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Example 20 - A manufacturer has 600 litres of a 12% acid

Example 20 - Chapter 6 Class 11 Linear Inequalities - Part 2
Example 20 - Chapter 6 Class 11 Linear Inequalities - Part 3 Example 20 - Chapter 6 Class 11 Linear Inequalities - Part 4


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Example 20 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 acid content in the resulting mixture will be more than 15% but less than 18%? Quantity of existing solution = 600 litres Amount of acid in existing solution = 600 × 12% = 600 × 12/100 = 72 litres Let the quantity of 30% acid solution to be added be x Amount of acid in the added solution = 30% of x = 30/100 × 𝑥 = 0.3x Given that, Amount of acid in total mixture should more than 15% i.e. Amount of acid in total mixture > 15% of total 72 + 0.3x > 15/100 (x + 600) 72 + 0.3x > 0.15(x + 600) 72 + 0.3x > 0.15x + 0.15 × 600 72 + 0.3x > 0.15x + 90 0.3x – 0.15x > 90 – 72 0.15x > 18 15/100 x > 18 x > 18 × 100/15 x > 120 Also, Given that, amount of acid in total mixture should be less than 18% i.e. Amount of acid in total mixture < 18% of total 72 + 0.3x < 18/100 (x + 600) 72 + 0.3x < 0.18(x + 600) 72 + 0.3x < 0.18x + 0.18 × 600 72 + 0.3x < 0.18x + 108 0.3x – 0.18x < 108 – 72 0.12x < 36 12/100 x < 36 x < 36 × 100/12 x < 300 Thus , x > 120 & x < 300 120 < x < 300 Thus, the number of litres of the 30% solution of acid will have to be more than 120 litres but less than 300 litres.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.