10% of salty water contained in the flask was poured out in the beaker. After this , a part of the water?

Question

contained in the beaker was vapourized by heating and due to this, the percentage of salt in the beaker increased m times. if it is known that after the content of the beaker was poured into the flask, the percentage of salt in beaker increased by x%, find the orignal % of salt in the flask.

ans is
(9m-1x%)/(m+1)

but how to solve ,
and in answer % sign is with x, or ? i dont know how to get.

Answer 1

Let's call y the original amount of salt in the flask, and v the original volume of liquid in the flask.

When 1/10 of the liquid is poured in the beaker, you have 0.9y salt and 0.9v liquid remaining in the flask. You have 0.1y and 0.1v in the beaker.

In increasing the salinity of the beaker m times, is done by dividing the liquid by m. At the end of this process, the beaker contains the same amount (0.1y) of salt, but 0.1v/m liquid.

When this is poured back into the flask, you add the beaker's and flask's contents: 0.9y+0.1y salt (0.9y + 0.1y = y), and 0.1v/m liquid (0.9v + 0.1v/m = (0.9m + 0.1)v/m.

Now, this represents an increase of x% in salinity:

new_salinity - old_salinity = x/100
(new_salt / new_liquid) - (old_salt / old_liquid) = x/100
( y / [(0.9m + 0.1)v/m]) - (y/v) = x/100

Now we solve for y/v, the original concentration:

( y / [(0.9m + 0.1)v/m]) - (y/v) = x/100
(y/v) (m/(0.9m + 0.1) - 1) = x/100
(y/v) ((m - 0.9m - 0.1)/(0.9m + 0.1)) = x/100
(y/v) ((0.1m - 0.1)/(0.9m + 0.1)) = x/100
(y/v) = (x/100) * (0.9m + 0.1)/(0.1m - 0.1)
(y/v) = (x/100) * (9m + 1)/(m - 1)

We'd multiply by 100 to convert the y/v value as a percentage, which cancels out the divisor of 100 on the right:

(y/v)% = x(9m + 1)/(m - 1)

That's not quite the same as your own equation, though.

Answer 2

you use x