Find all School-related info fast with the new School-Specific MBA Forum

It is currently 13 Jul 2014, 17:51

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Collection of work/rate problems?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 04 Feb 2010
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 4

Re: Collection of work/rate problems? [#permalink] New post 04 Jun 2010, 08:11
srivas wrote:
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


Slightly different approach:

After 24 minutes, A will have completed 24/60=40% of the task
Therefore, the portion completed by B = 100%-40%=60%
Therefore, B's rate is 60%/40%=1.5 times A's rate
Given B's rate is 5 pages more than A's rate by the problem, 0.5 times A's rate is 5 ppm => A's rate = 10 ppm, B's rate is 15 ppm
15*24+10*24=5*24+20*24=120+2*10*24=600
Intern
Intern
avatar
Joined: 10 Jun 2010
Posts: 8
Concentration: Marketing, General Management
GPA: 3.8
WE: Web Development (Investment Banking)
Followers: 0

Kudos [?]: 4 [0], given: 12

GMAT ToolKit User
Re: Collection of work/rate problems? [#permalink] New post 30 Jul 2010, 11:45
srivas wrote:
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,
....
.....
.....

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


How in the world am I going to solve that quadratic equation quickly in the actual exam? One of the answers is -1.20. I know a negative answer is not required but is there a faster way of doing this problem? :shock:

Thank you in advance.
Intern
Intern
avatar
Joined: 10 Jun 2010
Posts: 8
Concentration: Marketing, General Management
GPA: 3.8
WE: Web Development (Investment Banking)
Followers: 0

Kudos [?]: 4 [0], given: 12

GMAT ToolKit User
Re: Collection of work/rate problems? [#permalink] New post 01 Aug 2010, 10:08
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

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



You mean D: 3:30 right? Here is a different approach.

R = d/t
We know the the size of the pool (d) is the same.

So R1t1 = R2D2

or 1/4*10 = 1/5*t2
t2 = 50/4

Subtract this from 10: 10-50/4 = -2.5 (Ignore the negative)

Add this to 1 and gives you 3:30.

Does this method make sense?
Manager
Manager
avatar
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 73
WE 1: 6 years - Consulting
Followers: 3

Kudos [?]: 22 [0], given: 27

Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2010, 05:08
8. Answer is 3:30 PM. The drain pipe has been opened at 3:30 pm..let me know if u need a solution
_________________

Consider giving Kudos if my post helps in some way

Senior Manager
Senior Manager
User avatar
Joined: 18 Jun 2010
Posts: 303
Schools: Chicago Booth Class of 2013
Followers: 20

Kudos [?]: 123 [0], given: 194

GMAT Tests User Reviews Badge
Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2010, 05:56
snipertrader 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.

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


This question is solved wrongly. The correct answer is "B".

Statement 2 will be sufficient, since 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. In other words, if X worked for 2 hours, X would do the same work that Y did in 3 hours. Since we know that 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, then X working alone would take 6 hours to do the job.
Manager
Manager
avatar
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 73
WE 1: 6 years - Consulting
Followers: 3

Kudos [?]: 22 [0], given: 27

Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2010, 06:01
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?

My Solution:

This Question is based on Mandays fundamental.

6 machines work for 12 days to complete 1 job= 6*12 mandays (machine=man)

Now if we reduce days in this formula, machines will increase to compensate the reduction in days :

so we have :

6 mac*12 days = x mac * 8 days
x= 9 machines - a delta of 3 machines in addition ANS = 3 machines
_________________

Consider giving Kudos if my post helps in some way

Manager
Manager
avatar
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 73
WE 1: 6 years - Consulting
Followers: 3

Kudos [?]: 22 [0], given: 27

Re: Collection of work/rate problems? [#permalink] New post 16 Aug 2010, 06:14
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

My Approach - Supposedly the shortest and the best to solve such problems:

whenever we see such problems where time taken by pipes a (2), b(3) ,c(6) are given , take out the LCM of these values (the number which will be divisible of all these 3 values):
LCM of 2,3,6 = 12
Now this figure 12 is the total unit of Work (work done to fill full tank):
A does work -> 12/2= 6units /hour
B does work -> 12/3= 4units /hour
C does work -> 12/6= 2units /hour

for half tank -> total units of work=6 units
the equation can be written as -> x(-A-B+C) =6 (x is no of hours required and (-A-B+C) is work done in 1 hour - also '-' bcos A and B pumps out)

x(-6-4+2) = 6
x=-6/8 = 3/4

The explanation looks big, but if put inpractice, we can solve such questions without a pen-paper
_________________

Consider giving Kudos if my post helps in some way

Intern
Intern
avatar
Joined: 20 Jan 2010
Posts: 7
Followers: 0

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

Re: Collection of work/rate problems? [#permalink] New post 08 Sep 2010, 22:36
Hi where can I get the full list of the answers. Am confused with different responses to same qs!
Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 170
Followers: 1

Kudos [?]: 14 [0], given: 7

Re: Collection of work/rate problems? [#permalink] New post 06 Nov 2010, 15:14
srivas wrote:
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


Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you!
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18523
Followers: 3197

Kudos [?]: 21432 [0], given: 2547

Re: Collection of work/rate problems? [#permalink] New post 06 Nov 2010, 21:08
Expert's post
gettinit wrote:
srivas wrote:
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


Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you!


