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If 18 identical machines required 40 days to complete a job, how many

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If 18 identical machines required 40 days to complete a job, how many  [#permalink]

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New post 05 Mar 2019, 10:15
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  45% (medium)

Question Stats:

54% (01:12) correct 46% (02:00) wrong based on 59 sessions

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If 18 identical machines required 40 days to complete a job, how many fewer days would have been required to do the job if 54 additional machines of the same type had been used since the beginning?

A. 10 days
B. 40/3 days
C. 16 days
D. 80/3 days
E. 30 days
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Re: If 18 identical machines required 40 days to complete a job, how many  [#permalink]

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New post 05 Mar 2019, 21:50
18 machines did (1/40)th work in a day, then 72 ie (54+18) machines did x work in a day
x = 1/10 , ie 10 days. ( Use proportionals logic )
So difference is 30 days
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Re: If 18 identical machines required 40 days to complete a job, how many  [#permalink]

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New post 05 Mar 2019, 22:52
18 Machines - Rate 1/40 hence 1 machine takes - 1/(40*18).
72 Machines takes 1/(40*18)*72 = 10 days. So 30 days less
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Re: If 18 identical machines required 40 days to complete a job, how many  [#permalink]

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New post 06 May 2019, 23:23
mangamma wrote:
If 18 identical machines required 40 days to complete a job, how many fewer days would have been required to do the job if 54 additional machines of the same type had been used since the beginning?

A. 10 days
B. 40/3 days
C. 16 days
D. 80/3 days
E. 30 days

Potential traps: the question asks about additional machines and how many fewer days.

Just change the traditional (R * T) = W formula.
Add one more variable, # of machines, to the left hand side, this way
(# of machines * R * T) = W

Manipulate that formula in the same way that the "regular" RT=W gets manipulated

Step 1: Find the rate of an individual machine
# of machines = 18
R = ??
T = 40
W = 1

# of machines * R * T = W
\(18 * R * 40 = 1\)
\(R = \frac{1}{18*40} = \frac{1}{720}\)


Step 2: Use that individual machine rate to find the time that
54 additional machines would have needed to finish the work
# of machines = (18 + 54) = 72
R = \(\frac{1}{720}\)
T = ??
W = 1

# of machines * R * T = W
\(72 * \frac{1}{720} * T = 1\)
\(T * \frac{72}{720} = 1\)
\(T = \frac{720}{72} = 10\) days


72 machines (i.e., 54 additional machines) would have required 10 days to finish the work.

Step 3: How many fewer days would have been required?
-- 18 machines required 40 days.
-- 72 machines would have required 30 days.

(40 - 30) = 10 days

Answer E
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Re: If 18 identical machines required 40 days to complete a job, how many   [#permalink] 06 May 2019, 23:23
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