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Re: Painting the wall. [#permalink]
20 Nov 2009, 10:04

6

This post received KUDOS

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.

Re: Painting the wall. [#permalink]
20 Nov 2009, 10:15

16

This post received KUDOS

Bunuel wrote:

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

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 \(\frac{J+B}{J*B} = \frac{5}{24}\)

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.

Re: Painting the wall. [#permalink]
20 Nov 2009, 10:43

43

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

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 \(8\).

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
14 Sep 2012, 20:15

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.

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
14 Sep 2012, 20:18

@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.

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
15 Sep 2012, 02:20

1

This post received KUDOS

Expert's post

kartik222 wrote:

@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

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.

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
30 May 2013, 05:30

1

This post received KUDOS

(1) They're going to finish at 4:48, I broke this down into a fraction so I could try to make the even numbers match. The ratio that could hit this is in two minute intervals, you'll hit the hour but also the 48 minute mark on the final hour).

For each hour you have 30 units - so 144 units altogether (30 for the 12:00 hour, 30 for 1:00, 30 for 2:00, 30 for 3:00 and 24 for the 48 minutes in the 4:00 hour - each two minute span in the time period). This means that together they're painting 30/144 wall per hour. 30/2 is 15 - therefore these two numbers are not equal if J and B are even numbers.

A is sufficient.

(2) (J+B)^2=400 J+B= 20

There aren't any limiting parameters here and no finish time. If you decide to try it anyway - make it fit the rules of the problem.

10+10=20

Both parties are working at 1/10 wall per hour. Which means that together they're working at 2/10 wall per hour. It will take them 5 hours to paint the wall.

But since there's nothing that limits you, find another set of even numbers that aren't equal.

16+4=20. One party is working at 1/16 wall per hour, the other is working at 1/4 wall per hour. Together they're working at 5/16 wall per hour.It will take them 3 1/5 hours to paint the wall. These numbers give you an answer as well, this option isn't sufficient enough to give you an answer.

B is insufficient. The answer is (A). _________________

"You may not realize it when it happens, but a kick in the teeth may be the best thing in the world for you." - Walt Disney

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
30 May 2013, 09:22

1

This post received KUDOS

Expert's post

Bunuel wrote:

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

We are supposed to find out whether J=B? Assume that it is. Thus, the question can be interpreted as this : If Jane/Bill working alone take J=B hrs, how much time will they take if they work together? The time taken will be exactly half of what they took individually--> J/2 = H/2. Now as J=H=even=2k, thus the new time taken(in hours) must be an integral value(k). However, from F.S 1, we can see that the hours taken for them is not an an integral value. Thus \(J\neq{H}\).Sufficient.

From F.S 2, we know that (J+B) = 20. Thus, J=B=10 is a valid solution; and so is J=14,B=6. Insufficient.

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
10 Jun 2014, 16:44

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Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
28 Aug 2014, 10:04

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,

Two questions:

1) I was with you through the whole process until you said that "n" needed to be an integer. I just don't see that. Why do they have to be whole hours? Why can't N be .5 hours or .4 hours?

2) In atish' example -- it states "since J & B are even J+B can only be multiples of 10." Why does J+B have to be a multiple of 10? Can't it be 2+2 = 4 etc?

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
28 Aug 2014, 12:35

Bunuel wrote:

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

Case 1: 100/( 24/5) will give the efficiency of J+B which is equal to 20.83. Hence the efficiencies are never equal. Sufficient

Case 2: Not sufficient due to multiple solutions _________________

[color=#007ec6]Would love to get kudos whenever it is due.. Though the mountain is steep, nevertheless once i reach the finishing point, i would enjoy the view [/color]....

Jane can paint the wall in J hours, and Bill can paint the [#permalink]
10 Oct 2014, 23:06

I thought I'd jump in -

Quote:

2) In atish' example -- it states "since J & B are even J+B can only be multiples of 10." Why does J+B have to be a multiple of 10? Can't it be 2+2 = 4 etc?

J and B are even. So E+E=E always BUT we also just reached that J+B must be a multiple of 5 The only multiples of 5 that are even are the multiples of 10 too (like 10, 20...) (2+2=4 but 4 isnt a multiple of 5)

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
17 Oct 2014, 05:45

can't we just see from stmt 1. that if A and B are integers and represent number of hours to paint the wall, then for a = b and combined paint the wall in 4:48 minutes, then for either a or b do complete it on their own, it would take twice that amount of time (or, one working alone would do half the job in 4:48). This tells us directly that a and b are not integers, so sufficiently no?

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]
26 Nov 2014, 01:25

russ9 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,

Two questions:

1) I was with you through the whole process until you said that "n" needed to be an integer. I just don't see that. Why do they have to be whole hours? Why can't N be .5 hours or .4 hours?

2) In atish' example -- it states "since J & B are even J+B can only be multiples of 10." Why does J+B have to be a multiple of 10? Can't it be 2+2 = 4 etc?

"n" will be an integer because fractions are not even numbers or odd numbers, as they are not whole numbers. Hope it clears your doubt.

gmatclubot

Re: Jane can paint the wall in J hours, and Bill can paint the
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26 Nov 2014, 01:25

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