Let x = amount (liters) of 68% solution

then

460-x = amount (liters) of 22% solution

.68x + .22(460-x) = .61(460)

.68x + 101.2-.22x = 280.6

.68x - .22x = 179.4

.46x = 179.4

x = 390 liters

Say A liters of 68% alcohol solution must be used.

So it will have a total alcohol contents of Ax68/100=68xA/100 liters

Therefore (460-A) liters of 22% alcohol is used with total alcohol contents of (460-A)x22/100 liters

Total alcohol contents in 460 liter of 68% alcohol solution is

68xA/100+22x(460-A)/100=(68A+22x460-22A)/100=(46A+10120)/100 liters....(1)

61% alcohol content in 460 liters = 460x61/100 liters of alcohol....(2)

Since alcohol contents from equation (1) and (2) must equal

460x61/100 = (46A+10120)/100

28060/100=(46A+10120)/100.....multiply both sides by 100

100x28060/100=100x(46A+10120)/100

28060=46A+10120

28060-10120=46A

17940=46A...divide both sides by 46

A=17940/46=390

So 390 liters of 68% solution must be used.

Let's call the quantity of 68% solution x and the quantity of 22% solution y.

We know that x + y = 460 liters

Now we have to consider the concentrations of the solutions. We do this by taking the concentration and multiplying by the volume:

.68 (x) + .22 (y) = .61 (460)

From the first equation, we can solve for y: y = 460 - x
and replace that in for y in the second equation:
.68(x) + .22 (460 - x) = .61 (460)

Simplify and combine like terms:

.68x + 101.2 - .22x = 280.6

.46x = 179.4

x = 390 liters

So the answer is C