Question

A manufacturer has 600 litres of a 12% solution of acid. How many litres of 30% acid solution must be added to it so that acid content in the resulting mixture is more than 15% but less than 18%.

Answer 1

Hint: Assume the required amount of acid to be added be x. Hence form two linear equalities in x. Solve the linear inequations to get the range of values of x. Use the fact that if C and V are the concentration and volume of a solution, respectively, then the amount of solute is equal to CV.

Complete step-by-step answer:
Let the amount of acid to be added be x litres.
Initial concentration of acid = 12%.
Initial volume = 600 litres.
Hence the amount of solute present initially $=600\times {\displaystyle \frac{12}{100}}=72$ units.
Amount of solute added $={\displaystyle \frac{30}{100}}x=0.3x$ units.
Hence the final amount of solute present in the solution =0.3x
The final volume of the solution = 600+x.
Hence the final concentration of the solution $={\displaystyle \frac{72+0.3x}{600+x}}\times 100={\displaystyle \frac{7200+30x}{600+x}}$
Now we have
Final concentration is greater than 15%.
Hence $${\displaystyle \frac{7200+30x}{600+x}}>15$$
Since x>0, multiplying both sides by 600+x would not change the inequality
7200+30x>9000+15x
Subtracting 15x from both sides, we get
7200+15x>9000
Subtracting 7200 from both sides, we get
15x>1800
Dividing both sides by 15, we get
x>120
Also, we have
Final concentration is less than 18%.
Hence, we have
${\displaystyle \frac{7200+30x}{600+x}}<18$
Since x>0, multiplying both sides by 600+x would not change the inequality
We have
7200+30x<10800+18x
Subtracting 18x from both sides, we get
7200+12x<10800
Subtracting 7200 from both sides, we get
12x<3600
Dividing both sides by 12, we get
x<300
Hence we have 120Hence the manufacturer must add more than 120 litres but less than 300 litres to the solution.

Note: [1] Here any number between 120 and 300 is the solution of the given problem.
[2] In the question, we have assumed concentration to units of solute per unit volume of solutions.
[3] We have assumed the solute units are additive, i.e. a units +b units = a+b units.
[4] Example of such a concentration is weight/ volume percent.