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

1 2 3 4

Oppenheimer1945

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26 Dec 2024, 10:24

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Let total work be 300
A rate=6/hr
B rate=5/hr
C rate=4/hr
A+B+C= 15/hr
for 10 hr, 150 work done
remaining work=150
B stopped 15hr before completion
Let total time be T,
B+C worked for T-15 hrs : work done =9(T-15)=9T-225
C worked for 15 hrs: work done=60
90 units completed by B+C: 9(T-15)=90 => T=25hrs
Total time=10+25=35hrs
C)

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Let's define 1 the total amount of work and x the total hours required to finish the task.
Then, productivity of machine A is 1/50 and so on.

$\frac{1}{50} * 10 + \frac{1}{60} * (x - 15) + \frac{1}{75} * x = 1$,
$3x = 105$
$x = 35$

Answer: C

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The stem tells us that;

Machine A, working alone, can complete a task in 50 hours
Machine B, working alone, can complete the same task in 60 hours
Machine C, working alone, can complete it in 75 hours
All three machines started working together on the task
Machine A stopped working after 10 hours
Machine B stopped 15 hours before the task was completed

And we have to select;

How many total hours were required to finish the task

Work=Rate*Time

W=A*50 or B*60 Or C*75

Let the total time be T.

Also,W=(A+B+C)*10 + (B+C)(T-15-10) + C*15

Solving above equation;

W = 10A - 15B + T(B+C)

W = 10(W/50) - 15(W/60) + T { (W/60) +(W/75)}

1 = 10(1/50) - 15(1/60) + T { (1/60) +(1/75)}

Solving this;

T=35

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26 Dec 2024, 10:37

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The answer is 35.

Using the LCM approach, take LCM for 60,50 and 75

We have LCM = 300

Rate of A = 300/50 = 6units/hour ; Rate of B = 300/60 = 5units/hour ; Rate of C = 300/75 = 4units/hr

we have Total work 300 = 6*10 + 5(t-15) + 4t

solving we get t = 35

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26 Dec 2024, 10:40

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C

efficiency of A = W/50
B = W/60
C = W/75

For first 10 hrs
W/50 + W/60 + W/75 = x1/10 (x1 = work done in first 10 hrs)
x1 = W/2

Last 15 hrs
W/75 = x3/15 (x3 = work done in last 15hrs)
x3 = W/5

Work left =
W - W/2 - W/5 = 3W/10

Time taken for the remaining work
as we know both B and C were working until the last 15 hrs
W/60 + W/75 = (3W/10)/t
t = 10hrs

hence the total time taken to finish the task is 10 + 10 + 15 = 35

Bunuel

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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26 Dec 2024, 10:41

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Bunuel

This question was provided by GMAT Club
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The work rates of the machines are as follows:

Machine A completes 1/50 of the task per hour.
Machine B completes 1/60 of the task per hour.
Machine C completes 1/75 of the task per hour.

Machine A works for 10 hours before stopping. In that time, it completes:
10×(1/50)=10/50=1/5
So, 1/5 of the task is completed by Machine A.

Machine B works until 15 hours before the task is completed. Let T be the total time required to complete the task. Then, Machine B works for T−15 hours and completes:
(T−15)×(1/60)

Machine C works the entire time T and completes:
T×(1/75)

The sum of all contributions equals 1 (the entire task):
(1/5)+(T−15)×(1/60)+T×(1/75)=1
Simplify terms:
1/5 + (T−15)/60 + T/75 = 1

On solving the above equation T = 35,

The total time required is T=35 hours.
The final answer is C. 35.

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26 Dec 2024, 10:57

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Let total time taken be t
(10/50)+((t-15)/60)+(t/75)=1
t=35

Answer C

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26 Dec 2024, 11:35

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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?

