Question

At what time between \[5\] and \[6\] O’ clock, will the hands of a clock be at an angle of \[{62^ \circ }\]?
a). \[5\text{hours }{17}^{\dfrac{2}{11}}\text{minutes} \]
b). \[5\text{hours } {38}^{\dfrac{6}{11}}\text{minutes}\]
c). \[5\text{hours }16\text{minutes}\]
d). Both (B) and (C)

Answer 1

Hint: We know that in an hour or in \[60\]minutes the minute hand covers an angle of \[360\] degrees. Secondly, we also have from our previous knowledge that the hour hand covers an angle of \[30\] degrees.

Complete step-by-step solution:
So from our previous knowledge of clocks and its properties we know that in an hour or in \[60\]minutes the minute hand covers an angle of \[360\] degrees. Secondly, the hour hand covers an angle of \[30\] degrees.
So in every \[60\] minute the minute hand covers \[330\] degrees more than that of the hour hand.
So \[330\] degrees in \[55\] minutes, therefore we see that in every \[60\] minute the minute hand is \[55\] minute ahead.
And hence to be \[1\]minute ahead, time taken by minute hand is \[\dfrac{{60}}{{55}} = \dfrac{{12}}{{11}}\].
Now we observe different scenarios where,
Firstly the minute hand is behind so the distance when the time is \[5\] o’ clock is equal to \[25\] minute which is equal to \[\dfrac{{330}}{{55}} \times 25\] which is equal to \[150\]degrees.
So to reach \[62\] degrees the distance needed is \[150 - 62\] degrees which is \[88\]degrees or \[\dfrac{{44}}{3}\] minute
So the time taken after \[5\]is equal to \[\dfrac{{44}}{3} \times \dfrac{{12}}{{11}} = 16\] minute
Therefore the exact time when the hands will be together is \[3:16\]
Secondly we look when the minute hand is ahead so the distance when the time is \[5\]o’ clock is equal to \[25\] minute which is equal to \[\dfrac{{330}}{{55}} \times 25\] which is equal to \[150\] degrees.
So to reach \[62\] degrees the distance needed is \[150 + 62\] degrees which is \[212\] degree or \[\dfrac{{106}}{3}\] minute
So the time taken after \[5\]is equal to \[\dfrac{{106}}{3} \times \dfrac{{12}}{{11}}\] which is \[38{\kern 1pt} {\kern 1pt} (\dfrac{6}{{11}})\]minute
Therefore the exact time when the hands will be together is \[3:38{\kern 1pt} {\kern 1pt} (\dfrac{6}{{11}})\]
Hence the correct option from both the cases is (d).

Note: It is very important that we know the basic computation and evaluation of degrees to minutes to seconds to hours and vice versa while computing such solutions to the problems. It is also important to be familiar with the time each hand of the clock covers in any given time.