Worker A can complete a work in 24 days, Worker B can complete it in 36 days, and Worker C can complete it in 48 days. * They work in the order A → B → C, each working one day at a

Given

A = 24 days
B = 36 days
C = 48 days
Work cycle = 3 people, 1 day each
After 3 cycles, C leaves
A efficiency +50% afterwards

---

Solution

Step 1: Daily work rates

$$\text{A's 1-day work} = \frac{1}{24}, \quad \text{B's 1-day work} = \frac{1}{36}, \quad \text{C's 1-day work} = \frac{1}{48}$$

---

Step 2: Work done in 1 full cycle (3 days)

LCM of 24, 36, 48 = 144 units

$$\text{A's work} = 6 \text{ units}$$ $$\text{B's work} = 4 \text{ units}$$ $$\text{C's work} = 3 \text{ units}$$

Total per cycle:

$$6 + 4 + 3 = 13 \text{ units per 3 days}$$

---

Step 3: Work done in 3 cycles (9 days)

$$3 \times 13 = 39 \text{ units}$$

Total work = 144 units

Remaining work:

$$144 - 39 = 105 \text{ units}$$

---

Step 4: Worker C leaves → A efficiency +50%

• New A daily work:

$$\text{New A rate} = \frac{1}{24} \times 1.5 = \frac{1}{16}$$

• Convert to 144-unit system:

$$\text{A new daily work} = 9 \text{ units/day}$$

• B's daily work remains = 4 units/day

---

Step 5: Alternate A & B

• Remaining work = 105 units

• Work in 2-day cycles: 9 + 4 = 13 units per 2 days

• Number of full 2-day cycles:

$$105 ÷ 13 ≈ 8 cycles → 104 units$$

• Remaining work = 105 - 104 = 1 unit → next day A works

---

Step 6: Total days

• First 9 days (3 full cycles)
• Next 16 + 1 = 17 days

$$\text{Total days} = 9 + 17 = 26 \text{ days}$$

---

✅ Answer

26 days (Option A)