Guest Posted November 9, 2008 Report Share Posted November 9, 2008 (edited) You have to buy exactly 100 eggs You have exactly 100 coins There are 3 kinds of eggs: A costs 7 coins for 1 egg B costs 3 coins for 1 egg C costs 1 coin for 3 eggs Edit: there's one condition: you have to buy from all 3 kinds How many eggs of each type do you need to buy in order to have spend exactly 100 coins? Edited November 9, 2008 by Maggi Quote Link to post Share on other sites

0 Izzy 1 Posted November 9, 2008 Report Share Posted November 9, 2008 0 of A, 0 of B, 300 of C. Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 25 of B and 25 of C B = 75 coins and 25 eggs C = 25 coins and 75 eggs 100 coins spent, and 100 eggs bought Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 0 of A, 0 of B, 300 of C. Uhm please read carefully, you have to buy exactly 100 eggs Else it would have been way too easy Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 0 of A, 0 of B, 300 of C. But that would mean that you would buy 300 eggs. You have to buy 100 exactly Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 25 of B and 25 of C B = 75 coins and 25 eggs C = 25 coins and 75 eggs 100 coins spent, and 100 eggs bought Sorry forgot to put the 1 condition on there, I edited it, else it would have been right though, but then it would have been very easy Quote Link to post Share on other sites

0 Izzy 1 Posted November 9, 2008 Report Share Posted November 9, 2008 Oh, haha, no wonder it was so easy. I just read the last bit.. Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 A costs 7 coins for 1 egg B costs 3 coins for 1 egg C costs 1 coin for 3 eggs buy 2 eggs from A - 14 coins and 2 egg buy 20 eggs from B - 60 coins and 20 eggs buy 26 eggs from C - 26 coins and 78 eggs Quote Link to post Share on other sites

0 Prime 15 Posted November 9, 2008 Report Share Posted November 9, 2008 Still, there are many answers x = number of 7-coin eggs y = number of 3-coin eggs z = number of 1/3-coin eggs Then we have a system of two equations with three unknowns. 7x + 3y + z = 100 x + y + z = 100 The whole number solutions for (x, y, z) are: (8 ,5, 87); (6, 10, 84); (4, 15, 81); and (2, 20,78). Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 Still, there are many answers x = number of 7-coin eggs y = number of 3-coin eggs z = number of 1/3-coin eggs Then we have a system of two equations with three unknowns. 7x + 3y + z = 100 x + y + z = 100 The whole number solutions for (x, y, z) are: (8 ,5, 87); (6, 10, 84); (4, 15, 81); and (2, 20,78). Nice!! You found them ^^ There's only one tiny little mistake: 7x+3y+1/3z=100 Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 8 A eggs 5 B eggs 87 C eggs Solved through Excel Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 8 A eggs 5 B eggs 87 C eggs Solved through Excel You found one of the four possible answers Quote Link to post Share on other sites

0 Prime 15 Posted November 9, 2008 Report Share Posted November 9, 2008 Nice!! You found them ^^ There's only one tiny little mistake: 7x+3y+1/3z=100 Actually, the typo was in another equation. I meant it as x + y + 3z = 100 (That's how I had it on paper, but typed it wrong.) It is just my strange way of constructing the equations. Just because I don't like fractions and prefer whole numbers. There were x eggs of the first kind, y eggs of the second kind and 3z eggs of the third kind. Correspondingly the number of coins spent: 7x, 3y, and z. Quote Link to post Share on other sites

0 Guest Posted November 9, 2008 Report Share Posted November 9, 2008 Actually, the typo was in another equation. I meant it as x + y + 3z = 100 (That's how I had it on paper, but typed it wrong.) It is just my strange way of constructing the equations. Just because I don't like fractions and prefer whole numbers. There were x eggs of the first kind, y eggs of the second kind and 3z eggs of the third kind. Correspondingly the number of coins spent: 7x, 3y, and z. You're right, if you mathematically look at it since it does not matter whether you put 1/3z in the first equation or 3z in the second, but if you take a look at it in a different way (non-mathematical ) it's quite weird: in your first equation you were working with the coins, which would give: (price/each egg) x, (price/each egg) y, (price/each egg) z If x,y and z are equal to the amount of eggs of type A,B and C, then your equation cannot be right I think.. Look at 1 of the solutions for example: x=8 y=5 z=87 In your equation for the price, you'd get: 7x+3y+z=100 => 7*8+3*5+1*87=100 => 56+15+87 = 158 Which is wrong, because it should be equal to 100 However, your equations would be right if: x= amount of eggs of type A y= amount of eggs of type B z= 1/3 of the amount of eggs of type C And since you haven't really said what your x y and z are equal to, you're also right I guess Either way, you'll have to divide the amount of the type C eggs sometime.. Quote Link to post Share on other sites

## Question

## Guest

You have to buy exactly 100 eggs

You have exactly 100 coins

There are 3 kinds of eggs:

A costs 7 coins for 1 egg

B costs 3 coins for 1 egg

C costs 1 coin for 3 eggs

Edit: there's one condition: you have to buy from all 3 kinds

How many eggs of each type do you need to buy in order to have spend exactly 100 coins?

Edited by Maggi## Link to post

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