Why did the seller lose money by mixing eggs from two different pricing schemes?

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quark
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Stupid Seller?

Two men A and B go to the market to sell eggs. Each one has 30 eggs. A sells 2 for one unit(insert your own monetary unit) and B sells 3 for one unit. Just before they start to sell, A gets an urgent call from his home, he requests B to sell his stuff and goes home. B, the wise man, mixes both the eggs and sells 5 for 2 units. When finally he counts the money, he gets 1 unit short(60 = 12*5 and 12*2 = 24units)

Can we suggest him the trick so that he won't repeat the mistake again?
 
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use the price-per-uinit of A in al the eggs, including he's.
 
After B sold 10 mixtures of 5 eggs each (2 from A, 3 from B), he had sold all of his own eggs. From then on, he continued to sell A's eggs at the reduced price.
 
another possibility is that B didn't expect any reward or took in interests from him selling A's eggs...
 
Why can't anybody just answer it without manipulating the story or the whole point of this disscussion? :approve:
 
Jimmy!

Correct and a big egg for you. Can you put it mathematically(in white, ofcourse)
 
----No need to "white out"
Price of A eggs= 1/2 per egg
Price of B = 1/3 per egg
New Price for AB mix 2/5 per egg
Discount on A = 1/2 – 2/5 = 5/10 -4/10 =1/10
Discount on B = 1/3 - 2/15 = 5/15 -6/15 = -1/15 “Gain”
Loss on A's = 30*(1/10)= 3
Gain on B's = 30*(1/15)= 2
Net loss = 1

Expected was 30*1/2 plus 30*1/3 total 25 not 24
Thus Should Sell 60 for 25
Or price at 5 units to buy a dozen.
 
Mathematically you are correct. But your answer doesn't show at what point of time B went wrong. Here is a lead/mislead. Suppose A didn't go home and both started selling separately. I go there to buy eggs and take two from A and three from B and I get 5 for two units and they didn't loose(1/12 = 0.083units)anything.
 
B doesn't know how to add fractions or average (I'm assuming he wanted to average the price).
His work:
1/2 + 1/3 = 2/5

Actual work:
(1/2 + 1/3)/2 = (3/6 + 2/6)/2 = 5/6/2 = 5/12
 
quark said:
...but averaging did work in my above example.

Yes, averaging was the way to go, but the guy in the story seemed to think adding numerators and denominators would aveage it for him.
 
quark said:
...but averaging did work in my above example.
Sure it was a problem in your example as well,
Have the buyer keep buy'n 5 at a time and expect the same cost till he buys all the eggs.
At the end he'll need to get a $1 discount from A or pay him $1 more than his plan.
RB
 
Randall got it right. There won't be any problem as long as B's eggs are totally not sold out(or in that proportion, atleast). Once B's thirty eggs are over then loss for A is 10*(50-40) = 1 unit.