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, 260Solution
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
More Series Questions
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