Question on Faulty clock 
Clock A loses 4 minutes every hour, clock B always shows the correct
time and clock C gains 3 minutes every hour. On a Monday, all the
three clocks showed the same time, 8 PM.  On the following Wednesday,
when the clock C shows 2 PM, what time will clock A show?


*

*7:20 am

*8:40 am

*9:20 am

*10:40 am

Kindly suggest how to approach this problem.
My approach
There is a lapse of 42 hours from 8 pm of Monday to 2 pm of Wednesday. As clock C gains 3 minutes every hour, the actual lapse should be 42 hours - (42×3) minutes. Further clock A gains loses 4 minutes every hour, so the lapse of time in clock A should be 42 hours - ((42×3)-(42×4)) minutes.
I couldn't proceed , also I am not sure if it's a right approach or if there's a better way to do this.
Kindly help, thanks in advance.
 A: We have three "times" we can observe, one for each clock. If we knew the time on clock B, we could work out what time clocks A and C would show, but instead we're told the time on clock C and need to work out what time clock A shows.
As you note, if clock C reads 2pm then that means it is claiming that 42 hours have passed. So how many hours have actually passed?

*

*If 1 true hour passed, then C will show 1 hour 3 minutes as having passed.


*If 2 true hours passed, then C will show 2 hours 6 minutes as having passed.


*If $t$ true hours passed, then C will show $t$ hours and $3t$ minutes having passed. Or in other words, C will show $t \times 1 \frac{3}{60}$ hours passing.
Then if C showed 42 hours passing, then that means $t \times 1 \frac{3}{60} = 42$. Can you take this and rearrange it to find what $t$ is equal to?
Once you've done so, you can apply the same logic to finding the time on A - every hour that passed, A claims that only 56 minutes have passed. So if you know that $t$ hours have passed, A should show a time that is $56t$ minutes later.
A: Let the actual time elapsed since $8 \text{ pm} $ Monday be $ t $.
The time reported by clock $A$ is $\dfrac{56}{60} t $ while the time reported by clock $C$ is $\dfrac{63}{60} t$.
Hence, from the time on clock $C$, we know that
$\dfrac{63}{60} t = 42 $
Hence,  $ t = \dfrac{(60)(42)}{63}  = 40 $
Thus the time reported by clock $A$ is
$ \dfrac{56}{60}(40) = \dfrac{112}{3} \text{Hours}$
Deducting $24$ gives a reported time difference of $\dfrac{40}{3} = 13 \dfrac{1}{3} \text{ Hours } $
Finally, $13 + 8 \text{p.m.} = 9 \text{ a.m. }$
Therefore, the time appearing on clock $A$ is $9:20 \text{ a.m.}$
A: In $1$ hour, clocks A and C complete $\displaystyle\frac{56}{60}$ and $\displaystyle\frac{63}{60}$ revolutions, respectively.
Thus, they complete $1$ revolution in $\displaystyle\frac{60}{56}$ and $\displaystyle\frac{60}{63}$ hours, respectively.
When clock C displays $2$pm on the Wednesday, it has completed $42$ revolutions; this has taken $\displaystyle42\times\frac{60}{63}=40$ hours; concurrently, clock A has completed $\displaystyle40\div\frac{60}{56}=37\frac13$ revolutions; clock A then displays $8$pm$+\displaystyle37\frac13$ hours$=9{:}20$am.
