Ann's watch runs 2 min. per hour too slow. beth's watch runs 1 minute per hour too fast. for example, if both watches were correct at 12: 00 p. m. on a given day, then at 1: 00 p. m., ann's watch would read 12: 58 p. m. and beth's watch would read 1: 01 p. m. ann and beth set there watches to the correct time at noon on sunday. the next time they met, one of the watches was one hour ahead of the other. what was the earliest time this could have been?
(p. s.: what type of question is this and where can you find more like it? )

At 8:00 am on next day, Beth's watch was one hour ahead of Ann's watch

Step-by-step explanation:

Proportions

When two magnitudes have a linear dependency where one of them is the other by a constant quantity, we can say they are proportional. Like distance and time when moving at a constant speed, double time means double distance, triple time means triple distance and so on.

The problem explains how two watches run too slow or too fast in such a way that every hour they are 3 more minutes apart. If Ann and Beth set their watches to the correct time at noon on Sunday, by 1 pm their watches are 3 minutes apart. By 2 pm their watches are 6 minutes apart and so on.

When will their watches be 1 hour (60 minutes) apart? We'll use proportions to find it out.

If every hour they get 3 minutes apart, we need 60/3 = 20 hours to have their watches 1 hour apart.

At 8:00 am on next day, Beth's watch was one hour ahead of Ann's watch

answer: a.   one is a factor of every whole number since every number is divisible by itself.