MCQ Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 4

State the following statement is True or False

$p, q, r$ are statements with truth values T, F, T respectively, then the truth value of

$(\sim p\vee q)\wedge \sim r \rightarrow p$ , is T

0%
True
0%
False

Explanation

$p$	$q$	$r$	$\sim p\vee q$	$\left( \sim p\vee q \right) \wedge \sim r$	$\left( \sim p\vee q \right) \wedge \sim r\rightarrow p$
T	F	T	F	F	T

The contrapositive of

$(p \vee q)\Rightarrow r$ is

0%
$r \Rightarrow (p \vee q)$
0%
$\sim r \Rightarrow (p \vee q)$
0%
$\sim r \Rightarrow \sim p \wedge \sim q$
0%
$R \Rightarrow (q \vee r)$

Explanation

contrapositive of

$a \implies b$ is

$\sim b \implies \sim a$
contrapositive of

$p \lor q \implies r$ is

$\sim r \implies \sim ( p \lor q )$

$\sim r \implies (\sim p \land \sim q )$

Which of the following is/are a statement?

0%
Give me a glass of water.
0%
Asia is continent.
0%
The earth revolves round sun.
0%
The number $6$ has two prime factor $2,3$ .

State the following statement is True or False

$p, q, r$ are statements, with truth values T, F, T respectively, then the truth value of

$(\sim p \vee q) \wedge \sim r \Rightarrow p$ is T

0%
True
0%
False

Explanation

$p,q,r$ are statements with truth values

$T,F,T$ respectively

Consider

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

$=(F \vee F) \wedge F \Rightarrow T$

$=(F \wedge F) \Rightarrow T$

$=F \Rightarrow T=\sim(F) \vee T=T\vee T=T$
Clearly, option A.

$p\rightarrow (q \vee r)$ is false, then the truth values of

$p,q,r$ are respectively

0%
T, F, F
0%
F, F, F
0%
F, T, T
0%
T, T, F

Explanation

$p \rightarrow q$ is false only when

$p$ is true and

$q$ is false.

$\therefore p\rightarrow (q \vee r)$ is false when

$p$ is true and

$(q \vee r)$ is false, and

$q \vee r$ false when both

$q, r$ are false.

Hence T,F,F

$x = 5$ and

$y = -2$ , then

$x-2y = 9$ . The contrapositive of this statement is/are

0%
If $x-2y$ $\neq 9$ , then $x \neq$ $5$ or $y \neq 2$ .
0%
If $x-2y$ $\neq 9$ , then $x \neq 5$ and $y \neq -2$ .
0%
If $x-2y = 9$ , then $x = 5$ and $y = -2$ .
0%
none of these.

Explanation

Let

$p,q,r$ be the three statement.

$p : x = 5,\>q : y = 2,\>r : x - 2y = 9$
Here given statement is

$(p \wedge q) \rightarrow r$ and its contrapositive is

$\sim r \rightarrow \sim (p \wedge q)$
i.e.

$\sim r \rightarrow (\sim p \vee \sim q)$
i.e. If

$x - 2y \neq 9$ , then

$x \neq 5$ or

$y \neq 2$

The negation of the statement: "If I become a teacher, then I will open a school" is

0%
I will become a teacher and I will not open a school.
0%
Either I will not become a teacher or I will not open a school.
0%
Neither I will become a teacher nor I will open a school.
0%
I will not become a teacher or I will open a school.

Explanation

Let

$p:$ I become a teacher

$q:$ I will open a school

The given statement is

$p\rightarrow q=\left( \sim p \right) \vee q$

It negation is

$(~ \left( \text{~ p} \right) \vee q)=p\wedge \left( \text{~ q} \right)$

Thus negation of the given statement is 'I will become a teacher and I will not open school'.

$x = 5$ and

$y = 2$ then

$x - 2y = 9$ . The contrapositive of this statement is

0%
If $x -2y \neq 9$ then $x \neq 5$ or $y \neq 2$
0%
If $x -2y \neq 9$ then $x \neq 5$ and $y \neq - 2$
0%
If $x - 2y = 9$ then $x = 5$ and $y = -2$
0%
none of these.

Explanation

given three statements p:

$x=5$
q :

$y=2$
r:

$x+2y=9$

$(p \land q ) \implies r$ contrapositive of this is

$\sim r \implies \sim (p \land q)$

$\therefore x+2y \ne 9 \implies x \ne 5$ or

$y\ne 2$

Choose the conclusion of given statements:

All scientists working in America are talented. Some Indian scientists are working in America. Therefore, "Some Indian scientists are talented."

