A liquid P is 1 `3/7` times as heavy as water and water is 1 `2/5` times as heavy as another liquid Q. The amount of liquid 'P that must be added to 7 litres of the liquid 'Q' so that the mixture may weigh as much as an equal volume of water, will be

Options

A. 7litres
B. 5 `1/6` litres
C. 5litres
D. 4 `2/3` litres

Answer

To solve the problem step by step, we will follow the information given about the weights of liquids P, Q, and water.

### Step 1: Understand the weights of the liquids
- Liquid P is \(1 \frac{3}{7}\) times as heavy as water. This can be converted to an improper fraction:
\[ 1 \frac{3}{7} = \frac{10}{7} \]
So, the weight of liquid P compared to water is \(\frac{10}{7}\).

- Water is \(1 \frac{2}{5}\) times as heavy as liquid Q. This can also be converted to an improper fraction:
\[ 1 \frac{2}{5} = \frac{7}{5} \]
Thus, the weight of water compared to liquid Q is \(\frac{7}{5}\).

### Step 2: Express the weight of liquid Q in terms of water
From the relationship we have:
\[ \text{Weight of Q} = \frac{5}{7} \times \text{Weight of Water} \]

### Step 3: Set up the equation for the mixture
Let \(X\) be the amount of liquid P that we need to add to 7 liters of liquid Q. The total weight of the mixture must equal the weight of an equal volume of water (which is \(X + 7\) liters).

The weight of liquid P in terms of water is:
\[ \text{Weight of P} = \frac{10}{7} \times X \]

The weight of 7 liters of liquid Q in terms of water is:
\[ \text{Weight of Q} = \frac{5}{7} \times 7 = 5 \text{ (since the 7s cancel)} \]

### Step 4: Set up the equation for the total weight
Now we can set up the equation:
\[ \frac{10}{7}X + 5 = X + 7 \]

### Step 5: Solve for \(X\)
To solve for \(X\), first eliminate the fractions by multiplying the entire equation by 7:
\[ 10X + 35 = 7X + 49 \]

Now, rearranging gives:
\[ 10X - 7X = 49 - 35 \]
\[ 3X = 14 \]
\[ X = \frac{14}{3} \text{ or } 4 \frac{2}{3} \text{ liters} \]

### Conclusion
The amount of liquid P that must be added to 7 liters of liquid Q so that the mixture weighs as much as an equal volume of water is:
\[ \boxed{4 \frac{2}{3} \text{ liters}} \]