Time at camp?
Jimmy leaves his house in the morning to go to day camp.
Just as he is leaving his house he looks at an analog clock reflected in the mirror.
There are no numbers on the clock, so Jimmy makes an error in reading the time since it is a mirror image. Jimmy assumes there is something wrong with the clock and rides his bike to day camp.
He gets there in 20 minutes and finds that just as he gets there the day camp clock has a time that is 2 hours (2 hours and 30 minutes) later than the time that he saw in the mirror image of his clock at home.
What was the time when he got to day camp?
(The clock at camp and the clock at home were both set to the correct time.)
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
First subtract 20 minutes from 2 2 1 hours to compensate for his 20 minute bike ride to give a difference of 2 hours and 10 minutes.
To be a "Mirror Effect" it must be mirrored around 12 o'clock (when the hands are straight up), or around 6 o'clock (when the hands are pointing up and down), as we know he left in the morning, it must be 6 o'clock.
So, divide that 2 hours and 10 minutes by 2 and this will give you the center-point (65 minutes) for compensating for the mirror.
By adding that 65 minutes to 6 o'clock you get the time he left home 7 : 0 5 , and the time he saw in the mirror 4 : 5 5
Furthermore, by re-adding the 20 minutes from when he left 7 : 0 5 , you get what time he got to camp 7 : 2 5 .
Therefore, he got to day camp at 7 : 2 5