Problem help...

[Bigj]

Not sure if this is the right forum but can someone please posts a step by step solution to the following.....

A Farmer has to buy 100 animals (at least 1 of each) = £1000. The animals prices are:
1 cow is £50
1 sheep is £10
1 rabbit is 50p
E.g. 10 cows = £500
40 sheep = £400
50 rabbits = £25
So he's bought 100 animals but it only totals £925. Can you work out 100 animals for exactly £1000?

thanks in advance.

 

[Deleted member 4993]

Not sure if this is the right forum but can someone please posts a step by step solution to the following.....

A Farmer has to buy 100 animals (at least 1 of each) = £1000. The animals prices are:
1 cow is £50
1 sheep is £10
1 rabbit is 50p
E.g. 10 cows = £500
40 sheep = £400
50 rabbits = £25
So he's bought 100 animals but it only totals £925. Can you work out 100 animals for exactly £1000?

thanks in advance.

This can only be done through trial and error!

You started the first step - We need to make up 75 pounds.

Now suppose we buy 12 cows and 40 sheeps we get 1000 pounds - but not enough animals!!

So continue...

 

[Dr.Peterson]

Not sure if this is the right forum but can someone please posts a step by step solution to the following.....

A Farmer has to buy 100 animals (at least 1 of each) = £1000. The animals prices are:
1 cow is £50
1 sheep is £10
1 rabbit is 50p
E.g. 10 cows = £500
40 sheep = £400
50 rabbits = £25
So he's bought 100 animals but it only totals £925. Can you work out 100 animals for exactly £1000?

thanks in advance.

Algebra can reduce the trial and error to a triviality.

Let x = number of cows, y = number of sheep, z = number of rabbits.

Then x + y + z = 100, and 50x + 10y + 0.50z = 1000.

This is called a Diophantine system of equations.

Double the second equation to get 100x + 20y + z = 2000, so we have only integers. Then solve the first equation for z in terms of x and y, and substitute in the second. After simplifying, you will see that, in order for everything to be a whole number, it is obvious that one variable must be a multiple of ...

I'll let you finish. There's another trick that will make the final result particularly easy.