How many more machines..

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How many more machines..

by adi_800 » Sat Sep 04, 2010 5:47 am
14) Six machines working at the same rate constant rate complete a job in 12 days. How many additional machines working at same rate will be needed to complete the job in 8 days?
a. 3
b. 6
c. 9
d. 12
e. 15


My approach : 12 days - 1 job - 6 machines
8 days - 1 job - how many machines

= 8 * 6 /12 = 4 machines...

So additional machines = 4 -1 = 3, which is indeed the OA...

Is my approach right?

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by shibal » Sat Sep 04, 2010 6:33 am
adi_800 wrote:14) Six machines working at the same rate constant rate complete a job in 12 days. How many additional machines working at same rate will be needed to complete the job in 8 days?
a. 3
b. 6
c. 9
d. 12
e. 15


My approach : 12 days - 1 job - 6 machines
8 days - 1 job - how many machines

= 8 * 6 /12 = 4 machines...

So additional machines = 4 -1 = 3, which is indeed the OA...

Is my approach right?
not really.....

you know that it takes 6 machines to complete the job in 12 days. so you can deduce that one machine makes 1/72 of the job in a day (6*12 = 72).

now, use the formula rate*time=work; from our previous analysis and from what is given in the statement, we know that work=72 and time=8. plug them in the formula to find the rate, which is equal to 9. thus you need 9 machines working at the rate given by the exercise to complete the job in 8 days. if you have 6 today, then 3 are necessary to complete the job. hope it helps.

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by selango » Sat Sep 04, 2010 8:19 am
Another easy approach..

6*12=x*8

x=6

Additional machines=3
--Anand--

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by selango » Sat Sep 04, 2010 8:22 am
adi_800 wrote:14) Six machines working at the same rate constant rate complete a job in 12 days. How many additional machines working at same rate will be needed to complete the job in 8 days?
a. 3
b. 6
c. 9
d. 12
e. 15


My approach : 12 days - 1 job - 6 machines
8 days - 1 job - how many machines

= 8 * 6 /12 = 4 machines...

So additional machines = 4 -1 = 3, which is indeed the OA...

Is my approach right?
the below formula can be used

M1*D1*H1/W1=M2*D2*H2/W2

M-Men or machines or persons

D-days

H-Number of hours

W=work
--Anand--

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by arora007 » Sat Sep 04, 2010 11:16 am
adi_800 wrote:14) Six machines working at the same rate constant rate complete a job in 12 days. How many additional machines working at same rate will be needed to complete the job in 8 days?
a. 3
b. 6
c. 9
d. 12
e. 15


My approach : 12 days - 1 job - 6 machines
8 days - 1 job - how many machines

= 8 * 6 /12 = 4 machines...

So additional machines = 4 -1 = 3, which is indeed the OA...

Is my approach right?
If 12 days are taken by 6 machines, it should logically stem that.. 8 days would only be possible when more than 6 machines are employed, how many more has to be figured out...


6 * 12 = 8 * (6+X)
Now X = 9 - 6 = 3
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by GMATGuruNY » Sat Sep 04, 2010 11:19 am
Approach #1:

When the job is undefined, try plugging in your own number for the job.

Let's say each machine produces one unit each day. Then the 6 machines together would produce 6 units each day. Over 12 days, 6 * 12 = 72 units would be produced.

If we want to produce the 72 units in only 8 days, we'll need 72/8 = 9 units to be produced each day.

Since we'll need 3 more units to be produced each day, and each machine produces 1 unit per day, we'll need 3 more machines.

The correct answer is B.

Approach #2:

(number of machines) x (number of days) always has to yield the same amount of work.

So we could set up this equation:

(number of machines) x (number of days) = (number of machines) x (number of days)

6 * 12 = x * 8
72 = 8x
x = 9

Since we'll need 9 machines altogether, and we currently have 6, we'll need 9-6=3 more machines.

For all you math lovers: In math terms, the number of machines is inversely proportional to the number of days. When two values are inversely proportional, as one value goes up, the other must go down, so that their product is always the same. In the problem above, as the number of machines goes up, the number of days must go down, so that the product of the 2 values stays the same and we're always getting the same amount of work done.
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by Keeto702 » Sat Sep 04, 2010 1:21 pm
6 machines do the job in 12 days
thus 1 machine will do the job in 72 days (6*12)
Now we want the job to be done in 8 days so 72/8 = 9
which is 3 more than given 6 machines (9-6 = 3)


Let me know if my approach is crisp and simple