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GMAT Club Olympics 2025 (Day 2): 20 machines, each working at 01 Jul 2025, 22:27
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Option B should be the answer.
There are 20 machines.
Each machine will pair with the other 19 machines .
So if , pairs are formed , total 19 days' work is done.
So if all 20 machines work together for 6 additional days to complete the job,
total (19+6)= 25 days will be taken to complete the job.
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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Let A, B, C,D ..... be the working rates of each machine.

From the question stem, we can deduce each item A,B...would have worked with every other working rate 19 times.

It follows that

19 (A+B+C+......) + 6(A+B+C+....) is the total work

Combining the above equation, we can deduce that all machines working together can finish the total work in 19+6 = 25 days.

Therefore, Option B
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Quote:
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196
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Total unique combinations = 20C2 = 190, meaning 190 days of work done.

We have to find t = total work/rateAll

For the first 190 days, each machine worked with each other one, making 19 combinations in total.

Work done in first 190 days = (r1+r2+...r20)*19 = Let this be x1

Work done in last 6 days = (r1+r2+....r20)*6 = Let this be x2

Total work done = x1 + x2 => (r1+r2+....r20)*25

Now, to find t = total work/rateAll = (r1+r2+....r20)*25/(r1+r2+....r20) = 25

Therefore, (B) is the answer
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ANSWER IS B

Let the Rate of the machines be R
Now According to Question
20 Machines work in Pairs
Which means 19 machines work in days= 19(R1,R2,.......R20.............X
Additional Days for all 20 machines=6(R1,R2........R20)........... Y
Total work done in Days=A+B=25(R1,R2.....R20)

Time =[Work][/Rate]
Time taken by 20 Machines =25[R1,R2.....R20]/[R1,R2.....R20]
Time taken by 20 machines together is 25.
HENCE ANSWER IS B
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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Using the slot method, the unique pairs of 20 machines will be \(\frac{(20*19)}{(2)}\), accounting for the pairs that only work together once. (for example: pair AB is the same as pair BA, which is why we divide by 2).

Let's assume the total units of work that needs to be completed is 250, and assume that a pair of any two machines with different constant rates adds up to completing 1 unit a day.
so, 190 pairs complete 190 days * 1 unit = 190 units of work.

20 machines now work together, and as 2 machines add up to 1 unit of work, 20 machines will add up to 10 units a day.
so 20 machines complete 6 days * 10 units per day = 60 units, completing 250 units of work in total.

Now what if 20 machines were working together from day 1, to complete 250 units of work?

days * 10 = 250, thus days = 25. Option B.
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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While working in pairs, each machine works for 19 days (as one particular machine can form 19 unique pairs) and 6 more days, along with the other 20 machines.
So the total number of days each of the 20 machines works is 25 days to complete 1 full job.
sum of their rates * total days worked = 1 full job
Sum of their rates = 1/25

Now, 20 machines took 6 days to complete 6/25 th of the work(work = 1/25 * 6). To complete the whole work, they will take 25 days (1 = 1/25 * time).
Option B.
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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B. You have 20 machines, 2 exactly work on each day, so total pairs which can be made is 20c2= 190, therefore 190 days, + 6 days where all work together, here btw options D and E get eliminated as if 20 machines work altogether, naturally days will have to be less than that. But for finding the exact value , we need to solve. Moving on, if you think logically every machine will work with every other machine once, so each work with 19 others, let the total rate here be R,( combined rate of all the machines)=19R and then 6 days all work together=6R. adding=25R. The question asks if all worked together, 25 R is the work, if all work togther each day R will be done, so 25 Days.
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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Total days for all such unique pairs of machines to have worked once = 20C2 = (20 * 19)/2 = 190 days

Work done by 2 machines in a total of 190 days = 380 MW (MW = Machine Work as work is constant)

Work done by 20 machines to complete rest of work in 6 days = 20 * 6 = 120 MW

Total work = 120 + 380 = 500 MW

Now considering the question, if all 20 machines worked together from the beginning, they would have to complete 500 MW; That is the total days it would take = 500/20 = 25 days

Correct option is option B
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Rate of all 20 machines = R
Rate of individual machine = ri where I <=20

Days two machines work together = 20c2=190
In this we have each rate 19 times
Thus 19(r1+r2+....+r20)=19R
Thus total work= 19R+6R=25R
Number of days to complete work if rate had been R from beginning 25R/R=25

Ans B
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Since there are 20 machines that initially work in distinct pairs for 1 day each, we need to find the number of distinct pairs and, correspondingly, the number of days until each pair has worked once, since it is given no one pair works together more than once.

The total number of such pairs that can be formed from 20 machines is given by the combination \(20C_2\), which gives us 190 distinct pairs.

Since each pair works for 1 day, the total work done by all pairs over 190 days is calculated as the sum of each pair's work, which is given by:
=> \(1\times(R_1 + R_2) + 1\times(R_1 + R_3) +\dots+ 1\times(R_{20} + R_{19})\) ; where \(R_1, R_2, ..., R_{20}\) represent the constant work rates of each of the 20 machines.

