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GMAT Club Olympics 2025 (Day 2): 20 machines, each working at

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01 Jul 2025, 18:08

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Bunuel

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Answer: B
Since there are 20 machines and each workday requires a unique pair, we can have a total of 190 unique pairs: 19+18+17+16...+1 (1st machine can work with all remaining 19 machines; next machine can work with only 18 remaining machines (because it already paired with the 1st machine); and so on until sencond-to-last machine only has one machine to work with. As such, this part of the work lasted for 190 days

From the 190 days of unique pairings, each machine only worked for 19 days, as every one of the 20 machines had 19 unique pairings. If all 20 machines had worked from the beginning, the job would have been completed in 19 days + the 6 additional days

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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?

20 unique pairs = 20!/2!18! = 190 days of work = 1 day for each unique pair of 2 machines.

Every pair has to work 1 day => 190 days (1 day per pair machines) + 6days (work of all 20 machines) = 196 days total

IF all 20 machines did work from the beginning = (190/10) + 6 days (work of all 20 machines) = 25 days

Answer B

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The possible arranges are:
r1r2
r1r3 r2r3
r1r4 r2r4 r3r4
. r3r5
.
.
r1r20 r2r20 r3r20....r19r20
The sum of all this possibles arranges has an outcome of work equal to = 19(r1+r2+r3+...+r20)
Addition of 6 days working all together = 6(r1+r2+r3+...+r20)
Work = 25(r1+r2+r3+...+r20)
So, the days neede if all machines worked together is 25 days.

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Number of arrangement for 2 machines out of 20 machines: 20x19=380 (arrangements)

Number of unique pairs of machines: 380/2=190 (pairs)

Exactly 2 machines/day work, no pair works more than once

=> Total days that all unique pairs have worked once are the total number of unique pairs: 190 pairs = 190 days

Let rate of work done by 20 machines/day = R

=> R = Rate/day of Machine 1 + Rate/day of Machine 2 +... + Rate/day of Machine 20

During 190 days, each machines has to work in pair exactly 19 days (not 20 because they cannot work with themself)

=> During 190 days, Rate/day of each has been multiplied by 19
=> Total work of 20 machines during 190 days = 19 x (Rate/day of Machine 1 + Rate/day of Machine 2 +... + Rate/day of Machine 20) = 19R

In the next 6 days, all 20 machines work together = 6R

Total work = 19R + 6R = 25R

=> Total number of days if all 20 machines working together from the beginning: 25R/R = 25 (days)

Answer: B. 25

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Bunuel

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The machines have worked in pairs, and no pair works more than once. Hence total pairs are
20C2 = 190

Let the rates of work of each machine be denoted by r(n), where n refers to the nth machine

Let's write the work done by the machines
Day 1 = r(1) + r(2)
Day 2 = r(3) + r(2)
Day 3 = r(3) + r(4)
.......
.......
Day 190 = r(19) + r(20)

Now Adding up the work would give us
Initial work = r(1) + r(2) + r(3) + r(2) + .............................. + r(19) + r(20)
=> Initial work = 19 *( r(1) + r(2) + r(3) + ................ + r(20)) --> As each machine would partner with the other 19 machines once

Post this work, all the machine together worked for 6 days
Additional work = 6 *( r(1) + r(2) + r(3) + ................ + r(20))

Total work = Initial Work + Additional work
=> Total work = 19 *( r(1) + r(2) + r(3) + ................ + r(20)) + 6 *( r(1) + r(2) + r(3) + ................ + r(20))
=> Total work = 25 *( r(1) + r(2) + r(3) + ................ + r(20))

If all machines work together to complete the task of "Total Work" then we have

Rates of All machines working together $X$ number of days = Total work

=> r(1) + r(2) + r(3) + ................ + r(20) $X$ number of days = 25 *( r(1) + r(2) + r(3) + ................ + r(20))

=> number of days = $\frac{25 *( r(1) + r(2) + r(3) + ................ + r(20))}{r(1) + r(2) + r(3) + ................ + r(20)}$

=> number of days = 25

IMHO Option B

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Bunuel

So we know that there is a total of 20 machines.
Now, since they worked in pairs for the first part, each machine will be used 19 times (you can check it mentally; each machine gets paired with 19 other machines, so total times would be 19)
It is also said that over this paired working, all machines were used together for 6 additional days; therefore, if we add both together, a machine will be used 25 times in the total duration to complete work.
Therefore, if all the machines were working together from day 1, the work would be completed in 25 days.

