Question 1100328: Shortly after 5 O’clock, when the minute hand and the hour hand of the clock made a 90 degree angle, Ian went out to walk the dog. On his return after 5:30 the two hands again made a 90 degree angle. For how many minutes was Ian out with his dog?
Found 2 solutions by richwmiller, greenestamps:
Answer by richwmiller(17219)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The hour hand makes 1 revolution every 12 hours; the minute hand makes 1 revolution every hour. So the minute hand rotates 12 times as fast as the hour hand.
Let x be the fraction of an hour that the hour hand moved while Ian was out with his dog. Then 12x is the fraction of an hour that the minute hand moved.
The angle between the minute hand and hour hand was 90 degrees when Ian went out and again when he got back. That means the hour and minute hands were one-quarter of a revolution apart at both those times.
The minute hand went from 1/4 revolution behind the hour hand to 1/4 of a revolution ahead of the hour hand in the time that the hour hand rotated the fraction of an hour x. So the fraction of a revolution that the minute hand made in the time that Ian was out is 
But we know that the minute hand moves 12 times as fast as the hour hand. So
The hour hand made 1/22 of a revolution in the time Ian was out with his dog. Again since the minute hand moves 12 times as fast as the hour hand, the minute hand made 12/22 = 6/11 of a revolution.
And since the minute hand makes one revolution each hour, Ian was out with his dog for 6/11 of an hour, or 360/11 minutes.
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Now here is another way to solve problems like this....
The hour hand makes 1 revolution every 12 hours; the minute hand makes 12 revolutions in 12 hours.
That means that the minute hand will pass the hour hand 11 times in 12 hours. And since both hands rotate at constant rates, the interval between the times when the minute hand passes the hour hand is 12/11 hours or 1 and 1/11 hours.
The interval will be the same 12/11 of an hour for any particular angle between the two hands.
For example, for this problem, the minute hand is 90 degrees behind the hour hand at 3:00. So the minute hand will be 90 degrees behind the hour hand at these hours in a 12-hour cycle:
3, 4 1/11, 5 2/11, 6 3/11, 7 4/11, 8 5/11, 9 6/11, 10 7/11, 11 8/11, 12 9/11, and 1 10/11.
And the minute hand is 90 degrees ahead of the hour hand at 9:00, so the minute hand will be 90 degrees ahead the hour hand at these hours in a 12-hour cycle:
9, 10 1/11, 11 2/11, 12 3/11, 1 4/11, 2 5/11 3 6/11, 4 7/11, 5 8/11, 6 9/11, and 7 10/11.
Looking at those lists, we see that Ian started his walk at 5 2/11 hours and finished at 5 8/11 hours; that means his walk lasted 6/11 of an hour, which of course is the same result we got earlier.
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