1955 AHSME Problems/Problem 16

Revision as of 13:18, 23 July 2020 by Angrybird029 (talk | contribs) (Created page with "== Problem 16== The value of <math>\frac{3}{a+b}</math> when <math>a=4</math> and <math>b=-4</math> is: <math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 16

The value of $\frac{3}{a+b}$ when $a=4$ and $b=-4$ is:

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{3}{8}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \text{any finite number}\qquad\textbf{(E)}\ \text{meaningless}$

Solution

As the denominator is $4 - (-4) = 0$, the value of the fraction $\frac{3}{4 - (-4)}$ is $\textbf{(E)}\ \text{meaningless}$