Ch 1 · Q–

Class 10 Mathematics Exercise 1.1 NCERT Solutions Olympiad Board Exam

Chapter 1 — REAL NUMBERS

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

📋7 questions

⏱Ideal time: 20-25 min

📍Now at: Q1

Theory: Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of prime numbers, and this factorization is unique (except for the order of factors).

Key Concepts:

A prime number has exactly two factors: 1 and itself.
A composite number has more than two factors.
Prime factorisation is done using repeated division by prime numbers.

Solution Roadmap

Start dividing the number by the smallest prime (2).
Continue dividing until it is no longer divisible.
Move to next prime (3, 5, 7, ...).
Repeat until quotient becomes 1.
Write result in exponential form.

140

Step 1: Divide by 2
\[140 \div 2 = 70\]
Step 2: Divide again by 2
\[70 \div 2 = 35\]
Step 3: Divide by 5
\[35 \div 5 = 7\]
Step 4: Divide by 7
\[7 \div 7 = 1\]
Final Answer:
\[140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7\]
156

Step 1: Divide by 2
\[156 \div 2 = 78\]
Step 2: Divide by 2
\[78 \div 2 = 39\]
Step 3: Divide by 3
\[39 \div 3 = 13\]
Step 4: Divide by 13
\[13 \div 13 = 1\]
Final Answer:
\[156 = 2^2 \times 3 \times 13\]
3825

Step 1: Divide by 3
\[3825 \div 3 = 1275\]
Step 2: Divide by 3
\[1275 \div 3 = 425\]
Step 3: Divide by 5
\[425 \div 5 = 85\]
Step 4: Divide by 5
\[85 \div 5 = 17\]
Step 5: Divide by 17
\[17 \div 17 = 1\]
Final Answer:
\[3825 = 3^2 \times 5^2 \times 17\]
5005

Step 1: Divide by 5
\[5005 \div 5 = 1001\]
Step 2: Divide by 7
\[1001 \div 7 = 143\]
Step 3: Divide by 11
\[143 \div 11 = 13\]
Step 4: Divide by 13
\[13 \div 13 = 1\]
Final Answer:
\[5005 = 5 \times 7 \times 11 \times 13\]
7429

Step 1: Divide by 17
\[7429 \div 17 = 437\]
Step 2: Divide by 19
\[437 \div 19 = 23\]
Step 3: Divide by 23
\[23 \div 23 = 1\]
Final Answer:
\[7429 = 17 \times 19 \times 23\]

Why this question is important?

Board Exams: Direct questions on prime factorisation and its application in HCF/LCM are very common.
Foundation for Higher Maths: Used in algebra, number theory, and simplification problems.
Conceptual Strength: Builds clarity on primes, which is essential for advanced mathematics.

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Q2 →

Theory: LCM and HCF using Prime Factorisation

LCM (Least Common Multiple): Product of highest powers of all prime factors.
HCF (Highest Common Factor): Product of lowest powers of common prime factors.
Important Identity:

\[\text{LCM} \times \text{HCF} = \text{Product of the two numbers}\]

Solution Roadmap

Find prime factorisation of both numbers.
Write all prime factors with powers.
For LCM → take highest power of each prime.
For HCF → take lowest power of common primes.
Verify: LCM × HCF = product of numbers.

26 and 91

Prime Factorisation:
\[26 = 2 \times 13\] \[91 = 7 \times 13\]
LCM:
\[= 2^1 \times 7^1 \times 13^1 = 182\]
HCF:
\[= 13\]
Verification:
\[\text{LHS} = 182 \times 13 = 2366\] \[\text{RHS} = 26 \times 91 = 2366\] \[\text{LHS} = \text{RHS} \Rightarrow \text{Verified}\]
510 and 92

Prime Factorisation:

510:
\[510 \div 2 = 255\] \[255 \div 3 = 85\] \[85 \div 5 = 17\] \[17 \div 17 = 1\] \[510 = 2 \times 3 \times 5 \times 17\]
92:
\[92 \div 2 = 46\] \[46 \div 2 = 23\] \[23 \div 23 = 1\] \[92 = 2^2 \times 23\]
LCM:
\[= 2^2 \times 3 \times 5 \times 17 \times 23\] \[= 4 \times 3 \times 5 \times 17 \times 23\] \[= 23460\]
HCF:
\[= 2^1 = 2\]
Verification:
\[\text{LHS} = 23460 \times 2 = 46920\] \[\text{RHS} = 510 \times 92 = 46920\] \[\text{LHS} = \text{RHS} \Rightarrow \text{Verified}\]
336 and 54

Prime Factorisation:
\[336 \div 2 = 168\] \[168 \div 2 = 84\] \[84 \div 2 = 42\] \[42 \div 2 = 21\] \[21 \div 3 = 7\] \[7 \div 7 = 1\] \[336 = 2^4 \times 3 \times 7\] \[54 \div 2 = 27\] \[27 \div 3 = 9\] \[9 \div 3 = 3\] \[3 \div 3 = 1\] \[54 = 2 \times 3^3\]
LCM:
\[= 2^4 \times 3^3 \times 7\] \[= 16 \times 27 \times 7 = 3024\]
HCF:
\[= 2^1 \times 3^1 = 6\]
Verification:
\[\text{LHS} = 3024 \times 6 = 18144\] \[\text{RHS} = 336 \times 54 = 18144\] \[\text{LHS} = \text{RHS} \Rightarrow \text{Verified}\]

Why this question is important for Class 10 Boards?

