# Misc 13 - Chapter 5 Class 11 Linear Inequalities

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1

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Misc 3

Misc 4 Important

Misc 5 Important

Misc 6

Misc 7 Important Deleted for CBSE Board 2025 Exams

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Misc 11

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Misc 14

Chapter 5 Class 11 Linear Inequalities

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Last updated at April 16, 2024 by Teachoo

Misc 13 How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content? Volume of existing solution = 1125 litres Amount of acid in it = 45% of 1125 Hence, amount of water in it = 55% of 1125 = 55/100 × 1125 Let amount of water added be x litres So, Volume of new solution = 1125 + x Now, given that the new solution will contain more than 25% acid content Hence, 25 % of (1125 + x) = 506.25 25 % × (1125 + x) = 506.25 25/100 × (1125 + x) = 506.25 1/4 × (1125 + x) = 506.25 x = 1687.5 – 1125 x = 562.5 Hence, acid content should be between 25% and 30% i.e. amount of water added should be between 900 & 562.5 litres i.e. 562.5 < x < 900