A person goes to sleep between 1am and 2 am and he wakes up when his watch shows such a time that the two hands interchange their respective places. He wakes up between 2am and 3am, how long does he sleep?

Consider only hour hand & minute hand.

A person goes to sleep between 1am and 2 am and he wakes up when his watch shows such a time that the two hands interchange their respective places. He wakes up between 2am and 3am, how long does he sleep?

Consider only hour hand & minute hand.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThe answer is \(\boxed {\frac{12}{13}}\) hours.

The positions of hour-hand and minute-hand have a certain relationship. They are together at 00:00h or 12:00AM. Let 12:00AM position be \(0^\circ\), the time the person goes to sleep at \(t_1\) and wakes up at \(t_2\), the angle made by the hour-hand at \(t_1\) and minute-hand at \(t_2\) be \(\alpha\) and that by the hour-hand at \(t_2\) and minute-hand at \(t_1\) be \(\beta\). Note that the hour-hand makes \(30^\circ\) on the dial in \(1\) hour while the minute-hand makes \(360^\circ\) in \(1\) hour.

Therefore, we have \(\quad 30t_1 = \alpha \space, \quad 360t_1 = 360 + \beta\) (as \(t_1 > 1\) hour) \(\quad \Rightarrow t_1 = \dfrac {\alpha}{30} = 1 + \dfrac {\beta}{360} \)

Similarly, we have \(\quad t_2 = \dfrac {\beta}{30} = 2 + \dfrac {\alpha}{360}\)

The time that the person sleep, \(t = t_2-t_1 = \dfrac {\beta-\alpha} {30} = 1 - \dfrac {\beta - \alpha}{360}\)

\( \dfrac {\beta-\alpha} {30} \left( 1 + \dfrac {1}{12} \right) = 1\quad \Rightarrow \dfrac {\beta-\alpha} {30} = t = \boxed{\frac {12}{13}}\) hours. – Chew-Seong Cheong · 2 years, 11 months ago

Log in to reply

– Surya Prabha · 10 months, 3 weeks ago

Why you add 10 to y in the equation 10+y/60Log in to reply

– Aneesh Kundu · 2 years, 11 months ago

thanx a lot!!Log in to reply

The book is correct. Let \(x\) and \(y\) be the minutes past \(1 am\) and \(2 am\) where the hands could be. Then we solve this system of equations, for start and end times, the difference being the time of sleeping:

\(\dfrac { 10+y }{ 60 } =\dfrac { x }{ 5 } \)

\(\dfrac { 5+x }{ 60 } =\dfrac { y }{ 5 } \)

We get \(x=\dfrac { 125 }{ 143 } \) and \(y=\dfrac { 70 }{ 143 } \), from which we can work out the time of sleep

\((120+5+x)-(60+10+y)=\dfrac { 720 }{ 13 } =55+\dfrac { 5 }{ 13 } \) – Michael Mendrin · 2 years, 11 months ago

Log in to reply

– Surya Prabha · 10 months, 3 weeks ago

Why u add 10 to y in the equation (10+y)/60Log in to reply

– Aneesh Kundu · 2 years, 11 months ago

Thanx a lot!!Log in to reply

A similar question once appeared in INMO And I really think this is a gud one So pls reshare and write solutions – Aneesh Kundu · 2 years, 11 months ago

Log in to reply

he slept at 1:10 and woke up at 2:05 – Mehul Arora · 2 years, 11 months ago

Log in to reply

– Aneesh Kundu · 2 years, 11 months ago

According to the book the answer should be \[55\dfrac{5}{13}\text{ mins}\]Log in to reply

– Mehul Arora · 2 years, 11 months ago

but the question says that consider minute and hour hand onlyLog in to reply

– Aneesh Kundu · 2 years, 11 months ago

yes but that doesn't mean that time can't be fractionalLog in to reply