Question

At what time, between 10 o’clock and 11 o’clock are two hands of the clock symmetric with respect to the vertical line (give the answer to the nearest second)?
A. 10 h 9 min 13 sec
B. 10 h 9 min 14 sec
C. 10 h 9 min 22 sec
D. 10 h 9 min 50 sec

Answer 1

Hint: In this question take the time as a variable in minutes between 10 o'clock and 11 o'clock where the two hands of the clock are in symmetric with respective to the vertical line. The speed of the hour hand to minute hand is \[1:12\]. Then equate the distance moved by the minutes and the distance moved by the hours hand which will lead to the solution of the problem.

Complete step-by-step answer.

Let \[x\]minutes past 10 o'clock and 11 o’clock hands become symmetric.
Here in a clock each number is separated by \[{30^0}\]. Therefore, in one hour, the hour hand moves by \[{30^0}\], while the minute hand moves by \[{360^0}\]. So, the speed of hour hand to minute hand is \[1:12\]
The diagram of the clock will be as shown in the following figure:

Therefore, when the minute hand moves \[x\] minute distance, then the hour hand moves \[\dfrac{x}{{12}}\] minutes distance.
\[
   \Rightarrow \dfrac{x}{{12}} = 10 - x \\
   \Rightarrow \dfrac{x}{{12}} + x = 10 \\
   \Rightarrow \dfrac{{x + 12x}}{{12}} = 10 \\
   \Rightarrow \dfrac{{13x}}{{12}} = 10 \\
  \therefore x = \dfrac{{120}}{{13}}{\text{ minutes}} = 9{\text{ minutes 13}}{\text{.8 seconds}} \\
\]
Hence the required time is 10 h 9 min 13.8 sec which is approximately equals to 10 h 9 min 14 sec
Thus, the correct option is B. 10 h 9 min 14 sec.

Note: After obtaining the answer in minutes and seconds we have added it to the 10 hours as, we have to find the clock symmetric with respect to the vertical line between 10 o'clock and 11 o'clock. In a clock one hour is equal to sixty minutes and one minute is equal to sixty seconds.