Clock Angle

Geometry Level 4

When I left my home between 6:00 and 7:00, the hour and minute hands of my analog clock formed a $110^{\circ}$ angle. When I returned home, the hour and minute hands of the clock also formed a $110^{\circ}$ angle. If I left my home for more than five minutes and returned home before 7:00, how many minutes was I away from my home?

The answer is 40.

2 solutions

Abhay Tiwari
Jun 14, 2016

When the clock strikes 6:00 Hrs or 18:00hrs, the difference between Minute hand and the Hour hand is $180°$ .

Now, we know that an hour hand travels ${\dfrac{1}{2}}^{°}$ in one minute and a Minute hand travels $6°$ in one minute.

So, for the angle between them to be ${110}^{°}$ , $(180°+{\dfrac{x}{2}}^{°})-{6x}^{°}=110°.......\boxed{1}$ .

On solving this we get $x=\dfrac{140}{11} \space minutes$ .

Now, calculating the time when they both are overlapping, $(180°+{\dfrac{y}{2}}^{°})-{6y}^{°}=0°.......\boxed{2}$

On solving we get $y=\dfrac{360}{11} \space minutes$ .

Now, the question says that the time when Aaron Tsai came back they were again making, the surprising 110°.So for them to be again in that position, will be after the minute hand crosses the hour hand. And that is only after the time they overlap.

Calculating, the time for which they make 110°, $({0°+{\dfrac{z}{2}}}^{°})-6z°=110°......\boxed{3}$ , we get $z=20 \space Minutes$

Now, total time for which Aaron Tsai were out $=y-x+z=\dfrac{360}{11}-\dfrac{140}{11}+20=\color{#D61F06}{\boxed{40}}$

Nice approach. I solved this using the clock angle formula, which is $\text{Angle}=|30H-5.5M|$ , where $H$ is the hour of the time being displayed, and $M$ is the minute of the time being displayed. We know that $A=110$ and that $H=6$ . So we have

$110=180-5.5M \text{ or } -110=180-5.5M$

$\implies M=12.\overline{72} \text{ or } 52.\overline{72}$

The problem asks for the difference between the two possible values of $M$ . So our answer is $\boxed{40}$ .

Aaron Tsai - 5 years ago

This is pretty interesting Sir(+1). Never knew of this formula. But did you observe that the equation that you are getting is exactly the same as my equations. The reason for $30H$ is that each hour occupies $30°$ . And the reason for $-5.5M$ is that since the hour hand traverses ${\dfrac{1}{2}}^{°}$ for each minutes, whereas the minute hand traversed $6°$ for each minute. The hour hand will add the angle to $180°$ by ${\dfrac{1}{2}}^{°}$ each minute while the minutes hand reduces the angle by $6°$ each minute. And so the equation $Angle = |(30H+.5M)-6M|=|30H-5.5M|$ .

Abhay Tiwari - 5 years ago

Niranjan Khanderia
Jun 17, 2016

Hour arm rotates 360 degrees in 12 * 60 minutes. Minute arm rotates 12 * 360 degrees in 12 * 360 minutes.
So the two arms rotates towards or away from one another at 12 * 360 -360=11 * 360 degrees in 12 * 60 minutes.
So ONE degree in {12 * 60}/{11 * 360}=2/11 minutes.
So for 110 degrees it takes 20 minutes first to come together and another 20 minutes to go out at 110 degrees.
Total 20+20=40 minutes.

Nice approach using relative velocities! Just for additional twist, another solution can be had for the time slot 5:30 to 6:30 instead of 6:00 to 7:00

Ujjwal Rane - 4 years, 10 months ago

Clock Angle

The answer is 40.

2 solutions

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