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# Working simultaneously at their respective constant rates, M

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Joined: 07 Apr 2014
Posts: 130
Re: Working simultaneously at their respective constant rates, M [#permalink]

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11 Sep 2014, 06:18
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

1/a - rate of machine A for an hour
1/b-rate of machine B for an hour

1/x = 1/a +1/b for an hour

if working alone A 's rate is 1/a = 1/y for an hour

1/y+1/b = 1/x

b = xy/y-x
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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13 Oct 2014, 10:31
I did not do the long way or the bunuel's method
I knew that machine B must be the difference between y and x
the only solution that has y-x is E
therefore I picked E without even solving anything.
is this method ok to use? :D
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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13 Oct 2014, 13:57
What we are really looking for is what number multiplied by B(rate) equals 800. In short, Brate*t=800, because rate multiplied by time equals work
we need to find t in terms of x and y and we are given 2 clues.
1) "Machines A and B produce 800 nails in x hours" in mathematical notation: (Arate+Brate)*x=800 nails
2) "Machine A produces 800 nails in y hours" i.e. Arate*y=800 nails

We can substitute 2 into 1 by dividing both sides of 1 by x giving us: (Arate+Brate)=800/x
Then dividing both sides of 2 by y and isolating Arate, giving us Arate=800/y

putting 2 into 1, we get 800/y+Brate=800/x
lets put the 800s together by subtracting 800/y from both sides and getting Brate =(800/x)-(800/y)
we can simplify the right side of the equation and get (800y-800x)/xy
multiply both sides of the equation by xy and you get Brate*xy=800(y-x)
then divide both sides by y-x and you get Brate*xy/(y-x)=800
This means that t= xy/(y-x) (E), because Brate*t=800
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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14 Oct 2014, 05:34
These are the type of questions that make me feel like i wont be able to get a good score on the GMAT. I couldnt solve this at all, let alone solve it in 2 minutes. Still quite confused about the smart numbers in this case.
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Working simultaneously at their respective constant rates, M [#permalink]

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17 Aug 2015, 04:44
Hi, here is one more approach with smart numbers
Rate A=100 ->Ty=800/100=8
Rate B=100 ->Tb=800/100=8
Rate A+B=200 ->Tx=800/200=4
Plug in this numbers in the answer choices and you'll see that only answer (E) is the correct answer ->XY/(Y-X)=(8*4)/(8-4)=8
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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13 May 2016, 10:56
W= AxB/A+B

3x6/3+6 = 18/9 = 2 (together A and B produce 800 nails in 2 hours, x=2)
(A produce 800 nails in 3 hours, y=3)
(B produce 800 nails in 6 hours, find a match in the answer choice=6)

A) 2/5 not a match
B) 3/5 not a match
C) 6/5 not a match
D) 6/-1=-6 not a match
E) 6/1 = 6 correct answer

HINT: When it comes to work formula and you're dealing with variables, the perfect numbers to plug in this formula are 3 and 6.

Hope it helps
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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01 Jun 2016, 07:41
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Expert's post
This problem is what we call a combined worker problem, where

Work (of machine 1) + Work (of machine 2) = Total Work Completed

In this case,

Work (of Machine A) + Work (of Machine B) = 800

We know that Machines A and B produce 800 nails in x hours. Thus, the TIME that Machine A and B work together is x hours. We are also given that Machine A produces 800 nails in y hours. Thus, the rate for Machine A is 800/y. Since we do not know the rate for Machine B, we can label its rate as 800/B, where B is the number of hours it takes Machine B to produce 800 nails.

