Multiple choice

Find the average of the three original numbers, if three different numbers are chosen such that when each of the numbers is added to the average of the remaining two, it gives 35, 15 and 45 as result.

  1. 15.83

  2. 18.83

  3. 25.83

  4. 35.93

  5. None of these
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A Correct answer
Explanation

Let the numbers be x, y, z and their average be A = (x+y+z)/3. The equations are x + (y+z)/2 = 35, y + (x+z)/2 = 15, and z + (x+y)/2 = 45. Substituting (y+z) = 3A - x into the first gives x + (3A-x)/2 = 35, which simplifies to x + 3A = 70. Summing the three such equations gives 3(x+y+z) + 3A = 190, or 9A + 3A = 190, so 12A = 190, A = 15.833.

AI explanation

If the three numbers are a, b and c, the problem gives the equations a + (b+c)/2 = 35, b + (a+c)/2 = 15 and c + (a+b)/2 = 45. Multiplying each equation by 2 gives 2a + b + c = 70, a + 2b + c = 30 and a + b + 2c = 90. Adding all three new equations yields 4a + 4b + 4c = 190, which simplifies to a + b + c = 47.5. Dividing this sum by 3 gives the average of the three original numbers, which is 15.83.

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