[Guest]

90 oranges were divided among 3 men.

One took 50 oranges

Another took 30 oranges

And the third took 10 oranges.

Using two prices, and using the same two prices each. At what prices and how many oranges for each price would they have to sell them to receive 10 dollars each?

Maybe this one is hard. I do not have an equation, I do have a solution.

Edited October 8, 2009 by aliz8

[Guest]

One trivial answer is that everyone sells 10 oranges at 1 dollar each and gives the rest away - but I suppose that isn't what you want!

[Guest]

More thoughts on this problem.

There is a whole range of solutions, not just one. If this is to be narrowed down to one solution, more information is needed.

If the last one sells his oranges at one dollar each, then there are several ways the others can make $10. For example, A sells all his at 20 cents each, and B sells 5 of his at $1 each, and the remaining 25 at 20 cents each. Alternatively, the two prices could be $1 each and 6 for a dollar, leaving A selling 2 at a dollar each, and the remaining 48 at 6 for a dollar. B then sells 24 oranges at 6 for a dollar, and the other 6 at a dollar each, and so on.

:thumbsup:Donjar

Edited October 8, 2009 by donjar

[Guest]

50 for 20c & 0 for $1, 25 for 20c & 5 for $1, 0 for 20c & 10 for $1

Oap1+Oap2=50, Obp1+Obp2=30, Ocp1+Ocp2=10,P1*Oap1+P2*Oap2=P1*Obp1+P2*Obp2=P1*Ocp1+P2*Ocp2=10

keeping it discrete or at least rational.

Edited October 8, 2009 by Dawid

[Guest]

two prices are 0.2 dollar and 1 dollar.

Person A sells all 50 in 0.2. Totalling 10 dollar

Person B sells all 25 in 0.2 and 5 in 1. Totalling 10 dollar

Person A sells all 10 in 1. Totalling 10 dollar

So total are $0.2 count is 75 and $1 count is 15

[Guest]

If all three men sell oranges at both prices, then the solution is as follows:

Man A sells 1 orange at $3, and 49 oranges at 7 for $1.

Man B sells 2 oranges at $3 each, and 28 oranges at 7 for $1.

Man C sells 3 oranges at $3 each, and 7 oranges at 7 for $1.

Why anyone would pay $3 for an orange, I do not know.

[Guest]

Well Sarahbeara, it was a bit before my time, but Nell Gwyn (the mistress of Charles II of England) sold oranges, and had no difficulty in selling her wares for any price she wanted.

[Guest]

I just noticed that you have made your first post, Sarahbeara. Even though I am not a member of the hierarchy here, may I welcome you to a most engaging club.

Edited October 8, 2009 by jerbil

[Guest]

The first person sells 5 @ $1.10 and 45 @ $.10 The second person sells 7 @ $1.10 and 23 @ $.10 The third person sells 9 @ $1.10 and 1 @ $.10

[Guest]

So you have 2 prices: the more expensive one and the cheaper one.

The high price will be that at which the guy with less oranges will sell in order to get the full $10 profit.So that's 10/10 = 1 buck per orange.

The low price will be that at which the guy with most oranges will sell, that is 50/10 = 0.20. So 20 cents per orange.

Finally the last guy, has to make a balance between these 2 prices, hence the 2 equations with 2 variables:

0.2x+1y=10

x+y=30

Being x the number of oranges sold at a low price and y the number of oranges sold at a high price.

The answer, as stated in some posts above, is that he has to sell 5 oranges at 1 buck and 25 at 20 cents.

[Guest]

More thoughts on this problem.

There is a whole range of solutions, not just one. If this is to be narrowed down to one solution, more information is needed.

If the last one sells his oranges at one dollar each, then there are several ways the others can make $10. For example, A sells all his at 20 cents each, and B sells 5 of his at $1 each, and the remaining 25 at 20 cents each. Alternatively, the two prices could be $1 each and 6 for a dollar, leaving A selling 2 at a dollar each, and the remaining 48 at 6 for a dollar. B then sells 24 oranges at 6 for a dollar, and the other 6 at a dollar each, and so on.

:thumbsup:Donjar

IF the last sells his oranges at 1 dollar each then he is only using one price. Condition is, they use two prices. They all use the same two prices.

[Guest]

One trivial answer is that everyone sells 10 oranges at 1 dollar each and gives the rest away - but I suppose that isn't what you want!

Each sells their oranges and each make 10 dollars.

[Guest]

50 for 20c & 0 for $1, 25 for 20c & 5 for $1, 0 for 20c & 10 for $1

Oap1+Oap2=50, Obp1+Obp2=30, Ocp1+Ocp2=10,P1*Oap1+P2*Oap2=P1*Obp1+P2*Obp2=P1*Ocp1+P2*Ocp2=10

keeping it discrete or at least rational.

No zeros involved.

[Guest]

two prices are 0.2 dollar and 1 dollar.

Person A sells all 50 in 0.2. Totalling 10 dollar

Person B sells all 25 in 0.2 and 5 in 1. Totalling 10 dollar

Person A sells all 10 in 1. Totalling 10 dollar

So total are $0.2 count is 75 and $1 count is 15

They use two prices each. They used the same two prices each.

[Guest]

If all three men sell oranges at both prices, then the solution is as follows:

Man A sells 1 orange at $3, and 49 oranges at 7 for $1.

Man B sells 2 oranges at $3 each, and 28 oranges at 7 for $1.

Man C sells 3 oranges at $3 each, and 7 oranges at 7 for $1.

Why anyone would pay $3 for an orange, I do not know.

You solved it. Either you knew it of it wasn't that hard.

[Guest]

The first person sells 5 @ $1.10 and 45 @ $.10 The second person sells 7 @ $1.10 and 23 @ $.10 The third person sells 9 @ $1.10 and 1 @ $.10

well seems like there's more then one solution. Your's seems one of them.

[Guest]

So you have 2 prices: the more expensive one and the cheaper one.

The high price will be that at which the guy with less oranges will sell in order to get the full $10 profit.So that's 10/10 = 1 buck per orange.

The low price will be that at which the guy with most oranges will sell, that is 50/10 = 0.20. So 20 cents per orange.

Finally the last guy, has to make a balance between these 2 prices, hence the 2 equations with 2 variables:

0.2x+1y=10

x+y=30

Being x the number of oranges sold at a low price and y the number of oranges sold at a high price.

The answer, as stated in some posts above, is that he has to sell 5 oranges at 1 buck and 25 at 20 cents.

Using two prices, and using the same two prices each. Yours does not meet the condition. Two people already solved it with two different solutions. sarahbara and Tonybz.