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==
Let's say that we will have <math>3</math> slips for every number not exceeding <math>100</math> but bigger than <math>50.</math> This is to account for the <math>3p</math> probability part. Let's now say that we will only have one slip for each number below or equal to <math>50.</math> The probability(or <math>p</math>) will then be <math>\frac{1}{200}.</math> Now let's have all the squares under <math>50,</math> which are <math>1,4,9,16,25,36,49.</math> The probability for these are <math>\frac{7}{200}.</math> The numbers above <math>50</math> that are squares are <math>64,81,100.</math> We then need to multiply the probability by <math>3</math> so the probability of these are <math>\frac{9}{200}.</math> The answer is <math>\frac{7}{200}+\frac{9}{200}=0.008\implies\boxed{\textbf{(C).}}</math>
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Let's say that we will have <math>3</math> slips for every number not exceeding <math>100</math> but bigger than <math>50.</math> This is to account for the <math>3p</math> probability part. Let's now say that we will only have one slip for each number below or equal to <math>50.</math> The probability(or <math>p</math>) will then be <math>\frac{1}{200}.</math> Now let's have all the squares under <math>50,</math> which are <math>1,4,9,16,25,36,49.</math> The probability for these are <math>\frac{7}{200}.</math> The numbers above <math>50</math> that are squares are <math>64,81,100.</math> We then need to multiply the probability by <math>3</math> so the probability of these are <math>\frac{9}{200}.</math> The answer is <math>\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{\textbf{(C).}}</math>
  
 
~volkie thangy
 
~volkie thangy
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==See Also==
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{{AHSME box|year=1978|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 16:24, 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 $\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{\textbf{(C).}}$

~volkie thangy

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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