Question
'X' can complete 75% of a work in 15 days, whereas 'Y' can
complete 25% of the same work in 10 days. They both started working together, but 'Y' left the job 'n' days before the work finished. If the entire work is completed in 14 days, then find the value of [0.6n + (n²/5)].Solution
ATQ,
Time taken by 'X' to complete the entire work = (15/75) × 100 = 20 days
Time taken by 'Y' to complete the entire work = (10/25) × 100 = 40 days
Let the total work be 40 units (LCM of 20 and 40)
Efficiency of 'X' = 40/20 = 2 units/day
Efficiency of 'Y' = 40/40 = 1 unit/day
ATQ:
(2 × 14) + [1 × (14 - n)] = 40
28 + (14 - n) = 40
42 - n = 40
n = 2
Required value = 0.6n + (n²/5) = (0.6 × 2) + {(2 × 2)/5} = 1.2 + 0.8 = 2
The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represente...
Rows of Matrix I are numbered 0 to 4 and that of matrix II are numbered from 5 to 9. A letter from these matrices can be represented first by its row an...
The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represente...
The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represente...
