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

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01 Jul 2025, 08:25

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lets say total work = 1 unit
and as there are diff rates lets say R = R1+R2...+R20
Now every machine will get paired with every other machine this is like handshake so every machine will have to match up 19 times right 19R1+19R2+19R3....+19R20
so total = 19R
after that 6 days work so now Total = 19R+6R = 25R
so 25R=1 => R= 1/25
Days = 1/(1/25) => 25
Hence Ans : B

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Information given:
- 20 machines, each working at a different constant rate
- Each day, exactly 2 machines work, no pair repeats
- After all unique pairs have worked once, the 20 machines then work together for 6 more days to finish the job

Question:
- If all 20 machines had worked together from the beginning, how many days total would the job have taken?

Solution:
- First, calculate how many unique pairs you can make with 20 machines
- Number of pairs: (20 * 19) / 2 = 380 / 2 = 190
- Therefore, the 'pairs phase' lasts 190 days (190 pairs, 1 pair/day)

- Suppose each machine's rate = 1 unit
- Any pair working = 2 units/day
- All 20 machines together = 20 units/day

- During pairs phase: 190 x 2 = 380 units
- During final phase: 6 x 20 = 120 units
- Total units = 380 + 120 = 500

- If all 20 machines had worked together from the start, total days: 500 / 20 = 25 days

Answer: B, 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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Bunuel

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Assigning 1 pair of machines each day for 20 machines implies 19 days, hence total days to complete job is 19+6=25 days.

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Count how many unique pairs of machines can be formed
We are choosing 2 machines out of 20 without repetition, so:
Number of unique pairs = C(20, 2) = (20 × 19) / 2 = 190 pairs

So, for 190 days, 2 machines worked per day.

Let’s define 1 “machine-day” as the work done by 1 machine in 1 day.

For the first 190 days: 2 machines worked each day → 190 × 2 = 380 machine-days
Then, all 20 machines worked for 6 days → 20 × 6 = 120 machine-days

Total work done = 380 + 120 = 500 machine-days

Let the number of days be D. Then:
20 × D = 500
→ D = 500 / 20 = 25

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Let us consider machine 1 finishes alone in m1 days, machine 2 in m2 days and so on...
efficiency: 1/m1, 1/m2, 1/m3, ..... 1/m20

in simple terms:

job done in day1: (1/m1 +1/m2)
job done in day2: (1/m2+1/m3)
job done in day3: (1/m3+1/m4)
....
....
job done in day190 (why 190? "Unique pairs can work once hence 20C2 = 190): (1/m19+1/m20)

Now, think carefully, each machine will appear 19 times if you combine the total work done since day 1 to 190 apparently the equation formed by this approach will represent the expression of job done in one day if all the machines would have worked together , hence -

19*(1/m1 + 1/m2 + 1/m3 + 1/m4 + 1/m5 ....... + 1/m19 + 1/m20) + 6*(ALL) = 1
=> 19*(ALL) + 6*(ALL) = 1

Hence it would take 19+6= 25 days to finish if all the machines would have worked together.

*Suggestions or any rectification to this approach is highly appreciated

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Bunuel

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Let's call the rates at which each machine work as r1,r2......

If 2 machine groups work... So for 1st machine there will be 19 pairs, similarly for 2nd 19 pairs. ....

Total work = 19(r1+r2+.....r20) + 6(r1+r2+...r20) [ This is for the additional 6 days mentioned].
=> total work =25(r1+r2....r20).

So if from the start, if all 20 work together each day the work completed would be (r1+r2+r3....r20).

Hence to complete the total work we require 25 days.

Hence IMO B

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The number of ways to choose 2 machines from 20 is given by the combination formula:
C(n,k)= n!/ (k!(n−k)!)

Number of unique pairs = C(20,2)= 20!/(2!(20−2)!)=10×19=190 pairs.

now let's consider the contribution of each individual machine to this sum. Each machine will work with 19 other machines.
sum of all unique pairs (R i)=19×(R 1 +R 2 +...+R 20 )

The first phase lasts for 190 days (since each of the 190 unique pairs works for 1 day).
Let R_total =R 1 +R 2 +...+R 20

Work done in the first phase (W 1 ) = 19×R_total ×1 day=19R_total
Work done in the second phase (for 6 days)(W 2 ) = R_total ×6 days=6R_total

The total job W = W 1 +W2

W=19R_total + 6R_total = 25R_total

Now, lets calculate days to complete the job

Let D be the number of days it would take.
W=R_total ×D
Substitute W=25R_total
25R_total =R_total ×D
D=25 days

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Ans: B (25)

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.
So Total number of pairs = 20C2 = 190 ==> 190 days of pair work
In these 190 days, each machine has worked 19 times with another machine, so in total, there were 19 days of work done when all the machines work together.

