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12 Days of Christmas GMAT Competition - Day 9: In a factory, Machine A

1 2 3 4

Elite097

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27 Dec 2024, 00:31

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Time taken by machine A=50 hrs
Time taken by machine B= 60 hrs
Time taken by machine C=75 hrs
(LCM of 50,60 and 75)-Work =300 units

Rate of A=6 unts/hr
Rate of B=5 unts/hr
Rate of C=4 unts/hr

A stopped working after 10 hrs,
300 - 10*(4+5+6)=150

B stopped working before 15 hrs before the work is completed,
Remaining time needed to complete = 150/(5+4)=16.67 hrs
B & C work together for 1-2 hrs

Remaining work = 141(or 132) units
Time taken by C to complete = 141 (or 132)/4 ~35 (or 33 ) hrs.
Ans is E (35+1+10 ~45 hrs)

Bunuel

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In a factory, Machine A, working alone, can complete a task in 50 hours, Machine B, working alone, can complete the same task in 60 hours, and Machine C, working alone, can complete it in 75 hours. All three machines started working together on the task. However, Machine A stopped working after 10 hours, and Machine B stopped 15 hours before the task was completed. How many total hours were required to finish the task?

A. 25
B. 30
C. 35
D. 40
E. 45

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27 Dec 2024, 01:06

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Hi All,

work done per day by A,B,C = 1/50+1/60+1/75 -->1/20

They worked together for 10 hours, therefore 10*(1/20)= 1/2 work completed

Work left =1-1/2-->1/2

Work done per day by B,C=1/60+1/75= 3/100

Time take to complete 1/2 work = (1/2)/(3/100) (from the work done formula)

Time = 16.67 hours.

But B left before 15 hours of completetion, therfore B and C worked for only 1.67 hours

therefore work done= (3/100)*1.67= 1/20

Now C alone completes the remaining work i.e 1/2-1/20 = 9/20

Time taken = (9/20)/(1/75) =33.7 hours

Calculating total time = 10 + 1.67 + 33.7 = 45 approx

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Bunuel

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This question is Tricky, especially because there are 2 common types of mistakes that you can make that can hamper your progress:
1) Not understanding the statement " Machine B stopped 15 hours before the task was completed"
2) Not being able to plot the question appropriately

I follow this method (Note: create a common denominator to solve faster) (Bold is what you have to solve for)

	Rate	Time	Work
A	1/50 (6/300) Units of Work per hour	50 Hours	1 Units
B	1/60 (5/300) Units of Work per hour	60 Hours	1 Units
C	1/75 (4/300) Units of Work per hour	75 Hours	1 Units
A+B+C	(6+5+4)/300	10 Hours	150/300 Work done
C	Mentions that C worked alone since B worked for for total -15 Hours after A left 4/300	15 Hours	60/300 Work done
B+C	(5+4)/300 Units of Work Per Hour	Calculate from Work and Rate 10 Hours	300-(150+60)/300 = 90/300 Units Pending

Therefore, total Units of work = 10+15+10 = 45 Hours (E)

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27 Dec 2024, 01:13

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Time taken by each machine to complete the task :
A = 50; B = 60 & C = 75
Let the total work be 300..... (LCM of 50, 60 & 75)
Work per hour by each machine :
A = 6; B = 5 & C = 4

Let the total time taken be 't'
(6*10) + (5(t-15)) + 4t = 300
60 + 5t - 75 + 4t = 300
9t = 315
t = 45 hours

IMO E

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27 Dec 2024, 01:22

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Rate of Machine A $=\frac{1}{50}$

Rate of Machine B $= \frac{1}{60}$

Rate of Machine C $= \frac{1}{75}$

Total time taken for work, t

sum of work done by individual machines = 1

Work done by $A=\frac{10}{50}=\frac{1}{5}$

Work done by $B=\frac{t-15}{60}$

Work done by $C=\frac{t}{75}$

$\frac{1}{5}+\frac{t-15}{60}+\frac{t}{75}=1$

$\frac{5t-75+4t}{300}=\frac{4}{5}$

$9t=315$

$t=35$

Answer: C

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27 Dec 2024, 01:42

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Please find the attached solution

Attachments

WhatsApp Image 2024-12-27 at 13.49.55_8b9b9191.jpg [ 635.88 KiB | Viewed 754 times ]

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27 Dec 2024, 01:52

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Ans: 25 Hrs
Lets say total hrs required = t

then 10(A+B+C) + (B+C) * (t-15) + 15C = 1
first simplify this in ABC terms and make it in t = ###/### form

this will give t = 25

Or Another solution:

