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I came across this problem some years ago in a book (which I won't mention, and request others to not mention if they know). The solution is quite elegant and difficult to arrive at by mere trial and error. And I couldn't think of a proper way to solve it. Still, give it a try.

Once, there was an egg seller. She had $90$ eggs which she wanted to sell in the market. So, she sent her $3$ girls to the market. She gave the eldest and the cleverest girl $10$ eggs, the second girl $30$ eggs and the youngest one $50$. She told them, "You better decide for the price of the eggs among yourselves. But the three of you should stick to one price. You may change the price more than once, but the three of you must change it together. Also, I expect that the three of you earn the same amount of money each by selling your respective eggs. And you must bring home not less than $90$ units of money."

The egg seller told them they should not bring home less than $90$ units. The eldest girl thought of an ingenious way and explained her sisters her idea. In the evening, when they came home, each girl had earned $30$ units by selling her respective lot of eggs, and they together had brought home exactly $90$ units.

Can you think of what idea the eldest girl used?

Edit: Let me clarify the situation a bit more. The three girls occupy different places in the market. Since the eldest girl told them the plan, they have no connection whatsoever till evening. So, they cannot go and meet or sell their eggs to each other. It is not necessary to have a price per egg. Eg, $5$ units per $3$ eggs is a totally acceptable price. With their respective transactions, each of them earns $30$ units. All the eggs are sold in the end.

Edit 2: Also, if the girls decide to sell in bundles, they must sell until their bundles are finished. Eg, if they set the price to $5$ per $9$ eggs, the eldest will sell her $9$ eggs, the middle one will sell her $27$ eggs, the youngest will sell her $45$ eggs before changing the price again. This is one thing I should have told earlier. Sorry. O:-)

If any more doubt, please add in the comments.

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    $\begingroup$ Voted to close add unclear because all the answers make different assumptions about what's allowed. Please edit to clarify. $\endgroup$
    – xnor
    Feb 16, 2015 at 20:33
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    $\begingroup$ Even after the two edits, there still is something missing. The current formulation of the problem (Feb 17; 22 hours after the puzzle was posted) allows the following trivial side solution: In the beginning, the girls sell eggs at 30 units per bundle of 50; the youngest gets rid of her eggs and earns 30 units. Then the price is changed to 30 units per bundle of 30; the second girl earns 30 units. Finally, the price is changed to 30 units per bundle of 10; the eldest girl earns 30 units. $\endgroup$
    – Gamow
    Feb 17, 2015 at 12:07
  • $\begingroup$ @Gamow, that does fit the conditions. But, like you said, it's a trivial one. I didn't think so many loop holes could be found in that puzzle. :p $\endgroup$
    – Tejas
    Feb 18, 2015 at 11:01
  • $\begingroup$ I will keep this unanswered. Because many people gave solutions which fit the current state of the problem. And maybe in future, we might get more interesting solutions. $\endgroup$
    – Tejas
    Feb 18, 2015 at 11:07

11 Answers 11

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Edit: Someone else already had this answer, my apologies. I didn't notice it because it looked more complex than it actually is. Still, here's a simple way of reading it.

If we imagine that the eggs are sold in dozens, then:

perhaps the eggs are sold at a rate instead of a price. I suggest 6 units per dozen eggs, and 3 units for each individual egg. This rate allows the following:
- The youngest daughter has 50 eggs = 4 dozen + 2 individual. 4*6 + 2*3 = 24+6 = 30 units.
- The middle daughter has 30 eggs = 2 dozen + 6 individual. 2*6 + 6*3 = 12+18 = 30 units.
- The eldest daughter has 10 eggs = 0 dozen + 10 individual. 0*6 + 10*3 = 30 units.

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  • $\begingroup$ This is an acceptable answer! (+1) $\endgroup$
    – Tejas
    Feb 17, 2015 at 3:09
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I'm gonna have to make some assumptions

1- The girls are allowed to sell eggs to each other

2- since they have to sell eggs at the same price, customers will buy eggs uniformly from all 3 that have eggs available

one answer is:

step 1

they each sell 10 eggs at $1/piece

step 2

they briefly set the price to $0.33, so that the eldest can buy 30 eggs from the youngest

step 3

they sell all remaining eggs at $1/piece

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  • $\begingroup$ This is a good answer! But the girls cannot sell the eggs to each other. Sorry for the late edit. $\endgroup$
    – Tejas
    Feb 17, 2015 at 2:48
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Well, with the edits there is now a really simple answer: They set the starting price at 30 units per 50 eggs Then they drop it for 30 units per 30 eggs Then they drop it again to 30 units per 10 eggs

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Here is a solution with a single change of price:

  • In the morning, the girls sell the eggs at a price of 3 cent per bundle of 7 eggs.
    The eldest sells 1 bundle (and keeps 3 eggs),
    the second sells 4 bundles (and keeps 2 eggs),
    and the youngest sells 7 bundles (and keeps 1 egg).

