NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The World of Numbers PDF Download (2026-27)

The NCERT Solutions for Class 9 Maths Chapter 3 - The World of Numbers help students understand the fundamentals of number systems through detailed explanations and step-by-step solutions. Prepared according to the latest CBSE syllabus for the 2026-27 academic session, these solutions cover all exercises and questions from the chapter in a simple and student-friendly manner.

Students looking for Class 9 Maths Chapter 3 Solutions can use these answers to strengthen their concepts, complete assignments, and prepare effectively for examinations. The solutions are designed to make learning easier and help students develop confidence in solving mathematical problems.

Class 9 Maths Chapter 3 – The World of Numbers

Class 9 Chapter 3 Maths introduces students to the fascinating world of numbers and their properties. The chapter focuses on understanding different types of numbers, number patterns, and mathematical operations that form the foundation of higher mathematics. In Maths Chapter 3 Class 9, students learn to apply logical reasoning and numerical techniques to solve various mathematical problems.

Topics Covered in The World of Numbers

• Number systems
• Rational and irrational numbers
• Real numbers
• Properties of numbers
• Decimal expansions
• Representation of numbers on the number line
• Operations involving real numbers
• Applications of number concepts

Download NCERT Solutions for Class 9 Maths Chapter 3 PDF

Students can access the NCERT Solutions for Class 9 Maths Chapter 3 PDF for quick revision and offline study. The PDF contains complete solutions to all exercises, including Class 9 Maths Chapter 3 Exercise 3.1, making it a valuable resource for exam preparation.

Benefits of Downloading the PDF

• Access solutions anytime and anywhere
• Study offline without internet access
• Revise concepts quickly before exams
• Improve problem-solving skills
• Understand step-by-step methods

Class 9 Maths Chapter 3 The World of Numbers Solutions

Class 9 maths chapter 3 exercise 3.1

1. A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?

Solution: Given:

2 bags of spices = 15 copper ingots

12 bags of spices = ?

Calculation:

12 ÷ 2 = 6

6 × 15 = 90

Therefore, the merchant will leave with 90 copper ingots.

2. Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.

Solution: The numbers are:

11, 13, 17, 19

These are all prime numbers.

Next three prime numbers after 19 are:

23, 29, 31

Therefore, the next three numbers are 23, 29, and 31.

3. We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Solution: No, natural numbers are not closed under subtraction.

A set is said to be closed under subtraction if the result of subtracting any two numbers in the set is always a number from the same set. This is not true for natural numbers.

Example 1: 8 − 5 = 3

Since 3 is a natural number, the result belongs to the set of natural numbers.

Example 2: 5 − 8 = −3

Since −3 is not a natural number, the result does not belong to the set of natural numbers.

Example 3: 4 − 4 = 0

Since 0 is not considered a natural number in the traditional definition, the result is not a natural number.

Therefore, natural numbers are not closed under subtraction because subtracting two natural numbers does not always give a natural number.

4. Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?

Solution: On one hand, the four fingers (index, middle, ring, and little finger) each have 3 joints.

Number of joints = 4 × 3 = 12

By using the thumb to point to each joint, you can count from 1 to 12 on a single hand.

Calculation:

4 fingers × 3 joints per finger = 12 joints

Therefore, you can count up to 12 on one hand.

This method is believed to be one reason why some ancient civilizations used a base-12 (duodecimal) counting system instead of the base-10 system we use today. Since one hand can count 12 units, people could use the fingers of the other hand to keep track of how many groups of 12 had been counted. This made counting larger numbers easier and contributed to the development of base-12 number systems.

Class 9 Maths chapter 3 exercise 3.2

1. The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?

Solution:  Given,

• Temperature at noon = 4°C
• Temperature drops by = 15°C

Calculation:

4°C − 15°C = −11°C

Therefore, the temperature at midnight is −11°C. 

2. A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.

Solution: Given,

• Debt = −₹850
• Profit = +₹1,200
• Loss = −₹450

Equation:

−850 + 1200 − 450

Calculation:

= 350 − 450

= −100

Therefore, the trader's final financial standing is −₹100, which means he is still in debt of ₹100. 

