# How many liters of water will have to be added to $1125$ liters of the $45\%$ solution of acid so that the resulting mixture will contain more than $25 \%$ but less than $30\%$ acid content ?

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
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