📌 Given

(i) 3 50

(ii) 2 9

🎯 To Find

Convert each rational number into its decimal form using the process of long division and identify whether the decimal expansion is terminating or non-terminating repeating.

💡 Concept Reminder

• Convert every rational number into decimal form by long division.
• If the remainder becomes 0, the decimal expansion is terminating.
• If the same remainder repeats, the decimal expansion is non-terminating repeating.

✍ Step-by-Step Solution

(i) Convert 3 50 into decimal form.

Divide 3 by 50 using the long division method.

             0.06
50 ) 3.000
     0
     ----
      30
       0
     ----
      300
      300
     ----
        0

Therefore, 3 ÷ 50 = 0.06

The remainder becomes 0. Hence, 0.06 is a terminating decimal.

(ii) Convert 2 9 into decimal form.

Divide 2 by 9 using the long division method.

             0.22222...
9 ) 2.00000
    18
    ----
     20
     18
    ----
      20
      18
    ----
       20
       ...

Therefore, 2 ÷ 9 = 0.22222… (0.2)

The remainder repeats continuously. Hence, 0.22222… is a non-terminating repeating decimal.

✅ Final Answer

📘 Key Concept Used

• Every rational number can be converted into decimal form using the long division method.
• If the remainder becomes 0, the decimal expansion is a terminating decimal.
• If the remainder repeats continuously, the decimal expansion is a non-terminating repeating decimal.

⚠️ Common Mistake

• Stopping the long division before identifying the repeating pattern.
• Writing 2 ÷ 9 = 0.2 instead of 0.22222….
• Forgetting to mention whether the decimal is terminating or non-terminating repeating.

🎯 CBSE Exam Tip

In CBSE examinations, always show the complete long division steps and clearly mention whether the decimal expansion is terminating or non-terminating repeating. This helps you earn full step marks.

🧠 Memory Trick

Remainder becomes 0 → Terminating Decimal

Same remainder repeats → Non-Terminating Repeating Decimal

📌 Given

The number is √5.

🎯 To Find

Prove that √5 is an irrational number.

✅ Final Answer

Our assumption that √5 is a rational number leads to a contradiction because both p and q become divisible by 5, which is not possible as they are assumed to be co-prime.

Hence, our assumption is false.

∴ √5 is an irrational number.

📘 Key Concept Used

• Proof by contradiction.
• If the square of a number is divisible by a prime number, then the number itself is also divisible by that prime.
• Co-prime integers have no common factor other than 1.

⚠️ Common Mistake

• Not mentioning that p and q are co-prime before starting the proof.
• Writing p² = 5q² but not explaining why p is divisible by 5.
• Forgetting to prove that q is also divisible by 5.
• Writing the final conclusion without mentioning the contradiction.

🎯 CBSE Exam Tip

For proof questions, always write the steps in the correct order:

Assumption → Square Both Sides → Use Divisibility → Show Contradiction → Write Final Conclusion

Do not skip any mathematical step, as CBSE awards marks for the complete logical proof.

🧠 Memory Trick

Assume ↓ Square ↓ Both divisible ↓ Contradiction ↓ Irrational

Remember this five-step sequence whenever you prove that the square root of a non-perfect square is irrational.

📌 Given

Nine decimal numbers (both terminating and recurring decimals).

🎯 To Find

Convert each decimal number into the form p/q and write the fraction in its lowest terms.

(i) 0.532

(ii) 1.15

📌 Given

(i) 0.532

(ii) 1.15

🎯 To Find

Locate the given rational numbers on the number line.

Locate 0.532

Since

0.532 = 532 1000

it lies between 0 and 1 on the number line.

Step 1: Mark the interval from 0 to 1.

Step 2: Divide this interval into 10 equal parts and locate 0.5.

Step 3: Divide the interval from 0.5 to 0.6 into 10 equal parts and locate 0.53.

Step 4: Divide the interval from 0.53 to 0.54 into 10 equal parts. Mark the 2nd division after 0.53.

Hence, the required point represents 0.532.

Figure 4.1: Step-by-step construction showing how to locate 0.532 on the number line.

Locate 1.15

The decimal 1.15 means

1.15555…

It lies between 1.1 and 1.2 on the number line.

Step 1: Mark the interval from 1 to 2.

Step 2: Divide the interval from 1 to 2 into 10 equal parts and locate 1.1.

Step 3: Divide the interval from 1.1 to 1.2 into 10 equal parts and locate 1.15.

