Mathematical Reasoning Test

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Mathematical Reasoning Test - 17

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Question 1
1 / -0

The inverse proposition of $$(p \wedge \sim q)\Rightarrow r$$, is

A
$$\sim r \Rightarrow \sim p\vee q$$

B
$$\sim p\vee q \Rightarrow \sim r$$

C
$$r \Rightarrow p\wedge \sim q $$

D
none of these.

Solution

Inverse proportion of $$p \rightarrow q $$ is $$ \sim p \rightarrow \sim q$$
$$\therefore $$ inverse proportion of $$p \land \sim q \rightarrow r$$ is
$$\sim (p\ \wedge \sim q ) \rightarrow \sim r$$
$$\implies \sim p \lor q \rightarrow \sim r $$
Question 2
1 / -0

The contrapositive of $$(p \vee q)\Rightarrow r$$ is

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

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

C
$$\sim r \Rightarrow \sim p \wedge \sim q$$

D
$$R \Rightarrow (q \vee r)$$

Solution

contrapositive of $$a \implies b $$ is $$ \sim b \implies \sim a$$
contrapositive of $$ p \lor q \implies r $$ is
$$\sim r \implies \sim ( p \lor q )$$
$$ \sim r \implies (\sim p \land \sim q )$$
Question 3
1 / -0

Which of the following is the inverse of the proposition "If a number is prime, then it is odd"?

A
If a number is not prime, then it is odd.

B
If a number is not a prime, then it is not odd.

C
If a number is not odd, then it is not a prime.

D
If a number is not odd, then it is a prime.

Solution

p : A number is prime
q : It is odd
We have $$p\rightarrow q$$
The inverse of $$p\rightarrow q$$ is $$\sim p\rightarrow\sim q$$
i.e. If a number is not a prime then it is not odd.
Question 4
1 / -0

If $$x = 5$$ and $$ y = 2$$ then $$x - 2y = 9$$. The contrapositive of this statement is

A
If $$x -2y \neq 9$$ then $$x \neq 5$$ or $$y \neq 2$$

B
If $$x -2y \neq 9 $$ then $$x \neq 5$$ and $$y \neq - 2$$

C
If $$x - 2y = 9$$ then $$x = 5$$ and $$y = -2$$

D
none of these.

Solution

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 $$ y\ne 2 $$
Question 5
1 / -0

If $$x = 5$$ and $$y = -2$$, then $$x-2y = 9$$. The contrapositive of this statement is/are

A
If $$x-2y$$ $$\neq 9$$, then $$x \neq$$ $$5$$ or $$y \neq 2$$.

B
If $$x-2y$$ $$\neq 9$$, then $$x \neq 5$$ and $$y \neq -2$$.

C
If $$x-2y = 9$$, then $$x = 5$$ and $$y = -2$$.

D
none of these.

Solution

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 $$y \neq 2$$
Question 6
1 / -0

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

A
T, F, F

B
F, F, F

C
F, T, T

D
T, T, F

Solution

$$p \rightarrow q$$ 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
Question 7
1 / -0

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."

A
True

B
May be true

C
False

D
May be false

Solution

The statement is true as it's already given that All scientists working in America are talented.
Question 8
1 / -0

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) ?

A
(i) and (ii)

B
(ii) and (iii)

C
(iii) and (iv)

D
All of these

Solution

We know, If a statement is true then its truth value is T and if statement is false then F.
Hence all the statements (i),(ii),(iii) and (iv) are true hence their truth value is T
Question 9
1 / -0

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

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

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

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

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

Solution

The contrapositive of $$p \rightarrow ( \sim q \rightarrow \sim r)$$ is
$$ \equiv \sim (\sim q \rightarrow \sim r) \rightarrow \sim p $$
$$ \equiv \sim ( q \vee \sim r) \rightarrow p \equiv (\sim q \wedge r) \rightarrow \sim p $$
Question 10
1 / -0

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

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

B
$$q:$$ Circle is a particular case of an ellipse.

C
$$r:\>\displaystyle \sqrt{13}$$ is a rational number.

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

Solution

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

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Mathematical Reasoning Test - 17

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Mathematical Reasoning Test - 17

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Question 1

$$\sim r \Rightarrow \sim p\vee q$$

$$\sim p\vee q \Rightarrow \sim r$$

$$r \Rightarrow p\wedge \sim q $$

none of these.

Solution

Question 2

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

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

$$\sim r \Rightarrow \sim p \wedge \sim q$$

$$R \Rightarrow (q \vee r)$$

Solution

Question 3

If a number is not prime, then it is odd.

If a number is not a prime, then it is not odd.

If a number is not odd, then it is not a prime.

If a number is not odd, then it is a prime.

Solution

Question 4

If $$x -2y \neq 9$$ then $$x \neq 5$$ or $$y \neq 2$$

If $$x -2y \neq 9 $$ then $$x \neq 5$$ and $$y \neq - 2$$

If $$x - 2y = 9$$ then $$x = 5$$ and $$y = -2$$

none of these.

Solution

Question 5

If $$x-2y$$ $$\neq 9$$, then $$x \neq$$ $$5$$ or $$y \neq 2$$.

If $$x-2y$$ $$\neq 9$$, then $$x \neq 5$$ and $$y \neq -2$$.

If $$x-2y = 9$$, then $$x = 5$$ and $$y = -2$$.

none of these.

Solution

Question 6

T, F, F

F, F, F

F, T, T

T, T, F

Solution

Question 7

True

May be true

False

May be false

Solution

Question 8

(i) and (ii)

(ii) and (iii)

(iii) and (iv)

All of these

Solution

Question 9

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

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

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

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

Solution

Question 10

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

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

$$r:\>\displaystyle \sqrt{13}$$ is a rational number.

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

Solution

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