Aptitude - Time and Work - Discussion

In the case of A,

80 %. = 20 days
So
100% = ?

Cross multiplication
= 100 *20 /80
= 25 days.

So, A has the power to do daily;
100/25 = 4 % of the work.

As given
80 % of the work is done by A,
for the Remaining 20% of work, he calls B to finish work in 3 days.

In 3 Days, 'A' worked by using his power of 4% and did 4* 3 = 12

The remaining 20 -13 = 8 % will be completed by 'B' in 3 days
B' will work,
8 %. = 3 Day
100% =. ?

Crossing
= 100* 3/8
= 75/2,
= 37.5
And the answer is 37 and a half means i.e. 37 1/2.

A completes 80% of the work in 20 days.
This means A does 4% of the work each day (80% ÷ 20 days).
In 3 days, A would complete 12% of the work (4% × 3 days).
The remaining work, 8%, is done by B in 3 days.

To find out how long B would take to complete the entire work (100%), we calculate it as follows:

If B does 8% of the work in 3 days, then B's work rate is 8% ÷ 3 days = 2.67% per day.
Therefore, to complete 100% of the work, B would require:
100% ÷ 2.67% per day = 37.5 days.
So, B would take 37.5 days to complete 100% of the work.

@All.

Here A do 80% work in 20 days.

That mean remaining work = 20%.
On diving 80 by 20 = A do 4% work in 1 day
So the remaining 20 % has to be done in 20/4 =5 day.

Now, total days were taken by A = 20 + 5 = 25.
Now, ACC to question;

A+B Takes 3 days to complete the remaining work so 5 * 3 = 15 days.
Now;
A= 25,
A+B=15.

On taking LCM we get 75.
So now by dividing,
A's efficiency = 75/25 = 3,
A+B efficiency= 75/15=5.

Now B efficiency = 5-3 = 2 bcz A+ B= 5 nd A=3.
Now dividing B efficiency from total working days = 75/2 = 37.5.

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.

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.

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.

Solution by LCM Method:

A does 80% of the work in 20 days => 4/5th of the work in 20 days
=> (4/5)x = 20 => x=25 days (No of days A takes to complete the work)

Remaining work = 1/5 which is completed by A+B in 3 days.
(1/5)x = 3 => x=15 days (No of days A+B takes to complete the work).

Therefore,
A - 25 days - 3 units.
A+B - 15 days - 5 units.
LCM - 75.
A+B = 5 => B=2 units.
Therefore, total work done by B = 75/2.
= 37 1/2.

A can do 80% of work in 20 Days.

So, 20 days = 80%,
1 day = ?
= 4%,

A and B worked for 3 days so 4% A can do in 1 day and in 3 days it is 3*4=12%.
A and B together worked and completed the remaining 20% of the work in 3 days which A work is 12%.

So, A+B =20%,
12% +B =20,
B = 8%,
B worked 8% in 3 days,
3 days = 8%,
1 day = ?
=2.66%.

The whole work is 100% in itself,
B 100/2.66 ==37.5 or 37 1/2.

80% = 20 days.
100% = 20x100/80 = 25 days.

i.e. A alone can do whole work in 25days, A's one day work = 1/25.

But A worked 20 days (it mean 20/25 work completed)remaining 5/25.

A+B worked for 3 days to complete the remaining 5/25 part of work.

A's 3 days work = 1/25+ 1/25+ 1/25 = 3/25.

Remaining 2/25 part of work is done by B in 3 days.

3 days----2/25.
1 day-----?.
1 day----2/(25*3) = 2/75.

Total work done by B is 75/2 = 37 1/2.

(no of days required A to complete 100% work + no of days required B to complete 100% work) * no of days
= no of days required complete whole work by A+B -----> 1
A per-day work = 20/80 = 1/ 4 meaning 4 units/day.

Hence, no of days required A to complete 100% work = 100/4 = 25 days.
No of days are required to complete the whole work by A+B = 100/20 = 5.
The sub in 1.
(1/25 +1/B) *3 = 1/5.
3/B = 2/25.
B = 25*3/2
B = 37.5