Troublesome Clock

A person goes to a market between 4 to 5 PM and returned between 5 to 6 PM,when he comes back he finds that hour hand & minute hand have interchanged their positions . So at what time did he leave and what time did he come back !?

Note by Jaimin Pandya
3 years, 4 months ago

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Deleted my previous comment, here's my new answer:

LEAVE: $$4 : 26 \space \text{PM} \space$$ plus $$\frac {122}{143}$$ of a minute.

LEAVE: $$5 : 22 \space \text{PM} \space$$ plus $$\frac {34}{143}$$ of a minute.

Draw a clock shows that

The person leave between $$4 : 25 \space \text{PM} \space$$ and $$4 : 30 \space \text{PM} \space$$, and

The person arrive between $$5 : 20 \space \text{PM} \space$$ and $$5 : 25 \space \text{PM} \space$$

Denote $$0 < x,y < 5$$ such that $$4 : (25 + x)$$ and $$5 : (20 + y)$$ are the time we need to determine.

Then we have $$\frac {25+x}{60} \times 5 = y$$ and $$\frac {20+y}{60} \times 5 = x$$, solving them gives the above answer.

- 3 years, 4 months ago

Will you please write in detail I didn't get the last line of your solution

- 3 years, 4 months ago

Minute hand of leave = Hour hand of arrive

Minute hand of arrive = Hour hand of leave

Minute hand of leave = 25 + x

Notice that for every 60 minutes pass, the hour hand moves by 1, equivalently, for every minute pass, the hour hand moves by $$\frac {1}{60}$$

Which means $$\frac {25 + x}{60}$$ is the units the hour hand has moves after 4PM

Since $$1,2,3, \ldots, 12$$ on the clock denote 5 minute intervals

$$\frac {25 + x}{60} \times 5 =$$ Hour hand of arrive.

Likewise, we get the second equation.

- 3 years, 4 months ago

Thank you

- 3 years, 4 months ago

Comment deleted Feb 19, 2015

Answer will be in mixed fractions It's not that easy as you did

- 3 years, 4 months ago

Are you sure? It seems to me that the minute hand stays the name, namely at the 25 minute mark.

Staff - 3 years, 4 months ago