Question
A mixture is prepared by combining three ingredients M,
N, and O in the ratio of 5:4:3. Ingredient M contains 60% sugar, N contains 50% sugar, and O contains 40% sugar. Find the percentage of sugar in the mixture.Solution
Let the volumes of M, N, and O be 5x, 4x, and 3x liters respectively. Total volume = 5x + 4x + 3x = 12x liters Sugar in M = 5x × 0.60 = 3.00x liters Sugar in N = 4x × 0.50 = 2.00x liters Sugar in O = 3x × 0.40 = 1.20x liters Total sugar = 3.00x + 2.00x + 1.20x = 6.20x liters Percentage of sugar = (6.20x / 12x) × 100 = 51.67% The percentage of sugar in the mixture is approximately 52%. Correct Option: b
What value should come in place of (?) in the given expression.
450 ÷ 9 + 75% of 160 − 64 ÷ 4 = ?
2/5 of 3/4 of 7/9 of 14400 = ?
7/3 of 4/5 of 15/56 of ? = 83
[564 + 32 of 18 × 9 ÷ 12 + 162 ] ÷ 4 = ?
What should come in place of (?) question mark in the given expression.
(25% of 320) + (3/8 of 400) − 30 = ?
45% of 360 - 160 + ? = √324
187 ÷ 5 ÷ 0.4 = ? – 24 × 2.4
Find the simplified value of the following expression:
[{12 + (13 × 4 ÷ 2 ÷ 2) × 5 – 8} + 13 of 8]
18 × 15 + 86 – 58 =? + 38
√(24²+285-8²-172) = ?²
