ibps-rrb-os1 2017 Β· Question 41 of 74
Average
Let a consecutive odd numbers = π₯ β 2, π₯ and π₯ + 2 and consecutive even numbers = π¦ β 2, π¦, π¦ + 2 So, π¦ β 2 = 9 + π₯ + 2 π¦ β π₯ = 13 β¦ (i) and (π₯)2 + 507 = (π¦)2 π¦2 β π₯2 = 507 (π₯ + π¦)(π¦ β π₯) = 507 (π₯ + π¦) = 507 13 β π₯ + π¦ = 39 β¦ (π) Solving (i) and (ii) π¦ = 26 and π₯ = 13 so smallest odd numbers = π₯ β 2 = 13 β 2 = 11
Source: IBPS RRB PO Prelims 2017 Memory Based Paper (Adda247) Β· memory-based
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