3. Calculate the following using Brahmagupta’s laws:

(i) (−12) × 5

(ii) (−8) × (−7)

(iii) 0 − (−14)

(iv) (−20) ÷ 4

Solution: 

(i) (−12) × 5

= −60

(According to Brahmagupta's law, a negative number multiplied by a positive number gives a negative number.)

(ii) (−8) × (−7)

= 56

(According to Brahmagupta's law, a negative number multiplied by a negative number gives a positive number.)

(iii) 0 − (−14)

= 0 + 14

= 14

(Subtracting a negative number is the same as adding the corresponding positive number.)

(iv) (−20) ÷ 4

= −5

(A negative number divided by a positive number gives a negative number.)

4. Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 − (−5) = 15).

Solution: Consider that you have ₹10 in your pocket. A friend had previously borrowed ₹5 from you, so he owes you ₹5. This debt can be represented as −₹5 from your friend's side.

Now, if that ₹5 debt is removed (that is, your friend repays the money), your total amount increases by ₹5.

Mathematically:

10 − (−5)

= 10 + 5

= 15

So, subtracting a negative number means removing a debt or loss, which has the same effect as gaining that amount. Therefore, subtracting a negative number is equivalent to adding a positive number.

Hence, 10 − (−5) = 15.

Class 9 Maths chapter 3 exercise 3.3

1. Prove that the following rational numbers are equal:

(i) 2/3 and 4/6

Solution:  To prove that 2/3 and 4/6 are equal, multiply the numerator and denominator of 2/3 by 2.

2/3 = (2 × 2)/(3 × 2)

= 4/6

Therefore, 2/3 = 4/6.

We can also verify this using cross multiplication:

2 × 6 = 12

3 × 4 = 12

Since both products are equal, the two rational numbers are equal.

Hence, 2/3 and 4/6 are equivalent rational numbers.

(ii) 5/4 and 10/8

Solution: To prove that 5/4 and 10/8 are equal, multiply the numerator and denominator of 5/4 by 2.

5/4 = (5 × 2)/(4 × 2)

= 10/8

Therefore, 5/4 = 10/8.

We can also verify this using cross multiplication:

5 × 8 = 40

4 × 10 = 40

Since both products are equal, the two rational numbers are equal.

Hence, 5/4 and 10/8 are equivalent rational numbers.

(iii) -3/5 and -6/10

Solution: To prove that −3/5 and −6/10 are equal, multiply the numerator and denominator of −3/5 by 2.

−3/5 = (−3 × 2)/(5 × 2)

= −6/10

Therefore, −3/5 = −6/10.

We can also verify this using cross multiplication:

(−3) × 10 = −30

5 × (−6) = −30

Since both products are equal, the two rational numbers are equal.

Hence, −3/5 and −6/10 are equivalent rational numbers. 

(iv) 9/3 and 3

Solution: To prove that 9/3 and 3 are equal, simplify the fraction 9/3.

9/3 = 9 ÷ 3

= 3

Therefore,

9/3 = 3

We can also verify this using multiplication:

3 × 3 = 9

Since the numerator is obtained by multiplying the denominator by 3, the value of the fraction is 3.

Hence, 9/3 and 3 are equal. 

2. Find the sum:

(i) 2/5 + 3/10

Solution: To add 2/5 and 3/10, first make the denominators the same.

The LCM of 5 and 10 is 10.

Convert 2/5 into an equivalent fraction with denominator 10:

2/5 = (2 × 2)/(5 × 2)

= 4/10

Now add the fractions:

4/10 + 3/10 = 7/10

Therefore, 2/5 + 3/10 = 7/10. 

(ii) 7/12 + 5/8

Solution: To add 7/12 and 5/8, first find the LCM of 12 and 8.

LCM of 12 and 8 = 24

Convert the fractions to equivalent fractions with denominator 24:

7/12 = (7 × 2)/(12 × 2)

= 14/24

5/8 = (5 × 3)/(8 × 3)

= 15/24

Now add the fractions:

14/24 + 15/24 = 29/24

Therefore, 7/12 + 5/8 = 29/24. 

(iii) – 4/7 + 3/14

Solution: To add −4/7 and 3/14, first find the LCM of 7 and 14.

