Question

In what proportion should 10% ethanol be mixed with 65% ethanol to obtain 50% ethanol?

Answer 1

3 : 8

Explanation:

Let x quantity of 10% ethanol is mixed with y quantity of 65% ethanol to obtain 50% ethanol mixture,

Thus, the total quantity of resultant mixture = x + y

Also, ethanol in 10% ethanol mixture + ethanol in 65% ethanol mixture = ethanol in resultant mixture,

⇒ 10% of x + 65% of y = 50% of (x+y)

`\implies (10x)/(100)+(65y)/(100)=(50(x+y))/(100)`

⇒ 10x + 65y = 50(x+y)

⇒ 10x + 65y = 50x+50y

⇒ 10x - 50x = 50y - 65y

⇒ -40x = -15y

`\implies (x)/(y)=(15)/(40)=(3)/(8)`