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I have two clocks - table clock and wall clock

A table clock gains 2 mins every 12 hours and a wall clock loses 1 min every 12 hours both are set at 12 noon on tuesday(date is not known ) we need to tell after how many days would they be togeather (just like they were at 12 noon on tuesday) and what would that time be?

Thanks,

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Please share you idea how start/ your solution (partial/incomplete) –  Ram Jan 13 '13 at 4:30
    
This was asked in SSC exam? –  iostream007 May 16 '13 at 23:45

2 Answers 2

Let the two clocks meet after $x$ hours on a proper clock. The faster clock will show that $x + \dfrac{x}{360}$ hours have elapsed, while the slower clock will show that $x - \dfrac{x}{720}$ hours. Hence, for them to meet together, if both are a $24$ hour clock (say like a digital clock showing AM/PM), we need $$\dfrac{x}{360} + \dfrac{x}{720} = 24 \implies \dfrac{x}{240} = 24 \implies x = 5760 \text{ hours} = 240 \text{ days}$$ If both are $12$ hour clocks (say like an analog clock), we need $$\dfrac{x}{360} + \dfrac{x}{720} = 12 \implies \dfrac{x}{240} = 12 \implies x = 2880 \text{ hours} = 120 \text{ days}$$

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the time when they will meet would be –  user1838130 Jan 13 '13 at 4:25
    
They both will meet at $12$ noon. –  user17762 Jan 13 '13 at 4:27
    
x+x/360 the logic isnt clear here –  user1838130 Jan 13 '13 at 4:30
    
The first clock gains $2$ minutes for every $12$ hours i.e. gains $1$ hour every $30 \times 12 = 360$ hours, hence after $x$ hours, it gains $x/360$ hours. Similarly, for the second clock. –  user17762 Jan 13 '13 at 4:32

[You have to be careful with what it means to gain 2 mins every 12 hours. I'm assuming the 12 hours is in normal time, and not in the table clock time.]

They will meet again when their change in time gained/loss is equal to 12 hours. This means that the table clock has gained 8 hours, and the wall clock has lost 4 hours. This happens in $ 4 \times 60 \times 12$ hours, which is 120 days exactly. It will be a Wednesday (since $120 \equiv 1 \pmod{7}$).

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