Step 4: The number 1.15555… lies between 1.15 and 1.16. Mark this point on the number line.

Hence, the required point represents 1.15.

Figure: Step-by-step construction to locate 1.15 on the number line.

✅ Final Answer

• 0.532 is located between 0.53 and 0.54.
• 1.15 is located between 1.15 and 1.16.

📘 Key Concept Used

• Every rational number can be represented on the number line.
• Successively divide intervals into 10 equal parts to locate decimal numbers accurately.
• Recurring decimals are also rational numbers and can be represented on the number line.

⚠️ Common Mistake

• Locating 0.532 directly between 0 and 1 without successive subdivisions.
• Treating 1.15 as exactly 1.15.
• Skipping intermediate divisions on the number line.

🎯 CBSE Exam Tip

In CBSE examinations, always show the successive divisions used to locate the decimal number on the number line. Draw a neat and properly labelled number line to score full marks.

🧠 Memory Trick

Whole Number ↓ Tenths ↓ Hundredths ↓ Thousandths

Always move from bigger intervals to smaller intervals while locating decimals on the number line.

📌 Given

The given rational numbers are 3 and 4.

🎯 To Find

Find six rational numbers lying between 3 and 4.

First, express both numbers with denominator 1.

Since we need 6 rational numbers between them, multiply the numerator and denominator of each fraction by 7.

Therefore,

The numerators lying between 21 and 28 are 22, 23, 24, 25, 26 and 27.

Hence, the required six rational numbers are

✅ Final Answer

The six rational numbers lying between 3 and 4 are

📘 Key Concept Used

• Express the given numbers as equivalent fractions having the same denominator.
• Choose a denominator large enough to obtain the required number of rational numbers between the given numbers.
• The fractions formed by the intermediate numerators are the required rational numbers.

⚠️ Common Mistake

• Using different denominators for the two fractions.
• Choosing a denominator that does not produce enough rational numbers.
• Writing only five numbers instead of the required six rational numbers.
• Forgetting to write the fractions in their correct order.

🎯 CBSE Exam Tip

In CBSE examinations, do not write the final answer directly. Always show how you convert the given numbers into equivalent fractions with the same denominator, explain why the denominator is chosen, and then write the required rational numbers. This stepwise method helps you score full marks.

🧠 Memory Trick

Whenever you are asked to find rational numbers between two given numbers, first make the denominators the same, count the numerators lying between them, and then write the corresponding fractions.

📌 Given

The given rational numbers are 2 5 and 3 5

🎯 To Find

Find five rational numbers lying between the given fractions.

The given fractions already have the same denominator.

There is no integer between 2 and 3. Therefore, convert both fractions into equivalent fractions having a larger denominator.

Since we need 5 rational numbers, multiply the numerator and denominator of each fraction by 6.

Thus,

The integers between 12 and 18 are 13, 14, 15, 16 and 17.

Hence, the required five rational numbers are

📌 Given

The given rational numbers are 1 6 and 2 5 .

🎯 To Find

Find five rational numbers lying between the given fractions.

First, convert the given fractions into equivalent fractions having the same denominator.

The LCM of 6 and 5 is 30.

Thus, the given fractions become

The integers lying between 5 and 12 are 6, 7, 8, 9, 10 and 11.

Hence, any five of the following fractions are the required rational numbers.

✅ Final Answer

The required rational number is

📘 Key Concept Used

• Take the LCM of the denominators before adding rational numbers.
• Add like fractions by keeping the denominator the same.
• Solve the resulting linear equation to find the value of the unknown.

⚠️ Common Mistake

• Adding the denominators directly.
• Using an incorrect LCM.
• Forgetting to divide both sides by 8 after obtaining 8x = 16.

🎯 CBSE Exam Tip

In CBSE examinations, write every algebraic step clearly: LCM → Addition of Fractions → Simplification → Solve for x. Showing each step helps you score full marks.

🧠 Memory Trick

Take LCM ↓ Add Numerators ↓ Remove Denominator ↓ Solve for x

📌 Given

• a and b are non-zero rational numbers.
• a + 1 b = 0

🎯 To Find

Determine whether ab is positive or negative and justify your answer.

Given,

Subtract 1 b from both sides.

Multiply both sides by b.

Since b ≠ 0, the factor b cancels.

Since −1 is a negative rational number, ab is negative.

✅ Final Answer

We have obtained

Since −1 is a negative rational number, ab is negative.

