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    Question

    There are three numbers 'a', 'b' and 'c' (a < b < c)

    such that 'b' is equal to the average of 'a' and 'c'. 70% of 'c' is equal to 'a' and the difference between 'b' and 'c' is 21. Find the sum of all three numbers.
    A 338 Correct Answer Incorrect Answer
    B 420 Correct Answer Incorrect Answer
    C 310 Correct Answer Incorrect Answer
    D 357 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    ATQ, 'b' = (1/2) × (a + c) Or, 2b = a + c And, 0.7c = 'a' Or, c = 10:7 Let 'c' = '10d' and 'a' = '7d' So, 2b = 7d + 10d Or, 'b' = (17d/2) Now, c - b = 21 Or, 10d - (17d/2) = 21 Or, (3d/2) = 21 Or, 'd' = 14 Required sum = 7d + {(17/2) × 14} + (10 × 14) = 98 + 119 + 140 = 357

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