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Re: Collection of work/rate problems?
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04 Jun 2010, 09:11
srivas wrote: 3.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ? 1:600 2:800 3:1000 4:1200 5:1500
Soln: Printer A alone finishes the task in 60 mins Hence work done in 1 min is (1/60) Let printer B alone finish the task in B mins. Hence work done in 1 min is (1/B) Together they finish in 24 mins. Hence work done in 1 min is (1/24)
Now, we get equation, (1/60) + (1/B) = (1/24) Solving, we get B = 40 mins.
Number of papers that A prints per minute be = X Hence number of papers that B prints per minute be = (X + 5)
Total papers printer in 24 mins will be = 24X + 24(X+5)  eq(1)
If A were to work alone it will finish task in 60 mins. Hence papers printed in 60 mins will be = 60X  eq(2)
Equating eq(1) & eq(2) we get, 24X + 24(X+5) = 60X Solving for X, we get X = 10.
Hence the task contains 60X pages to be printed = 60 * 10 = 600 pages. Ans: 1 Slightly different approach: After 24 minutes, A will have completed 24/60=40% of the task Therefore, the portion completed by B = 100%40%=60% Therefore, B's rate is 60%/40%=1.5 times A's rate Given B's rate is 5 pages more than A's rate by the problem, 0.5 times A's rate is 5 ppm => A's rate = 10 ppm, B's rate is 15 ppm 15*24+10*24=5*24+20*24=120+2*10*24=600



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Re: Collection of work/rate problems?
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30 Jul 2010, 12:45
srivas wrote: 2.Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
A. 4 B. 6 C. 8 D. 10 E. 12
Soln: To produce w widgets, .... ..... .....
Therefore, (1/D) + 1/(2 + D) = (5/12)
Solving, we get D = 4.
X takes 6 days to do w widgets. Therefore it will take 12 days to produce 2w widgets. Ans:E How in the world am I going to solve that quadratic equation quickly in the actual exam? One of the answers is 1.20. I know a negative answer is not required but is there a faster way of doing this problem? Thank you in advance.



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Re: Collection of work/rate problems?
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01 Aug 2010, 11:08
h2polo wrote: 8. A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened? *at 2:00 pm * at 2:30 pm * at 3:00 pm * at 3:30 pm * at 4:00 pm
Here is how I solved the problem:
let A = the number of hours the filling valve is open by itself let B = the number of hours the filling and draining valve are open together
Filling rate: 1 pool / 4 hrs Draining rate: 1 pool / 5 hrs
therefore,
(1 pool / 4 hrs)*(A+B hrs)  (1 pool / 5 hrs)*(B hrs) = 1 pool
and we know that the pool was filled in 10 hours:
A + B = 10
so now we have two equations and two unknowns; solve for A:
(A+B)/4  B/5 = 1 5*(A+B)  4*5 = 20 5*A + B = 20
substitute B for 10A:
5*A + 10  A = 20 A = 2.5
So the pool was filled 2 and half hours after 1 PM or 3:30 PM
ANSWER: E. 3:30 PM You mean D: 3:30 right? Here is a different approach. R = d/t We know the the size of the pool (d) is the same. So R1t1 = R2D2 or 1/4*10 = 1/5*t2 t2 = 50/4 Subtract this from 10: 1050/4 = 2.5 (Ignore the negative) Add this to 1 and gives you 3:30. Does this method make sense?



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Re: Collection of work/rate problems?
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16 Aug 2010, 06:08
8. Answer is 3:30 PM. The drain pipe has been opened at 3:30 pm..let me know if u need a solution



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Re: Collection of work/rate problems?
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16 Aug 2010, 06:56
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 This question is solved wrongly. The correct answer is "B".Statement 2 will be sufficient, since it gives us a comparison of X and Y's rates. We know: the amount of work Y did in 3 hours was half what X did in 4 hours. In other words, if X worked for 2 hours, X would do the same work that Y did in 3 hours. Since we know that 4 hours of X and 3 hours of Y is enough to finish the job, and 3 hours of Y is equivalent to 2 hours of X, then X working alone would take 6 hours to do the job.



