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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?


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Solution

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%of'X'=90100×'X'=90X100

Quantity of acid present in ‘Y’ liters of the solution ‘B’ = 97%of'Y'=97100×'Y'=97Y100

Quantity of acid present in 21 liters of the solution (A + B) = 95%of'21'=95100×'21'=95×21100=1995100

As per question 90X100+97Y100=1995100

90X+97Y=1995…………………….Equation(2)

On Multiplying equation(1) by 90

90X+90Y=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.


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