Aptitude - Time and Work - Discussion

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

Substitute C in eq 1.
A=B+ (1/50).
From eq 2 we get A= (1/10) -B.
Substitute this in above eq.

(1/10)-B=B+ (1/50).

Solve it.

Thank you @Awoke your method was so easy.

How come 1/10, Can you please explain it?

This problem we can do it in a simple way.

(A+B)'s 1 day's work = 1/10;
C's 1 day's work = 1/50;
A's 1day's work = (B+C)'s 1day's work;
i.e. A+B = 1/10;

(B+C)+B = 1/10;
2B+C = 1/10;
2B = (1/10) - (1/50);
2B = 4/50;
B = 2/50;
B = 1/25;

So, B alone can do it in 25 days.

Thanks @Sneha.

You explained it well.

A = B+C( A work is equal to B and C work)---> 1
A+ B = 10 days (A and B can do a work in 10 days)--> 2
C=50 days( C alone do work in 50 days)----> 3.

Let's assume work as a 50 parts.(take LCM of 10,50).

A+B=5parts(A anb B can do 5 parts a day work)--> 4.
C=1 part(C can do work 1 part a day I,e 50 parts 50 days)---> 5.

From equation 1: A=B+C
A = B + 1(from eq 5).
3 + 2 = 5(from eq 4).
B = 2 parts.
50 parts of work can do in 25 days. B can do parts a day.

A = B + C,
A+B = 10 or A and B's one day work A+B=1/10.
C=50 or C=1/50.

A+B=1/10 (Since A=B+C),
B+C+B=1/10,
2B+1/50=1/10,
2B=1/10-1/50,
2B=2/25,
B=2/25*1/2.
B=1/25 (Which is one day's work).
B would take 25 days to complete certain work.

They gave A = B+C, A+B = 10Days, C=50 Days;
A+B = 1/10;
since A = B+C;
B+C+B = 1/10;
2B+C = 1/10;
2B+1/50 = 1/10;
therefore B = 1/25;
B can complete in 25 days.

1/A+1/B=1/10 ---> EQN1.

Given that,
1/C = 1/50,
1/A=1/B+1/C ---> EQN2.

From EQN1 and EQN2 we get,
1/B + 1/B+1/C = 1/10,
2/B + 1/C = 1/10,
2/B + 1/50 = 1/10,
2/B = 2/25,
1/B = 1/25.

Hence B alone can do it in 25 Days.

Total work is = 50
C's efficiency = 50/50=1.
A+B's Efficiency = 50/10=5.
A+B+C's Efficiency = 5 + 1 = 6.

A = B+C,
B+C+B+C=6,
2(B+C)=6,
B+C= 6/2 =3,
A's Efficiency = 3,
So B's Efficiency =5-3 =2.
Hence B alone do it in 50/2 = 25 Days.