Collection of work/rate problems? : GMAT Quantitative Section - Page 4
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 09 Dec 2016, 12:24

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Collection of work/rate problems?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 140 [0], given: 3

Re: Collection of work/rate problems? [#permalink]

### Show Tags

06 Sep 2009, 05:41
1
This post was
BOOKMARKED
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
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 140 [0], given: 3

Re: Collection of work/rate problems? [#permalink]

### Show Tags

06 Sep 2009, 05:49
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

Soln: Initial Height of tree = 4 feet
Let It increase by X feet each year.

Then the height of tree at end of 4th year is = 4 + 4X
Then the height of tree at end of 6th year is = 4 + 6X

Since the tree was 1/5 taller than it was at end of the 4th year,
= 4 + 6X = (4 + 4X) * 6/5

Solving we get,
X = 2/3.

Ans: 2/3
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 140 [0], given: 3

Re: Collection of work/rate problems? [#permalink]

### Show Tags

06 Sep 2009, 09:48
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

Soln: From the initial statements we get the price P of the jacket to be
P = 4x = 5y

Now considering statement 1 alone, it says Tom makes 20% less per hour than Jim does. This is already obtained from the initial statements of 4x = 5y.
Therefore y = .8x . This statement is not enough to find the price of the jacket.
Hence Statement 1 alone is INSUFFICIENT

Now considering statement 2 alone, we have an equation in x and y. Since we also have another equation from the initial statements, the two equations are,

4x = 5y => eq(1)
x + y = 43.75 => eq(2)

Since there are two equations we can solve for x or y and hence find the price of jacket.
Hence Statement 2 alone is SUFFICIENT

Manager
Joined: 22 Jul 2009
Posts: 191
Followers: 4

Kudos [?]: 258 [1] , given: 18

Re: Collection of work/rate problems? [#permalink]

### Show Tags

15 Sep 2009, 15:54
1
KUDOS
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
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

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.

Manager
Joined: 22 Jul 2009
Posts: 191
Followers: 4

Kudos [?]: 258 [0], given: 18

Re: Collection of work/rate problems? [#permalink]

### Show Tags

15 Sep 2009, 16:15
h2polo wrote:
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

Nice method.

I did it differently:

.........V1+V2.......V1.........V2
R........C/48......C/120....(C+50*120)/120
T........48...........120......
W........C..............C......

C/48 = C/120 + (C+6000)/120 => 5C/240 = 2C/240 + (2C+12000)/240 => 5C = 2C + 2C + 12000 => C = 12000
_________________

Please kudos if my post helps.

Manager
Joined: 22 Jul 2009
Posts: 191
Followers: 4

Kudos [?]: 258 [0], given: 18

Re: Collection of work/rate problems? [#permalink]

### Show Tags

15 Sep 2009, 16:28
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

This question has also been solved here: ps-pool-77282.html
_________________

Please kudos if my post helps.

Manager
Joined: 22 Jul 2009
Posts: 191
Followers: 4

Kudos [?]: 258 [2] , given: 18

Re: Collection of work/rate problems? [#permalink]

### Show Tags

15 Sep 2009, 16:48
2
KUDOS
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
_________________

Please kudos if my post helps.

Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

Re: Collection of work/rate problems? [#permalink]

### Show Tags

13 Nov 2009, 02:45
h2polo 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.

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

Anyone have the OA for this question?

I was looking at it again... I am thinking that the answer may be B.

Based purely on logic, if Machine X worked for 4 to produce twice as many bottles as Machine Y did in 3 hrs. Then we know that Machine X had completed 2/3 of the job already. So Machine X needs to work only for two more hours to complete the job by itself.

Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

Re: Collection of work/rate problems? [#permalink]

### Show Tags

13 Nov 2009, 02:52
powerka wrote:
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
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

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.

Yep, that was definitely a typo.

The equation should have been:

40/(B+4) + 40/B = 50/6

Thanks!
Intern
Joined: 23 Nov 2009
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Collection of work/rate problems? [#permalink]

### Show Tags

23 Nov 2009, 23:30
Also for problem number 4 here is a simpler solution

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

20 * 25 = x * 25 + 37.5 * 10 = 5
Manager
Joined: 22 Sep 2009
Posts: 222
Location: Tokyo, Japan
Followers: 2

Kudos [?]: 22 [0], given: 8

Re: Collection of work/rate problems? [#permalink]

### Show Tags

29 Nov 2009, 23:20
h2polo wrote:
h2polo 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.

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

Anyone have the OA for this question?

I was looking at it again... I am thinking that the answer may be B.

Based purely on logic, if Machine X worked for 4 to produce twice as many bottles as Machine Y did in 3 hrs. Then we know that Machine X had completed 2/3 of the job already. So Machine X needs to work only for two more hours to complete the job by itself.

I am voting for B as well.
Intern
Joined: 21 Oct 2009
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Collection of work/rate problems? [#permalink]

### Show Tags

11 Dec 2009, 09:18
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

Ans............

rate of A = 40/x+4
rate of B = 40/x
rate of A+B = 50/6

We need to find x first, before going for the printer A 80 pages

(40/(x+4))+(40/x) = (50/6)......i find this equation will take more time (even if we use the formula) to calculate x.

