Question

After 25 kg of water has been evaporated from solution of salt and water, which had 20% salt, the remaining solution had 30% salt. find the weight of the original solution by percentage method

Answer 1

Let x be the original wt of solution.

wt of salt = 20x/100

After evaporation, the new wt of solution = x-25

wt of salt = 30(x-25)/100

Now equate the wt of salt

20x/100 = 30(x-25)/100

20x = 30x - 750

10x = 750 or x = 75 Kg ,

The wt of original solution is 75 Kg

Answer 2

Answer:

The weight of the original solution, $x$, calculated is $75kg$.

Explanation:

Given data,

Before evaporation of water,

Let the weight of the original solution = $xkg$

The weight of the salt in $xkg$ solution, w₁ = $20\% x$ = $\frac{20x}{100}$ = $\frac{x}{5}kg$

The weight of water in $xkg$ solution, w₂ = $x - \frac{x}{5}$ = $\frac{4x}{5}kg$

After evaporation of water,

The weight of the solution becomes, w₃ = $(x-25)kg$

The weight of the remaining salt in the solution, w₄ = $30\%$ = $\frac{30}{100}$

The original solution's weight =?

As given in the solution, the equation becomes:

• $\frac{w_{1} }{w_{3} } = w_{4}$
• $\frac{\frac{x}{5} }{x-25} = \frac{30}{100}$
• $\frac{100x}{5} = 30 ( x-25)$
• $2x = 3 (x-25)$
• $2x = 3x-75$
• $x = 75kg$

Hence, the weight of the original solution = $75kg$.