Class 9 Maths Chapter 3 Exercise 3.5 Solutions Ganita Manjari NCERT
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Built to strengthen your understanding of real numbers, this exercise covers core mathematical concepts such as identifying terminating and repeating decimals without long division, classifying rational and irrational numbers, examining algebraic proofs for repeating decimals like \(0.\bar{9} = 1\), and exploring the fascinating properties of cyclic numbers in fractional reciprocals.
Our expert-crafted solutions mirror the exact format expected in your school notebooks, using clean mathematical notation and direct, logical problem-solving steps without filler. Whether you are checking your homework, preparing for your school assessments, or mastering decimals and irrational numbers, this guide gives you the clarity you need to succeed with confidence.
🔑 Key Concepts to Remember:
- Terminating Decimals: A rational number \(\frac{p}{q}\ln\) in lowest form has a terminating decimal expansion if the prime factorization of its denominator \(q\) is of the form \(2^n 5^m\), where \(n\) and \(m\) are non-negative integers.
- Repeating Decimals: If the prime factors of the denominator contain any prime other than 2 or 5, the decimal expansion is non-terminating repeating.
- Rational vs Irrational: Rational numbers have terminating or repeating decimal expansions, whereas irrational numbers have non-terminating, non-repeating expansions.
NCERT Class 9 Maths Ganita Manjari Chapter 3 Exercise 3.5 Solutions
Question 1
Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: \(\frac{7}{20}\), \(\frac{4}{15}\) and \(\frac{13}{250}\). Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
Solution:
Part 1: Without Long Division (Checking Denominators)
- \(\frac{7}{20}\): Denominator \(20 = 2^2 \times 5^1\). Since prime factors contain only 2 and 5, it is a terminating decimal.
- \(\frac{4}{15}\): Denominator \(15 = 3^1 \times 5^1\). Since it contains a prime factor 3 (other than 2 or 5), it is a repeating decimal.
- \(\frac{13}{250}\): Denominator \(250 = 2^1 \times 5^3\). Since prime factors contain only 2 and 5, it is a terminating decimal.
Part 2: Verification via Long Division
- \(\frac{7}{20} = 7 \div 20 = \mathbf{0.35}\) (Terminating)
- \(\frac{4}{15} = 4 \div 15 = \mathbf{0.2666...} = \mathbf{0.2\bar{6}}\) (Repeating)
- \(\frac{13}{250} = 13 \div 250 = \mathbf{0.052}\) (Terminating)
Question 2
Perform the long division for \(\frac{1}{13}\). Identify the repeating block of digits. Does it show cyclic properties if you evaluate \(\frac{2}{13}\)? Now compute \(\frac{3}{13}\), \(\frac{4}{13}\), etc. What do you notice?[cite: 11]
Solution:
Step 1: Long Division for \(\frac{1}{13}\)
$$\frac{1}{13} = 0.076923076923... = \mathbf{0.\overline{076923}}$$
The repeating block consists of 6 digits: 076923.
Step 2: Evaluating \(\frac{2}{13}\), \(\frac{3}{13}\), etc.
- \(\frac{1}{13} = 0.\overline{076923}\)
- \(\frac{2}{13} = 2 \times 0.076923... = \mathbf{0.\overline{153846}}\)
- \(\frac{3}{13} = 3 \times 0.076923... = \mathbf{0.\overline{230769}}\)
- \(\frac{4}{13} = 4 \times 0.076923... = \mathbf{0.\overline{307692}}\)
Observation: We notice that all these fractions generate the exact same cyclic permutation of the digits `076923`. They are cyclic numbers because multiplying the base pattern by different integers shifts the cyclic order of the same digit group.
Question 3
Classify the following numbers as rational or irrational:
(i) \(\sqrt{81}\)
(ii) \(\sqrt{12}\)
(iii) \(0.33333 \dots\)
(iv) \(0.123451234512345 \dots\)
(v) \(1.01001000100001 \dots\) (Notice the pattern: Is it repeating a single block?)
