Aptitude - Time and Work - Discussion

Given : AB=10; C=50; find B=?;

LCM of 50, 10 is "50"(total work),
Efficiency of 'AB' =total work : 50/10 = 5,
Efficiency of 'C' =total work : 50/50 = 1.

From efficiency of 'AB & C' is ---> A+B+C : 5 +1 = 6.
Question itself given that : A=B+C;
We know A+B+C = 6--->(1) , in question they said A=B+C -----> (2) ;
Substitute (2) in (1):
We get (1).
B+C+B+C =6;
2B+2C=6; (w.k.t efficiency of C = 1)
2B+2(1)=6;
2B+2=6;
2B=6-2 ; ---> 2B=4; ---> B=2;
Hence, total B alone worked : (50)/2 = '25' days.

Great, Thanks everyone for explaining.

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.

a = b+ c ---> (1)
a+ b = 10 = 1/10 for 1 day work ---> (2)
c = 50 = 1/50 for 1 day work
b = 1/10 - a ---> (3)---from eqn(1)
c = 1/50.
a = b +c ---from (1)
a = 1/10 -a + 1/50 ---from (3)
a = 3/50.
b = 1/10 - a = 1/10 -3/50 = 1/25 .
b = 25 days.

A = B+C.
A + B = 1/10,
2B + C = 1/10,
C = 1/50,
B = 1/25,
B = 25hrs.

A+B can finish work in 10 days so its efficiency is 5 unit per day.
c can finish work in 50 days so its efficiency will be 1 unit per day.
So from here, we can calculate the total work, which will be 50 unit.
now according to the question,

Efficiency of A =efficiency of B+C
Adding B both sides,
A+B=2B+C
5=2B+1.

Therefore the efficiency of B will be,2 unit per day.

Now since the total work is 50 unit,we can calculate the time,
Time = total work/efficiency
Time=25 days.

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.

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.

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.

1/A = 1/B + 1/ C -->1
1/A + I/B = 1/10 -->2
1/C = 1/50 -->3

Subs 1 in 2.

1/B + 1/C + 1/B = 1/10 -->4

Subs 3 in 4.
2/B = 1/10 - 1/50.

2/B = 4/50,
B = 25.