Correct Answer - Option 2 : 2 : 3

**Calculation:**

**Let volume of alcohol solutions are V**_{1 }and V_{2 }respectively.

⇒ Volume of alcohol in the resulting mixture = 15% of V_{1 }+ 20% of V_{2}

⇒ ((15/100) × V_{1}) + ((20/100) × V_{2})

⇒ (15V_{1}+ 20V_{2 })/100

which is also equal to 18% of (V_{1 }+ V_{2}).

⇒ (15V1+ 20V2 )/100 = (18 × (V1 + V2))/100

⇒ **2V**_{2 }= 3V_{1}

**The ratio of volume is 2 : 3 .**

If a mixture of Q1 quantity C1% content is mixing with Q2 quantity C2% mixture, and the concentration of the new mixture is Cnew% , then the following equation can be used

C1% of Q1 + C2% of Q2 = Cnew% of (Q1 + Q2)

⇒ C1Q1 + C2Q2 = Cnew(Q1 + Q2)