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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 2 years, 8 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.

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

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 – Thank you

Comment deleted Feb 19, 2015

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

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

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TopNewestDeleted 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.

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

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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.

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Comment deleted Feb 19, 2015

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

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

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