Question

At what time between 10 O’ clock and 11 O’ close are the two hands of the clock symmetric with respect to the vertical line (give the answer to the nearest second)?
A) 10h9m13s
B) 10h9m14s
C) 10h9m22s
D) 10h9m50s

Answer 1

Hint: In this question we have to find the minutes after 10 hours, at which the hands of the clock become symmetric. To find this first we are going to find minutes. For this we are going to equate two terms and after adding minutes to hours we will get the final answer.

Complete step by step solution:
After x minutes the two hands of the clock become symmetric to each other.
As we already know that the ratio of angular speed of hour to minute hand is $1:12$.
If the minute hand moves by x minute distance, the hour hand will move by ${\displaystyle \frac{x}{12}}$ minute distance.
Now to find the value of x, we will equate $${\displaystyle \frac{x}{12}}$$ with $$10-x$$.
$$\Rightarrow {\displaystyle \frac{x}{12}}=10-x$$
$$\Rightarrow {\displaystyle \frac{x}{12}}+x=10$$
$$\Rightarrow {\displaystyle \frac{13x}{12}}=10$$
$$\Rightarrow x={\displaystyle \frac{120}{13}}$$
$$\Rightarrow x=9.23\text{ minutes}$$
Hence, from the above calculation we have found out the value of x or the value of minutes $$x=9.23\text{ minutes}$$. Hence the time at which two hands of the clock will be symmetric is$$\text{10h 9m 22s}$$.

Hence option (C) is correct.

Note: The ratio of angular speed of hour hand and minute hand is $1:12$. Now, we are going to discuss how we get this ratio. As we know that the minute hand of the clock covers ${360}^{\circ }$ in 60 minutes. The hour hand of a clock covers ${30}^{\circ }$ in 60 minutes. We can say it in other way that the minute hand of the clock covers $6\text{ deg/minute}$ and the hour hand of the clock covers $\text{0}\text{.5 deg/minute}$. If we find the ratio of the angular speed of hour to minute hands, it comes out to be $1:12$. This information of the ratio of angular speed is must to solve such problems.