Mathematical Reasoning – JEE Main Previous Year Question with Solutions

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Q. Statement-1: $\sim(p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$

Statement-2 $: \sim(p \leftrightarrow \sim q)$ is a tautology.

(1) Statement–1 is true, Statement–2 is false.

(2) Statement–1 is false, Statement–2 is true.

(3) Statement–1 is true, Statement–2 is true ;

Statement–2 is a correct explanation for Statement–1.

(4) Statement–1 is true, Statement–2 is true ;

Statement–2 is not a correct explanation for statement–1.

[AIEEE-2009]

Sol. (1)

Q. Let $S$ be a non-empty subset of $R$. Consider the following statement:

$p:$ There is a rational number $x \in S$ such that $x>0$

which of the following statements is the negation of the statement p?

(1) There is a rational number $x \in$ S such that $x \leq 0$

(2) There is no rational number $x \in S$ such that $x \leq 0$

(3) Every rational number $x \in S$ satisfies $x \leq 0$

(4) $x \in S$ and $x \leq 0 \Rightarrow x$ is not rational.

[AIEEE-2010]

Sol. (3)

Given $\mathrm{S} \subseteq \mathrm{R}$ and

$\mathrm{p}=$ There is a rational number $\mathrm{x} \in \mathrm{S}$ such that $\mathrm{x}>0$

then $\sim \mathrm{p}:$ Any rational number $\mathrm{x} \in \mathrm{S}$ such that $\mathrm{x}$ $\not>$ 0

i.e. $\sim \mathrm{p}:$ Every rational number $\mathrm{x} \in \mathrm{S}$ satisfy $\mathrm{x} \leq 0$

Q. Consider the following statements

p : Suman is brilliant

q : Suman is rich

r : Suman is honest

The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as :-

( 1)$\sim \mathrm{q} \leftrightarrow \sim \mathrm{p} \wedge \mathrm{r}$

( 2)$\sim(\mathrm{p} \wedge \sim \mathrm{r}) \leftrightarrow \mathrm{q}$

( 3)$\sim \mathrm{p} \wedge(\mathrm{q} \leftrightarrow \sim \mathrm{r})$

( 4)$\sim(\mathrm{q} \leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$

[AIEEE-2011]

Sol. (2,4)

Given Statement :

$(\mathrm{p} \wedge \sim \mathrm{r}) \Leftrightarrow \mathrm{q}$

Negations of $\mathrm{p} \Leftrightarrow \mathrm{q}$ are

$\sim(\mathrm{p} \Leftrightarrow \mathrm{q}), \sim(\mathrm{q} \Leftrightarrow \mathrm{p})$

$\sim \mathrm{p} \Leftrightarrow \mathrm{q}$ and $\sim \mathrm{q} \Leftrightarrow \mathrm{p}$

Hence negations of given statement

are $\sim(\mathrm{q} \Leftrightarrow(\mathrm{p} \wedge \sim \mathrm{r}))$

and $\sim(\mathrm{p} \wedge \sim \mathrm{r}) \Leftrightarrow \mathrm{q}$

Q. The only statement among the followings that is a tautology is :

( 1) $\mathrm{q} \rightarrow[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})]$

(2) $\mathrm{p} \wedge(\mathrm{p} \vee \mathrm{q})$

(3) $\mathrm{p} \vee(\mathrm{p} \wedge \mathrm{q})$

(4) $[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$

[AIEEE-2011]

Sol. (4)

$[\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q})] \rightarrow \mathrm{q}$

$[\mathrm{p} \wedge(\sim \mathrm{p} \vee \mathrm{q})] \rightarrow \mathrm{q}$

$[(\mathrm{p} \wedge \sim \mathrm{p}) \vee(\mathrm{p} \wedge \mathrm{q})] \rightarrow \mathrm{q}$

$[\mathrm{c} \vee(\mathrm{p} \wedge \mathrm{q})] \rightarrow \mathrm{q}$

$\left\{\begin{array}{c}{\mathrm{p} \wedge \sim \mathrm{p} \equiv \mathrm{c} \equiv \mathrm{contradiction}} \\ {\because \mathrm{c} \vee \mathrm{p} \equiv \mathrm{p}}\end{array}\right.$

$\Rightarrow(\mathrm{p} \wedge \mathrm{q}) \rightarrow \mathrm{q}$

$\Rightarrow \sim(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{q}$

$\Rightarrow(\sim \mathrm{p} \vee \sim \mathrm{q}) \vee \mathrm{q}$

$\Rightarrow \sim \mathrm{p} \vee(\mathrm{q} \vee \sim \mathrm{q})$

$\Rightarrow \sim \mathrm{p} \vee(\mathrm{t}) \equiv$ tautology

Q. The negation of the statement

“If I become a teacher, then I will open a school”, is :

(1) I will not become a teacher or I will open a school.

(2) I will become a teacher and I will not open a school.

(3) Either I will not become a teacher or I will not open a school.

(4) Neither I will become a teacher nor I will open a school.

[AIEEE-2012]

Sol. (2)

Q. Consider :

Statement-I: $(\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\sim \mathrm{p} \wedge \mathrm{q})$ is a fallacy.