  • In the afternoon, the girls sell the eggs at a price of 9 cent per single egg.
    The eldest sells 3 bundles,
    the second sells 2 bundles,
    and the youngest sells 1 bundle.

  • Altogether, the eldest receives $1\cdot3+3\cdot9=30$ cent,
    the second receives $4\cdot3+2\cdot9=30$ cent, and
    the youngest receives $7\cdot3+1\cdot9=30$ cent.

(There are plenty of other solutions that use two or three or four changes of price.)

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  • $\begingroup$ This is correct! $\endgroup$
    – Tejas
    Feb 18, 2015 at 11:04
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My suggestion is without price changes.

An assumption based on the riddle:

The riddle does not explicitely tell that all eggs need to be sold, only that they should earn the same amount of money.

The solution:

The girls set their price to 3 units money pr. egg. Girl 1 sells all of her (10) eggs, but the second and third only sells 10 of their lots. The rest of the eggs (20+40) they simply throw away. Each girl will return with 30 units of money.

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  • $\begingroup$ Well, the egg seller might get happy by knowing that they earned exactly $90$ units. But she's going to get very angry on knowing how they earned it. $\endgroup$
    – Tejas
    Feb 17, 2015 at 2:51
  • $\begingroup$ @Tejas I understand you have found it in a book, but you could easily improve the question to stop my solution, by adding a simply "all of", in this part: "..,earn the same amount of money each by selling all of your respective eggs." $\endgroup$
    – Enigma
    Feb 17, 2015 at 5:49
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Set the price of an egg at 3 eggs. The youngest girl should then buy 10 eggs for the price of 30 from the oldest girl.

Set the price at 1 unit per egg. At the end of the day, each girl will have made 30 units, or 90 units in total.

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  • $\begingroup$ This is a good answer! But the girls cannot sell their eggs to each other. Sorry for the late edit. $\endgroup$
    – Tejas
    Feb 17, 2015 at 2:49
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There are probably multiple solutions where we assume the 3 sisters are at different locations in the market such that different people could buy from all three at the same time and buying from each other would mean leaving your eggs behind and you just don't do that.

The one I have come up with involve selling them in bundles and then selling the remaining eggs per sister.

Solution:

sell 12 eggs for 6 dollars and the remaining for 3 each

The price change would happen once all bundles available per sister have been sold. As a result, clearly there is a pattern.

Let $x$, $y$, and $z$ represent egg count per sister.

Then, you need to find the bundle size $b$ such that bundle price $p$ and single price $s$ fulfill the following:

$$ \begin{align} & \left\lfloor\frac xb*p \right\rfloor + \mod(x,b)*s \\ = &\left\lfloor\frac yb*p\right\rfloor+\mod(y,b)*s \\ = &\left\lfloor\frac zb*p\right\rfloor + \mod(z,b)*s\\ = &30 \end{align}$$

Therefore, the profit per sister is:

$\$30$, since sister with 10 can only sell singles at 3. sister with 30 sells 2 bundles of 12 at 6 and 6 eggs as singles at 3 and sister with with 50 sells 4 bundles at 6 and 2 singles at 3.

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  • $\begingroup$ nice answer. selling eggs by the 12 makes a lot of sense, and it adds up to the correct answer. but it only works if we assume enough people are willing to pay 3 units/piece. $\endgroup$
    – user3453281
    Feb 16, 2015 at 18:00
  • $\begingroup$ This is an acceptable answer! (+1) $\endgroup$
    – Tejas
    Feb 17, 2015 at 3:12
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A simple and stupid solution:

1. They set the price to \$3, the eldest sells her 10 eggs
2. They set the price to \$1, the second sells her 30 eggs
3. They set the price to \$0.6, the youngest sells her 50 eggs

Or, minimizing the number of price changes:

They set the price to \$3, the eldest sells 10 eggs, the second 5
They set the price to \$0.6, the second sells 25 eggs, the youngest 50

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Let there be 2 different prices, $A$ and $B$,

And call the numbers sold under the first price $X_a$, $Y_a$, and $Z_a$.

