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Collection of work/rate problems?

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Collection of work/rate problems? [#permalink] New post 27 Mar 2009, 23:15
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Guys, does anyone have a collection of work/rate problems??
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Re: Collection of work/rate problems? [#permalink] New post 15 Aug 2009, 10:53
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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
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Re: Collection of work/rate problems? [#permalink] New post 15 Aug 2009, 11:09
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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
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Re: Collection of work/rate problems? [#permalink] New post 15 Aug 2009, 11:35
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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.
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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 10:30
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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!
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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 07:01
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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
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Re: Collection of work/rate problems? [#permalink] New post 30 Mar 2009, 11:37
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I think

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


Q3 A(600)
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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 09:01
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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!
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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:13
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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
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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:13
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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
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Re: Collection of work/rate problems? [#permalink] New post 15 Aug 2009, 12:27
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This one was tough... because I kept mis-reading the question:

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

Here is the base equation that we want to set up:

let A = number of days where 25 people are working
let B = number of days where only 10 people are working (ie. after the 15 people leave)

(Rate of 25 people working)*(A)+(Rate of 10 people working)*(B) = 1 job

The trick is the figure out the rate of 10 people working:

If it takes 25 people to finish 1 job in 20 days, then it would take one person 20*25 days to finish 1 job or...

X number of people "X/(20*25)" days to complete 1 job.

So...

(Rate of 25 people working)*(A)+(Rate of 10 people working)*(B) = 1 job
(1/20)*A + (10/(20*25))*B = 1

We already know that B = 37.5

(1/20)*A + (10/(20*25))*37.5 = 1
A = 5

ANSWER A. 5
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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 04:52
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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
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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 07:04
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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
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 03:33
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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
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 06:52
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Okay... that's it. I answered every problem in the original post. Please do not take the answers as the official answers, because they are not.

If there are questions on them, please don't hesitate to ask...

But more importantly if you find an answer that is incorrect please post the correct solution for the benefit of others.

If you also find a faster way to do them, those shortcuts are definitely welcome as well.

I know for several of these problems, I wouldn't be able to do them in less than two mins.

Last edited by h2polo on 13 Nov 2009, 03:02, edited 1 time in total.
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Re: Collection of work/rate problems? [#permalink] New post 15 Sep 2009, 15:54
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h2polo wrote:
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.
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Re: Collection of work/rate problems? [#permalink] New post 15 Sep 2009, 16:48
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h2polo wrote:
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?


I like using RTW charts.

...........25m.......1m........10m
R ........1/20.....1/500......1/50
T..........20........500........50
W..........1..........1...........1

1/20*T + 1/50*37.5 = 1 => 1/20*T = 12.5/50 => T = 5
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Re: Collection of work/rate problems? [#permalink] New post 24 Mar 2010, 13:26
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At 28 I also believe it is B :

Here is my reasoning:
Let´s say that X produced A bottles in 4h => Y produced A/2 products in 3h
 X produced A/4 bottles in 1h
 So, for X to produce all the amount alone: A/4 * W hours = A + A/2 =>
W=6 hours
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Re: Collection of work/rate problems? [#permalink] New post 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] New post 08 Nov 2012, 10:49
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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] 08 Nov 2012, 10:49
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