From the above, we observe that each machine’s work rate appears 19 times, so we can factor out 19 from sum of all the machines' work rates, giving us:
=> \(19\times(R_1 + R_2 +\dots+ R_{20})\) - (1)

Next, the total work done by all 20 machines working together over an additional 6 days is:
=> \(6\times(R_1 + R_2 +\dots+ R_{20})\) - (2)

Adding (1) and (2) together and equating to the total work, we get:
=> \(19\times(R_1 + R_2 +\dots+ R_{20}) + 6\times(R_1 + R_2 +\dots+ R_{20}) = 1\)
=> \(25\times(R_1 + R_2 +\dots+ R_{20}) = 1\)
=> \((R_1 + R_2 +\dots+ R_{20}) = \frac{1}{25}\) units of work/day

Thus, the combined work rate of all 20 machines working together is \(1/25\) units of work per day, and so if 20 machines together complete \(1/25\) of the job each day, it will take them 25 days to complete the entire job, had they worked together from the beginning
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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There are (20,2)=190 unique pairs, so 190 days pass with 2 machines working per day.

The total work done by all pairs equals 19R, since each machine appears in 19 pairs and the total machine rate is
R
R. With 6 more days of all 20 machines working together at rate
R
R, total work is 19R+6R=25R, so with all 20 from the start, the job takes 25 days.

Option B.
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Very interesting Q Ans is 26.
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Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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Lets assume each machine works at rate x1, x2, x3......x20 per day.
No. of differrent unique pair possible from 20 Machines = (20*19)/2 = 190
Work done by 190 pairs + 6 * total work by all 20 machines = Total Work (W)

(xi+xj) for all pairs + 6 (x1+x2+x3.....x20) =W { each machine appear 19 times }
19 (x1+x2+x3...x20) +6 (x1+x2+x3.....x20) =W
W= 25(x1+x2+x3.....x20)

Days = Total Work / rate of all machines working together
= 25 (x1+x2+x3.....x20)/( x1+x2+x3.....x20)=25 (B)
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Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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First, we need to find how many unique pairs of 2 machines that we can have out of a total of 20 machines.
We can solve it using combinations, 20C2 = \(\frac{20!}{18!*2}\) = 190 pairs.
This means that they've worked for 190 days so far with 2 machines each day.
If x is the average working speed of a single machine, \(\frac{1}{2x}=190\)
For all 20 machines working together, this amount of work will take \(\frac{1}{20x}=\frac{190}{10}=19\) days.
Adding the extra 6 days where the 20 machines worked together, we get 19 + 6 = 25 days of work if they all worked together from the start.
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Assume that the machines work rate (work/day) is 1,2,3,....20 (ensuring different rates)

Number of possible pairs = Number of days of work (exc the last 6 days) = 20C2 = 190
Work done in these 190 days = 19 x 1 + 19 x 2 + ...19 x 20 (because each machine has 19 pairs, hence each machine works on 19 days). = 19 (1 + 2 + ....20)
6 days in which all machines work. Work done = 6 x (1+2+...20)

Total work done = (19+6) (1+2+....20) (A)

If all machines worked from the start, work done/day = (1+2+..20) (B)
So, Days to complete the work if all machines had been working = A/B = 25 days

Option (B)

Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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Total unique pairs are 20C2 = 190 and each machine works with the other 19 once since it cannot work with itself
Assume the total combined rate of the machines is R and each machine works for 19 days
That means total work done = 19xR
During the 6 days the total work done was = 6xR
Total work is (19+6)R= 25R
To find number of days for all the machines working together you take the total work done divide by the combined rate so 25R/R = 25
ANS B
Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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During the period in which exactly 2 machines work together, each machine works in pairs with exactly 19 other machines in 19 days. Assume the productivity per day of each machine is Y(i)

Total productivity during the period in which the machines work in pairs: 19 x Y(1) + ... + 19 x Y(20) = 19 x (Daily Productivity of 20 machines).

Combined with the last period of 6 days in which the machines work together: Total Work = 25 x (Daily Productivity of 20 machines)

So if the machines work together, in 25 days they would have completed the work.

Answer: B
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first, how many unique pairs can we make from 20 machines? thats C(20,2) = 190 pairs.

so 190 pairs work for 190 days (one pair per day), then all 20 machines work togeather for 6 more days.
let me think in terms of "machine-days" of work:

pairs phase: 190 pairs × 2 machines per pair = 380 machine-days
all machines phase: 20 machines × 6 days = 120 machine-days
total work = 380 + 120 = 500 machine-days

if all 20 machines worked from the begining:
500 machine-days ÷ 20 machines = 25 days

the answer is B. 25
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20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


first: calculate the number of unique pairs: using combination method of picking 2 from 20: 20!/(2!18!) = 190 pairs
the work rate of one pair, on average, is just 1/10 of the whole 20 machines (rate of 2 machines/ rate of 20 machines on average).
=> So 1 pair (2 machines) working 190 days equal to 20 machines working 19 days

Then we know the 20 machines need to work together for 6 additional days
=> 19 + 6 = 25. Choose B
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We established that there are 190 unique pairs of machines. Each pair works for 1 day. If you look at how many times each individual machine's rate appears in this grand sum, you'll find that each machine's rate is paired with every other of the 19 machines.
Work done by all 20 machines together at the end: After the 190 days of pairs, all 20 machines work together for 6 more days.
The total work to complete the job is the sum of the work from pairs and the work from all 20 machines is 25 days (19 days plus the additional 6 days). Thus, Answer B.

Regards,
Lucas

Bunuel
20 machines, each working at a different constant rate, work to complete a certain job. Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job. If all 20 machines had worked together from the beginning, in how many days would the job have been completed?

A. 19
B. 25
C. 26
D. 190
E. 196


 


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