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:

$14\dfrac{14}{23}$

$14\dfrac1{14}$

$13\dfrac{101}{115}$

$13\dfrac{83}{115}$

$13\dfrac{13}{23}$

Answer: A

Concepts:clock date and time rate fraction

Difficulty rating: 1670

Small Hint:

The watch advances $\frac{575}{576}$ as fast as real time

Big Hint:

From the displayed $1$ P.M. on March $15$ to $9$ A.M. on March $21$ is $8400$ displayed minutes

Solution:

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.

1964 AMC 12 Problem 40

Answer: A

Difficulty rating: 1670

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years