# Problem of the week #38 - December 17th, 2012

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#### Jameson

Staff member
A clock is observed. The hour hand is exactly at the minute mark, and the minute hand is six minutes ahead of it. Later, the clock is observed again. This time, the hour hand is exactly on a different minute mark, and the minute hand is seven minutes ahead of it. How much time elapsed between the first and second observations?

Hint:
The hour hand is exactly on a minute mark five times per hour -- on the hour, twelve minutes past the hour, twenty four minutes past, thirty six minutes past, and forty eight minutes past.

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#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) MarkFL
2) Sudharaka

Solution (from Sudharaka):
Let the hour hand be on the $$x$$ minute mark when the minute hand is six minutes ahead of it and let the hour hand be on the $$y$$ minute mark when the minute hand is seven minutes ahead of it. Then $$x+6=12,24,36,48$$ and $$y+7=12,24,36,48$$. Hence,

$x=6,18,30,42\mbox{ and }y=5,17,29,41$

Note that the hour hand should be one minute past the hour when the minute mark is twelve minutes past the hour. The hour hand should be two minutes past the hour when the minute mark is twenty four minutes past the hour. The hour hand should be three minutes past the hour when the minute mark is thirty six minutes past the hour. The hour hand should be four minutes past the hour when the minute mark is forty eight minutes past the hour.

Therefore the only possibility among the listed values for $$x$$ and $$y$$ is, $$x=6$$ and $$y=17$$. That is initially the time is, $$1.12$$ and in the second observation it's $$3.24$$. Time that is elapsed in between is therefore, 2 hours and 12 minutes (in fact it could be $$(12n+2)$$ hours and $$12$$ minutes where $$n\in\mathbb{N}=\mathbb{Z}\cup\{0\}$$).

A second solution from MarkFL:
Label the minutes on the clock from 0-59. Let $\displaystyle m=12k$ where $\displaystyle k\in\{0,1,2,3,4\}$ be the minute at which the minute hand is located, and $\displaystyle h=5H+k$ be the minute at which the hour hand is located, where $\displaystyle1\le H\le12$ is the hour.

Let $\displaystyle m-h=d$

$\displaystyle 12k-(5H+k)=d$

$\displaystyle 11k-5H=d$

The integer solution is:

$\displaystyle (k,H)=(d+5n,2d+11n)$ where $\displaystyle n\in\mathbb{Z}$

For the first observation, we are told:

$\displaystyle d=6$

The only solution satisfying the given restrictions is $\displaystyle (k,H)=(1,1)$, i.e., $\displaystyle n=-1$.

For the second observation, we are told:

$\displaystyle d=7$

The only solution satisfying the given restrictions is $\displaystyle (k,H)=(2,3)$, i.e., $\displaystyle n=-1$.

And so we find the elapsed time between observations is (3 - 1) = 2 hours and (2 - 1)·12 = 12 minutes.

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