0%
True
0%
May be true
0%
False
0%
May be false

Consider the statements
(i)Two plus three is five.
(ii) Every square is a rectangle.
(iii) Sun rises in the east.
(iv) The earth is not a star.
Which of the above statements have truth value (T) ?

0%
(i) and (ii)
0%
(ii) and (iii)
0%
(iii) and (iv)
0%
All of these

The contrapositive of

$p\rightarrow (\sim q\rightarrow \sim r)$ is equivalent to

0%
$(\sim q\wedge r)\rightarrow \sim p$
0%
$(q\wedge \sim r)\rightarrow \sim p$
0%
$p\rightarrow (\sim r\vee q)$
0%
$p\wedge (q\vee r)$

Explanation

The contrapositive of

$p \rightarrow ( \sim q \rightarrow \sim r)$ is

$\equiv \sim (\sim q \rightarrow \sim r) \rightarrow \sim p$

$\equiv \sim ( q \vee \sim r) \rightarrow p \equiv (\sim q \wedge r) \rightarrow \sim p$

Which one of the following statements is not a false statement?

0%
$p:$ Each radius of a circle is a chord of the circle.
0%
$q:$ Circle is a particular case of an ellipse.
0%
$r:\>\displaystyle \sqrt{13}$ is a rational number.
0%
$s:$ The centre of a circle bisects each chord of the cirlce.

Explanation

We know that equation of an ellipse is given by

$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
If we take

$\displaystyle a=b$ then we get

$\displaystyle x^{2}+y^{2}=a^{2}$ which satisfies all the conditions of circle

$\displaystyle \therefore$ circle is the particular case of an ellipse.

Ans: B

Consider the. following compound statement
(i) Mumbai is the capital of Rajasthan or Maharashtra,
(ii)

$\displaystyle \sqrt{3}$ is a rational number or an irrational number,
(iii)

$125$ is a multiple of

$7$ or

$8$
(iv) A rectangle is a quadrilateral or a regular hexagon.
Which of the above statements is not true?

0%
(i)
0%
(ii)
0%
(iii)
0%
(Iv)

Explanation

(i) The component statements of " Mumbai is the capital of Rajasthan or Maharashtra" are

$p :$ Mumbai is the capital of Rajasthan.

$q :$ Mumbai is the capital of Maharashtra.
We note that

$p$ is false and

$q$ is true, so the compound statement is true

(ii) The component statements of

$\displaystyle \sqrt{3}$ is a rational or an irrational are

$p :$

$\displaystyle \sqrt{3}$ is a rational number.

$q:$

$\displaystyle \sqrt{3}$ is an irrational number.
We note that

$p$ is false and

$q$ is true, so the compound statement is true.

(iii) The component statements of

$125$ is a multiple of

$7$ or

$8$ are

$p:125$ is a multiple of

$7.$

$q:125$ is a multiple of

$8.$
We note that

$p$ and

$q$ both are false statements, so compound statement is false.

(iv) The component statements of "A rectangle is a quadrilateral or a regular hexagon." are

$p: A$ rectangle is a quadrilateral.

$q: A$ rectangle is a regular hexagon.
We note that

$p$ is true and

$q$ is false, so the compound statement is true.

The contrapositive of

$(\sim p\wedge q)\rightarrow \sim r$ is equivalent to

0%
$(p\wedge q)\rightarrow r$
0%
$(p\wedge q)\vee r$
0%
$r\rightarrow (p\vee \sim q)$
0%
none of these

Explanation

The contrapositive of

$(\sim p\wedge q)\rightarrow \sim r$ is,

$\equiv \sim (\sim r) \rightarrow \sim (\sim p\wedge q)$

$\equiv r \rightarrow ( p \vee \sim q)$

The contrapositive of the statement "If you believe in yourself and are honest then you will get sucess" is

0%
If you do not believe yourself and are dishonest then you will not get success.
0%
If you do not believe yourself and are dishonest then you will get success.
0%
If you get success then you are honest and you also believe in yourself.
0%
If you will not get success then you don't not believe in yourself or are not honest

The converse of

$p \rightarrow (q \rightarrow r)$ is

0%
$(q \wedge \sim r) \vee p$
0%
$(\sim q \vee r)\vee p$
0%
$(q \wedge \sim r) \wedge \sim p$
0%
$(q \wedge \sim r) \wedge p$

Explanation

The converse of

$p \rightarrow (q \rightarrow r)$ is,

$\equiv (q \to r) \to p \equiv (\sim q \vee r) \to p \equiv (q\wedge \sim r) \to p$

In the following letter sequence, some of the letters are missing. These are given in order as one of the alternatives below. Choose the correct alternative.