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20 machines can form 20×19÷2=190 unique pairs → 190 days.

In those 190 days every machine shows up 19 times (once with each of the other 19 machines).

So the pair phase does the same amount of work as 19 days with all 20 machines running together.

After that, all 20 machines really do work together for 6 more days.

Total work for the whole job = Equivalent to 19+6=25 “full-team” days

They’d need exactly those 25 full-team days.

Answer is B

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01 Jul 2025, 19:29

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20c2 combinations exist and these are 190 days . 6/196 work is done in 6 days , then total work is done in 196 days

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Answer is B. 25

It is given that each of the machines have a different constant rate.
Let us suppose $w_i$ is the work done by machine $i$ in a day, where $i$ ranges from 1 to 20

If any 2 machines can work on a single day and the pairs cannot have repeated values, each machine works with the other 19 machines exactly once. so there are 19 combinations for 1 single machine. Each machine will work on 19 days.

So the amount of total work done by all machines = $w_1 * 19 + w_2*19 + w_3*19 + .... + w_{20}* 19 $
$ = 19 (w_1 + w_2 + w_3 + .....+ w_{20} ) -> I$

Then for 6 days, all machines work together, so here each machine is working for 6 days.
Total work done by all machines in 6 days = $ w_1 * 6 + w_2*6 + w_3*6 + .... + w_{20} * 6$
$= 6 (w_1 + w_2 + w_3 + .....+ w_{20} ) -> II$

From I and II, total work done in all days
$= I + II$
$= 25 (w_1 + w_2 + w_3 + .....+ w_{20} )$

If all the machines were working together since day 1,
>>each day they would finish $(w_1 + w_2 + w_3 + .....+ w_{20} )$ of the work

So we can see that to achieve the total work, they would have to work for 25 days.

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Total machines = 20
Each machine individually works for 19 days(one day with each of the other machine in the group). This can be imagined as all machines working together for 19 days.
So total work done till now = 19x(all machines)

Also it is given that 6 additional days all machines worked together to complete the whole job.
So remaining work completed in 6 days = 6x(all machines)

Combining both the above, Total Work Done = (19+6)x(all machines) = 25x(all machines)
So answer is 25 days for all machines working together to finish the whole job.

Answer B

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The number of ways to choose 2 machines out of 20 is 20C2 = 190.
So, there are 190 unique pairs, meaning the first phase lasts 190 days.

Each day, a pair (i,j) works, contributing wi+wj to the job.
Total work in the first phace = Sum over all pairs (i,j) of (wi+wj).

Because we use each machine 19 times:
sum (all pairs wi+wj) = sum (19 * wi) = 19 * sum(wi)

The second phase is 6 * sum(wi)

First pahse + second phase = 19 * sum(wi) + 6 * sum(wi)

Total job is done in 19+6=25 days using all the machines.

Answer B

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Gieven,
There are 20 machines, each with a different constant work rate.
Each day, exactly 2 machines work together.
No pair works together more than once.
After all unique pairs have worked once, all 20 machines work together for 6 more days to finish the job.

