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VizzyFanboy
Joined: 14 Jul 2023
Last visit: 17 Nov 2025
Posts: 32
Given Kudos: 31
Location: India
Schools: IIMA '26 (A) EPGP '26 (A)
GMAT Focus 1: 675 Q88 V84 DI79 
01 Jul 2025, 10:13
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Not a 100% sure about this, since it says each of the 20 machines has a different work rate.
But, 20 machines selected in a pair comes up to 20C2 = 20!/2!18! = 190 days.
Hence the total number of days taken to complete the task = 190+6 = 196.
If 2 machines at a time take 190 days to complete a certain fraction of the task, 1 machine at a time takes 190*2 days to complete same fraction of the task. Similarly, 20 machines would take = 190*2/20 = 19 days.
Total number of days taken by all 20 machines to complete the task = 19+6 = 25.
Answer = B
But, 20 machines selected in a pair comes up to 20C2 = 20!/2!18! = 190 days.
Hence the total number of days taken to complete the task = 190+6 = 196.
If 2 machines at a time take 190 days to complete a certain fraction of the task, 1 machine at a time takes 190*2 days to complete same fraction of the task. Similarly, 20 machines would take = 190*2/20 = 19 days.
Total number of days taken by all 20 machines to complete the task = 19+6 = 25.
Answer = B
01 Jul 2025, 10:21
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Each of the \(20\) machines are paired up with another machine only once, and the pair works for \(1\) day, until all unique pairs are exhausted.
Lets say the \(20\) machines are named \(a, b, c, d.........t \) and their rate of work is also denoted by the same alphabet.
Pairs including machine \(a\) would be \((a,b), (a,c), (a,d).............(a,t)\). Machine \(a\) will make \(19\) pairs (because there are \(20\) machines and \((a,a)\) is not a valid pair).
The same would be true for all other machines. They will make \(19\) pairs with every remaining machine.
The work done by pair \((a,b)\) in \(1\) day will be \(a+b\)
The work done by pair \((a,c)\) in \(1\) day will be \(a+c\)
Similarly, each pair will complete a fraction of work until all pairs are exhausted. If we add all the work done by pairs, we should get - >
\((a+b)+(a+c)...........(t+r)+(t+s)= 19(a+b+c.......t)\) (We can factor out a \(19\) because each machine's rate of work shows up \(19\) times in this)
Let \(a+b+c.......t = R\)
The total work done by all the pairs \(= 19(R)\)
The remaining work was done by all \(20\) machines in \(6\) days.
So the rate of work of all machines when working together is \(= a+b+c.......t \)
In \(6\) days, they completed \(6(a+b+c.......t) = 6(R)\) units of work.
The total work done can be assumed to be 1 unit, which is the sum of the work done by the pairs and the work done by all the machines working together for \(6\) days.
\(19(R) + 6(R) = 1\)
\(25(R) = 1\)
\(R = \frac{1}{25}\)
So, all machines working together can complete \(1\) unit of work in \(25\) days.
Lets say the \(20\) machines are named \(a, b, c, d.........t \) and their rate of work is also denoted by the same alphabet.
Pairs including machine \(a\) would be \((a,b), (a,c), (a,d).............(a,t)\). Machine \(a\) will make \(19\) pairs (because there are \(20\) machines and \((a,a)\) is not a valid pair).
The same would be true for all other machines. They will make \(19\) pairs with every remaining machine.
The work done by pair \((a,b)\) in \(1\) day will be \(a+b\)
The work done by pair \((a,c)\) in \(1\) day will be \(a+c\)
Similarly, each pair will complete a fraction of work until all pairs are exhausted. If we add all the work done by pairs, we should get - >
\((a+b)+(a+c)...........(t+r)+(t+s)= 19(a+b+c.......t)\) (We can factor out a \(19\) because each machine's rate of work shows up \(19\) times in this)
Let \(a+b+c.......t = R\)
The total work done by all the pairs \(= 19(R)\)
The remaining work was done by all \(20\) machines in \(6\) days.
So the rate of work of all machines when working together is \(= a+b+c.......t \)
In \(6\) days, they completed \(6(a+b+c.......t) = 6(R)\) units of work.
The total work done can be assumed to be 1 unit, which is the sum of the work done by the pairs and the work done by all the machines working together for \(6\) days.
\(19(R) + 6(R) = 1\)
\(25(R) = 1\)
\(R = \frac{1}{25}\)
So, all machines working together can complete \(1\) unit of work in \(25\) days.
01 Jul 2025, 10:23
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20 machines
Since each machine will work with each other, to cover all possible pair, we initially have:
(20*19)/2 = 190 -> 20 machines, working each one with all others. It's divided by two to remove duplicated pais (Machine 1 with Machine 2 = Machine 2 with Machine 1).
So, each machine works 19 days from the pairs + 6 days from all the set. Therefore, if all machines worked at the same time since the begining, it would be required 19 + 6 = 25 days of work.
Since each machine will work with each other, to cover all possible pair, we initially have:
(20*19)/2 = 190 -> 20 machines, working each one with all others. It's divided by two to remove duplicated pais (Machine 1 with Machine 2 = Machine 2 with Machine 1).
So, each machine works 19 days from the pairs + 6 days from all the set. Therefore, if all machines worked at the same time since the begining, it would be required 19 + 6 = 25 days of work.
