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

Let the total time be t, therefore A has worked for 10 hours, B has worked for t - 15, and C has worked for t
10/50 + (t-15)/60 + t/75 = 1
Solving gives t = 35

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?

Let's assume that total x hours were required to finish the task.

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

.2 + x/60 - .25 + x/75 = 1
x/60 + x/75 = 1.05 = 21/20
9x/300 = 21/20
x = 35 hours

IMO C

Step 1: Machine Work Rates
Machine A: 1 task in 50 hours → Rate = 1/50
Machine B: 1 task in 60 hours → Rate = 1/60
Machine C: 1 task in 75 hours → Rate = 1/75
Step 2: Work Done in First 10 Hours
All three machines work together for 10 hours.
Combined rate of all 3 machines:
1/50 + 1/60 + 1/75 = 1/20 (after finding the common denominator).
In 10 hours, work done = 10 * (1/20) = 1/2 of the task.
Step 3: Work Done After 10 Hours
After 10 hours, Machine A stops. Machines B and C continue.
Combined rate of B and C:
1/60 + 1/75 = 9/300 = 3/100.
Remaining work = 1/2 of the task.
Time to finish = (1/2) ÷ (3/100) = 50/3 ≈ 16.67 hours.
Step 4: When Does Machine B Stop?
Machine B stops 15 hours before the task is finished.
So, Machine B works for 16.67 - 15 = 1.67 hours after 10 hours.
Step 5: Total Time to Finish Task
Machine A worked for 10 hours.
Machine B worked for 1.67 hours.
Machine C worked for 16.67 hours.
Total time = 10 + 16.67 = 26.67 hours ≈ 30 hours.

Answer:
30 hours.

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

Rate*time = work

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

on simplification we get t=35

Rate for all 3 together = 1/50 + 1/60 + 1/75 = 1/20
Rate for B and C together = 1/60 + 1/75 = 3/100

Work by all 3 = 1/20*10 = 1/2, i.e. 50% is done by all 3 in 10hrs

Let total number of hours to complete the remaining work be x, then B and C worked together for x-15 hrs

1/2 = (x-15)*(3/100) + 15* 1/75
Solving this we get x = 25 hrs

Total therefore is 35hrs

Let total hours taken be T

Then,

10/50 + (T - 15)/60 + T/75 = 1
\(\frac{10 *6 + (T - 15)*5 + 4T}{300} = 1\)
60 + 5T - 75 + 4T = 300
9T = 315
T = 35

Answer: C

A:
w in 50 hours
R(A) = w/50 in 1 hr

B:
W in 60 hours
R(B) = w/60

C:
R(C) = w/75

Lets say the work was completed in t hours, then B worked for t-15 hours
W = 10*(w/50) + (t-15)*(w/60) + (w/75)*t
1 = 1/5 + (t-15)/60 + (t/75)

4/5 = 1/15 ((t-15)/4 + (t)/5)
4 = 1/3*20 (5t-75 +4t)
4*20*3 = 9t-75
(240+75)/9 = t
315/9 = t = 35 hours

better take eg of number of units= 3000

A/hr=60
B/hr= 50
C/hr=40

(A+B+C)/hr= 150

Total t hrs

A= 10*60= 600
B= (T-15)*50= 50T-750
C=T*40= 40T

600+50T-750+40T=3000
90T= 3000+150
T= 35 hrs

Ans C

	Time taken for completion alone	Work done in 1 hr
A	50	1/50
B	60	1/60
C	75	1/75

First 10 hours all machine worked.
work completed = work completed by A + B + C
= 10(1/50 + 1/60 + 1/75)
=2(1/10 + 1/12 +1/15)
=2 (18+15+12)/180
= 0.5 = 50%
Last 15 hours, only machine C worked
work completed in last 15 hours = work done by C
= 15 (1/75) = 0.2 = 20 %
Remaining work = 30%, completed by B & C together
Work done in an hour by B & C together = 1/60 + 1/75 = 3/100 = 3 %
Time taken to complete 30% of work = 30/3% = 10 hours

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

Initial rates (per hour):
A: 1/50
B: 1/60
C: 1/75

Phase 1 (0-10h): All machines
Work done = (1/50 + 1/60 + 1/75) × 10

Phase 2: B and C only
Until B stops, rate = 1/60 + 1/75

Phase 3: C only
Last 15h at rate 1/75

Calculate total work = 1
10(1/50 + 1/60 + 1/75) + t(1/60 + 1/75) + 15(1/75) = 1
where t = time with B and C

t = 10 hours
Total time = 10 + 10 + 15 = 35 hours

Answer: C. 35

Total units of work = LCM (50, 60 , 75 ) = 300

A can complete 300/50 = 6 units of work in one hour
B can complete 300/60 = 5 units of work in one hour
C can complete 300/75 = 4 units of work in our hour

In 10 hours, total work completed = 60 + 50 + 40 = 150

Work remaining = 150

Let's assume it took t more hours to complete the work after machine A stopped working

5(t-15) + 4t = 150

5t - 75 + 4t = 150

t = 25

Total Hours Needed = 25 + 10 = 35

Option C

Let the entire work be 300 units. (LCM of all 3)

So A can do 6 u, B can do 5 u, & C can do 4 u.

