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

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

20 Nov 2009, 09:37

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Question Stats:

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65% (02:34) wrong

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Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
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Re: Painting the wall. [#permalink]

20 Nov 2009, 11:04

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If both J & B work they will finish the job in J*B/(J+B) hrs.
Also given both J&B are even.

Lets start with statement B
(J+B)^2 = 400
If J & B were equal then J=B=10
They will finish the work in 5 hrs i.e 5 PM, statement A is different hence we know J & B are not equal. Sufficient but before we jump to C lets look at statement A alone

J & B finish the job by 4:48p.m they take 4 48/60 hrs = 4 4/5 = 24/5

So we know J*B/(J+B) = 24/5
since J and B are even they can't be fractions, hence
J*B has to be a multiple of 24
J+B has to be a multiple of 5
since J & B are even J+B can only be multiples of 10.
Lets assume J=B and start plugging in values
when J*B/(J+B) = 48/10
J+B = 10, J=B=5 so J*B = 25 too small
when J*B/(J+B) = 96/20
J+B = 10, J=B=10 so J*B = 100 too big
when J*B/(J+B) = 144/30
J+B = 30, J=B=15 so J*B = 225 too big

J*B values keep increasing and will never equal the value derived by assuming j&B are equal.

Hence answer should be A.
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Re: Painting the wall. [#permalink]

20 Nov 2009, 11:15

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Bunuel wrote:

Question Stem : In one hour Jane can paint \frac{1}{J} of the wall and Bill can paint \frac{1}{B} of the wall.
Together, in one hour, they can paint \frac{1}{J} + \frac{1}{B} of the wall.
Starting time : 12:00 p.m.
If J and B are even, can we tell if J = B ?

St. (1) : Finish time : 4:48 p.m.
This tells us that when they worked together, they finished painting the wall in 288 minutes or \frac{288}{60} hours.
Therefore, in one hour, working together, they finished painting \frac{60}{288} of the wall.

Therefore we get the equation : \frac{1}{J} + \frac{1}{B} = \frac{60}{288}

This can be reduced to \frac{J+B}{J*B} = \frac{5}{24}

Now let us assume that A and B are equal.

Therefore, \frac{J+J}{J*J} = \frac{5}{24} ---> \frac{2*J}{J^2} = \frac{5}{24} ---> J = \frac{48}{5}

Thus it is obvious that if J is equal to B, in order to satisfy statement 2, J and B cannot be even. Since we are given that J and B have to be even, it is possible to conclude that J is not equal to B.

Hence this statement is Sufficient.

St. (2) : (J + B)^2 = 400
This one is easy to discard by just plugging in some values.
The statement holds true for J = B = 10 as well as for J = 12 and B = 8.
Since it gives contradicting solutions for the question stem (J=B in case and J#B in the other), this statement is Insufficient.

Answer : A
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Re: Painting the wall. [#permalink]

20 Nov 2009, 11:43

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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.
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Re: Painting the wall. [#permalink]

20 Nov 2009, 12:03

very elegant solution Bunuel. And thanks for the awesome question. +1
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Jane and Bill paint together [#permalink]

08 Jun 2010, 13:55

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[Reveal] Spoiler:

OA is A

total hours worked together = JB/(J+B) hrs

If J=B,
then the hours worked together would be = J*J/2J = J/2 hrs

since J is even, total hours should be an integer.

(1) total hours worked = 4.75
this is not an integer, so J≠B

SUFFICIENT

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

14 Sep 2012, 21:15

Hi all,

Adding to Bunel's solution

Once you get :
T= JB/(J+B) then if you want you can just skip J=B step.
All you have to do is know that both J and B are even i.e. both J=B=2n, where is an integer.

So,

T = (2n)(2n)/(2n+2n)= 4(n^2)/4n = n
hence, now using options all you have to prove is T=n i.e. T is a integer.

thanks,

-Kartik

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

14 Sep 2012, 21:18

@bunel,

I have a question regarding option a)
4hr 48 mins if you look at it from hours prespective then yes, T is not an integer BUT:
4 hr 48 mins == 288 mins. Now, 288 is a integer.

thoughts?

I feel the question is not clear on this!

-Kartik

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

15 Sep 2012, 03:20

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kartik222 wrote:

J and B are given in hours and we are told that both are even numbers, so the number of hours they need together to paint (T) must be even number too.

Hope it's clear.
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Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]

15 Sep 2012, 03:22

hey thanks Bunuel!

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink] 15 Sep 2012, 03:22

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

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

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