After all such unique pairs have worked once, all 20 machines work together for 6 additional days to complete the job.
Including this, a total of 25 days (19+6) of work was done when all the machines worked together.

If all 20 machines had worked together from the beginning, in how many days would the job have been completed?
So if all the machines have worked together total of 25 days would have been the work days to complete the job.

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How many unique pairs of 20 machines?
Number of unique pairs of 20 machines = 20C2 =. 20! = 20×19 = 190 (20-2)! 2! 2 So, for 190 days, 2 machines work each day.
Each machine is in 19 different pairs (with each of the other 19 machines).
So, every machine worked for 19 days in those 190 days.
If a machine has a rate of work (like machine A does 1 unit/day), then in 19 days, it does 19 units.
So total work done in first 190 days =Sum of all machines’ rates × 19 = 19 × total rate (R) = 19R
In the last 6 days, all 20 machines work together, so work done = 6 × R
So total work = 19R + 6R = 25R
If all 20 machines worked together from the start, daily rate = R
So number of days = 25R = 25
R

Final Answer: 25 days

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Let's call machines X1 to X20, each having work rates of 1/X1, 1/X2, 1/X2, ...... , 1/X19, 1/X20. Since all 20 Machines work in pairs, we will have 19 unique pairs for X1, 18 for X2, ....... 1 unique pair for X19. Or in other words, each machine would have worked for 19 days, aprt from that all the machines worked for additional 6 days together, that means total work is completable in 19 + 6 days. i.e 25 days.

To visualise total work rate was equal to

(1/X1 + 1/X2) + (1/X1 + 1/X3) + ..... 19 terms ..... + (1/X1 + 1/X20 ) + (1/X2 +1/X3) +.... 18 terms ... + (1/X2 +1/X20) +...similarly.......+ (1/X19 +1/X20) + ( last 6 days all machines worked together =) 6( 1/X1 + 1/X2 + 1/X3 + 1/X4 + ....... 1/X19 + 1/X20) = 1.

If we closely examine the above equation, we can see each term is repeated 19 times in total apart from the last 6 days. So we can rewrite this as,

19 ( 1/X1 + 1/X2 + 1/X3 + ...... + 1/X20) + 6 ( 1/X1 + ....... + 1/X20) = 25 ( 1/X1 + ..... + 1/X20) = 1. This shows if all machines work together from the beginning, the work could have been completed by 25 days.

Bunuel

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In this problem, 20 machines each work at a different constant rate. For the first part of the job, only 2 machines work each day, and every possible unique pair works together exactly once. Since there are C (20, 2) = 190 unique pairs, this part takes 190 days.
After that, all 20 machines work together for 6 more days to finish the job, making the total time taken 196 days.

If all 20 machines had worked together from the beginning, they would have completed more work per day. In the first 190 days, each machine appears in 19 different pairs, so the total work done during that time is equivalent to 19 days of all machines working together.
Adding the 6 final days, the total job equals 25 full days of work by all machines. Therefore, if all 20 machines had worked together from the start, the entire job would have been completed in 25 days.

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We have to choose 2 machines out of 20 machines.
So, 20C2 =190

Let Ri and Rj be the rate of the i-th and j-th machines respectively.
So, the work done by both in one day is (Ri + Rj).

Notice that each machine will appear in 19 pairs
So, the sum can be rewritten as: 19∗(R1+R2+...+R20).
This work is completed over 190 days

Work done by all 20 machines:
Let Rtotal be the sum of the rates of all 20 machines: Rtotal=R1+R2+...+R20.
The work done by all 20 machines working together for 6 days is 6∗Rtotal.

Total job completions:
W=19∗Rtotal+6∗Rtotal
W=25∗Rtotal

If all 20 machines worked together from the beginning, the time taken would be W/Rtotal.
Days = (25∗Rtotal)/Rtotal=25 days.