LCM of 7 and 14 = 14

Convert −4/7 into an equivalent fraction with denominator 14:

−4/7 = (−4 × 2)/(7 × 2)

= −8/14

Now add the fractions:

−8/14 + 3/14 = −5/14

Therefore, −4/7 + 3/14 = −5/14. 

3. Find the difference:

(i) 5/6 – 1/4

Solution:  To find 5/6 − 1/4, first find the LCM of 6 and 4.

LCM of 6 and 4 = 12

Convert the fractions to equivalent fractions with denominator 12:

5/6 = (5 × 2)/(6 × 2)

= 10/12

1/4 = (1 × 3)/(4 × 3)

= 3/12

Now subtract:

10/12 − 3/12 = 7/12

Therefore, 5/6 − 1/4 = 7/12. 

(ii) 11/8 – 3/4

Solution: To find 11/8 − 3/4, first find the LCM of 8 and 4.

LCM of 8 and 4 = 8

Convert 3/4 into an equivalent fraction with denominator 8:

3/4 = (3 × 2)/(4 × 2)

= 6/8

Now subtract:

11/8 − 6/8 = 5/8

Therefore, 11/8 − 3/4 = 5/8. 

(iii) -7/9 – (-2/3)

Solution: To find −7/9 − (−2/3), first change subtraction of a negative number into addition:

−7/9 − (−2/3) = −7/9 + 2/3

Now find the LCM of 9 and 3.

LCM of 9 and 3 = 9

Convert 2/3 into an equivalent fraction with denominator 9:

2/3 = (2 × 3)/(3 × 3)

= 6/9

Now add the fractions:

−7/9 + 6/9 = −1/9

Therefore, −7/9 − (−2/3) = −1/9. 

4. Find the product:

(i) 2/3 × 3/10

Solution: To find 2/3 × 3/10, multiply the numerators and denominators:

(2 × 3)/(3 × 10)

= 6/30

Simplify the fraction by dividing the numerator and denominator by 6:

6/30 = 1/5

Therefore, 2/3 × 3/10 = 1/5. 

(ii) 7/11 × 5/8

Solution: To find 7/11 × 5/8, multiply the numerators and denominators:

(7 × 5)/(11 × 8)

= 35/88

Since 35 and 88 have no common factor other than 1, the fraction is already in its simplest form.

Therefore, 7/11 × 5/8 = 35/88. 

(iii) -4/7 × 5/14

Solution:  To find 7/11 × 5/8, multiply the numerators and denominators:

(7 × 5)/(11 × 8)

= 35/88

Since 35 and 88 have no common factor other than 1, the fraction is already in its simplest form.

Therefore, 7/11 × 5/8 = 35/88. 

5. Find the quotient:

(i) 2/3 ÷ 3/10

To divide fractions, multiply the first fraction by the reciprocal of the second fraction.

2/3 ÷ 3/10

= 2/3 × 10/3

= (2 × 10)/(3 × 3)

= 20/9

Therefore, 2/3 ÷ 3/10 = 20/9. 

(ii) 7/11 ÷ 5/8

= 7/11 × 8/5

= (7 × 8)/(11 × 5)

= 56/55

Therefore, 7/11 ÷ 5/8 = 56/55. 

(iii) -4/7 ÷ 5/14

= -4/7 × 14/5

= (-4 × 14)/(7 × 5)

= -56/35

Simplify by dividing numerator and denominator by 7:

= -8/5

Therefore, -4/7 ÷ 5/14 = -8/5. 

6. Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3

Solution: 

LHS = (1/2 + 3/4) × 8/3

First, add the fractions:

1/2 = 2/4

So,

1/2 + 3/4 = 2/4 + 3/4 = 5/4

Now multiply:

5/4 × 8/3

= (5 × 8)/(4 × 3)

= 40/12

= 10/3

Therefore,

LHS = 10/3

RHS = 1/2 × 8/3 + 3/4 × 8/3

First term:

1/2 × 8/3

= (1 × 8)/(2 × 3)

= 8/6

= 4/3

Second term:

3/4 × 8/3

= (3 × 8)/(4 × 3)

= 24/12

= 2

Now add:

4/3 + 2

= 4/3 + 6/3

= 10/3

Therefore,

RHS = 10/3

Since,

LHS = RHS = 10/3

Hence,

(1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3. 

