Mathematical Reasoning JEE Main previous year questions with solutions

5 solved JEE Main questions on Mathematical Reasoning, free to read — no sign-in needed. The full chapter has 135 questions; sign in to attempt the remaining 130 in the exam simulator.

1. Q1JEE Main 2023 (13 Apr, Shift 1)
   The negation of the statement $((A∧(B∨C))\Rightarrow (A∨B))\Rightarrow A$ is
   1. A.equivalent to $~C$
   2. B.equivalent to $B∨~C$
   3. C.a fallacy
   4. D.equivalent to $~A$
   Show answer & solution

   Answer: (D)

   Given, $((A∧(B∨C))\Rightarrow (A∨B))\Rightarrow A$ We know that, $X\Rightarrow Y\equiv ~X∨Y$, Now using the above formula we get, $\equiv \left(~(A∧(B∨C))∨(A∨B)\right)\Rightarrow A$ $\equiv ~(~(A∧(B∨C))∨(A∨B))∨A$ $\equiv (A∧(B∨C))∧~(A∨B)∨A$ Now to solve $(A∧(B∨C))∧~(A∨B)$ let us assume a universal set $\left\{1,2,3,4,5,6,7,8\right\}$, set $A=\left\{1,2,3\right\}$, set $B=\left\{3,4,5\right\}$ and set $C=\left\{5,6,7\right\}$ So, $A∧\left(B∨C\right)$ will be $\left\{3\right\}$ and $~(A∨B)$ will be $\left\{6,7,8\right\}$, So, $(A∧(B∨C))∧~(A∨B)=\emptyset$ $\equiv F∨A$ $\equiv A$ $∴$ Negation of statement$=~A$
2. Q2JEE Main 2023 (12 Apr, Shift 1)
   Among the two statements $\left({S}_{1}\right):\left(p\Rightarrow q\right)∧\left(p∧\left(~q\right)\right)$ is a contradiction and $\left({S}_{2}\right):\left(p∧q\right)∨\left(\left(~p\right)∧q\right)∨\left(p∧\left(~q\right)\right)∨\left(\left(~p\right)∧\left(~q\right)\right)$ is a tautology
   1. A.only $({S}_{2})$ is true
   2. B.only $({S}_{1})$ is true
   3. C.both are false
   4. D.both are true
   Show answer & solution

   Answer: (D)

   Given, ${S}_{1}:\left(p\Rightarrow q\right)∧\left(p∧~q\right)$ $\equiv \left(~p∨q\right)∧\left(p∧~q\right)$ $\left\{\text{as}A\Rightarrow B\equiv ~A∨B\right\}$ $\equiv \left(~p∧\left(p∧~q\right)\right)∨\left(q∧\left(p∧~q\right)\right)$ $\equiv \left(f∧~q\right)∨\left(f∧p\right)$ $\equiv f∨f\equiv f$ $\left({S}_{2}\right):\left(p∧q\right)∨\left(\left(~p\right)∧q\right)∨\left(p∧\left(~q\right)\right)∨\left(\left(~p\right)∧\left(~q\right)\right)$ $\equiv \left(p∧q\right)∨\left(~p∧q\right)∨\left(p∧~q\right)∨\left(~p∧~q\right)$ $\equiv \left(\left(p∨~p\right)∧q\right)∨\left(\left(p∨~p\right)∧~q\right)$ $\equiv \left(t∧q\right)∨\left(t∧~q\right)\equiv q∨~q\equiv t$ Hence, both statement are true.
3. Q3JEE Main 2023 (11 Apr, Shift 2)
   The converse of $\left(\left(~p\right)∧q\right)\Rightarrow r$ is
   1. A.$\left(\left(~p\right)∨q\right)\Rightarrow r$
   2. B.$\left(~r\right)\Rightarrow p∧q$
   3. C.$\left(~r\right)\Rightarrow \left(\left(~p\right)∧q\right)$
   4. D.$\left(p∨\left(~q\right)\right)\Rightarrow \left(~r\right)$
   Show answer & solution

   Answer: (D)

   We know that, Converse of $A\Rightarrow B$ is $B\Rightarrow A$ So, using the above formula we get, Converse of $\left(~p∧q\right)\Rightarrow r$ $\equiv r\Rightarrow (~p∧q)$ $\equiv \left(~r\right)∨\left(~p∧q\right)$ $\equiv \left(~p∧q\right)∨\left(~r\right)$ $\equiv ~\left(p∨~q\right)∨\left(~r\right)\left\{\text{as}~\left(A∨B\right)\equiv ~A∧~B\right\}$ $\equiv \left(p∨\left(~q\right)\right)\Rightarrow \left(~r\right)$
4. Q4JEE Main 2023 (11 Apr, Shift 1)
   The number of ordered triplets of the truth values of $p,q$ and $r$ such that the truth value of the statement $\left(p∨q\right)∧\left(p∨r\right)\Rightarrow \left(q∨r\right)$ is True, is equal to
   Show answer & solution

   Answer: 7

   Given, Expression $\left(p∨q\right)∧\left(p∨r\right)\Rightarrow \left(q∨r\right)$ Now by truth table we get, $p$ $q$ $r$ $p∨q$ $p∨r$ $\left(p∨q\right)∧\left(p∨r\right)$ $q∨r$ $\left(p∨q\right)∧\left(q∨r\right)\Rightarrow \left(q∨r\right)$ T T T T T T T T T T F T T T T T T F T T T T T T T F F T T T F F F T T T T T T T F T F T F F T T F F T F T F T T F F F F F F F T Now from above truth table we can say that there are total $7$ triplet are possible.
5. Q5JEE Main 2023 (10 Apr, Shift 2)
   The statement $~\left[p∨\left(~\left(p∧q\right)\right)\right]$ is equivalent to
   1. A.$~\left(p∧q\right)$
   2. B.$\left(p∧q\right)∧\left(~p\right)$
   3. C.$\left(~\left(p∧q\right)\right)∧q$
   4. D.$~\left(p∨q\right)$
   Show answer & solution

   Answer: (B)

   Given, $~\left[p∨\left(~\left(p∧q\right)\right)\right]$ Now we know that, By demorgan law, $~\left(A∨B\right)\equiv ~A∧~B$ Now using above formula we get, $~\left[p∨\left(~\left(p∧q\right)\right)\right]$ $\equiv ~p∧\left(p∧q\right)$ $\left[\text{as}~\left(~A\right)\equiv A\right]$ $\equiv \left(p∧q\right)∧\left(~p\right)$