A large watermelon weighs 20 kg. with  96% of its weight being water. It is allowed to stand in the sun and some of the water evaporates so that only of its 95% weight is water. Its reduced weight will be :

Correct option is A

Given:

1. The initial weight of the watermelon is 20 kg .
2. Initially, $$\frac{4}{5}$$​ of its weight is water.
3. After evaporation, $$\frac{3}{5}$$​ of its weight is water.
4. We need to find the reduced weight of the watermelon.

Formula Used:

The weight of non-water content remains constant before and after evaporation.

Let:
​$$w_1$$​ = initial weight of watermelon,
​$$f_1$$​ = initial fraction of water,
​$$f_2$$​ = final fraction of water,
​$$w_2$$​ = final weight of watermelon.

​$$\text{Non-water weight} = (1 - f_1) \times w_1 = (1 - f_2) \times w_2$$​

Solution:

1. **Initial non-water weight:**

$$\text{Non-water weight} = \left(1 - \frac{4}{5}\right) \times 20 = \frac{1}{5} \times 20 = 4 \, \text{kg}$$​

2. **After evaporation:**
The fraction of non-water content is:

$$1 - \frac{3}{5} = \frac{2}{5}$$​

Using the formula for non-water weight:

$$\frac{1}{5} \times 20 = \frac{2}{5} \times w_2$$​​

Simplify:

$$4 = \frac{2}{5} \times w_2$$​​

Multiply through by 5:

$$20 = 2 \times w_2$$​​

Divide by 2:

$$w_2 = 16 \, \text{kg}$$​​

Final Answer:

The reduced weight of the watermelon is 16 kg
**Option A: 16 kg**