Difference between revisions of "1975 AHSME Problems/Problem 14"

(Created page with "== Problem 14 == If the <math>whatsis</math> is <math>so</math> when the <math>whosis</math> is <math>is</math> and the <math>so</math> and <math>so</math> is <math>is \cdot...")
 
 
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Taking a look at the answer choices, we see that our answer is <math>\boxed{\textbf{(E)}\ so\text{ and }so}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369]
 
Taking a look at the answer choices, we see that our answer is <math>\boxed{\textbf{(E)}\ so\text{ and }so}</math>. ~[https://artofproblemsolving.com/wiki/index.php/User:Jiang147369 jiang147369]
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==See Also==
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{{AHSME box|year=1975|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 16:13, 19 January 2021

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

See Also

1975 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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