Clock word problemIn 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

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.
d) Use each value of h between 1 and 12 to determine all of the times (up
to 2 decimal places) when both hands of a clock are pointing in exactly the
same direction.

selmasharafaz | College Teacher | (Level 3) Adjunct Educator

Posted on

On finding part c, we can express m as a function of h, by the equation,

m = h x 5.4545 or m = h x (60/11)

So, the times at which the two hands of the clock coincide and point towards the same direction are,

time = h + m , where m = h x (60/11)

or time = h + [h x (60/11)]

Therefore, the 12 times with value of h from 1 to 12 are,

h = 1 ; time = 1 h + [1 x (60/11)] m ; time = 01 h : 5.45 min

h = 2 ; time = 2 h + [2 x (60/11)] m ; time = 02 h : 10.91 min

h = 3 ; time = 3 h + [3 x (60/11)] m ; time = 03 h : 16.36 min

h = 4 ; time = 4 h + [4 x (60/11)] m ; time = 04 h : 21.82 min

h = 5 ; time = 5 h + [5 x (60/11)] m ; time = 05 h : 27.27 min

h = 6 ; time = 6 h + [6 x (60/11)] m ; time = 06 h : 32.73 min

h = 7 ; time = 7 h + [7 x (60/11)] m ; time = 07 h : 38.18 min

h = 8 ; time = 8 h + [8 x (60/11)] m ; time = 08 h : 43.64 min

h = 9 ; time = 9 h + [9 x (60/11)] m ; time = 09 h : 49.09 min

h = 10 ; time = 10 h + [10 x (60/11)] m ; time = 10 h : 54.55 min

h = 11 ; time = 11 h + [11 x (60/11)] m ; time = 12 h : 0.00 min

h = 12 ; time = 12 h + [12 x (60/11)] m ; time = 1 h : 5.45 min (the value is same as when h = 1 and the cycle continues)

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted on

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 or 1h 05.4545m

d) All such times are:

1*(1h 05.4545m) = 01h 05.45m = 01:05.45

2*(1h 05.4545m) = 02h 10.91m = 02:10.91

3*(1h 05.4545m) = 03h 16.36m = 03:16.36

4*(1h 05.4545m) = 04h 21.82m = 04:21.82

5*(1h 05.4545m) = 05h 27.27m = 05:27.27

6*(1h 05.4545m) = 06h 32.73m = 06:32.73

7*(1h 05.4545m) = 07h 38.18m = 07:38.18

8*(1h 05.4545m) = 08h 43.64m = 08:43.64

9*(1h 05.4545m) = 09h 49.09m = 09:49.09

10*(1h 05.4545m) = 10h 54.55m = 10:54.55

11*(1h 05.4545m) = 11h 60.00m = 12:00.00