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

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Bunuel

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Bunuel

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Bunuel

GMAT Club Official Explanation:

Each machine worked 1 day in a pair with each of the remaining 19 machines. Hence, each machine worked for 19 days in pairs.

After all unique pairs had worked once, all 20 machines worked together for 6 more days.

So, each machine worked a total of 19 + 6 = 25 days.

Thus, if all machines had worked together from the beginning, they would have completed the job in 25 days.

Answer: B.

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Let R be the combined work rate of all 20 machines.

Phase 1: Machines work in unique pairs.
The number of unique pairs from 20 machines is "20 choose 2", which is 190. So this phase lasts 190 days.
Crucially, each individual machine is part of 19 different pairs. Therefore, each machine works for exactly 19 days during this phase.
The total work done in Phase 1 is equivalent to all machines working together for 19 days, which is 19 * R.

Phase 2: All machines work together.
This phase lasts for 6 additional days.
The work done in Phase 2 is 6 * R.

Total Work:
The total work to complete the job is the sum of the work from both phases.
Total Work = (19 * R) + (6 * R) = 25 * R.

Conclusion:
The total job is equivalent to all 20 machines working together for 25 days.

The correct answer is (B) 25.

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Number of pairs = 20C2 = 190

Each day a unique pair works. So there are 10 pairs working each day.
Total days = 190/10=19 days

In addition, all 20 machines work together for 6 additional days to complete the job
This makes the total number of days = 19+6=25

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[size=85][size=100]Answer[/size][/size]

[size=85][size=100](B) 25[/size][/size]

Explanation

Let's break this down logically. The key is to figure out the total amount of "work" done in terms of the machines' combined daily output.

Let R be the combined rate of all 20 machines working together for one day.

Step 1: The "Pairs" Phase

First, how many days does this phase last? We need to find the number of unique pairs of 2 machines from a group of 20. This is a combination:
- C(20, 2) = (20 × 19) / 2 = 190 days.
Now, how much work is done? Over these 190 days, each machine is paired with the other 19 machines. This means each machine works for a total of 19 days.
So, the total work done in this phase is equivalent to all 20 machines working together for 19 days.
- Work in pairs phase = 19 × R.

Step 2: The "All Together" Phase

After the pairs phase, all 20 machines work together for 6 additional days.
The work done here is simply 6 × R.

Step 3: Calculate the Total Work

To complete the job, the total work is the sum of the work from both phases.
- Total Work = (19 × R) + (6 × R) = 25 × R.
This means the entire job is equivalent to all 20 machines working together for 25 days.

Step 4: Answer the Question

The question asks: "If all 20 machines had worked together from the beginning, in how many days would the job have been completed?"
Since the total work is 25 × R, it would take exactly 25 days.

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25 days.

20 machines. / 2 per day never the same pair.

190 unique pairs in a group a group of 20 so 190 days. (19*20 / 2)
6 days at the end needed.

(190*2 ) + (6*20) = 500 machine working days. 500/20 = 25

25 days

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From 20 machines, the number of unique 2-machine combinations would be 20C2 = 190

Let x1, x2, x3 and so on be work rates per day of 20 machines
So over the 190 days, the work completed would be sum of (xi + xj) for all the pairs of i and j

Since each machine works 19 days with other machine, 19* (sum of xi, for all other 19 machines)
Let R be (sum of xi, for all other 19 machines)
So total work for 2 machines at a time would be 19*R

For 6 days, the work done was 6*R
So total work = 19*R + 6*R = 25R

but if they all began to work at same time it would have take 25 days to do so.

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Bunuel

Each Machine would work with the other 19 machines for 19 days, and then all the machines would simultaneously work for 6 days, hence each machine would work for (19+6) = 25 days;

Hence, when all the machines work simultaneously, they have to work for 25 days.

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The ans is 196.
so, each pair of machines work once so total combinations = 20C2 = 190.
6 additional days extra were taken . so, 190+6 = 196 days.

ans is E

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Let 1 machine can complete the task in x hrs
Rate of 1 machine = 1/x

Total number of pairs = 20C2 = 190

Total work done = work done by 190 machines taken 2 at a time + work done by 20 machines for 6 days
Total work = (1/x + 1/x) * 190 + (20/x)*6 = 500/x

When 20 machines work together since beginning
No of days = Total work/work done by 20 machines
No of days = (500/a)/(20/a) = 25 days

