A manufacturer has 200 liters of acid solution which has 15% acid content. How many liters of acid solution with 30% acid content may be added so that acid content in the resulting mixture will be more than 20% but less than 25% ?
Answers
Let x liters of 30% acid solution be added to 600 liters of 12% solution of acid. Then,
Total quantity of mixture = (600 + x) liters
Total acid content in the (600 + x) liters of mixture
= 30x/100 + 12/100 X 600
It is given that acid content in the resulting mixture must be more than 15% and less than 18%.
Therefore, 15% of (600 + x) < (30x/100 + 12/100 X 600) < 18% of (600 + x)
=> 15/100 X (600 + x) < 30x/100 + 12/100 X 600 < 18/100 X (600 + x)
=> 15(600 + x) < 30x + 12 X 600 < 18(600 + x) [Multiplying throughout by 100]
=> 9000 + 15x < 30x + 7200 < 10800 + 18x
=> 9000 + 15x < 30x + 7200 and 30x + 7200 < 10800 + 18x
=> 9000 – 7200 < 30x – 15x and 30x – 18x < 10800 – 7200
=> 1800 < 15x and 12x < 3600
=> 15x > 1800 and 12x < 3600
=> x > 120 and x < 300
Hence, the number of liters of the 30% solution of acid must be more than 120 but less than 300.
Proportion of Acid content in the resulting mixture = (30+0.3x)/(200+x)
20/100 < (30+0.3x)/(200+x) < 25/100
=> 1/5 < (30+0.3x)/(200+x) < ¼
We split this inequality into two parts.
1/5 < (30+0.3x)/(200+x) => 200+x < 150+1.5x
=> 50 < 0.5x => x>100 ---(1)
(30+0.3x)/(200+x) < ¼ => 120+1.2x < 200+x
=> 0.2x < 80 => x < 80/0.2 => x < 400 ---(2)
From (1) and (2), 100 < x < 400