MCQ Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 9

Check the validity of the following statement:

$p:60$ is a multiple of

$3$ and

$5$

0%
True
0%
False

Explanation

$r:60$ is a multiple of

$3\quad (3\times 20) \rightarrow True$

$s:60$ is a multiple of

$5\quad (5\times 12) \rightarrow True$

$True \wedge True \implies True$

$r \wedge s \implies True$

Hence

$p$ is true

Check whether the following statements are true or not :

$p:$ If

$x,y$ are integers such that

$xy$ is even, then at least one of

$x$ and

$y$ is an even integer.

0%
True
0%
False

Explanation

$p:$ If

$x$ and

$y$ are integers such that

$xy$ is even then atleast one of

$x$ and

$y$ is an even integer.

Let

$q$ and

$r$ be two statements.

Here,

$q:xy$ is an even integer.

$r:$ Atleast one of

$x$ and

$y$ is an even integer.

Let

$r$ be not true.

It is false that atleast one of

$x$ and

$y$ is an even integer.

Now,

$x$ and

$y$ are odd integers.

$x=2m+1$ and

$y=2n+1$ for same integers

$m$ and

$n$

$\Rightarrow xy=(2m+1)(2n+1)$

$\Rightarrow xy=2(2mn+m+n)+1$

so,

$xy$ is not an even integer. Then,

$xy$ is not true.

Hence, the statement is true.

So, the option

$A$ is correct.

Check whether the following statements are true or not :

$p:$ If

$x$ and

$y$ are odd integers, then

$x+y$ is an integer.

0%
True
0%
False

Explanation

Let

$p : x$ and

$y$ are odd integers .

$q : x+y$ is an integer .

Then , given statement is : if

$p$ , then

$q$ .

Let

$p$ is true .

Hence

$x$ and

$y$ are odd integers .

Let

$x = 2m+1$ and

$y =2n+1$ then ,

$x+y = 2m+1 + 2n + 1 = 2(m+n) + 2$ .

Which is an integer.

Hence,

$q$ is true.

That is ,

$p$ is true implies

$q$ is true

"If

$p$ then

$q$ " is true .

The given statement is true.

Check the validity of the following statement:

$p:100$ is a multiple of

$4$ and

$5$

0%
True
0%
False

Explanation

$True \wedge True \implies True$

$r:100$ is a multiple of

$4\rightarrow$ TRUE

$(4\times 25)$

$s:100$ is a multiple of

$5\rightarrow$ TRUE

$(5\times 20)$

$r \wedge s \implies T$

Hence

$p$ is true

Determine whether the argument used to check the validity of the following statement is correct.

$p:$ If

$x^{2}$ is irrational, then

$x$ is rational'
The statement is true because the number

$x^{2}=\pi^{2}$ is irrational, therefore

$x=\pi$ irrational.

0%
True
0%
False

Explanation

Here, the argument used is,

$x^2=\pi^2$ is irrational, therefore

$x=\pi$ is irrational and,

$p:$ " If

$x^2$ is irrational, then

$x$ is rational.

Let us take an irrational number given by

$x=\sqrt n$ ,

where

$n$ is a rational number.

Now, square both sides, we get,

$x^2=k$

Therefore,

$x^2$ is a rational number, which contradicts our statement.

Hence, the argument used to check validity of given statement is false.

Which of the following is NOT equivalent to

$p\rightarrow q$ ?

0%
p only if q
0%
q is necessary for p
0%
q only if p
0%
p is sufficient for q

Explanation

$q$ only if

$p$ is a statement that says if

$q$ is true then

$p$ is true.

This is not equivalent to

$p \rightarrow{} q$ . This non-equivalency can be checked by letting

$p$ be a true statement and

$q$ be a false statement. With these values, the statement

$p \rightarrow{} q$ is false but the statement

$q$ only if

$p$ is true.

Negation of the statement:

$\sqrt{5}$ is an integer or

$5$ is irrational is?

0%
$\sqrt{5}$ is not an integer or $5$ is not irrational
0%
$\sqrt{5}$ is not an integer and $5$ is not irrational
0%
$\sqrt{5}$ is irrational or $5$ is an integer
0%
$\sqrt{5}$ is an integer and $5$ is irrational

