IBPS PO Mains 12 Oct 2025 Shift 1 Paper

Concept:Check whether the Column I statement forces the Column II statement to be true. If the first statement is impossible, that pair is invalid.Explanation:Let Set $A = \{a,\ a+5,\ a+10,\ a+15,\ a+20,\ a+25\}$, where $a$ is a two-digit multiple of 5.Let Set $B = \{b,\ b+6,\ b+12\}$, where $b$ is a two-digit multiple of 6.For statement P, take the difference as B minus A.Smallest difference $= b - a$.Third-smallest difference $= (b+12) - (a+10) = b - a + 2$.Equality would give $b - a = b - a + 2$, which is impossible.So P is always false, and combinations I and IV are not correct.For Q, Set B contains 36, so $b \in \{24,\ 30,\ 36\}$.For X, second-smallest terms are equal: $a+5 = b+6 \Rightarrow a = b+1$.This works only for $b=24$, giving $a=25$. For $b=30$ or $b=36$, $a=31$ or $37$, which are not multiples of 5.Thus Q does not always force X, so II is not definite.For Y, sum of largest terms is 86: $(a+25)+(b+12)=86 \Rightarrow a+b=49$.If $b=30$, Q is true, but choosing $a=20$ makes Y false. So V is not definite.For R, $a+25 > b+12 \Rightarrow a - b > -13$.Take $a=10,\ b=12$; R is true, but X is false and Y is false. So III and VI are not definite.Thus no listed combination is definitely correct.Answer:Option E — None of these.