Robert has two watches, one which loses 6 seconds every 24 hours and one which gains 1 second
per hour. He sets both of them to the correct time at 6 : 00 p.m. How many hours will pass
before the positive difference between the time shown on both watches is 4 hours?
i hit my head on this question for a long time ! finally got the answer. total difference : 6seconds+24seconds - 30seconds they lose in a day. now calculate the total seconds in four hours . which is 14400. now divide 14400 by 30 seconds. which will be 480. now multiple that number by 24 hours. the final and correct sanswer will be 11520. :)
Let's analyze the situation step by step.
1. The watch that loses 6 seconds every 24 hours:
- This means it loses 6/24 = 1/4 seconds per hour.
- Since it was set to the correct time at 6:00 p.m., the time this watch shows after 'h' hours will be 6:00 p.m. minus (h/4) seconds.
2. The watch that gains 1 second per hour:
- This means it gains 1 second per hour.
- Since it was set to the correct time at 6:00 p.m., the time this watch shows after 'h' hours will be 6:00 p.m. plus 'h' hours.
3. The positive difference between the two watches:
- The positive difference between the two watches is 4 hours.
- This means the watch that loses time is behind by 4 hours from the watch that gains time.
- So, we need to find the value of 'h' where the difference between the two watches is 4 hours.
Now, let's set up an equation using the information above:
(6:00 p.m. - (h/4) seconds) - (6:00 p.m. + h hours) = 4 hours
Breaking it down further:
(18:00 - (h/4) seconds) - (18:00 + h) = 4 hours
Simplifying:
- (h/4) seconds - h = 4 hours
Now, let's solve for 'h':
- (h/4) seconds - h = 4
Multiply the equation by 4 to remove the fraction:
- h seconds - 4h = 16
Combining like terms:
- 3h = 16
Divide both sides by -3:
h = 16 / -3
Approximately, h ≈ -5.33 hours
Since time cannot be negative, we discard this solution.
Therefore, there is no positive difference of 4 hours between the two watches.
To find out the number of hours that will pass before the positive difference between the time shown on both watches is 4 hours, we need to analyze how the time on each watch changes over time.
First, let's consider the watch that loses 6 seconds every 24 hours. We need to convert the given rate to hours. There are 60 seconds in a minute and 60 minutes in an hour, so there are 60 * 60 = 3600 seconds in an hour. Therefore, the watch loses 6/3600 = 1/600 of an hour every 24 hours.
Now, let's consider the watch that gains 1 second per hour. Since this rate is already given in hours, there is no need to convert it.
Since both watches were set correctly at 6:00 p.m., the initial time on both watches is the same. Let's say after x hours, the positive difference between the time shown on both watches is 4 hours.
Therefore, the time shown on the watch that loses 6 seconds every 24 hours would be (6:00 p.m. + x hours) - (1/600 * x) hours.
The time shown on the watch that gains 1 second per hour would be 6:00 p.m. + x hours.
To find the positive difference between the time shown on both watches, we subtract the time shown on the watch that gains 1 second per hour from the time shown on the watch that loses 6 seconds every 24 hours:
[(6:00 p.m. + x hours) - (1/600 * x) hours] - [6:00 p.m. + x hours] = 4 hours
Now, we can solve the equation to find the value of x (the number of hours that will pass before the positive difference between the time shown on both watches is 4 hours).