Q.
T1, T2, T3 and T4 can do a job together in 8/3 days. Time taken by T2 alone to complete the job is 2 days less than T4, while T2 is 87.5% more efficient than T1. Time taken by T3 and T4 together to complete the work is 60/11 days.M = Find time taken by T2 and T3 together to complete the job.N = Find time taken by T5 to complete the job alone, if T2 and T5 together complete the job in 3 days.L= If T4 worked at twice of its efficiency and T1 worked at 75% of its efficiency, then find time taken by both together to complete the work.A.M > N > LB.M = N < LC.M < N = LD.M = N > LE.M < N > L
Ans: M = N > L
Sol:
Let the total work = 240 units
Efficiency of (T1 + T2 + T3 + T4) = [240 / (8/3)] =
90 units/day
Efficiency of (T3 + T4) = [240 / (60/11)] = 44
units/day
Efficiency of (T1 + T2) = 90 – 44 = 46 units
Ratio of efficiency of T2 and T1 respectively =
15:8
So, efficiency of T2 = 46/23 x 15 = 30 units/day
Efficiency of T1 = 46 – 30 = 16 units/day
Time taken by T2 to complete the work = 240/30
= 8 days
Time taken by T1 to complete the work = 240/16
= 15 days
Time taken by T4 to complete the work = 8 + 2 =
10 days
Efficiency of T4 = 240/10 = 24 units/day
Efficiency of T3 = 44 – 24 = 20 units/day
For M,
Time taken by T2 and T3 together to complete
the work = 240/ (50) = 4.8 days n
For N,
Efficiency of T2 = 30 units/day
Efficiency of (T2 + T5) = 240/3 = 80 units/day
Efficiency of T5 = 80 – 30 = 50 units/day
Required time = 240/50 = 4.8 days
For L,
New efficiency of T4 = 2 x 24 = 48 units/day
New efficiency of T1 = 16 x 3/4 = 12 units/day
Required time = 240/ (48 + 12) = 4 days
So, M = N > L
Hence the answer is option D
