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08 Dec 2021, 03:15
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:28) correct
60%
(02:27)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
The hands of a strange clock move such that they would meet twice as frequently if they run in opposite directions than if they run in the same direction. How many times would the faster hand meet the slower hand in the time that the slower hand completes 20 rotations, given that the hands run in opposite directions?
A. 60
B. 40
C. 80
D. 120
E. 150
Are You Up For the Challenge: 700 Level Questions
A. 60
B. 40
C. 80
D. 120
E. 150
Are You Up For the Challenge: 700 Level Questions
29 Dec 2021, 12:36
Kudos
Bookmarks
Say the time taken by the hands to meet when they are running in the same direction and when they are running in the opposite direction be 2t and t.
For every 1/4 round of the hour hand, the minute hand will meet it once when they move in the opposite direction.
So when the hour hand completes 20 rounds, the minute hand should meet it 20 * 4 = 80 times. Ans (A)
For every 1/4 round of the hour hand, the minute hand will meet it once when they move in the opposite direction.
So when the hour hand completes 20 rounds, the minute hand should meet it 20 * 4 = 80 times. Ans (A)
29 Dec 2021, 13:03
Kudos
Bookmarks
Bunuel
Breaking Down the Info:
We may set two variables x and y, with the units degrees per hour. (units can be whatever as long as you are consistent).
Let's assume the hands meet once every hour if they run in the same direction, which means they meet every half hour if they run in the opposite direction.
Then \((y - x)*1 = 360\) and \((x + y)*0.5 = 360\)
Solving this gives \(y = 540\) and \(x = 180\), again in degrees per hour. Thus x would take 40 hours to complete 20 rotations. We assumed they would meet every half hour running in the opposite direction, so they meet 80 times in 40 hours.
Answer: C
