Aptitude - Time and Work - Discussion

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.

A = B+C ,
A - C = B ---> (i).

Then,

A + B + C = 1/10 + 150 = 3/25.
From (i) putting a value of B
A+B+C= 3/25,
A+A - C + C = 3/25.
2A = 3/25.
A = 3/50.
Now, A+B = 1/10,
Putting A = 3/50,
B = 1/10-3/50.
B = 1/25 (B one day work) then total days is 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.

Time taken by A and B =10 days and Time taken by C =50 days

Total work=50 units

Efficiency of A and B = 5
The efficiency of C =1

A=B+C is given

A+B=5 => B+C+B=5 => 2B+C=5
2B+1=5 =>2B=4
B=2 is the efficiency

Time taken by B= work /efficiency of B =>50/2 =25.

Therefore, B can complete the work in 25days.

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.

If A = B+C.
So A-B =C.

i,e A-B = 1/50 ----> eq 1
And
A+B = 1/10 ----> eq 2.
From eq 2 - eq 1 we will get the answer.
.

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.

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.

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.

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.