1964 AMC 12 Problem 40

Question

A watch loses

2\dfrac12

minutes per day. It is set right at

1

P.M. on March

15.

Let

n

be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows

9

A.M. on March

21,

n

equals:

14\dfrac{14}{23}

14\dfrac1{14}

13\dfrac{101}{115}

13\dfrac{83}{115}

13\dfrac{13}{23}

The watch advances

\frac{575}{576}

as fast as real time

From the displayed

1

P.M. on March

15

to

9

A.M. on March

21

is

8400

displayed minutes

The watch runs at

\frac{575}{576}

of the correct rate. The displayed elapsed time is

5

days

20

hours, or

8400

minutes. Thus the real elapsed time is

8400(\frac{576}{575}),

and the correction is

\begin{aligned} n&=8400\left(\frac{576}{575}-1\right)\\ &=\frac{8400}{575} =\frac{336}{23}\\ &=14\frac{14}{23}. \end{aligned}

Therefore, the correct answer is A.

Problem 40 in other years