Statement-II : $(\mathrm{p} \rightarrow \mathrm{q}) \leftrightarrow(\sim \mathrm{q} \rightarrow \sim \mathrm{p})$ is a tuatology

(1) Statement-I is true, Statement-II is true; statement-II is a correct explanation for Statement-I.

(2) Statement-I is true, Statement-II is true; statement-II is not a correct explanation for Statement-I.

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

[JEE-MAINS-2013]

Sol. (2)

Given statement is $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$

As we know $\sim(\mathrm{p} \leftrightarrow \mathrm{q}) \equiv \sim \mathrm{p} \leftrightarrow \mathrm{q}$ or $\mathrm{p} \leftrightarrow \sim \mathrm{q}$

$\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$

Q. The statement $\sim(p \leftrightarrow \sim q)$ is :

(1) equivalent to $p \leftrightarrow q$

(2) equivalent to $\sim p \leftrightarrow q$

(3) a tautology

(4) a fallacy

[JEE(Main)-2014]

Sol. (1)

Given statement is $\sim(\mathrm{p} \leftrightarrow \sim \mathrm{q})$

As we know $\sim(\mathrm{p} \leftrightarrow \mathrm{q}) \equiv \sim \mathrm{p} \leftrightarrow \mathrm{q}$ or $\mathrm{p} \leftrightarrow \sim \mathrm{q}$

$\therefore \sim(\mathrm{p} \leftrightarrow \sim \mathrm{q}) \equiv \mathrm{p} \leftrightarrow \mathrm{q}$

Q. The negation of $\sim \mathrm{s} \vee(\sim \mathrm{r} \wedge \mathrm{s})$ is equivalent to :

( 1) $\mathrm{s} \vee(\mathrm{r} \vee \sim \mathrm{s})$

(2) $\mathrm{s} \wedge \mathrm{r}$

(3) $\mathrm{s} \wedge \sim \mathrm{r}$

( 4) $\mathrm{s} \wedge(\mathrm{r} \wedge \sim \mathrm{s})$

[JEE(Main)-2015]

Sol. (2)

$\square \mathrm{s} \vee(\square \mathrm{r} \wedge \mathrm{s})$

$(\square \mathrm{s} \vee \sim \mathrm{r}) \wedge(\square \mathrm{s} \wedge \mathrm{s})$

$(\square \mathrm{s} \vee \sim \mathrm{r}) \wedge \mathrm{t}$

$(\square \mathrm{s} \vee \sim \mathrm{r})$

$\sim(\square \mathrm{s} \vee \sim \mathrm{r})$

$\mathrm{s} \wedge \mathrm{r}$

Q. The Boolean Expression (p\wedge\simq) Vq\vee(\simp\wedgeq) is equivalent to :-

(1) $\mathrm{pv} \sim \mathrm{q}$

(2) $\sim \mathrm{p} \wedge \mathrm{q}$

(3) $\mathrm{p} \wedge \mathrm{q}$

(4) $\mathrm{p} \vee \mathrm{q}$

[JEE(Main)-2016]

Sol. (4)

Given boolean expression is

$(\mathrm{p} \wedge \sim \mathrm{q}) \vee \mathrm{q} \vee(\sim \mathrm{p} \wedge \mathrm{q})$

$(\mathrm{p} \wedge \sim \mathrm{q}) \mathrm{Vq}=(\mathrm{p} \vee \mathrm{q}) \wedge(\sim \mathrm{q} \vee \mathrm{q})=(\mathrm{p} \vee \mathrm{q}) \wedge \mathrm{t}=(\mathrm{p} \vee \mathrm{q})$

Now,

$(\mathrm{pVq}) \vee(\sim \mathrm{p} \wedge \mathrm{q})=\mathrm{p} \vee \mathrm{q}$

Q. The following statement $(\mathrm{p} \rightarrow \mathrm{q}) \rightarrow[(\sim \mathrm{p} \rightarrow \mathrm{q}) \rightarrow \mathrm{q}]$ is :

(1) a fallacy

(2) a tautology

(3) equivalent to $\sim \mathrm{p} \rightarrow \mathrm{q}$

(4) equivalent to $\mathrm{p} \rightarrow \sim \mathrm{q}$

[JEE(Main)-2017]

Sol. (2)

Q. The Boolean expression $\sim(p \vee q) \vee(\sim p \wedge q)$ is equivalent to :

(1) p (2) q (3) $\sim \mathrm{q}$ ( 4)$\sim \mathrm{p}$

[JEE(Main)-2018]

Sol. (4)

$\sim(p \vee q) \vee(\sim p \wedge q)$

$(\sim p \wedge \sim q) \vee(\sim p \wedge q)$

$\Rightarrow \sim p \wedge(\sim q \vee q)$

$\Rightarrow \sim p \wedge t \equiv \sim p$

• May 21, 2020 at 1:06 pm

Chill bro

• May 16, 2020 at 9:21 am

Excellent

• May 12, 2020 at 4:31 pm

nice questions

• April 9, 2020 at 3:03 pm

Very good problem

• April 9, 2020 at 1:18 pm