IBPS PO Mains 12 Oct 2025 Shift 1 Paper

Concept:Determine the values of $k$, $p$, and $s$, then compute the series elements and compare the three quantities.Explanation:The smallest two-digit perfect square is $16$, so $k = \sqrt{16} = 4$.The two-digit prime numbers are $11, 13, 17, \dots$, so the second smallest is $p = 13$.Since $2^5 = 32 < 40$ and $2^6 = 64 > 40$, the maximum natural number is $s = 5$.Now compute the series elements:First element $= k + 1 = 4 + 1 = 5$.Second element $= 2^s - 2 = 2^5 - 2 = 32 - 2 = 30$.Third element $= p^2 - 19 = 13^2 - 19 = 169 - 19 = 150$.Fourth element $= 120(k + 1) = 120 \times 5 = 600$.The pattern is $\times 6, \times 5, \times 4, \times 3$: $5 \times 6 = 30$, $30 \times 5 = 150$, $150 \times 4 = 600$, and $600 \times 3 = 1800$.Hence, fifth element $C = 1800$.Now compare the quantities:Quantity 1 $= 4 \times 150 = 600$.Quantity 2 $= \frac{1800}{3} = 600$.Quantity 3 $= k \times s^3 + 100 = 4 \times 5^3 + 100 = 4 \times 125 + 100 = 600$.All three quantities are equal.Answer:Quantity 1 $=$ Quantity 2 $=$ Quantity 3.Correct option: E. $Quantity 1 = Quantity 2 = Quantity 3$.