Question

a solution of 165 litres contain 80% of acid and the rest of water how much water must be added to the above solutions such that the resulting mixture contain 25% of water

Answer 1

(80/100)×165=132

water=33 litres

(25/100)×165
=41.25

41.25-33=8.25

Answer 2

Answer: 11 liters of water must be added to the above solutions such that the  resulting mixture contain 25% of water.

Step-by-step explanation:

Since we have given that

Total quantity of a solution = 165 liters

Percentage of acid = 80%

Quantity of acid is given by

$\frac{80}{100}\times 165\\\\=132\ ltr$

Percentage of  water = 20%

Quantity of water is given by

$\frac{20}{100}\times 165\\\\=33\ ltrs$

Amount of water must be added be 'x', so that the resulting mixture contain 25% of water.

According to question, it becomes,

$\frac{33+x}{165+x}=\frac{25}{100}\\\\\frac{33+x}{165+x}=\frac{1}{4}\\\\4(33+x)=165+x\\\\132+4x=165+x\\\\4x-x=165-132\\\\3x=33\\\\x=\frac{33}{3}=11\ ltrs$

Hence, 11 liters of water must be added to the above solutions such that the  resulting mixture contain 25% of water.