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

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AviNFC

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01 Jul 2025, 09:03

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Let machines be A,B...T
each machines pairs with 19 others.
so per day work..rA+rB...rB+rT..

Thus total work 19rA+19rB+19rC+....19rT+6(rA+rB...+rT)
or total work =25*(rA+rB+...rT) = time*(rA+rB+...rT)
time=25

Ans B

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B. 25

We know that the total work(w) done is = work done by unique pairs + work done by all 20 machines together
Let their individual rates be x1, x2......x20
w = work done when they are paired together + (x1+x2.......+x20) * 6 days

since each machine will be paired exactly 19 times
work done when they are paired together = 19*(x1+x2.......+x20)

so,
w = (x1+x2.......+x20) * 19 + (x1+x2.......+x20) * 6
w = 25 (x1+x2.......+x20) --- (1)

If they worked together from the beginning then days taken = w/collective rates
Hence, from (1) days = [25 (x1+x2.......x20)] / (x1+x2.......x20)
final answer = 25

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01 Jul 2025, 09:08

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Total unique pairs = 20C2 = 190

M1 = x1 work/day
M2 = x2 work/day
.
.
M20 = x20 work /day

Adding work of 190 days = 19*(x1+x2+....+x20) work done total in 190 days

M1 to M20 work together = (x1+x2+...+x20) amount of work done in one day

x1+x2+...+x20 = X work/day

19X + 6X = 25X = total work

When 20 machines work together they complete X amount of work per day, therefore they will take 25 days to complete 25X of work when working together from the beginning

Bunuel

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Total number of pairs = 20C2=20*19/2=190
If we assume the number of days it takes one machine to do the said work W to be D days we can formulate the equation as -

(W/D)*2*190(Since 2 machines work each day as a pair) + 20*(W/D)*6 (all machines work for 20 days) = W

380W/D+120W/D=W
500W/D=W
D=500

So one machine on its own takes 500 days to complete the work. If all 20 machines work together every single day & if we assume D to be total days taken,

(W/500)*20*D=W
(W/25)*D=W
D=25

Therefore the answer to this question is option (B) 25 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?

Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once. This process takes 20C2 days or 190 days.
Additionally, 20 machines work together for 6 more days to complete the job.
If all 20 machines worked together on other 190 days instead of 2 then the 190 day job would have taken 19 days.
Total days to complete the job if all machines worked together from the beginning= 6+19=25 days

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We are given 20 machines, each with a different constant rate.
- In the first phase, each possible pair of machines works together for exactly one day. The number of unique pairs is C(20,2) = 190. So, this phase lasts 190 days.
- Then, in the second phase, all 20 machines work together for 6 days to complete the job.
Let the sum of the rates of all 20 machines be R (jobs per day).
In the first phase, each pair (i,j) works for one day. The work done by a pair is (r_i + r_j).
The total work done in the first phase is the sum over all pairs of (r_i + r_j).
Note that each machine is paired with every other machine exactly once. Since there are 19 other machines, each machine's rate appears in 19 pairs. Therefore, the total work in the first phase is:
Total work Phase 1 = 19 * (r_1 + r_2 + ... + r_20) = 19R.
In the second phase, all 20 machines work together for 6 days, so the work done is 6 * R.
Therefore, the total work for the job is:
W = 19R + 6R = 25R.
If all 20 machines worked together from the beginning, their combined rate is R. The time T required to complete the job is:
T = W / R = 25R / R = 25 days.
Thus, the job would have been completed in 25 days.

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01 Jul 2025, 09:18

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The number of ways to choose 2 machines out of 20 is:
20C2 = 20*19/2 = 190

Total days = 190 + 6 = 196 days

Days worked by each machine = 6 (wherein each machine worked) + 19 (wherein it paired with other machines) = 25

Since each machine worked for 25 days effectively

Answer B

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Bunuel

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Rate of each machine (units / day)= r1, r2 ... r20

Imagine that there are just 4 machines

A, B , C, D

Number of pairs

AB
AC
AD

BC
BD

CD

Hence, each machine works for 3 times.

For 20 machines, each machine will work for 19 times.

