Anushka and Anirudh working together can complete a piece of work in 20 days. They started the work together, but Anushka left af ter x days and Anirudh finished the remaining work in the next x/2 days. Had Anushka left after 3x/4 days, Anirudh would have taken x days to finish the remaining work. Find the ratio of the efficiency of Anushka to that of Anirudh.

Options

A. `2:3`
B. `3:2`
C. `2:1`
D. `1:1`

Answer

To solve the problem step by step, we will denote the work done by Anushka in one day as \( A \) and the work done by Anirudh in one day as \( B \).

### Step 1: Determine the combined work rate of Anushka and Anirudh
Since Anushka and Anirudh together can complete the work in 20 days, their combined work rate is:
\[ A + B = \frac{1}{20} \quad \text{(work done in one day)} \]

### Step 2: Work done by Anushka and Anirudh when Anushka leaves after \( x \) days
In \( x \) days, the work done by both Anushka and Anirudh is:
\[ \text{Work done in } x \text{ days} = (A + B) \cdot x = \frac{x}{20} \]
After \( x \) days, Anushka leaves, and Anirudh finishes the remaining work in \( \frac{x}{2} \) days. The work done by Anirudh in \( \frac{x}{2} \) days is:
\[ B \cdot \frac{x}{2} \]
The total work is 1 (the whole work), so we can set up the equation:
\[ \frac{x}{20} + B \cdot \frac{x}{2} = 1 \]

### Step 3: Solve for \( B \)
Rearranging the equation gives:
\[ B \cdot \frac{x}{2} = 1 - \frac{x}{20} \]
Multiplying through by 20 to eliminate the fraction:
\[ 10Bx = 20 - x \]
Thus,
\[ 10Bx + x = 20 \]
Factoring out \( x \):
\[ x(10B + 1) = 20 \]
So,
\[ x = \frac{20}{10B + 1} \quad \text{(1)} \]

### Step 4: Work done when Anushka leaves after \( \frac{3x}{4} \) days
If Anushka leaves after \( \frac{3x}{4} \) days, the work done by both in that time is:
\[ \text{Work done in } \frac{3x}{4} \text{ days} = (A + B) \cdot \frac{3x}{4} = \frac{3x}{80} \]
Anirudh would then take \( x \) days to finish the remaining work:
\[ B \cdot x \]
Setting up the equation for total work gives:
\[ \frac{3x}{80} + B \cdot x = 1 \]

### Step 5: Solve for \( B \) again
Rearranging this equation gives:
\[ B \cdot x = 1 - \frac{3x}{80} \]
Multiplying through by 80:
\[ 80Bx = 80 - 3x \]
Thus,
\[ 80Bx + 3x = 80 \]
Factoring out \( x \):
\[ x(80B + 3) = 80 \]
So,
\[ x = \frac{80}{80B + 3} \quad \text{(2)} \]

### Step 6: Equate the two expressions for \( x \)
From equations (1) and (2):
\[ \frac{20}{10B + 1} = \frac{80}{80B + 3} \]
Cross-multiplying gives:
\[ 20(80B + 3) = 80(10B + 1) \]
Expanding both sides:
\[ 1600B + 60 = 800B + 80 \]
Rearranging gives:
\[ 800B = 20 \implies B = \frac{1}{40} \]

### Step 7: Find \( A \)
Substituting \( B \) back into the equation for \( A + B \):
\[ A + \frac{1}{40} = \frac{1}{20} \]
Thus,
\[ A = \frac{1}{20} - \frac{1}{40} = \frac{2}{40} - \frac{1}{40} = \frac{1}{40} \]

### Step 8: Calculate the ratio of efficiencies
The efficiencies of Anushka and Anirudh are given by \( A \) and \( B \):
\[ \text{Efficiency of Anushka} : \text{Efficiency of Anirudh} = A : B = \frac{1}{40} : \frac{1}{40} = 1 : 2 \]

### Final Answer
The ratio of the efficiency of Anushka to that of Anirudh is:
\[ \text{Ratio} = 2 : 1 \]