Book: RD Sharma - Mathematics
Chapter: 15. Linear Inequations
Subject: Maths - Class 11th
Q. No. 11 of Exercise 15.4
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A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If there are 640 liters of the 8% solution, how many liters of 2% solution will have to be added?
Volume of the 8% solution = 640 litres
Boric acid present in the 8% solution = 8% of 640 …(i)
And the rest 92% of 640 litres is water in the 8% solution.
Let volume of 2% solution added to 640 liters be x.
Boric acid present in 2% solution = 2% of x …(ii)
New volume of 8% solution = 640 + x …(iii)
Boric acid present in the new solution (that is, after adding x litres of 2% solution to 8% solution) = Boric acid present in the 8% solution + Boric acid present in the 2% solution [from (i) & (ii)]
⇒ Boric acid present in the new solution = 8% of 640 + 2% of x
⇒Boric acid present in the new solution = 
⇒Boric acid present in the new solution = 
…(iv)
According to the question,
The resulting mixture is to be more than 4% but less than 6% boric acid.
That is, the boric acid content in the resulting mixture must be more than 4% but less than 6% boric acid.
So, first let us take boric acid content in the resulting mixture to be more than 4%.
⇒ Boric acid present in the new solution > 4% of the new volume of 8% solution
[from (iii) & (iv)]
⇒ 2x + 5120 > 2560 + 4x
⇒ 5120 – 2560 > 4x – 2x
⇒ 2560 > 2x
⇒ 2x < 2560
⇒ x < 1280
Now, let us the take boric acid in the resulting mixture to be less than 6%.
⇒ Boric acid present in the new solution < 6% of the new volume of 8% solution
[from (iii) & (iv)]
⇒ 2x + 5120 < 3840 + 6x
⇒ 5120 – 3840 < 6x – 2x
⇒ 1280 < 4x
⇒ 4x > 1280
⇒ x > 320
We have
x < 1280 & x > 320
⇒ 320 < x < 1280
Hence, the required liters of 2% solution to be added to 8% of the solution is between 320 liters and 1280 liters.