$\alpha \beta$ _

$\alpha \alpha$ _

$\beta \beta \beta$ _

$\alpha \alpha \alpha \alpha$ _

$\beta \beta \beta ...$

0%
$\alpha \beta \beta \alpha$
0%
$\beta \alpha \beta \alpha$
0%
$\alpha \alpha \alpha \beta$
0%
$\alpha \beta \alpha \beta$

Explanation

This follows the following pattern

$\beta\alpha\beta\alpha$ is the missing part.
Hence, option 'B' is correct..

The negation of the statement

$q \vee (p \wedge \sim r)$ is equivalent to

0%
$\sim q \wedge(p \rightarrow r)$
0%
$\sim q \vee \sim(p \rightarrow r)$
0%
$q \wedge (\sim p \wedge r)$
0%
None of these.

Explanation

The negation of the statement

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

$\equiv \sim (q \vee (p \wedge \sim r) ) \equiv \sim q \wedge (\sim p \vee \sim(\sim r))\equiv \sim q \wedge (\sim p \vee r)\equiv \sim q \wedge (p\to r)$

The converse of

$p \rightarrow (q \rightarrow r)$ is

0%
$(q \wedge \sim r) \vee p$
0%
$(\sim q \vee r) \vee p$
0%
$(q \wedge \sim r) \wedge \sim p$
0%
$(q \wedge \sim r) \wedge p$

Explanation

The converse of

$p \rightarrow (q \rightarrow r)$ is,

$\equiv (q \to r) \to p \equiv (\sim q \vee r) \to p \equiv \sim (\sim q \vee r) \vee p\equiv (q \wedge \sim r) \vee p$

The contrapositive of

$\sim p \rightarrow ( q \rightarrow \sim r)$ is

0%
$(~q \wedge r) \rightarrow ~p$
0%
$(q \rightarrow r) \rightarrow~p$
0%
$(q \vee ~r) \rightarrow ~ p$
0%
None of these

Explanation

The contrapositive of

$\sim p \rightarrow ( q \rightarrow \sim r)$ is

$\sim ( q \rightarrow \sim r) \rightarrow \sim (\sim p )$

$\equiv \sim (\sim q \vee \sim r) \rightarrow p \equiv (q \wedge r) \rightarrow p$

The negative of the statement "If a number is divisible by

$15$ then it is divisible by

$5$ or

$3$ "

0%
If a number is not divisible by $15$ , then it is not divisible by $5$ and $3$
0%
A number is divisible by $15$ and it is not divisible by $5$ or $3$
0%
A number is not divisible by $15$ or it is not divisible by $5$ and $3$
0%
A number is divisible by $15$ and it is not divisible by $5$ and $3$

Explanation

Let

$p, q, r$ be three statements defined as

$p$ : a number

$N$ is divisible by

$15$

$q$ : number

$N$ is divisible by

$5$

$r$ : number

$N$ is divisible by

$3$
Here given statement is

$p \rightarrow (q \vee r)$
Here negative of above statement is

$\sim (p \rightarrow (q \vee r))\equiv p \wedge (\sim (q \vee r)$

$\equiv p \wedge ( \sim q \wedge \sim r)$
i.e. A number is divisible by

$15$ and it is not divisible by

$5$ and

$3$ .

If statements

$p, q, r$ have truth values T, F, T respectively then which of the following statement is true

0%
$(p \rightarrow q) \wedge r$
0%
$(p \rightarrow q) \vee \sim r$
0%
$(p \wedge q)\vee (q \wedge r)$
0%
$(p \rightarrow q) \rightarrow r$

Explanation

$(p \rightarrow q) \wedge r \equiv (T \rightarrow F) \wedge T \equiv F \wedge T \equiv F$
B.

$(p \rightarrow q) \vee \sim r\equiv (T \rightarrow F) \vee F \equiv F \vee F \equiv F$
C.

$(p \wedge q) \vee (q \wedge r)\equiv(T \wedge F) \vee (F \wedge T)\equiv F\vee F \equiv F$
D.

$(p \rightarrow q) \rightarrow r \equiv (T \rightarrow F) \rightarrow T\equiv F\rightarrow T \equiv T$

The contrapositive of

$p \rightarrow (\sim q \rightarrow \sim r)$ is

0%
$(\sim q\wedge r) \rightarrow \sim p$
0%
$(q \rightarrow r) \rightarrow \sim p$
0%
$( q \vee \sim r) \rightarrow \sim p$
0%
none of these.

