Mathematical Reasoning Test

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Mathematical Reasoning Test - 31

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Question 1
1 / -0

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

A
$$q \vee r$$

B
$$q \wedge r$$

C
$$q \rightarrow r$$

D
none of these

Solution

$${\textbf{Step -1: After applying associative and distributive law}}$$
  $$\text{Given, } \left( {p \wedge q} \right) \vee \left( { \sim p \wedge q} \right) \vee \left( { \sim q \wedge r} \right)$$
$$ \equiv \left[ {\left( {p \vee \sim p} \right) \wedge q} \right] \vee \left( { \sim q \wedge r} \right)$$ $$[\because\boldsymbol{(a\vee b)\wedge c} \boldsymbol{\equiv (a\wedge c)\vee (b\wedge c)}]$$
$${\textbf{Step -2: Apply complement law}}$$
$$ \equiv \left( {T \wedge q} \right) \vee \left( { \sim q \wedge r} \right)$$ $$[\because\boldsymbol{(a\vee\sim a)\equiv T}]$$
$${\textbf{Step -3:Apply identity law}}$$
               $$ \equiv q \vee \left( { \sim q \wedge r} \right)$$   $$[\because\boldsymbol{(T\wedge a)\equiv q}]$$
$${\textbf{Step -4: Apply distributive law}}$$
               $$ \equiv \left( {q \vee \sim q} \right) \wedge \left( {q \vee r} \right)$$   $$[\because\boldsymbol{(a\wedge b)\vee c} \boldsymbol{\equiv (a\vee c)\wedge (b\vee c)}]$$
$${\textbf{Step -5: Apply complement law}}$$
                $$ \equiv T \wedge \left( {q \vee r} \right)$$   $$[\because\boldsymbol{(a\vee\sim a)\equiv T}]$$
$${\textbf{Step -6: Apply identity law}}$$
  $$ \equiv q \vee r$$ $$[\because\boldsymbol{(T\wedge a)\equiv q}]$$
$${\textbf{Hence, }}$$$$\mathbf{\left( {p \wedge q} \right) \vee \left( { \sim p \wedge q} \right) \vee \left( { \sim q \wedge r} \right) \equiv q \vee r}$$
Question 2
1 / -0

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:

A
You are not wet when you are in river or it rains.

B
You are wet when you are in river and it rains.

C
You are wet when it rains and you are in a river

D
You are wet when it rains or you are in a river.

Solution
Question 3
1 / -0

$$∼(p⇒q)⟺∼p\vee ∼q \, is$$

A
a tautology

B
a contradiction

C
neither a tautology nor a contradiction

D
cannot come to any conclusion

Solution
Question 4
1 / -0

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.

A
Counter example

B
Contrapositive method

C
Direct method

D
By Contradiction

Solution

Given : The given statement is in the form "if p then q" we have to show that this is false. If p then $$\sim q$$.
If $$n = 99$$ 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
if $$n$$ 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.
Question 5
1 / -0

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:

A
The statement p is true and the statement q is not true

B
The statement p is false and the statement q is true.

C
The statement p is true and the statement q is false

D
The statement p is true and the statement q is true

Solution
Question 6
1 / -0

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

A
$$(p\vee q)$$

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

C
$$p\wedge(q)$$

D
$$p\vee(q)$$

Solution

$$\Rightarrow [(P)^q]$$
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.
Question 7
1 / -0

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

A
Counter example

B
Direct method

C
By Contradiction

D
Contrapositive method

Solution
Question 8
1 / -0

An electrical circuit for a set of 4 lights depends on a system of switches A, B, C and D. When these switches work they have the following effect on the lights: They each toggle the state of two lights (i.e. on becomes off and off becomes on). The lights that each switch controls are as follows.
A B C D
1 and 2 2 and 4 1 and 3 3 and 4
In configuration I shown below, switches C-B-D-A are asserted in order, resulting in configuration 2. One switch did not work and had no effect at all. Which was that switch?

A
A

B
B

C
C

D
D

Solution
Question 9
1 / -0

The converse of $$\sim p \rightarrow q$$ is equivalent to

A
$$p\rightarrow \sim q$$

B
$$p\rightarrow q$$

C
$$\sim q\rightarrow p$$

D
$$q\rightarrow \sim p$$
Question 10
1 / -0

Let P(n) denote the statement that $$n^2+n$$ is odd. It is seen that $$P(n)\Rightarrow P(n+1), P(n)$$ is true for all.

A
$$n > 1$$

B
n

C
$$n > 2$$

D
None of these

Solution

$$P(n)=n^2+n$$. It is always odd (by statement) but square of any odd and
also, sum of two odd number is always even. So, for no any 'n' for which its statement is true.