Baker’s Percentage Exercises, Part 3, With Answers

Answers to exercises for Part 3 of the Baker’s Percentage Tutorial.

Exercise 1

Consider this formula:

  • White flour 50%
  • Whole wheat flour 20%
  • Semolina (flour) 30%
  • Water 70%
  • Instant yeast 1.5%
  • Salt 2%

How much of each ingredient would you use if you wanted to make 1400 g of dough?

Exercise 1 Answer

The total % is 173.5.

Total Flour Weight is (1400/173.5) x 100 = 807 g.

Use the TFW to calculate the amount of each ingredient:

  • White flour: 404 g (50% x 807)
  • Whole wheat flour: 161 g (20% x 807)
  • Semolina: 242 g (30% x 807)
  • Water: 565 g (70% a 807)
  • Instant yeast: 12 g (1.5% x 807)
  • Salt: 16 g (2% x 807)

Exercise 2

Consider this formula:

  • Flour 75%
  • Whole wheat flour 20%
  • Whole rye flour 5%
  • Water 50%
  • Milk 10%
  • Instant yeast 1%
  • Salt 2%

If you only have 50 g of milk left in your refrigerator, how much dough can you make? How much of each of the other ingredients would you need?

Exercise 2 Answer

The 50 g of milk accounts for 10 parts of the dough, so each part is 5 g (50/10).

The total number of parts (total %) is 163, so 815 g of dough can be made (5 x 163).

The flour is 100 parts, so the Total Flour Weight is 500 g (5 x 100).

Use the TFW to calculate each ingredient:

  • White flour: 375 g
  • Whole wheat flour: 100 g
  • Whole rye flour: 25 g
  • Water: 250 g
  • Milk: 50 g
  • Instant yeast: 5 g
  • Salt: 10 g

Post a comment » One Comment

  1. Thank you for the tutorial on the baker’s math! Very helpful stuff. As with all math, there is more than one way to solve a problem. Regarding problem 2, it seems more straightforward to answer it as follows:

    Since we know we have 50 grams of milk, which = 10% of the loaf, that means that 1 gram of x = 1% of the loaf.

    From there, you need only multiply 5 x the %# for each ingredient (e.g., White Flour grams = 5×75, Salt = 5×2. (and if you want to get the weight of the loaf, you sum up all of the calculated ingredients).

    Your solution is more complicated. I do recognize, however, that since your solution is a bit more complex, it also gives one a deeper understanding of the problems. For example, one might forget how to get to the % calculations if they only have the weights of the ingredients — but your solution keeps all of the relevant calculations at the forefront of our minds so that we can do all the math regardless of the problem.

    My comment was only meant for those who always have the % figures and just want to solve this particular problem: I have the % and I have this much of an ingredient, how much do I need for everything else.

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