Aptitude - Time and Work - Discussion

A's 80%=20days.

So we want 100% of A's Work
Therefore
A*(80/100) =20
A*(4/5) =20
A =(20*5) /4
A =100/4=25
So A's 100% of work finished in 25 days. Therefore A=1/25 ...... (1)

Now remaining (100,.-80%) =20% work done by (A+B) in 3 days

(A+B) 's 20%=3days

So we want 100% of (A+B) 's work

(A+B) *(20/100) =3
(A+B) *(1/5) =3
(A+B) =3*5
=15 days

so (A+B) finished in 15 days
therefore
(A+B) =1/15...........(2)

From equation (1) & (2)
(A+B) =1/15
(1/25) +B=1/15 [A=1/25]
B=[1/15]-[1/25]
B=[5-3]/75 [75/15=5 & 75/25=3]
B=75/2
B=37(1/2) days [remain 1/2]

So do it in notebook otherwise you can't understand.
Take the left side of the notebook & derive A's 100% of work and right side of the notebook derive (A+B) 's 100% of work.

Then the middle side of the notebook solves these A's and (A+B)'s. You get B's 100% of work. that is the answer.

now check it
(1/25) +(2/75) =1/15

5/75=1/15 [(75/25) =3 & (75/75)=1]
1/15=1/15.
LHS=RHS.

Thanks.

Thanks @Karan.

Good, Thanks @Keerthan G.

Here the answer in simple steps:

A 's work in 20 days = 80/100 = 4/5

Then, A's 1 day work is = 4/5*20 =4/100=1/25.

(Here we simply multiply the 20 in the denominator)

So,

A+B 's 3 days work = 1/5
A+B's 1 day work = 1/5 * 3 = 1/15.

(Here the 1/5 comes as, in Factional total work is 1 so when we subtract the A s 4/5 from 1 we get the remaining work which is going to complete by A and B is 1/5).

So we got A's work and A+B 's work when we subtract the both we got B's time to complete it
1/15-1/25 = 2/75 it is an efficiency when you reciprocal it, we got 75/3 that is in 35.5 or 35 1/2 days.

Finally, I understood the concept, Thanks @Rajendra.

@Rajendra.

Thank you for explaining.

This can also be solved in this way.
A - 80% work in 20 days .
=>1/A=1/20 (1 day work of A) ---> 1

Also Given remaining 20% work done in 3 days by help of B.
Let us calculate days if they work for other 80%.

20% = 3 days.
80% = x.
x = 12 days.
=> 1/A + 1/B = 1/12.

Substitute value of 1/A from eq 1
then B = 30

=> 80% = 30 days,
100% = x.
By calculating x = 37.5.
So, B alone can finish work in 37.5 days.

ELIMINATION METHOD (SUPER FAST TRICK).

A completes 80% of the work in 20 days.
A's 1-day work = 80% ÷ 20 = 4%
A's 3-day work = 4% × 3 = 12%
Remaining 20% is shared after A's contribution:
B's 3-day work = (100% − 80%) − 12% = 8%
B's 1-day work = 8% ÷ 3 = 2.67%

Checking total days for B:
2.67 × 40 = 106.8% (Not possible)
2.67 × 37 = 98.79% (Not exactly 100%)
2.67 × 23 = 61.41%
2.67 × 37.5 ≈ 100%
Therefore, B alone can complete the work in approximately 37.5 days.

Sorry I can't get it why we should multiply with that 5/4 ?

A does 80% in 20 days..so
A does 100% in 25 days..
A's 1 day work=1/25
A+B does 20% in 3 days..so
A+B does 100% in 15 days
A+B's 1 day work=1/15
1/25+B=1/15
B=1/15-1/25
B,s 1 day work=2/75
so B,s work=75/2 i.e 37 1/2 days..