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D01-29

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Re D01-29  [#permalink]

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New post 18 Oct 2018, 05:58
I think this is a high-quality question and I agree with explanation. Great question as always.
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Re: D01-29  [#permalink]

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New post 31 Oct 2018, 23:01
Bunuel wrote:
Official Solution:


Mac can finish the job in \(M\) days: the rate of Mac is \(\frac{1}{M}\) job/day;

Jack can finish the job in \(J\) days: the rate of Jack is \(\frac{1}{J}\) job/day;

After working together for \(T\) days, Mac left and Jack alone worked to complete the remaining work in \(R\) days, hence Mac worked for \(T\) days only and did \(\frac{T}{M}\) part of the job while Jack worked for \(T+R\) days and did \(\frac{T+R}{J}\) part of the job;

Since Mac and Jack completed an equal amount of work (so half of the job each) then \(\frac{T}{M}=\frac{1}{2}\) and \(\frac{T+R}{J}=\frac{1}{2}\). So, \(T=\frac{M}{2}\) and \(T=\frac{J}{2}-R\), therefore \(\frac{M}{2}=\frac{J}{2}-R\), which gives \(J=M+2R\).

(1) \(M = 20\) days. Not sufficient, we still need the value of \(R\).

(2) \(R = 10\) days. Not sufficient, we still need the value of \(M\).

(1)+(2) \(J=M+2R=20+2*10=40\). Sufficient.


Answer: C




You made this so easy.

Can you please suggest,what was your line of thinking ..I took a longer approach and got confused at the end...

Here is my approach,

Mac's Work =1/M
Jacks' = 1/J

When both work for T days work done = ((1/M) + (1/J))*T
Remaining work = 1 - {(M+J)/MJ}*T

Now work done by jack= T+R days..
So I would make acomplex equution equation work done by Jack and Mac and this would really confuse me spotting R and T
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Re: D01-29  [#permalink]

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New post 31 Oct 2018, 23:06
prabsahi wrote:
Bunuel wrote:
Official Solution:


Mac can finish the job in \(M\) days: the rate of Mac is \(\frac{1}{M}\) job/day;

Jack can finish the job in \(J\) days: the rate of Jack is \(\frac{1}{J}\) job/day;

After working together for \(T\) days, Mac left and Jack alone worked to complete the remaining work in \(R\) days, hence Mac worked for \(T\) days only and did \(\frac{T}{M}\) part of the job while Jack worked for \(T+R\) days and did \(\frac{T+R}{J}\) part of the job;

Since Mac and Jack completed an equal amount of work (so half of the job each) then \(\frac{T}{M}=\frac{1}{2}\) and \(\frac{T+R}{J}=\frac{1}{2}\). So, \(T=\frac{M}{2}\) and \(T=\frac{J}{2}-R\), therefore \(\frac{M}{2}=\frac{J}{2}-R\), which gives \(J=M+2R\).

(1) \(M = 20\) days. Not sufficient, we still need the value of \(R\).

(2) \(R = 10\) days. Not sufficient, we still need the value of \(M\).

(1)+(2) \(J=M+2R=20+2*10=40\). Sufficient.


Answer: C




You made this so easy.

Can you please suggest,what was your line of thinking ..I took a longer approach and got confused at the end...

Here is my approach,

Mac's Work =1/M
Jacks' = 1/J

When both work for T days work done = ((1/M) + (1/J))*T
Remaining work = 1 - {(M+J)/MJ}*T

Now work done by jack= T+R days..
So I would make acomplex equution equation work done by Jack and Mac and this would really confuse me spotting R and T




My main question is how did you take the hint from this question while reading that you have to use this approach..you line of thought
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Re: D01-29  [#permalink]

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New post 31 Oct 2018, 23:20
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Bunuel wrote:
Mac can finish a job in \(M\) days and Jack can finish the same job in \(J\) days. After working together for \(T\) days, Mac left and Jack alone worked to complete the remaining work in \(R\) days. If Mac and Jack completed an equal amount of work, how many days would have it taken Jack to complete the entire job working alone?


(1) \(M = 20\) days

(2) \(R = 10\) days


Here is my understanding now after screwing up the question

We see Mac works for T days and Jack works for T+R days and another important thing to notice is they complete half of the work each.Now Rate * Time = Work done..This means for Jack 1/J (T+R) = 1/M * (T)=1/2 Now total number of days which jack will take is (T+R)..If you see equation in red you can calculate T=2M and the same u can substitute in 1/J(2M+R)


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Re: D01-29  [#permalink]

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New post 14 Nov 2018, 08:58
Hi Leonsandcastle332

I believe we are using the W=RT formula. As Bunuel said:
Mac takes M days to do the job so in one day he completes 1/M part of the job. This is the rate.
He works for T days (Time). Therefore, the amount of work he completes in T days is : T*1/M (Time* Rate)

Hope I am rightly addressing your concern.
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Re D01-29  [#permalink]

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New post 15 Nov 2018, 07:19
I think this is a high-quality question and I agree with explanation.
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Re D01-29  [#permalink]

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New post 30 Jan 2019, 07:21
I think this is a high-quality question and I agree with explanation.
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Re: D01-29  [#permalink]

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New post 11 Mar 2019, 10:43
They both do equal amount of work, so lets take 40 being total work.

Mac and Jack both do equal amount of work so "20" work by Mac and "20" work by Jack.

Now, Mac does total work "40" in 20 days, so 2 work in one day so, for to do 20 work he worked for 10 days, so, T = 10 days.
Now, Jack worked for rest of 10 days alone, and T days (10 days) and he completed 20 work, so, he works at pace of 1 work a day.

Therefore, total work being 40, it will take him 40 days.

Sorry, my approach is similar to PS problem, hope it helps.
I am using the EGMAT LCM method.
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Re D01-29  [#permalink]

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New post 18 Apr 2019, 06:37
I think this is a high-quality question and I agree with explanation. great
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Re: D01-29  [#permalink]

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New post 06 May 2019, 02:10
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I think this is a high-quality question and I agree with explanation.
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Re: D01-29  [#permalink]

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New post 14 Jun 2019, 23:19
This is a great question
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Re: D01-29  [#permalink]

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New post 26 Jun 2019, 09:38
I think this is a high-quality question and I agree with explanation.
When I tried to solve by finding combined work of these two, it became a complicated formula.
Then it was not possible to eliminate T and R variables, so I concluded as E.
But after seeing the explanation, I can clearly see how it can be equated separately.
Still, I don't know when to use the combined work formula like 1/A + 1/B and when otherwise.
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Re: D01-29   [#permalink] 26 Jun 2019, 09:38

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