Multiple choice

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

  1. 120 < x < 300

  2. 100 < x < 270

  3. 150 < x < 320

  4. 180 < x < 370
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A Correct answer
Explanation

Initial: 600L at 12% = 72L acid. Added: xL at 30% = 0.3xL acid. Total: (72 + 0.3x) / (600 + x). We need 0.15 < (72 + 0.3x)/(600 + x) < 0.18. Solving 0.15(600+x) < 72+0.3x gives 90 + 0.15x < 72 + 0.3x -> 18 < 0.15x -> x > 120. Solving (72+0.3x) < 0.18(600+x) gives 72 + 0.3x < 108 + 0.18x -> 0.12x < 36 -> x < 300.

AI explanation

Let x be the liters of the 30% acid solution added to the 600 liters of 12% acid solution. For the lower limit of 15% acid, the equation is 0.12(600) + 0.30x = 0.15(600 + x), which yields x = 120. For the upper limit of 18% acid, the equation is 0.12(600) + 0.30x = 0.18(600 + x), which yields x = 300. Therefore, the required volume of the 30% acid solution must be more than 120 liters but less than 300 liters.

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