Mathematical Reasoning Test

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Mathematical Reasoning Test - 27

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Question 1
1 / -0

If p, q, r have truth values T, F, T respectively, then which of the following is $$\text{True}$$?

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

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

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

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

Solution

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.
Question 2
1 / -0

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

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

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

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

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

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

Solution

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

By distributive property

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

So option $$D$$ is correct.
Question 3
1 / -0

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

A
If X is not is prime number, then x is not odd

B
If X is not odd, then x is not a prime number

C
If X is a prime number, then x is not odd

D
If X is not a prime number, then x is odd

Solution

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
Question 4
1 / -0

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?

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

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

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

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

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

Solution

$$p:57$$ 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

$$\sim p\vee (q\wedge r)$$
$$=\sim F\vee (T\wedge T)=T\vee (T)=T$$
Question 5
1 / -0

The converse of the contrapositive of the conditional $$p\rightarrow \sim q$$ is.

A
$$p\rightarrow q$$

B
$$\sim p \rightarrow \sim q$$

C
$$\sim q \rightarrow p$$

D
$$\sim p \rightarrow q$$

Solution

The contrapositive of the statement $$p\longrightarrow \sim q\\ $$ is $$q\longrightarrow \sim p$$

The converse of the statement $$q\longrightarrow \sim p$$ is $$\sim p \longrightarrow q$$
Question 6
1 / -0

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.

A
Only the converse is true

B
Only the inverse is true

C
Both are true

D
Neither is true

E
The inverse is true, but the converse is sometimes true

Solution

The converse is: If a quadrilateral is a rectangle, then it is a square. The inverse is: If a quadrilateral is not a square, then it is not a rectangle. Both statements are false; or use Venn diagrams to picture the sets mentioned.
Question 7
1 / -0

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

A
$$p\vee q$$

B
$$p\leftrightarrow q$$

C
$$\sim p\vee \sim q$$

D
$$q\vee \sim q$$

Solution
Question 8
1 / -0

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 $$1$$.)

A
$$1, 2$$

B
$$1, 3$$

C
$$1, 4$$

D
$$2, 3$$

E
$$2, 4$$

Solution

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.
Question 9
1 / -0

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:

A
(1)

B
(2)

C
(3)

D
(4)

E
(5)

Solution

Statement: All men are good drivers.
Negation of the above statement is:
Statement is false that all men are good drivers, that is, at least one man is bad driver.
Question 10
1 / -0

Which one of the following statements is not true for the equation
$$i{ x }^{ 2 }-x+2i=0$$.
where $$i\equiv \sqrt { -1 } $$?

A
The sum of the roots is $$2$$

B
The discriminant is $$9$$

C
The roots are imaginary

D
The roots can be found by using the quadratic formula

E
The roots can be found by factoring, using imaginary numbers

Solution

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Mathematical Reasoning Test - 27

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Mathematical Reasoning Test - 27

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Question 1

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

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

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

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

Solution

Question 2

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

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

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

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

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

Solution

Question 3

If X is not is prime number, then x is not odd

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

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

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

Solution

Question 4

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

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

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

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

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

Solution

Question 5

$$p\rightarrow q$$

$$\sim p \rightarrow \sim q$$

$$\sim q \rightarrow p$$

$$\sim p \rightarrow q$$

Solution

Question 6

Only the converse is true

Only the inverse is true

Both are true

Neither is true

The inverse is true, but the converse is sometimes true

Solution

Question 7

$$p\vee q$$

$$p\leftrightarrow q$$

$$\sim p\vee \sim q$$

$$q\vee \sim q$$

Solution

Question 8

$$1, 2$$

$$1, 3$$

$$1, 4$$

$$2, 3$$

$$2, 4$$

Solution

Question 9

(1)

(2)

(3)

(4)

(5)

Solution

Question 10

The sum of the roots is $$2$$

The discriminant is $$9$$

The roots are imaginary

The roots can be found by using the quadratic formula

The roots can be found by factoring, using imaginary numbers

Solution

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