IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Concept:Use the given equations and alternating series patterns to find unknown variables, then calculate the 7th term of Series 2.Explanation:From Note III, $2x$ and $2y$ are roots of $d^2 - 18d + 32 = 0$.Solving: $d = \frac{18 \pm \sqrt{324 - 128}}{2} = \frac{18 \pm 14}{2}$, so $d = 16$ or $d = 2$.Thus $\{2x, 2y\} = \{16, 2\}$, giving $(x,y) = (1,8)$ or $(8,1)$.Since all elements are positive, $y - 7.5 > 0$, so $y = 8$ and $x = 1$.Series 1: $15, B, 18, (4x)^2, 21, y+11$ becomes $15, B, 18, 16, 21, 19$.Odd positions: $15, 18, 21$ increase by 3.Even positions: $B, 16, 19$ also increase by 3, so $B + 3 = 16$, giving $B = 13$.Series 2: $A, 22, (A-B+23), 32, (A+B+7), 42$.Substitute $B = 13$: $A, 22, A+10, 32, A+20, 42$.Odd positions: $A, A+10, A+20$; even positions: $22, 32, 42$.So the 7th term will be $A + 30$.Now determine $A$. Given $A, B, Z$ are distinct primes less than 20 and their sum is even.Since $B = 13$ is odd, $A + Z$ must be odd, so one of them is the even prime $2$.Series 3: $(y-7.5), x, 28, 4, Z^3, 16$ becomes $0.5, 1, 28, 4, Z^3, 16$.Testing $Z = 2$ gives $Z^3 = 8$, which fits the series well, so $Z = 2$.Hence $A$ cannot be 2, so $A$ is an odd prime from $\{3, 5, 7, 11, 17, 19\}$.Then $A + 30$ can be $33, 35, 37, 41, 47, 49$.Among the options, $47$ is possible.Answer: C. 47