Aptitude - Time and Work - Discussion

80% = (80/100).

so, A's (80/100) work done in 20 days.

Total Work of A = (100/80 * 20) = 25 //
Now if we consider total Work as 1.
4/5 part was already done by A remaining 1/5 he completed with the help of B.

So, (1/5) of work done by (A+B) in 3 days,
by calculating we get (A+B) = 3*(5/1) = 15 days,

Now to find B's Work we have A+B=15.
substitute A value and in above equation i.e (A+B = 15).

so
1/25 + 1/B = 1/15.
1/B = (1/15) - (1/25) ---- (LCM = 75).
1/B = (5-3) / 75,
1/B = 2/75,
B = 75 / 2 = 37 1/2.

Please, anyone, help me to get it clearly.

For A:
80% work = 20 days.
4/5 work = 20 days.
cross multiply.
1 work = 5/4 * 20 days.
1 work = 25 days.
A's 25 days = 1 work ------>1.

A+B remain work done in 3 days.
total work = 1,
80% work = 4/5,
100% work = 1,
100% - 80% = 1 - 4/5 = 1/5.

For A+B:
Work = days.
1/5 work = 3 days.
Cross multiply
1 work = 15 days.


And A+B's 15 days = 1 work ---->2

Now we know A's and A+B's work
Now find B's work
Apply A's in A+B's.

A+B 1 Days= 1/15 works
A's 1 days = 1/25 works---->from 1..

1/25 + B = 1/15 works.
B's days = 1/15 - 1/25 Works.
B's days= 2/75 works.
we have to calc works.
cross multiply.
B's work = 75/2 days = 37 1/2 days.

Whole work done by A is?

In 20 days 80%
In how many days 100%
80% in 20 days.

100% in x days ---- cross multiply.
100 * 20/80 = 25 days.
A completes work in 25 days.

Now A+B
20% in 3 days
100% in x days ---- cross multiply.

100 * 3/20= 15 days.
A+B complete work in 15 days.
now A's one-day work = 1/25.
A+B one day's work = 1/15.
now we want B,
B=(A+B)-A,
=(1/15)-(1/25)=>2/75............................B's one day's work,

Work done by B is (reciprocal of B's one day work) 75/2 days

now divide 75/2.

75/2=37 and the remainder is 1.
hence answer is 37*(1/2)days.

A=80%
Remain 100 - 80 = 20%.

A = (1/20) * (80/100) = 1/25.
A+b = (1/3)+(20/100) = 1/15.
A+b.
(1/25) + b = 1/15.
B = 2/75,
B = 75/2,
B = 37(1/2).

Here;
20 is 80 %,
25 is 100%,
A+B = 3 days this is 20 %.
20% * 5 = 3 * 5 for 100% work,
B= A+B- A.
B=1/15- 1/25,
B = 2/ 75 = 37.5.

A finish 80% work in 20 days.
A finish 100% work in 20/80*100 = 25 days,
A+B finish the remaining 20% of work in 3 days,
A+B finish 100% work in 3/20*100 = 15 days.

Now A finishes the whole work in 25 days and A+B finishes the same work in 25 days.
Now total work is LCM of 15 and 25 =75

A's work of one day =75/25 = 3.
A+B work of one day=75/15 = 5.
B's work of one day= (A+B)-A = 2 - 3 = 2.

Now the time is taken by B to complete the whole work= total work/B's per day work =75/2 = 37.5.

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.

Hey Guys! Let's view this differently,

|80% = 80/100 = 4/5|
|20% = 20/100 = 1/5|

As/Q,
4/5 A = 20 => A = 25 --> 3 <-- Efficiency of A {as, TW/25=3}
LCM (25, 15) = 75 <-- Total Work, TW(say).

1/5 (A+B) = 3 => (A+B) = 15 --> 5 <-- Efficiency of A+B {as, TW/5=5}.

Since the Efficiency of A is 3 and A+B is 5, this implies, the efficiency of B is 2.

Now,
TW=75 and B's efficiency is 2,
So, no. of days needed by B is, 75/2 = 35.5 <--answer.

Your explanation is short and good.

Thanks @Ajith.