If the wrong number of the series i ii and iii are represented as P Q and R respectively then which ...........

Quantitative Aptitude(IBPS PO) Number series

Given below there are three number series (I, II, and III). One of the numbers in each series (I, II, and III) is wrong or (odd one out).

Series I: 590, 615, 651, 711, 795, 927

Series II: 64, 164, 285, 510, 546, 770

Series III: 18, 25.5, 40.5, 63, 93, 131

If the wrong number of the series (i), (ii), and (iii) are represented as (P, Q, and R) respectively, then which of the following equations is/are correct?

This questions was previously asked in

IBPS PO Mains (26th November 2022)

View All IBPSPO Previous Year Papers

Q - (P + R) = 40

P < Q > R

(R + P) < Q

Both 1 and 3

Both 2 and 3

छोटे Courses बड़े Results

Master Number series

Prepare through Micro-courses

Try Now

Explanation:

Correct option:5
Solution:
Series I: 590, 615, 651, 711, 795, 927
Logic used: 591 + (12 x 2) = 615
615 + (12 x 3) = 651
651 + (12 x 5) = 711
711 + (12 x 7) = 795
795 + (12 x 11) = 927
The wrong number in the series is (590).

Series II: 64, 164, 285, 510, 546, 770
Logic used: 64 +(6 +4)²= 164
164+(1+6+4)²=285
285 +(2 +8 + 5)²= 510
510 + (5 + 1 + 0)² = 546
546 +(5 + 4 +6)²= 771
The wrong number in the series is (770).

Series III: 18, 25.5, 40.5, 63, 93, 131
18 + 7.5 = 25.5
25.5 + 15 = 40.5
40.5 + 22.5 = 63
63 + 30 = 93
93 + 37.5 = 130.5
The wrong number in the series is (131).
Value of P, Q, and R are 590, 771, and 131 respectively.
In option (1): Q - (P + R) = 40 771 - (590 + 131) = 40 771 - 721 = 50 So, option (1) is incorrect
In option (2): P < Q > R 590 < 771 > 131 So, option 2 is correct.
In option (3): (ft + P) < Q (131 + 590) < 771 721 < 771 So, option 3 is correct.
Hence, the correct answer is both 2 and 3.