Explanation

The contrapositive of

$p \rightarrow (\sim q \rightarrow \sim r)$ is

$\equiv \sim (\sim q \rightarrow \sim r) \rightarrow \sim p$

$\equiv \sim(q \vee \sim r) \to \sim p$

$\equiv (\sim q \wedge r) \to \sim p$

Negation of

$\displaystyle q \vee \sim \left ( p\wedge r \right )$ is

0%
$\displaystyle \sim q\wedge \sim \left ( p\wedge r \right )$
0%
$\displaystyle \sim q\wedge \left ( p\wedge r \right )$
0%
$\displaystyle \sim q\vee \left ( p\wedge r \right )$
0%
None of these

Explanation

$\displaystyle \sim \left ( q\vee \sim \left ( p\wedge r \right ) \right ) \equiv \sim q\wedge \left ( \sim \left ( \sim (p\wedge r \right ) \right ) \equiv \sim q\wedge \left ( p\wedge r \right )$

Find the truth value of the compound statement, 4 is the first composite number and

$2+5=7$ .

0%
T
0%
F
0%
Neither T nor F
0%
cannot be determined

Explanation

The given statement is True, as

$4$ is the first composite number. Also

$2 + 5 = 7$

The negation of the statement "

$2 + 3 = 5$ " and "

$8 < 10$ " is

0%
$2 + 3 \neq 5$ and $8\nless 10$
0%
$2 + 3$ \neq 5 or $8 > 10$
0%
$2 + 3 \neq 5$ or $8 \ngeqslant 10$
0%
None of these

Explanation

Take

$p:2+3=5$ and

$q:8<10$

So the given conjunction is

$p\wedge q$

Now

$\sim p:2+3\neq 5$ and

$\sim q:8\nless 10$

Now Negation of the given conjunction

$p\wedge q$ is

$\sim \left( p\wedge q \right)$

$\sim \left( p\wedge q \right) :2+3\neq 5$ or

$8\nless 10$

Find the quantifier which best describes the variable of the open sentence

$x^2+2\ge0$

0%
Universal.
0%
Existential.
0%
Neither (a) nor (b).
0%
Does not exist.

Explanation

An universal quantifier is a symbol or a logic to denote that the statement is true for all values under the scope.

We know that

$x^2$ is always

$\ge 0$
Hence,

$x^2 + 2 \ge 0$ holds true for all values of

$x$ .

Thus, the quantifier to describe the variable of the given sentence is universal.

The inverse of the statement

$(p \wedge \sim q) \rightarrow r$ is

0%
$\sim (p \vee \sim q) \rightarrow \sim r$
0%
$(\sim p \wedge \sim q) \rightarrow \sim r$
0%
$(\sim p \vee q) \rightarrow \sim r$
0%
None of these.

Explanation

$\textbf{Step-1: Apply the concept of logical reasoning. }$

$\text{We have,}$

$(p \wedge \sim q) \rightarrow r$

$\therefore$

$\text{The inverse of the statement}$

$(p \wedge \sim q) \rightarrow r$

$\text{is,}$

$\equiv \sim (p \wedge \sim q) \rightarrow \sim r \equiv (\sim p \vee \sim (\sim q)) \to \sim r$

$\equiv (\sim p \vee q) \to \sim r$

$\textbf{Hence, option C}$

What is the truth value of the statement 'Two is an odd number iff 2 is a root of

$x^2+2=0$ '?

0%
T
0%
F
0%
Neither T nor F
0%
Cannot be determined

Explanation

Both the statements are false as Two is not an odd number and

$2$ is not the root of

$x^2 + 2 = 0$

We know that two False statements together make a True statement., Hence, truth value of the given statement is True.

The negative of the statement "If a number is divisible by 15 then it is divisible by 5 or 3"

0%
if a number is divisible by 15 then it is not divisible by 5 and 3
0%
a number is divisible by 15 and it is not divisible by 5 or 3
0%
a number is divisible by 15 or it is not divisible by 5 and 3
0%
a number is divisible by 15 and it is not divisible by 5 and 3

Explanation

Let

$p, q, r$ be three statements defined as

$p$ : a number N is divisible by 15

$q$ : a number N is divisible by 5

$r$ :a number N is divisible by 3

Here given statement is

$p\rightarrow (q\vee r)$

Here negative of above statement is

$\sim (p \rightarrow (q \vee r))$

$=\sim [(\sim p) \vee (q \vee r)]$ (

$\because p\rightarrow q=\sim p \vee q$ )

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

$= p \wedge (\sim q \wedge \sim r)$ (

$\because \sim (\sim p)=p, \sim (p\vee q=\sim p \wedge \sim q$ )

i.e. a number is divisible by 15 and it is not divisible by 5 and 3.

CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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