Author 
Message 
TAGS:

Hide Tags

SVP
Status: It's near  I can see.
Joined: 13 Apr 2013
Posts: 1702
Location: India
Concentration: International Business, Operations
GPA: 3.01
WE: Engineering (Real Estate)

Re D0129
[#permalink]
Show Tags
18 Oct 2018, 05:58
I think this is a highquality question and I agree with explanation. Great question as always.
_________________
"Do not watch clock; Do what it does. KEEP GOING."



Senior Manager
Joined: 09 Jun 2014
Posts: 353
Location: India
Concentration: General Management, Operations

Re: D0129
[#permalink]
Show Tags
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



Senior Manager
Joined: 09 Jun 2014
Posts: 353
Location: India
Concentration: General Management, Operations

Re: D0129
[#permalink]
Show Tags
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



Senior Manager
Joined: 09 Jun 2014
Posts: 353
Location: India
Concentration: General Management, Operations

Re: D0129
[#permalink]
Show Tags
31 Oct 2018, 23:20
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)Press Kudos if it helps!!



Intern
Joined: 03 Jun 2018
Posts: 1

Re: D0129
[#permalink]
Show Tags
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.



Intern
Joined: 24 Feb 2018
Posts: 31
Location: India
GPA: 3.35
WE: Military Officer (Military & Defense)

Re D0129
[#permalink]
Show Tags
15 Nov 2018, 07:19
I think this is a highquality question and I agree with explanation.



Manager
Status: Victory is never a one time thing.
Joined: 14 Jan 2018
Posts: 57
Location: Oman
Concentration: Marketing, Entrepreneurship
GMAT 1: 590 Q49 V21 GMAT 2: 650 Q47 V33
GPA: 3.8

Re D0129
[#permalink]
Show Tags
30 Jan 2019, 07:21
I think this is a highquality question and I agree with explanation.



Manager
Status: Studying 4Gmat
Joined: 02 Jan 2016
Posts: 234
Location: India
Concentration: Strategy, Entrepreneurship
GPA: 4
WE: Law (Manufacturing)

Re: D0129
[#permalink]
Show Tags
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.



Intern
Joined: 10 Aug 2014
Posts: 10

Re D0129
[#permalink]
Show Tags
18 Apr 2019, 06:37
I think this is a highquality question and I agree with explanation. great



Intern
Joined: 26 Sep 2017
Posts: 2

Re: D0129
[#permalink]
Show Tags
06 May 2019, 02:10
I think this is a highquality question and I agree with explanation.



Intern
Joined: 31 May 2019
Posts: 2

Re: D0129
[#permalink]
Show Tags
14 Jun 2019, 23:19
This is a great question



Intern
Joined: 07 Jun 2019
Posts: 7

Re: D0129
[#permalink]
Show Tags
26 Jun 2019, 09:38
I think this is a highquality 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.



VP
Joined: 14 Feb 2017
Posts: 1314
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35
GPA: 3
WE: Management Consulting (Consulting)

Re: D0129
[#permalink]
Show Tags
03 Aug 2019, 22:37
I initially interpreted that since Jack and Mac completed the same amount of work T/M = (T+R)/J However this yields inconclusive info for Jack. Why is this incorrect Bunuel ?



Intern
Joined: 13 Sep 2019
Posts: 4

Re D0129
[#permalink]
Show Tags
03 Dec 2019, 11:01
I think this is a highquality question and I agree with explanation.







Go to page
Previous
1 2
[ 34 posts ]



