A container holds 5 liters of a mixture that is 20% alcohol. How much pure alcohol must be added to make the mixture 50% alcohol?

Jawaban: 2.5 L

Initially, alcohol = 20% of 5 = 1 L. Let x be pure alcohol added. New amount = 1+x, total volume = 5+x. Need (1+x)/(5+x)=0.5 → 1+x = 2.5+0.5x → 0.5x = 1.5 → x = 3. But wait, 1+3=4, 5+3=8, 4/8=0.5. So x=3. But option D is 3 L. So answerIndex should be 3. I'll correct that.