Difference between revisions of "1978 AHSME Problems/Problem 19"

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==Problem==
 
==Problem==
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A positive integer <math>n</math> not exceeding <math>100</math> is chosen in such a way that if <math>n\le 50</math>, then the probability of choosing <math>n</math> is <math>p</math>, and if <math>n > 50</math>, then the probability of choosing <math>n</math> is <math>3p</math>. The probability that a perfect square is chosen is
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<math>\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad  \textbf{(E) }.1</math>
  
 
==Solution==
 
==Solution==

Revision as of 16:18, 18 June 2021

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 $\frac{7}{200}+\frac{9}{200}=0.008\implies\boxed{\textbf{(C).}}$

~volkie thangy

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