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 !?

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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. – Pi Han Goh · 2 years, 3 months ago

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– Jaimin Pandya · 2 years, 3 months ago

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

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 · 2 years, 3 months ago

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– Jaimin Pandya · 2 years, 3 months ago

Thank youLog in to reply

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– Jaimin Pandya · 2 years, 3 months ago

Answer will be in mixed fractions It's not that easy as you didLog in to reply

– Calvin Lin Staff · 2 years, 3 months ago

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