Question

90%and97% pure acid Sol.. is mixed together to obtain 21 liters 95%pure solution. howany liters of each solution are need.?

Answer 1

let's take x litres from 90% pure and (21-x) litres from 97% pure solution. then
$x \times \frac{90}{100} + (21 - x) \frac{97}{100} = 21 \times \frac{95}{100} \\ 90x + 2037 - 97x = 1995 \\ 7x = 42 \\ x = 6$
that is take 6 litres from 90%pure and 15 litres from 97% pure solutions

Answer 2

Step-by-step explanation:

\huge {\mathcal{\purple{H}\green{e}\pink{y}\blue{!}}}Hey!

Let the given solutions be A and B respectively.

Let X litres of A be mixed with Y litres of B .

Then,

X + Y = 21 ---------(1)

★Quantity of acid in X litres of A = (90% of X ) Litres

=> ( 90/100 × X ) Litres

=> 9X / 10 Litres

★ Quantity of acid in Y litres of B = (97% of Y ) litres

=> ( 97/100 × Y ) Litres

=> (97Y/100) litres

★ Quantity of acid in 21 litres of (A + B) = ( 95% of 21 litres )

=> ( 95/100 × 25) litres

=> ( 399/20 ) litres

Therefore,

9X / 10 + 97Y/100 = 399/20

=> 90X + 97Y = 1995 --------(2)

Multiply equation (1) by 90 we get,

90X + 90Y = 1890 -----------(3)

Subtract equation (3) from equation (2) we get,

90X + 97Y = 1995

90X + 90Y = 1890

----------------------------

7Y = 105.

Y = 105/7 = 15

Putting the value of Y in equation (1)

X + Y = 21

X + 15 = 21

X = 21-15

X = 6

So,

6 litres of 90% solution is mixed with 15 litres of 97% solution.

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Thank you so much for the question