Author 
Message 
Senior Manager
Joined: 23 Oct 2010
Posts: 318
Location: Azerbaijan
Concentration: Finance

Re: Collection of work/rate problems?
[#permalink]
Show Tags
01 Feb 2012, 00:41
does anyone have all these problems in one document? if yes,please share
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Intern
Joined: 04 Apr 2012
Posts: 1

Re: Collection of work/rate problems?
[#permalink]
Show Tags
29 May 2012, 10:54
I'm having some trouble with #28. I'm having some difficulty understanding this problem (28) I got an answer B, or that statement 2 is sufficient alone to determine how many hours it would have taken machine X to fill out everything. Machine X: xrate/ hour, 4 hours, so output = 4x Machine Y: y rate/hour, 3 hours, so output = 3y. (1) Insuf (2) Suff... because: 4x = (3y) * 2 4x = 6y y = 4/6x, = 2/3x. New rate table: Machine X: xrate/ hour, 4 hours, so output = 4x Machine Y: 2/3x rate/hour, 3 hours, so output = 2x. we know that the entire lot = 4x + 3y, so the new total output is 6x. If machine X is to operate on its own, at rate X... x (rate) * time = 6x. Time = 6 hours to operate the entire thing. Is this logic flawed? Quote: ****28.Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?
(1) Machine X produced 30 bottles per minute. (2) Machine X produced twice as many bottles in 4hours as machine Y produced in 3 hours.
let X = time it takes Machine X to fill lot let Y = time it takes Machine Y to fill lot
4*(1/X) + 3*(1/Y) = 1
From Statement 1 we know the total number of bottles made in four hours by Machine X
NOT SUFFICIENT
From Statement 2 we know that Machine Y produced half the total mentioned above in three hours
Using this info, we can determine how long it would take Machine X to fill the lot:
Total # of bottles * (1 hr / (30*60) bottles) = Number of Hours it would take Machine X to fill the lot
ANSWER: C. Both statements together are sufficient***



Manager
Joined: 08 Apr 2012
Posts: 114

Re: Collection of work/rate problems?
[#permalink]
Show Tags
29 May 2012, 11:17
jsnkwok wrote: I'm having some trouble with #28. I'm having some difficulty understanding this problem (28) I got an answer B, or that statement 2 is sufficient alone to determine how many hours it would have taken machine X to fill out everything. Machine X: xrate/ hour, 4 hours, so output = 4x Machine Y: y rate/hour, 3 hours, so output = 3y. (1) Insuf (2) Suff... because: 4x = (3y) * 2 4x = 6y y = 4/6x, = 2/3x. New rate table: Machine X: xrate/ hour, 4 hours, so output = 4x Machine Y: 2/3x rate/hour, 3 hours, so output = 2x. we know that the entire lot = 4x + 3y, so the new total output is 6x. If machine X is to operate on its own, at rate X... x (rate) * time = 6x. Time = 6 hours to operate the entire thing. Is this logic flawed? Quote: ****28.Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?
(1) Machine X produced 30 bottles per minute. (2) Machine X produced twice as many bottles in 4hours as machine Y produced in 3 hours.
let X = time it takes Machine X to fill lot let Y = time it takes Machine Y to fill lot
4*(1/X) + 3*(1/Y) = 1
From Statement 1 we know the total number of bottles made in four hours by Machine X
NOT SUFFICIENT
From Statement 2 we know that Machine Y produced half the total mentioned above in three hours
Using this info, we can determine how long it would take Machine X to fill the lot:
Total # of bottles * (1 hr / (30*60) bottles) = Number of Hours it would take Machine X to fill the lot
ANSWER: C. Both statements together are sufficient***
Hi jsnkwok, You are absolutely correct. Regards, Shouvik.



Intern
Joined: 11 Oct 2012
Posts: 37

Re: Collection of work/rate problems?
[#permalink]
Show Tags
08 Nov 2012, 11:18
h2polo wrote: 11.6 machines each working at the same constant rate together can complete a job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the job in 8 days?
I think I saw this one in the first PowerPrep Test:
This is a simple two equation problem:
let x = the number of additional machines needed to complete the job in 8 days let y = the rate for one machine to complete the job
Equation 1:
Rate * Time = Work 6*y*12 = 1 job
Equation 2:
(6+x)*y*8 = 1 job
so... y = 1/(6*12) (6+x)*(1/(6*12))*8 = 1 x = 3 In case anyone cares, I solved this a different way, which I thought was more intuitive. We know from the problem that 6 machines work at a rate of 1/12 (since it takes 12 hours to do "1 work" so to speak). Therefore, 1 machine works at a rate of 1/72 (1/12 divided by 6). Letting W = 1, we can use R*T=W to solve where R = M/72 (M machines working at 1/72 rate), T = 8, and W = 1. After solving, we get M = 9, therefore, additional = 96 3



