This set of Class 9 Maths Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Representing Real Numbers on the Number Line & Real Numbers Operations”.

1. Summation or Subtraction of two non-zero rational numbers is a/an __________

a) natural number

b) whole number

c) rational Number

d) irrational number

View Answer

Explanation: Let’s take two non-zero rational numbers to understand this.

Consider two numbers, 1/2 and 3/4

Adding these two, we get 1/2 + 3/4 = {4 + (2*3)}/2*4

= 10/8

= 5/4

Which is rational number since its result is terminating (5/4 = 1.25).

Subtracting these two, we get 3/4 – 1/2 = {(2*3) – 4}/2*4

= 2/8

= 1/4

Which is rational number since its result is terminating (1/4 = 0.25).

Hence, result of two terminating numbers would also be terminating so we can say that Summation or Subtraction of two rational number is always rational number.

However, as we saw in this example, it is not necessary that the result of summation or subtraction of two non-zero rational number would be Integer. So

**the result may or may not be Natural number or Whole number**.

2. If we multiply or divide two rational numbers, we get a/an __________

a) natural number

b) whole number

c) rational Number

d) irrational number

View Answer

Explanation: Let’s take two non-zero rational numbers to understand this.

Consider two numbers, 1/2 and 3/4

Multiplying these two, we get 1/2 * 3/4 = 3/8

Which is rational number since its result is terminating (3/8= 0.875).

Subtracting these two, we get (3/4) / (1/2) = (3*2) / 4

= 6/4

Which is rational number since its result is terminating (6/4= 1.5).

Hence, result of two terminating numbers would also be terminating so we can say that Summation or Subtraction of two rational number is always rational number.

However, as we saw in this example, it is not necessary that the result of product or division of two non-zero rational number would be Integer. So

**the result may or may not be Natural number or Whole number**.

3. If we add or subtract an irrational number and a rational number (non-zero), then we get a/an __________ number.

a) natural

b) rational

c) irrational

d) rational or irrational

View Answer

Explanation: Let’s take two non-zero numbers to understand this. One rational and one irrational number.

Consider two numbers, 3 and \(\sqrt{2}\)

Adding these two, we get 3 + \(\sqrt{2}\) Now, \(\sqrt{2}\) = 1.414213…

Expansion of √2 is non-terminating, hence result of 3 + \(\sqrt{2}\) would be non-terminating and non-recurring.

The same will happen while subtracting this.

Hence, we can say that if we add or subtract an irrational number and a rational number (non-zero) then we get an irrational number.

4. If we add or subtract two irrational numbers, we get __________ number.

a) natural

b) rational

c) irrational

d) rational or irrational

View Answer

Explanation: Let’s take two irrational numbers to understand this.

Consider \(\sqrt{2}\) and –\(\sqrt{2}\) By adding these, we get \(\sqrt{2}\) + (-\(\sqrt{2}\)) = 0

Which is

**rational**number since it is terminating.

Now, consider \(\sqrt{5}\) and \(\sqrt{2}\) By adding these, we get \(\sqrt{5}\) + \(\sqrt{2}\) Which is

**irrational**since the expansion of √5 and √2 is non-terminating and non-recurring.

Now, let’s take two numbers for subtraction.

Consider \(\sqrt{2}\) and \(\sqrt{2}\).

By subtracting these, we get \(\sqrt{2} – \sqrt{2}\) = 0

Which is

**rational**number since it is terminating.

Consider \(\sqrt{5}\) and \(\sqrt{2}\) By subtracting these, we get \(\sqrt{5}\) – \(\sqrt{2}\) Which is irrational since the expansion of \(\sqrt{5}\) and \(\sqrt{2}\) is non-terminating and non-recurring.

Hence, we can say that if we add or subtract two irrational numbers, we get rational or irrational number.

5. 3 + \(\sqrt{2}\) is a/an __________ number.

a) natural

b) irrational

c) rational

d) whole

View Answer

Explanation: \(\sqrt{2}\) = 1.414213…

Since the expansion of √2 is non-terminating and non-recurring, result of 3 + \(\sqrt{2}\) would be also non-terminating and non-recurring. So we can say that 3 + \(\sqrt{2}\) is an irrational number.

6. π – π is __________ number.

a) natural

b) irrational

c) rational

d) negative

View Answer

Explanation: π – π = 0

Which is rational number since it is terminating.

7. By simplifying \((\sqrt{7}+\sqrt{5}) * (\sqrt{7}-\sqrt{5})\), we get __________

a) 24

b) 2

c) 12

d) 74

View Answer

Explanation: We know that \((\sqrt{a}+\sqrt{b}) * (\sqrt{a}-\sqrt{b})\) = a-b

By applying that rule here, we get = \((\sqrt{7}+\sqrt{5}) * (\sqrt{7}-\sqrt{5})\)

= 7-5

= 2.

