MCQ Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 8

P: he studies hard, q: he will get good marks. The symbolic form of " If he studies hard then he will get good marks "is_____

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$\sim q\Rightarrow p$
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$p\Rightarrow q$
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$\sim p\vee q$
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$p\Leftrightarrow q$

Consider the following two statements :
P: If 7 is odd number, then is divisible by

$2$ .
Q: If 7 is a prime number, Then is an odd number.
If

${ V }_{ 1 }$ is the value of the contra-positive of

$p$ and

${ V }_{ 2 }$ Is the truth value of contra-positive of Q, then the orders pair

$({ V }_{ 1 },{ V }_{ 2 })$ equals :

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$(T,T)$
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$(T,F)$
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$(F,T)$
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$(F,F)$

Explanation

P= odd number cannot be divisible by 2,

${V_1}$ is contra positive value of P

${V_1} = T$

Q=Every prime number is an odd number but not every odd number . is a prime number

${V_2}$ is the contra positive value of Q

$\left( {{V_1},{V_2}} \right) = \left( {T,F} \right)$

Negation of compound proposition : If the examination is difficult, then I shall pass if I study hard

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The examination is difficult and I study hard and I shall pass
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The examination is difficult and I study hard but I shall not pass
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The examination is not difficult and I study hard and I shall pass
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All of these

The cost of an article including the sales tax is

$Rs\ .616$ . The rate of sales tax is

$10\%$ ,if the shopkeeper has made a profit of

$12\%$ , then the cost price of the article is?

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$Rs\ .350$
0%
$Rs\ .400$
0%
$Rs\ .500$
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$Rs\ .800$

Explanation

Let the cost price $=Rs100$

Then, including tax $=100+\dfrac{11}{100}=110\%$

So,

$110\%$ of S.P $=Rs\,616$

Then,

$S.P=\dfrac{616}{110}\times 100$

$S.P=Rs\,560$

Now, Cost price of $12\%$ $=\dfrac{100\times560}{100+12}=\dfrac{100\times 560}{112}$

$=Rs\,500$

Hence, this is the answer.

Consider the following statement
p:you want to success
q:you will find a away,
then the negation of

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

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you want success or you will find a way
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you want of success and you do not find a way
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if you want of success then you cannot find a way
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if you do not want to succeed then you will not find a way

$p, q, r$ are three propositions, then the negation of

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

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

If statement

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

$p,q,r$ respectively, can be :

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$FTF$
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$TTT$
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$FFF$
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$FTT$

Explanation

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

$\therefore$ If p=T or p=F

q=T q=T

r=F r=F

Then the statement

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

1167974_1277597_ans_2a02148a64a3451e8514aaec2d4ecec3.JPG

Negation of

$(\sim p\rightarrow q)$ is ________________.

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$\sim { p }{ \wedge }\sim q$
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$\sim \left( p\vee q \right) \vee \left( p\vee \left( \sim p \right) \right)$
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$\sim \left( p\vee q \right) \wedge \left( p\vee \left( \sim p \right) \right)$
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$\left( \sim p\vee q \right) \wedge \left( p\vee \sim q \right)$

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

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$p \vee q$
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$p \wedge q$
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$p \rightarrow q$
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none of these

Explanation

We have,

$P \wedge (q \, \vee \sim p) = ?$

From truth table,

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

From above truth table, we get

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

Hence, option B is correct answer.

Consider the statement "if it is raining, then sky is not filled with clouds which among the Q 7 following is true

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contrapositive of the given statement "if sky is filled with clouds then it is not raining"
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converse of the given statement is "If sky is filled with clouds then it is raining"
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"If it is not raining then sky is filled with clouds" is the inverse of the given statement
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Both 1 , 2 and 3 are true

The statement

$P \rightarrow ( q \vee r )$ is not equivalent to

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$( p \rightarrow q ) \vee ( q - r )$
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$p \wedge ( \sim q ) \rightarrow r$
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$P \wedge ( \sim r ) \rightarrow q$
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$p \wedge q \rightarrow ( p \wedge r ) \vee ( q \wedge r )$

The false statement in the following is

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$p \wedge ( \sim p )$ is a contradiction.
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$( p \rightarrow q ) \leftrightarrow ( \sim q \rightarrow \sim p )$ is a contradiction
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$\sim ( \sim p ) \leftrightarrow p$ is a tautology
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$p \vee ( \sim p )$ is a tautology

Consider the following statements

$P:$ Suman is brilliant

$Q:$ Suman is rich

$R:$ Suman is honest.
The negative of the statement."Suman is brilliant and dishonest, if and only if Suman is rich" can be expressed as

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$\sim [Q \leftrightarrow (P \wedge \sim R)]$
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$\sim Q \leftrightarrow P \wedge R$
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$\sim (P \wedge \sim R) \leftrightarrow Q$
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$\sim P \wedge (Q \leftrightarrow \sim R)$

Explanation

Suman is brilliant and dishonest

$\Rightarrow P \wedge \sim~ R$

Suman is brilliant and dishonest if and only if suman is rich

$\Rightarrow Q \leftrightarrow \left( P \wedge \sim R \right)$

Negative of the above statement will be-

$\sim \left[ Q \leftrightarrow \left( P \wedge \sim R \right) \right]$

Consider the following two statements:

$P:$ if

$7$ is an odd number, then

$7$ is divisible by

$2$ .

