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 90 × 4 - 4 = 356 356 × 3 - 3 = 1065 1065 × 2 - 2 = 2128 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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