Can you calculate the total elapsed time – [HARD]
An observation is made on a clock and it is found that the hour hand is placed exactly at the minute mark while the minute hand is six minutes ahead it. After some time, the clock is observed again and it is found that the hour hand is exactly on another minute mark while the minute hand is seven minutes ahead of it.
Analyzing this data, can you calculate the total time that has elapsed between the two observations?
ANSWER- 2hr 11min
This question cannot be solved unless you know a fact i.e. the hour hand is exactly on the minute mark five times every hour – on the hour, 12 minutes past hour, 24 minutes past hour, 36 minutes past hour and 48 minutes past hour.
Denoting the number of hours as X and the number of minutes as Y, and lets begin with the calculation.
When we have the hour hand on the minute mark,
The position of hour hand: 5X + Y/12
And the position of the minute hand = Y
When the first observation was taken,
Y = Y = 5X + Y/12 + 6
Or, 60X = 11Y – 72.
With the given facts, we know for sure that Y can only be 0, 12, 24, 36 or 48.
According to which, the possible values for X and Y can be 1 and 12 respectively. Thus the time must be 1:12 when the first observation was taken.
Similarly, in the second observation, let’s make the equation:
60X = 11Y – 84
Here, the possible values for X and Y here are 3 and 24 respectively. According to this, the time is 3:24 for sure.
From 1:12 to 3:24, two hours and eleven minutes have passed.
2 hours and 12 minutes.
This question can actually be answered with less information–it’s not necessary to know how far ahead the minute hand is, only how much further ahead it is the second time. So this answer considers the question to be:
An observation is made on a clock and it is found that the hour hand is placed exactly at the minute mark. After some time, the clock is observed again and it is found that the hour hand is exactly on another minute mark while the minute hand is one minute further ahead of the hour hand than the minute hand was ahead of the hour hand when the first observation was made.
Every 12 minutes, or five times an hour, the hour hand is exactly on a minute mark. Every 12 minutes, the minute hand advances 12 minutes and the hour hand advances one minute, so the distance between the hour and minute hand increases by 11, modulo 60.
If the number of 12-minute periods that pass is p, then we want to solve for 11p modulo 60 = 1. So starting with 1, keep adding 60 until we find the first multiple of 11:
1, 61, 121 <– BINGO
121 = 11 * 11, so the minimum number of 12 minute periods that have passed is p = 11. 11 * 12 = 132 minutes = 2 hours and 12 minutes.
Since 11 and 60 are relatively prime, we know that in a 12 hour period, the hour hand will be on a minute mark 60 different times and the amount the minute hand is ahead of the hour hand will be different on each occasion, with the difference assuming each of the values from 0 to 59. This realization is what allows the problem to be uniquely solved with only the information about how much the difference between the hour and minute hand changed.
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