Machine A alone takes 50 hours
Machine B alone takes 60 hours
Machine C alone takes 75 hours

Machine A stopped working after 10 hours, and Machine B stopped 15 hours before the task was completed

Work done = Rate * Time
Rate of A = 1/50
Rate of B = 1/60
Rate of C = 1/75

When working together, rates are added

For the first 10 hours work done is,
10 * (1/50 + 1/60 + 1/75) = 1/2

Therefore, half the work was done in 10 hours

For the remaining half,
1/2 = (1/60 + 1/75) * (t - 15) + 15/75
Solving,
t = 25

Total time taken = 10 + 25 = 35

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26 Dec 2024, 11:35

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t=hours to finish the task

10*(1/50)+(t-15)*(1/60)+t*(1/75)=1

60+5t-75+4t=300
9t=315
t=35

IMO C

Prakhar9802

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A in 50 hrs
B in 60 hrs
C in 75 hrs

Let total work to be completed be 300 units (took 300 as it is the LCM of all three)
Efficiency-

A does 6 units of work in 1 hr
B does 5 units of work in 1 hr
C does 4 units of work in 1 hr

Total units of work in 1 hr = 15

Total units of work to be done = 300

Total time taken if all three work together = 300/15 = 20 hrs

In the first 10 hrs, half of the work is done,
that is, 150 units of work is done in 10 hrs.

after 10 hrs,
only B and C work.

B stops BEFORE 15 HOURS FROM COMPLETION, that means C works alone in the last 15 hrs.

C does 4 units of work in 1 hr, so,
in 15 hrs, C does 60 units of work.

Remaining work = 150-60 = 90

This much work would be done by B and C together.

B and C in 1 hr does 9 units of work.
So 90 units will be complete in 10 hrs.

Therefore, total time taken for completion = 10 + 10 + 15 = 35 hrs

FINAL ANSWER - Option C

Bunuel

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OmerKor

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26 Dec 2024, 13:24

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Hi everyone

This is a RTD question.
I will build a table in order to illustrate the way to the solution.

	Rate	Time	Work
A		50h
B		60h
C		75h

In order to make the question easier, we will find a LCM of 50-60-75.
I will take the number 300. 300 is "1" "Work".

	Rate	Time	Work
A	6	50h	300
B	5	60h	300
C	4	75h	300

Now we have the "constraints":

	Rate	Time	Work
A+B+C	15	10h	150
C	4	15h	60
B+C	9	10h	300-210=90

I filled the date in the table due to information in the question.

Time: 10h+15h+10h=35h
Answer Choice C

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Answer is C. 35

For the first 10 hours, A + B + C = 10/50 + 10/60 + 10/75 = 225/450 = 1/2 complete
For the last 15 hours, C = 15/75 = 1/5 complete
For the middle X hours, B + C = 1-(1/2+1/5) = 3/10 to complete --> X/60 + X/75 = 3/10 --> X = 10

Task takes 10 + 15 + 10 = 35 hours in total

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Let the rates of A, B and C be Ra, Rb & Rc

Ra= 1/50 ; Rb = 1/60 & Rc = 1/75

Let the total time to complete the work combing A, B & C be x

Now work done (W1) combining A, B & C

W1= (1/50 + 1/60 + 1/75) * 10

Now work done (W2) only by C, considering C only worked for last 15 hours

W2= (1/75) * 15

Now work done (W3) combining B & C

W3= (1/60 + 1/75) * (x - 10 - 15)

W3 = (1/60 + 1/75) * ( x-25)

Also, we know that W1+ W2 + W3 = 1

Now,

((1/50 + 1/60 + 1/75) * 10) + ( (1/75) * 15) + ( (1/60 + 1/75) * ( x-25)) = 1

Solving for x gives x= 35

So, total time to complete the work is 35 hours

Answer is Option (C)

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26 Dec 2024, 14:54

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Let the total work be 300 units.

