There are three numbers 'm', 'n' and 'o' (m < n < o) such that 'n' is equal to the average of 'm' and 'o'. 60% of 'o' is equal to 'm' and the difference between 'n' and 'o' is 18. Find the sum of all three numbers.
ATQ, 'n' = (1/2) × (m + o) Or, 2n = m + o And, 0.6o = 'm' Or, o = 5:3 Let 'o' = '5c' and 'm' = '3c' So, 2n = 3c + 5c Or, 'n' = (8c/2) Now, o - n = 18 Or, 5c - (8c/2) = 18 Or, (2c/2) = 18 Or, 'c' = 9 Required sum = 3c + {(8/2) × 9} + (5 × 9) = 27 + 36 + 45 = 108
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