If I have 1 1/2 ounces of a 60% juice, 40% water solution and 1 1/4 ounces of a 0% juice, 100% water solution, what would be the final percentage of juice in the 2 3/4 ounce solution?
2007-08-06 05:24:33 · 3 answers · asked by forshizzle992 2 in Science & Mathematics ➔ Mathematics
1st: .6*1.5 = .9 oz. of juice.
.4*1.5 = .6 oz. of water
2nd: 0*1.25 = 0 oz of juice
1*1.25 = 1.25 oz of water
Mix: 1.5 + 1.25 = 2.75 oz total
.9 oz. of juice/2.75 oz total = 0.32727 oz. of juice
.32727*100 = 32.7272% juice.
2007-08-06 05:32:24 · answer #1 · answered by miggitymaggz 5 · 1⤊ 0⤋
Amount of juice in the first solution:
1 1/2 * 60% = 3/2 * 6/10 = 18/20
Amount of juice in the second solution:
1 1/4 * 0% = 0
Answer: (18/20) / (11/4) = 56/220 = 25.(45)%
2007-08-06 05:30:37 · answer #2 · answered by Alexander 6 · 0⤊ 1⤋
1.5 ounces and 60 percent juice
1.5 * 0.6 = 0.9 ounces of juice
Since no new juice is added, the final percentage is:
(0.9/2.75)*100 = 32.73 percent
2007-08-06 05:33:11 · answer #3 · answered by Captain Mephisto 7 · 1⤊ 0⤋