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02 Jul 2025, 06:15
Kudos
Bookmarks
there are 2 parts in the question:
1st part , machines are working in pairs & 2nd part 20 machines working all together.
1st part , machines are working in pairs:
total combinations=(20*19)/2=190 days
each m/c has a unique rate so total rate R= r1+r2....+r20
let work for day1,W1= (r1+r2)*1
let work for day2W2= (r1+r3)*1
let work for day3W3=(r1+r4)*1
...... every machines rate appears 19 times in the 190 days of work.
total work done in 190 days = 19*(r1+r2+.....r20)= 19R
similarly work done in 6 days=2*6
total work done = 19R+6R=25R
2ND PART
time taken by 20 m/c = W/R= 25R/R= 25...... AND (B)
1st part , machines are working in pairs & 2nd part 20 machines working all together.
1st part , machines are working in pairs:
total combinations=(20*19)/2=190 days
each m/c has a unique rate so total rate R= r1+r2....+r20
let work for day1,W1= (r1+r2)*1
let work for day2W2= (r1+r3)*1
let work for day3W3=(r1+r4)*1
...... every machines rate appears 19 times in the 190 days of work.
total work done in 190 days = 19*(r1+r2+.....r20)= 19R
similarly work done in 6 days=2*6
total work done = 19R+6R=25R
2ND PART
time taken by 20 m/c = W/R= 25R/R= 25...... AND (B)
Bunuel
02 Jul 2025, 06:41
Kudos
Bookmarks
The key sentence is "Each day, exactly 2 machines are assigned to work, and no pair of machines works together more than once.", indicating the combinatoric:
- 20!/ (2!18!) = 190 choices of two machines working together, meaning 190 days of work for 20 machines
- With the 6 additional days of work for all machines, we have a total of 196 days
- 20!/ (2!18!) = 190 choices of two machines working together, meaning 190 days of work for 20 machines
- With the 6 additional days of work for all machines, we have a total of 196 days
02 Jul 2025, 07:16
Kudos
Bookmarks
Let x1,x2,........x20 be the number of days each of the machines 1....20 take individually to complete the work.
Since, only a distinct pair of two machines can work on a day and each machine will form 19 distinct pairs with other machines.
So, work completed by the time all distinct pairs are exhausted will be:
W1 = 19 * (1/x1 + 1/x2.....+ 1/x20)
Work done in last 6 days = W2 = 6 * (1/x1 + 1/x2 + ............ + 1/x20)
W1/W2 = 19/6
Total Work done W = W1+W2
From the above ratio
Time taken to do 6/25 of total work = 6 days
So, time taken to do total work = 6 * 25/6 = 25 days
Therefore Option B is correct answer.
Since, only a distinct pair of two machines can work on a day and each machine will form 19 distinct pairs with other machines.
So, work completed by the time all distinct pairs are exhausted will be:
W1 = 19 * (1/x1 + 1/x2.....+ 1/x20)
Work done in last 6 days = W2 = 6 * (1/x1 + 1/x2 + ............ + 1/x20)
W1/W2 = 19/6
Total Work done W = W1+W2
From the above ratio
Time taken to do 6/25 of total work = 6 days
So, time taken to do total work = 6 * 25/6 = 25 days
Therefore Option B is correct answer.
Bunuel
