Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Apr 2015, 20:02

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40361 [8] , given: 5420

Jane can paint the wall in J hours, and Bill can paint the [#permalink]  20 Nov 2009, 08:37
8
KUDOS
Expert's post
23
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

35% (02:51) correct 65% (01:58) wrong based on 1103 sessions
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
[Reveal] Spoiler: OA

_________________
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 10

Kudos [?]: 117 [6] , given: 3

Re: Painting the wall. [#permalink]  20 Nov 2009, 10:04
6
KUDOS
If both J & B work they will finish the job in J*B/(J+B) hrs.
Also given both J&B are even.

(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.

_________________

Rock On

Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
Followers: 78

Kudos [?]: 797 [16] , given: 18

Re: Painting the wall. [#permalink]  20 Nov 2009, 10:15
16
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 $$\frac{J+B}{J*B} = \frac{5}{24}$$

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.

_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40361 [43] , given: 5420

Re: Painting the wall. [#permalink]  20 Nov 2009, 10:43
43
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 $$8$$.

_________________
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 10

Kudos [?]: 117 [0], given: 3

Re: Painting the wall. [#permalink]  20 Nov 2009, 11:03
very elegant solution Bunuel. And thanks for the awesome question. +1
_________________

Rock On

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 326
Followers: 11

Kudos [?]: 364 [1] , given: 20

Jane and Bill paint together [#permalink]  08 Jun 2010, 12:55
1
KUDOS
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

[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

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Manager
Joined: 27 Dec 2011
Posts: 59
Followers: 1

Kudos [?]: 15 [0], given: 12

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

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
Manager
Joined: 27 Dec 2011
Posts: 59
Followers: 1

Kudos [?]: 15 [0], given: 12

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.

thoughts?

I feel the question is not clear on this!

-Kartik
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40361 [1] , given: 5420

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  15 Sep 2012, 02:20
1
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.

Hope it's clear.
_________________
Manager
Joined: 27 Dec 2011
Posts: 59
Followers: 1

Kudos [?]: 15 [0], given: 12

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  15 Sep 2012, 02:22
hey thanks Bunuel!
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40361 [0], given: 5420

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  30 May 2013, 03:55
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66

_________________
Current Student
Joined: 08 Jun 2012
Posts: 108
Location: United States
Concentration: Marketing, Other
GMAT 1: 550 Q30 V36
GMAT 2: 660 Q41 V40
GMAT 3: 670 Q44 V38
GPA: 3.27
WE: Corporate Finance (Mutual Funds and Brokerage)
Followers: 3

Kudos [?]: 29 [1] , given: 70

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  30 May 2013, 05:30
1
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

Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
Followers: 54

Kudos [?]: 702 [1] , given: 135

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  30 May 2013, 09:22
1
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.

A.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4667
Followers: 290

Kudos [?]: 52 [0], given: 0

Re: Jane can paint the wall in J hours, and Bill can paint the [#permalink]  10 Jun 2014, 16:44
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 331
Followers: 0

Kudos [?]: 20 [0], given: 23

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 $$8$$.

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?
Intern
Joined: 18 Mar 2014
Posts: 37
Followers: 0

Kudos [?]: 1 [0], given: 68

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]....

Intern
Joined: 23 Aug 2014
Posts: 43
GMAT Date: 11-29-2014
Followers: 0

Kudos [?]: 7 [0], given: 28

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)

Hope that helps

Cheers
Dee
Intern
Joined: 07 Sep 2014
Posts: 24
Location: United States (MA)
Concentration: Finance, Economics
Followers: 0

Kudos [?]: 7 [0], given: 6

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?
Intern
Joined: 16 Feb 2014
Posts: 3
Location: India
Concentration: Technology, Marketing
GMAT 1: 630 Q46 V30
WE: Engineering (Computer Software)
Followers: 0

Kudos [?]: 0 [0], given: 8

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 $$8$$.

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.
Re: Jane can paint the wall in J hours, and Bill can paint the   [#permalink] 26 Nov 2014, 01:25
Similar topics Replies Last post
Similar
Topics:
1 If Sally can paint a house in 4 hours, and John can paint 7 02 Nov 2012, 09:46
If Sally can paint a house in 4 hours, and John can paint 1 10 Jul 2009, 13:35
Tom, working alone, can paint a room in 6 hours. Peter and 11 12 Oct 2008, 19:08
Tom, working alone, can paint a room in 6 hours. Peter and 1 14 Feb 2008, 08:19
Tom, working alone, can paint a room in 6 hours. Peter and 5 20 Aug 2007, 18:34
Display posts from previous: Sort by