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Will project Alpha be completed by the deadline?

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Will project Alpha be completed by the deadline?

by bhumika.k.shah » Tue Jan 26, 2010 5:22 am
Andy, George and Sally are a team of consultants working on Project Alpha. They have an eight hour deadline to complete the project. The team members work at constant rates throughout the eight hour period. If the team of three has to begin work now and no one else can work on this project, will Project Alpha be completed by the deadline?

1.Sally can finish the project alone in 4k + 7 hours, where k is a positive integer with a minimum value of 1 and a maximum value of 5.
2.Working alone, George will take 2k + 1 hours and Andy will take 3 + 2k hours, where k is a positive integer with a minimum value of 1 and a maximum value of 5


OA B

I know A alone is not sufficent ... what next?
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by thephoenix » Tue Jan 26, 2010 7:01 am
are u sure its B
bcoz without knowing the eff of sally we can not solve it

i am getting it as D
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by bhumika.k.shah » Tue Jan 26, 2010 7:05 am
The oA is B

i dont know why though...

And about ur answer as D i know it is wrong for sure..
As A along cant b sufficient since we need to knw the ratio of work done by the remaining two...

If we need to check whether sally herself completes the entire work we will have to take k as 1
which would mean that sally alone takes 11 hours to complete the plan...

that is far beyond the schedule time!

so A is insufficient!

dont know about B though
thephoenix wrote:are u sure its B
bcoz without knowing the eff of sally we can not solve it

i am getting it as D
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by Osirus@VeritasPrep » Tue Jan 26, 2010 7:19 am
it is B. Its a maximization problem.

With statement (2) if you assume that k is 5 which would mean that the two guys are working the slowest they can possibly work you would add these two rates together to get their maximum slowest rate together.

you get: 1/11 + 1/13 = 13/143 + 11/143 = 24/143

Next, set up the Work= Rate X Time

1 = 24/143 X 8


24/143 * 8 is greater than one so regardless of what Sally does, the two guys working at their slowest rates will finish the project in 8 hours.
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by thephoenix » Tue Jan 26, 2010 7:25 am
bhumika.k.shah wrote:The oA is B

i dont know why though...

And about ur answer as D i know it is wrong for sure..
As A along cant b sufficient since we need to knw the ratio of work done by the remaining two...

If we need to check whether sally herself completes the entire work we will have to take k as 1
which would mean that sally alone takes 11 hours to complete the plan...

that is far beyond the schedule time!

so A is insufficient!

dont know about B though
thephoenix wrote:are u sure its B
bcoz without knowing the eff of sally we can not solve it

i am getting it as D
sorry i meant C

but still i am not satisfied with OA let sme one else cum up with soln , or else i need to scratch my head a bit more to digest the ans
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by bhumika.k.shah » Tue Jan 26, 2010 7:31 am
great explanation osirus. the answer indeed is B

thanks!

how would i know how to approach this question in this manner ...by maximising and blah!

thephoenix hope u got it now ????
osirus0830 wrote:it is B. Its a maximization problem.

With statement (2) if you assume that k is 5 which would mean that the two guys are working the slowest they can possibly work you would add these two rates together to get their maximum slowest rate together.

you get: 1/11 + 1/13 = 13/143 + 11/143 = 24/143

Next, set up the Work= Rate X Time

1 = 24/143 X 8


24/143 * 8 is greater than one so regardless of what Sally does, the two guys working at their slowest rates will finish the project in 8 hours.
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by Osirus@VeritasPrep » Tue Jan 26, 2010 7:40 am
If you ever see a problem where they give you a range of what a variable could be instantly think that it is probably a max/min problem. If you are unsure whether to maximize or minimize it and if time permits, simply try both.
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by bhumika.k.shah » Tue Jan 26, 2010 7:50 am
But what made u directly take k as 5 in this case ?
i mean how were u so sure about it ???

sowree for asking such silly questions but i m too weak at math! :(
osirus0830 wrote:If you ever see a problem where they give you a range of what a variable could be instantly think that it is probably a max/min problem. If you are unsure whether to maximize or minimize it and if time permits, simply try both.
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by Osirus@VeritasPrep » Tue Jan 26, 2010 7:55 am
bhumika.k.shah wrote:But what made u directly take k as 5 in this case ?
i mean how were u so sure about it ???

