% fat-content chocolate (but we could use any 2 letters or symbols)

So we know that x + y = 200

Let's state the percentages as decimals:  20% = 0.20,    40% = 0.40,    and 37% = 0.37

We also know that the 200KG will be a 37% blend, so that's (0.37)(200) = 74

So now we can add the components together to get that blend:  0.20X + 0.40y = 74

Now we have two equations, each using the same x and y

We can solve for either x or y in one of these and substitute back into the other equation.  x + y = 200 looks easier, so let's solve it for x:    x = 200 - y  

Now put that x into the other equation:  (0.20)(200 - y) + (0.40y) = 74

40 - 0.20y + 0.40y = 74

40 + 0.20y = 74;   0.20y = 34;  y = 170kg

x + y = 200;  x + 170 = 200;  x = 200 - 170;  x = 30kg

So you need only 30KG of 20% fat-content chocolate, but 170KG of 40% fat-content chocolate.  A common-sense look at the original question shows that makes sense:  the 37% result we want is much closer to 40% than to 20%


For these kind of "mixing" problems, it is essentially a system of equations. Let's first think about this intuitively. For one, we know that there are two batches of chocolate (say x and y) and that the sum of thier content must total 200 kg. So, individual amounts must sum to some total amount. 

Amount of x + amount of y = total of 200

Next, we want to think of the actual amount of fat contained in each mixture. For example, if I ask "how much fat is in a 5 gal chocolate mixture if the concentration, C, is 20%? Well, that would be 0.20*5=1. If you didn't know it was 5 gal, it would be 0.20*X. The total amount of fat for each batch must also sum to something as well. So let's build the second equation by applying that principle; the concentration times the amount in batch 1 plus the concentration times the amount in batch 2 must equal the final concentration times the total amount of the mixture.

C1(x)+C2(y)=C3(total), or

So we have two equations and two unknows (a system of eqautions).

Eqn. 1) x + y = 200 (the sum of the mixtures)
Eqn. 2) 0.20x+0.40y=0.37(200) (the sum of their fat (or fill-in-the-blank) content)

Let's solve by substitution. First, solve eqn 1 for either x or y (doesn't matter).

Next, substitute for the variable x in eqn 2.
0.20(200-y)+0.40y=74; distribute, collect like terms, and solve for y.

y=170 kg

Now that we know y, substitute into eqn 1 to find x.
x=30 kg 

So we need 30 kg of 20% fat content added to 170 kg of 40% fat content to give 200 kg of chocolate with a 37% fat content.