B) 25

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01 Jul 2025, 08:50

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Step 1: How many unique pairs of machines?
Choose 2 out of 20:
(202)=20×192=190 days(220)=220×19=190 days
So, this phase lasts 190 days, with 2 machines working per day.
[hr]
Step 2: Total work done
Let’s say the total job = 1 unit.
Let the total rate of all 20 machines together be RR.

In the 190-day phase, each machine works with 19 others, so each machine’s rate is counted 19 times.
So total work in 190 days = 19R19R
Then, all 20 machines work together for 6 days → work = 6R6R
So total work = 19R+6R=25R19R+6R=25R

But total job = 1, so:
25R=1⇒R=12525R=1⇒R=251
[hr]
Step 3: If all 20 worked together from the start
They would work at rate R=125R=251, so time to finish:
11/25=25 days1/251=25 days
[hr]
✅ Final Answer: B. 25

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Number of pairs of 20 machines: 20C2 = (20x19)/2 = 190 => Need 190 days for all pairs work. Each day, 2 machines work => In 190 days those pairs work: 190x2=380. Then, 20 machines work together for 6 additional days => In 6 days, those machines work: 20x6=120. If all 20 machines had worked together from the beginning, days would the job have been completed: (390+120)/20 = 500/2 = 25 => B.

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Please correct me if my reasoning is wrong.

In 20 machines
exactly 2 machines are assigned to work, and no pair of machines works together more than once
each machine works 19 days out of all days.
19[m1+m2+m3...+m20] =
later extra 6 days of working together
6[m1+m2...+m20]

total effort= 25[m1+m2+m3...+m20]

So,if they start working together from start, then can finish in 25 days.

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Here they said 2 machines work each day,and there are 20 machines in total.So how many unique pairs are there?
That will give us the number of days those machines worked in the first phase.

20C2 will give us the unique pairs and that is 190.That means the machines worked for 190 days in the first phase.

Then I got stuck.

The question asks us how many days would the job have been completed

rate=work/time

so somehow we need to find rate and work to get the time

in second phase we can calculate work

work=rate*time
let rate be R which adds up all the rates of the machines and time is given
work(phase 2)=R*6=6R

in first phase,we need to find total work

work=rate*time
let rate be R which adds up all the rates of the machines
work(phase 1)=R*days.
We need to find the days

Each machine will work with 19 other machines(not 20 because it cant pair with itself),so that means each machine works for 19 days.
So it will be like 19r1+19r2+19r3........(where r1,r2,r3 are different rates for the different machines)

so work(phase2)=19R

total work(phase1+phase2)=25R

Finally we can get the answer(phew)

rate=work/time

time=work/rate=25R/R=25

The answer is B.

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Let ri be the constant rate of work for the i-th machine, where i=1,2,...,20 and Let the total amount of work to be completed be W

Now since each machine works with one another in day only once so total work days 20-1=19

total work in with pair of two machines = 19×(r1+r2+...+r20)=19×Rtotal.
Total work, W=19×Rtotal+6×Rtotal = 19+6(Rtotal)

If they would have worked together from the first day then Total work done, W= Rate of all 20(Rtotal)×Time

19+6×(Rtotal)= Time×(Rtotal)
Time =25

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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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There are a total of 20 machines. If these machines are to work in pairs, and each machine works with every other machine just once, then we have the total number of days it would take for these machines to work with each other = 19 + 18 + ... + 1.

Sum of the AP => 19/2 * (1+19) = 190

If the total number of days = 190, then we can say that each machine has worked for 19 days because each machine has worked with each of the other 19 machines.
Therefore, if from the beginning these machines were to be working together, the work which took 190 days to complete using the pairing method, would have actually taken 19 days.
The extra 6 days at the end was worked upon by all the machines together so it will be added to the 19 days which would have been taken.

Hence, total number of days = 19 + 6 = 25.

Answer = B.

Note : To calculate that the pairing method would take 190 days, we could also use 20C2 = 20!/(18!*2!) = 20*19/2 = 190

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Total = a
Cybersecurity = b
Cloud = c
Neither = d

So,
Both = b + c + d − a

Fraction who attended both =
(both/a) = (b+c+d−a)/a

Answer: C

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