If $$x+4=8$$, then $$x=4$$, Inverse of the statement is-

If $$x+4\neq8$$, then $$x\neq4$$

MCQ Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 7

The inverse of statement "If a dog is barking, then it will not bite" is

0%
If a dog is not barking, then it will bite.
0%
If a dog is biting, then it will not bark.
0%
If a dog is not biting, then it will not bark.
0%
None of the above

Which of the following statements is the inverse of
"Our pond floods whenever there is a thunderstorm."?

0%
If there is a thunderstorm, then our pond floods.
0%
If we do not get a thunderstorm, then our pond does not flood.
0%
If our pond does not flood, then we did not get a thunderstorm.
0%
None of the above

Which of the following statements is the inverse of
"If you do not understand geometry, then you do not know how to reason deductively." ?

0%
If you reason deductively, then you understand geometry.
0%
If you understand geometry, then you reason deductively.
0%
If the do not reason deductively, then you understand geometry.
0%
None of the above

$x+4=8$ , then

$x=4$ , Inverse of the statement is-

0%
If $x+4=8$ , then $x\neq4$
0%
If $x+4\neq8$ , then $x\neq4$
0%
If $x+4\neq8$ , then $x=4$
0%
none of the above

Explanation

$x+4\neq8$ , then

$x\neq4$

this is the inverse obtained by negating hypothesis and conclusion condition

Which statement represents the inverse of the statement "If it is snowing, then Skeeter wears a sweater."?

0%
If Skeeter wears a sweater, then it is snowing.
0%
If Skeeter does not wear a sweater, then it is not snowing.
0%
If it is not snowing, then Skeeter does not wear a sweater.
0%
If it is not snowing, then Skeeter wears a sweater.

Explanation

To find the inverse we need to negate the hypothesis and conclusion, and hence on negating it is snowing we get it is not snowing and on negating skeeter wears a sweater we get skeeter does not wear a sweater.

$C$

Write negation of:
All natural numbers are integers and all integers are not natural numbers.

0%
All natural numbers are not integers and all integers are not natural numbers.
0%
All natural numbers are integers and all integers are natural numbers.
0%
Some natural numbers are integers and all integers are not natural numbers.
0%
All natural numbers are not integers and some integers are natural numbers.

Explanation

the negation of

All natural numbers are integers and all integers are not natural numbers. is All natural numbers are not integers and some integers are natural numbers. or Some natural numbers are not integers and some integers are natural numbers

more or less option d is correct.

$D$

$p, q, r$ are simple proportions with truth values

$T, F, T$ , then the truth value of

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

0%
True
0%
False
0%
True, if $r$ is false
0%
True, if $q$ is true

Explanation

$\sim p\vee q$ means

$F\vee F = F, \sim r$ means

$F$

$\therefore [(\sim p\vee q)\wedge \sim r)=F\wedge F = F$ which imply

$p = T$ means

$F\Rightarrow T = T$ .

Hence, the truth value of the given statement is True.

$\sim \left[ p\wedge (\sim q) \right] =$

0%
$\sim p \ \wedge \sim q$
0%
$\sim p \ \vee \sim q$
0%
$\sim p \ \wedge \ q$
0%
$\sim p\vee q$

Explanation

$\sim \left [p\wedge(\sim q)\right]=\sim p\vee\sim (\sim q)$ (by demorgan's law)

$=\sim p\vee q$ (

$\because \sim(\sim p)=p$ )

Option

$D$ is correct.

The inverse of the propositions

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

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

Explanation

In logic, an inverse is a type of conditional sentence which is an immediate interference made from another conditional sentence. Any conditional sentence has an inverse. The inverse of

$P\rightarrow Q\quad is\quad \neg P\rightarrow \neg Q$

Hence, answer is (∼r)

$\rightarrow$ (∼p) ∨ q

$p \leftrightarrow q$ is equivalent to

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

Explanation

Consider truth table of

$p\leftrightarrow q$

$($ Let

$T$ denotes true ans

$F$ denotes false

$)$

$p=T$ and

$q=T$ then

$p\leftrightarrow q = T$

$p=T$ and

$q=F$ then

$p\leftrightarrow q = F$

$p=F$ and

$q=T$ then

$p\leftrightarrow q = F$

$p=F$ and

$q=F$ then

$p\leftrightarrow q = T$

Now consider truth table of

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

$p=T$ and

$q=T$ then

$(p\rightarrow q)\wedge (q\rightarrow p) = T$

$p=T$ and

$q=F$ then

$(p\rightarrow q)\wedge (q\rightarrow p) = F$

$p=F$ and

$q=T$ then

$(p\rightarrow q)\wedge (q\rightarrow p) = F$

$p=F$ and

$q=F$ then

$(p\rightarrow q)\wedge (q\rightarrow p) = T$

Observe that both truth tables are equal , therefore the correct option is

$D$

$p$ is

$T$ and

$q$ is

$F$ , then which of the following have the truth value

$T$ ?