19 (r1 + r2 + r3 ..+ r20)

The additional total work = 6 * (r1 + r2 + .. r20)

Total work = 19 + 5 (r1 + r2 + .. +r20)

= 25 (r1+ r2+ ... +r20)

If the machines work simultaneously, each day the machine's output will be (r1+r2+..r20)

Number of days = 25 (r1+ r2+ ... +r20) / (r1+r2+..r20)

= 25

Option B

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Total =20 machines. 2 machines each day. No.of days for 20 machines is 20/2=10 days. Additional 6 days leads to 16 days (10+6) . 20 machines took 16 days to complete the job. Therefore 1 machine took 20/16=1.25 days to complete the job. For all 20 machines to work together, it will take 25 days (20*1.25) to complete the job.

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Number of unique 2 machine pairs = 20C2 = (20 x 19)/2 = 190.
So we have 190 days of 2-machine work.

Let each machine's rates be r1, r2, r3, ..., r20.
Combined rate of all machines, R = r1 + r2 + ... + r20

Now, each machine appears in 19C1 = 19 pairs, so over the 190 days, each machine contributes its rate 19 times.
So, total work done = 19R

All 20 machines work for 6 more days
Work done = 6R

Total work = 19R+6R=25R

If all 20 machines had worked together from the start,

Time = 25R/R = 25 days.

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First off, let’s label 20 machines as a, b, c, d, ..., t, and their total work is X.
We know there are C(20,2) ways to pick 2 machines as a pair, which is (20*19)/2=190.
1901 + 6X = total work, let’s call it T.
190*1 actually means 380 letters of work.
380/20=19 pairs.
So, that’s 19X.
19X + 6X = T.
X = T/25.

T/(X) = T/(T/25) = 25.
Answer is B.

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01 Jul 2025, 09:31

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total machines 20
2 machines each day work will take 20c2 = 190 days to complete work

After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job.
1 particular machine can have total of 19 pairs , i.e 19 days
and also the case that 6 additional days each machine work together ; 6 days
total days will be 25
OPTION B , 25

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We’re told:

There are 20 machines, each working at a different constant rate.

Every day, exactly 2 machines work.

No pair works together more than once.

After all unique pairs of machines have worked once, then all 20 machines work together for 6 days to finish the job.

The question: If all 20 machines had worked together from the beginning, how many total days would it have taken to finish the same job?

Step 1: Count the number of unique pairs of 20 machines.
This is a standard combination:
(20,2) = (20×19)/2 =190
So for 190 days, 2 machines worked each day, each pair only once.

Step 2: Let’s define the total amount of work done as 1 full job.
Let each machine Mi have a work rate ri (e.g., jobs per day).

The total job is of size 1.

So:During the first 190 days, each day 2 machines worked. The total work done is the sum of the rates of those 2 machines, for 1 day.

So for all 190 days, the total work done =
∑ (ri + rj) {i.e for all unique pairs (i,j)} = ∑ (ri + rj ) (i.e for each pair)

Let’s compute that more carefully.

Step 3: Sum over all pairs: total work in first 190 days
We observe:

Each machine is in exactly (20 – 1) = 19 unique pairs (because there are 19 other machines).

So each machine works for 19 days, each time paired with a different machine.

Therefore, each machine contributes its rate ri for 19 days in the first phase.

So total work done in first 190 days:

∑i= 19*ri = 19*(r1 +r2 +⋯+r20 )=19R

Let’s define:
R= (r1 +r2 +⋯+r20 ) (i.e total rate of all 20 machines)

Step 4: Work done in last 6 days
After those 190 days, all 20 machines work together for 6 days.

