View source for 1964 AHSME Problems/Problem 14

== Problem ==

A farmer bought <math>749</math> sheep. He sold <math>700</math> of them for the price paid for the <math>749</math> sheep. 
The remaining <math>49</math> sheep were sold at the same price per head as the other <math>700</math>. 
Based on the cost, the percent gain on the entire transaction is:

<math>\textbf{(A)}\ 6.5 \qquad
\textbf{(B)}\ 6.75 \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 7.5 \qquad
\textbf{(E)}\ 8  </math>   

== Solution ==

Let us say each sheep cost <math>x</math> dollars.  The farmer paid <math>749x</math> for the sheep.  He sold <math>700</math> of them for <math>749x</math>, so each sheep sold for <math>\frac{749}{700} = 1.07x</math>.

Since every sheep sold for the same price per head, and since every sheep cost <math>x</math> and sold for <math>1.07x</math>, there is an increase of <math>\frac{1.07x - 1x}{1x} = 0.07</math>, or <math>7\%</math>, which is option <math>\boxed{\textbf{(C}}</math>

==See Also==
{{AHSME 40p box|year=1964|num-b=13|num-a=15}}

[[Category:Introductory Algebra Problems]]

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