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If a + b + c = 2013 then find maximum possible value of √(3a+12) + √(3b+12) + √(3c+12) - 100
Let a = b = c = x
Then, x + x + x = 2013
3x = 2013
x = 671
Now substituting the values of a, b, c
= √(3(671)+12) + √(3(671)+12) + √(3(671)+12) - 100
= √(2025) + √(2025) + √(2025) - 100
= 45 + 45 + 45 - 100
= 35
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