Explanation

Let

$p=\sqrt5$ is an integer

$\Rightarrow$

$\sim p=$

$\sqrt5$ is not an integer

$q= 5$ is irrational

$\Rightarrow$

$\sim q=$

$5$ is not an irrational

$\sqrt{5}$ is not an integer and

$5$ is not an irrational number

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

The equivalent form of the statement

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

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

Explanation

We have

$a \rightarrow b = \sim a \vee b$

$\therefore\ p \rightarrow \sim q = \sim p \vee \sim q$

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

$= \sim (\sim (p \wedge q))$ .... De Morgan's Law

$= p \wedge q$ .... Complement Law

_________________________ [OR] ____________________________

We can also solve it using Truth Table

$p$	$q$	$\sim p$	$\sim q$	$p \wedge q$	$p \wedge \sim q$	$p \vee \sim q$	$\sim p \vee q$	$p \rightarrow \sim q$	$\sim (p \rightarrow \sim q)$
$0$	$0$	$1$	$1$	$0$	$0$	$1$	$1$	$1$	$0$
$0$	$1$	$1$	$0$	$0$	$0$	$0$	$1$	$1$	$0$
$1$	$0$	$0$	$1$	$0$	$1$	$1$	$0$	$1$	$0$
$1$	$1$	$0$	$0$	$1$	$0$	$1$	$1$	$0$	$1$

So from column

$5$ and

$10$ we can write that,

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

The statement pattern

$(p\wedge q)\wedge [\sim r\vee (p\wedge q)]\vee (\sim p\wedge q)$ is equivalent to _________.

0%
r
0%
q
0%
$p\wedge q$
0%
p

Explanation

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

We will now use absorption law

$[(a\vee b)\wedge a=a]$ for first 2 bracket.

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

Now by using disrtibutive law we get

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

$1\wedge q$

$q$

Tell if the following statement is true or false. In case give a valid reason for saying so

$p:$ If

$x$ and

$y$ are integers such that

$x>y$ . then

$-x<-y$ .

0%
True
0%
False

Explanation

Given

$x>y$

Multiply both sides by

$-1$

$-x<-y$

$\therefore$ both statements true.

Tell if the following statement is true or false. In case give a valid reason for saying so

$p:$ Each radius of a circle is a chord of the circle.

0%
True
0%
False

Explanation

This statement is false.

By definition the chord,

It should intersect the circle in two points.

$OB$ and

$OC$ are the radius of circle.

But they dont intersect the circle at two points.

Hence

$OB$ &

$OC$ are not the chord of a circle.

So, option

$B$ is correct.

1434867_1668009_ans_792b7f2210ac4be48847fbf7daeec965.png

The negation of the compound proposition

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

0%
$(p\wedge ∼q)\wedge ∼p$
0%
$(p\wedge ∼q)\vee ∼p$
0%
$(p\wedge ∼q)\vee ∼p$
0%
none of these

Given, "If I have a Siberian Husky, then I have a dog." Identify the converse

0%
If I do not have a Siberian Husky, then I do not have a dog.
0%
If I have a dog, then I have a Siberian Husky.
0%
If I do not have a dog, then I do not have a Siberian Husky.
0%
If I do not have a Siberian Husky, then I have a dog.

The component statements are:

p: You are wet when it rains.

q: You are wet when you are in river.

The compound statement of these component statements using appropriate connective is:

0%
You are not wet when you are in river or it rains.
0%
You are wet when you are in river and it rains.
0%
You are wet when it rains and you are in a river
0%
You are wet when it rains or you are in a river.

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

0%
$q \vee r$
0%
$q \wedge r$
0%
$q \rightarrow r$
0%
none of these

Explanation

${\textbf{Step -1: After applying associative and distributive law}}$

$\text{Given, } \left( {p \wedge q} \right) \vee \left( { \sim p \wedge q} \right) \vee \left( { \sim q \wedge r} \right)$

$\equiv \left[ {\left( {p \vee \sim p} \right) \wedge q} \right] \vee \left( { \sim q \wedge r} \right)$ $[\because\boldsymbol{(a\vee b)\wedge c} \boldsymbol{\equiv (a\wedge c)\vee (b\wedge c)}]$

${\textbf{Step -2: Apply complement law}}$

$\equiv \left( {T \wedge q} \right) \vee \left( { \sim q \wedge r} \right)$ $[\because\boldsymbol{(a\vee\sim a)\equiv T}]$

${\textbf{Step -3:Apply identity law}}$

$\equiv q \vee \left( { \sim q \wedge r} \right)$ $[\because\boldsymbol{(T\wedge a)\equiv q}]$

${\textbf{Step -4: Apply distributive law}}$

$\equiv \left( {q \vee \sim q} \right) \wedge \left( {q \vee r} \right)$ $[\because\boldsymbol{(a\wedge b)\vee c} \boldsymbol{\equiv (a\vee c)\wedge (b\vee c)}]$

${\textbf{Step -5: Apply complement law}}$

$\equiv T \wedge \left( {q \vee r} \right)$ $[\because\boldsymbol{(a\vee\sim a)\equiv T}]$

