Worksheet: Number System – Mixed Problems

📄 Worksheet 1: Number System – Mixed Problems

Class: 9 CBSE
Chapter: Number System
Topic: Simplification, Rational/Irrational, Rationalisation, Exponents
Total Marks: 40
Time: 60 Minutes

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

Choose the correct option and write the answer.

1. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5})$ equals:

   • A. 2

   • B. $\sqrt{2}$

   • C. 12

   • D. $\sqrt{35}$

2. Which of the following is a rational number?

   • A. $\sqrt{2}$

   • B. $\frac{1}{\sqrt{3}}$

   • C. $\frac{3}{4}$

   • D. $\pi$

3. $(2\sqrt{3})^2 =$

   • A. 12

   • B. 6

   • C. $4\sqrt{3}$

   • D. 9

4. $16^{3/4}$ is equal to:

   • A. 4

   • B. 8

   • C. 16

   • D. 2

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

   • A. 5

   • B. 25

   • C. $\frac{1}{25}$

   • D. $-25$

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

Answer in one step wherever possible.

6. Rationalise: $\frac{1}{\sqrt{3} + 1}$

7. Classify as rational or irrational: $3 - \sqrt{11}$

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

9. Represent $\sqrt{2}$ geometrically on a number line. (Sketch or describe method)

10. Find:
    a) $27^{1/3}$
    b) $81^{3/4}$

🔹 Section C: Short Answer Questions (3 marks each)

Show working where required.

11. Simplify and express in simplest form:
    $\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

12. Evaluate:
    a) $(3^2 \cdot 3^3) \div 3^4$
    b) $(2^3)^2 \cdot 2^{-4}$

13. Without actual calculation, state whether the following are rational or irrational. Justify:

    • a) $\sqrt{49}$

    • b) $\frac{2}{\sqrt{7}}$

    • c) $\sqrt{3} + \sqrt{7}$

🔹 Section D: Long Answer Questions (4 marks each)

Explain each step clearly.

14. Simplify the expression:

$$(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2$$

15. Rationalise and simplify:

$$\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}}$$

16. Prove that $\sqrt{2}$ is irrational.
    (Use contradiction method)

17. Evaluate:

$$\left( \frac{8}{27} \right)^{2/3}$$

Then use exponent laws to show:

$$8^{2/3} \div 2^{4/3}$$

🔹 Section E: Challenge Question (5 marks)

This tests higher-order thinking.

18. Rohit says that since π is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of π using a circle of radius 7 cm (Use $\pi \approx \frac{22}{7}$) and check the accuracy of this approximation.

✅ Instructions for Students:

• Solve each section step by step.

• For geometry-related questions, draw figures neatly.

• Highlight final answers.

• Use separate sheets if necessary.

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Worksheet: Number System – Mixed Problems

📄 Worksheet: Number System – Mixed Problems

Class: 9 CBSE

Chapter: Number System

Topic: Simplification, Rational/Irrational, Rationalisation, Exponents

Total Marks: 40

Time: 60 Minutes

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

Choose the correct option and write the answer.

1. \( (\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) \) equals:

• 2
• \( \sqrt{2} \)
• 12
• \( \sqrt{35} \)

2. Which of the following is a rational number?

• \( \sqrt{2} \)
• \( \frac{1}{\sqrt{3}} \)
• \( \frac{3}{4} \)
• \( \pi \)

3. \( (2\sqrt{3})^2 = \)

• 12
• 6
• \( 4\sqrt{3} \)
• 9

4. \( 16^{3/4} \) is equal to:

• 4
• 8
• 16
• 2

5. \( \left( \frac{1}{5^2} \right)^{-1} = \)

• 5
• 25
• \( \frac{1}{25} \)
• -25

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

Answer in one step wherever possible.