Check the solution here: work-problem-98599.html#p759876

Similar questions:
a-good-one-98479.html?hilit=point%20quadratic
hours-to-type-pages-102407.html?hilit=point%20quadratic

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 170
Followers: 1

Kudos [?]: 14 [0], given: 7

Re: Collection of work/rate problems? [#permalink] New post 08 Nov 2010, 19:21
Bunuel wrote:
gettinit wrote:
srivas wrote:
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

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


Can someone please explain to me how we solved to get D=4? I am having a tough time getting to this number. Thank you!


Check the solution here: work-problem-98599.html#p759876

Similar questions:
a-good-one-98479.html?hilit=point%20quadratic
hours-to-type-pages-102407.html?hilit=point%20quadratic

Hope it helps.


Thanks Bunuel, this was very helpful! appreciate your responsiveness.
Intern
Intern
avatar
Affiliations: CFA Charterholder
Joined: 25 Oct 2010
Posts: 24
Schools: EWMBA Haas
Followers: 0

Kudos [?]: 2 [0], given: 5

GMAT Tests User
Re: Collection of work/rate problems? [#permalink] New post 29 Nov 2010, 10:43
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
Can somebody please break down this section:

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

Im having difficulty doing the math...Much appreciated.

Thank you...

Casey
Manager
Manager
avatar
Joined: 04 May 2009
Posts: 68
Location: Astoria, NYC
Followers: 1

Kudos [?]: 7 [0], given: 1

Re: Collection of work/rate problems? [#permalink] New post 02 Dec 2010, 09:57
i am in the same boat as casey.....
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

Im having difficulty doing the math...Much appreciated.
i know the 40/(B+4) + 40/B should be in the form XY/x+y...but i am not getting the final equation.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18523
Followers: 3197

Kudos [?]: 21432 [0], given: 2547

Re: Collection of work/rate problems? [#permalink] New post 02 Dec 2010, 10:09
Expert's post
anish319 wrote:
i am in the same boat as casey.....
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

Im having difficulty doing the math...Much appreciated.
i know the 40/(B+4) + 40/B should be in the form XY/x+y...but i am not getting the final equation.


First of all: in case of any question or doubt about specific problem please post this problem in PS or DS subforum.

As for this problem:

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?
A. 12
B. 18
C. 20
D. 24
E. 30

Let the time needed to print 40 pages for printer A be a minutes, so for printer B it would be a-4 minutes.

The rate of A would be rate=\frac{job}{time}=\frac{40}{a} pages per minute and the rate of B rate=\frac{job}{time}=\frac{40}{a-4} pages per minute.

Their combined rate would be \frac{40}{a}+\frac{40}{a-4} pages per minute. Also as "two printers can print 50 pages in 6 minutes" then their combined rate is rate=\frac{job}{time}=\frac{50}{6}, so \frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}.

\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6} --> \frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}. At this point we can either try to substitute the values from answer choices or solve quadratic equation. Remember as we are asked to find time needed for printer A to print 80 pages, then the answer would be 2a (as a is the time needed to print 40 pages). Answer D works: 2a=24 --> a=12 --> \frac{1}{12}+\frac{1}{8}=\frac{5}{24}.

Answer: D.

Similar problems:
work-problem-98599.html#p759876
hours-to-type-pages-102407.html?hilit=answer%20choices%20or%20solve%20quadratic%20equation.%20R

Theory about work problems:
word-translations-rates-work-104208.html?hilit=printer#p812628

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 29 Dec 2010
Posts: 33
Followers: 0

Kudos [?]: 2 [0], given: 3

Re: Collection of work/rate problems? [#permalink] New post 11 Jan 2011, 14:56
hi does anyone have the answers to 4, 5, 8, 9, 10, 11, 12, 17,18,19 ?

regards
Intern
Intern
avatar
Joined: 06 Jul 2011
Posts: 2
Followers: 0

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

Re: Collection of work/rate problems? [#permalink] New post 16 Sep 2011, 05:29
snipertrader wrote:
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

Solution

2/3 were basic so 1/3 will be deluxe
Assuming it took 1 hr to complete Basic stereo ; deluxe will take 7/5 X 1 hr

Total time taken (basic radio + deluxe radio) = 2/3 + (1/3 x 7/5) = 17/15

Ratio of hrs needed for deluxe ratio to total hrs = (7/15)/(17/15) = 7/17


I can't seem to figure out why you multiplied 1/3 by 7/5. I assume you are right because the answer I get isn't one of the choices, so I'm hoping someone can explain to me why this was done. I try to solve these problems using a small chart: (Total=Rate*Time)

As you stated, let's assume it takes 1 hr to make a B stereo.