$(i)p\vee q$

$(ii)\sim p\vee q$

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

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

0%
$(i),(ii),(iii)$
0%
$(i),(ii),(iv)$
0%
$(i),(iii),(iv)$
0%
$(ii),(iii),(iv)$

Explanation

$(i)p\vee q = T\vee F =T$

$(ii)\sim p\vee q=(\sim T)\vee F = F\vee F=F$

$(iii)p\vee (\sim q)=T\vee (\sim F)=T\vee T= T$

$(iv)p\wedge (\sim q)=T\wedge (\sim F)=T\wedge T=T$

Hence option 'C' is correct

$p:3$ is a prime number and

$q:$ one plus one is three, then the compound statement "It is not that

$3$ is a prime number or it is not that one plus one is three" is

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

Explanation

$p:3$ is a prime number

$q:$ one plus one is three

Compound statement: "It is not that

$3$ is a prime number or it is not that one plus one is three" is

$\sim p\vee \sim q$

$p$ 's truth value is

$T$ and

$q$ 's truth value is

$F$ , then which of the following have the truth value

$T$ ?
(i)

$p\vee q$
(ii)

$\sim p\vee q$
(iii)

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

$p\wedge (\sim q)$

0%
(i), (ii), (iii)
0%
(i), (iii), (iv)
0%
(i), (ii), (iv)
0%
(ii), (iii), (iv)

Explanation

Given

$p's$ truth value is

$T$ and

$q's$ truth value is

$F$ .

We have to determine which of the given have the truth value

$T$ .

$(i) p\vee q$

$p$	$q$	$p \vee q$
$T$	$F$	$T$

$(ii) \sim p\vee q$

$p$	$\sim p$	$q$	$\sim p \vee q$
$T$	$F$	$F$	$F$

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

$p$	$q$	$\sim q$	$p \vee (\sim q)$
$T$	$F$	$T$	$T$

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

$p$	$q$	$\sim q$	$p \wedge (\sim q)$
$T$	$F$	$T$	$T$

From the above tables we see that,

$(i), (iii), (iv)$ have truth value

$T$ .

Which of the following is not a logical statement?

0%
$8$ is less than $6$
0%
Every set is a finite set
0%
Kashmir is far from here
0%
The sun is a star

Explanation

$8$ is less than

$6$

$\rightarrow$ logical

B) Every set is a finite set

$\rightarrow$ logical

C) Kashmir is far from here

$\rightarrow$ not a logical statement

D) The sun is a star

$\rightarrow$ logical

The truth values of

$p, q$ and

$r$ for which

$(p \wedge q) \vee (\sim r)$ has truth value F are respectively

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

Explanation

option	$p$	$q$	$r$	$p\wedge q$	$\sim r$	$(p\wedge q)\vee (\sim r)$
A	$F$	$T$	$F$	$F$	$T$	$T$
B	$F$	$F$	$F$	$F$	$T$	$T$
C	$T$	$T$	$T$	$T$	$F$	$T$
D	$T$	$F$	$F$	$F$	$T$	$T$
E	$F$	$F$	$T$	$F$	$F$	$F$

Option

$E$ gives truth value

$F$ .

So option

$E$ is correct.

$\left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right)$ is a :

0%
A tautology
0%
A contradiction
0%
Both a tautology and a contradiction
0%
Neither a tautology nor a contradiction

Explanation

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

$=f\wedge f=f$
(By using associative laws and commutative laws)

Therefore,

$\left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right)$ is a contradiction.

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

$\text{True}$ ?

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

Explanation

Since, p is true and q is false.

$\therefore p\rightarrow q$ has truth value F.
Statement r has truth value T.

$\therefore (p\rightarrow q)\wedge r$ has truth value F.

Also,

$(p\rightarrow q)\wedge \sim r$ has truth value F.

Now,

$p\wedge q$ has truth value F and

$p\vee r$ has truth value T.

$\therefore (p\wedge q)\wedge (p\vee q)$ has truth value F.
As,

$p\wedge r$ has truth value T.

Therefore,

$q\rightarrow (p\wedge r)$ has truth value T.

$\sim [(\sim p) \wedge q]$ is logically equivalent to

0%
$\sim (p \vee q)$
0%
$\sim [p \wedge (\sim q)]$
0%
$p \wedge (\sim q)$
0%
$p \vee (\sim q)$
0%
$(\sim p) \vee (\sim q)$

Explanation

$\sim [(\sim p)\wedge q]$

By distributive property

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

So option

$D$ is correct.

