Aptitude - Time and Work - Discussion

@Ankita.

Follow this method.
Here total work is 12 hours as lcm of 2-4-3.
A does 3 work in 1 hour ND he finishes 12 work in four hours.
In the same way, abt A and C you know A does 3 work in 1 hour thus C also does 3 work in 1 hour so total work 6 in 1 hour thus 12 work in 2 hour is completed by A and C.

And now C does 3 work in 1 hour so b does 1 work in 1 hour and thus the both do 4 work in 1 hour and finishes in 3 hours.

As a result, B takes 1 hour to do 1 work so B Will take 12 hours to do 12 work.

You explained well @Akshay.

How it will 7/12-1/2=1/12?

Thank you so much @Himanshi Mishra.

I was looking for this LCM method. This is way too easy. Thanks.

Guys!.

I simply subtracted the (A's) work - (B's+C's)work
so I got,
1/4-1/3=-1/12 for 1hour.
hence for b to do alone his work 12 hour.

A in 4hrs.

B+C _ 3.
A+C _ 2.
Total work by taking lcm of 4,3,2 is 12units.
A in one day do = 12/4= 3.
A+C_ 6.
So C=3.
In the same way,
B+C= 12/3 > 4.
So B=1.
At last B =12/1=12. That's the answer!.

Clear and detailed explanation. Thank you all.

A=B+C(given) ----> (1)
one day work of A+B=1/10 --> (2)
one day work of C=1/50 ---> (3).

So for total (A+B+C)'s 1 day work, we can assume from 1--> C=A-B=1/50----> (4)
So adding equation 2 and 4 we get A=3/50.

Substituting above A in equation 2 we get B=1/25.
Hence 25 days because(If A's 1 day's work =1/n then A can finish the work in n days.)).

Well said @Hindu.

Thanks for your answer @Bits.