6. Rationalise: \( \frac{1}{\sqrt{3} + 1} \)
7. Classify as rational or irrational: \( 3 - \sqrt{11} \)
8. Simplify: \( (5 + \sqrt{2})^2 \)
9. Represent \( \sqrt{2} \) geometrically on a number line. (Sketch or describe method)
10. Find:
    • a) \( 27^{1/3} \)
    • b) \( 81^{3/4} \)

🔹 Section C: Short Answer Questions (3 marks each)

11. Simplify and express in simplest form: \( \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \)
12. Evaluate:
    • a) \( (3^2 \cdot 3^3) \div 3^4 \)
    • b) \( (2^3)^2 \cdot 2^{-4} \)
13. Without actual calculation, state whether the following are rational or irrational. Justify:
    • a) \( \sqrt{49} \)
    • b) \( \frac{2}{\sqrt{7}} \)
    • c) \( \sqrt{3} + \sqrt{7} \)

🔹 Section D: Long Answer Questions (4 marks each)

14. Simplify the expression:
    \( (2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 \)
15. Rationalise and simplify:
    \( \frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \)
16. Prove that \( \sqrt{2} \) is irrational. (Use contradiction method)
17. Evaluate:
    • \( \left( \frac{8}{27} \right)^{2/3} \)
    • Then use exponent laws to show: \( 8^{2/3} \div 2^{4/3} \)

🔹 Section E: Challenge Question (5 marks)

18. Rohit says that since \( \pi \) is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of \( \pi \) using a circle of radius 7 cm (Use \( \pi \approx \frac{22}{7} \)) and check the accuracy of this approximation.

✅ Instructions for Students:

• Solve each section step by step.
• For geometry-related questions, draw figures neatly.
• Highlight final answers.
• Use separate sheets if necessary.

Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

 This lecture includes simplification of expressions with square roots, classification of numbers, rationalisation of denominators, and laws of exponents.

📘 Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

🔹 Topic 1: Simplification Involving Square Roots

Let’s begin with examples that show how to simplify algebraic expressions containing square roots.

📍 Example 15: Simplify

1. $(5 + \sqrt{7})(2 + \sqrt{5})$
   Use distributive law (FOIL method):

   $$= 5 \cdot 2 + 5 \cdot \sqrt{5} + \sqrt{7} \cdot 2 + \sqrt{7} \cdot \sqrt{5} = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$$

2. $(5 + \sqrt{5})(5 - \sqrt{5})$
   Identity: $(a + b)(a - b) = a^2 - b^2$

   $$= 25 - 5 = 20$$

3. $(\sqrt{3} + \sqrt{7})^2$
   Identity: $(a + b)^2 = a^2 + 2ab + b^2$

   $$= 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}$$

4. $(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})$
   Identity: $(a - b)(a + b) = a^2 - b^2$

   $$= 11 - 7 = 4$$

🔹 Topic 2: Classifying Numbers as Rational or Irrational

A number is irrational if it cannot be expressed as a fraction $\frac{p}{q}$, where p and q are integers and $q \ne 0$.

📍 Classify the following:

1. $2 - \sqrt{5}$ → Irrational
   (Rational - Irrational = Irrational)

2. $(3 + \sqrt{23}) - \sqrt{23} = 3$ → Rational

3. $\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$ → Rational

4. $\frac{1}{\sqrt{2}}$ → Irrational

5. $2\pi$ → Irrational

🔹 Topic 3: Simplifying Square Root Expressions

Use standard identities and combine like terms.

Examples:

1. $(3 + \sqrt{3})(2 + \sqrt{2})$

2. $(3 + \sqrt{3})(3 - \sqrt{3})$

3. $(\sqrt{5} + \sqrt{2})^2$

4. $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$

👉 Use formulas like:

• $(a + b)^2 = a^2 + 2ab + b^2$

• $(a + b)(a - b) = a^2 - b^2$

🔹 Topic 4: Think and Discuss

📌 Why is π irrational, even though it's defined as a ratio (C/d)?

