Aptitude - Time and Work - Discussion

A = B+C.
A+B->10.
C->50.

We can write C as A-B.
Therefore, A-B->50.
LCM of 10 and 50 is 50.

As we know that efficiency x time = work ------> (1)
Here work = lcm of 10 and 50 = 50

Therefore the efficiency of A+B is 50/10 = 5(use formula 1)
The efficiency of C i.e A-B is 50/50 = 1,
Solving both equations we get A=3 B = 2.

Now time is taken by B is=50/2 =25 (by using formula 1).
So, the time taken by B alone is 25days.

Easiest method:

A = B+C ---> (1)
A+B = 1/10 ---> (2)
C=1/50 ---> (3).

From (2) A = 1/10-B.

By substituting (2) & (3) in (1), we get;

1/10-B = B + 1/50.
Solving for B.

2B = 1/10 - 1/50 = 1/25.

a+b, +c can finished 50 work.
a+b finished in 5 days,
c finished in 1 day,
can say one day work (a+b+c) in 6 days,
a= b+c put,
a+(a) =6,
a=3.

Suppose A's one day work=1/x=(B+C) one day work.
A+B one day work=1/10,
C's one day work=1/50,
B's work=?
B's one day work= (A+B)'s one-day work-A's one day work.
=1/10-1/x.

Again, for B's one day work=(B+c)'s one-day work- c's one day work.
= 1/x-1/50.
by comparing these, we get;
1/10-1/x = 1/x-1/50,
by solving this we get x=50/3,
Hence A's one day work=3/50,
B's one day work=1/10-3/50,
B's one day work= 1/25,
So, B alone can do it in 25 days.

@Saurabh.

Thanks for the solution.

Awesome method thanks @Awoke.

We know A+B= 10 ----> 1
C=10 ----> 2
GIVEN;
A=B+C ----> 3.

JUST PUT 3, 2 IN 1
B + C + B = 10
2B + 60 = 10
B = -50/2
B = 25 (IT BECAME POSITIVE BECAUSE DAY IS NOT In NEGATIVE).

A = b+c.

A+b = 1/10,c=1/50
B+C+B = 1/10.

2b = (1/10)-(1/50).
2b = 2/25,
b = 1/25,
So,b = 25.

Thank you @Awoke your method was easy.

Thank you @Sneha you made the given solution easier to understand.