Difference between revisions of "1993 AJHSME Problems/Problem 2"

Latest revision as of 23:10, 4 July 2013

When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be

$\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133$

$\begin{align*} \dfrac{49}{84} &= \dfrac{7^2}{2^2\cdot 3\cdot 7} \\ &= \dfrac{7}{2^2\cdot 3} \\ &= \dfrac{7}{12}. \end{align*}$

The sum of the numerator and denominator is $7+12=19\rightarrow \boxed{\text{C}}$ .

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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
All AJHSME/AMC 8 Problems and Solutions

@@ Line 1: / Line 1: @@
-== Problem ==
+==Problem==
 When the fraction <math>\dfrac{49}{84}</math> is expressed in simplest form, then the sum of the numerator and the denominator will be
 <math>\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133</math>
-== Solution ==
+==Solution==
-The fraction is already in simplest form.
-The prime factorization of <math>49</math> is <math>7^2</math> and for <math>84</math> it is <math>2*37</math>, so the greatest common factor is 1.
+<cmath>\begin{align*}
+\dfrac{49}{84} &= \dfrac{7^2}{2^2\cdot 3\cdot 7} \\
+&= \dfrac{7}{2^2\cdot 3} \\
+&= \dfrac{7}{12}.
+\end{align*}</cmath>
+The sum of the numerator and denominator is <math>7+12=19\rightarrow \boxed{\text{C}}</math>.
+==See Also==
-Then we add <math>49+84</math> and get <math>133 \Rightarrow E</math>.
+{{AJHSME box|year=1993|num-b=1|num-a=3}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}