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Question

Question: At what time are the hands of the clock together between 5 and 6?...

At what time are the hands of the clock together between 5 and 6?

Explanation

Solution

Take 5 O’clock in an initial situation. At this point, the hands of the clock are 25 minutes apart from each other. Also, we need to use the information that 1 hour is equal to 60 minutes. In a clock, 60 minutes means 1 circular round of minute hand. The point is when the minute hand moves 60 steps at the same time hour hand moves only 5 steps (from the perspective of 1 minute as 1 step)

Complete step-by-step answer :
The Idea to solve this question is to observe the obvious thing that is 1 hour = 60 minutes
It means as the minute hand moves 60 steps, the hour hand moves 5 steps but slower than minute hand. Which can also be seen as
60 minute-steps = 5 hour-steps
So, 1 minute-steps = 560=112\dfrac{5}{{60}} = \dfrac{1}{{12}} hour-steps

Let us assume it takes y minute-steps.
So, this y minute step should be equal to steps from 12 to 5 which is 25 steps plus the steps moved by hour hand. Which is m×112m \times \dfrac{1}{{12}}. This can also be written as
m=25+m×112 m=25+m12 \begin{gathered} m = 25 + m \times \dfrac{1}{{12}} \\\ \Rightarrow m = 25 + \dfrac{m}{{12}} \\\ \end{gathered}
We can solve this linear equation with unknown m to get our required answer.
Multiplying the whole equation by 12 and taking m-involving terms together.
12m=12×25+m 12mm=300 11m=300 m=30011 \begin{gathered} 12m = 12 \times 25 + m \\\ \Rightarrow 12m - m = 300 \\\ \Rightarrow 11m = 300 \\\ \Rightarrow m = \dfrac{{300}}{{11}} \\\ \end{gathered}
On dividing 300 by 11 we’ll the minutes at which two hands of the watch coincide.
m=30011=27.27m = \dfrac{{300}}{{11}} = 27.\overline {27}
Hence, two hands of the watch coincide between 5 to 6 at 27.3 minutes approximately.

Note : The hack in this question is to observe that as the minute hand moves and travels the distance till 5 the hour hand also moves. So, to get the answer we have to find a time where both meets. One can also solve this by the time and distance concept.