I have been having real problems with work problems for some reason, so I am going to solve each and post my solutions... feel free to correct any mistakes!
1.
Rate of X: 1 Job / 12 Hrs
Rate of Y: 1/10*Rate of X = (1/10)(1/12) = 1 Job / 120 Hrs
Therefore,
(1/3 Job)*Rate of Y = (1/3 Job)* (120 Hrs / 1 Job) = 40 Hrs
ANSWER: 40 Hrs
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Sat Aug 15, 2009 12:09 pm
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
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
let Y = number of days for machine Y to produce w widgets
Rate of Y = w widgets / Y days Rate of X = w widgets / (Y+2) days
Formula: Rate (together) = Rate of Machine X + Rate of Machine Y
(5/4)*w/3 = w/Y+w/(Y+2) or (5/4)/3 = 1/Y+1/(Y+2)
simplify:
5*Y^2-14Y-24=0
then I used the quadratic formula to get Y = 4
therefore,
2w widgets * 1/Rate of X 2w widgets * (Y+2) days / w widgets 2w widgets * 6 days / w widgets = 12 days
ANSWER: E. 12 days
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Sat Aug 15, 2009 12:35 pm
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
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 ? A:600 B:800 C:1000 D:1200 E:1500
let B = time it takes for B to complete 1 job
Standard Work Formula: 1/(time it takes for A and B to complete together) = 1/(time it takes for A to complete) + 1/(time it takes for B to compelte)
1/24 = 1/60 +1/B B = 40 min
let A = # of pages printer A can print in 1 min
Pages per min equation for Printer A: (A pages / 1 min)*(60 min) = 1 job
Pages per min equation for Printer B: (A+5 pages / 1 min)*(40 min) = 1 job
therefore,
(A pages / 1 min)*(60 min) = (A+5 pages / 1 min)*(40 min) A = 10 pages / 1 min
(10 pages / 1 min)*(60 min) = 600 pages!
ANSWER: E. 600 pages
Again... let me know if you guys agree with the answer or have a faster way of doing them.
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Sun Aug 16, 2009 11:30 am
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
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!
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Mon Aug 17, 2009 8:01 am
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
14.John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?
Rate of John + Rate of Jane = Rate Together 1/20+1/12=1/T T=7.5
let X = days working together
X*(1/7.5) + 4*(1/20) = 1 job
X = 6
therefore,
6+4 = 10 days
nitya34
Re: Collection of work/rate problems? [#permalink]
Posted: Mon Mar 30, 2009 12:37 pm
Director
Affiliations: CFA Level 1 Joined: Fri Jan 04, 2008 Posts: 954 Location: India Schools: One Year MBA Followers: 18
1.Working, independently X takes 12 hours to finish a certain work. he finishes 2/3 of the work . The rest of the work is finished by Y whose rate is 1/10 th of X. In how much time does Y finsh his work?
Q1->is it 40 hrs?
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
Q2 E(12)and
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 ? A:600 B:800 C:1000 D:1200 E:1500
14. John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?
Ans 10
Solution
Let the total no of days worked = n So jane worked n-4 days
Work rate of John = 1/20 Work rate of John = 1/12
setting up the equation R x t = w janes+ johns work + Johns work = total work
(1/20+1/12) x (n-4) + (1/20 *4) = 1 Solve for N
Sorry, i dont know how to use the proper symbols in the post..so its kinda crude!
_________________ If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.
snipertrader
Re: Collection of work/rate problems? [#permalink]
Posted: Thu Aug 06, 2009 12:13 am
Senior Manager
Affiliations: ACA, CPA Joined: Sun Apr 26, 2009 Posts: 453 Location: Vagabond Schools: BC WE 1: Big4, Audit WE 2: Banking Followers: 3
25.A certain car averages 25 miles per gallon of gasoline when driven in the city and 40 miles per gallon when driven on the highway. According to these rates, which of the following is closest to the number of miles per gallon that the car averages when it is driven 10 miles in the city and then 50 miles on the highway?
28 30 33 36 38
Ans 36 (approx)
Formula = Total distance traveled/total Gallons of fuel used for the same
_________________ If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.
snipertrader
Re: Collection of work/rate problems? [#permalink]
Posted: Thu Aug 06, 2009 12:13 am
Senior Manager
Affiliations: ACA, CPA Joined: Sun Apr 26, 2009 Posts: 453 Location: Vagabond Schools: BC WE 1: Big4, Audit WE 2: Banking Followers: 3
23.Micheal and Adam can do together a piece of work in 20 days. After they have worked together for 12 days Micheal stops and Adam completes the remaining work in 10 days. In how many days Micheal complete the work separately.
80 days 100 days 120 days 110 days 90 days
_________________ If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Sat Aug 15, 2009 1:27 pm
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
13.Working together at their respective rates, machine A, B, and C can finish a certain work in 8/3 hours. How many hours will it take A to finish the work independently? (1) Working together, A and B can finish the work in 4 hours. (2) Working together, B and C can finish the work in 48/7 hours.
let A = time it takes machine A to finish the job by itself let B = time it takes machine B to finish the job by itself let C = time it takes machine C to finish the job by itself
From the orginal statement we know that: 1/A+1/B+1/C = 1/(8/3)
From Statement 1 we know that: 1/A + 1/B = 1/4
INSUFFICIENT
From Statement 2 we know that 1/B + 1/C = 1/(48/7)
SUFFICIENT
ANSWER: B. Statement 2 alone is sufficient
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Mon Aug 17, 2009 8:04 am
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
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
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Tue Aug 18, 2009 4:33 am
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
26.Working together, John and Jack can type 20 pages in one hour. They will be able to type 22 pages in one hour if Jack increases his typing speed by 25%. What is the ratio of Jack's normal typing speed to that of John? 1/3 2/5 1/2 2/3 3/5
let N = rate of John typing let K = rate of Jack typing
N + K = 20 N + 1.25*K = 22
Solve the system of equations: N = 12 K = 8
K/N = 8/12
ANSWER: D. 2/3
h2polo
Re: Collection of work/rate problems? [#permalink]
Posted: Tue Aug 18, 2009 7:52 am
Manager
Joined: Thu Aug 13, 2009 Posts: 214 Schools:Sloan Followers: 2
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.
_________________ Please kudos if my post helps.
powerka
Re: Collection of work/rate problems? [#permalink]
Thanks for the url there are 30 probs lets solve 3 at a time ==================== 1.Working, independently X takes 12 hours to finish a certain work. he finishes 2/3 of the work . The rest of the work is finished by Y whose rate is 1/10 th of X. In how much time does Y finsh his work?
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
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 ===============
=========Next 3 Qs are========= 4.25 men reap a field in 20 days . when should 15 men leave the work.if the whole field is to be reaped in 37-1/2 days after they leave the work?
A 5 days B 10 days
5.Read the question from the diagram above and (WE CAN AVOID THIS) C 7 days D 7-1/2 days
6.when a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year? 3/10 2/5 1/2 2/3 6/5 ============