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Re: Collection of work/rate problems?
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16 Aug 2010, 07:01
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?
My Solution:
This Question is based on Mandays fundamental.
6 machines work for 12 days to complete 1 job= 6*12 mandays (machine=man)
Now if we reduce days in this formula, machines will increase to compensate the reduction in days :
so we have :
6 mac*12 days = x mac * 8 days x= 9 machines  a delta of 3 machines in addition ANS = 3 machines



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Re: Collection of work/rate problems?
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16 Aug 2010, 07:14
12.At their respective rates, pump A, B, and C can fulfill an empty tank, or pumpout the full tank in 2, 3, and 6 hours. If A and B are used to pumpout water from the halffull tank, while C is used to fill water into the tank, in how many hours, the tank will be empty? A. 2/3 B. 1 C. 3/4 D. 3/2 E. 2
My Approach  Supposedly the shortest and the best to solve such problems:
whenever we see such problems where time taken by pipes a (2), b(3) ,c(6) are given , take out the LCM of these values (the number which will be divisible of all these 3 values): LCM of 2,3,6 = 12 Now this figure 12 is the total unit of Work (work done to fill full tank): A does work > 12/2= 6units /hour B does work > 12/3= 4units /hour C does work > 12/6= 2units /hour
for half tank > total units of work=6 units the equation can be written as > x(AB+C) =6 (x is no of hours required and (AB+C) is work done in 1 hour  also '' bcos A and B pumps out)
x(64+2) = 6 x=6/8 = 3/4
The explanation looks big, but if put inpractice, we can solve such questions without a penpaper



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Re: Collection of work/rate problems?
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08 Sep 2010, 23:36
Hi where can I get the full list of the answers. Am confused with different responses to same qs!



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Re: Collection of work/rate problems?
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06 Nov 2010, 16:14
srivas wrote: 2.Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
A. 4 B. 6 C. 8 D. 10 E. 12
Soln: To produce w widgets, Time taken by machine Y is = D days Then, Time taken by machine X is = D + 2 days
Now, Since both machines working together produce (5/4)w widgets in 3 days, the number of days in which they will produce w widgets working together is,
(5/4)w => 3 days w => K days
By direct variation, k = (4/5) * 3 = (12/5) days
Now, Since machine Y takes D days to finish the work, Hence work done in one day = (1/D) Similarly work done by X in one day = 1/(2 + D) Together they do (5/12) of work in one day
Therefore, (1/D) + 1/(2 + D) = (5/12)
Solving, we get D = 4.
X takes 6 days to do w widgets. Therefore it will take 12 days to produce 2w widgets. Ans:E Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you!



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Re: Collection of work/rate problems?
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06 Nov 2010, 22:08
gettinit wrote: srivas wrote: 2.Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
A. 4 B. 6 C. 8 D. 10 E. 12
Soln: To produce w widgets, Time taken by machine Y is = D days Then, Time taken by machine X is = D + 2 days
Now, Since both machines working together produce (5/4)w widgets in 3 days, the number of days in which they will produce w widgets working together is,
(5/4)w => 3 days w => K days
By direct variation, k = (4/5) * 3 = (12/5) days
Now, Since machine Y takes D days to finish the work, Hence work done in one day = (1/D) Similarly work done by X in one day = 1/(2 + D) Together they do (5/12) of work in one day
Therefore, (1/D) + 1/(2 + D) = (5/12)
Solving, we get D = 4.
X takes 6 days to do w widgets. Therefore it will take 12 days to produce 2w widgets. Ans:E Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you! Check the solution here: workproblem98599.html#p759876Similar questions: agoodone98479.html?hilit=point%20quadratichourstotypepages102407.html?hilit=point%20quadraticHope it helps.
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Re: Collection of work/rate problems?
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08 Nov 2010, 20:21
Bunuel wrote: gettinit wrote: srivas wrote: 2.Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?
A. 4 B. 6 C. 8 D. 10 E. 12
Soln: To produce w widgets, Time taken by machine Y is = D days Then, Time taken by machine X is = D + 2
Now, Since both machines working together produce (5/4)w widgets in 3 days, the number of days in which they will produce w widgets working together is,
(5/4)w => 3 days w => K days
By direct variation, k = (4/5) * 3 = (12/5) days
Now, Since machine Y takes D days to finish the work, Hence work done in one day = (1/D) Similarly work done by X in one day = 1/(2 + D) Together they do (5/12) of work in one day
Therefore, (1/D) + 1/(2 + D) = (5/12)
Solving, we get D = 4.
X takes 6 days to do w widgets. Therefore it will take 12 days to produce 2w widgets. Ans:E Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you! Check the solution here: workproblem98599.html#p759876Similar questions: agoodone98479.html?hilit=point%20quadratichourstotypepages102407.html?hilit=point%20quadraticHope it helps. Thanks Bunuel, this was very helpful! appreciate your responsiveness.



