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Jane can paint the wall in J hours, and Bill can paint the

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Re: Jane can paint the wall in J hours, and Bill can paint the  [#permalink]

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New post 03 Dec 2016, 08:56
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 ill-defined or I miss something?
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Re: Jane can paint the wall in J hours, and Bill can paint the  [#permalink]

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New post 03 Dec 2016, 11:36
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.

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Re: Jane can paint the wall in J hours, and Bill can paint the  [#permalink]

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New post 08 Feb 2018, 23:33
Is noon always = 12pm?
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Re: Jane can paint the wall in J hours, and Bill can paint the  [#permalink]

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New post 09 Feb 2018, 04:49
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  [#permalink]

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New post 28 Aug 2018, 03:50
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  [#permalink]

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New post 28 Aug 2018, 03:57
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  [#permalink]

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New post 28 Aug 2018, 04:28
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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Jane can paint the wall in J hours, and Bill can paint the &nbs [#permalink] 28 Aug 2018, 04:28

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