Basic
Total = 2/3
Rate = x
time = 1hr
therefore rate (x) = (2/3) / 1 = 2/3

Deluxe
Total = 1/3
Rate = y
time = 7/5 *1hr = 7/5
therefore rate (y) = (1/3) / (7/5) = 5/21

so 1/x + 1/y = 1/total time (t)
1/(2/3) + 1/(5/21) = 1/t
t = 3/2 + 21/5 = 15/10 + 42/10 = 57/10

so to find the fraction of Deluxe Time/Total time = (7/5)/(57/10) = not the right answer

I can see that in order to match the snipertrader's solution, I need to rearrange my 'total', 'rate' & 'time' to be:

Basic
Total = x
rate = 1
time = 2/3

Deluxe
total = y
rate = 7/5
time = 1/3

in which case you would get 7/17 as the answer, but I can't make the leap as to why 2/3 and 1/3 need to be the TIME and not the TOTAL? I'm hoping someone can walk me through this.

Thank you!!
Senior Manager
Senior Manager
User avatar
Joined: 03 Mar 2010
Posts: 448
Schools: Kelley,Tepper,Carlson,Kenan Flager,Broad
Followers: 4

Kudos [?]: 101 [0], given: 22

GMAT Tests User
Re: Collection of work/rate problems? [#permalink] New post 21 Sep 2011, 08:12
Answers to first 15.
1.40
2.C (8)
3.A(600)
4.8 1/3
6.2/3
7.C (97.2)
8.3:30pm
9.12,000
10. 24
11. 3
12. 3/4
13. D (1/8)
14. 10
15. C


Rest tomorrow. :lol:
_________________

My dad once said to me: Son, nothing succeeds like success.

Senior Manager
Senior Manager
User avatar
Joined: 03 Mar 2010
Posts: 448
Schools: Kelley,Tepper,Carlson,Kenan Flager,Broad
Followers: 4

Kudos [?]: 101 [0], given: 22

GMAT Tests User
Re: Collection of work/rate problems? [#permalink] New post 22 Sep 2011, 10:21
16. C x-3/3x
17. A 18
18. B 3
19. A 6
20. C 6
21. A 24
22 D 66 2/3
23.B 100
24 D 25
25 D 36
26 D 2/3
27.E 4/9
Missing Answer choices for Q.27 are: (A) 1/9 (B) 1/6 (C) 1/3 (D) 7/18 (E) 4/9
28. B
29. D 12/7
30. 7/17 A
_________________

My dad once said to me: Son, nothing succeeds like success.

Intern
Intern
avatar
Joined: 09 Oct 2011
Posts: 10
Location: United States
Concentration: Leadership, Entrepreneurship
GMAT Date: 12-27-2011
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 6

GMAT Tests User
Re: Collection of work/rate problems? [#permalink] New post 22 Oct 2011, 06:47
Need help understanding this
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

The answer is B, but I want to understand how I can figure this out without solving the whole problem.

From the question we have 4x = 5y

From (1) we have y = 0.8x (or y - 0.8x = 0)
From (2) we have x + y = 43.75 (or y = 43.75 - x)

At this point I see I have a new equation from both (1) and (2) and my initial response is that both are individually sufficient (D).

How can I figure out at this point that (1) is not sufficient and (2) is sufficient without spending much time in solving all the equations.

thanks,
Vinay
Intern
Intern
avatar
Joined: 25 Oct 2011
Posts: 3
Followers: 0

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

Re: Collection of work/rate problems? [#permalink] New post 16 Dec 2011, 08:13
No need to use any algebra here.

A = tap that is filling the pool
B = leak in the pool

A is running for 10 hrs, which means it can fill 2.5 times the pool in that time.
However, it is able to fill just 1 pool, which means that B leaks 1.5 times the pool in that time. Now, B can empty 1 pool in 5 hrs, so it will take 7.5 hrs to empty 1.5 times the pool.

Therefore, leak started 2.5 hrs after A started = which is 3.30pm.

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

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
Re: Collection of work/rate problems?   [#permalink] 16 Dec 2011, 08:13
    Similar topics Author Replies Last post
Similar
Topics:
22 Collections of work/rate problem with solutions Baten80 13 14 Aug 2011, 11:49
1 Work-Rate Problem Safiya 9 29 Aug 2010, 19:43
Work/Rate Problem consultinghokie 6 24 Apr 2006, 21:00
work-rate problem FN 5 03 Oct 2005, 11:59
Display posts from previous: Sort by

Collection of work/rate problems?

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   3   4   5   6    Next  [ 109 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.