# Difference between revisions of "1954 AHSME Problems/Problem 50"

## Problem

The times between $7$ and $8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $84^{\circ}$ are:

$\textbf{(A)}\ \text{7: 23 and 7: 53}\qquad \textbf{(B)}\ \text{7: 20 and 7: 50}\qquad \textbf{(C)}\ \text{7: 22 and 7: 53}\\ \textbf{(D)}\ \text{7: 23 and 7: 52}\qquad \textbf{(E)}\ \text{7: 21 and 7: 49}$

## Solution

At $7$ o'clock, the hour hand is at the position $\tfrac{7}{12}\cdot 360^{\circ}=210^{\circ}$ clockwise from the $12$ o'clock position, and the minute hand is exactly at the $12$ o'clock position. Thus the minute hand is $360^{\circ}-210^{\circ}=150^{\circ}$ ahead of the hour hand, while it is also $210^{\circ}$ behind the hour hand. So, when the minute hand first makes an $84^{\circ}$ angle with the hour hand, the minute hand will be $84^{\circ}$ behind the hour hand, and the second time they make and $84^{\circ}$ angle, the minute hand will be $84^{\circ}$ ahead the hour hand.

The minute hand moves clockwise at a rate of $360^{\circ}/\text{hr}$, while the hour hand moves at $(1/12)\cdot 360^{\circ}/\text{ hr}=30^{\circ}/\text{hr}$. Therefore the minute hand catches up to the hour hand at a rate of $360^{\circ}-30^{\circ}=330^{\circ}$ per hour. Therefore it will take $(210^{\circ}-84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{126}{330}\text{hr}$ for the two hands of the clock to make an $84^{\circ}$ angle. It will also take $(210^{\circ}+84^{\circ})/(330^{\circ}/\text{hr})=\tfrac{294}{330}\text{hr}$ after $7$ o'clock for the hands to make an $84^{\circ}$ angle for the second time. Converting these values to minutes, we see that it will take $22\tfrac{10}{11}$ minutes, and $53\tfrac{5}{11}$ minutes, for the two hands to make $84^{\circ}$ angles. Thus the times, correct to the nearest minute, at which the hands of the clock will form an $84^{\circ}$ angle are $\boxed{\textbf{(A)}\ \text{7: 23 and 7: 53}}$.

## Solution 2

Using the formula \$(60h-11m)/2