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Re: Collection of work/rate problems? [#permalink]
29 May 2012, 09:54
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.
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***
Re: Collection of work/rate problems? [#permalink]
29 May 2012, 10:17
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.
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***
Re: Collection of work/rate problems? [#permalink]
08 Nov 2012, 10:18
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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 = 9-6 3 _________________
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Re: Collection of work/rate problems? [#permalink]
08 Nov 2012, 10:49
1
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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. _________________
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Re: Collection of work/rate problems? [#permalink]
09 Nov 2012, 04:20
Expert's post
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.
Re: Collection of work/rate problems? [#permalink]
09 Nov 2012, 07:41
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.
Thanks 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. _________________
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Re: Collection of work/rate problems? [#permalink]
15 Nov 2012, 16:43
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
Re: Collection of work/rate problems? [#permalink]
16 Nov 2012, 04:17
Expert's post
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.
Re: Collection of work/rate problems? [#permalink]
10 Nov 2014, 17:19
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Re: Collection of work/rate problems? [#permalink]
29 Dec 2014, 07:30
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..?
Re: Collection of work/rate problems? [#permalink]
31 Dec 2014, 03:20
Expert's post
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..?
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