Is there any short cut or alternate method to answer this problem? I am struck on this pls help
Intern
Joined: 23 Aug 2009
Posts: 16
Followers: 0

Kudos [?]: 13 [0], given: 8

Re: Collection of work/rate problems? [#permalink]

### Show Tags

29 Dec 2009, 00:23
Answers for 2, 3 , 4 , 8

2. Work done by machine Y in one day to produce w widgets = 1/x---->eq1
Work done by machine X in one day to produce w widgets = 1/(x+2)--->eq2
Machine X and Y together produce 5/4 w widgets in three days. Therefore in one
day they produce 5/12 widgets.

hence the equation:
[1/(x+2) + 1/x] = 5/12
Solving we get
5x^2-14x-24=0
(5x+6)(x-4)
x=-5/6 or x=4

Work done in one day to produce w widgets is 1/(4+2).
Therefore machine X takes 6 days to produce w widgets.
and 12 days to produce 2w widgets.

3. 1/A + 1/B =1/24
1/A =1/60

Therefore 1/B = [(1/24)- (1/60)] = 1/40

In one minute the amt of work done between a and b is 5 pages.

hence 1/40 - 1/60 = 1/120
If 1/120 is the amt work done to produce 5 pages, then a total of 600 pages is produced.

4. 25 men's 1 day's work =1/20
1 man's 1 day's work = 1/(25 x 20)
10 men's 1 day's work =10/(25x20) = 1/50

When 10 men's work in one day is 1/50, then the work completed by 10 men in 37-1/2 days is = (75x2)X(1/50)= 3/4

Remaining work is 1-3/4 =1/4

Hence 25 men completed 1/4 work in 5 days.

8. A pool can be filled in 4 hrs and can be drained in 5 hours.

If the pool is filled and emptied at the same time then the total work done in 1 hr to fill the pool will be: 1/4 - 1/5 = 1/20

Let the valve A fill the pool and valve B empty the pool.
Let x hours be the time when only valve A was open.
Total time taken to fill the pool is from 1:00 pm to 11:00 pm ie 10 hrs.

Part filled by A in x hrs + part filled by (A+B) in (10-x) hrs =1
1/4(x) + 1/20(10-x) =1
Solving we get
4/20 (x) = 1/2
Therefore x= 5/2 = 2.5 hours. So the Valve B or the drain was opened at 3.30pm
_________________

Peace

Intern
Joined: 12 Mar 2010
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Collection of work/rate problems? [#permalink]

### Show Tags

22 Mar 2010, 03:36
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

I think the correct answer is B

Statement 1 is insufficient, since it gives no information about Y. So we cannot calculate the total number of bottles produced by A&B respectively

Statement 2 will be sufficient, as 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.Thus if X worked for 2 hours he would do the same work that Y did in 3 hours. Hence 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,
thereforeX working alone would take 6 hours to do the job.
Intern
Joined: 17 Mar 2010
Posts: 39
Followers: 2

Kudos [?]: 64 [0], given: 26

Re: Collection of work/rate problems? [#permalink]

### Show Tags

24 Mar 2010, 07:32
sorry, just saw that my question was answered
_________________

If my post helped you : KUDOS KUDOS KUDOS

Intern
Joined: 17 Mar 2010
Posts: 39
Followers: 2

Kudos [?]: 64 [1] , given: 26

Re: Collection of work/rate problems? [#permalink]

### Show Tags

24 Mar 2010, 13:26
1
KUDOS
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
_________________

If my post helped you : KUDOS KUDOS KUDOS

Intern
Joined: 12 Mar 2010
Posts: 10
Followers: 0

Kudos [?]: 10 [0], given: 1

Re: Collection of work/rate problems? [#permalink]

### Show Tags

01 Apr 2010, 06:22
powerka wrote:
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
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

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.

This took me quite a long time to solve though my approach was correct. Is there a faster way to solve this or did I just take a long time to get the roots of the quadratic?
Manager
Joined: 24 Mar 2010
Posts: 104
Followers: 1

Kudos [?]: 30 [0], given: 12

Re: Collection of work/rate problems? [#permalink]

### Show Tags

06 Apr 2010, 06:21
great material...thanks
_________________

Please do consider giving kudos if you like my posts

Intern
Joined: 01 May 2010
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Collection of work/rate problems? [#permalink]

### Show Tags

05 May 2010, 08:17
h2polo wrote:
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

I think you got it wrong right there...
if John and Jane continued to work together, the task would have been completed in 4 more days
but since Jane left, John needs more time to finish the task as now he is working alone..
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 277 [0], given: 16

Re: Collection of work/rate problems? [#permalink]

### Show Tags

08 May 2010, 04:59
leilak wrote:
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

i agree that B alone is sufficient. BUT A alone is again sufficient:
X: R*4= x%
-> R=x/400
Y: R*3= 1-x%
-> R= (1-x/100)/3
if x/400 = 30 then we can find rate for X then find rate for Y thus sufficient (????)

please comment as not sure if this is corect
Re: Collection of work/rate problems?   [#permalink] 08 May 2010, 04:59

Go to page   Previous    1   2   3   4   5   6    Next  [ 112 posts ]

Similar topics Replies Last post
Similar
Topics:
6 Basic work/rate problems and algebra 4 31 Jan 2013, 10:29
46 Collections of work/rate problem with solutions 15 14 Aug 2011, 11:49
3 Problems with Work/Rate questions 8 11 Feb 2011, 07:20
7 Machines X and Y produce bottles at their respective constant rates. 12 29 Aug 2010, 19:43
137 Collection of remainder problems in GMAT 62 13 Jan 2009, 10:57
Display posts from previous: Sort by