7. Simplify the following using the distributive property: (7/9)(6/7 − 3/4).

Solution:  Given,

(7/9)(6/7 − 3/4)

Using the distributive property:

(7/9)(6/7) − (7/9)(3/4)

Now simplify each term:

(7 × 6)/(9 × 7) = 6/9 = 2/3

and

(7 × 3)/(9 × 4) = 21/36 = 7/12

So,

2/3 − 7/12

Convert to a common denominator:

2/3 = 8/12

Therefore,

8/12 − 7/12 = 1/12

Hence, (7/9)(6/7 − 3/4) = 1/12. 

8. Find the rational number x such that: (5/6)(x + 3/5) = (5/6)x + 1/2

Solution: Given,

(5/6)(x + 3/5) = (5/6)x + 1/2

Using the distributive property on the left-hand side:

(5/6)x + (5/6 × 3/5) = (5/6)x + 1/2

Simplify:

(5/6)x + 15/30 = (5/6)x + 1/2

(5/6)x + 1/2 = (5/6)x + 1/2

Subtract (5/6)x from both sides:

1/2 = 1/2

This is true for every value of x.

Therefore, x can be any rational number. 

Class 9 Maths chapter 3 exercise 3.4

1. Represent the rational numbers 2/3, -5/4 and 1½ on a single number line.

Solution: To represent the rational numbers −5/4, 2/3, and 1½ on a number line, first convert them into easily understandable forms:

• −5/4 = −1¼ = −1.25
• 2/3 ≈ 0.67
• 1½ = 3/2 = 1.5

Number Line:

Steps:

• −5/4 lies between −2 and −1, one-fourth unit to the left of −1.
• 2/3 lies between 0 and 1, about two-thirds of the distance from 0 to 1.
• 1½ lies exactly halfway between 1 and 2.

Therefore, the points −5/4, 2/3, and 1½ can be marked on the number line as shown above.

2. Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Solution: Given rational numbers: −1/2 and 1/4

First, find the LCM of the denominators 2 and 4.

LCM(2, 4) = 4

Convert −1/2 into an equivalent fraction with denominator 4:

−1/2 = −2/4

Now the given rational numbers are −2/4 and 1/4.

The rational numbers lying between −2/4 and 1/4 include:

−1/4, 0, and 1/8

Therefore, three rational numbers between −1/2 and 1/4 are −1/4, 0, and 1/8.

3. Simplify the expression: (-1/4) + (5/12)

Solution: To add −1/4 and 5/12, first find the LCM of the denominators 4 and 12.

LCM(4, 12) = 12

Convert −1/4 into an equivalent fraction with denominator 12:

−1/4 = (−1 × 3)/(4 × 3)

= −3/12

Now add the fractions:

−3/12 + 5/12

= 2/12

= 1/6

Therefore, (−1/4) + (5/12) = 1/6. 

4. A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?

Solution: First, convert the mixed fractions into improper fractions:

15¾ = (15 × 4 + 3)/4 = 63/4

2¼ = (2 × 4 + 1)/4 = 9/4

The total silk cloth available is 15¾ metres (63/4 metres).

If 2¼ metres of silk is required to make 1 kurta, then the number of kurtas that can be made from 15¾ metres of silk is:

Number of kurtas = (63/4) ÷ (9/4)

= (63/4) × (4/9)

= 63/9

= 7

Therefore, 15¾ metres of fine silk can be used to make exactly 7 kurtas.

5. Find three rational numbers between 3.1415 and 3.1416.

Solution: To find rational numbers between 3.1415 and 3.1416, we can extend the decimal places and choose numbers that lie between them.

3.1415 < 3.14151 < 3.14152 < 3.14153 < 3.1416

Thus, the numbers 3.14151, 3.14152, and 3.14153 all lie between 3.1415 and 3.1416.

Therefore, three rational numbers between 3.1415 and 3.1416 are 3.14151, 3.14152, and 3.14153.

