D01-30

Official Solution:

Painter A can paint a house alone in 20 days, while Painter B can do it in 30 days. They began painting a house together, but at some point, A left for a number of days before rejoining B to finish the job. B worked alone for 5 days, and then A and B together completed the remaining work in 4 days. How many days did A and B work together before A initially left B?

A. 4
B. 5
C. 6
D. 7
E. 8


Painter A has a rate of \(\frac{1}{20}\) job/day, while Painter B has a rate of \(\frac{1}{30}\) job/day.

The combined rate of A and B working together is \(\frac{1}{20}+\frac{1}{30}=\frac{1}{12}\) job/day.

During the 5 days that B worked alone (second stage), B completed \(5*\frac{1}{30}=\frac{2}{12}\) of the job.

In the 4 days that A and B worked together (third stage), they completed \(4*\frac{1}{12}=\frac{4}{12}\) of the job.

Therefore, during the first stage of work, A and B together completed \(1-(\frac{2}{12}+\frac{4}{12})=\frac{6}{12}\) of the job. At their combined rate of \(\frac{1}{12}\) job/day, it took them 6 days to complete this portion of the job.


Answer: C
_________________

I prefer assuming total work before solving such questions. Helps me to avoid fractions.

ALONE
A finishes work in 20 days
B finishes work in 30 days

Taking LCM of 20 and 30 (Or any convenient common mulitple) --> 120
So if total work is 120, then to finish 120 units of work as per their current rate
A finishes work 120/20 = 6 units per day
B finishes work 120/30 = 4 units per day
Both together finish 6 + 4 = 10 units per day.

Coming back to what's asked -
B alone worked for 5 days, so 4*5 = 20 units of total 120 units is done
Together they worked for 4 hours before finishing the work = 10*4 = 40 units.
We've got 40 + 20 = 60 units done.
120-60 = 60 units pending.
This work is nothing but the work that A & B did together before A left B alone to suffer.
Hence 60/10 = 6 days!

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. Great question. I freaked out when I first faced it but was able to solve it within 2 mins! Thanks

I will go with this approach,

rate of A= 1/20 job per day
rate of B = 1/30 job per day

when they work together the no of days required to complete the task = 1/20 + 1/30 = 3+2/60 = 5/60 = 1/12

when A and B were working together let us assume that after X days A left the working while ,
so the work completed in X days by both of them = x * 1/12= x/12

so the work remaining = 1-x/12 = 12-x/12

Now B worked for 5 days , so total work done by B in 5 days = 5*1/30 = 1/6

now the total work remaining = total work remaining after x days - work done alone by B in 5 days
= 12-x/12 - 1/6
= 10-x/12

now A rejoined the work and they both completed the remaining work in 4 days
remaining work = 4 days of work for A and B
10-x/12 = 4/12
10-x = 4
x= 6

so the no of days after which A left working together is 6

let for x days A worked with B before he left
Now
A worked for X+ 4 days
B worked for X+4+5 ie. X+9 days.

(X+4)/20 + (x+9)/30 =1
solve this you will get X=6

Given from the prompt:

A & B worked together + B worked alone + A& B joined again to complete work

Combined rate = 1/20+ 1/30 = (3+2)/60=5/60= 1/12 job/day

Let T= time for both A & B together before A leaving B

To solve the question:

Use the standard formula

Work = rate * time

Following the sequence stated by prompt

T *1/12 + 5 *1/30 + 4 *1/12 = 1 .............( the work here is one house)

T /12 + 1/6 + 1/3 = 1

T /12 + 3/6= 1

T /12 = 1/2 .......T=6

Answer: C

I think this is a high-quality question.

I think this is a high-quality question and I agree with explanation.

Let's pick up a Number which is the LCM of both of A and B's rates, for eg. 60.
Here, let 60 be the total units of work to be done.

So, if A takes 20 days to finish 60 units, it means he accomplishes 60/20= 3 units per day.
Similarly, if B takes 30 days to finish 60 units, he accomplishes 60/30= 2 units per day

Now, the total work done by them in 1 day= 3+2= 5 units.

For the 5 days B worked by himself, he gets 5*2 units= 10 units done.
The 4 days A and B work together in the end, they get 4* 5 units= 20 units done.
Total work done= 30 units.
Leftover work= 60-30 units= 30 units. This is the work that A and B must have done together, before A left.

So, given the combined rate, A and B together must have done 30 units in 30/5= 6 days!

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
_________________

I think this is a high-quality question and I agree with explanation. I agree with the explanation just I want to point out that the work done in second stage, can you explain why you considered 5*1/30= 2/12

In the second stage, B worked alone for 5 days at the rate of \(\frac{1}{30}\) job/day, hence completed \(5*\frac{1}{30}=\frac{2}{12}\) of the job.
_________________

I think this is a high-quality question and I agree with explanation.
_________________