GeeksforGeeks App
Open App
Browser
Continue

# Class 11 NCERT Solutions- Chapter 14 Mathematical Reasoning – Miscellaneous Exercise on Chapter 14

### (i) p: For every positive real number x, the number x – 1 is also positive.

Solution:

~p: There exists atleast a positive real number x, such that x – 1 is not positive.

### (ii) q: All cats scratch.

Solution:

~q: There exists cats that do not scratch.

### (iii) r: For every real number x, either x > 1 or x < 1.

Solution:

~r: There exists a real number x, such that neither x > 1 nor x < 1.

### (iv) s: There exists a number x such that 0 < x < 1.

Solution:

~s: There does not exist a number x, such that 0 < x < 1.

### (i) p: A positive integer is prime only if it has no divisors other than 1 and itself.

Solution:

Statement p can be understood as follows.

If a positive integer is prime, then it has no divisors other than 1 and itself.

The converse of the statement is as follows.

If a positive integer has no divisor other than 1 and itself, then it is prime.

The contrapositive of the statement is as follows.

If positive integer has divisor other than 1 and itself, then it is not prime.

### (ii) q: I go to a beach whenever it is a sunny day.

Solution:

The given statement can be understood as follows.

If it is a sunny day, then I go to a beach.

The converse of the statement is as follows.

If I go to a beach, then it is a sunny day.

The contrapositive of the statement is as follows.

If I do not go to a beach, then it is not a sunny day.

### (iii) r: If it is hot outside, then you feel thirsty.

Solution:

The converse of statement r is as follows.

If you feel thirsty, then it is hot outside.

The contrapositive of statement r is as follows.

If you do not feel thirsty, then it is not hot outside.

### (i) p: It is necessary to have a password to log on to the server.

Solution:

p: If you have a password, then you can log on to the server.

### (ii) q: There is traffic jam whenever it rains.

Solution:

q: If it rains, then there is a traffic jam.

### (iii) r: You can access the website only if you pay a subscription fee.

Solution:

r: If you pay the subscription fee, then you can access the website.

### (i) p: If you watch television, then your mind is free and if your mind is free, then you watch television.

Solution:

p: You watch television if and only if your mind is free.

### (ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly.

Solution:

q: You get an A grade if and only if you do all the homework regularly.

### (iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.

Solution:

r: A quadrilateral is equiangular if and only if it is a rectangle.

### Write the compound statements connecting these two statements with ‘And’ and ‘Or’. In both cases check the validity of the compound statement.

Solution:

The compound statement with ‘And’ is ‘25 is a multiple of 5 and 8′.

This statement is not valid, because 25 is not a multiple of 8.

The compound statement with ‘Or’ is ‘25 is a multiple of 5 or 8’.

This statement is valid, because although 25 is not a multiple of 8, it is a multiple of 5.

### (i) p: The sum of an irrational number and a rational number is irrational (by contradiction method).

Solution:

The given statement is,

p: The sum of an irrational number and a rational number is irrational.

Let us assume that the given statement, p, is false. That is, we assume that the sum of an irrational number and

a rational number is rational.

Therefore, √a + b/c = d/e is irrational where a, b, c, d and e are integers.

d/e – b/c is a rational number and √a is an irrational number.

This is a contradiction. So, our assumption is wrong.

Therefore, the sum of an irrational number and a rational number is rational.

Hence, the given statement is valid.

### (ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method).

Solution:

The given statement is,

q: If n is a real number with n > 3, then n2 > 9.

Let’s assume that n is a real number with n > 3, but n2 > 9 is false,i.e., n2 < 9.

Then, n > 3 where n is a real number.

Squaring both the sides,

n2 > (3)2

⇒ n2 > 9, which is a contradiction to our assumption that is n2 < 9.

Therefore, the given statement is valid.

### p: If a triangle is equiangular, then it is an obtuse-angled triangle.

Solution:

The given statement can be written in the following five different ways:

(i) A triangle is equiangular implies that it is obtuse-angled.

(ii) A triangle is equiangular only if it is an obtuse-angled.

(iii) For a triangle to be equiangular, it is necessary that the triangle is obtuse-angled.

(iv) For a triangle to be obtuse-angled, it is sufficient that the triangle is equiangular.

(v) If a triangle is not obtuse-angled, then the triangle can not be equiangular.

My Personal Notes arrow_drop_up
Related Tutorials