Six machines, each working at the same constant rate

[Guest JK]

GMATers- Please note: The following question appears in the PowerPrep program CAT test.

I would like confirmation on the method I used to solve the following question. Since PowerPrep only provides answers, I want to see if the method I used is correct or if I was just lucky.

"Six machines, each working at the same constant rate, together can complete a certain 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?"

A) 2
B) 3
C) 4
D) 6
E) 7

Answer is B. Below is the method I used:

If 6 machines work at constant rate (1/x) and take 12 days to complete 1 job, then:
Step 1: Find individual rate of machine
R*T=D
6(1/x)(12)=1 and so rate of 1 machine is 1/72 (translated in english this means 1 machine can do 1/72 of the job in 1 day)

Step 2: Find how many extra machines are necessary to complete job in 8 days
(6/72+x/72)(8)=1
(6+x)/9=1
x=3 or 3 extra machines are necessary.
Is this method legitimate or is there another method that may be wholly applicable to these type of problems? Thanks :D

[JadranLee]

Your method looks great, JK. To make things a bit easier for yourself, you could also have made a table like this to help you keep track of all the quantities and how they are related:

| 6 machines | 6+x machines
Rate |

Time | 12 | 8

Work |


(Sorry, the formatting is messed up, but I think you can see what I mean.)

I put 12 and 8 in the table because those numbers are given in the problem. You're free to pick any number you like for the size of the job (i.e. the number that goes in the work row). For this problem, you could have picked 72 instead of 1 to represent the size of the job.

-Jad

[minu]

Jad cld u please explain the problem again..due to the formatting i can't understand what u did and i am stumped by this problem....thanx!

[blue_lotus]

Let us convert the given information into a genal unit (machiune days)
The number of machine days required for the job = 12 * 6 = 72 machine days

Now we have to do the same work(72 unit) in 8 days .
Therefore number of machines required = 72/8 = 9

The question asks how many additional machines required.
As we already had 6 machines , we would require 9 - 6 = 3 machines.

[StaceyKoprince]

nice, blue lotus

The chart would look something like this:

------------6 machines-----6+x machines
Rate-----------y---------------y----------(rate is same)

Time --------12----------------8---------(in days)

Work--------1 job-----------1 job

Remember that rt=w. Then, fill in what you know and get to work! Jad suggested picking the setting the size of the job to 72 units to make things easier eg, 6 machines do 12y work and 6+x machines do 8y work. You can get the rate from the first equation (the 6 machines column): if 6 machines do 12y work (and complete the job), then each machine does 2y work (or 1/6 of the total job) over that 12 day period, which also translates to (1/6) * (1/12) or 1/72 of the job per machine per day. Well, that's an annoying number! So say the whole job is 72 widgets and each machine produces one widget (or 1/72 of the job) per day. Then continue on with the original poster's method, but use these nicer numbers instead.

You could also use blue lotus's suggested approach - a very elegant solution to this problem. You can use that if the rates are both constant for all of the machines throughout the problem and the SAME rate for all of the machines.

[benkriger]

It does not need to be so complicated.

1) Each machine has the same constant rate, so just choose 1 as the rate. (or 2 if you do not like picking 1 as a number).

2) Calculate the number of units produced by the 6 machines in 12 days at their constant rate of 1: 6*12*1=72units!

3) Calculate the number of units produced by the X machines in 8 days at the same constant rate of 1, setting it equal to the 72 units, because the problem tells us that the same number of units are made: X*8*1=72

4) Solve for X: X=9

5) Answer the problem: 9machines - 6 machines = 3machines!

------- Just to make it easier for those who dont want to read----

1)pick number for constant rate (i used 1)
2)find units by 6 machines in 12 days: 6*12*1=72 units
3)Set up equation to find x machines in 8 days: X*8*1=72
4)Find the difference between X in step 3 and 6. Which is 3.

