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    Question

    Given below are three series I, II and III. Each series

    consist an odd one out number. Odd one out number in series I, II and III are 'P', 'Q' and 'R', respectively. You have to find the values of 'P','Q' and 'R' and establish a relation among them. I. 20, 100, 356, 1065, 2128 II. 224, 220, 195, 128, 10, -186 III. (208/3), (104/3), (104/3), 52, 108, 260
    A P = Q, Q > R Correct Answer Incorrect Answer
    B P < Q, Q > R Correct Answer Incorrect Answer
    C P = Q = R Correct Answer Incorrect Answer
    D P > Q, Q > R Correct Answer Incorrect Answer
    E P < Q, Q < R Correct Answer Incorrect Answer

    Solution

    For series I: 20 × 5 - 5 = 95 95 × 4 - 4 = 376 376 × 3 - 3 = 1125 1065 × 2 - 2 = 2248 So, P = 100 For series II: 224 - 22 = 220 220 - 52 = 195 195 - 82 = 131 131 - 112 = 10 10 - 142 = -186 So, Q = 128 For series III: (208/3) × 0.5 = (104/3) (104/3) × 1 = (104/3) (104/3) × 1.5 = 52 52 × 2 = 104 104 X 2.5 = 260 So, R = 108 Therefore, P < Q and Q > R

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