📘 Key Concept Used

• Use transposition to isolate the required term.
• Multiply both sides of an equation by the same non-zero quantity.
• If b ≠ 0, then b cancels from both sides.
• Determine the sign from the final value obtained.

⚠️ Common Mistake

• Forgetting that b ≠ 0 before cancelling b.
• Writing ab = 1 instead of ab = −1.
• Giving only the answer without proper justification.

🎯 CBSE Exam Tip

Always justify your conclusion by showing every algebraic step. Write: Given → Transpose → Multiply → Simplify → Conclusion to score full marks in CBSE examinations.

🧠 Memory Trick

Transpose ↓ Multiply ↓ Cancel ↓ Identify the Sign

📌 Given

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place.

🎯 To Find

• Show that the number can be written in the form p 10⁴ where p is an integer not divisible by 10.
• State whether the denominator in the lowest form must be divisible by 2⁴ or 5⁴.

Since the last non-zero digit occurs in the 4th decimal place, the decimal has exactly four decimal places after removing any trailing zeros.

Therefore, the given rational number can be written as

where p is an integer.

Since the last non-zero digit occurs in the fourth decimal place, the numerator p cannot end with the digit 0.

Now write the denominator in prime factor form.

When the fraction is reduced to its lowest form, any common factors of the numerator and denominator are cancelled.

If p is divisible by 2 or 5, the corresponding factors are cancelled from the denominator.

Therefore, after simplification, the denominator may no longer contain the factors 2⁴ or 5⁴.

✅ Final Answer

The given rational number can always be written in the form

where p is an integer not divisible by 10.

However, it is not necessary that the denominator in its lowest form remains divisible by 2⁴ or 5⁴, because common factors of the numerator and denominator are cancelled while reducing the fraction to its lowest form.

📘 Key Concept Used

• A terminating decimal having four decimal places can be written with denominator 10⁴.
• 10⁴ = 2⁴ × 5⁴.
• While reducing a fraction to its lowest form, common factors of the numerator and denominator are cancelled.
• Therefore, the denominator in lowest form may not contain 2⁴ or 5⁴.

⚠️ Common Mistake

• Thinking that the denominator always remains 10⁴ after simplification.
• Forgetting to express 10⁴ as 2⁴ × 5⁴.
• Ignoring the cancellation of common factors while reducing the fraction.

🎯 CBSE Exam Tip

In proof-based questions, always write the argument in sequence: Decimal Form → Fraction Form → Prime Factorisation → Cancellation → Conclusion. This systematic approach helps you score full marks in CBSE examinations.

🧠 Memory Trick

Decimal ↓ 10⁴ ↓ 2⁴ × 5⁴ ↓ Cancel Common Factors ↓ Lowest Form

🎉 Chapter 3 End Exercise Completed

You have successfully completed all solutions of the Class 9 Maths Chapter 3 End Exercise (Ganita Manjari 2026). Practice these questions regularly to strengthen your understanding of rational numbers and score excellent marks in your CBSE examinations.

📌 Given

The rational number is 18 125

🎯 To Find

Determine whether its decimal expansion is terminating or non-terminating. If it is terminating, find the number of decimal places.

A rational number has a terminating decimal expansion if, after writing the fraction in its lowest form, the denominator has only the prime factors 2 and/or 5.

The given fraction is

Since 18 and 125 have no common factor other than 1, the fraction is already in its lowest form.

Now find the prime factorisation of the denominator.

The denominator contains only the prime factor 5.

Therefore, the decimal expansion of 18 125 is terminating.

Since the highest power of 5 in the denominator is 3, the decimal expansion will have 3 decimal places.

✅ Final Answer

The denominator of the fraction 18 125 contains only the prime factor 5.

Therefore, its decimal expansion is terminating.

Since 125 = 5³, the decimal expansion contains 3 decimal places.

📘 Key Concept Used

• Write the rational number in its lowest form.
• Find the prime factors of the denominator.
• If the denominator contains only the prime factors 2 and/or 5, the decimal expansion is terminating.
• The number of decimal places is equal to the greater power of 2 or 5 in the denominator.

⚠️ Common Mistake

• Performing long division even though the question clearly says “without performing division”.
• Not reducing the fraction to its lowest form before checking the denominator.
• Thinking that only denominators with factor 2 terminate. A denominator containing only 5 also gives a terminating decimal.
• Writing the wrong number of decimal places.