Solution:
- (i) \(\sqrt{81}\): Simplifies to \(9\), which can be written as \(\frac{9}{1}\). Therefore, it is a Rational Number.
- (ii) \(\sqrt{12}\): Simplifies to \(2\sqrt{3}\). Since \(\sqrt{3}\) is irrational, it is an Irrational Number.
- (iii) \(0.33333 \dots\) (\(0.\bar{3}\)): It is a non-terminating repeating decimal, which can be expressed as \(\frac{1}{3}\). Therefore, it is a Rational Number.
- (iv) \(0.123451234512345 \dots\) (\(0.\overline{12345}\)): It is a repeating decimal block. Therefore, it is a Rational Number.
- (v) \(1.01001000100001 \dots\): The number of zeros between 1 increases successively, meaning it is non-terminating and non-repeating. Therefore, it is an Irrational Number.
Question 4
The number \(0.\bar{9}\) (which means \(0.99999 \dots\) ) is a rational number. Using algebra (let \(x = 0.\bar{9}\), multiply by 10, and subtract), explain why \(0.\bar{9}\) is exactly equal to 1.
Solution:
Let \(x = 0.99999 \dots\) — (Equation 1)
Multiplying both sides by 10:
\(10x = 9.99999 \dots\) — (Equation 2)
Subtracting Equation 1 from Equation 2:
$$10x - x = 9.99999 \dots - 0.99999 \dots$$
$$9x = 9$$
$$x = \frac{9}{9} = 1$$
Final Answer: Through algebraic subtraction, \(0.\bar{9}\) simplifies precisely to \(\mathbf{1}\).
Question 5
We have seen that the repeating block of \(\frac{1}{7}\) is a cyclic number. Try to find more numbers (\(n\)) whose reciprocals (\(\frac{1}{n}\)) produce decimals with repeating blocks that are cyclic.
Solution:
Reciprocals of certain prime numbers generate pure repeating decimals with cyclic blocks equal to \(p - 1\) digits (known as full reptend primes). Examples include:
- \(\frac{1}{7}\): \(0.\overline{142857}\) (6-digit cyclic block)
- \(\frac{1}{13}\): \(0.\overline{076923}\) (6-digit cyclic block)
- \(\frac{1}{17}\): \(0.\overline{0588235294117647}\) (16-digit cyclic block)
- \(\frac{1}{19}\): \(0.\overline{052631578947368421}\) (18-digit cyclic block)
Answer: Other numbers/primes like \(7, 13, 17,\) and \(19\) produce reciprocals with cyclic repeating blocks.
Frequently Asked Questions (FAQs)
Q1: How can you tell if a rational number has a terminating decimal without dividing?
A: Check the denominator in its lowest form. If its prime factors contain only 2, 5, or both, the decimal terminates.
Q2: What is the difference between rational and irrational numbers in decimal form?
A: Rational numbers have terminating or non-terminating repeating decimals, while irrational numbers have non-terminating and non-repeating decimals.
Q3: What textbook do these solutions belong to?
A: These step-by-step answers follow the latest Ganita Manjari Part I textbook issued for NCERT Class 9 Mathematics.
Q4: Why does \(0.\bar{9}\) equal 1 algebraically?
A: When you set \(x = 0.999...\) and compute \(10x - x\), the infinite repeating decimal parts cancel out completely, leaving \(9x = 9\), which results in \(x = 1\).
Q5: What are cyclic numbers in decimal fractions?
A: They are repeating decimal blocks generated by unit fractions where multiplying the sequence by integers shifts the digits in a cyclic pattern.
Q6: Are square roots of all non-square numbers irrational?
A: Yes, square roots of numbers that are not perfect squares (like \(\sqrt{12}\) or \(\sqrt{3}\)) always yield non-terminating, non-repeating decimals, making them irrational.
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