Machine A completes 300/50 = 6 units per hour
Machine B completes 300/60 = 5 units per hour
Machine C completes 300/75 = 4 units per hour

Machines A, B, and C work together for 10 hours:
1. Total work done by all three machines in 10 hours = (6 + 5 + 4) * 10 = 150 units.
2. Remaining work = 300 - 150 = 150 units.
Machine B stops 15 hours before completion:
1. So, Machine C works alone for those final 15 hours.
2. Work done by Machine C in 15 hours = 4 * 15 = 60 units.
3. Remaining work = 150 - 60 = 90 units.
Machines B and C work together on the remaining 90 units:
1. Together, B and C complete 5 + 4 = 9 units per hour.
2. Time to finish the remaining 90 units = 90 / 9 = 10 hours.

Total time to complete the task:

10 hours (initial work by A, B, and C)
15 hours (C working alone)
10 hours (B and C working together)

So the total time is 10 + 15 + 10 = 35 hours.

Answer is C.

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Let the total hours required to finish the task be t. We have:

(1/50) x 10 + (1/60) x (t-15) + (1/75) x t = 1
1/5 + t/16 - 1/4 + t/75 = 1
t = 35

Answer: C

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1. Analyze given info:
- tA= 50 hours
- tB = 60 hours
- tC = 75 hours

{start}|---10 hours---|---x hours---|---15 hour---|{done}
|A + B + C work |B + C work |only C work

2. Find total hour of work if they work together
- 1/t* = 1/tA + 1/tB + 1/tC = 1/50+1/60+1/75 => t* = 20 hours
=> the first 10 hours, A & B C has completed 1/2 total job

- find total hour to finish 1/2 job:
W/2 = Rb*(T-15) + Rc*(T-15) + Rc*15
=>1/2 = (1/60)*(T-15) + (1/75)*(T-15) + (1/75)*15
=> T = 25 hours
=> Total time = first 10 hours + next 25 hours = 35 hrs.

Ans C

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26 Dec 2024, 21:30

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To solve this question, it is given that machine A can do its work alone for 50 hours & it did its work for 10 hours on the work in question.

Hence if the total amount of work can be taken to be equal to W, the amount of work done by machine A would be equal to W/50*10=W/5

The remaining work would hence be equal to W-W/5=4W/5. This would be equal to the total amount of work done by both B & C combined.

If we assume the total time taken to complete the work be equal to T, the equation for the total work done would be hence equal to:

4W/5=W*(T-15)/60+W*T/75
4W/5=5W*(T-15)+4*W*T/300
On cancelling out W on both sides,
4*300/5=5T-75+4T
240+75=9T
315=9T
T=35

Hence the total numbe rof hours spent to complete the work is equal to 35 hours

Hence the answer to this question is option (C) 35

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A= 50 hr./task=> efficiency of A= 1/50 task per hr.
B= 60 hr./ task=> efficiency of B= 1/60 task per hr.
C= 75 hr/ task=> efficiency of C= 1/75 task per hr.

We know, efficiency of A+ efficiency of B+ efficiency of C= 1

Given, A worked for only 10 hrs., and B stopped 15 hrs. before the completion of work. Let t be the time taken by all 3 of them for total completion of work.

So, 10/50+ (t-15)/60+ t/75= 1

Solving the above eqn. we get,
t= 35

Ans.- (C)

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Bunuel

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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Machine A work in one Hour: 1/50 = rate
Machine B work in one Hour: 1/60 = rate
Machine C work in one Hour: 1/75 = rate

Total work = 1
Work = rate *time

T be the total time taken to complete the work when all 3 machines were working
eqn as per given info
10/50 + (T-15)/60 + T/75 = 1
T/60 +T/75 = 21/20
(5T+4T)/300 = 21/20
T = 21*300/(20*9)
T = 35
Total time is 35 Hours

IMO 35

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26 Dec 2024, 23:48

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All machines combine rate = 1/50 + 1/60 + 1/75 = 1/20
Machine A works 10 hours = 10 * 1/20 = 1/2

Machine B works T-15 hours
Machine C works T hours
Combined rate of B and C : 1/60 + 1/75 = 3/100
Equation : 3(t-15)/100 + 3T/100 = 1/2
(6T - 45)/100 = 1/2
6T - 45 = 50
T = 15.83 hours
Total time = 10 + T = 10 + 15.83 = 25.83 hours