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

Revision as of 20:37, 16 June 2011

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

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

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

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

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

1984 AHSME (Problems • Answer Key • Resources)
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

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 ==Solution==
-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>.
-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> 1000, \boxed{\text{B}} </math>.
 ==See Also==
 {{AHSME box|year=1984|before=First Problem|num-a=2}}