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Re: Collection of work/rate problems?
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29 Nov 2010, 11:43
10.It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages? * 12 * 18 * 20 * 24 * 30
Rate of printer A and printer B printing 40 pages together:
40 pages * 6 min / 50 pages
let B = # of mins for printer B to print 40 pages
Rate of Printer A + Rate of Printer B = Rate of Printer A and B 40/(B+4) + 40/B = 40*6/50 simplify: 5B^2 + 28*B + 96 = 0 (I had to use the quadratic formula here) B = 8
Rate of Printer A = 40 pages / (8+4) min
therefore
Rate of printer A to print 40 pages:
80 pages * (12 min / 40 pages) = 24 mins
ANSWER: D. 24 mins
This one took me a long time to solve... way more than 2 mins. If anyone finds a quicker way to solve this, please post!
There is a mistake in the explanation above.
The answer is indeed 24 mins, which is derived from the equation 5B^2 + 28*B + 96 = 0, just as h2polo said.
BUT>
Rate of Printer A + Rate of Printer B = Rate of Printer A and B = 40/(B+4) + 40/B = 40*6/50  this is wrong; solve, and you get 24B^2  304*B  800 = 0, which is not equal to 5B^2 + 28*B + 96 = 0.
Rate of Printer A and B = 40/(B+4) + 40/B = 50/6  this is correct; from there you get 5B^2 + 28*B + 96 = 0, and then experience the enjoyment of finding its roots.
Positive root is 8 > 8+4=12 is the time it takes A to print 40 pages > 12*2=24 is the time it takes A to print 80 pages Can somebody please break down this section:
Rate of Printer A and B = 40/(B+4) + 40/B = 50/6  this is correct; from there you get 5B^2 + 28*B + 96 = 0
Im having difficulty doing the math...Much appreciated.
Thank you...
Casey



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Re: Collection of work/rate problems?
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02 Dec 2010, 10:57
i am in the same boat as casey..... Rate of Printer A and B = 40/(B+4) + 40/B = 50/6  this is correct; from there you get 5B^2 + 28*B + 96 = 0
Im having difficulty doing the math...Much appreciated. i know the 40/(B+4) + 40/B should be in the form XY/x+y...but i am not getting the final equation.



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Re: Collection of work/rate problems?
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02 Dec 2010, 11:09
anish319 wrote: i am in the same boat as casey..... Rate of Printer A and B = 40/(B+4) + 40/B = 50/6  this is correct; from there you get 5B^2 + 28*B + 96 = 0
Im having difficulty doing the math...Much appreciated. i know the 40/(B+4) + 40/B should be in the form XY/x+y...but i am not getting the final equation. First of all: in case of any question or doubt about specific problem please post this problem in PS or DS subforum.As for this problem: It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages?A. 12 B. 18 C. 20 D. 24 E. 30 Let the time needed to print 40 pages for printer A be \(a\) minutes, so for printer B it would be \(a4\) minutes. The rate of A would be \(rate=\frac{job}{time}=\frac{40}{a}\) pages per minute and the rate of B \(rate=\frac{job}{time}=\frac{40}{a4}\) pages per minute. Their combined rate would be \(\frac{40}{a}+\frac{40}{a4}\) pages per minute. Also as "two printers can print 50 pages in 6 minutes" then their combined rate is \(rate=\frac{job}{time}=\frac{50}{6}\), so \(\frac{40}{a}+\frac{40}{a4}=\frac{50}{6}\). \(\frac{40}{a}+\frac{40}{a4}=\frac{50}{6}\) > \(\frac{1}{a}+\frac{1}{a4}=\frac{5}{24}\). At this point we can either try to substitute the values from answer choices or solve quadratic equation. Remember as we are asked to find time needed for printer A to print \(80\) pages, then the answer would be \(2a\) (as \(a\) is the time needed to print \(40\) pages). Answer D works: \(2a=24\) > \(a=12\) > \(\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\). Answer: D. Similar problems: workproblem98599.html#p759876hourstotypepages102407.html?hilit=answer%20choices%20or%20solve%20quadratic%20equation.%20RTheory about work problems: wordtranslationsrateswork104208.html?hilit=printer#p812628Hope it helps.
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Re: Collection of work/rate problems?
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11 Jan 2011, 15:56
hi does anyone have the answers to 4, 5, 8, 9, 10, 11, 12, 17,18,19 ?
regards



