enigma123 wrote:
A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15). Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day)?
A)5:00
B)5:34
C)5:42
D)6:00
E)6:24
Guys - again the official answer is not provided. But I have done the Maths to get D. Do you think D is the right answer? Can someone explain to me if you think D is not the right answer?
Here is my approach to such problems:
1. Loses 15 mins every hour (i.e. 60 mins) means the clock's speed is a fourth less than the normal speed. So the clock's speed is (3/4)th the normal speed. It covers only 45 mins in the time in which a correct clock covers 60 mins.
2. Gains 20 mins every hour means the clock's speed is a third more than normal speed. So the clock's speed is (4/3) times the normal speed. It covers 1 hr 20 mins in the time in which a correct clock covers 1 hr.
Let speed of a correct clock = s
Speed of clock 1 = (3/4)s
Speed of clock 2 = (3/4)s * (5/4) = (15/16)s
Speed of clock 3 = (15/16)s * (2/3) = (5/8)s
Speed of clock 4 = (5/8)s * (4/3) = (5/6)s
Speed of the fourth clock is (5/6)th the normal speed. If a correct covers 6 hrs, the fourth clock will cover 5 hrs. Hence, the time shown will be 5:00 pm
This means that if the speed of clock 1 is s1, the speed of clock 2 is s1 + s1/4 (i.e. clock 2 covers 15 extra mins for every 60 mins covered by clock 1). This gives us that speed of clock 2 is 5s1/4.
Since the speed of clock 1 itself is 3s/4, the speed of clock 2 will be (5/4)*(3s/4) = 15s/16