# Misc 13 - Chapter 6 Class 11 Linear Inequalities (Term 2)

Last updated at Feb. 17, 2020 by Teachoo

Last updated at Feb. 17, 2020 by Teachoo

Transcript

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

Chapter 6 Class 11 Linear Inequalities (Term 2)

Serial order wise

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