Read the given information carefully and answer the following questions.
A boat covers certain distance of it's journey in three parts i.e. upstream, downstream and in still water. Ratio of distance covered in downstream to upstream is 7:3 and total distance covered is 375 km. When boat goes downstream it consume 25% less fuel per km and while moving in upstream it consumes `12(1)/(2)%` more fuel per km than that of in still water and it cover 175km in still water. Now, after reaching its destination, boat returns to initial point covering the same path and it takes `(10)/(3)` lit more fuel in return journey.
What is the rate of consumption of fuel of boat in upstream?

Options

A. 7 km in 1 lit
B. 8 km in 1 lit
C. 10 km in 1 lit
D. 12 km in 1 lit

Answer

To solve the problem step by step, we will break down the information given and calculate the required values systematically.

### Step 1: Understand the distances covered
The total distance covered by the boat is 375 km. The distance ratio of downstream to upstream is given as 7:3.

Let the distance covered downstream be \(7x\) and the distance covered upstream be \(3x\).

From the total distance:
\[ 7x + 3x + 175 = 375 \]
This simplifies to:
\[ 10x + 175 = 375 \]
Subtracting 175 from both sides gives:
\[ 10x = 200 \]
Thus,
\[ x = 20 \]

### Step 2: Calculate the distances
Now we can find the distances:
- Downstream distance = \(7x = 7 \times 20 = 140\) km
- Upstream distance = \(3x = 3 \times 20 = 60\) km
- Distance in still water = 175 km (given)

### Step 3: Determine fuel consumption rates
Let the fuel consumption in still water be \(x\) liters/km.

- **Downstream consumption**: Since it consumes 25% less fuel, the consumption rate is:
\[ \text{Downstream} = x \times \left(1 - \frac{25}{100}\right) = x \times \frac{75}{100} = \frac{3x}{4} \text{ liters/km} \]

- **Upstream consumption**: Since it consumes 12.5% more fuel, the consumption rate is:
\[ \text{Upstream} = x \times \left(1 + \frac{12.5}{100}\right) = x \times \frac{112.5}{100} = \frac{9x}{8} \text{ liters/km} \]

### Step 4: Calculate fuel consumed in the initial journey
The total fuel consumed in the initial journey can be calculated as follows:

1. **Fuel consumed downstream**:
\[ \text{Fuel}_{\text{downstream}} = \text{Distance}_{\text{downstream}} \times \text{Consumption}_{\text{downstream}} = 140 \times \frac{3x}{4} = 105x \]

2. **Fuel consumed upstream**:
\[ \text{Fuel}_{\text{upstream}} = \text{Distance}_{\text{upstream}} \times \text{Consumption}_{\text{upstream}} = 60 \times \frac{9x}{8} = 67.5x \]

3. **Fuel consumed in still water**:
\[ \text{Fuel}_{\text{still water}} = 175 \times x = 175x \]

Adding these together gives:
\[ \text{Total Fuel}_{\text{initial}} = 105x + 67.5x + 175x = 347.5x \]

### Step 5: Calculate fuel consumed in the return journey
In the return journey, the distances switch:
1. **Fuel consumed downstream** (now upstream):
\[ \text{Fuel}_{\text{return upstream}} = 60 \times \frac{3x}{4} = 45x \]

2. **Fuel consumed upstream** (now downstream):
\[ \text{Fuel}_{\text{return downstream}} = 140 \times \frac{9x}{8} = 157.5x \]

3. **Fuel consumed in still water** remains the same:
\[ \text{Fuel}_{\text{still water}} = 175x \]

Adding these gives:
\[ \text{Total Fuel}_{\text{return}} = 45x + 157.5x + 175x = 377.5x \]

### Step 6: Set up the equation based on the problem statement
According to the problem, the return journey consumes \(\frac{10}{3}\) liters more fuel:
\[ 347.5x + \frac{10}{3} = 377.5x \]

### Step 7: Solve for \(x\)
Rearranging gives:
\[ \frac{10}{3} = 377.5x - 347.5x \]
\[ \frac{10}{3} = 30x \]
Thus,
\[ x = \frac{10}{3} \div 30 = \frac{10}{90} = \frac{1}{9} \text{ liters/km} \]

### Step 8: Calculate the rate of consumption of fuel in upstream
The rate of consumption in upstream is:
\[ \text{Consumption}_{\text{upstream}} = \frac{9x}{8} = \frac{9 \times \frac{1}{9}}{8} = \frac{1}{8} \text{ liters/km} \]

### Final Answer
The rate of consumption of fuel of the boat in upstream is \(\frac{1}{8}\) liters/km.

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