Question 281733
Let {{{c}}} = number of cows
Let {{{p}}} = number of pigs
Let {{{k}}} = number of chickens
given:
{{{10c + 3p + (1/2)*k = 100}}} dollars
{{{c + p + k = 100}}}
There are 2 equations and 3 unknowns, so it
can't be solved directly. but you can get there
There can't be all cows- that would be
only 10 animals for $100
There can't be all pigs- that would be
33 animals for $99
Say there are 50 chickens and 50 pigs:
{{{k = 50}}} {{{(1/2)*50 = 25}}}
{{{p = 50}}} {{{3*50 = 150}}}
That's $175 for 100 animals. There must be more
chickens than pigs
Let's say:
{{{k = 60}}} {{{(1/2)*60 = 30}}}
{{{p = 40}}} {{{3*40 = 120}}}
That's $150 for 100 animals
I have to run, but keep going and you'll get it
You must have at least 1 cow, too