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Collection of work/rate problems?

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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
_________________

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.

Senior Manager
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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:14
22.Working alone at its own constant rate, a machine seals k cartons in 8 hours,
and working alone at its own constant rate, a second machine seals k cartons in
4 hours. If the two machines, each working at its own constant rate and for the
same period of time, together sealed a certain number of cartons, what percent
of the cartons were sealed by the machine working at the faster rate?

25%
33 1/3%
50%
66 2/3%
75%
_________________

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.

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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:17
21.One hour after Yolanda started walking from X to Y, a distance of 45 miiles, Bob started walking along the same road from Y to X. If Yolanda's walking rate was 3 miles /hour and Bob's was 4 miles / hour, how many miles had Bab walked when they met?

24
23
22
21
19.5

Solution - using KISS approach - Speed of both persons are added 3+4 =7

Yolanda already traveled 3 miles ( 3 m x 1 hr)
So remaining distance = 45-3 = 42 miles

So bobs time = 42/7 = 6 hrs

So bob walked = 4 x 6 = 24 miles
_________________

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.

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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:17
20.A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

a) 3

b) 4

c) 6

d) 9

e) 12
_________________

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Re: Collection of work/rate problems? [#permalink] New post 05 Aug 2009, 23:18
19.Machince A and B are each used to manufacture 660 sprockets. It takes A 10 hours longer to produce 660 sprockets than machine B. B produces 10 percent more sprockets per hour than A. How many sprockets per hours does machine A produce?

A. 6
B. 6.6
C. 60
D. 100
E 110
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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 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 16 Aug 2009, 02:53
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

let R = the growth of the tree each year

height in four years = 4+4*R
height in six years = 4+6*R

therefore,

4+6*R=(6/5)*(4+4*R)

R = 2/3
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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 02:59
thanks h2polo

I was looking for some of the solutions that i could not solve.
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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 03:35
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

From problem statement we can get:
4*x = 5*y

From Statement 1 we get:
y = .80*x

At first glance, it looks like we can solve the problem with these two equations, but upon closer inspection we see that they are the same equation:
y = .80*x
y = 8/10*x
10*y = 8*x
5*y = 4*x

Statement 1 - NOT SUFFICIENT

From Statement 2 we get:
x + y = $43.75

SUFFICIENT

ANSWER: B. Statement 2 alone is sufficient
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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 04:14
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 10-A:

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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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 09:09
9. With both valves open, the pool will be filled with water in 48 minutes. The first valve alone would fill the pool in 2 hours. What is the capacity of the pool if every minute the second valve admits 50 cubic meters of water more than the first?
* 9000 cubic meters
* 10500 cubic meters
* 11750 cubic meters
* 12000 cubic meters
* 12500 cubic meters

let B = # of mins for the second valve alone to fill the pool

1/(time it takes first valve to fill the pool) + 1/B = 1/(total time its takes to fill the pool)
1/120 + 1/B = 1/48
B = 80

let C = the capacity of the pool
C/80 - C/120 = 50
C = 12000

ANSWER: D. 12000
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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 16 Aug 2009, 10:48
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
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Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2009, 11:08
12.At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full 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

let X = # of hours for the tank to empty

X*Rate A + X*Rate B - X*Rate C = 1/2 tank

X*(1/2) + X*(1/3) - X*(1/6) = 1/2
X = 3/4

It takes 3/4 hrs to empty the tank

ANSWER: C. 3/4

Let me know if the logic here is flawed...
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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, 07:01
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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 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
Re: Collection of work/rate problems?   [#permalink] 17 Aug 2009, 07:04
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