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RD Chapter 31- Mathematical Reasoning Ex-31.6 Interview Questions Answers

Question 1 :
Check the validity of the following statements:
(i) p: 100 is a multiple of 4 and 5.
(ii) q: 125 is a multiple of 5 and 7.
(iii) r: 60 is a multiple of 3 or 5.

Answer 1 :

(i) p: 100 is a multiple of 4 and 5.
We know that 100 is a multiple of 4 as well as 5. So, the given statement is true.
Hence, the statement is true.
(ii) q: 125 is a multiple of 5 and 7
We know that 125 is a multiple of 5 and not a multiple of 7. So, the given statement is false.
Hence, the statement is false.
(iii) r: 60 is a multiple of 3 or 5.
We know that 60 is a multiple of 3 as well as 5. So, the given statement is true.
Hence, the statement is true.

Question 2 :
Check whether the following statement is true or not:
(i) p: If x and y are odd integers, then x + y is an even integer.
(ii) q : if x, y are integer such that xy is even, then at least one of x and y is an even integer.

Answer 2 :

(i) p: If x and y are odd integers, then x + y is an even integer.
Let us assume that ‘p’ and ‘q’ be the statements given by
p: x and y are odd integers.
q: x + y is an even integer
the given statement can be written as :
if p, then q.
Let p be true. Then, x and y are odd integers
x = 2m+1, y = 2n+1 for some integers m, n
x + y = (2m+1) + (2n+1)
x + y = (2m+2n+2)
x + y = 2(m+n+1)
x + y is an integer
q is true.
So, p is true and q is true.
Hence, “if p, then q “is a true statement.”
(ii) q: if x, y are integer such that xy is even, then at least one of x and y is an even integer.
Let us assume that p and q be the statements given by
p: x and y are integers and xy is an even integer.
q: At least one of x and y is even.
Let p be true, and then xy is an even integer.
So,
xy = 2(n + 1)
Now,
Let x = 2(k + 1)
Since, x is an even integer, xy = 2(k + 1). y is also an even integer.
Now take x = 2(k + 1) and y = 2(m + 1)
xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)
So, it is also true.
Hence, the statement is true.

Question 3 :
Show that the statement
p : “If x is a real number such that x3 + x = 0, then x is 0” is true by
(i) Direct method
(ii) method of Contrapositive
(iii) method of contradiction

Answer 3 :

(i) DirectMethod:

Let us assume that ‘q’ and ‘r’ be the statements given by

q: x is a real number such that x+ x=0.

r: x is 0.

The given statement can be written as:

if q, then r.

Let q be true. Then, x is a real number such that x+x = 0

x is a real number such that x(x+ 1) = 0

x = 0

r is true

Thus, q is true

Therefore, q is true and r is true.

Hence, p is true.

(ii) Methodof Contrapositive:

Let r be false. Then,

R is not true

x ≠ 0, xR

x(x2+1)≠0, xR

q is not true

Thus, -r = -q

Hence, p : q and r is true

(iii) Methodof Contradiction:

If possible, let p be false. Then,

P is not true

-p is true

-p (p => r) is true

q and –r is true

x is a real number such that x3+x = 0and x≠ 0

x =0 and x≠0

This is a contradiction.

Hence, p is true.

Question 4 :
Show that the following statement is true by the method of the contrapositive
p: “If x is an integer and x2 is odd, then x is also odd.”

Answer 4 :

Let us assume that ‘q’ and ‘r’ be the statements given

q: x is an integer and x2 is odd.

r: x is an odd integer.

The given statement can be written as:

p: if q, then r.

Let r be false. Then,

x is not an odd integer, then x is an even integer

x = (2n) for some integer n

x2 = 4n2

x2 is an even integer

Thus, q is False

Therefore, r is false and q is false

Hence, p: “ if q, then r” is a true statement.

Question 5 :

Show that thefollowing statement is true
“The integer n is even if and only if n2 iseven”

Answer 5 :

Let the statements,

p: Integer n is even

q: If n2 is even

Let p be true. Then,

Let n = 2k

Squaring both the sides, we get,

n2 = 4k2

n2 = 2.2k2

n2 is an even number.

So, q is true when p is true.

Hence, the given statement is true.

Question 6 :

By giving a counterexample, show that the following statement is not true.
p: “If all the angles of a triangle are equal, thenthe triangle is an obtuse angled triangle.”

Answer 6 :

Let us consider a triangle ABC with all angles equal.

Then, each angle of the triangle is equal to 60.

So, ABC is not an obtuse angle triangle.

Hence, the statement “p: If all the angles of a triangle areequal, then the triangle is an obtuse angled triangle” is False.

Question 7 :

Which of thefollowing statements are true and which are false? In each case give a validreason for saying so
(i) p: Each radius of a circle is a chord of thecircle.
(ii) q: The centre of a circle bisect each chord ofthe circle.
(iii) r: Circle is a particular case of an ellipse.
(iv) s: If x and y are integers such that x > y,then – x < – y.
(v) t: √11 is a rational number.

Answer 7 :

(i) p: Eachradius of a circle is a chord of the circle.

The Radius of the circle is not it chord.

Hence, this statement is False.

(ii) q:The centre of a circle bisect each chord of the circle.

A chord does not have to pass through the center.

Hence, this statement is False.

(iii) r:Circle is a particular case of an ellipse.

A circle can be an ellipse in a particular case when the circlehas equal axes.

Hence, this statement is true.

(iv) s:If x and y are integers such that x > y, then – x < – y.

For any two integers, if x – y id positive then –(x-y) isnegative.

Hence, this statement is true.

(v) t:√11 is a rational number.

Square root of prime numbers is irrational numbers.

Hence, this statement is False.

Question 8 :

Determine whetherthe argument used to check the validity of the following statement is correct:
p: “If x2 is irrational, then x isrational.”
The statement is true because the number x2 =π2 is irrational, therefore x = π is irrational.

Answer 8 :

Argument Used: x2 = π2 isirrational, therefore x = π is irrational.

p: “If x2 is irrational, then x is rational.”

Let us take an irrational number given by x = √k, where k is arational number.

Squaring both sides, we get,

x2 = k

x2 is a rational number and contradicts ourstatement.

Hence, the given argument is wrong.


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RD Chapter 31- Mathematical Reasoning Ex-31.6 Contributors

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