6. Can you think of other way(s) to find a rational number between any two rational numbers?

Solution: Yes, there are several methods to find rational numbers between two given rational numbers:

1. Using the Average Method

If a and b are two rational numbers, then their average,

(a + b) / 2

is always a rational number that lies between a and b.

2. Using a Common Denominator

Convert the given rational numbers into equivalent fractions with the same denominator. Then choose any fraction whose numerator lies between the two numerators.

3. Using Decimal Expansion

Express the rational numbers in decimal form and select one or more decimal numbers that lie between them.

Therefore, the average method, common denominator method, and decimal expansion method can all be used to find rational numbers between two given rational numbers.

Class 9 Maths chapter 3 exercise 3.5

1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Solution:  A rational number p/q in its simplest form has a terminating decimal expansion if the prime factors of the denominator q are only 2 and/or 5. If the denominator contains any prime factor other than 2 or 5, the decimal expansion is non-terminating recurring (repeating).

(i) 7/20

Prime factorisation of 20:

20 = 2² × 5

Since the denominator contains only the prime factors 2 and 5, the decimal expansion of 7/20 is terminating.

Converting into decimal:

7/20 = 0.35

Therefore, 7/20 is a terminating decimal.

(ii) 4/15

Prime factorisation of 15:

15 = 3 × 5

Since the denominator contains the prime factor 3, which is neither 2 nor 5, the decimal expansion of 4/15 is repeating.

Converting into decimal:

4/15 = 0.2666...

Therefore, 4/15 is a non-terminating recurring decimal.

(iii) 13/250

Prime factorisation of 250:

250 = 2 × 5³

Since the denominator contains only the prime factors 2 and 5, the decimal expansion of 13/250 is terminating.

Converting into decimal:

13/250 = 0.052

Therefore, 13/250 is a terminating decimal.

2. Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?

Solution: 

Answer: First, let us observe the decimal expansion of 1/13:

1/13 = 0.076923076923...

The repeating sequence is 076923.

Now consider the decimal expansions of the next fractions:

2/13 = 0.153846153846...

Repeating sequence: 153846

3/13 = 0.230769230769...

Repeating sequence: 230769

4/13 = 0.307692307692...

Repeating sequence: 307692

5/13 = 0.384615384615...

Repeating sequence: 384615

6/13 = 0.461538461538...

Repeating sequence: 461538

From these expansions, we can observe that:

• Each fraction produces a repeating decimal.
• The repeating digits contain the same six-digit pattern arranged in different orders.
• Every repeating block is a cyclic rotation of the others.

Therefore, the decimal expansions of the fractions with denominator 13 exhibit a cyclic pattern, demonstrating an interesting cyclic property of repeating decimals. 

3. Classify the following numbers as rational or irrational:

(i) √81

√81 = 9

Since 9 can be written as 9/1, it is a rational number.

(ii) √12

√12 = 2√3

Since √3 is irrational, √12 is also irrational.

(iii) 0.33333...

This is a non-terminating recurring decimal.

0.33333... = 1/3

Therefore, it is a rational number.

(iv) 0.123451234512345...

The block 12345 repeats indefinitely.

Since it is a repeating decimal, it is a rational number.

(v) 1.01001000100001...

This decimal is non-terminating and non-repeating.

Therefore, it is an irrational number.

(vi) 23.560185612239874790120

This is a terminating decimal.

All terminating decimals are rational numbers because they can be expressed as fractions.

Therefore, it is a rational number

4. The number 0.9̅ (which means 0.99999…) is a rational number. Using algebra (let x = 0.9̅, multiply by 10, and subtract), explain why 0.9̅ is exactly equal to 1.

Solution: Let x = 0.99999…

Multiply both sides by 10:

10x = 9.99999…

Now subtract:

10x – x = 9.99999… – 0.99999…

9x = 9

x = 1

But x = 0.99999…

Therefore, 0.99999… = 1.

5. We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.

Solution: Some examples of numbers whose reciprocals produce cyclic repeating decimals are:

1/7 = 0.142857142857...

1/17 = 0.0588235294117647...

1/19 = 0.052631578947368421...

In each case, the decimal expansion is non-terminating and repeating. The repeating digits form a cyclic pattern, where the sequence of digits repeats in a regular order.

