Question

How many gallons of 30% alcohol solution
and how many gallons of 60% alcohol
solution must be mixed to produce 18
gallons of 50% solution?​

Answer 1

Step-by-step explanation:

please tell the answer

Answer 2

Solution:-

Assume,

x = gallons of 30% alcohol

x = gallons of 30% alcoholy = gallons of 60% alcohol

x gallons of 30% alcohol and y gallons of 60% alcohol solution must be mixed to produce 18 gallons of 50% solution.

x+y = 18 ......(1)

x+y = 18 ......(1)0.30x +0.60y = 0.50(x+y) ......(2)

From eqn (1) we get ,

x+y = 18

or, y= (18-x) ......(3)

if we put eqn(3) in the eqn(2) we get,

0.30x +0.60y = 0.50(x+y)

or, 0.30x +0.60(18-x) = 0.50(x+18-x)

or,0.30x+ 10.8-0.60x = 0.50*(18)

or, -0.30x + 10.8 = 9

or, -0.30x = 9 - 10.8

or, -0.30x = -1.8

or, x = (1.8/0.30) gallons

or, x = 6 gallons

if x = 6 gallons then the value of y is,

x + y = 18

or, 6 + y = 18

or, y = 18 - 6

or, y = 12 gallons

Hence, 6 gallons of a 30% alcohol solution and 12 gallons of a 60% alcohol solution must be mixed to produce 18 gallons of a 50% alcohol solution.(Ans)