Given two ingredients, one with 42% fat and one with 3% fat, how can I calculate how much I will need of each to get 9% fat?

  • 4
    I’m voting to close this question because it's really just math.
    – Erica
    Commented Aug 27, 2020 at 22:36

2 Answers 2


As a matter of arithmetic, what you want is for the weighted average fat content to be 9%. We can say

42X + 3Y = 9

where X and Y are the fractions of the whole that are from each of the two components.


as the parts add up to the whole, so


leading to


This means that a mixture of 2/13 42% fat plus 11/13 3% fat will make 9%.

It's important to be consistent. If your fat percentage of both ingredients is by weight, then work in weight throughout, if by volume, work in volume.

To put this into practice, let's say both are by weight, and we want 100g in total.

2/13×100g = 15.4g

11/13×100 = 84.6g

You can't measure that accurately, so round to 15g of 42% fat and 85g of 3%, and you'll have 100g of 9%


This is really more of a math question than it is a cooking question. However, I love math questions, so here we go.

If we have two ingredients, A and B, which contain a% and b% (respectively) of something (here, fat), a mixture of the two consisting of x% A and y% B will contain (xa + yb)% total fat. Note that this works because a, b, x and y are all percentages. To find out the exact quantities of A and B you need, we use x + y = 1 (assuming two ingredients only and 100% = 1, i.e., we work with proportions/fractions rather than percentages).

Thus, to obtain an end result with z% fat, substituting y = 1 - x we find xa + (1 - x) b = z, which simplifies to x = (z - b) / (a - b).

For your specific example, we have a = 0.42, b= 0.03 and z = 0.09, which gives x = 0.06 / 0.39 = 0.154. In other words, about 15.4% of your end product should be A (the high-fat ingredient).

PS. It seems that math formatting is not enabled on this site; but perhaps I am mistaken. If I figure out how to format the above using MathJax, I will update the answer.

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