8. By rationalising the denominator of \(\frac{1}{\sqrt{3}}\), we get __________

a) \(\frac{\sqrt{3}}{3}\)

b) \(\sqrt{3}\)

c) \(\frac{3}{\sqrt{3}}\)

d) \(3\sqrt{3}\)

View Answer

Explanation: When the denominator of an expression contains a term with a square root, the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator.

By multiplying \(\frac{1}{\sqrt{3}}\) by \(\frac{\sqrt{3}}{\sqrt{3}}\), we will get same expression since \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1.

Therefore, \(\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * (\frac{\sqrt{3}}{\sqrt{3}}) = \frac{\sqrt{3}}{3}\).

9. By simplifying \((\sqrt{3}+\sqrt{2}) * (\sqrt{3}+\sqrt{2})\), we get __________

a) 5 + 4√6

b) 5 + 2√6

c) 1

d) 5

View Answer

Explanation: We know that \((\sqrt{a}+\sqrt{b}) * (\sqrt{a}+\sqrt{b}) = a + 2\sqrt{a}\sqrt{b} + b.\)

By applying that rule here, we get \((\sqrt{3}+\sqrt{2}) * (\sqrt{3}+\sqrt{2}) = 3 + 2\sqrt{3}\sqrt{2} + 2

= 5 + 2\sqrt{6}\) since \(\sqrt{a}\sqrt{b} = \sqrt{(ab)}\),

\(\sqrt{3}\sqrt{2} = \sqrt{(3*2)} = \sqrt{6}\).

10. By rationalising the denominator of \(\frac{1}{(6-\sqrt{3})}\), we get __________

a) \(\frac{6-\sqrt{3}}{3}\)

b) \(\frac{6-\sqrt{3}}{33}\)

c) \(\frac{6+\sqrt{3}}{3}\)

d) \(\frac{6+\sqrt{3}}{33}\)

View Answer

Explanation: When the denominator of an expression contains a term with a square root, the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator.

By multiplying \(\frac{1}{(6-\sqrt{3})}\) by \(6+\sqrt{3}\), we will get same expression since \(\frac{6+\sqrt{3}}{6+\sqrt{3}}\) = 1.

Therefore, \(\frac{1}{(6-\sqrt{3})} = \frac{1}{(6-\sqrt{3})} * \frac{6+\sqrt{3}}{6+\sqrt{3}} = \frac{6+\sqrt{3}}{(6*6) – (\sqrt{3}*\sqrt{3})}\)

= \(\frac{6+\sqrt{3}}{(36-3)}\)

= \(\frac{6+\sqrt{3}}{33}\).

11. By rationalising the denominator of \(\frac{1}{5+\sqrt{7}}\), we get __________

a) \(\frac{5-\sqrt{7}}{5+\sqrt{7}}\)

b) \(\frac{5-\sqrt{7}}{18}\)

c) \(\frac{5+\sqrt{7}}{18}\)

d) \(\frac{5+\sqrt{7}}{5-\sqrt{7}}\)

View Answer

Explanation: When the denominator of an expression contains a term with a square root, the process of converting it to an equivalent expression whose denominator is a rational number is called rationalising the denominator.

By multiplying \(\frac{1}{5+\sqrt{7}}\) by \(5-\sqrt{7}\), we will get same expression since \(\frac{5-\sqrt{7}}{5-\sqrt{7}}\) = 1.

Therefore, \(\frac{1}{5+\sqrt{7}} = {\frac{1}{5+\sqrt{7}}} * (\frac{5-\sqrt{7}}{5-\sqrt{7}})\)

= \(\frac{5-\sqrt{7}}{(5*5)-(\sqrt{7}*\sqrt{7})}\)

= \(\frac{5-\sqrt{7}}{25-7}\)

= \(\frac{5-\sqrt{7}}{18}\).

12. Summation or Subtraction of rational number and irrational number is always a/an __________ number.

a) natural

b) whole number

c) rational number

d) irrational number

View Answer

Explanation: Let’s take one rational number and one irrational number for example.

Consider 3 and \(\sqrt{2}\)

We know that 3 is rational number and \(\sqrt{2}\) is irrational number.

By adding 3 and \(\sqrt{2}\), we get 3+\(\sqrt{2}\)

We know that \(\sqrt{2}\) = 1.414213…

Since the expansion of \(\sqrt{2}\) is non-terminating and non-recurring, expansion of 3+\(\sqrt{2}\) will also be non-terminating and non-recurring. Therefore result of summation will be irrational number.

The same will happen if we subtract \(\sqrt{2}\) from 3.

Hence we can say that Summation or Subtraction of rational number and irrational number is always irrational number.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 9**.

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