Intern
Joined: 11 Oct 2012
Posts: 37

Re: Collection of work/rate problems?
[#permalink]
Show Tags
08 Nov 2012, 11:49
h2polo wrote: 15. working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B? 1) A's constant rate is 25LTS / min 2) the tanks capacity is 1200 lts.
From the original statement we know that: 1/A + 1/B = 1/(1/2)
From Statement 1 we can plug in and find B: SUFFICIENT
Statement 2 is completely unnecessary:
ANSWER: A. Statement 1 alone is sufficient I don't think this is correct. We know that the rate of A and B = 1/30 of a tank per minute (given in the stem). If A had said that pump A fills X OF THE TANK per minute, then it'd be sufficient alone since A + B = rate of A + rate of B and can solve for rate of B given that we know A. However, this gives it to us in units other than what the stem gave us, so we need to know the tank's capacity. Statement 2 gives us the capacity, so if we use them in conjunction, we can solve the problem. Should be C, not A, IMO.



Math Expert
Joined: 02 Sep 2009
Posts: 59622

Re: Collection of work/rate problems?
[#permalink]
Show Tags
09 Nov 2012, 05:20
brooksbrahs wrote: h2polo wrote: 15. working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B? 1) A's constant rate is 25LTS / min 2) the tanks capacity is 1200 lts.
From the original statement we know that: 1/A + 1/B = 1/(1/2)
From Statement 1 we can plug in and find B: SUFFICIENT
Statement 2 is completely unnecessary:
ANSWER: A. Statement 1 alone is sufficient I don't think this is correct. We know that the rate of A and B = 1/30 of a tank per minute (given in the stem). If A had said that pump A fills X OF THE TANK per minute, then it'd be sufficient alone since A + B = rate of A + rate of B and can solve for rate of B given that we know A. However, this gives it to us in units other than what the stem gave us, so we need to know the tank's capacity. Statement 2 gives us the capacity, so if we use them in conjunction, we can solve the problem. Should be C, not A, IMO. Yes, the answer to this question is C not A.Working together at their constant rates, A and B can fill an empty tank to capacity in 1/2 hours. What is the constant rate of pump B?\(rate*time=job\). We are told that \((A+B)*30=C\), where \(A\) is the rate of pump A in lts/min, \(B\) is the rate of pump B in lts/min and C is the capacity of the tank in liters. Question: \(B=?\) (1) A's constant rate is 25 LTS / min > \(A=25\) > \((25+B)*30=C\) > clearly insufficient (two unknowns), if \(C=1200\), then \(B=15\) but of \(C=1500\), then \(B=25\). (2) The tanks capacity is 1200 lts. > \(C=1200\). \((A+B)*30=1200\) > \(A+B=40\). Also insufficient. (1)+(2) \(A=25\) and \(A+B=40\) > \(B=15\). Sufficient. Answer: C. Discussed here: workingtogetherattheirconstantratesaandbcanfill97316.html
_________________



Intern
Joined: 11 Oct 2012
Posts: 37

Re: Collection of work/rate problems?
[#permalink]
Show Tags
09 Nov 2012, 08:41
Bunuel wrote: brooksbrahs wrote: h2polo wrote: 15. working together at their constant rates , A and B can fill an empty tank to capacity in1/2 hr.what is the constant rate of pump B? 1) A's constant rate is 25LTS / min 2) the tanks capacity is 1200 lts.
From the original statement we know that: 1/A + 1/B = 1/(1/2)
From Statement 1 we can plug in and find B: SUFFICIENT
Statement 2 is completely unnecessary:
ANSWER: A. Statement 1 alone is sufficient I don't think this is correct. We know that the rate of A and B = 1/30 of a tank per minute (given in the stem). If A had said that pump A fills X OF THE TANK per minute, then it'd be sufficient alone since A + B = rate of A + rate of B and can solve for rate of B given that we know A. However, this gives it to us in units other than what the stem gave us, so we need to know the tank's capacity. Statement 2 gives us the capacity, so if we use them in conjunction, we can solve the problem. Should be C, not A, IMO. Yes, the answer to this question is C not A.Working together at their constant rates, A and B can fill an empty tank to capacity in 1/2 hours. What is the constant rate of pump B?\(rate*time=job\). We are told that \((A+B)*30=C\), where \(A\) is the rate of pump A in lts/min, \(B\) is the rate of pump B in lts/min and C is the capacity of the tank in liters. Question: \(B=?\) (1) A's constant rate is 25 LTS / min > \(A=25\) > \((25+B)*30=C\) > clearly insufficient (two unknowns), if \(C=1200\), then \(B=15\) but of \(C=1500\), then \(B=25\). (2) The tanks capacity is 1200 lts. > \(C=1200\). \((A+B)*30=1200\) > \(A+B=40\). Also insufficient. (1)+(2) \(A=25\) and \(A+B=40\) > \(B=15\). Sufficient. Answer: C. Discussed here: workingtogetherattheirconstantratesaandbcanfill97316.htmlThanks for the detailed answer, it was useful, but did my logic in my previous post make sense? I've found that for a lot of these data sufficiency questions, the best way to solve them is to figure out what the stem has given and what information needs to be found to solve it. From there, one can just analyze what the two statements give independently and then figure out if each or both together are sufficient together without necessarily solving it. Obviuosly, there will need to be calculation done to figure out what it's giving you, but I have noticed that I can save A LOT of time by not formally solving them and just conjecturing logically as to whether or not it's useful, particularly if I'm in a time crunch. Therefore, I just wanted to check if what I stated in my previous post made sense logically, because that's how I solved it and I want to make sure my method seems somewhat sound.



