Question

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?

Answer 1

Let x litres of water is required to be added.
Then, total mixture = (x + 1125) litres 
It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres.
This resulting mixture will contain more than 25% but less than 30% acid content. 
∴ 30% of (1125 + x) > 45% of 1125 
And, 25% of (1125 + x) < 45% of 1125 
30% of (1125 + x) > 45% of 1125 
\(⇒ \frac{30}{100}(1125+x) > \frac{45}{100 }×1125\)
⇒ 30(1125+x) > 45x1125
⇒ 30×1125 + 30x > 45x1125
⇒ 30x > 45x1125 - 30x1125
⇒30x > (45-30)×1125
⇒ x > 15×1125/30
⇒ x > 562.5
25% of (1125 + x) < 45% of 1125 
\(⇒ \frac{25}{100}(1125+x) < \frac{45}{100}×1125\)
⇒ 25 (1125+x) > 45×1125
⇒25×1125+25x > 45×1125
⇒25x > 45x1125 - 25×1125
⇒25x > (45-25)×1125
\(⇒ x >\frac{ 20×1125}{25}\)
⇒ x > 900
∴ 562.5 < x < 900 
Thus, the required number of litres of water that is to be added will have to be more than 562.5 but less than 900.

Answer 2

Let's add x liters of water to the 1125 liters of a 45% acid solution. 
The total quantity of the mixture will be (1125+x) liters. 
The total acid content in the initial 1125 liters of the mixture is 45% of 1125. 
Given that the resulting mixture must have an acid content between 25% and 30%. 
Multiplying by 100: 
\(28125 + 25x < 50625 < 33750 + 30x \)
So, \(x < \frac{50625}{28125} \ and \ x >\frac{ 50625}{33750 }\)
Thus, the water to be added should be more than \(562.5\) liters but less than 900 liters.