Aptitude - Time and Work - Discussion

Thanks for the answer @Bala.

Since we are given with the condition -
A=B+C ------1 (eqn).
C's 1-day work = 1/50.
(A+B)'s 1-day work = 1/10.

Therefore, using eqn 1.
A+B = 2B+C,
1/10 = 2B+ 1/50,
2B=40/500,
B=1/25.
B alone can finish work in 25 days.

Hope this method will help you understand this question easily.

Perfect @Shweta. Thanks.

A + B = 10.
C = 50.

Take lcm of 10, 50 which will give total work.
So A+B=50/10=5 ------> (1)
C=50/50=1 -----> (2)
We also know B+C=A ----> (3)
On solving 1,2 & 3.
B=2 unit.
So work done b alone = 50/2 = 25.

1/A+1/B=1/10 ------------> 1

GIVEN
1/A = 1/B + 1/C -------------> 2
1/A-1/B = 1/C from eq2.
1/A-1/B = 1/50.

Now solve 1 and
1/A=6/100 put this value in eqn 1.
We get 1/B = 4/100.
So, B can do in 25 days.

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.

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).