[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 November 9, 2008 by Maggi

[Izzy]

0 of A, 0 of B, 300 of C.

[Guest]

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

[Guest]

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

[Guest]

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

[Guest]

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

[Izzy]

Oh, haha, no wonder it was so easy. I just read the last bit..

[Guest]

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

[Prime]

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).

[Guest]

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

[Guest]

8 A eggs

5 B eggs

87 C eggs

Solved through Excel

[Guest]

8 A eggs

5 B eggs

87 C eggs

Solved through Excel

You found one of the four possible answers

[Prime]

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.

[Guest]

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..