[Ben Ku]

benkriger, Your approach is correct and works well. Basically, you're selecting the "smart" number of 1 to represent the rate (specifically 1 job / day). As you can see, there are many correct approaches to solve a rate problem. As you prepare for the GMAT, try to approach a problem in multiple ways, and determine which way works best for you in a particular problem.

[warren.margolin]

6 * 12 = y * 8
6 * 12 / 8 = y
9 = y

therefore additional 3 required

[mschwrtz]

That's exactly right Warren. It is possible to set the two sets of factors equal without worrying about what their product is. I don't think that it saves much effort in this case, but it could if the number were even a bit more complicated.

Here's a made-up question (well, I guess that they're all made up by somebody):

24 identical machines take 12 days to complete a work order. How many days fewer would 32 such machines require?

Warren's method allows us to say:

Solve for 12-x

24*12=32x
(24*12)/32=x
cancel three 2s out of 24, two 2s out of 12, and all five 2s out of 32, and we have x=9. It would take 12-9 fewer days.

Notice that this avoids some of the computation involved in
288=32x
.
.
.

[its4christian]

HI Michael,

Warren method looks like a simple proportion, but inverted....can you explain why it works and if we can create a "general rule" for the next similar problem. (instead of using RTD chart, consider this problem as a disguised ratio problem...?)

This was my simple approach ( I felt it was too simple....)

days/machines=days/machines

6/12= x/8
x=4

Is there an extra step to get to 3?

Thanks a lot,

Chris

[mschwrtz]

Hey Chris,

I'm not 100% I followed your question. Does this help:

Normally in a work-rate problem, the relevant formula is rt=w.
In this case, rather than t we have the product machine days, or md.
So in this case the relevant formula is a special case of the general formula, rmd=w.
We can rewrite this as md=w/r.
In this case, r is treated as a constant, and so is w.
So md is also a constant.
So where you have "days/machines=days/machines" you should have "days*machines=days*machines"

Maybe you could see even in "rmd=w" that if r and w are constants then md is fixed.

[sadat.sathak]

Hey Guys,

First post...and even though I note there are multiple approaches, just wanted to make sure I could articulate a method that made sense to me(after all the best learning happens when you teach)

Rephrase question:
6 machines complete a job in 12 days
x machines complete the job in 8 days

Therefore the x machines complete the job in 8/12= 2/3 the time that 6 machines do. (we can already intuitively note that 6 more machines-double of what it currently is-would reduce the time by half, therefore the additional machines should be less than 6)

Keeping in mind the 'inverse proportionality rule (2/3 becomes...3/2)

3/2*6= Voila... 9 machines in total.
Hence 9-6 machines= 3 additional machines.

[tim]

cool..

[jkewalramani]

I think we can also look at percentages as an option . This method would only work if you are good at calculations . It takes 12 hours to do 100% , around 8.3% every hour . 6 machines do 8.3% which is roughly 1.3% per machine .If we need the job to be completed in 8 hours which is around 12.5 % leaves a gap of 4.2 % which can be estimated to 3 machines i.e 3 * 1.3 =4% .

[RonPurewal]

I think we can also look at percentages as an option . This method would only work if you are good at calculations . It takes 12 hours to do 100% , around 8.3% every hour . 6 machines do 8.3% which is roughly 1.3% per machine .If we need the job to be completed in 8 hours which is around 12.5 % leaves a gap of 4.2 % which can be estimated to 3 machines i.e 3 * 1.3 =4% .

percentages aren't a very good option here, because the number 100 is not compatible with the numbers in the problem (6 or 12).
in other words, you are basically just using the strategy of "pick your own numbers" (variously known as "VIC" or "smart numbers" in our books) -- but with ill-chosen numbers.

this method will work better if you pick numbers that are less awful.
let's say that the "constant rate" is 1 unit per machine per day.
then, since 6 machines can do a whole job in 12 days, one whole job is equivalent to 72 units.
if you want to do this job (72 units) in 8 days instead, you need to do 9 units of work per day. for that, you'll need 9 machines, which is 3 more than you originally had.
done.