The contrapositive statement of statement "If x is prime number, then x is odd" is

0%
If X is not is prime number, then x is not odd
0%
If X is not odd, then x is not a prime number
0%
If X is a prime number, then x is not odd
0%
If X is not a prime number, then x is odd

Explanation

If p, then q means

$p\rightarrow q$

$p\rightarrow q$ is equivalent to its contrapositive

$p\rightarrow q=\bar { p } +q=\sim p+q$
Contrapositive:

$\sim q\rightarrow \sim p=\sim (\sim q)+\sim p$

$=q+\sim p$

$=\sim p+q$

$\therefore$ Contrapositive of "If x is prime,then x is odd" is
If x is not odd, then x not a prime number

Let

$p:57$ is an odd prime number

$q:4$ is a divisor of

$12$

$r:15$ is the LCM of

$3$ and

$5$
be three simple logical statements. Which one of the following is true?

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

Explanation

$p:57$ is an odd number

$q:4$ is a divisor of

$12$

$r:15$ is the LCM of

$3$ and

$5$

Then,

$p$ is

$F$

$q$ is

$T$

$r$ is

$T$ By the given options

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

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

The converse of the contrapositive of the conditional

$p\rightarrow \sim q$ is.

0%
$p\rightarrow q$
0%
$\sim p \rightarrow \sim q$
0%
$\sim q \rightarrow p$
0%
$\sim p \rightarrow q$

Explanation

The contrapositive of the statement

$p\longrightarrow \sim q\\$ is

$q\longrightarrow \sim p$

The converse of the statement

$q\longrightarrow \sim p$ is

$\sim p \longrightarrow q$

Given the true statement: If a quadrilateral is a square, then it is a rectangle. It follows that, of the converse and the inverse of this true statement.

0%
Only the converse is true
0%
Only the inverse is true
0%
Both are true
0%
Neither is true
0%
The inverse is true, but the converse is sometimes true

$p\vee q$ is true and

$p\wedge q$ is false, then which of the following is not true?

0%
$p\vee q$
0%
$p\leftrightarrow q$
0%
$\sim p\vee \sim q$
0%
$q\vee \sim q$

Given are three positive integers

$a, b,$ and

$c$ . Their greatest common divisor is

$D$ ; their least common multiple is

$M$ . Then, which two of the following statements are true?

$(1)$ The product

$MD$ cannot be less than

$abc$

$(2)$ The product

$MD$ cannot be greater than

$abc$

$(3)$

$MD$ equals

$abc$ if and only if

$a, b, c$ are each prime

$(4)$

$MD$ equals

$abc$ if and only if

$a, b, c$ are relatively prime in pairs
(This means : no two have a common factor greater than

$1$ .)

0%
$1, 2$
0%
$1, 3$
0%
$1, 4$
0%
$2, 3$
0%
$2, 4$

Explanation

Represent

$a, b, c$ in terms of their prime factors. Then

$D$ is the product of all the common prime factors, each factor taken as often as it appears the least number of times in

$a$ or

$b$ or

$c$ .

$M$ is the product of all the non-common prime factors, each factor taken as often as it appears the greatest number of times in

$a$ of

$b$ or

$c$ .
Therefore,

$MD$ may be less than

$abc$ , but it cannot exceed

$abc$ .
Obviously,

$MD$ equals

$abc$ when there are no common factors.

Given the following six statements:
(1) All women are good drivers
(2) Some women are good drivers
(3) No men are good drivers
(4) All men are bad drivers
(5) At least one man is a bad driver
(6) All men are good drivers.
The statement that negates statement (6) is:

0%
(1)
0%
(2)
0%
(3)
0%
(4)
0%
(5)

Which one of the following statements is not true for the equation

$i{ x }^{ 2 }-x+2i=0$ .
where

$i\equiv \sqrt { -1 }$ ?

0%
The sum of the roots is $2$
0%
The discriminant is $9$
0%
The roots are imaginary
0%
The roots can be found by using the quadratic formula
0%
The roots can be found by factoring, using imaginary numbers

The moment of inertia of the plate about the

$x-$ axis is

0%
$\dfrac{{M{L^2}}}{8}$
0%
$\dfrac{{M{L^2}}}{32}$
0%
$\dfrac{{M{L^2}}}{24}$
0%
$\dfrac{{M{L^2}}}{6}$

Negation of

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

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

The negation of

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

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

Explanation

$q \vee \sim (p \wedge r)\,\,\,\,\,\,\,\,\,\,\left[ {\because \sim ( \sim p) = p} \right]$

$\sim q \wedge \sim (p \wedge r)\,\,\,\,\,\left[ {\because \sim \left( {p \vee q} \right) = \sim p \wedge \sim q} \right]$ [by De Morgans law]

Dual of

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

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

CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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