Three professors Dr. Gupta, Dr. Sharma, Dr. Singh are evaluating Answer script of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr, Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work?

Options

A. 7.2 days
B. 9.5 days
C. 11.5 days
D. None of these

Answer

To solve the problem step by step, we will first determine the efficiencies of the three professors and then calculate how much work they complete in the given time.

### Step 1: Determine the efficiencies of Dr. Gupta, Dr. Sharma, and Dr. Singh.

Let the efficiency of Dr. Singh be \( S \).

- Dr. Sharma is 20% more efficient than Dr. Singh:
\[ \text{Efficiency of Dr. Sharma} = S + 0.2S = 1.2S \]

- Dr. Gupta is 40% more efficient than Dr. Sharma:
\[ \text{Efficiency of Dr. Gupta} = 1.2S + 0.4(1.2S) = 1.2S + 0.48S = 1.68S \]

### Step 2: Determine the time taken by each professor to complete the evaluation work.

Let the total work be \( W \).

- Time taken by Dr. Singh to complete the work:
\[ T_S = \frac{W}{S} \]

- Time taken by Dr. Sharma to complete the work:
\[ T_{Sh} = \frac{W}{1.2S} \]

- Time taken by Dr. Gupta to complete the work:
\[ T_G = \frac{W}{1.68S} \]

According to the problem, Dr. Gupta takes 10 days less than Dr. Sharma:
\[ T_{Sh} - T_G = 10 \]

Substituting the expressions for \( T_{Sh} \) and \( T_G \):
\[ \frac{W}{1.2S} - \frac{W}{1.68S} = 10 \]

### Step 3: Solve for \( W \) in terms of \( S \).

To solve the equation, find a common denominator:
\[ \frac{W \cdot 1.68 - W \cdot 1.2}{1.2 \cdot 1.68S} = 10 \]
\[ \frac{W(1.68 - 1.2)}{2.016S} = 10 \]
\[ \frac{W(0.48)}{2.016S} = 10 \]
\[ W = \frac{10 \cdot 2.016S}{0.48} \]
\[ W = 42S \]

### Step 4: Calculate the time taken by each professor.

Now we can find the time taken by each professor:
- Dr. Singh:
\[ T_S = \frac{W}{S} = \frac{42S}{S} = 42 \text{ days} \]
- Dr. Sharma:
\[ T_{Sh} = \frac{W}{1.2S} = \frac{42S}{1.2S} = 35 \text{ days} \]
- Dr. Gupta:
\[ T_G = \frac{W}{1.68S} = \frac{42S}{1.68S} = 25 \text{ days} \]

### Step 5: Calculate the work done by Dr. Gupta and Dr. Sharma.

- Work done by Dr. Gupta in 10 days:
\[ \text{Work}_{G} = 10 \cdot 1.68S = 16.8S \]

- Remaining work after Dr. Gupta:
\[ \text{Remaining Work} = W - \text{Work}_{G} = 42S - 16.8S = 25.2S \]

- Work done by Dr. Sharma in 15 days:
\[ \text{Work}_{Sh} = 15 \cdot 1.2S = 18S \]

- Remaining work after Dr. Sharma:
\[ \text{Remaining Work} = 25.2S - 18S = 7.2S \]

### Step 6: Calculate the time taken by Dr. Singh to complete the remaining work.

- Time taken by Dr. Singh to complete the remaining work:
\[ T_{Singh} = \frac{7.2S}{S} = 7.2 \text{ days} \]

### Final Answer:
Dr. Singh can complete the remaining evaluation work in **7.2 days**.

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