In a village, there lived a kindly baker who sold many items (cakes, biscuits, buns, bread loaves, croissants, etc) in his small shop. He made enough profits for running his family.

Once, he made a dozen loaves of a special bread each of which cost him 2 U (U is some unit of currency) to make. He usually sold it for 3 U each, when the buyer was rich; but he kindly sold it for 1 U each, when he knew that the buyer was poor. After closing-time, he found that he was sold-out and had made a profit of 3 U on the bread alone.

Now Dear Reader, something for you to think over :

How is this situation possible ?
How many rich buyers bought the special bread ?

Hint :

Depending on how you usually think, you may treat it as one unknown or 2 unknowns.
If you think there is one unknown, then there are actually 2 unknowns.
If you think there are two unknowns, then there are actually 3 unknowns.
Basically, there is one EXTRA unknown.

  • $\begingroup$ I will be adding the exact expected answer to all my puzzles, so that the clues are explained, but if somebody else has got the correct answer, then that will be the "Accepted Answer". $\endgroup$
    – Prem
    May 9, 2015 at 16:49
  • 1
    $\begingroup$ The "cost" bit could be better phrased. You say a dozen loaves cost the baker 2 U, but you meant that a dozen loaves cost the baker 2 U each. $\endgroup$ Feb 13, 2018 at 21:32
  • $\begingroup$ @QuantumRipple , Done. $\endgroup$
    – Prem
    Feb 15, 2018 at 13:24

5 Answers 5


The answer is:

yes, it is possible and the baker had 8 rich buyers and 5 poor buyers.

This is because:

he makes a profit of 1U for each rich buyer and -1U (i.e. a loss of 1U) for each poor buyer. So the parity (odd/even) of his profit changes for each buyer. If the total number of loaves were 12 (an ordinary 'dozen'), his profit would be even and the answer would be no, it's not possible. But because he's a baker, we can assume he makes a baker's dozen, i.e. 13 loaves. Now since his profit is (number sold to rich buyers) - (number sold to poor buyers) = 3, while (number sold to rich buyers) + (number sold to poor buyers) = 13, he must have had 8 rich buyers and 5 poor buyers.

  • 1
    $\begingroup$ @randal'thor TIL the baker's dozen :) thanks $\endgroup$
    – A.D.
    May 9, 2015 at 14:13
  • 2
    $\begingroup$ @randal'thor , there was one word missing in the question, "Is this situation possible ?" should be "How is this situation possible ?" which implies that Baker + Dozen is the core of this puzzle. $\endgroup$
    – Prem
    May 9, 2015 at 16:18
  • $\begingroup$ Yeah, I'm pretty sure the "baker's dozen" is the whole trick to this riddle. $\endgroup$
    – user88
    May 9, 2015 at 16:37
  • $\begingroup$ @randal'thor , Request you to remove the first part ("not possible") and high-light only the second part ("it is possible") so that I can mark it as the accepted answer. $\endgroup$
    – Prem
    May 10, 2015 at 5:59
  • $\begingroup$ @Prem I've edited as you asked, though I included a reference to the first part so that people can follow the thought process I had. $\endgroup$ May 10, 2015 at 11:23

Doesn't the "baker's dozen" solution, or even the idea that the puzzle has a single solution, rely on the assumption that all the buyers are either rich or poor? There's nothing in it to say that there aren't also buyers who are neither, and are possibly sold the bread at cost, in which case the number of rich buyers could be:

between 3 and 7,


0 to 4 poor buyers


1, 3, 5, 7 or 9 buying at cost.

  • $\begingroup$ Welcome! Please remember to put your answers in spoiler tags. Happy puzzling. $\endgroup$ Jun 13, 2017 at 2:45
  • 1
    $\begingroup$ +1 , Perspective is totally unique !! It still matches the given hint "Basically, there is one EXTRA unknown.", which is the number of buyers who are neither !! $\endgroup$
    – Prem
    Jun 13, 2017 at 8:49

Specifically Expected Answer :

One-unknown way of thinking :

If R is the number of rich buyers, then "3R + (12-R) = 2 times 12 + 3"
which gives fractional R.
Here the second variable is "D = Dozen" (either 12 [Common] or 13 [Bakers]) so the correct equation is "3R + (D-R) = 2 times D + 3"
which gives integral R, when D is 13.

Two-unknowns way of thinking :

If R is the number of rich buyers and P is the number of poor buyers, then "3R + P = 2 times 12 + 3" & "R + P = 12"
which gives fractional R.
Here the third variable is "D = Dozen" (either 12 [Common] or 13 [Bakers]) so the correct equations are "3R + P = 2 times D + 3" & "R + P = D"
which gives integral R, when D is 13.

Bonus Clue was :

In the last line, "Basically, there is one EXTRA unknown." is redundant, pointing to "one EXTRA" or 12+1.

Core of the puzzle :

Baker + Dozen = Bakers Dozen : refer http://en.wikipedia.org/wiki/Dozen and follow.

Hope all clues were highly visible , yet hidden !!!!


Not the intended solution, but one that fits the hint:

12 Loaves baked (cost 24U) 8 sold to the Rich (income 24U) 3 sold to the Poor (income 3U) 1 stolen (income 0U)

Add it all up:

Income = 24U + 3U = 27U Cost = 24U Profit = 27U - 24U = 3U Unknowns: Sold to Rich, Sold to Poor, Stolen.


Another way to see it :

There are many answers. Even not accounting for the baker's dozen


It is stated that he sells 3U when he knows the buyer is rich and 1U for the poor. Therefore, he can sell hopefully his bread at the normal price of 2U, if it is indeed the normal price

An answer could be

4 rich buying at 3U 1 poor buying at 1U the rest (7) buying at 2U (cost price)

But could be anything else

depending on the normal selling price


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