🎯 CBSE Exam Tip

In CBSE examinations, never perform division when the question asks you not to. Simply write the fraction in its lowest form, find the prime factors of the denominator, and use the rule of terminating decimals. This method is faster and helps you score full marks.

🧠 Memory Trick

Lowest Form ↓ Prime Factors of Denominator ↓ Only 2 and/or 5 ? ↓ Yes → Terminating ↓ Highest Power = Decimal Places

📌 Given

The denominator of the rational number in its lowest form is 2³ × 5.

🎯 To Find

Find the number of decimal places in its terminating decimal expansion and justify the answer.

The denominator is already in its lowest form.

A rational number has a terminating decimal expansion if its denominator contains only the prime factors 2 and/or 5.

Here, the denominator contains only the prime factors 2 and 5. Therefore, the decimal expansion is terminating.

The number of decimal places is equal to the greater of the powers of 2 and 5.

The greater power is 3.

Hence, the decimal expansion will have 3 decimal places.

✅ Final Answer

The denominator is 2³ × 5, which contains only the prime factors 2 and 5.

Therefore, the decimal expansion is terminating.

The greater power of 2 and 5 is 3. Hence, the decimal expansion will have 3 decimal places.

📘 Key Concept Used

• Write the denominator in its prime factorised form.
• If the denominator contains only the prime factors 2 and/or 5, the decimal expansion is terminating.
• The number of decimal places is equal to the greater power of 2 or 5.

⚠️ Common Mistake

• Adding the powers of 2 and 5 (3 + 1 = 4), which is incorrect.
• Using the smaller power instead of the greater power.
• Forgetting that this rule applies only when the fraction is in its lowest form.

🎯 CBSE Exam Tip

Always remember this important CBSE rule: Decimal Places = Greater power of 2 or 5 in the denominator (after reducing the fraction to its lowest form). Write this rule before giving your answer to score full marks.

🧠 Memory Trick

Lowest Form
↓ Prime Factors
↓ Only 2 & 5
↓ Take the Greater Power
↓ Decimal Places

📌 Given

a = 7 12     b = 5 6

🎯 To Find

• Express both fractions with the same denominator m.
• Write exactly five rational numbers between them.
• Explain why k₂ − k₁ > n + 1 is required.

First, write both fractions with the same denominator. The LCM of 12 and 6 is 12.

Thus,

Here,

Since 3 ≤ 6, there are not enough integers between 7 and 10 to write five rational numbers. Therefore, we choose a larger common denominator.

Multiply both numerator and denominator of each fraction by 3.

Now,

Since 30 − 21 = 9 > 6, there are enough integers between 21 and 30 to obtain exactly five rational numbers.

The integers lying between 21 and 30 are

Any five of these numerators will give rational numbers lying between a and b. Choosing the first five,

These are the required five distinct rational numbers lying between 7 12 and 5 6 .

The condition k₂ − k₁ > n + 1 is necessary because there must be at least n integers between k₁ and k₂ to obtain n distinct rational numbers having the same denominator.

✅ Final Answer

Five rational numbers between 7 12 and 5 6 are

The condition k₂ − k₁ > n + 1 ensures that there are enough integers between k₁ and k₂ to obtain exactly n rational numbers.

📘 Key Concept Used

• Convert the given fractions into equivalent fractions having the same denominator.
• Choose a denominator large enough so that sufficient integers lie between the numerators.
• The intermediate numerators give the required rational numbers.
• For finding n rational numbers, the condition k₂ − k₁ > n + 1 must be satisfied.

⚠️ Common Mistake

• Using the LCM denominator directly without checking whether enough numerators are available.
• Writing fewer or more than the required five rational numbers.
• Not explaining why a larger denominator is chosen.
• Ignoring the condition k₂ − k₁ > n + 1.

🎯 CBSE Exam Tip

In CBSE examinations, always show these steps: LCM → Equivalent Fractions → Difference of Numerators → Required Rational Numbers → Reason for Choosing a Larger Denominator. This complete method helps you secure full marks.

🧠 Memory Trick

Same Denominator
↓ Enough Numerators
↓ Choose the Middle Fractions
↓ Write the Answer

📌 Given

x + y + z = 0

xy + yz + zx = 0

🎯 To Find

Show that x = y = z = 0.

From the first equation,

Therefore,

Substitute this value of z into the second equation.

Expand the brackets.

Simplify the equation.

Now use the identity

Multiplying both sides of x² + xy + y² = 0 by 4, we get

Since both terms are squares (or positive multiples of squares), each term must be zero.

