Aptitude - Time and Work - Discussion

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

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 ---> (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.

Given:

A = B + C.
A + B = 10,
C = 50.

Solution :

A + B = 10.
(A + B)'s one day way is 1/10.
A + B = 1/10 ---> 1
from A = B + C ---> 2

Substitute eqn 2 in 1;
B + C + B = 1/10,
2B + 1/50 = 1/10,
2B = 1/10 - 1/50,
2B = 4/50,
B = 1/25.
B alone can do the work in 25 days.

1/A + 1/B = 1/10
As A and B + C IS SAME

1/A = 1/B+1/C. (1/C=1/50 AS GIVEN IN QUES)

Equate both equations
1/A + 1/B = 1/10.
(1/B + 1/C) + 1/B = 1/10,
2/B + 1/50 = 1/10,
B = 25.

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.

A = B + C.

Lcm of 10 and 50 = 50 ( Total Work)
Eff of A + B = 5 --->1.
Eff of C = 1.
A = 5- B from 1.
A = B + C.
5 - B = B + 1.
B = 2.
Time= TW/Eff of B.
= 50/2 = 25.

A = B + C.
LCM of a+b and c =50,
Efficiency of A+B =5 c=1 than A+B+C=6.

A + B + C = 6 BUT A = B + C
A + A = 6
A = 3.
Then B = 2.

Ans = 50/2 = 25.

A = B+C,
A+B = 10,
C = 50.


A = B+C=> A = B+1/50.

A+B = 1/10 => B+1/50+B = 1/10.
2B + 1/50 = 1/10.
2B = 4/50.
B = 1/25.
25 DAYS.

Short and Simple:

a+b=10, c=50, total work 50.
perday a + b = 5, c = 1.
a+b = 5,
b+c+b = 5,
2b+c = 5,
b = 2/d.
b = 50/2 = 25.