Thus, 7, 17, and 19 are examples of numbers whose reciprocals generate cyclic recurring decimals.

Therefore, numbers such as 7, 17, 19, and others with similar properties can produce repeating decimal expansions that exhibit cyclic behaviour. 

End-of-Chapter Exercises

1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) 3/50

Solution:

Divide 3 by 50 using long division:

3 ÷ 50 = 0.06

Therefore,

3/50 = 0.06

Since the decimal expansion ends after two decimal places, it is a terminating decimal.

(ii) 2/9

Solution:

Divide 2 by 9 using long division:

2 ÷ 9 = 0.22222...

Therefore,

2/9 = 0.22222...

The digit 2 repeats indefinitely, so it is a non-terminating repeating (recurring) decimal.

2. Prove that √5 is an irrational number.

Solution: We prove that √5 is irrational using the method of contradiction.

Assume that √5 is a rational number. Then it can be expressed in the form:

√5 = p/q

where p and q are integers, q ≠ 0, and the fraction p/q is in its simplest form (that is, p and q have no common factor other than 1).

Squaring both sides:

5 = p²/q²

Multiplying by q²:

p² = 5q²

This implies that p² is divisible by 5. Therefore, p must also be divisible by 5.

Let:

p = 5k

for some integer k.

Substituting this value into the equation p² = 5q²:

(5k)² = 5q²

25k² = 5q²

Dividing both sides by 5:

5k² = q²

This shows that q² is also divisible by 5, which means q is divisible by 5.

Thus, both p and q are divisible by 5. This means they have a common factor of 5, which contradicts our original assumption that p/q was in lowest terms.

Since our assumption leads to a contradiction, it must be false.

Therefore, √5 is an irrational number. 

3. Convert the following decimal numbers in the form of p/q.

(i) 12.6

12.6 = 126/10

= 63/5

Therefore, 12.6 = 63/5.

(ii) 0.0120

Solution:

0.0120 = 120/10000

= 3/250

Therefore, 0.0120 = 3/250.

(iii) 3.05̅2̅ = 3.05252525...

Solution:

Let x = 3.05252525...

100x = 305.252525...

10000x = 30525.252525...

Subtracting:

10000x − 100x = 30525.252525... − 305.252525...

9900x = 30220

x = 30220/9900

= 1511/495

Therefore, 3.05̅2̅ = 1511/495.

(iv) 1.23̅5̅ = 1.23535353...

Solution:

Let x = 1.23535353...

100x = 123.535353...

10000x = 12353.535353...

Subtracting:

10000x − 100x = 12353.535353... − 123.535353...

9900x = 12230

x = 12230/9900

= 6115/4950

= 1223/990

Therefore, 1.23̅5̅ = 1223/990.

(v) 0.2̅3̅ = 0.23232323...

Solution:

Let x = 0.23232323...

100x = 23.232323...

Subtracting:

100x − x = 23.232323... − 0.232323...

99x = 23

x = 23/99

Therefore, 0.2̅3̅ = 23/99.

(vi) 2.05̅ = 2.055555...

Solution:

Let x = 2.055555...

10x = 20.55555...

100x = 205.55555...

Subtracting:

100x − 10x = 205.55555... − 20.55555...

90x = 185

x = 185/90

= 37/18

Therefore, 2.05̅ = 37/18.

(vii) 2.125̅ = 2.125555...

Solution:

Let x = 2.125555...

1000x = 2125.5555...

10000x = 21255.5555...

Subtracting:

10000x − 1000x = 21255.5555... − 2125.5555...

9000x = 19130

x = 19130/9000

= 1913/900

Therefore, 2.125̅ = 1913/900.

(viii) 3.125̅ = 3.125555...

Solution:

Let x = 3.125555...

1000x = 3125.5555...

10000x = 31255.5555...

Subtracting:

10000x − 1000x = 31255.5555... − 3125.5555...

9000x = 28130

x = 28130/9000

= 2813/900

Therefore, 3.125̅ = 2813/900.

(ix) 2.1̅6̅2̅5̅ = 2.162516251625...

Solution:

Let x = 2.162516251625...

10000x = 21625.16251625...

Subtracting:

10000x − x = 21625.16251625... − 2.16251625...

