1978 AHSME Problems/Problem 19

Problem

A positive integer $n$ not exceeding $100$ is chosen in such a way that if $n\le 50$ , then the probability of choosing $n$ is $p$ , and if $n > 50$ , then the probability of choosing $n$ is $3p$ . The probability that a perfect square is chosen is

$\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad \textbf{(E) }.1$

Solution

Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$ ) will then be $\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\frac{9}{200}.$ The answer is $\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{\textbf{(C).}}$

~volkie thangy

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1978 AHSME Problems/Problem 19

Problem

Solution