Worker A can complete a work in 20 days, Worker B in 30 days, and Worker C in 60 days. All three work together for 6 days. Then Worker B leaves. After that, Worker A works every da

Given

A = 20 days
B = 30 days
C = 60 days
Together for 6 days
After that, B leaves
A works daily
C works alternate days (starting next day)

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Solution

Step 1: Daily work rates

$$A = \frac{1}{20}, \quad B = \frac{1}{30}, \quad C = \frac{1}{60}$$

LCM of 20, 30, 60 = 60 units

$$A = 3 \text{ units/day} B = 2 \text{ units/day} C = 1 \text{ unit/day}$$

Total work = 60 units

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Step 2: Work done in first 6 days (all together)

Daily combined work:

$$3 + 2 + 1 = 6 \text{ units/day}$$

In 6 days:

$$6 \times 6 = 36 \text{ units}$$

Remaining work:

$$60 - 36 = 24 \text{ units}$$

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Step 3: After B leaves

Now:

• A works daily = 3 units/day
• C works alternate days = 1 unit every 2 days

So in 2 days cycle:

Day 1: A + C = 4 units Day 2: A only = 3 units

Total in 2 days:

$$4 + 3 = 7 \text{ units}$$

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Step 4: Finish remaining 24 units

Number of full 2-day cycles:

$$24 ÷ 7 = 3 \text{ cycles} (21 units)$$

Work done in 3 cycles:

$$3 \times 7 = 21$$

Remaining work:

$$24 - 21 = 3 \text{ units}$$

Next day = A + C = 4 units → enough to finish.

Days after B leaves:

$$3 cycles = 6 days + 1 extra day = 7 days$$

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Step 5: Total Days

$$6 + 7 = 13 \text{ days}$$

Closest option:

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✅ Answer

14 days (Option C)