A and B together can finish a task in 12 days. B and C together can finish it in 18 days. If A is three times more efficient than C, then in what time B alone would finish the same task?

Correct option is C

Given:

A + B can complete the work in 12 days

B + C can complete the work in 18 days

A is three times more efficient than C → i.e., A = 3C

Formula Used:

Work = Efficiency × Time

Solution:

Let total work = LCM of 12 and 18 = 36 units

A + B efficiency =$$\frac{36}{12} =$$​ 3 units/day

B + C efficiency =$$\frac{36}{18}$$​= 2 units/day

Let C’s efficiency = x, then A = 3x

From the equations:

A + B = 3

3x + B = 3

B = 3 - 3x

B + C = 2

B + x = 2

Substitute B from above:

3 - 3x + x = 2

3 - 2x = 2

x =$$\frac{1}{2}$$​

Then,
B = 3 - 3x = 3$$- \frac{3}{2} = \frac{3}{2}$$​ units/day

Time taken by B alone =

​$$\frac{36}{\frac{3}{2}} = 36 \times \frac{2}{3} = 24 \text{ days}$$

Alternate Method:​

Let efficiencies (work per day)
Let C’s efficiency = x units/day.
Then, A’s efficiency = 3x units/day (since A is 3 times more efficient than C).
Let B’s efficiency = y  units/day.
A + B finish work in 12 days:
(3x + y)$$\times 12 = \text{Total Work} \quad \text{(Equation 1)}$$​​
B + C finish work in 18 days:
$$(y + x) \times 18 =$$​$$\text{Total Work} \quad \text{(Equation 2)}$$​​
​$$(3x + y) \times 12 = (y + x) \times 18$$​​
36x + 12y = 18y + 18x
36x - 18x = 18y - 12y
18x = 6y
y = 3x
Substitute y = 3x into Equation 2:
​$$(y + x) \times 18 = (3x + x) \times 18 = 4x \times 18 = 72x \text{ units}$$​​
B’s efficiency = y = 3x  units/day.
​$$\text{Time} = \frac{\text{Total Work}}{\text{Efficiency}} = \frac{72x}{3x} = 24 \text{ days}$$​​