And under the second price $X_b$, $Y_b$, and $Z_b$.

We have:

$AX_a + BX_b = AY_a + Y_b = AZ_a + BZ_b$
$X_a + X_b = 50$
$Y_a + Y_b = 30$
$Z_a + Z_b = 10$

Eliminating the $b$ amounts,

$AX_a + B(50 - X_a) = AY_a + B(30 - Y_a) = AZ_a + B(10 - Z_a)$

or

$(A - B)X_a + 50B = (A - B)Y_a + 30B = (A - B)Z_a + 10B$

Subtracting $10B$ and dividing by $(A - B)$,

$X_a + \frac{40B}{A - B} = Y_a + \frac{20B}{A - B} = Z_a$

$Z_a = n$
$Y_a = n + \frac{20B}{B - A}$
$X_a = n + \frac{40B}{B - A}$

which gives us a solution for any given $A$, $B$, and $n$.

Taking $A = \frac{3}{7}$ and $B = 9$ just as a guess from the problem's examples, we get

$\frac{B}{B - A} = \frac{9}{\frac{60}{7}} = \frac{21}{20}$

$X_a = n + 42$
$Y_a = n + 21$
$Z_a = n$

Clearly, we can choose any value for $n$ from $0$ to $8$ and get valid numbers for $X_a$ and $Y_a$; since we need to sell a whole number of eggs at the second price, and thus need the differences between the numbers sold at the first price to be a whole number (actually, just a multiple of $\frac{1}{3}$), we can choose either $0$ or $7$ for $n$. So either

$A = \frac{3}{7}$
$X_a = 49$
$Y_a = 28$
$Z_a = 7$

and

$B = 9$
$Xb = 1$
$Yb = 2$
$Zb = 3$

Hence, each earns $\$30$

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  • $\begingroup$ This is a really great answer. (+1) However, please see edit 2. $\endgroup$
    – Tejas
    Feb 17, 2015 at 3:20
  • $\begingroup$ My answer still fits, The strategy is that all of them first sell their eggs at a price of 1 unit per 7 eggs, then when they can no more sell the eggs at this price (i.e they dont have bundles of 7), they then change the price to 3 units per egg. $\endgroup$
    – The Dragonista
    Feb 17, 2015 at 3:27
  • $\begingroup$ But you got $Z_a=0$, meaning the eldest girl doesn't sell any egg at the first price, while she does have one bundle of $7$. $\endgroup$
    – Tejas
    Feb 17, 2015 at 3:28
  • $\begingroup$ whoops!!... i missed that...my bad! @Tejas $\endgroup$
    – The Dragonista
    Feb 17, 2015 at 3:29
  • $\begingroup$ The answer is really great. I'll just say that you're pretty close to the answer which I have in mind. ;) $\endgroup$
    – Tejas
    Feb 17, 2015 at 3:31
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I have a solution with two price changes necessary (assuming that people will buy the eggs for these prices).

Initial Price:

Set the price at 1 unit per egg.

First sale

The youngest sells 30 of her eggs, leaving her with 30 units of money and 20 eggs left.

Price change:

They then all change the price to 3 units per egg.

And now some sneakiness on the part of the kids:

The youngest now buys the 10 eggs from the eldest for the 30 monetary units. She now has 30 eggs, but no money. The eldest has 30 units of money, but no eggs. The middle child still has 30 eggs.

Price change #2:

Change the price back to 1 unit per egg

Final sales:

The middle child and youngest now sell each of their 30 eggs for 1 unit each, leaving everyone with 30 units of money. 90 units total, no eggs left.

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  • $\begingroup$ I don't think this works. if the price is set to $1, then by the time the youngest has sold 30 eggs, the middle one has also sold 30 eggs, and the eldest one is out of eggs. $\endgroup$
    – user3453281
    Feb 16, 2015 at 15:39
  • $\begingroup$ @user3453281 they don't have to sell them at the same rate $\endgroup$
    – Holloway
    Feb 16, 2015 at 15:51
  • $\begingroup$ I think they do have to sell them at the same rate. it's what makes the answer non trivial. otherwise the answer would simply be: they set the price at 3 units and the eldest sells hers, when she is out of eggs they switch to 1 unit. then middle one sells hers at 1$, when she is out of eggs they switch to 0.6 unit and the youngest one sells hers. $\endgroup$
    – user3453281
    Feb 16, 2015 at 15:56
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The girls, who love each other dearly, redistribute the eggs between themselves on the way to the market so they have 30 each. They then sell them at 1 unit per egg, no price change needed.

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