sowree for asking such silly questions but i m too weak at math! :(
osirus0830 wrote:If you ever see a problem where they give you a range of what a variable could be instantly think that it is probably a max/min problem. If you are unsure whether to maximize or minimize it and if time permits, simply try both.
No question is silly. This is actually shocking me because I remember when I first started my prep, I couldn't answer any of the questions posted here and now I can actually offer help (which is good since my test is Friday...lol). I took 5 right away because 5 would maximize how long it would take each to finish the project. If k =5 then it would take George 11 hours. Without knowing Sally's rate in statement (2) in order to answer the question you needed to know that the two guys could finish the project without her working at their absolute slowest rates. The since the rate is 1/time, to get the slowest rate you would want to get the maximum amount of time. You would find the maximum amount of time it would take each to finish by making the variable as high as it can be. If you make it smaller you would minimize the time, which is the best case scenario, but the best case scenario isn't true in all situations. If they could finish the project within 8 hours working with their worst case scenario rates, then they don't even need Sally and you can answer the question. I hope this helps.
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by Ashish8 » Tue Jan 26, 2010 8:47 am
I thought the formula was 1/a + 1/b = 1/c. If we use k = 5, this would be 1/11 + 1/13 = 1/c --> 24/143 = 1/c --> c = 143/24 --> 5.9 < 8. Is my formula wrong?
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by Osirus@VeritasPrep » Tue Jan 26, 2010 8:50 am
Ashish8 wrote:I thought the formula was 1/a + 1/b = 1/c. If we use k = 5, this would be 1/11 + 1/13 = 1/c --> 24/143 = 1/c --> c = 143/24 --> 5.9 < 8. Is my formula wrong?


You're right you just have to finish it. 1/c or 24/143 is their combined rate. To discover whether or not they will finish the job in 8 hours use the work = rate x time forumula. In this case the rate is 24/143, and time equals 8 hours, and work equals 1 (one job). when you multiply the rate by 8 you get a number greater than 1 (one job) so it is sufficient.
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by bhumika.k.shah » Tue Jan 26, 2010 8:54 am
Thanks osirus :D
it indeed helped :)

all the best for your exam!!
do share your GMAT D day experience with all of us later on :-)


osirus0830 wrote:
bhumika.k.shah wrote:But what made u directly take k as 5 in this case ?
i mean how were u so sure about it ???

sowree for asking such silly questions but i m too weak at math! :(
osirus0830 wrote:If you ever see a problem where they give you a range of what a variable could be instantly think that it is probably a max/min problem. If you are unsure whether to maximize or minimize it and if time permits, simply try both.
No question is silly. This is actually shocking me because I remember when I first started my prep, I couldn't answer any of the questions posted here and now I can actually offer help (which is good since my test is Friday...lol). I took 5 right away because 5 would maximize how long it would take each to finish the project. If k =5 then it would take George 11 hours. Without knowing Sally's rate in statement (2) in order to answer the question you needed to know that the two guys could finish the project without her working at their absolute slowest rates. The since the rate is 1/time, to get the slowest rate you would want to get the maximum amount of time. You would find the maximum amount of time it would take each to finish by making the variable as high as it can be. If you make it smaller you would minimize the time, which is the best case scenario, but the best case scenario isn't true in all situations. If they could finish the project within 8 hours working with their worst case scenario rates, then they don't even need Sally and you can answer the question. I hope this helps.
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by Ashish8 » Tue Jan 26, 2010 9:55 am
osirus0830 wrote:
Ashish8 wrote:I thought the formula was 1/a + 1/b = 1/c. If we use k = 5, this would be 1/11 + 1/13 = 1/c --> 24/143 = 1/c --> c = 143/24 --> 5.9 < 8. Is my formula wrong?


You're right you just have to finish it. 1/c or 24/143 is their combined rate. To discover whether or not they will finish the job in 8 hours use the work = rate x time forumula. In this case the rate is 24/143, and time equals 8 hours, and work equals 1 (one job). when you multiply the rate by 8 you get a number greater than 1 (one job) so it is sufficient.
stupid mistake on my part. If 2 out of 3 people can do the job in less than 8 hours, you don't care how long the 3rd person takes. It can only make the time faster.
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by sreak1089 » Tue Jan 26, 2010 10:27 am
I think you need not go any further once you discover 5.9 < 8 coz you know deadline is 8 hours. Two of the folks working at their slowest speed itself is enough to finish the project in less than 8 hours. So, it is over-whelmingly obvious that the Alpha project deadline will be met :)
osirus0830 wrote:
Ashish8 wrote:I thought the formula was 1/a + 1/b = 1/c. If we use k = 5, this would be 1/11 + 1/13 = 1/c --> 24/143 = 1/c --> c = 143/24 --> 5.9 < 8. Is my formula wrong?


You're right you just have to finish it. 1/c or 24/143 is their combined rate. To discover whether or not they will finish the job in 8 hours use the work = rate x time forumula. In this case the rate is 24/143, and time equals 8 hours, and work equals 1 (one job). when you multiply the rate by 8 you get a number greater than 1 (one job) so it is sufficient.
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by sreak1089 » Tue Jan 26, 2010 10:30 am
Absolutely, but in reality it does not quite happen that way though. lol :)
It can only make the time faster.
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