# Clock word problem3. In this problem we will ﬁgure out all the times when the two hands of aclock are pointing in exactly the same direction and all of the times whenthe two hands are exactly...

Clock word problem

3. In this problem we will ﬁgure out all the times when the two hands of a

clock are pointing in exactly the same direction and all of the times when

the two hands are exactly 180

◦

apart.

a) According to a certain clock it is currently 2:16. At 2:16, the hour hand

of the clock is not pointing exactly at the 2, rather it is pointing at part way

between the 2 and the 3. Write decimals between 0 and 12 to express in hour

units exactly where each of the hands of the clock is pointing.

b) Suppose the current time is h:m o’clock. That is, it is m minutes after the

hour of h. So h is a whole number between 1 and 12, and m is a real number

between 0 and 60. Write decimals between 0 and 12 to express in hour units

exactly where each of the hands of the clock is pointing.

c) For particular values of h and m, when the time is h:m o’clock, the two

hands of the clock are pointing in exactly same direction. Write an equation

expressing m as a function of h.

Please answer part C the other questions are provided as background

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In 12 hours, the two hands of the clock meets exactly 13 times pointing in the same direction. Of these 13 times, the first time and the last time are at 12. That means, the hands starts at 12 and reaches back at 12 in a period of 12 hours.

Other than these, there are 11 times when the two handles of the clock are pointing in the same direction. This happens at intervals of 1 hour and 5.45 minutes. ie., after 12, again the two handles meet pointing in the same direction when the time is 1:05:45 (1 hour, 5 min and 45 seconds). This happens again at 2 hours and 11 minutes.

So the relation between the coincidence and time is that the hands meet at increments of hour + (hour x 5.45).

The value 5.45 is obtained by dividing 60 with the 11 times.

So,

When minute = m

hour = h,

The handles meet at times, when **m = h x 5.45**

So , m can be expressed as a function of h in the equation,

**m = h x 5.45**

c) The two hands are exactly in the same direction for a 12 hour clock only 12-1=11 times, hence interval between each such time = 12/11 hours = 1h+(60/11)m

The minutes lapsed in terms of hour are therefore 60/11 times hour

if m is the function of h for the two hands of the clock pointing in the same direction, then **minutes lapsed**

** m = 60h/11**