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Re D0129
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15 Sep 2014, 23:12
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
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Re: D0129
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14 Sep 2015, 14:00
Hi Bunuel 
What was your thought process in not setting T/M = 1/2 and T+R/J = 1/2 equal to each other? That is, T/M = T+R/J? I was stuck after I did such (algebraic mess, and leads you to the wrong conclusion). Would be great to learn your thought process.



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Re: D0129
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25 Dec 2015, 03:19
One option is to go by options and see check with the values.
It gives an idea that both values are required to find out the value of J.



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Re: D0129
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07 Mar 2016, 21:10
mystseen wrote: Hi Bunuel 
What was your thought process in not setting T/M = 1/2 and T+R/J = 1/2 equal to each other? That is, T/M = T+R/J? I was stuck after I did such (algebraic mess, and leads you to the wrong conclusion). Would be great to learn your thought process. Here the idea is to find the value of 'J'. Therefore, the focus should be to eliminate the extra variables where ever you can. I was also stuck with the same bit as you did, but i realized that one can eliminate the variable 'T' by substituting the variable 'T' in the other equation (Jack's equation) which kinda did the trick as I was left with an equation with only M, R and J. Hope this helps a little bit.
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Re D0129
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04 Jul 2016, 09:02
I think this is a highquality question and I agree with explanation.



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10 Jul 2016, 02:20
I think this is a highquality question and I agree with explanation.



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Re: D0129
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14 Jul 2016, 02:29
This is definitely a high quality question. I solve it using a number of variable approach and i want to ask to some math expert whether it is correct.
===R================T===========W *==(1/m)+(1/t)=========t==========(t/m)+(t/j) => in time = t M&J working together complete \((t/m)+(t/j)\)
**==1/j===========((mj/m+jt)====== 1((t/m)+(t/j)) J works at his constant rate \(1/j\) for the remaining time which is equal to \(((1/m)+(1/j)) t\)
We are asked to find j: from ** we can multiply R*T=W and find a single equation in 3 unknowns therefore we need stat 1 and stat 2 which gives us 2 unknown to find the third one. Is the process correct ?



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24 Sep 2016, 04:43
I think this is a highquality question and I agree with explanation.



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06 Nov 2016, 19:22
I think this is a highquality question and I agree with explanation.



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Maybe I don't understand this completely. I understand that T/M = (R+T)/J. If W is the total work done, then both of these equal W/2. Thus, it's clear that T/M=(R+T)/J=W/2.
How can you say that T/M=1/2, when you don't know how much work was done? Which part of this question allows us to say that W=1, thus T/M=(R+T)/J=1/2?? Ie, which part of the prompt restrict the variables such that we know W=1?



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Re: D0129
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28 Dec 2016, 05:39



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Re: D0129
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13 May 2018, 11:00
I think this is a high quality question and l agree with the explanation



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Re: D0129
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18 Jun 2018, 10:55
Basic question regarding ratios since my ratios grip is weak
If instead of saying that work was equal, the question had said, Mac did 30% of the work that Jack did,
this would mean Mac work/ Jack work = 3/ 10
In case of above ratio, it would be correct to say that Mac did 3/ 13 and Jack did 10/13 work right?
Hence T/M = 3/ 13 right?
Just want to solidify my grip on ratios, that is why asking



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Re: D0129
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18 Jun 2018, 19:36



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Re: D0129
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13 Aug 2018, 18:06
When you say "T/M" part of the job...why are we putting time over rate? W=RT...so I don't understand why we are putting the number of days, T (time), over the rate (M).
Same thing for the other equation.
Thank you



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13 Aug 2018, 23:48



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Re: D0129
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14 Aug 2018, 17:30
I think I'm almost there, but not quite there.
We are setting Time to a value...so we're not really using the W=RT formula with this problem?



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14 Aug 2018, 22:54



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Re: D0129
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15 Aug 2018, 17:38
Bunuel wrote: leonsandcastle332 wrote: I think I'm almost there, but not quite there.
We are setting Time to a value...so we're not really using the W=RT formula with this problem? We are solving using ratios of work done. Don't we need a variable for rate in this ratio?







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