Question 1116863
.
A farmer bought some sheep for $2100. Two sheep died, and he sold the rest for $30 more a head than he paid for them. 
If he made a profit of $60, how many sheep did he buy?
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<pre>
Let n be the number of sheep he bought.

Then he paid  {{{2100/n}}} dollars for each.     (*)


Later he sold  (n-2) sheep  for 2100 + 60 = 2160 dollars,  and we know that  {{{2160/(n-2)}}}  is 30 dollars more than (*).


It gives you an equation

{{{2160/(n-2)}}} - {{{2100/n}}} = 30.


Divide both sides by 30 to simplify. You will get

{{{72/(n-2)}}} - {{{70/n}}} = 1.


Multiply both sides by n*(n-2). You will get


    72n - 70*(n-2) = n*(n-2)

    2n + 140 = n^2 - 2n

    n^2 - 4n - 140 = 0


    (n-14)*(n+10) = 0


The solution is  n = 14  sheeps.


<U>Answer</U>. &nbsp;&nbsp;14 sheeps were originally.
</pre>

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