9999x = 21623

x = 21623/9999

Therefore, 2.1̅6̅2̅5̅ = 21623/9999. 

4. Locate the following rational numbers on the number line.

(i) 0.532

Solution:

The number 0.532 lies between 0.53 and 0.54 on the number line.

To locate it:

• Divide the interval from 0.53 to 0.54 into 10 equal parts.
• Mark the 2nd division after 0.53.

Therefore, 0.532 is located at the second subdivision between 0.53 and 0.54.

(ii) 1.15̅

Solution:

Let

1.15̅ = 1.15555...

This number lies between 1.15 and 1.16 on the number line.

To locate it:

• Divide the interval from 1.15 to 1.16 into 10 equal parts.
• Since 1.15555... is slightly greater than 1.155, mark a point just to the right of 1.155.

Therefore, 1.15̅ (1.15555...) is located between 1.15 and 1.16, slightly beyond the midpoint of the interval.

5. Find 6 rational numbers between 3 and 4.

Solution: To find 6 rational numbers between 3 and 4, write them with a common denominator.

3 = 30/10

4 = 40/10

Now choose any six fractions between 30/10 and 40/10:

31/10, 32/10, 33/10, 34/10, 35/10, 36/10

In decimal form, these are:

3.1, 3.2, 3.3, 3.4, 3.5, 3.6

Therefore, six rational numbers between 3 and 4 are 3.1, 3.2, 3.3, 3.4, 3.5, and 3.6. 

6. Find 5 rational numbers between 2/5 and 3/5.

Solution: To find 5 rational numbers between 2/5 and 3/5, first convert them into equivalent fractions with a larger common denominator.

Multiply both numerator and denominator by 6:

2/5 = 12/30

3/5 = 18/30

Now choose five fractions between 12/30 and 18/30:

13/30, 14/30, 15/30, 16/30, 17/30

Therefore, five rational numbers between 2/5 and 3/5 are:

13/30, 14/30, 15/30, 16/30, and 17/30.

7. Find 5 rational numbers between 1/6 and 2/5.

Solution: To find 5 rational numbers between 1/6 and 2/5, first express them with a common denominator.

LCM of 6 and 5 = 30

So,

1/6 = 5/30

2/5 = 12/30

Now choose five fractions between 5/30 and 12/30:

6/30, 7/30, 8/30, 9/30, 10/30

Simplifying where possible:

6/30 = 1/5

7/30 = 7/30

8/30 = 4/15

9/30 = 3/10

10/30 = 1/3

Therefore, five rational numbers between 1/6 and 2/5 are:

1/5, 7/30, 4/15, 3/10, and 1/3.

8. If x/3 + x/5 = 16/15, find the rational number x.

Solution: Given: x/3 + x/5 = 16/15

Taking x common, we have:

x(1/3 + 1/5) = 16/15

Now, 1/3 + 1/5 = 5/15 + 3/15 = 8/15

So, x × 8/15 = 16/15

Therefore, x = (16/15) ÷ (8/15)

= (16/15) × (15/8)

= 16/8

= 2

Thus, the rational number x is 2.

9. Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.

Solution: Given,

a + 1/b = 0

So, a = -1/b

Multiply both sides by b:

ab = -1

Since ab = -1, which is negative,

Therefore, ab is negative.

10. A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.

Solution: If the last non-zero digit of a decimal number is in the fourth decimal place, then the number has exactly four decimal places. Such a number can be expressed as:

p/10⁴

where p is an integer.

Since the final non-zero digit appears in the fourth decimal place, p cannot be a multiple of 10. Otherwise, the decimal representation would terminate before the fourth place.

Thus, the number can be written in the form:

p/10⁴, where p is an integer not divisible by 10.

Now,

10⁴ = 10000 = 2⁴ × 5⁴

When the fraction is simplified to its lowest terms, any common factors shared by p and 10⁴ are cancelled. As a result, the denominator in the reduced form may no longer contain 2⁴ or 5⁴.

For example:

0.1250 = 1250/10000

= 1/8

Here, the denominator in lowest form is 8 = 2³, which is not divisible by 2⁴ or 5⁴.