Hence B is correct

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First, I need to determine how many unique pairs can be formed from 20 machines. This is a combination problem where we select 2 machines from 20:
Number of unique pairs = C(20,2) = (20 × 19)/(2 × 1) = 190 pairs
So there were 190 days where 2 machines worked together, followed by 6 days where all 20 machines worked together.
Let's denote the rate of machine i as r_i (in terms of job/day).
When 2 machines work together for a day, their combined rate is (r_i + r._]) job/day.
When all 20 machines work together, their combined rate is (r_1+r_2 + ... + r_20) job/day.
The total work done equals 1 job, which can be written as:
Sum of all pairs working for 1 day each + All 20 machines working for 6 days = 1 job
For the sum of all pairs, each machine appears in exactly 19 pairs (it can pair with each of the other 19 machines). So each machine's rate r_i appears 19 times in the sum.
Therefore:
19(г 1+ г_2+ ... + г_20) + 6(г_1+ г_2+ ... + r_20) = 1 job
(19 + 6)(г_1+r_2+.. + r_20) =1 job
25(г_1+r_2+.. + r_20) =1 job
This means that the combined rate of all 20 machines is 1/25 job/day.
Therefore, if all 20 machines had worked together from the beginning, they would have completed the job in 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?

Let the efficiency (part of work done per day) of the machines be e1, e2, ..... e20

e1 + e2 + e1 + e3 + e1 + e4 + ... + e1 + e20 + e2 + e3 + e2 + e4 + ...+e2 + e20 + e3 + e4 + ...e3 + e20 + ... e19 + e20 + 6 (e1 + e2+ ... + e20) = 1

19 (e1 + e2+ ... + e20 ) + 6 (e1 + e2 + ... + e20) = 25 (e1 + e2 + ... + e20) = 1
e1 + e2 + .... + e20 = 1/25

Efficiency of all machines working together = 1/25

If all 20 machines had worked together from the beginning, the number of days in which the job would have been completed = 25 days

IMO B

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You can pair 2 machines out of 20 in
(202)=20×192=190 ways
So, for 190 days, 2 machines work each day — each pair only once.

Each machine ends up working with 19 others, once each. So every machine contributes to 19 days of work.
That means, across all those 190 days, each machine works for 19 days total.
So, total work in 190 days =
Same as if all 20 machines had worked individually for 19 days each.
Which is:
19 × (sum of all machine rates)
Let’s call this total combined rate = R
So work done in first phase = 19R

6 days × R = 6R
So total work = 19R + 6R = 25R

If all machines worked together from the start (rate = R),
time needed =
Total work ÷ total rate =
25R ÷ R = 25 days

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We have 20 machines and they work in pairs
So unique paris are 20C2 = 20*19/2 =190
Each machine has worked 19 days in these 190 days, so if amounts to 19 days of work for every machine working together
Then in the end all machines work together for 6 days
Total day = 19+6 =25
Hence B

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20C2 * 2t + 6* 20t = 1
t = 1/500

d*20t= 1 => d(no. days for 20 machines) = 500/20 = 25

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As we are taking unique pairs, machine 1 works for 19 days, machine 2 works for 19 days, machine 3 works for 19 days, and so on.

So, each machine works for 19 days.

After that, each machine for 6 days.

So in total each machine works for 25 days. Total work is 25 * rate of each machine

Now, if the machines are started all in one go right from the start, each day every machine will run and it will take 25 days to complete. Because the amount of work that has to be done is the same.

Option B

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The correct answer is 25. Below attached is the detailed solution for this.

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Bunuel

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20C2 = 190
It takes 190days for all unique pairs to run

Every 10 days, each unique machine has worked one of the days.

therefore 190 days of unique pairs = 19 days of all 20 machines working together

19+6 = 25,

It would take 25 days of all machines working together

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20 machines can form 10 unique pair but no where it is mentioned that the pair are unique which means a pair can be formed in 20C(2) ways which is 190 pair but as we know unique pairs are 10 so it means each machine shows up 19 times

work done in first 190 days is:
let rate of each machine be a, b,c and so on
the rate by 19 machine would be 19(a+b+c....)

similarly work done in 6 days is:
6(a+b+c...)

Let sum of all rates be R THEN total work = 19R + 6 R=25R

Days= Total Work/ Total Rate = 25R/R = 25 Days

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The number of possible unique pairings of machines can be found using the handshake formula where
(number of machines x number of other machines each machine can work with)/2 = 190

Because each pair works on one day, I know it took the first phase 190 days before the final 6 day phase begun.

Setting up the rate equation, I don't know the individual rates and have no way of finding out so I'll try and express it in terms of their combined rate = R. Because each machine works only on a day that it is paired with a new machine, it will work for 19 days (to make up a total of 19 different pairs). This is of course the case for all machines so 19 x (rate machine 1) + 19 x (rate machine 2)....... = 19 (rate machine 1 + r machine 2 +.....) = 19R

For the second phase, they all work together for 6 days so the work done = 6R

Adding the work done during both phases means that the job is worth 25R, or 25 days of all machines working together.

Answer: B) 25

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