$Q:$ if

$7$ is a prime number, then

$7$ is an odd number If

$V_{1}$ is the truth value of the contrapositive of

$P$ and

$V_{2}$ is the truth value of contrapositive of

$Q$ . Then the ordered pair

$\left(V_{1},V_{2}\right)$ equals:

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$\left(F,F\right)$
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$\left(T,T\right)$
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$\left(F,T\right)$
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$\left(T,F\right)$

Negation of

$p \to \left( {p \vee \sim q} \right)$ is

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$( \sim p) \to ( \sim p \vee q)$
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$p \wedge ( \sim p \wedge q)$
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$\left( { \sim p} \right) \vee \left( { \sim p \vee \sim q} \right)$
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$\left( { \sim p} \right) \leftarrow \left( { \sim p \to q} \right)$

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

$p,q$ and

$r$ are respectively

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$F,T,F$
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$T,F,T$
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$F,F,F$
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$T,T,T$

Consider the following three statements:
P : 5 is a prime number.
Q : 7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35.
Then the truth value of which one of thefollowing statements is true ?

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$(P \wedge Q)\vee (\sim R)$
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$(\sim P)\wedge (\sim Q \wedge R)$
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$(\sim P) \vee (Q\wedge R)$
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$P\vee (\sim Q \wedge R)$

Explanation

P is True

Q is False

R is True

Option 4)

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

State whether the statement

P: "if x is a real number such that

$x^3+2x=0$ , then

$x$ is

$0$ " is true/false

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True
0%
False

Explanation

Given that,

$P :$ if

$x$ is real number such that

$x^3 + 2x = 0$ then,

$x = 0$

$x^3 + 2x = 0$

$x (x^2 + 2) = 0$

$x = 0$

$x^2 + 2 = 0$

for

$x^2 + 2 = 0$

$x^2 = \, -2$

$x =\pm \sqrt{-2}$

$x$ is real number

$\therefore x = 0$

Hence, statement is true.

Which of the following is not a tautology?

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

Explanation

$(1)$

$p\rightarrow (p\vee q)$

$\sim p\vee (p\vee q)=t$

$(2)$

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

$(\sim p\vee \sim q)\vee p=t$

$(3)$

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

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

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

$=t\wedge \sim q=\sim q\neq t$

$(4)$
Alter:
(a)

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

(b)

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

(c)

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

The negation of

$p\wedge \left( q \rightarrow\ r \right)$ is

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

Which of the following is not true?

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If the last column of its truth table contains only $F$ then it is a contradiction
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Negation of a negation of a statement is the statement itself
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If $p$ and $q$ are two statements then $p \leftrightarrow q$ is a tautology
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If the last column of its truth table contains only $T$ then it is tautology

Explanation

Truth table

$p$	$q$	$p\leftrightarrow q$
$T$	$T$	$T$
$T$	$F$	$F$
$F$	$T$	$F$
$F$	$F$	$T$

Since the last column of the truth table has False and Truth.It cannot be tautology.

State whether the statement
P: "if x is a real number such that

$x^3+2x=0$ , then

$x$ is

$0$ " is true/false.

0%
True
0%
False

Let

$p\rightarrow (\sim q\vee r)$ is false, then truth values of p, q, r are respectively.

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

The negation of the boolean expression

$\sim s\vee \left( \sim r\wedge s \right)$ is equivalent to:

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$r$
0%
$s\wedge r$
0%
$s\vee r\quad$
0%
$\sim s\wedge \sim r\quad$

Explanation

$\sim \left( \sim s\vee (\sim r\wedge s) \right)$

$s\wedge (r\vee \sim s)$

$(s\wedge r)\vee (s\wedge \sim s)$

$(s\wedge r)\vee (F)$

$(s\wedge r)$

Determine whether the following compound statements are true or false:
Delhi is in India and

$2+2=4$

0%
True
0%
False

Explanation

Delhi is in India is TRUE :

$p$

$2+2=4$ is TRUE :

$q$

Therefore the compound statement is true.

Since the connective is 'and' and both the statements are true

So, the compound statement is true

Determine whether the following compound statements are true or false:
Delhi is in India and

$2+2=5$

0%
True
0%
False

Explanation

Delhi is in INDIA is TRUE :

$p$

$2+2=5$ is FALSE :

$q$

The connective word is 'and' and the second statement is false

Therefore the compound statement is FALSE.

Determine whether the following compound statement are true of false:
Delhi is in England and

$2+2=4$

0%
True
0%
False

Explanation

Delhi is in England is FALSE :

$p$

$2+2=4$ is TRUE :

$q$

Since the connecting word is 'and ' and one of the statement is false

Therefore the compound statement is FALSE

Determine whether the following compound statements are true or false:
Delhi is in England and

$2+2=5$

0%
True
0%
False

Explanation

Delhi is in ENGLAND is FALSE:

$p$

$2+2=5$ is FALSE :

$q$

Because one of the statement is False

Therefore the compound statement is FALSE.

Check the validity of the following statement:

$p:125$ is a multiple of

$5$ and

$7$

0%
True
0%
False

Explanation

$r:125$ is a multiple of

$5\rightarrow$ TRUE

$(5\times 25)$

$s:125$ is a multiple of

$7\rightarrow$ FALSE

Hence

$p$ is false.

State whether the statement

$p:$ If

$x$ is a real number such that

$x^{3}+19x=0$ , then

$x$ is

$0$ is true / False

0%
True
0%
False

Explanation

$x^3 + 19x= 0$

$x(x^2 + 19)=0$

$x = 0$ ,

$x^2 + 19 =0$

so given statement is true

CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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