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Clock related speed distance problem 2

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Clock related speed distance problem 2 [#permalink] New post 21 Sep 2009, 12:51
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Q.
A wall clock gains 2 mins in 12 hrs, while a table clock loses 2 mins in 36 hrs; both are set right at noon on Tuesday.The correct time when they both show the same time next would be
a. 12.30 night
b. 12 noon
c. 1.30 night
d. 12 night
Manager
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Re: Clock related speed distance problem 2 [#permalink] New post 22 Sep 2009, 13:04
got A – 12.30 night

My logic was: Minutes - distance, hours - time

So: wall clock speed: 2/12 or 1/6, table - 2/36 or 1/18.

Total distance they travel towards each other is limited by 60 min.

Need to know the number time to travel for each. 1/6x + 1/18x = 60 2/9x=60 x=540/2 x=270 Now divide 270 by 24 =>11,5 days, so 12.30 at night Not sure 100% in it though, so comments are appreciated :)
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Re: Clock related speed distance problem 2 [#permalink] New post 24 Sep 2009, 04:13
According to me
The wall clock gains 6 mins in 36 hrs;
The table clock loses 2 mins in 36hrs;
therefore, the time differential in 36 hrs = 8 mins..for the clocks to show same time time differential has to be 12 hrs
after 3240 hrs they will give time differential 12 hrs; and as 3240 is perfectly divisible by 24...ans is 12 noon

arkadiyua wrote:
got A – 12.30 night

My logic was: Minutes - distance, hours - time

So: wall clock speed: 2/12 or 1/6, table - 2/36 or 1/18.

Total distance they travel towards each other is limited by 60 min.

Need to know the number time to travel for each. 1/6x + 1/18x = 60 2/9x=60 x=540/2 x=270 Now divide 270 by 24 =>11,5 days, so 12.30 at night Not sure 100% in it though, so comments are appreciated :)
Manager
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Re: Clock related speed distance problem 2 [#permalink] New post 17 Oct 2009, 19:59
so what is the answer guys??

Rohit
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Re: Clock related speed distance problem 2 [#permalink] New post 17 Oct 2009, 23:31
:|
virtualanimosity wrote:
According to me
The wall clock gains 6 mins in 36 hrs;
The table clock loses 2 mins in 36hrs;
therefore, the time differential in 36 hrs = 8 mins..for the clocks to show same time time differential has to be 12 hrs
after 3240 hrs they will give time differential 12 hrs; and as 3240 is perfectly divisible by 24...ans is 12 noon



I may sound stupid here but if the time difference is 12hrs after 3240hrs (135days) then the answer should be 12night considering both clocks were set to 12noon Tuesday.
Can someone please explain this :?: :? :( :|
Re: Clock related speed distance problem 2   [#permalink] 17 Oct 2009, 23:31
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