${\textbf{Step -6: Apply identity law}}$

$\equiv q \vee r$ $[\because\boldsymbol{(T\wedge a)\equiv q}]$

${\textbf{Hence, }}$ $\mathbf{\left( {p \wedge q} \right) \vee \left( { \sim p \wedge q} \right) \vee \left( { \sim q \wedge r} \right) \equiv q \vee r}$

$∼(p⇒q)⟺∼p\vee ∼q \, is$

0%
a tautology
0%
a contradiction
0%
neither a tautology nor a contradiction
0%
cannot come to any conclusion

Disjunction of two statements p and q is denoted by

0%
$p \leftrightarrow q$
0%
$p \rightarrow q$
0%
$p \leftarrow q$
0%
$p \vee q$

If p and q are mathematical statements, then in order to show that the statement p and q is true, we need to show that:

0%
The statement p is true and the statement q is not true
0%
The statement p is false and the statement q is true.
0%
The statement p is true and the statement q is false
0%
The statement p is true and the statement q is true

An implication or conditional "if p then q "is denoted by

0%
$p \vee q$
0%
$p \rightarrow q$
0%
$p \leftarrow q$
0%
None of these

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

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

Explanation

$\Rightarrow [(P)^q]$

Here, '

$\wedge$ ' means 'and', if we remove the brackets, the sign gets reversed into '

$\vee$ ' which is 'or'.

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

Hence, option (C) is correct.

Name the technique used in the first step of the solution to the problem below :
Verify that 5 is irrational
Solution : Let us assume that 5 is rational

0%
Counter example
0%
Direct method
0%
By Contradiction
0%
Contrapositive method

Name the technique used in the solution of the problems below :

Question: Show that the following statement is false: If n is an odd integer, then n is prime.

Solution: The given statement is in the form “if p then q” we have to show that this is false, If p then ~q.

If n= 99 is odd integer which is not a prime number. Thus, we conclude that the given statement is false.

0%
Counter example
0%
Contrapositive method
0%
Direct method
0%
By Contradiction

Explanation

Given : The given statement is in the form "if p then q" we have to show that this is false. If p then

$\sim q$ .

$n = 99$ is odd integer which is not a prime number. Thus, we conclude that the given statement is false.

So, we need to justify that the statements is false. Now, compound statements are,

$p : n$ is an odd integer

$q : n$ is prime

Now, let assume

$n = 99$

So,

$n = 2m +1$ which states as odd integer

$n$ is

$99$ then

$n = 2 \times 4 +1$

We know that prime number is a number which is divisible by itself and unity.

Here, if we break

$99$ in multiples then it

$99 = 9 \times 11$ , which states that it has a multiple including the number itself and unity.

So,

$99$ is odd integer but not prime.

And we know that, a mathematical statement has two parts: a condition and a conclusion. However, showing that a mathematical statement is false only requires finding one example where the statement isn't true.

So, such example is called counter examples as it goes against, the statement's conclusion.

$\therefore$ Option A is correct.

Check whether the following statement is true or false? In each case give a valid reason for your answer.

$p: \sqrt {11}$ is an irrational number.

0%
True
0%
False

Check whether the following statement is true or false. Also give a valid reason for your answer.

$r:$ Each radius of a circle is a chord of the circle.

0%
True
0%
False

Check whether the following statement is true or false? Also give a valid reason for your answer.

$u:$ The quadratic equation

$x^2+x+1=0$ has no real roots.

0%
True
0%
False

Check whether the following statement is true or false? Also give a valid reason for your answer.

$s:$ The centre of a circle bisects each chord of the circle.

0%
True
0%
False

Which of the following statements are true and which are false? In each case give a valid reason for your answer.

$q:$ Circles is a particular case of an ellipse.

0%
True
0%
False

Explanation

True. The equation of an ellipse is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ .

Putting

$b=a$ , it becomes

$x^2+y^2=a^2$ , which is the equation of a circle.

Check whether the following statement is true or false? Also give a valid reason for your answer.

$t:$ If

$a$ and

$b$ are integers such that

$a < b$ , then

$-a > -b$

0%
True
0%
False

Explanation

Given statement is True.

when we multiply by a negative sign the equality sign changes.

since

$a<b \Rightarrow -a> -b$ is true.

Let

$A=\left\{ 2, 3, 5, 7\right\}$ . Examine whether the statements given below are true or false.

$\exists \ x\in A$ such that

$x$ is even.

0%
True
0%
False

Let

$A=\left\{ 2, 3, 5, 7\right\}$ . Examine whether the statements given below are true or false.

$\exists \ x\in A$ such that

$x+2=6$ .

0%
True
0%
False

CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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