Therefore, it is not necessary for the denominator in lowest form to be divisible by 2⁴ or 5⁴.

11. Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

Solution: Given:

18/125

The denominator 125 can be expressed as:

125 = 5³

Since the denominator contains only the prime factor 5, the decimal expansion of the fraction will be terminating.

To determine the number of decimal places, convert the denominator into a power of 10:

125 × 8 = 1000

Multiplying both the numerator and denominator by 8:

18/125 = (18 × 8)/(125 × 8)

= 144/1000

= 0.144

The decimal terminates after three decimal places.

Therefore, 18/125 is a terminating decimal, and its decimal representation is 0.144.

12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.

Solution: Consider the denominator:

2³ × 5 = 8 × 5 = 40

A rational number whose denominator (in simplest form) contains only the prime factors 2 and 5 always has a terminating decimal expansion**.

The number of decimal places in the terminating decimal is determined by the larger exponent of 2 or 5 in the denominator.

Here:

Exponent of 2 = 3

Exponent of 5 = 1

Therefore, the decimal expansion will have 3 decimal places.

For example:

1/40 = 0.025

This decimal terminates after three digits following the decimal point.

Hence, a rational number with denominator 40 will have a terminating decimal expansion with 3 decimal places.

13. Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ − k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ − k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Solution:  a = 7/12 and b = 5/6

First, express both fractions with the same denominator.

Since:

5/6 = 10/12

we have:

a = 7/12 and b = 10/12

The difference between the numerators is:

10 − 7 = 3

Since this is not enough to find five rational numbers between them, multiply both fractions by 3:

7/12 = 21/36

10/12 = 30/36

Now:

a = 21/36 and b = 30/36

The difference between the numerators is:

30 − 21 = 9

This provides enough integers between 21 and 30 to obtain five rational numbers.

Therefore, five rational numbers lying between 21/36 and 30/36 are:

22/36, 23/36, 24/36, 25/36, 26/36

Hence, five rational numbers between 7/12 and 5/6 are:

22/36, 23/36, 24/36, 25/36, and 26/36.

Why is the condition k₂ − k₁ > n + 1 important?

Suppose we want to find n rational numbers between k₁/m and k₂/m.

The integers strictly between k₁ and k₂ are:

k₁ + 1, k₁ + 2, ..., k₂ − 1

The total number of such integers is:

k₂ − k₁ − 1

To obtain n distinct rational numbers between the given fractions, we must have at least n integers available between the numerators.

Therefore:

k₂ − k₁ − 1 ≥ n

which gives:

k₂ − k₁ ≥ n + 1

In practice, the condition k₂ − k₁ > n + 1 is used to ensure that there are enough numerators available to construct the required number of distinct rational numbers.

Thus, this condition guarantees that we can find n rational numbers between the given fractions. 

14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.

Solution: Given,

x + y + z = 0 …(1)

xy + yz + zx = 0 …(2)

We use the identity:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

From (1), (x + y + z)² = 0² = 0

From (2), xy + yz + zx = 0

So, 0 = x² + y² + z² + 2(0)

Therefore, x² + y² + z² = 0

Now, x², y² and z² are all non-negative rational numbers.

The sum of three non-negative numbers is 0 only when each one is 0.

Hence, x² = 0, y² = 0, z² = 0

Therefore, x = 0, y = 0, z = 0

So, all the rational numbers x, y and z are simultaneously zero.

15. Show that the rational number (a + b) / 2 lies between the rational numbers a and b.

Solution: 

Assume that a < b. We need to prove that the rational number (a + b)/2 lies between a and b.

First, compare a and (a + b)/2.

Since a < b, adding a to both sides gives:

a + a < a + b

2a < a + b

Dividing both sides by 2:

a < (a + b)/2

Next, compare (a + b)/2 and b.

Since a < b, adding b to both sides gives:

a + b < b + b

a + b < 2b

Dividing both sides by 2:

(a + b)/2 < b

Combining the two results, we get:

a < (a + b)/2 < b

Therefore, the rational number (a + b)/2 lies between a and b.

If b < a, a similar argument shows that:

b < (a + b)/2 < a

Thus, regardless of which number is larger, (a + b)/2 always lies between a and b. 

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