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14 Jun 2021, 05:45
Kudos
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Difficulty:
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Question Stats:
39% (02:53) correct
61%
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wrong
based on 70
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Four clocks were set right at the same time. One clock gains ‘a’ minutes in a day, second clock loses ‘b’ minutes in a day, third clock gains ‘c’ minutes in a day and the fourth clock gains ‘d’ minutes. After how much time will all the clocks again show the correct time simultaneously if a + b + c + d = 1 hour, a : b = c : d = 2 : 1 and b : d = 3 : 1?
(A) 3 hours
(B) 12 hours
(C) 24 hours
(D) 24 days
(E) 144 days
(A) 3 hours
(B) 12 hours
(C) 24 hours
(D) 24 days
(E) 144 days
14 Jun 2021, 15:09
Kudos
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a:b:c:d=6:3:2:1
a+b+c+d=60
a=30; b=15; c=10; d=5
In order to show correct time, relative distance it has to travel wrt to correct clock = 24*60 mins [If the clock is 24-hour clock] or 12*60 [If the clock is 12-hour clock]
They will show correct time in = \(\frac{24*60}{gcd(30,15,10,5)} = \frac{24*60}{5}= 288days\) [ If all the clocks are 24-hour clock]
They will show correct time in = \(\frac{12*60}{gcd(30,15,10,5)} = \frac{12*60}{5}= 144days\) [ If all the clocks are 12-hour clock]
a+b+c+d=60
a=30; b=15; c=10; d=5
In order to show correct time, relative distance it has to travel wrt to correct clock = 24*60 mins [If the clock is 24-hour clock] or 12*60 [If the clock is 12-hour clock]
They will show correct time in = \(\frac{24*60}{gcd(30,15,10,5)} = \frac{24*60}{5}= 288days\) [ If all the clocks are 24-hour clock]
They will show correct time in = \(\frac{12*60}{gcd(30,15,10,5)} = \frac{12*60}{5}= 144days\) [ If all the clocks are 12-hour clock]
Bunuel
Four clocks were set right at the same time. One clock gains ‘a’ minutes in a day, second clock loses ‘b’ minutes in a day, third clock gains ‘c’ minutes in a day and the fourth clock gains ‘d’ minutes. After how much time will all the clocks again show the correct time simultaneously if a + b + c + d = 1 hour, a : b = c : d = 2 : 1 and b : d = 3 : 1?
(A) 3 hours
(B) 12 hours
(C) 24 hours
(D) 24 days
(E) 144 days
(A) 3 hours
(B) 12 hours
(C) 24 hours
(D) 24 days
(E) 144 days
19 Jun 2021, 19:48
Kudos
Bookmarks
nick1816
a:b:c:d=6:3:2:1
a+b+c+d=60
a=30; b=15; c=10; d=5
In order to show correct time, relative distance it has to travel wrt to correct clock = 24*60 mins [If the clock is 24-hour clock] or 12*60 [If the clock is 12-hour clock]
They will show correct time in = \(\frac{24*60}{gcd(30,15,10,5)} = \frac{24*60}{5}= 288days\) [ If all the clocks are 24-hour clock]
They will show correct time in = \(\frac{12*60}{gcd(30,15,10,5)} = \frac{12*60}{5}= 144days\) [ If all the clocks are 12-hour clock]
a+b+c+d=60
a=30; b=15; c=10; d=5
In order to show correct time, relative distance it has to travel wrt to correct clock = 24*60 mins [If the clock is 24-hour clock] or 12*60 [If the clock is 12-hour clock]
They will show correct time in = \(\frac{24*60}{gcd(30,15,10,5)} = \frac{24*60}{5}= 288days\) [ If all the clocks are 24-hour clock]
They will show correct time in = \(\frac{12*60}{gcd(30,15,10,5)} = \frac{12*60}{5}= 144days\) [ If all the clocks are 12-hour clock]
Hi nick1816, hope you are doing great!
Needless to say, kudos for the concise solution again! I couldn't solve this under timed conditions and had to take an analytical approach for this.
a:b:c:d = 6:3:2:1
Now, we can draw the analogy between the clocks and people going around a circular track. We need to determine the pair of clocks with the least number of overlaps. Since, clocks C and D are moving in the same direction and have the least ratio in terms of time difference, thus we need to find the points of overlaps for clocks C and D.
Number of overlap points = difference of their ratio = 2-1 = 1 => The starting point is the only overlap point i.e. mid-night or noon.
This is where it gets tricky. It's not only the minute hand that has to coincide, but also the hour hand. The minute hands of C and D will next coincide when clock D has gained an hour with respect to the correct time. Thus, the number of days = 60/5 = 12. But at the end of 12 days, the times in clock C and D will be 02:00 AM (clock C gains twice as much time as clock D) and 01:00 AM respectively. Now the hour hands need to coincide. Since, the ratio is same for the hour hands, they will coincide after 12 more days.
Thus, the total number of days after when all the clocks will show the same time = 12*12 = 144 days.
Regards
Sovan
