Let x be the liters of water added to 1125 liters of 45 % acid solution

total mixture $= (x+1125)$

The amount of acid content in the mixture is $45 \% \;of\;1125$

The resulting mixture will contain more than $25 \%$ acid content.

Hence $25\%$ of $(1125 +x) < 45 \%$ of $1125$

$=> \large\frac{25}{100} $$(1125 +x) < \large\frac{45}{100 } $$\times 1125$

Multiplying both sides by 100

$=> 25 (1125 +x) < 45 \times 1125$

$=> 25 x +25 \times 1125 < 45 \times 1125$

Subtracting $25 \times 1125$ on both sides.

=> $ 25 x < 45 \times 1125 -25 \times 1125$

=> $ 25 x < (45- 25) \times 1125$

=> $25 x < 20 \times 1125$

dividing by 25 on both sides

=> $ x < \large\frac{20 \times 1125}{25}$

=> $ x < 900$

The resulting solution must be less than $30 \%$ acid content.

$45 \%$ of $1125 < 30 \%$ of (1125+x)

$\large\frac{45}{100} $$ \times 1125 < \large\frac{30}{100} $$(1125 +x)$

Multiplying both sides by 100.

$45 \times 1125 - 30 \times 1125 < 30x$

Subtracting both sides by $30 \times 1125$

=> $ 45 \times 1125 -30 \times 1125 < 30 x$

=> $15 \times 1125 < 30x$

Dividing by 30 on both sides

$\large\frac{1125 \times 15}{30 } $$ < x$

$562.5 < x $------(2)

We see from (1) and (2) that x must be more than $562.5 \;litres$ and less than 900 litres

The solution set is $ 562.5 < x< 900$