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

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

Rate of 18 Identical Machines = (1 job) / (40 Days)


Rate of 1 Machine ------> Since the 1 Job remains CONSTANT ---- Rate of Work is INVERSELY Proportional to Time needed to compete the Work

Divide the Rate by 18 Machines -------> Results in 18 * the amount of TIME needed to complete the Job


Rate of 1 Machine = (1 job) / (18 * 40) days


If we Add an ADDITIONAL 54 Machines from the beginning:

Total Machines = 54 + 18 already used = 72 Total Machines


Let D = No. of Days it takes to complete the Work with the Additional 54 machines

#Machines.......... * .......Rate per 1 Mach....... * .........TIME = Work to be Completed
(72 Machines)...... * ....... (1)/(18*40) ........... * ............D = 1 Job


(72D) / (18*40) = 1

(8D) / (2 * 40) = 1

(D) / (2 * 5) = 1

D = 10 Days to complete the job with 72 machines



Originally took 40 Days

-

10 Days it would have taken had we used an additional 54 machines
___________________________________________________


30 Fewer Days

-E-


Edit: much easier Method 2

Given the Same, Constant Job:

The Number of Machines (each working at an Identical Rate) is INVERSELY Proportional to the Amount of Time necessary to complete the Job

If more machines are added———-> the amount of Time needed to complete the Job DECREASES Proportionally

18 machines————————> take 40 days to perform the job


(18 + 54) = 72 machines ————-> take X days to perform the job


The 18 machines Increase by a Factor of * (4) ——— 18 * (4) = 72 machines

Therefore, the amount of Time needed to complete the Same Job will Decrease by a Factor of * (1/ 4) ———in other words, it will take 1/4th the amount of Time

40 days * (1/4) = 10 days


It would have taken

30 Fewer Days

Given

• 18 identical machines required 40 days to complete a job.

To Find

• 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.

Approach and Working Out

• 18 machines in the first case and 54 + 18 = 72 machines in the second case.
• Ratio of machine 1: 4.

o Ratio of time = 4 : 1
o The First case it took 40 days, so the number of days now would be 10.
o 40 – 10 = 30 fewer days

.


Correct Answer: Option E

18 identical machines work for 40 days to complete 18 * 40 = 720 units of work.

Additional '54' same type machines: 18 + 54 = 72 will take \(\frac{720 }{ 72}\) = 10 days to complete the same work.

Fewer days: 40 - 10 = 30 days

Answer E

When it says fewer days shouldn’t it mean the difference of days required to do the job ?

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