# Benediktov Problem

**Tricky Problem**

The Great Russian poet Benediktov was very fond of mathematics and he collected and compiled a whole volume of trick brain teasers. Though this work was never published, the manuscript was found in 1924. An interesting problem contained in the manuscript, captioned ‘An ingenious Way of Solving a Tricky Problem’ goes as follows:

One woman made a living by selling eggs, had 90 eggs which she wanted her three daughters to sell. So she gave her eldest daughter 10 eggs, 30 to her second daughter and 50 to the youngest.

‘You better agree among yourselves’, she told them; ‘ about the price you’re going to ask fo you eggs, and stick to it. But I hope that the eldest will receive as much for her 10 eggs as the second will for her 30 eggs and the third for her 50 eggs. In other words each of you should bring back the same amount of money and keep in mind; the total for 90 eggs should not be less than 90 bucks.

I found this interesting puzzle from the book named More Puzzles by Shakuntala Devi.

Post your answers below as comments or discuss it on the QuickerMaths facebook page.

The story I saw published in Figures for Fun: Stories and Conundrums by Yakov Perelman (Translated in English from Russian by G. Ivanov-Mumjiev) and published by Foreign Languages Publishing House, Moscow (1957)

Since the problem states, the total should not be less than 90 bucks,then we can assume that the three daughters can can earn same amount that is more than 90 bucks. Given this, we can have 15, 5, and 3 (as prices) respectively. That is because, the common multiple for the number of eggs each daughter has is 150.

Daughter1: 10eggs * @$15= 150

Daughter2: 30eggs * @$05= 150

Daughter3: 50eggs * @$03= 150

JUST A THOUGHT, if you notice the prices, 15=5*3 or 15/5=3 or 15/3=5. 😀

hope my answer makes sense too..:)

the answer is in multiple of 3,1 and 0.6 i.e 3,1,0.6 are the minimum rates which they should ask for, to fullfill the criteria………

If each daughter sells@ 3 bucks / 7 eggs for 7 times ( or 7 customers) and 9 bucks/egg for the eigth time (or the eighth customer) then each can receive 30 bucks.

10 egg @6.00 per egg = 6*10 =60

(60/30=2, @2.00 for 30 eggs.)

30 egg @2.00 per egg =2*30 =60

(60/50=2, @1.20 for 50 eggs.)

50 egg @1.20 per egg = 1.20*50 =60

the answer may be.. multipliesof 3,1,0.6

Akshay is correct as the sum is atleast 90 bucks the price of eggs may be any multiple of 3, 1, 0.6

the answer is eldest-3,younger-1 and youngest-0.6…

but where is the solution given?

I think… answer can be anything in the multiples of 3, 1 & 0.6…. considering that total amount to be brought back is AT LEAST 90 bucks and not Exact 90 Bucks..

90/3= 30$ Each need to bring back this amount

Considering that y = amount to charge.

the eldest ==> 10y = 30$, thus y = 3,00$

the middle ==> 30y = 30$, thus y = 1,00$

the youngest ==> 50y = 30$, thus y = 0,60$

eldest- 15 bucks/egg

middle- 5 buck/egg

youngest- 3 buck/egg

eldest- 3 bucks/egg

middle- 1 buck/egg

youngest- 0.6 buck/egg

6.00,2.00,1.20

Firstly sailing price @3 bus for 7 egg

First receive 3 bus sailing 7 egg remaing

3 egg

Second receive 12 bus sailing 28 egg remain

2 egg

third receive 21 bus sailing 49 egg remain

1 egg

after that each one should sell remain egg

@9 bus for 1 egg