how many birds would buy from every species?

Question

A farmer sells chickens,geese and ducks.Every chicken costs $100$ dollars,every goose costs $200$ dollars and every duck costs $250$ dollars.A customer wants to buy $40$ of these birds and spending $4.400$ dollars.how many birds would buy from every species?

Could anyone give any hint to find it?

Answer 1

The fact that those numbers are integers means you can solve it without needing two more equations.

$100x + 200y + 250z = 4400$ becomes $$2x + 4y + 5z = 88$$

So $z$ has to be an even number. So $z=2k$, $k \leq 7$.

$$x+2y+5k = 44$$ Now any $k$ from $0$ to $7$ gives you a solution:

$k$ even: solve $x+2y = 44-5k$ (even), so $x=0$ and y= $(44-5k)/2$

$k$ odd: solve $x+2y = 44-5k$ (odd), so $x=1$ and y= $(43-5k)/2$

Now you just have to make $x+y+2k=40$, so do eight sums with the numbers above and you're done.

Answer 2

Solve the following system:

• $100c+200g+250d=4400$
• $c+g+d=40$

Since there are more variable ($3$) than equations ($2$), there are infinitely many solutions.

But not all of these solutions are necessarily non-negative integers, so let's try to solve it anyway:

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• $100c+200g+250d=4400$
• $c+g+d=40$

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• $100c+200g+250d=4400$
• $100c+100g+100d=4000$

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• $100g+150d=400$

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• $[g=1,d=2]\vee[g=4,d=0]$ trial and error

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• $c+g+d=40$
• $[c=37,g=1,d=2]\vee[c=36,g=4,d=0]$

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• $100c+200g+250d=4400$
• $[c=37,g=1,d=2]\vee[c=36,g=4,d=0]$

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• $[g=1,d=2,c=37]$

Answer 3

Thought: Here's a less algebraic approach.

If one buys $40$ chickens, one only spends $\$4000$. So we must swap out some chickens for the same number of geese and ducks, but so as to get the price up by $\$400$ (to $\$4400$).

If we swap out a chicken for a goose, our price goes up by $\$100$; and if we swap out a chicken for a duck, our price goes up by $\$150$.

You have to find combinations of these swaps to get to the extra $\$400$.