C worked for 15 hrs so completed 15/75 = 1/5 part of the work alone --> Remaining 4/5th
ABC together worked for 10 hrs so (1/50 +1/60+1/75) = 1/2 they completed half of the work --> Remaining 3/10
from this (1/60 +1/75) (t-15) = 3/10
(t-15) = 3/10 * (60*75)/135
t-15 = 10
t= 25 hrs

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In a factory, Machine A, working alone, can complete a task in 50 hours, Machine B, working alone, can complete the same task in 60 hours, and Machine C, working alone, can complete it in 75 hours. All three machines started working together on the task. However, Machine A stopped working after 10 hours, and Machine B stopped 15 hours before the task was completed. How many total hours were required to finish the task?

A, B, C together work for 10 hours,

(1/50+1/60+1/70)10
=1/2

C works for 15 hour alone,
1/75x15=1/5

Total work completed, 1/2+1/5= 3/5, remaining= 2/5 which done by B&C.

B & C completed work in 1 hour,

1/60+1/75
=3/100

To, complete 2/5 required hours,

100/3x2/5
=40/3
=13.++++

So, entire work will be completed in 10+15+13.+++
38+++

Answer: D. 40

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27 Dec 2024, 03:44

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Rate A = 1/50
Rate B= 1/60
Rate C= 1/75
In one hour the three will have completed 1/50+1/60+1/75= 1/2 of the work
Since Machine B stopped at some point leaving machine C to work alone for 15 hours, the work covered by C for the 15 hours was 15x1/75= 1/5
So the work that was done by both B&C was 1-(1/2+1/5)= 3/10
Divide the work by the combined rate of both C&B= (3/10)/3/100= 10 hours
Total time equals= 10+15+10= 35 hours
C

Bunuel

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the answer is 35, equivalent to answer C

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27 Dec 2024, 04:18

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A work=10/50
B work=(time-15)/60
C work=time/75

All together:
1/5+(time-15)/60+time/75=1

Solving:
9*time=315

time=35

Answer C

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In a factory, Machine A, working alone, can complete a task in 50 hours, Machine B, working alone, can complete the same task in 60 hours, and Machine C, working alone, can complete it in 75 hours. All three machines started working together on the task. However, Machine A stopped working after 10 hours, and Machine B stopped 15 hours before the task was completed. How many total hours were required to finish the task?

A. 25
B. 30
C. 35
D. 40
E. 45

Suppose, Task = 300 units

Time taken (hrs) Rate(units/hr)
Machine A -> 50 6
Machine B -> 60 5
Machine C -> 75 4

In first 10 hrs (all A, B and C work) ----> 10 * (6 + 5 + 4) = 150 units................(1)

In last 15 hrs (only C works) ----> 15 * (4) = 60 units.................(2)

thus, in middle T hrs (suppose), work completed = 300 - (150 + 60) = 90 units
in T hrs, only B & C work, so Rate = (5+4) units/hr = 9 units/hr

So, T = 90 / 9 = 10 hrs...................................................................................(3)

Using (1), (2) and (3)

Total time taken = 10 + 15 + 10 = 35 hrs

(C) is the CORRECT answer

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Bunuel

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If we consider the total unit of work to be 1.

Rate of A = 1/50
Rate of B = 1/60
Rate of C = 1/75

Given A worked for 10 hours = 10 * (1/50) = 1/5
Given B worked till 15 hours before the Total time = t-15 * (1/60)
Given C worked complete time = t * (1/75)
1/5 + (t-15)/60 + t/75 = 1

Least common denominator of 5, 60, 75 is 300.

By multiplying it to the equation and further solving it we get t=35

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27 Dec 2024, 05:28

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we know work=rate*time,
so if work =1 indicates the whole completed work, then rate =1/time

Rate of Machine A=1/50
Rate of Machine B=1/60
Rate of Machine A=1/75,

we know the total work they have to do is 1,
hence
Work done by A{for 10 hours}+ Work done by B{let the total time work done by C=t hours so B would do t-15}+ Work done by C =1
10*1/50+(t-15)*1/60+t*1/75=1
so t from here will be 35 hours,

Method 2
together they can do work will be equal to the rates of A and B and C
thus Rate of all together =Rate A+Rate B+Rate C
=1/50+1/60+1/70,
now A+B+C have done 10 hours of work so work completed till first 10 hours we can calculate
which is Rate{ABC}*Time {which will come out as 1/2 after calculation} ,
now the remaining work will be 1-1/2=1/2
for this remaining work Machine C has used let T hours {as machine C is doing till last}so Machine B must have done T-15 hrs.
so
Rate{BC}*T-15+Rate C* T=1/2
this will again give us T which is 35 Hrs.

