Question 264388
This is a mixture problem. we can set up a table based on given information
liquid . . . . . . % . . . . . . L . . . . . . .%L
orange. . . . . .60 . . . . . x . . . . . . .60x
water. . . . . . . 0 . . . . . . y. . . . . . . 0y 
mixture . . . . . 40 . . . . . x+y. . . . .40x + 40y
Looking at the third column, we get
 {{{60x = 40x + 40y}}} 
or
{{{20x = 40y}}}
dividing, we get
{{{x = 2y}}}
This means what ever the amount of orange juice is, the amount of water must be twice that.
we don't have enough information to answer with actual hard numbers.