• π is defined as the ratio of circumference to diameter, but it's not a ratio of two integers.

• So while it appears rational in form, it is not expressible exactly as a fraction.

• Its decimal form is non-terminating and non-repeating, making it irrational.

🔹 Topic 5: Number Line Representation

📍 Represent $\sqrt{3}$ on the number line using geometry (similar to the method in Lecture 5).

1. Mark 0 and 1 on the number line.

2. Construct segment 1.5 units from 0, form a right triangle.

3. Use Pythagoras’ Theorem to locate $\sqrt{3}$.

🔹 Topic 6: Rationalising the Denominator

To rationalise means to eliminate the square root from the denominator.

Examples:

1. $\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

2. $\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}$

3. $\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{3}$

4. $\frac{1}{\sqrt{7} - 2} = \text{Multiply numerator and denominator by } \sqrt{7} + 2$

🔹 Topic 7: Laws of Exponents

Let $a \ne 0$, and m, n are integers.

1. $a^m \cdot a^n = a^{m+n}$

2. $(a^m)^n = a^{mn}$

3. $\frac{a^m}{a^n} = a^{m-n}$

4. $a^m \cdot b^m = (ab)^m$

🔹 Exercise Practice (from Ex 1.5)

1. Find:

• $64^{\frac{2}{3}} = (4^3)^{2/3} = 4^2 = 16$

• $32^{\frac{3}{5}} = (2^5)^{3/5} = 2^3 = 8$

2. Simplify powers using laws of exponents:

• $2^3 \cdot 2^5 = 2^8$

• $\left( \frac{1}{3^7} \right)^{-1} = 3^7$

• $\frac{11^{\frac{1}{2}} \cdot 11^{\frac{1}{4}}}{11^{\frac{3}{4}}} = 11^{(1/2 + 1/4 - 3/4)} = 11^0 = 1$

📝 Homework

1. Rationalise: $\frac{1}{\sqrt{3} + \sqrt{2}}$

2. Classify: $\sqrt{49} + \pi$

3. Simplify: $(\sqrt{3} + 1)^2$

4. Prove: $\sqrt{2}$ is irrational.

Lecture 5: Geometrical Representation of Square Roots

📘 Lecture 5: Geometrical Representation of Square Roots & Properties of Square Roots

🔹 Topic 1: What does √a mean?

For any real number $a > 0$,
$\sqrt{a} = b \text{ means } b^2 = a \text{ and } b > 0$

This definition works for both:

• Natural numbers (e.g., √4 = 2, because 2² = 4)

• Real numbers (e.g., √3.5)

🔹 Topic 2: How to Represent √x Geometrically?

We now explore how to construct $\sqrt{x}$ geometrically when $x$ is a positive real number.

📐 Example: Constructing √3.5 geometrically

Refer to Fig. 1.11

Steps:

1. Draw a line segment AB = 3.5 units on a number line.

2. From point B, mark 1 unit to the right. Let that point be C.

3. Find midpoint of AC, mark it as O.

4. Draw a semicircle with center O and radius OC.

5. Draw a perpendicular line from point B to the semicircle. Let it intersect at point D.

6. Then, BD = √3.5

✅ This gives a geometric way to represent square root of 3.5.

📐 General Case: Represent √x geometrically

Refer to Fig. 1.12

Steps for any real number $x > 0$:

1. Mark AB = x units.

2. From B, mark 1 unit to the right, name the point as C.

3. Find the midpoint O of AC.

4. Draw a semicircle with diameter AC.

5. Draw a perpendicular from B to the semicircle and mark intersection point as D.

6. Then, BD = √x

🧠 Proof using Pythagoras Theorem

In △OBD, ∠OBD = 90°, so it’s a right triangle.

