A, B, and C are capable of completing a job in 30, 45, and 90 days, respectively. A works every day, while B and C join A every fifth day. How long will it take to finish the task?

Correct — A. Take the job as 90 units, the LCM of 30, 45 and 90.

A = 90 ÷ 30 = 3 units a day
B = 90 ÷ 45 = 2 units a day
C = 90 ÷ 90 = 1 unit a day

One 5-day cycle: A alone for 4 days = 4 × 3 = 12, then day 5 all three = 3 + 2 + 1 = 6
Work per cycle = 12 + 6 = 18 units

Cycles needed = 90 ÷ 18 = 5
Time = 5 cycles × 5 days = 25 days → option (a)

Why the others are wrong

• (b)
  18 days — 18 is the work in one cycle, in units, not a count of days. In 18 days A does 54 units and B and C add at most 12, well short of 90.
• (c)
  16 days — In 16 days A does 48 units, and B and C come at most 4 times, adding 3 units each: at most 60 of the 90 units.
• (d)
  12 days — 12 days is faster than all three working every day, which takes 90 ÷ 6 = 15 days. With B and C coming only one day in five, the job must take longer than 15 days.

Work questions with helpers who come and go are solved in cycles. Convert each worker to units a day on a common total, add up one full repeat of the pattern, then count how many repeats the job needs.

Here the pattern repeats every 5 days and does 18 units, and 90 is exactly 5 × 18, so the job ends at the close of a cycle.

When the total is not a whole number of cycles, finish the last part day by day in the order the pattern runs.

Key facts

• Units a day = total units ÷ days taken alone: on a 90-unit job A does 3, B 2 and C 1.
• One 5-day pattern here does 4 × 3 + 6 = 18 units.
• Bounds check: A alone takes 30 days and all three every day take 15, so the answer lies between.

Common traps

• Stopping at 18, the units done in one cycle, and marking it as the number of days.
• Assuming the job always ends on a helper day: here it does, but when the leftover is smaller than one cycle's work, A may finish alone earlier.

No directly related past PYQ was found.