Intern
Status: Preparing for GMAT
Joined: 30 Aug 2012
Posts: 1
Location: United States
GPA: 3.81
WE: Engineering (Computer Hardware)

Re: Collection of work/rate problems?
[#permalink]
Show Tags
15 Nov 2012, 17:43
I think B would be the correct answer. Let machine y produce = Y bottles in 3 hrs ( rate = y/3 bottles/hr) Let machine x produce = 2Y bottles in 4 hrs ( rate = 2y/4 = y/2 bottles/hr) Total no of bottles produced by both x and y = 3Y. So, time machine x takes to produce 3Y bottles is = 3Y/(Y/2) = 6hrs. snipertrader wrote: 28.Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?
(1) Machine X produced 30 bottles per minute. (2) Machine X produced twice as many bottles in 4hours as machine Y produced in 3 hours.
Ans C With Statement 1  we cant find out the total work With Statement II  we can only express the speed of one machine in terms of the other
Both statements are needed for the complete picture



Math Expert
Joined: 02 Sep 2009
Posts: 59622

Re: Collection of work/rate problems?
[#permalink]
Show Tags
16 Nov 2012, 05:17
ks4gsb wrote: I think B would be the correct answer. Let machine y produce = Y bottles in 3 hrs ( rate = y/3 bottles/hr) Let machine x produce = 2Y bottles in 4 hrs ( rate = 2y/4 = y/2 bottles/hr) Total no of bottles produced by both x and y = 3Y. So, time machine x takes to produce 3Y bottles is = 3Y/(Y/2) = 6hrs. snipertrader wrote: 28.Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?
(1) Machine X produced 30 bottles per minute. (2) Machine X produced twice as many bottles in 4hours as machine Y produced in 3 hours.
Ans C With Statement 1  we cant find out the total work With Statement II  we can only express the speed of one machine in terms of the other
Both statements are needed for the complete picture Yes, the answer to this question is B. Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?Let the rate of X be \(x\) bottle/hour and the rate of Y \(y\) bottle/hour. Given: \(4x+3y=job\). Question: \(t_x=\frac{job}{rate}=\frac{job}{x}=?\) (1) Machine X produced 30 bottles per minute > \(x=30*60=1800\) bottle/hour, insufficient as we don't know how many bottles is in 1 lot (job). (2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours > \(4x=2*3y\), so \(3y=2x\) > \(4x+3y=4x+2x=6x=job\) > \(t_x=\frac{job}{rate}=\frac{job}{x}=\frac{6x}{x}=6\) hours. Sufficient. Answer: B. Discussed here: machinesxandyproducedidenticalbottlesatdifferent104208.html
_________________



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 402
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Re: Collection of work/rate problems?
[#permalink]
Show Tags
29 Dec 2014, 08:30
Hello,
Can I have the solution for 23 please? I am not sure if I understand the question to be honest though. Are we looking for the number of days that it would take Michael to finish the whole project seperately of after the 12 days that they have been working together? Or neither of the two..?
Thank you.



Math Expert
Joined: 02 Sep 2009
Posts: 59622

Re: Collection of work/rate problems?
[#permalink]
Show Tags
31 Dec 2014, 04:20
pacifist85 wrote: Hello,
Can I have the solution for 23 please? I am not sure if I understand the question to be honest though. Are we looking for the number of days that it would take Michael to finish the whole project seperately of after the 12 days that they have been working together? Or neither of the two..?
Thank you. This question is discussed here: http://gmatclub.com/forum/michealanda ... 61136.html== Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
_________________



NonHuman User
Joined: 09 Sep 2013
Posts: 13733

Re: Collection of work/rate problems?
[#permalink]
Show Tags
02 Nov 2019, 12:38
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.
_________________




Re: Collection of work/rate problems?
[#permalink]
02 Nov 2019, 12:38



Go to page
Previous
1 2 3 4
[ 72 posts ]



