Art of Problem Solving

1986 AJHSME Problems/Problem 2: Difference between revisions

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We see that <math>\frac{1}{3} = \frac{5}{15}</math>, and <math>\frac{2}{5} = \frac{6}{15}</math>, so obviously <math>\frac{1}{3}</math> is smaller.
We see that <math>\frac{1}{3} = \frac{5}{15}</math>, and <math>\frac{2}{5} = \frac{6}{15}</math>, so obviously <math>\frac{1}{3}</math> is smaller.


<math>\boxed{\text{A}}</math>
<math>\boxed{\text{A) } \frac{1}{3} }</math>
 
==Alternate Solution==
 
Find the reciprocals of all numbers.
 
 
\( \frac{1}{3} = 3 \)
 
 
\( \frac{2}{5} = \frac{5}{2} = 2.5 \)
 
 
\( 1 = 1 \)
 
 
\( 5 = \frac{1}{5} = 0.2 \)
 
 
\( 1986 = \frac{1}{1986} = 0.0005035246727 \)
 
 
Therefore, just by comparing the options, we get to know that \( \frac{1}{3} \) has the largest reciprocal.
 
Thus, the answer is
 
<math>\boxed{\text{A) } \frac{1}{3} }</math>
 
~ Alternate solution by abcde26


==See Also==
==See Also==

Latest revision as of 01:16, 23 October 2025

Problem

Which of the following numbers has the largest reciprocal?

$\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{5} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 1986$

Solution

For positive numbers, the larger the number, the smaller its reciprocal. Likewise, smaller numbers have larger reciprocals.

Thus, all we have to do is find the smallest number.

But which one is it? $\frac{1}{3}$? or $\frac{2}{5}$? We see that $\frac{1}{3} = \frac{5}{15}$, and $\frac{2}{5} = \frac{6}{15}$, so obviously $\frac{1}{3}$ is smaller.

$\boxed{\text{A) } \frac{1}{3} }$

Alternate Solution

Find the reciprocals of all numbers.


\( \frac{1}{3} = 3 \)


\( \frac{2}{5} = \frac{5}{2} = 2.5 \)


\( 1 = 1 \)


\( 5 = \frac{1}{5} = 0.2 \)


\( 1986 = \frac{1}{1986} = 0.0005035246727 \)


Therefore, just by comparing the options, we get to know that \( \frac{1}{3} \) has the largest reciprocal.

Thus, the answer is

$\boxed{\text{A) } \frac{1}{3} }$

~ Alternate solution by abcde26

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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