# 1975 AHSME Problems/Problem 14

## Problem 14

If the $whatsis$ is $so$ when the $whosis$ is $is$ and the $so$ and $so$ is $is \cdot so$, what is the $whosis \cdot whatsis$ when the $whosis$ is $so$, the $so$ and $so$ is $so \cdot so$ and the $is$ is two ( $whatsis, whosis, is$ and $so$ are variables taking positive values)? $\textbf{(A)}\ whosis \cdot is \cdot so \qquad \textbf{(B)}\ whosis \qquad \textbf{(C)}\ is \qquad \textbf{(D)}\ so\qquad \textbf{(E)}\ so\text{ and }so$

## Solution

From the problem, we are given: $whatsis = so$ $whosis = is$ $so + so = is \cdot so$

We want to find what $whatsis \cdot whosis$ is when $whosis = so$, $so + so = so \cdot so$, and $is = 2$.

Since $is = 2$ and $so + so = is \cdot so = so \cdot so$ (from the given), that means $so = 2$. Now we know that $whatsis = 2$ and $whosis = 2$ as well because they are equal to $so$ and $is$, respectively.

Plugging in the values, we get $whatsis \cdot whosis = so \cdot is = 4$. This also means that $whatsis \cdot whosis = so + so$.

Taking a look at the answer choices, we see that our answer is $\boxed{\textbf{(E)}\ so\text{ and }so}$. ~jiang147369

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 