Given :
Volume of the existing solution = 1125 liters
Amount of acid in the existing solution = 45% of 1125 …(i)
And the rest 55% of 1125 liters is the amount of water in it, which need not be computed.
Let the water added (in liters) be x in 1125 liters of solution.
According to the question,
x liters of water has to be added to 1125 liters of the 45% solution.
We can say that,
Even if x liters of water is added to the 1125 liters of solution, acid content will not change.
Only water content and the whole volume of the solution will get affected.
So, the resulted solution will have acid content as follows :
The acid content in the solution after adding x liters of water = 45% of 1125 …(ii)
[∵ We know that the amount of acid content will not change after adding water to the whole solution. So, from equation (i), we have this conclusion]
Also,
According to the question,
This resulting mixture will contain more than 25% acid content.
So,
We have,
Acid content in the solution after adding x litres of water > 25% of new mixture
⇒ 45% of 1125 > 25% of (1125 + x)
[∵ from equation (ii)]
⇒ \(\frac{45}{100}\) \(\times\) 1125 > \(\frac{25}{100}\) \(\times\) (1125 + x)
⇒ 45 × 1125 > 25(1125 + x)
⇒ 9 × 1125 > 5(1125 + x)
⇒ 9 × 225 > 1125 + x
⇒ 2025 > 1125 + x
⇒ x < 2025 – 1125
⇒ x < 900
Also,
This resulting mixture will contain less than 30% acid content.
So,
We have Acid content in the solution after adding x litres of water < 30% of new mixture
⇒ 45% of 1125 < 30% of (1125 + x)
[∵ from equation (ii)]
⇒ \(\frac{45}{100}\)\(\times\)1125 >\(\frac{30}{100}\) \(\times\)(1125 + x)
⇒ 45 × 1125 < 30(1125 + x)
⇒ 9 × 1125 < 6(1125 + x)
⇒ 3 × 1125 < 2(1125 + x)
⇒ 3375 < 2250 + 2x
⇒ 2x + 2250 > 3375
⇒ 2x > 3375 – 2250
⇒ 2x > 1125
⇒ x > \(\frac{1125}{2}\)
⇒ x > 562.5
Thus,
We have got,
x < 900 and x > 562.5.
⇒ 562.5 < x < 900
Hence,
The required liters of water to be added to 1125 liters of solution is between 562.5 liters and 900 liters.