Let’s calculate:

• Radius of semicircle =
  $OC = \frac{x + 1}{2}$

• OB =
  $x - \frac{x + 1}{2} = \frac{x - 1}{2}$

Using Pythagoras:

$$BD^2 = OD^2 - OB^2 = \left( \frac{x+1}{2} \right)^2 - \left( \frac{x-1}{2} \right)^2 = \frac{(x+1)^2 - (x-1)^2}{4} = \frac{4x}{4} = x$$

Thus,

$$BD = \sqrt{x}$$

📘 Topic 3: Properties of Square Roots

Let $a, b > 0$ be real numbers.

1. $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

2. $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

3. $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$

4. $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$

5. $(\sqrt{a} + \sqrt{b})(\sqrt{c} + \sqrt{d}) = \sqrt{ac} + \sqrt{ad} + \sqrt{bc} + \sqrt{bd}$

6. $(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b$

📚 Exercise for Practice

1. Geometrically construct $\sqrt{2.5}$ using the method described.

2. Prove that $(\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3$

3. Use Pythagoras Theorem to verify that your geometric construction is correct.

4. Show that $\sqrt{8} \cdot \sqrt{2} = \sqrt{16}$

📌 Key Takeaway

We can visualize square roots of real numbers on a number line using semicircles and perpendicular lines, and use algebraic identities to simplify square root expressions.

Lecture 4: Converting Recurring Decimals to Rational Numbers

📘 Class 9 – CBSE Mathematics

Chapter 1: Number System

Lecture 4: Converting Recurring Decimals to Rational Numbers + Exercise 1.3

🔹 Examples: Converting Recurring Decimals to Rational Numbers

✅ Example 7

Convert: $0.3333\ldots = 0.\overline{3}$

Let $x = 0.3333\ldots$

Multiply both sides by 10:

$$10x = 3.3333\ldots$$

Now subtract:

$$10x - x = 3.3333\ldots - 0.3333\ldots = 3 \Rightarrow 9x = 3 \Rightarrow x = \frac{1}{3}$$

✅ Example 8

Convert: $1.272727\ldots = 1.\overline{27}$

Let $x = 1.272727\ldots$

Multiply both sides by 100 (since 2 digits repeat):

$$100x = 127.272727\ldots$$

Now subtract:

$$100x - x = 127.2727\ldots - 1.2727\ldots = 126 \Rightarrow 99x = 126 \Rightarrow x = \frac{126}{99} = \frac{14}{11}$$

✅ Example 9

Convert: $0.2353535\ldots = 0.2\overline{35}$

Let $x = 0.2353535\ldots$

Multiply by 100:

$$100x = 23.535353\ldots$$

Now subtract:

$$100x - x = 23.5353\ldots - 0.2353\ldots = 23.3 \Rightarrow 99x = 23.3 = \frac{233}{10} \Rightarrow x = \frac{233}{990}$$

📝 Exercise 1.3 – Solved

Q1. Convert to decimal form and classify the decimal expansion

Fraction	Decimal	Type
(i) 36/100	0.36	Terminating
(ii) 1/11	0.\overline{09}	Non-terminating recurring
(iii) 4/8	0.5	Terminating
(iv) 3/13	0.\overline{230769}	Non-terminating recurring
(v) 2/11	0.\overline{18}	Non-terminating recurring
(vi) 329/400	0.8225	Terminating

Q2. Predict the decimal expansions:

Given:
$\frac{1}{7} = 0.\overline{142857}$

Use cyclic patterns:

• $\frac{2}{7} = 0.\overline{285714}$

• $\frac{3}{7} = 0.\overline{428571}$

• $\frac{4}{7} = 0.\overline{571428}$

• $\frac{5}{7} = 0.\overline{714285}$

• $\frac{6}{7} = 0.\overline{857142}$

📌 These are cyclic permutations of the digits in $\frac{1}{7}$.

