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Working simultaneously at their respective constant rates, M

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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)
[Reveal] Spoiler: OA

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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)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = \(800/x\) ..... 1
a = \(800/y\) ..... 2


Use 2 in 1... we get

b = \(800 (y-x) / xy\)

Finally

Rate of B * time = Work done by B (we want time)

\(800 (y-x) / xy * t = 800\)

t = \(xy / (y-x)\)

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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)

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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)


Pick some smart numbers for x and y.

Say x=1 hour and y=2 hours (notice that y must be greater than x, since the time for machine A to do the job, which is y hours, must be more than the time for machines A and B working together to do the same job, which is x hours).

In this case, the time needed for machine B to do the job must also be 2 hours: 1/2+1/2=1.

Now, plug x=1 and y=2 in the options to see which one yields 2. Only option E fits.

Answer: E.
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Simplest solution Here-

RA + RB = 1/X

RA = 1/Y

RB = (1/X - 1/Y) = (Y-X)/XY

Time = 1/RB = XY/(X+Y)


I have treated 800 as equivalent to unity(= 1), as it's presence in final answer was trivial, as it will eventually cancel out, taking it unity has make the solution quite Un Complex..
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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:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate
solution-required-100221.html?hilit=work%20rate%20done
work-problem-98599.html?hilit=work%20rate%20done
hours-to-type-pages-102407.html?hilit=answer%20choices%20or%20solve%20quadratic%20equation.%20R

Theory on work/rate problems: work-word-problems-made-easy-87357.html

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66


Hope this helps
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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?


You can add rates when required to find simultaneous work done etc.

In distance / time / speed related problems, we cannot add up the speeds

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

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Jp27 wrote:
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)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = \(800/x\) ..... 1
a = \(800/y\) ..... 2


Use 2 in 1... we get

b = \(800 (y-x) / xy\)

Finally

Rate of B * time = Work done by B (we want time)

\(800 (y-x) / xy * t = 800\)

t = \(xy / (y-x)\)


Yeah, R*T=W is a lengthy way to solve these problems but, I have seen that it is almost a sure shot way to solve most of the problems on this concept. Picking up the smart numbers may be a neat way to solve these questions but it highly depends on the mental state when you are taking the exam.
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ronr34 wrote:
Jp27 wrote:
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)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = \(800/x\) ..... 1
a = \(800/y\) ..... 2


Use 2 in 1... we get

b = \(800 (y-x) / xy\)

Finally

Rate of B * time = Work done by B (we want time)

\(800 (y-x) / xy * t = 800\)

t = \(xy / (y-x)\)


I had a problem with this.
(1/A + 1/B) X = 800

(1/A)Y = 800

and when comparing both, I have too many unknowns....


That's because you stop on a halfway. Try to continue as suggested by Jp27 in the post you are quoting.
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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Another approach:

Rate(A) = 800/y
Rate(A+B) = 800/x

Rate A + Rate B = Rate(A+B)

=> Rate(B) = Rate(A+B) - Rate(A)
= 800(y-x)/xy

Then the GODFATHER equation Rate * Time = Work

800(y-x)/xy * Time = 800

Time = xy/(y-x)

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gciftci wrote:
HI Brunel I get it sir, just got confused with this "If an object moves the same distance twice, but at different rates, then the average rate will NEVER be the average of the two rates given for the two legs of the journey.(MGMAT)"


This is about completely different matter: it says that if an object covers 100 miles at 10 miles per hour and another 100 miles at 20 miles per hour, then the average speed for 200 miles won't be (10+20)/2=15 miles per hour.

(average speed) = (total distance)/(total time):

(total distance) = 100 + 100 = 200 miles.

(total time) = 100/10 + 100/20 = 15 hours.

(average speed) = (total distance)/(total time) = 200/15 miles per hour.

Notice here though that we can add or subtract rates (speeds) to get relative rate.

For example if two cars are moving toward each other from A to B (AB=100 miles) with 10mph and 15mph respectively, then their relative (combined) rate is 10+15=25mph, and they'll meet in (time)=(distance)/(rate)=100/25=4 hours;

Similarly if car x is 100 miles ahead of car y and they are moving in the same direction with 10mph and 15mph respectively then their relative rate is 15-10=5mph, and y will catch up x in 100/5=20 hours.

Hope it's clear.
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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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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, 10:23
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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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Re: Working simultaneously at their respective constant rates, M [#permalink]

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New post 20 Oct 2013, 16:48
A and B : 1/x = 1/800
A alone: 1/y = 1/800
B alone: A and B - A: 1/x - 1/y = 0

solving it: (y - x)/xy => total time it takes is the reciprocal therefore B alone = xy/(y-x)

Answer: E

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

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New post 02 Nov 2013, 08:20
Jp27 wrote:
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)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = \(800/x\) ..... 1
a = \(800/y\) ..... 2


Use 2 in 1... we get

b = \(800 (y-x) / xy\)

Finally

Rate of B * time = Work done by B (we want time)

\(800 (y-x) / xy * t = 800\)

t = \(xy / (y-x)\)


I had a problem with this.
(1/A + 1/B) X = 800

(1/A)Y = 800

and when comparing both, I have too many unknowns....

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

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New post 06 Apr 2014, 16:49
I managed to solve this via plugging in numbers but I went over 3 minutes! With these plug-in-numbers type of problems, I strive to pick simple numbers but there always seems to be one, usually the one i'm solving for, that ends up being a rather complicated number. My question is:

1) I plugged in Rate A and Rate A+B and then solved for time. Should I have plugged in numbers for time directly. Is there a general rule as to what number I should be plugging in?
2) In this case, to keep things super simple, I could have plugged the total Rate to be 8 and Ra and Rb to both be 4. Is it bad practice to choose the same numbers for the individual rates/times?

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

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New post 07 Apr 2014, 01:04
russ9 wrote:
I managed to solve this via plugging in numbers but I went over 3 minutes! With these plug-in-numbers type of problems, I strive to pick simple numbers but there always seems to be one, usually the one i'm solving for, that ends up being a rather complicated number. My question is:

1) I plugged in Rate A and Rate A+B and then solved for time. Should I have plugged in numbers for time directly. Is there a general rule as to what number I should be plugging in?
2) In this case, to keep things super simple, I could have plugged the total Rate to be 8 and Ra and Rb to both be 4. Is it bad practice to choose the same numbers for the individual rates/times?



Just refer to method of Bunuel; he did using plug-ins.

I used same variables available & got correct answer (Had taken 800 = 1 as done by honchos)
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Re: Working simultaneously at their respective constant rates, M [#permalink]

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New post 22 Apr 2014, 03:11
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?

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Re: Working simultaneously at their respective constant rates, M   [#permalink] 22 Apr 2014, 03:11

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