Working simultaneously at their respective constant rates, M

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)

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



Hope this helps

Machines A and B produce 800 nails in x hours.
Rate = ouput/time
So, their COMBINED rate = 800/x nails per hour

Working alone at its constant rate, Machine A produces 800 nails in y hours
Rate = ouput/time
So, Machine A's rate = 800/y nails per hour

We know that: (Machine A's rate) + (Machine B's rate) = (their combined rate)
Substitute to get: 800/y + (Machine B's rate) = 800/x
Rewrite as follows: Machine B's rate = 800/x - 800/y
Rewrite with common denominators: Machine B's rate = 800y/xy - 800x/yx
Combine: Machine B's rate = (800y - 800x)/xy

How many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?
Time = output/rate
= 800/((800y - 800x)/xy)
= (800)(xy/800y - 800x)
= 800xy/(800y - 800x)
= xy/(y-x)

Answer: E

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..

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)

Rgds,
TGC!

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?

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.

i am not able to comprehend ,

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


how come ??

A needs 2 hours to do a job and A and B together need 1 hour to do the same job.

In 1 hour A does 1/2 of the job thus another half is done by B in 1 hour. Half of the job in 1 hour = whole job in 2 hours.

Or: 1/2 + (rate of B) = 1 --> (rate of B) = 1/2 --> time of B = 2 hours.

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

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

Answer: E

The solutions in this thread show that there are multiple ways to answer this question. Below are tips that you can use for a couple of different approaches (the links go to pages with lists of questions that you can use to practice each tip):

Rate problems: Use D = R x T and W = R x T: In this question, it's probably most useful to use the variations R = W/T and T = W/R, because you are given 2 rates (A + B = 800/x and A = 800/y), and you really need to solve for B, because the question asks you for the time taken by Machine B, which is 800/B. You can subtract the second rate equation from the first rate equation to solve for B, then simplify this and plug it into 800/B to get the answer.

Set the amount of work equal to 1 for a single job: As pointed out by others on this thread, because the amount of work done (800 nails) is always the same, we can simplify our calculations even more by just calling this a single job and setting the amount of work equal to 1, instead of 800.

Add rates when they are simultaneous: When we are talking about two simultaneous rates, such as Machines A and B working together, it's important to realize that we can add the rates to get their combined rate. This is what allows us to say that the combined rate is A+B, if the rates for each machine are A and B.

Pick smart numbers to plug into variables in answer choices: When you see that the answer choices contain variables, an alternative approach is to choose smart numbers to plug in for each variable. You also need to know what value you need to get after you plug in the numbers, so that you can eliminate any answer choices that don't give you that value. One really easy scenario to use is one where each machine produces 400 nails per hour (this is equivalent to what Bunuel used in his smart numbers solution), in which case x = 1 and y = 2, and the final value needs to be 2 hours. When you plug x = 1 and y = 2 into the answer choices, E is the only one that gives an value of 2 hours, so that is the correct answer. Note that, if more than one answer choice gave you the correct value, you would need to choose another set of numbers and plug them into the remaining answer choices.

Please let me know if you have any questions, or if you want me to post a video solution!

Video solution from Quant Reasoning starts at 21:55
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Answer: Option E

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Video solution from Quant Reasoning:
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Let A's rate be r and B's rate be r'.

Thus we have
(1) rx + r'x = 800 (took them both x hours to create 800 nails).

The problem tells us A can produce 800 nails in y hours --> ry = 800. Therefore, r is (800/y).

Plug r from above into 1 to solve for r' (the rate for B). (800/y)x + r'x = 800 --> r'x = 800 - 800x/y --> r' = 800/x - 800/y

we are not done yet. we now have the rate but we need to figure out time (t) that it takes to produce 800 nails using only machine B.

Solve it for t (time that it takes) to produce 800 nails. r't =800 --> (800/x - 800/y) t = 800 --> (1/x-1/y)t = 1 --> t = xy/y-x.