Multiple choice

Ages of three men are in the ratio 3:6:8. If the thrice the sum of ages of eldest and youngest brothers is eighteen less than the six times of age of the other brother. Find the average age of the brothers.

  1. 28

  2. 34

  3. 51

  4. 58.5
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B Correct answer
Explanation

Ages are 3x, 6x, 8x. Thrice the sum of eldest and youngest: 3(8x + 3x) = 33x. Six times the other brother: 6(6x) = 36x. Equation: 33x = 36x - 18 => 3x = 18 => x = 6. Ages are 18, 36, 48. Average = (18+36+48)/3 = 102/3 = 34.

AI explanation

Let the ages of the three men be 3x, 6x, and 8x. The problem states thrice the sum of the eldest and youngest is eighteen less than six times the middle brother's age, giving the equation 3(8x + 3x) = 6(6x) - 18. Simplifying 3(11x) = 36x - 18 yields 33x = 36x - 18, so x = 6. The ages are 18, 36, and 48, and their average is (18 + 36 + 48) / 3, which equals 34.

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