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D01-30

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D01-30  [#permalink]

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New post 16 Sep 2014, 00:12
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Question Stats:

76% (02:25) correct 24% (02:44) wrong based on 66 sessions

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Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?

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

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Re D01-30  [#permalink]

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New post 16 Sep 2014, 00:13
Official Solution:

Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?

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


The rate of A is \(\frac{1}{20}\) job/day;

The rate of B is \(\frac{1}{30}\) job/day;

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

In 5 days that B worked alone (2nd stage) \(5*\frac{1}{30}=\frac{2}{12}\) of the job was done;

In 4 day that A and B worked together (3rd stage) \(4*\frac{1}{12}=\frac{4}{12}\) of the job was done;

So, in the 1st stage of the work \(1-\frac{2}{12}-\frac{4}{12}=\frac{6}{12}\) part of the job was done by A and B together, which at the rate of \(\frac{1}{12}\) job/day took them 6 days.


Answer: C
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Re: D01-30  [#permalink]

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New post 08 Dec 2015, 06:01
5
1
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!

+Kudos, if this helped!
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Re: D01-30  [#permalink]

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New post 15 Feb 2016, 06:24
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nishantdoshi wrote:
I think this the explanation isn't clear enough, please elaborate.



Hi,
Lets divide the work in two parts, one when B worked alone and second when both worked together..
Since B can finish work in 30 days, His 1 day work = 1/30 of total..
so in 5 days, he will finish= 5/30= 1/6 of work..

now second part, when both work together..
we should not confuse in two different time frames, that is one at the start and other in the end.. we just require how many days both worked together..
one day work of both= 1/20 + 1/30 = 5/60= 1/12..
so they have to finish 1-1/6 work= 5/6..
it will be done in (5/6)/(1/12)= 5*12/6= 10..
so both work for 10 days together..
In the second part they worked for 4 days to finish the work..
so they worked for 10-4=6 days initially..
ans 6

C
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Re D01-30  [#permalink]

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New post 19 Apr 2016, 23:32
I think this is a high-quality question and I agree with explanation.
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New post 16 Jul 2016, 10:56
I think this is a high-quality question and I agree with explanation.
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New post 22 Aug 2016, 12:00
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
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D01-30  [#permalink]

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New post 30 Nov 2016, 19:55
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
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Re D01-30  [#permalink]

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New post 29 Jan 2017, 07:35
I think this is a poor-quality question and I don't agree with the explanation. 1/20 + 1/30 is not 1/12
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Re: D01-30  [#permalink]

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New post 29 Jan 2017, 10:18
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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
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New post 29 Jan 2017, 10:20
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New post 30 Jan 2017, 09:15
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Bunuel wrote:
Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?

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



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
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New post 18 May 2017, 02:58
I think this is a high-quality question.
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New post 07 Jul 2017, 12:38
I think this is a high-quality question and I agree with explanation.
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Re: D01-30  [#permalink]

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New post 28 May 2018, 11:49
I solved by setting up a table and following the story in the problem.

W R T
Painter A 60 60 / 20 = 3 20
Painter B 60 60 / 30 = 2 30
combined 60 3 + 2 = 5 12

Painter B worked alone for 5 days at a rate of 2 (things) per day for a total of 10 (things)
Painters A and B worked together for 4 days at a rate of 5 (things) per day for a total of 20 (things)

1 whole house = 60 (things)

60 - 10 - 20 = 30 (things) that A and B worked on together at their combined rate

30 / 5 = 6 days
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Re: D01-30  [#permalink]

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New post 05 Oct 2018, 01:21
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!
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Re: D01-30  [#permalink]

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New post 30 May 2019, 01:14
How is the 5 days 5 days that B worked alone (2nd stage) \(5*\frac{1}{30}=\frac{2}{12}\)? Shouldn't 1/30*5=1/6?

Bunuel wrote:
Official Solution:

Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?

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


The rate of A is \(\frac{1}{20}\) job/day;

The rate of B is \(\frac{1}{30}\) job/day;

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

In 5 days that B worked alone (2nd stage) \(5*\frac{1}{30}=\frac{2}{12}\) of the job was done;

In 4 day that A and B worked together (3rd stage) \(4*\frac{1}{12}=\frac{4}{12}\) of the job was done;

So, in the 1st stage of the work \(1-\frac{2}{12}-\frac{4}{12}=\frac{6}{12}\) part of the job was done by A and B together, which at the rate of \(\frac{1}{12}\) job/day took them 6 days.


Answer: C
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Re: D01-30   [#permalink] 30 May 2019, 01:14
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