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 !?

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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.

Pi Han Goh - 6 years, 3 months ago

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

Jaimin Pandya - 6 years, 3 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.

Pi Han Goh - 6 years, 3 months ago

@Pi Han Goh – Thank you

Jaimin Pandya - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$