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Collection of work/rate problems?

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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 07:11
16.Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?

1/3x
3x/x-3
x-3/3x
x/x-3
x-3/x

let y = amount of room painted by Joseph in 20 min

Rate of Lindsay + Rate of Joseph = Rate Total

(1/x)/20 + y/20 = (1/3)/20
y = (x-3)/(3*x)

ANSWER: C.
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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 07:20
17.Machines X and Y run at different constant rates, and machine X can complete a certain job in 9 hours. Machine X worked on the job alone for the first 3 hours and the two machines, working together, then completed the job in 4 more hours. How many hours would it have taken machine Y, working alone, to complete the entire job?
18
13+1/2
7+1/5
4+1/2
3+2/3

First we need to find the rate together:

let T = time it takes for X and Y to complete one job together in hours

3 hrs * (1 job / 9 hrs) + 4 hrs * (1 job / T hours) = 1 job
T = 6

Rate of X + Rate of Y = Rate Together
1/9 + 1/Y = 1/T
1/9 + 1/Y = 1/6
Y = 18

ANSWER: A. 18 hours
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Re: Collection of work/rate problems? [#permalink] New post 17 Aug 2009, 11:41
18.Six machines, each working at the same constant rate, together can complete a certain 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?

A) 2
B) 3
C) 4
D) 6
E) 8

Rate of one machine = 1 job / (12*6) days

let X = number of machines needed to complete the job in 8 days

1/(6*12) * 8 * X = 1 job

X = 9

9-6 = 3

ANSWER: B. 3
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:36
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

let B = amount of time B takes to produce 660 sprockets in hours

Rate of Machine A:
660 / (10 +B)

Rate of Machine B:
660 / B

B produces 10% more sprockets per hour than A:
(660 / B) / (660 / (10 +B)) = 1.1
B = 100

therefore,

660/(10+100) = 6

ANSWER: A. 6
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:39
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

Rate of Type R * # of Machines + Rate of Type S * # of Machines = Rate Together

let M = # of machines

M/36 + M/18 = 1/2

M = 6

ANSWER: C. 6
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:41
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

ANSWER: A. 24
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:42
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%

ANSWER: D.
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:44
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

ANSWER: B. 100
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 01:49
24.Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
26days
27days
23days
25days
24 days

Rate Together * # of days working together + Rate of Peter * # of days working alone = 1 completed job

let P = 3 of hours Peter can complete one job alone

(1/20)*12 + (1/P)*10 = 1

P = 25
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 02:02
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

Total miles driven: 50 + 10 = 60

Total gallons used driving in the city:
10 miles * (1 gal / 25 gal) = 2/5

Total gallons used driving on the highway:
50 mi * (1 gal / 40 gal) = 5/4

60 / (2/5 + 5/4) = 36....

ANSWER: D. 36
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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, 04:03
27.Tom, working alone, can paint a room in 6 hours. Peter and John, working independently, can paint the same room in 3 hours and 2 hours, respectively. Tom starts painting the room and works on his own for one hour. He is then joined by Peter and they work together for an hour. Finally, John joins them and the three of them work together to finish the room, each one working at his respective rate. What fraction of the whole job was done by Peter?

let X = amount of time in hours for Tom, Peter, and John to complete the job

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

X = 1/3

Total work done by Peter:

1/3 + (1/3)*(1/3) = 4/9

ANSWER: 4/9
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 04:20
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
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 04:58
29.One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?

(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7

let S = time it takes a smurf to finish a treehouse
let E = time it takes an elf to finish a treehouse
let F = time it takes a fairy to finish a treehouse

1/S + 1/E = 1/2
1/2 + 2/F = 1/2
1/E + 1/F = 1/4

3 equations and three unkowns

solve for X:

1/S + 1/F + 1/E = 1/X

X = 12/7

ANSWER: D. 12/7
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Re: Collection of work/rate problems? [#permalink] New post 18 Aug 2009, 06:48
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

let T = total number of stereos produced
let B = number of hours it takes to produce 1 basic radio

total number of hours to produce delux radios, TD: (1/3*T)*(7/5*B)

total number of hours to produce basic radios, TB: (2/3*T)*B

TD / (TB + TD) = 7/17

ANSWER: A. 7/17
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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 02 Sep 2009, 18:33
h2polo, thank a lot, great post !!

could you explain #15 a bit more in detail. I don't see how A by itself is sufficient, I got C as the answer, thanks

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
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Re: Collection of work/rate problems? [#permalink] New post 04 Sep 2009, 04:07
Max,

Good catch! The equation I meant to use was:

Rate of A + Rate of B = Rate of A&B working together

Statement 1 gives us the Rate of A, but in order to figure out the Rate of A&B we need to know what the tank capacity is:

Rate of A&B = 1200 lts / 30 min
Rate of A = 25 lts / 1 min

The answer is therefore C. Both statements together are sufficient
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Re: Collection of work/rate problems? [#permalink] New post 06 Sep 2009, 05:23
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?

Soln: Since X finishes 2/3 of the work, the work left over for Y to finish is 1/3.

Since the rate of Y is 1/10th that of X, hence Y will take 10 times the TIME that X takes to finish the same work. Hence Y will finish the same work in
= 12 * 10 = 120 hours

Therefore, If Y takes 120 hours to finish the work of 1 unit, 1/3 units of work will be finished by Y in = (1/3) * 120 = 40 hours.
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Re: Collection of work/rate problems? [#permalink] New post 06 Sep 2009, 05:32
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
Re: Collection of work/rate problems?   [#permalink] 06 Sep 2009, 05:32
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