For first 10 hrs work completed = 15u * 10 = 150u.
Last 15 hrs only C worked = 4u*15 = 60u.

So for t hrs in b/w B & C worked. = 9u*t.

So we know total work is 300u=150u+60u+9tu

=> t=10 hrs.

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

IMO C

A's speed = 1/50;
B's speed = 1/60;
C's speed = 1/75;

Assume, t is the total time required to complete the task, working together.

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

(60 + 5t - 75 + 4t) / 300 = 1

9t = 315; t = 35;

Let X be the total no of hours. From the given info

A- worked for 10 hours
B - worked for X-15 hours
C- Worked for X hours

We know the rates of work of each machine as 1/50, 1/60 and 1/75 for A,B and C respectively.

Set up the equation as

10(1/50)+X(1/75)+((X-15)(1/60))=1
---> 60+4X+5X-75=300
--->9X=315---> X=35 hours.

1. First, let's calculate the rates of each machine:

Machine A: 1/50 task/hour

Machine B: 1/60 task/hour

Machine C: 1/75 task/hour

2. Let's say the total time to complete the task is x hours. We know that:

Machine A works for 10 hours

Machine B works for (x- 15) hours

Machine C works for x hours

3. Now we can set up an equation. The total work done by all machines must equal 1 task:

(1/50 × 10) + (1/60 × (x - 15)) + (175 × x) = 1

4. Let's solve this equation:

1/5 + (x - 15/60 + x/75 = 1

15/75 + (75x - 1125/4500 + 60x/4500 = 1

675 + 75x - 1125 + 60x= 4500

135x = 4500

135x= 4950

x=4950/135 = 36.67

5. The closest answer to 36.67 is 35 hours.

Therefore, the correct answer is C. 35 hours.

Step 1: Define the rates of work for each machine
Machine A's rate of work: 1/50 tasks per hour.
Machine B's rate of work: 1/60 tasks per hour.
Machine C's rate of work: 1/75 tasks per hour.

Step 2: Work done by Machine A
Machine A works for only the first 10 hours. The work done by Machine A is:
Work by A= 10×1/50 = 10/50 = 1/5

Step 3: Represent the total work
The total task requires 1 complete unit of work.

Remaining work after Machine A stops:
Remaining work=1−1/5 = 4/5​

Step 4: Represent Machine B's working time
Machine B stops working 15 hours before the task is completed.
Let T be the total time required to finish the task.
Then Machine B works for: T−15hours.

Step 5: Express total work done by all machines
The total work is completed by:
Machine A in 10 hours (1/5)
Machine B in T−15 hours.
Machine C in T hours.
The total work equation becomes:
1/5 + (1/60)(T−15)+ (1/75)T = 1

Step 6: Solve for T
Simplify each term:
1/60 (T−15)= T/60 − 15/60 = T/60 − 1/4
​
1/75 * T = T/75
Substitute these into the equation:
1/5 + (T/60) − 1/4 + (T/75) =1

Thus, −(1/20) + (T/60) + (T/75) = 1

Combine terms involving
T using the least common denominator of 60 and 75 (which is 300):
T/60 + T/75 = 5T/300 + 4T/300 = 9T/300 = 3T/100

The equation becomes:
−1/20 + 3T/100 = 1

Solve for T:
T= 105/3 = 35

Step 7: Verify the answer
The total time required to complete the task is: 35hours
​

Let the Total Work (W) be 300
Time taken by A = 50 hrs, A's rate of work = 6
Time taken by B = 60 hrs, A's rate of work = 5
Time taken by C = 75 hrs, A's rate of work = 4

Rate of work of A + B + C = 15

A stopped after 10 hours, work completed in 10 hrs = 15*10 = 150

B stopped 15 hours before the task was completed, therefore C worked alone for 15 hours, work completed by C alone = 4*15 = 60

Total work completed uptil is 150 + 60 = 210, it means B and C together completed 90 units of work. Rate of work of B and C = 9, time taken by them to complete 90 units = 10 hours.

Total time taken = Time taken by A, B and C + Time taken by B, and C + Time taken by C alone
Total time taken = 10 + 10 + 15 = 35

Answer C.