CAT 2024 Slot 3 · Quant · Question 9

Sam finishes alone in 20 days, so his rate is \frac{1}{20} of the job per day.

Mohit is twice as fast, so his rate is \frac{2}{20}=\frac{1}{10} per day.

Mohit is thrice as fast as Ayna, so Ayna's rate is \frac{1}{10}\div 3=\frac{1}{30} per day.

Take the whole job as 60 units, the LCM of the denominators: Sam does 3 units a day, Mohit 6 and Ayna 2.

Day 1 is Sam and Mohit, giving 3+6=9 units; day 2 is Sam and Ayna, giving 3+2=5; day 3 is Mohit and Ayna, giving 6+2=8.

One three-day cycle therefore produces 9+5+8=22 units.

Two full cycles cover 6 days and 44 units, leaving 60-44=16 units.

Day 7 repeats day 1 and adds 9 units, bringing the total to 53 with 7 units left.

Day 8 repeats day 2 and adds 5 units, bringing the total to 58 with 2 units left.

Day 9 is a Mohit-and-Ayna day, which Sam sits out, and the last 2 units are finished then.

Sam worked on days 1, 2, 4, 5, 7 and 8 — six days at 3 units each, so he did 18 units.

His share is \frac{18}{60}=\frac{3}{10}, the second option.

Shortcut. Sam works on two of every three days, so a careless answer would be \frac{2}{3}\times \frac{3}{22} per cycle. The real check is only whether the final partial day includes Sam — here it does not, so his tally stops at a whole number of cycles plus two days.