Q3. Express in the form $\frac{p}{q}$:

(i) $0.6 = \frac{6}{10} = \frac{3}{5}$
(ii) $0.47 = \frac{47}{100}$
(iii) $0.001 = \frac{1}{1000}$

Q4. Express $0.99999\ldots$ in the form $\frac{p}{q}$

Let $x = 0.99999\ldots$
Then $10x = 9.99999\ldots$
Subtract: $10x - x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1$

✅ So, $0.99999\ldots = 1$

Q5. Repeating block length in $\frac{1}{7}$

Do long division:
$\frac{1}{7} = 0.\overline{142857}$

🔹 Number of digits in the repeating block = 6

Q6. Property of q for rational number $\frac{p}{q}$ to be terminating

Decimal of $\frac{p}{q}$ terminates iff q (in lowest terms) has no prime factor other than 2 or 5.

✅ Example:

• $\frac{1}{8} = 0.125$: q = 8 = $2^3$

• $\frac{1}{6} = 0.16666...$: q = 6 = $2 \times 3$ → Not terminating

Q7. Three numbers with non-terminating, non-recurring decimals:

• π = 3.1415926535...

• √2 = 1.41421356...

• e = 2.718281828...

Q8. Three irrational numbers between $\frac{5}{7}$ and $\frac{9}{11}$:

Convert bounds to decimal:

• $\frac{5}{7} ≈ 0.714$

• $\frac{9}{11} ≈ 0.818$

Now pick irrational numbers in between:

• 0.718281828... (e/4)

• 0.75√2

• 0.8π/3

Q9. Classify as rational or irrational

Number	Type
(i) $\sqrt{23}$	Irrational
(ii) $\sqrt{225} = 15$	Rational
(iii) 0.3796	Rational
(iv) 7.478478...	Rational
(v) 1.10100100010000...	Irrational

Would you like me to create a printable worksheet and answer key for this exercise?

Lecture 3: Real Numbers and Their Decimal Expansions

 .

📘 Class 9 – CBSE Mathematics

Chapter 1: Number System

Lecture 3: Real Numbers and Their Decimal Expansions

🔹 Decimal Expansions of Rational Numbers

When we divide a rational number $\frac{p}{q}$ (with $p, q \in \mathbb{Z}, q \ne 0$), its decimal expansion can be of two types:

🔸 Case I: Terminating Decimal Expansion

• If, during division, the remainder becomes zero after a certain number of steps, the decimal expansion stops or ends.

• Such decimals are called terminating decimals.

✅ Examples:

• $\frac{7}{8} = 0.875$

• $\frac{1}{2} = 0.5$

• $\frac{639}{250} = 2.556$

Observation: The division process ends after a finite number of steps.

🔸 Case II: Non-Terminating Recurring Decimal Expansion

• If the remainder never becomes zero and starts repeating after a certain point, then the decimal goes on forever but with a repeating pattern.

• These are called non-terminating recurring (repeating) decimals.

✅ Examples:

• $\frac{10}{3} = 3.333... = 3.\overline{3}$

• $\frac{1}{7} = 0.142857142857... = 0.\overline{142857}$

• $3.5727272... = 3.57\overline{27}$

Note: The bar $\overline{\phantom{0}}$ indicates the digits that repeat.

🔹 Important Observations from Examples

1. The remainders either:

   • Become zero (Terminating)

   • Or start repeating (Non-Terminating Recurring)

2. In repeating cases, the number of distinct remainders before repetition is less than the divisor.

   • Example: $\frac{10}{3}$ → Divisor is 3 → Only 1 digit (3) repeats

   • Example: $\frac{1}{7}$ → Divisor is 7 → 6 digits (142857) repeat

3. The decimal expansions of rational numbers are either terminating or non-terminating recurring.

🔹 Conclusion

If a number:

• Terminates (e.g., 3.25), or

• Repeats (e.g., 1.272727...)

Then it is a rational number.

So,

• Decimal expansions help us identify rational numbers.

• All rational numbers will either terminate or have repeating decimals.

• If a number has a non-terminating non-repeating decimal (like π or $\sqrt{2}$), it is irrational.