Now,
Number of unique pairs of machine working together = 20C2 = 20!/18!*2! = 20*19/2 = 190 unique pairs i.e, 190 days of pair work

We know that, on a given day 2 machines work = wi + wj
Each machine works with other 19 pairs
=> Total pair work = ∑ (wi+wj) = 19 * (w1 + w2+ ... + w20)

Total work during the 6 days machine work together = 6*(w1 + w2+ ... + w20)
Therefore, Total work during the entire process = 19 * (w1 + w2+ ... + w20) + 6*(w1 + w2+ ... + w20) = 25 *(w1 + w2+ ... + w20) = 25W ----------------------------[lets consider (w1 + w2+ ... + w20) = W]

Now, if all the 20 machines worked together from the start,
Total work done in a day = w1 + w2+ ... + w20 = W
Total number of days taken to complete entire work = Total work / Total work done in a day
= 25W / W
= 25

B. 25 days

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Each will have to work 19 days for all combinations to be form (can form 19 pairs), so 19(work of all machines) + 6(work of all machines) = 25 day if all worked simultaneously

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When the pairs were working, each machine's work rate is part of 19 different pairs (it pairs with each of the other 19 machines once). So, over 190 days (20C2), each machine's total contribution is its work rate multiplied by 19 days.

In the next stage, all machines work together for 6 days, so each machine contributes its work rate for 6 more days. Therefore, each machine's total contribution is (19 + 6) = 25 days' worth of its work rate. The total work is the sum of all 20 machines' work rates, each multiplied by 25.

Correct answer is B

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M=20, 20C2=190
let total work=1, Total rate of all machines included=R
Work done in pairing days =(n−1)R=19R
Work done in last 6 days with all machines =6*R
Then:
19R+6R=1
=>25R=1
R=1/25

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First, every possible pair of machines works together for one day (that's 190 days total, since there are 190 unique pairs possible from 20 machines).
Then, all 20 machines work together for 6 more days to finish the job.

In the pairing phase, each machine works with each other machine exactly once. So each machine works for 19 days.
Then in the final 6 days, all machines work together, so each machine works 6 more days.
Total work per machine: 19 + 6 = 25 days of work.

If we had all 20 machines working together from the start, their combined power would complete the same total work (25 days from each machine) in just 25 days, because they're all working simultaneously.

The right answer is B

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There are 20 machines, and each day exactly 2 different machines work. No pair works more than once. So first, we count how many unique pairs of 2 machines can be formed from 20 machines.

Number of unique pairs = 20 choose 2 = 20 × 19 / 2 = 190 pairs
So for 190 days, 2 machines work each day, and each unique pair works once.

After that, all 20 machines work together for 6 more days.

Let the total work required be W.
Let the work done by machine i per day be Ri.
Let the total rate of all machines together be R = R1 + R2 + ... + R20

During the 190 days of unique pairs, each pair of machines i and j works once, so work done = Ri + Rj
So total work during 190 days = sum over all unique pairs of Ri + Rj

This total sum is equal to (20 - 1) times the total of all rates.
Because each machine appears in 19 different pairs, its rate is added 19 times.
So total work during pair days = 19 × (R1 + R2 + ... + R20) = 19R

Then, during the 6 days when all 20 machines work together, they work at rate R each day
So total work in those 6 days = 6R

So total work done = 19R + 6R = 25R

If all 20 machines had worked together from the beginning, they would do R work per day
So total work W = 25R
Total time = W / R = 25R / R = 25 days

Answer is B

I personally think this should be higher rated in terms of difficulty.

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What we want to find out is the total number of days these machines worked. Its given that the machines work in unique pairs, after all such unique pairs finish working, then all 20 machines work for 6 more days. So we need to find
= the number of days the unique pairs will work + 6
= the number of ways to arrange machines in unique pairs of two + 6
= $20_C_2 + 6$
= $(20 * 19)/2 + 6$
= $190 + 6$
= $196$
option E

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We can find all possible combination by the PnC formula which will get us to 20*19/(2*1)=190 Now if every machine has to work 6 additional days it would get us to 19+6(Total power of all machines)=25*Total power
So if all machines are working from the start we would get job done in 25 days ie Option B

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Quote:

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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