Question

A manufacturer has 600 litres of a 12-/- solution of acid . how many litres of a 30-/- acid solution must be added to it , so that the acid content in the resulting mixture will be more than 15-/-. But less than 18-/-?
PLEASE HELP ME FRIENDS.....

Answer 1

In 600 liter acid=72 liter. Adding x litre of 30% acid means adding acid content of 0.3x. New num=72+0.3x. new den=600+x. So (72+0.3x)/(600+x)=0.15 gives x=120 liter. similarly for making it 18% we add 300 liter of 30% acid.

Answer 2

AnswEr:

Let x litres of 30% acid solution be added to 600 litres of 12% solution of acid. Then,

$\tt \: Total \: quantity \: of \: mixture = (600 + x) \: litres$

• Total acid content in the (600+x) litres of mixture -

$\rightarrow \tt \frac{30x}{100} + \frac{12}{100} \times 600$

• It is given that acid content in the resulting mixture must be more than 15% and less than 18%.

$\therefore \tt \: 15\% \: of \: (600 + x) < ( \frac{30x}{100} + \frac{12}{100} \times 600) \: 18\% \: of \: (600 + x) \\ \\ \rightarrow \tt \frac{15}{100} \times (600 + x) < \frac{30x}{100} + \frac{12}{100} \times 600 < \\ \tt \frac{18}{100} \times (600 + x) \\ \\ \rightarrow \tt15(600 + x) < 30x + 12 \times 600 < 18(600 + x) \\ \\ \rightarrow \tt9000 + 15x < 30x + 7200 < 10800 + 18x \\ \\ \rightarrow \tt9000 + 15x < 30x + 7200 \: \: and \: \: \\ \tt \: 30x + 7200 < 10800 + 18x \\ \\ \rightarrow \tt \: 9000 - 7200 < 30x - 15x \: \: and \: \: \\ \tt \: 30x - 18x < 10800 - 7200 \\ \\ \rightarrow \tt1800 < 15x \: \: \: and \: \: \: 12x <3 600 \\ \\ \rightarrow \tt15x > 1800 \: \: \: and \: \: \: 12x < 3600 \\ \\ \rightarrow \tt \: x > 120 \: \: \: and \: \: \: x < 300 \\ \\ \rightarrow \tt120 < x < 300$

◕ Hence, the number of litres of the 30% solution of acid must be more than 120 but less than 300.