Bunuel

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27 Dec 2024, 06:21

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Solve this step by step in simple terms.
Given:

Machine A: Completes the task in 50 hours.
Machine B: Completes the task in 60 hours.
Machine C: Completes the task in 75 hours.
Machine A stops after 10 hours.
Machine B stops 15 hours before the task is finished.

Let's Find:
Total time (T) to complete the task.
Step-by-Step Solution:

Calculate Work Rates:
- Machine A: 1/50 of the task per hour.
- Machine B: 1/60 of the task per hour.
- Machine C: 1/75 of the task per hour.
Work Done in Different Periods:
- First 10 Hours (All three machines working):
  (1/50 + 1/60 + 1/75) × 10 = 15/300 × 10 = 1/20 × 10 = 0.5 tasks
- From 10 Hours to T−15T - 15 Hours (Machines B and C working):
  (1/60 + 1/75) × (T − 25) = 9/300 × (T − 25) = 3/100 × (T−25)
- Last 15 Hours (Only Machine C working):
  1/75 × 15 = 0.2 tasks
Total Work Done:
0.5 + 3/100(T − 25) + 0.2 = 1 task
0.7 + 3/100(T − 25) = 1
3/100(T − 25) = 0.3
3(T − 25) = 30
T − 25 = 10
T = 35 hours

Answer:
C. 35

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27 Dec 2024, 06:25

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Productivity of the machines is: $a=1/50, b=1/60$ and $c=1/75$

All machines together completed the tasks, and each worked for a certain period of time:
$1=t*c+10a+(t-15)b=t(c+b) + 1/5-1/4$
From there, $t(c+b)=t(1/75+1/60) = 9t/300=3t/100$

And finally $1+0.25-0.2=1.05=\frac{3t}{100}$
And $t=\frac{100*1.05}{3}=35$ and the answer is C.

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27 Dec 2024, 06:38

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Rate of A alone= $\frac{1}{50}$
Rate of B alone= $\frac{1}{60}$
Rate of C alone= $\frac{1}{75}$

Combined rate= $\frac{1}{20}$
Let the total time be $t$

Work done by A.B,C in 10 hours = $10*\frac{1}{20}= \frac{1}{2}$ ...so half the task was completed

B&C together worked for a total of $t-25$ hours, which is $(t-25)*\frac{3}{100}$

As per question,
Work done by A,B,C in 10 hours + Work done by B&C in t-25hours + Work done by C alone in 15 hours= 1
$\frac{1}{2} + (t-25)*\frac{3}{100} + 15*\frac{1}{75} = 1$
$t=35$

IMO C

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27 Dec 2024, 07:05

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Work rate of machine A/hr = 1/50
Work rate of machine B/hr = 1/60
Work rate of machine C/hr = 1/75

Work done in first 10 hrs together = 10 x (1/50+1/60+1/75) = 1/2 = 0.5
Machine A stops after 10hrs,
Combined work rate of B and C = 1/60 + 1/75 = 3/100

Now,
Total time required to finish the task = T
=> Machine B and C work together for the time = T - 10 - 15= T -25 (Since Machine C works for the final 15hrs alone)

Work done by C = 1/75 x 15 = 1/5
Total work done by B and C together = 1 - (0.5 +0.2) = 0.3
=> 3/100 x t = 3/10
=> t = 10hrs

C. Total time = 10+10+15 = 35hrs

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27 Dec 2024, 07:17

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Let efficiency of A, B and C be 'a', 'b', 'c' respectively
Assume work need to be done is 'W' and is completed in 'x' hrs

a * 50 = b * 60 = c * 75 = W

Given

a * 10 + b * (x - 15) + c * x = W
$\\
\\
\frac{W}{5} + \frac{W}{60 }(x - 15) + \frac{W}{75 }* x = W\\
\\
\frac{4}{5 }= \frac{x}{60 }- \frac{1}{4} + \frac{x}{75 }\\
\\
\frac{4}{5} + \frac{1}{4} = \frac{(5x + 4x)}{300 }\\
\\
\frac{21}{20 }= \frac{(9x)}{300 }\\
\\
$

7 * 5 = x = 35

OPTION C

Bunuel