To better organize our information we can set up a rate x time = work matrix:

We now can say:

Work (of Machine A) + Work (of Machine B) = 800

800x/y + 800x/B = 800

To cancel out the denominators, we can multiply the entire equation by yB. This gives us:

800xB + 800xy= 800yB

xB + xy = yB

xy = yB – xB

xy = B(y – x)

xy /(y – x) = B

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Re: Working simultaneously at their respective constant rates, M [#permalink]

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31 Jan 2017, 04:49
1) $$(A+B)*x=800; A+B=\frac{800}{x}$$
2) $$A*y=800; A=\frac{800}{y}$$
3) $$B=(A+B)-A=\frac{800}{x}-\frac{800}{y}=\frac{800}{(800y-800x)/xy}=\frac{800xy}{800(y-x)}=\frac{xy}{(y-x)}$$

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Re: Working simultaneously at their respective constant rates, M [#permalink]

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22 Apr 2017, 16:33
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

Solved in under 2 minutes.

First,
We should know what is the question? Asking B time.

Second,
Since Time = $$\frac{Work}{Rate}$$ and we already know that work = 800, so we must calculate Rate.

Third,
Rate B = Rate combined - Rate A = [$$\frac{800}{x}$$] - [$$\frac{800}{y}$$] = $$\frac{800 (y-x)}{(xy)}$$

Hence, we can calculate Time = 800 * $$\frac{(xy)}{800 (y-x)}$$.
We can eliminate 800 so answer is = $$\frac{xy}{(y-x)}$$, E.
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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03 Jun 2017, 06:44
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

Sollution : (A+B)*X=A*Y
B= (Y-X)/X (1)

B*Z=A*Y
B = (A*Y)/Z (2)

From (1) & (2) (A*Y)/Z = (Y-X)/X
Z = (Y*X)/(Y-X)
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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19 Sep 2017, 22:07
2
KUDOS
1
This post was
BOOKMARKED
here 800 nails can be taken as "one" work.
then together they are taking x hours
A alone taking Y hours
then B alone can be obtained by
1/B = 1/X-1/Y
1/B = Y-X/XY
b= XY/Y-X
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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21 Nov 2017, 07:05
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Bunuel wrote:
gciftci wrote:
Q: I get the algebra but I got confused with this question because I thaught adding and deviding rates was a NoNo? Why is it diffrent in this case?

No, we CAN easily sum the rates. For example:

If we are told that A can complete a job in 2 hours and B can complete the same job in 3 hours, then A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The combined rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$. For example when we are told that a man can do a certain job in 3 hours we can write: $$3*rate=1$$ --> $$rate=\frac{1}{3}$$ job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then $$5*(2*rate)=1$$ --> so rate of 1 printer is $$rate=\frac{1}{10}$$ job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then $$3*(2*rate)=12$$ --> so rate of 1 printer is $$rate=2$$ pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is $$rate_a=\frac{job}{time}=\frac{1}{2}$$ job/hour and B's rate is $$rate_b=\frac{job}{time}=\frac{1}{3}$$ job/hour. Combined rate of A and B working simultaneously would be $$rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$ job/hour, which means that they will complete $$\frac{5}{6}$$ job in one hour working together.

3. For multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}$$, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$t_1$$ and $$t_2$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}$$ hours.

Some work problems with solutions:
http://gmatclub.com/forum/time-n-work-p ... cal%20rate
http://gmatclub.com/forum/facing-proble ... reciprocal
http://gmatclub.com/forum/what-am-i-doi ... reciprocal
http://gmatclub.com/forum/gmat-prep-ps- ... cal%20rate
http://gmatclub.com/forum/questions-fro ... cal%20rate
http://gmatclub.com/forum/a-good-one-98 ... hilit=rate
http://gmatclub.com/forum/solution-requ ... ate%20done
http://gmatclub.com/forum/work-problem- ... ate%20done
http://gmatclub.com/forum/hours-to-type ... ation.%20R

Theory on work/rate problems: http://gmatclub.com/forum/work-word-pro ... 87357.html

All DS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=46
All PS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=66

Hope this helps

Bunuel hello, great explantation. Just trying to understand when you multiply 3 * rate =1 how do go get 1 ? also when you multiply 3 (2*rate) =12 how did you get 12 ? and what number is implied by RATE ? Big thanks for brilliant explanation in advance and have great day!
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Posts: 44400
Re: Working simultaneously at their respective constant rates, M [#permalink]

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21 Nov 2017, 07:27
dave13 wrote:
Bunuel wrote:
gciftci wrote:
Q: I get the algebra but I got confused with this question because I thaught adding and deviding rates was a NoNo? Why is it diffrent in this case?

No, we CAN easily sum the rates. For example:

If we are told that A can complete a job in 2 hours and B can complete the same job in 3 hours, then A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The combined rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$. For example when we are told that a man can do a certain job in 3 hours we can write: $$3*rate=1$$ --> $$rate=\frac{1}{3}$$ job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then $$5*(2*rate)=1$$ --> so rate of 1 printer is $$rate=\frac{1}{10}$$ job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then $$3*(2*rate)=12$$ --> so rate of 1 printer is $$rate=2$$ pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is $$rate_a=\frac{job}{time}=\frac{1}{2}$$ job/hour and B's rate is $$rate_b=\frac{job}{time}=\frac{1}{3}$$ job/hour. Combined rate of A and B working simultaneously would be $$rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$ job/hour, which means that they will complete $$\frac{5}{6}$$ job in one hour working together.

3. For multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}$$, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$t_1$$ and $$t_2$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}$$ hours.

Some work problems with solutions:
http://gmatclub.com/forum/time-n-work-p ... cal%20rate
http://gmatclub.com/forum/facing-proble ... reciprocal
http://gmatclub.com/forum/what-am-i-doi ... reciprocal
http://gmatclub.com/forum/gmat-prep-ps- ... cal%20rate
http://gmatclub.com/forum/questions-fro ... cal%20rate
http://gmatclub.com/forum/a-good-one-98 ... hilit=rate
http://gmatclub.com/forum/solution-requ ... ate%20done
http://gmatclub.com/forum/work-problem- ... ate%20done
http://gmatclub.com/forum/hours-to-type ... ation.%20R

Theory on work/rate problems: http://gmatclub.com/forum/work-word-pro ... 87357.html

All DS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=46
All PS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=66

Hope this helps

Bunuel hello, great explantation. Just trying to understand when you multiply 3 * rate =1 how do go get 1 ? also when you multiply 3 (2*rate) =12 how did you get 12 ? and what number is implied by RATE ? Big thanks for brilliant explanation in advance and have great day!

We are equating to the amount of job: 1 job, 12 pages, etc.

For more check:

17. Work/Rate Problems

On other subjects:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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23 Nov 2017, 09:36
khairilthegreat wrote:
RateA + RateB = 800/x

RateA = 800/y
RateB = 800/z

So --> 800/y + 800/z = 800/x

1/y + 1/z = 1/x --> 1/z = 1/x - 1/y
z = xy/(y-x)

how did you get from 1/z = 1/x - 1/y this one z = xy/(y-x) ? can you please explain step by step ?
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Joined: 02 Sep 2009
Posts: 44400
Re: Working simultaneously at their respective constant rates, M [#permalink]

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23 Nov 2017, 09:40
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Expert's post
dave13 wrote:
khairilthegreat wrote:
RateA + RateB = 800/x

RateA = 800/y
RateB = 800/z

So --> 800/y + 800/z = 800/x

1/y + 1/z = 1/x --> 1/z = 1/x - 1/y
z = xy/(y-x)

how did you get from 1/z = 1/x - 1/y this one z = xy/(y-x) ? can you please explain step by step ?

Basic algebraic manipulations:

$$\frac{1}{z} = \frac{1}{x} - \frac{1}{y}$$

$$\frac{1}{z} = \frac{y-x}{xy}$$

$$z=\frac{xy}{y-x}$$
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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23 Nov 2017, 09:52
Bunuel wrote:
gciftci wrote:
Q: I get the algebra but I got confused with this question because I thaught adding and deviding rates was a NoNo? Why is it diffrent in this case?

No, we CAN easily sum the rates. For example:

If we are told that A can complete a job in 2 hours and B can complete the same job in 3 hours, then A's rate is 1/2 job/hour and B's rate is 1/3 job/hour. The combined rate of A and B working simultaneously would be 1/2+1/3=5/6 job/hours, which means that the will complete 5/6 job in hour working together.

THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$. For example when we are told that a man can do a certain job in 3 hours we can write: $$3*rate=1$$ --> $$rate=\frac{1}{3}$$ job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then $$5*(2*rate)=1$$ --> so rate of 1 printer is $$rate=\frac{1}{10}$$ job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then $$3*(2*rate)=12$$ --> so rate of 1 printer is $$rate=2$$ pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is $$rate_a=\frac{job}{time}=\frac{1}{2}$$ job/hour and B's rate is $$rate_b=\frac{job}{time}=\frac{1}{3}$$ job/hour. Combined rate of A and B working simultaneously would be $$rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$ job/hour, which means that they will complete $$\frac{5}{6}$$ job in one hour working together.

3. For multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}$$, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$t_1$$ and $$t_2$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}$$ hours.

Some work problems with solutions:
http://gmatclub.com/forum/time-n-work-p ... cal%20rate
http://gmatclub.com/forum/facing-proble ... reciprocal
http://gmatclub.com/forum/what-am-i-doi ... reciprocal
http://gmatclub.com/forum/gmat-prep-ps- ... cal%20rate
http://gmatclub.com/forum/questions-fro ... cal%20rate
http://gmatclub.com/forum/a-good-one-98 ... hilit=rate
http://gmatclub.com/forum/solution-requ ... ate%20done
http://gmatclub.com/forum/work-problem- ... ate%20done
http://gmatclub.com/forum/hours-to-type ... ation.%20R

Theory on work/rate problems: http://gmatclub.com/forum/work-word-pro ... 87357.html

All DS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=46
All PS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=66

Hope this helps

it helps thanks!
one of the links to similar topic you shared is locked, so I could not post a question there: here it is https://gmatclub.com/forum/it-takes-6-d ... cal%20rate (I couldn't understand how m/3+5=w/9 to solve for M and W in your solution,- Also I couldn't understand why in the second equation you wrote - M/3 and not 3/m - As per formula Time = JOB / RATE and Rate = Job / time so in both cases in the numerator is JOB and not TIME - I wanted to ask a question but couldn't since the topic is locked )
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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23 Nov 2017, 10:14
Bunuel wrote:
dave13 wrote:
khairilthegreat wrote:
RateA + RateB = 800/x

RateA = 800/y
RateB = 800/z

So --> 800/y + 800/z = 800/x

1/y + 1/z = 1/x --> 1/z = 1/x - 1/y
z = xy/(y-x)

how did you get from 1/z = 1/x - 1/y this one z = xy/(y-x) ? can you please explain step by step ?

Basic algebraic manipulations:

$$\frac{1}{z} = \frac{1}{x} - \frac{1}{y}$$

$$\frac{1}{z} = \frac{y-x}{xy}$$

$$z=\frac{xy}{y-x}$$

Bunuel thanks! one question how after this $$\frac{1}{z} = \frac{y-x}{xy}$$ you got this $$z=\frac{xy}{y-x}$$[/quote] ? please say in words what you did
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Posts: 44400
Re: Working simultaneously at their respective constant rates, M [#permalink]

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23 Nov 2017, 10:23
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Expert's post
dave13 wrote:

Bunuel thanks! one question how after this $$\frac{1}{z} = \frac{y-x}{xy}$$ you got this $$z=\frac{xy}{y-x}$$? please say in words what you did

You really should brush-up fundamentals before attempting questions.

$$z=\frac{xy}{y-x}$$

Cross-multiple: $$z(y-x)=xy$$

Divide by y-x: $$z=\frac{xy}{y-x}$$
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Re: Working simultaneously at their respective constant rates, M   [#permalink] 23 Nov 2017, 10:23

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