Aptitude - Time and Work - Discussion

Here, 1 day's work of a and b is(50/10) = 5.
1 day work of c is(50/50) = 1
a+b+c=(5+1)=>6->(1).
Given that a's time=sum of b's and c's time so,a=b+c
we can write a=b+c as b=a-c and then substitute this in (1)
that is a+b+c=6.
a+a-c+c=6=>2a=6->a=3.
Then later in that (1).
a+b+c = 6.
b = 6 - a - c,
b = 6 - 3 - 1,
b = 2.
That is 50/2=>25days.

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.

A+B = 1/10 --> i
A = B+C --> ii
C = 1/50 --> iii

Subs (ii) in (i)
B + C + B = 1/10.
C + 2B = 1/10.
subs C = 1/50.
1/50 + 2B = 1/20.
2B = 1/20 - 1/50,
2B = 2/25,
B = (2/25) x (1/2),
B = 2/50,
B = 1/25.

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 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.

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.

A = B+C
A+B = C
10 50
LCM 50.
5 : 1.
A+B = 5 Put value A.
2B + C = 5.
2B = 4 , B = 2,
B = 50/2,
B = 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.