1954 AMC 12 Problem 50

Question

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

degrees are:

7{:}23

and

7{:}53

7{:}20

and

7{:}50

7{:}22

and

7{:}53

7{:}23

and

7{:}52

7{:}21

and

7{:}49

At

t

minutes after

7{:}00,

the signed separation of the hands is

210^\circ-5.5t^\circ

Solve

\lvert210-5.5t\rvert=84

for both values of

t

At

t

minutes after

7{:}00,

the hour hand is at

210^\circ+0.5t^\circ

and the minute hand is at

6t^\circ.

Thus

|210-5.5t|=84.

The two solutions are

\begin{aligned} t&=\frac{126}{5.5}\approx22.91,\\ t&=\frac{294}{5.5}\approx53.45. \end{aligned}

To the nearest minute, the times are

7{:}23

and

7{:}53.

Thus, the correct answer is A.

Problem 50 in other years