Joey 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 Joey makes an error in reading the time since it is a
mirror image. Joey 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

^{1}/_{2}hours (2 hours and 30 minutes) later than the time that he saw in the mirror image of his clock at home.
What time was it when
he got to day camp?

(The clock at camp
and the clock at home were both set to the correct time.)

**Solution**

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*****

*****

First subtract 20
minutes from 2 1/2 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 (7:05), and the time he saw
in the mirror (4:55).

Furthermore,
by re-adding the 20 minutes from when he left (7:05), you get what time he got
to camp (7:25).

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