Solved Worksheet 1: Number System

✅ Solved Worksheet: Number System – Mixed Problems

Class: 9 CBSE
Total Marks: 40 | Time: 60 Minutes

🔹 Section A: Multiple Choice Questions (1 mark each)

1. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) = 7 - 5 = \boxed{2}$ → Answer: A

2. Rational number: $\boxed{\frac{3}{4}}$ → Answer: C

3. $(2\sqrt{3})^2 = 4 \times 3 = \boxed{12}$ → Answer: A

4. $16^{3/4} = (2^4)^{3/4} = 2^3 = \boxed{8}$ → Answer: B

5. $\left( \frac{1}{5^2} \right)^{-1} = 5^2 = \boxed{25}$ → Answer: B

🔹 Section B: Very Short Answer (2 marks each)

6. Rationalise:

$$\frac{1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{3 - 1} = \boxed{\frac{\sqrt{3} - 1}{2}}$$

7. $3 - \sqrt{11}$: Irrational, because $\sqrt{11}$ is irrational and difference with rational number remains irrational. → \boxed{\text{Irrational}}

8. $(5 + \sqrt{2})^2 = 25 + 2 \times 5 \times \sqrt{2} + 2 = \boxed{27 + 10\sqrt{2}}$

9. Geometrical Representation of $\sqrt{2}$:
   Draw a right-angled triangle with both legs of 1 unit. The hypotenuse is $\sqrt{2}$.
   Using compass, draw an arc from origin with radius $\sqrt{2}$. The intersection point on number line is $\boxed{\sqrt{2}}$.

a) $27^{1/3} = \boxed{3}$
b) $81^{3/4} = (3^4)^{3/4} = 3^3 = \boxed{27}$

🔹 Section C: Short Answer (3 marks each)

$$\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{(\sqrt{5} + \sqrt{2})^2}{5 - 2} = \frac{5 + 2 + 2\sqrt{10}}{3} = \boxed{\frac{7 + 2\sqrt{10}}{3}}$$

a) $(3^2 \cdot 3^3) \div 3^4 = 3^{2+3-4} = 3^1 = \boxed{3}$

b) $(2^3)^2 \cdot 2^{-4} = 2^{6} \cdot 2^{-4} = 2^{2} = \boxed{4}$

a) $\sqrt{49} = 7$ → \boxed{\text{Rational}}
b) $\frac{2}{\sqrt{7}}$ → Denominator irrational → \boxed{\text{Irrational}}
c) $\sqrt{3} + \sqrt{7}$ → sum of irrationals not simplifiable → \boxed{\text{Irrational}}

🔹 Section D: Long Answer (4 marks each)

$$(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 = [4 + 4\sqrt{3} + 3] - [4 - 4\sqrt{3} + 3] = 7 + 4\sqrt{3} - (7 - 4\sqrt{3}) = \boxed{8\sqrt{3}}$$

$$\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \\ = \frac{3(\sqrt{5} + \sqrt{2}) + 2(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} \\ = \frac{3\sqrt{5} + 3\sqrt{2} + 2\sqrt{5} - 2\sqrt{2}}{5 - 2} \\ = \frac{5\sqrt{5} + \sqrt{2}}{3} = \boxed{\frac{5\sqrt{5} + \sqrt{2}}{3}}$$

16. Prove $\sqrt{2}$ is irrational – contradiction method:

Assume $\sqrt{2} = \frac{p}{q}$, where $p, q$ are integers, $\gcd(p, q) = 1$

Squaring:
$2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2$

So, $p^2$ even ⇒ $p$ even ⇒ $p = 2k$
Then: $p^2 = 4k^2 = 2q^2 ⇒ q^2 = 2k^2$ ⇒ $q$ also even

⇒ $p$ and $q$ both even ⇒ Contradiction to assumption that they are coprime.

∴ $\boxed{\sqrt{2} \text{ is irrational}}$

a)

$$\left( \frac{8}{27} \right)^{2/3} = \frac{8^{2/3}}{27^{2/3}} = \frac{(2^3)^{2/3}}{(3^3)^{2/3}} = \frac{2^2}{3^2} = \frac{4}{9} \Rightarrow \boxed{\frac{4}{9}}$$

b)

$$8^{2/3} \div 2^{4/3} = (2^3)^{2/3} \div 2^{4/3} = 2^2 \div 2^{4/3} = 2^{2 - 4/3} = 2^{2/3} = \boxed{2^{2/3}}$$

🔹 Section E: Challenge Question (5 marks)

Rohit's statement:
“π is defined as the ratio of circumference to diameter ⇒ it must be rational” is incorrect.

Explanation:

• Though $\pi = \frac{C}{d}$, both circumference and diameter are real numbers.

• The ratio is not always rational.

• π has a non-terminating, non-repeating decimal expansion, and cannot be expressed as a fraction ⇒ Irrational.

Approximation Check:

Radius $r = 7 \text{ cm}$
Circumference $C = 2\pi r \approx 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}$

True value using $\pi \approx 3.1416$:
$C = 2 \times 3.1416 \times 7 = 43.9824 \approx 44 \text{ cm}$

Hence, $\frac{22}{7}$ is a very good rational approximation of π, but π itself remains irrational.

Would you like a print-ready PDF version of this worksheet + solution set with your name/school/brand/logo at the top?