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Difference between revisions of "1987 AHSME Problems/Problem 26"

(Created page with "==Problem== The amount <math>2.5</math> is split into two nonnegative real numbers uniformly at random, for instance, into <math>2.143</math> and <math>.357</math>, or into <ma...")
 
(Added a solution with explanation)
 
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\textbf{(D)}\ \frac{3}{5}\qquad
 
\textbf{(D)}\ \frac{3}{5}\qquad
 
\textbf{(E)}\ \frac{3}{4}  </math>
 
\textbf{(E)}\ \frac{3}{4}  </math>
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== Solution ==
 +
The two parts may round to <math>0</math> and <math>2</math>; <math>1</math> and <math>2</math>; <math>1</math> and <math>1</math>; <math>2</math> and <math>1</math>; or <math>2</math> and <math>0</math>. By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from <math>0</math> to <math>2.5</math>, in which one of the parts is found, into five equal ranges of <math>0</math> to <math>0.5</math>, <math>0.5</math> to <math>1</math>, <math>1</math> to <math>1.5</math>, <math>1.5</math> to <math>2</math>, and <math>2</math> to <math>2.5</math>). As exactly two of the five cases give a sum of <math>1 + 2 = 3</math>, the answer is <math>\frac{2}{5}</math>, which is answer <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:01, 31 March 2018

Problem

The amount $2.5$ is split into two nonnegative real numbers uniformly at random, for instance, into $2.143$ and $.357$, or into $\sqrt{3}$ and $2.5-\sqrt{3}$. Then each number is rounded to its nearest integer, for instance, $2$ and $0$ in the first case above, $2$ and $1$ in the second. What is the probability that the two integers sum to $3$?

$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{2}{5} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{3}{4}$

Solution

The two parts may round to $0$ and $2$; $1$ and $2$; $1$ and $1$; $2$ and $1$; or $2$ and $0$. By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from $0$ to $2.5$, in which one of the parts is found, into five equal ranges of $0$ to $0.5$, $0.5$ to $1$, $1$ to $1.5$, $1.5$ to $2$, and $2$ to $2.5$). As exactly two of the five cases give a sum of $1 + 2 = 3$, the answer is $\frac{2}{5}$, which is answer $\boxed{B}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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