Lecture 7: Worksheet 2: – Number System

 

📘 Lecture 7: Worksheet 2: – Number System

Class: 9 CBSE
Chapter: Number System
Topic: Mixed Problems
Worksheet Title: Consolidation and Skill Practice
Total Marks: 40
Time: 1 Hour

🔹 Section A: Fill in the Blanks (1 mark each)

1. $\sqrt{50} = \_\_\_$

2. $(a^3)^2 = \_\_\_$

3. $\sqrt{2} \times \sqrt{8} = \_\_\_$

4. A number that cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, is called ___.

5. $\left(\frac{1}{4}\right)^{-1/2} = \_\_\_$

🔹 Section B: Match the Columns (2 marks each)

Match the expression in Column A with its simplified value in Column B.

Column A	Column B
a) $(5 - \sqrt{3})(5 + \sqrt{3})$	A. $8$
b) $\sqrt{12} \div \sqrt{3}$	B. $22$
c) $\frac{1}{\sqrt{7} - 2}$	C. $\frac{\sqrt{7} + 2}{3}$
d) $2^3 \cdot 2^{-2}$	D. $5^2 - 3$

🔹 Section C: Solve the Following (3 marks each)

6. Simplify:
   $(\sqrt{5} + 2)^2$

7. Rationalise the denominator and simplify:
   $\frac{2}{\sqrt{3} + \sqrt{2}}$

8. If $a = 2^{1/3}$ and $b = 2^{2/3}$, then evaluate $a \cdot b$

9. Classify the following as Rational or Irrational:
   a) $\sqrt{121}$
   b) $\frac{3\sqrt{2}}{7}$
   c) $\pi - 3$

🔹 Section D: Application-Based (4 marks each)

10. A rope is $\sqrt{75}$ metres long. Another rope is $\sqrt{27}$ metres long.
    a) Find the total length in simplest form.
    b) Classify the result as rational or irrational.

11. Simplify using exponent laws:
    a) $5^{1/2} \cdot 5^{3/2}$
    b) $\left( \frac{4}{9} \right)^{-3/2}$

🔹 Section E: Conceptual & Reasoning (5 marks each)

12. If a number is written as $2 + \sqrt{5}$, show that its conjugate is $2 - \sqrt{5}$.
    Now multiply the two and determine the result.
    What conclusion can you draw about the product of irrational conjugates?

13. A student claims:

“$\frac{22}{7}$ is the value of $\pi$, so it must be a rational number.”
Do you agree with this claim? Justify your answer and explain the difference between approximation and actual value of irrational numbers.

📌 Challenge Task (5 bonus marks)

Construct a proof that $\sqrt{3}$ is irrational using contradiction method (proof by assumption).

✅ Student Instructions:

• Read each question carefully.

• Use pencil and scale for diagrams.

• Try all sections for full understanding.

• Marks are mentioned next to each question.

📘 Lecture 7 Worksheet 2: – Number System: Solutions

🔹 Section A: Fill in the Blanks

1. $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$

2. $(a^3)^2 = a^{3 \cdot 2} = a^6$

3. $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$

4. Irrational number

5. $\left(\frac{1}{4}\right)^{-1/2} = (4)^{1/2} = \sqrt{4} = 2$

🔹 Section B: Match the Columns

Column A	Column B
a) $(5 - \sqrt{3})(5 + \sqrt{3})$	B. $22$
b) $\sqrt{12} \div \sqrt{3}$	A. $2$
c) $\frac{1}{\sqrt{7} - 2}$	C. $\frac{\sqrt{7} + 2}{3}$
d) $2^3 \cdot 2^{-2}$	D. $2^{3 - 2} = 2^1 = 2$

So matching answers:
a–B, b–A, c–C, d–D

🔹 Section C: Solve the Following

6. $(\sqrt{5} + 2)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$

7. $\frac{2}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{2(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}$
   Denominator: $3 - 2 = 1$
   Answer: $2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$

8. $a = 2^{1/3}, b = 2^{2/3} \Rightarrow a \cdot b = 2^{1/3 + 2/3} = 2^1 = 2$

a) $\sqrt{121} = 11$ – Rational
b) $\frac{3\sqrt{2}}{7}$ – Irrational
c) $\pi - 3$ – Irrational (since π is irrational)

🔹 Section D: Application-Based

a) $\sqrt{75} + \sqrt{27} = \sqrt{25 \cdot 3} + \sqrt{9 \cdot 3} = 5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$
b) Since $\sqrt{3}$ is irrational, $8\sqrt{3}$ is also irrational.

a) $5^{1/2} \cdot 5^{3/2} = 5^{1/2 + 3/2} = 5^2 = 25$
b) $\left( \frac{4}{9} \right)^{-3/2} = \left( \frac{9}{4} \right)^{3/2} = \left( \sqrt{9}/\sqrt{4} \right)^3 = (3/2)^3 = 27/8$

🔹 Section E: Conceptual & Reasoning

Given number = $2 + \sqrt{5}$, its conjugate is $2 - \sqrt{5}$
Product:
$(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1$
Conclusion: The product of conjugates $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$, always gives a rational number.

No, we do not agree.

• $\pi \approx \frac{22}{7}$, but this is only an approximation.

• Actual value of $\pi$ is non-terminating, non-repeating = irrational

• Rational approximations are used for calculations, but the true nature of $\pi$ remains irrational.

📌 Challenge Task: Proof $\sqrt{3}$ is irrational

Assume $\sqrt{3}$ is rational.
Then $\sqrt{3} = \frac{p}{q}$, where p, q are integers with no common factor and $q \neq 0$.
Squaring both sides:
$3 = \frac{p^2}{q^2} \Rightarrow p^2 = 3q^2$
So p² is divisible by 3 → p is divisible by 3 → p = 3k
Then $p^2 = 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2 \Rightarrow q$ is also divisible by 3.

So both p and q are divisible by 3, contradicting that they have no common factor.
Hence, $\sqrt{3}$ is irrational.

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