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Re: Collection of work/rate problems?
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16 Sep 2011, 06:29
snipertrader wrote: 30.Company S produces two kinds of stereos: basic and deluxe. Of the stereos produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe stereo as it does to produce a basic stereo, then the number of hours it took to produce the deluxe stereos last month was what fraction of the total number of hours it took to produce all the stereos?
A.7/17 B.14/31 C. 7/15 D.17/35 E.1/2
Solution
2/3 were basic so 1/3 will be deluxe Assuming it took 1 hr to complete Basic stereo ; deluxe will take 7/5 X 1 hr
Total time taken (basic radio + deluxe radio) = 2/3 + (1/3 x 7/5) = 17/15
Ratio of hrs needed for deluxe ratio to total hrs = (7/15)/(17/15) = 7/17 I can't seem to figure out why you multiplied 1/3 by 7/5. I assume you are right because the answer I get isn't one of the choices, so I'm hoping someone can explain to me why this was done. I try to solve these problems using a small chart: (Total=Rate*Time) As you stated, let's assume it takes 1 hr to make a B stereo. Basic Total = 2/3 Rate = x time = 1hr therefore rate (x) = (2/3) / 1 = 2/3 DeluxeTotal = 1/3 Rate = y time = 7/5 *1hr = 7/5 therefore rate (y) = (1/3) / (7/5) = 5/21so 1/x + 1/y = 1/total time (t) 1/(2/3) + 1/(5/21) = 1/t t = 3/2 + 21/5 = 15/10 + 42/10 = 57/10so to find the fraction of Deluxe Time/Total time = (7/5)/(57/10) = not the right answer I can see that in order to match the snipertrader's solution, I need to rearrange my 'total', 'rate' & 'time' to be: BasicTotal = x rate = 1 time = 2/3 Deluxetotal = y rate = 7/5 time = 1/3 in which case you would get 7/17 as the answer, but I can't make the leap as to why 2/3 and 1/3 need to be the TIME and not the TOTAL? I'm hoping someone can walk me through this. Thank you!!



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Re: Collection of work/rate problems?
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21 Sep 2011, 09:12
Answers to first 15. 1.40 2.C (8) 3.A(600) 4.8 1/3 6.2/3 7.C (97.2) 8.3:30pm 9.12,000 10. 24 11. 3 12. 3/4 13. D (1/8) 14. 10 15. C Rest tomorrow.
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Re: Collection of work/rate problems?
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22 Sep 2011, 11:21
16. C x3/3x 17. A 18 18. B 3 19. A 6 20. C 6 21. A 24 22 D 66 2/3 23.B 100 24 D 25 25 D 36 26 D 2/3 27.E 4/9 Missing Answer choices for Q.27 are: (A) 1/9 (B) 1/6 (C) 1/3 (D) 7/18 (E) 4/9 28. B 29. D 12/7 30. 7/17 A
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Re: Collection of work/rate problems?
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22 Oct 2011, 07:47
Need help understanding this 7.If Jim earns x dollars per hour, it will take him 4 hours to earn exactly enough money to purchase a particular jacket. If Tom earns y dollars per hour, it will take him exactly 5 hours to earn enough money to purchase the same jacket. How much does the jacket cost? (1) Tom makes 20% less per hour than Jim does. (2) x + y = $43.75
The answer is B, but I want to understand how I can figure this out without solving the whole problem.
From the question we have 4x = 5y
From (1) we have y = 0.8x (or y  0.8x = 0) From (2) we have x + y = 43.75 (or y = 43.75  x)
At this point I see I have a new equation from both (1) and (2) and my initial response is that both are individually sufficient (D).
How can I figure out at this point that (1) is not sufficient and (2) is sufficient without spending much time in solving all the equations.
thanks, Vinay



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Re: Collection of work/rate problems?
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16 Dec 2011, 09:13
No need to use any algebra here. A = tap that is filling the pool B = leak in the pool A is running for 10 hrs, which means it can fill 2.5 times the pool in that time. However, it is able to fill just 1 pool, which means that B leaks 1.5 times the pool in that time. Now, B can empty 1 pool in 5 hrs, so it will take 7.5 hrs to empty 1.5 times the pool. Therefore, leak started 2.5 hrs after A started = which is 3.30pm. h2polo wrote: 8. A pool can be filled in 4 hours and drained in 5 hours. The valve that fills the pool was opened at 1:00 pm and some time later the drain that empties the pool was also opened. If the pool was filled by 11:00 pm and not earlier, when was the drain opened? *at 2:00 pm * at 2:30 pm * at 3:00 pm * at 3:30 pm * at 4:00 pm
Here is how I solved the problem:
let A = the number of hours the filling valve is open by itself let B = the number of hours the filling and draining valve are open together
Filling rate: 1 pool / 4 hrs Draining rate: 1 pool / 5 hrs
therefore,
(1 pool / 4 hrs)*(A+B hrs)  (1 pool / 5 hrs)*(B hrs) = 1 pool
and we know that the pool was filled in 10 hours:
A + B = 10
so now we have two equations and two unknowns; solve for A:
(A+B)/4  B/5 = 1 5*(A+B)  4*5 = 20 5*A + B = 20
substitute B for 10A:
5*A + 10  A = 20 A = 2.5
So the pool was filled 2 and half hours after 1 PM or 3:30 PM
ANSWER: E. 3:30 PM




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