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Re: Jane can paint the wall in J hours, and Bill can paint the
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03 Dec 2016, 09:56
Looking at the first statement:
Ra = 1/a=1/(2k) Rb=1/b=1/(2n) convert 4:48 to hours: 4 + 48/60 = 4 + ⅘ = 20/5 + ⅘ = 24/5 24/5 (1/2k + 1/2n) = 1 24/5 (1/2) (1/k + 1/n) = 1 12/5 (k+n)/(kn) = 1 (k+n)/(kn) = 5/12 k+n=5 kn=12
However, there are no such positive integers k and n. Is the problem illdefined or I miss something?



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Re: Jane can paint the wall in J hours, and Bill can paint the
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03 Dec 2016, 12:36
Hi lexxus, This DS question is a bit more 'layered' than most DS questions. From the prompt, we know that J and B are both EVEN INTEGERS. We're asked if J and B are equal. This is a YES/NO question. Fact 1 tells us that it takes the two people 4 4/5 hours to paint the wall together. There are only 2 possible pairs of even integers that will lead to THAT result: 6 and 24 8 and 12 In both situations, the answer to the question is NO, so Fact 1 is SUFFICIENT. GMAT assassins aren't born, they're made, Rich
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Re: Jane can paint the wall in J hours, and Bill can paint the
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09 Feb 2018, 00:33
Is noon always = 12pm?



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Re: Jane can paint the wall in J hours, and Bill can paint the
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09 Feb 2018, 05:49
sucal000 wrote: Is noon always = 12pm? Yes, what else could it be



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Re: Jane can paint the wall in J hours, and Bill can paint the
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28 Aug 2018, 04:50
Bunuel wrote: Nice solutions atish and sriharimurthy, +1 to both of you.
Though it can be done easier.
Jane and Bill working together will paint the wall in \(T=\frac{JB}{J+B}\) hours. Now suppose that \(J=B\) > \(T=\frac{J^2}{2J}=\frac{J}{2}\), as \(J\) and \(B\) are even \(J=2n\) > \(T=\frac{2n}{2}=n\), as \(n\) is an integer, working together Jane and Bill will paint the wall in whole number of hours, meaning that in any case \(T\) must be an integer.
(1) They finish painting in 4 hours and 48 minutes, \(T\) is not an integer, > \(J\) and \(B\) are not equal. Sufficient.
(2) \(J+B=20\), we can even not consider this one, clearly insufficient. \(J\) and \(B\) can be \(10\) and \(10\) or \(12\) and \(8\).
Answer: A. Hi Bunuel, Correct me if I am wrong. Before jumping into the individual statements, it has been found that "T" must be an integer. Now in statement 2: 1st case :10 and 10 gives me T as an integer but 2nd case: 12 and 8 {(1/12+1/8)*T =1} does not give T as an integer. So is it not that J and B have to be 10 and 10. Please give your suggestion. Thank you.
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Re: Jane can paint the wall in J hours, and Bill can paint the
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28 Aug 2018, 04:57
Rishovnits wrote: Bunuel wrote: Nice solutions atish and sriharimurthy, +1 to both of you.
Though it can be done easier.
Jane and Bill working together will paint the wall in \(T=\frac{JB}{J+B}\) hours. Now suppose that \(J=B\)> \(T=\frac{J^2}{2J}=\frac{J}{2}\), as \(J\) and \(B\) are even \(J=2n\) > \(T=\frac{2n}{2}=n\), as \(n\) is an integer, working together Jane and Bill will paint the wall in whole number of hours, meaning that in any case \(T\) must be an integer.
(1) They finish painting in 4 hours and 48 minutes, \(T\) is not an integer, > \(J\) and \(B\) are not equal. Sufficient.
(2) \(J+B=20\), we can even not consider this one, clearly insufficient. \(J\) and \(B\) can be \(10\) and \(10\) or \(12\) and \(8\).
Answer: A. Hi Bunuel, Correct me if I am wrong. Before jumping into the individual statements, it has been found that "T" must be an integer. Now in statement 2: 1st case :10 and 10 gives me T as an integer but 2nd case: 12 and 8 {(1/12+1/8)*T =1} does not give T as an integer. So is it not that J and B have to be 10 and 10. Please give your suggestion. Thank you. From the stem we got that T is an integer IF J = B, not that T is an integer in all cases.
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Jane can paint the wall in J hours, and Bill can paint the
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28 Aug 2018, 05:28
Bunuel wrote: Rishovnits wrote: Bunuel wrote: Nice solutions atish and sriharimurthy, +1 to both of you.
Though it can be done easier.
Jane and Bill working together will paint the wall in \(T=\frac{JB}{J+B}\) hours. Now suppose that \(J=B\)> \(T=\frac{J^2}{2J}=\frac{J}{2}\), as \(J\) and \(B\) are even \(J=2n\) > \(T=\frac{2n}{2}=n\), as \(n\) is an integer, working together Jane and Bill will paint the wall in whole number of hours, meaning that in any case \(T\) must be an integer.
(1) They finish painting in 4 hours and 48 minutes, \(T\) is not an integer, > \(J\) and \(B\) are not equal. Sufficient.
(2) \(J+B=20\), we can even not consider this one, clearly insufficient. \(J\) and \(B\) can be \(10\) and \(10\) or \(12\) and \(8\).
Answer: A. Hi Bunuel, Correct me if I am wrong. Before jumping into the individual statements, it has been found that "T" must be an integer. Now in statement 2: 1st case :10 and 10 gives me T as an integer but 2nd case: 12 and 8 {(1/12+1/8)*T =1} does not give T as an integer. So is it not that J and B have to be 10 and 10. Please give your suggestion. Thank you. From the stem we got that T is an integer IF J = B, not that T is an integer in all cases. Okay. I missed it. Thanks
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Re: Jane can paint the wall in J hours, and Bill can paint the
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31 Aug 2019, 13:16
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Re: Jane can paint the wall in J hours, and Bill can paint the
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