Question

A certain 12 hour digital clock displays the hour and the minute of a day. Due to a defect in the clock whenever the digit 1 is supposed to be displayed it displays 7. What fraction of the day will the clock show the correct time?
(a) \[\dfrac{1}{2}\]
(b) \[\dfrac{5}{8}\]
(c) \[\dfrac{3}{4}\]
(d) \[\dfrac{5}{6}\]

Answer 1

Hint: In this question, the clock is given to be 12 hour type and digital and it displays time in hour and minute format. It is given that the clock displays 7 when the actual value is 1. Therefore, the time shown will be incorrect if either the hour or minute in the actual time has a value where 1 is a digit.

Complete step-by-step solution -
It is given that the clock displays 7 when the actual value is 1. Therefore, the clock will show incorrect time whenever the hour part of the time contains 1. Therefore, as time between 1 to 2 o’clock, 10 to 11 o’ clock, 11 to 12 o’ clock, 12 to 1 o’ clock contains 1, therefore time will be wrongly displayed at these times. Also, as it is a 12 hour clock, we should count the time twice for both day and night. Therefore, the total time for which the time is incorrectly shown by the clock is \[\begin{align}
  & =2\times \text{(time between 1 to 2 o’clock, 10 to 11 o’ }\!\!'\!\!\text{ clock, 11 to 12 o’ clock, 12 to 1 o'clock)} \\
 & =2\times 4\text{ hours}\times 60\min =480\min \\
 & =\dfrac{480}{60}hr=8hr..................(1.1) \\
\end{align}\]
Where in the last step we have used the fact that 1 hour contains 60mins and thus the number of hours can be obtained by dividing the total time in minutes by 60.
As, the total number of hours in one day is 24, in the remaining time \[24-8=16\text{ hours,}\]the hour hand shows the correct time. In one hour, the error in display of minutes will appear whenever there is 1 in the minute time, therefore in one hour there will be an error in minutes at (01,10,11,12,13,…19,21,31,41,51) which is 15 in number. Therefore, in the remaining 16 hours of the day, the minute hand will show an error 15 times. Therefore, the total time at which the clock shows incorrect time in 16 hours is \[16\times 15=240\min ...............(1.2)\]
Adding (1.1) and (1.2), we get the total incorrect time = \[480\text{ }+\text{ }240\text{ }=\text{ }720\text{ min}.......................\text{(1}\text{.3)}\]
However, as 1 hour contains 60 minutes, we obtain that
Total number of minutes in the day \[=24\times 60=1440\min ...............(1.4)\]
Therefore, from (1.3) and (1.4), the fraction of the time when time shown by clock is incorrect to that when time shown is correct is \[=\dfrac{720}{1440}=\dfrac{1}{2}\]
Therefore, as time shown can be either correct or incorrect therefore, the sum of fractions of correct and incorrect time should be 1. Therefore, the total fraction of correct time can be obtained as\[=1-\dfrac{1}{2}=\dfrac{1}{2}\]
Which matches option (a). Hence, option (a) is the right answer .

Note: We should note that the errors in the minute hand between (1 – 2, 10 – 11, 11 – 12, 12 – 1) should not be considered because at these times the hour hand already shows an error and thus, the errors in these times are accounted while calculating error in hours and we have to count the number of errors in shown time and not the number of times there is errors in the digits of the clock.