Question

90%acid solution (90%pure acid &10%water) and 97% acid solution are mixed to obtain 21 liters of 95% acid solution. How many liters of each solution are mixed?

Answer 1

Let 90% acid solution (90%pure acid &10%water) be the solution ‘A’ and 97% acid solution(97%pure acid & 3%water) be the solution ‘B’.

Let ‘X’ liters of the solution ‘A’ is mixed with ‘Y’ liters of the solution ‘B’ to obtain 21 liters of 95% acid solution(95%pure acid & 5%water).

Given that, X + Y = 21……………… Equation(1)

Quantity of acid present in ‘X’ liters of the solution ‘A’ = $90\%\mathrm{of}\text{'}\mathrm{X}\text{'}=\frac{90}{100}\times \text{'}\mathrm{X}\text{'}=\frac{90\mathrm{X}}{100}$

Quantity of acid present in ‘Y’ liters of the solution ‘B’ = $97\%\mathrm{of}\text{'}\mathrm{Y}\text{'}=\frac{97}{100}\times \text{'}\mathrm{Y}\text{'}=\frac{97\mathrm{Y}}{100}$

Quantity of acid present in 21 liters of the solution (A + B) = $95\%\mathrm{of}\text{'}21\text{'}=\frac{95}{100}\times \text{'}21\text{'}=\frac{95\times 21}{100}=\frac{1995}{100}$

As per question $\frac{90\mathrm{X}}{100}+\frac{97\mathrm{Y}}{100}=\frac{1995}{100}$

$90\mathrm{X}+97\mathrm{Y}=1995$…………………….Equation(2)

On Multiplying equation(1) by 90

$90\mathrm{X}+90\mathrm{Y}=1890$………………….Equation(3)

On Solving equations (1) and (3), we get X = 6 and Y = 15

Hence, ‘6’ liters of 90% acid solution are mixed with ‘15’ liters of 97% acid solution to obtain 21 liters of 95% acid solution.