A chemist has three solutions of acid and water. The first solution has 40 percent acid, the second has 60 percent acid, and the third has 80 percent acid. He mixes 2 liters of the first solution, 3 liters of the second solution, and 5 liters of the third solution. What is the percentage of acid in the final mixture?
ATQ, Amount of acid in the first solution is 0.40 × 2 = 0.8 liters. Amount of acid in the second solution is 0.60 × 3 = 1.8 liters. Amount of acid in the third solution is 0.80 × 5 = 4 liters. Total amount of acid = 0.8 + 1.8 + 4 = 6.6 liters. Total volume of the mixture = 10 liters. Percentage of acid in the mixture = 6.6 / 10 × 100 = 66%
31.98% of 449.99 = ? - 194.97 + 59.98% of 124.99
(15.98% of 399.99) - 6.998 = √?
25.31% of 5199.90 + (19.9 × 17.11) + 46.021 =? + 168.98
11.89 × 2.10 × 4.98 × 4.03 ÷ 7.98 of 15.03 = ?
[(343) 1/3 ÷ {(12.001)2 × (1 ÷ (4.03 × 2.97) 2 )}] = ?
? = {29.7% of (97.72 × 40.04)} ÷ 3.92
45.22% of (71.9 x 5.01) + 69.97 =?
12.5% of 6400 + (17 × 25) = ?% of 2200+ 125
(6859.01)1/3 × 10.11 × 14.47 ÷ 20.32 = ? + 45.022
(168.24 ÷ 23.98) x 19.81 + ? = 176.33
