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Working/Rate Problem

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Working/Rate Problem

by vladmire » Tue Oct 14, 2008 6:25 pm
Company C has a machine that, working alone at its constant rate, processess 100 units of a certain product in 5 hours. if company c plans to buy a new machine that will process this product at a constant rate and if the two machines working together at their respective constant rates are to process 100 units of this product in 2 hours what should be the constant rate in units per hour of the new machine?

50
45
30
25
20


Working/rate
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by cramya » Tue Oct 14, 2008 7:50 pm
The answer should be C) 30

Old machine produces 100 machines in 5 hours
In 1 hour its rate is 20 machines/hour

Let R2 be the rate of the new machine

When 2 machines work together the rates are added (i.e 20+R2)

Using work formula

Rate * no of machines * time = work

(20+R2) * 1 (since they work together they are treated as 1 machine) * 2(given) = 100

20+R2 = 100/2
20+R2 = 50
R2 = 30

CHECK:

OLD MACHINE WORKS FOR 2 HOURS AN PRODUCES 40 UNITS
NEW MACHINE WORKS FOR 2 HOURS AND PRODUCES 60 UNITS
TOGETHER THEY PRODUCE 100 UNITS IN 2 HOURS

Hope this helps!
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by raunekk » Wed Oct 15, 2008 2:03 am
old machine,

100 mac - 5 hrs
thus, 20 mac - 1 hrs

new machine,
let rate be x


combined,
should produce,
100 units - 2 hrs
50 units - 1hrs

Thus new machine rate is,
new + old = combined
x+ 20 = 50

x=30

thus C!!
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by msvmuthu » Thu Oct 16, 2008 2:34 am
One more way to solve the problem:

here the job is constant : 100 machines there fore we could omit the job for now.

formula : 1/r + 1/s = 1/h

where r and s are the no of hours required by machines r and s to complete a job separately and h is the no of hours required by both machines together for the same job.

here no of hours required by machine r = 5 and we dont know for machine s and total time taken by both machines would be 2.

Applying the values to the above said formula:

1/5 + 1/s = 1/2

solving for s:
2s+10=5s
3s=10
s=10/3

so time taken for the machine s to produce 100 units is 10/3 hrs.
But the question asked is rate of s! we need to find the no of units per hour!

10/3 hrs -> 100 units
1 hr -> (100 * 1) / (10/3)

solving would give => 30
so the rate of new machine s is 30
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