Since

both terms on the left-hand side are non-negative because they are squares (or positive multiples of squares). The sum of two non-negative numbers can be zero only when each term is zero.

Therefore,

Hence,

Substituting y = 0 into 2x + y = 0, we get

From the first given equation,

Substituting x = 0 and y = 0,

Hence,

✅ Final Answer

The only rational numbers satisfying

are

📘 Key Concept Used

• Substitute one variable using the first equation.
• Simplify the second equation.
• Use the identity 4(x² + xy + y²) = (2x + y)² + 3y².
• The sum of two non-negative numbers can be zero only if each is zero.

⚠️ Common Mistake

• Expanding xy + yz + zx incorrectly after substitution.
• Not using the identity correctly.
• Concluding directly that x = y = 0 without proving each squared term is zero.
• Forgetting to substitute the values back to find z.

🎯 CBSE Exam Tip

For proof questions, write the complete sequence: Substitution → Simplification → Identity → Sum of Squares → Conclusion. Never skip the reasoning that a sum of non-negative numbers can be zero only when each number is zero.

🧠 Memory Trick

Substitute
↓ Simplify
↓ Use Identity
↓ Squares = 0
↓ x = y = z = 0

📌 Given

Let a and b be rational numbers such that

🎯 To Find

Show that

To prove that

consider the difference

Taking LCM,

Since a < b, we have b − a > 0. Therefore,

Hence,

Now, we shall prove that a + b 2 < b in Part 2.

Now, we prove that

Consider the difference

Taking LCM,

Since a < b, we have b − a > 0. Therefore,

Hence,

Combining the results obtained in Part 1 and Part 2,

Thus, a + b 2 lies between the rational numbers a and b.

✅ Final Answer

Therefore,

Hence, the rational number a + b 2 lies between a and b.

📘 Key Concept Used

• To prove one number lies between two numbers, prove both inequalities separately.
• If a < b, then b − a > 0.
• A positive quantity divided by a positive number remains positive.
• The average of two rational numbers always lies between them.

⚠️ Common Mistake

• Proving only one inequality and forgetting the other.
• Making mistakes while taking the LCM.
• Using a − b instead of b − a.
• Not writing the final combined inequality.

🎯 CBSE Exam Tip

For proof-based questions, always show both parts separately: First prove the number is greater than the lower limit, then prove it is smaller than the upper limit. Finally, combine both inequalities to complete the proof.

🧠 Memory Trick

Greater than a
↓ Less than b
↓ Combine Both
↓ Average Lies Between

📌 Given

A square root spiral is formed by joining several right triangles.

Each new triangle has one perpendicular side of length 1 unit.

🎯 To Find

Find the length of the hypotenuse of each right triangle using the Pythagoras Theorem.

According to the Pythagoras Theorem,

Now, calculate the hypotenuse of each triangle one by one.

Triangle 1

Base = 1 unit
Perpendicular = 1 unit

Triangle 2

Base = √2
Perpendicular = 1

Triangle 3

Base = √3
Perpendicular = 1

Triangle 4

Base = 2
Perpendicular = 1

Triangle 5

Base = √5
Perpendicular = 1

Similarly, continue the same process for the remaining triangles.

Triangle 6

Base = √6
Perpendicular = 1

Triangle 7

Base = √7
Perpendicular = 1

Triangle 8

Base = √8
Perpendicular = 1

Triangle 9

Base = 3
Perpendicular = 1

Triangle 10

Base = √10
Perpendicular = 1

Triangle 11

Base = √11
Perpendicular = 1

✅ Final Answer

The lengths of the hypotenuses of the successive right triangles are:

📘 Key Concept Used

• Each new triangle is a right triangle.
• One side is always 1 unit.
• The previous hypotenuse becomes the base of the next triangle.
• Apply the Pythagoras Theorem repeatedly.

⚠️ Common Mistake

• Using the wrong base for the next triangle.
• Forgetting that the previous hypotenuse becomes the next base.
• Writing √8 as 8 or √12 as 12.
• Not applying the Pythagoras Theorem at every step.

🎯 CBSE Exam Tip

Always calculate the hypotenuse of one triangle before moving to the next. In the Square Root Spiral, the previous hypotenuse becomes the base of the next right triangle. Showing each step using the Pythagoras Theorem helps you score full marks in CBSE examinations.

🧠 Memory Trick

Previous Hypotenuse
↓ Add 1²
↓ Apply Pythagoras
↓ Next Square Root