Difference between revisions of "1984 AHSME Problems/Problem 1"

(created links)
(Solution)
Line 8: Line 8:
 
We can use [[difference of squares]] to factor the [[denominator]], yielding:
 
We can use [[difference of squares]] to factor the [[denominator]], yielding:
  
<math> \frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{1000} </math>.
+
<math> \frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{2000} </math>.
  
We see that the <math> 1000 </math> in the denominator cancels with one of the <math> 1000 </math>s in the [[numerator]], yielding <math> 1000, \boxed{\text{B}} </math>.
+
We see that the <math> 1000 </math> in the denominator cancels with one of the <math> 1000 </math>s in the [[numerator]], yielding <math> 500, \boxed{\text{C}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|before=First Problem|num-a=2}}
 
{{AHSME box|year=1984|before=First Problem|num-a=2}}

Revision as of 10:26, 2 October 2011

Problem

$\frac{1000^2}{252^2-248^2}$ equals

$\mathrm{(A) \  }62,500 \qquad \mathrm{(B) \  }1,000 \qquad \mathrm{(C) \  } 500\qquad \mathrm{(D) \  }250 \qquad \mathrm{(E) \  } \frac{1}{2}$

Solution

We can use difference of squares to factor the denominator, yielding:

$\frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{2000}$.

We see that the $1000$ in the denominator cancels with one of the $1000$s in the numerator, yielding $500, \boxed{\text{C}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions