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Working simultaneously at their respective constant rates, M

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

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New post 11 Sep 2014, 06:18
Walkabout wrote:
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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New post 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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New post 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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New post 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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New post 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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New post 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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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:

Image

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

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

The correct answer is E

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

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New post 22 Apr 2017, 16:33
Walkabout wrote:
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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New post 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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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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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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Re: Working simultaneously at their respective constant rates, M [#permalink]

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New post 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 ! ! !
Ultimate GMAT Quantitative Megathread

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

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

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

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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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