So the total rate is R, and for 6 days:
Work done=6R

Step 5: Total work = Work done in both phases = 1 job
Total work=19R+6R=25R=1
-> So total rate of all 20 machines: R= 1/25

Step 6: If all 20 machines had worked together from the beginning...
Their total rate is R = 1/25 jobs/day
So to do 1 full job, time required = 1/R = 1/ (1/25) = 25 days

Final Answer: B

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01 Jul 2025, 09:42

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Suppose 20 machines work in 1 day be a1, a2, a3, .....a20 for 20 different machines

Now 2 machines are used daily and that 2 machines are not used further

So 2 machines pair can be selected from 20 pairs in 20C2 ways = 20*19/2= 190

So 190 pairs work will be like

a1+a2, a3+a4, a5+a6.....a19+a20 and so on

a1+a3, a1+a4, a1+a5,a1+a6, a1+a7, a1+a8, a1+a9....a1+a19, a1+a20

Thus, a1 will come 19 times in similar fashion others will come 19 times

so we can write 19*(a1+a2+a3+a4......a20) as total work for 190 days

and for 6 days all 20 machines will work simultaneously

So Total work will be = 19*(a1+....a20)+6*(a1+....a20)
= 25*(a1+a2+....a20)

Thus if all machines work from start they will complete work in 25 days

Answer is B

so if we add

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Bunuel

This question was provided by GMAT Club
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Total number of pairs = 20C2 = 190

Hence 2 pair machines work for 190 days after that all 20 machines start.

Let r be the rate of work for each machine, then the equationn is as follows:
(2/r)*190 + (20/r)*6 = 1 from which r is equal to => r = 1/500

Hence the number of days it will take 20 machines = 500/20 = 25 days. Hence option B

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No of unique pairs = 20C2 = 190 days
Let: $r_{1}, r_{2}, r_{3}, ..., r_{20} $ be individual machine rates
Total work done in first 190 days: each machine will work with 19 others each once, so each machine operated in 19 days = $19*(r_{1}+ r_{2}+ r_{3} +...+ r_{20})=19*R_{(all)}$
Total work done in last 6 days: $6*R_{(all)}$
Total work: $19*R_{(all)}+ 6*R_{(all)} = 25*R_{(all)}$

If all 20 machines had worked together from the beginning:
Time = $\frac{25*R_{(all)}}{ R_{(all)}}$=25 days

Answer: B

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01 Jul 2025, 10:06

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Bunuel

This question was provided by GMAT Club
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20 machines working at a different constant rates. Let the rates be r1, r2, r3, r4...........r18,r19,r20.

Each day exactly two people work = 20 C 2 = (20*19)/(1*2) = 190 cases.

Each rate repeats for 19 times. If Machine 1 is chosen, then machine 1 can pair with exactly 19 other machines.

So the summation of rates is 19*(r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20) = 19* A

Where, r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20 = Rate = A

Given, 20 machines work for another 6 days.

6* (r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20) = 6*A

Total work = 19*A + 6*A = 25*A

The number of days when all 20 people work

= (25*A ) / (Total rates )

= (25*A) / (r1+ r2+ r3+ r4+ .........+ r17+ r18+ r19+ r20)

= (25*A)/ (A)

= 25

Option B

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There are 20 machines , we can make n(n-1)/2 unique pairs out of 20

That is 20(20-1)/2 = 190 pairs.

Lets assume there is 190 unit of work has to be done, and 1 pair (2 machines) making 1 unit daily.

So 20 machines will make 10 units daily. Making 10 units daily it will take 19 days to make 190 units.

Hence option A is correct.

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Bunuel

This question was provided by GMAT Club
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There is a simple trick to this question.

We know that for 20 machines each of them worked with every other machine, So let us take the first machine as an example:

Machine 1 (M1) would work with M2, M3, ..., M20 ==> 19 pairs
So M1 worked for 19 days

Similarly, the same occurs for all 20 machines.
So, each of the 20 machines work for 19 days in total.

We know that 20 machines worked together for 6 days after the unique pair of machine so we will add that too

-> 19 + 6
= 25

Answer:
(B)

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Possible pairs among 20 machines = 19+18+...+1 = 20*9+10 = 190 (b/c, Machine 1 pairs with 2-20 (19 pairs), Machine 2 pairs with 1, 3-20 but the pair (2,1) should be excluded to prevent double conting (18 pairs)...)

If work done by all machines 1-20 together = t = a1+a2+...+a20,
Work done by pairs = 190 * t
For the last six days = 6 * t

Total work = 190t + 6t = 196t, so will take 196 days to complete if all machines work together

Answer: E