Direct application of prime factorisation method.
Verification step is frequently asked in board exams.
Strengthens accuracy in long calculations.
Forms base for algebraic and number system problems.

← Q1

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Q3 →

Theory: HCF and LCM of Three Numbers

HCF: Product of lowest powers of only those prime factors which are common in all numbers.
LCM: Product of highest powers of all prime factors present in any number.
If no prime factor is common in all numbers, then HCF = 1.

Solution Roadmap

Find prime factorisation of each number.
List all prime factors.
HCF → take only common primes in all numbers with smallest power.
LCM → take all primes with greatest power.

12, 15 and 21

Prime Factorisation:
\[\begin{aligned}12 \div 2 &= 6,\\ 6 \div 2 &= 3,\\ 3 \div 3 &= 1\end{aligned}\] \[12 = 2^2 \times 3\] \[\begin{aligned}15 \div 3 &= 5,\\ 5 \div 5 &= 1\end{aligned}\] \[15 = 3 \times 5\] \[\begin{aligned}21 \div 3 &= 7,\\ 7 \div 7 &= 1\end{aligned}\] \[21 = 3 \times 7\]
HCF:

Common prime in all three numbers = 3
\[HCF = 3^1 = 3\]
LCM:
\[= 2^2 \times 3^1 \times 5^1 \times 7^1\] \[= 4 \times 3 \times 5 \times 7 = 420\]
17, 23 and 29

Prime Factorisation:

All numbers are prime:
\[\begin{aligned}17 &= 17,\\ 23 &= 23,\\ 29 &= 29\end{aligned}\]
HCF:

No common prime factor
\[HCF = 1\]
LCM:
\[= 17 \times 23 \times 29\] \[= 11339\]
8, 9 and 25

Prime Factorisation:
\[\begin{aligned}8 \div 2 &= 4,\\ 4 \div 2 &= 2,\\ 2 \div 2 &= 1\end{aligned}\] \[8 = 2^3\] \[\begin{aligned}9 \div 3 = 3,\\ 3 \div 3 = 1\end{aligned}\] \[9 = 3^2\] \[\begin{aligned}25 \div 5 = 5,\\ 5 \div 5 = 1\end{aligned}\] \[25 = 5^2\]
HCF:

No common prime factor in all three numbers
\[HCF = 1\]
LCM:
\[= 2^3 \times 3^2 \times 5^2\] \[= 8 \times 9 \times 25 = 1800\]

Why this question is important for Class 10 Boards?

Tests understanding of HCF when more than two numbers are involved.
Common mistake area: identifying “common in all” vs “common in some”.
Directly asked in exams in similar format.
Strengthens accuracy in prime-based reasoning.

← Q2

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Q4 →

Theory: Relation between HCF and LCM

For any two positive integers:

\[\text{HCF} \times \text{LCM} = \text{Product of the numbers}\]

This identity helps to find LCM when HCF and numbers are known.

Solution Roadmap

Write the identity: HCF × LCM = product of numbers.
Substitute the given values.
Rearrange to find LCM.
Simplify step by step.

Given:

\[\text{HCF}(306, 657) = 9\]

Using identity:

\[\text{HCF} \times \text{LCM} = 306 \times 657\]

Substitute value:

\[9 \times \text{LCM} = 306 \times 657\]

Step 1: Divide both sides by 9

\[\text{LCM} = \frac{306 \times 657}{9}\]

Step 2: Simplify the fraction

\[306 \div 9 = 34\] \[\text{LCM} = 34 \times 657\]

Step 3: Multiply step-by-step

\[657 \times 34 = 657 \times (30 + 4)\] \[= 657 \times 30 + 657 \times 4\] \[= 19710 + 2628\] \[= 22338\]

Final Answer:

\[\text{LCM}(306, 657) = 22338\]

Why this question is important for Class 10 Boards?

Direct application of HCF–LCM identity.
Frequently asked as a short-answer or 2–3 mark question.
Tests algebraic manipulation and simplification skills.
Helps avoid lengthy prime factorisation when HCF is given.

← Q3

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Q5 →

!-- THEORY -->

Theory: Condition for a number to end with 0

A number ends with digit 0 if and only if it is divisible by 10.
Since \(10 = 2 \times 5\), the number must contain both factors 2 and 5.
If any number lacks factor 5, it cannot end with 0.

Solution Roadmap

Express \(6^n\) in prime factor form.
Check whether factor 5 is present.
Use divisibility condition of 10.
Conclude about last digit.

Step 1: Express \(6^n\) in prime factor form

\[6 = 2 \times 3\] \[6^n = (2 \times 3)^n\] \[= 2^n \times 3^n\]

Step 2: Check required condition

For a number to end with 0, it must contain both factors 2 and 5.

Step 3: Analyze factors of \(6^n\)

\(6^n\) contains factor \(2^n\) → factor 2 is present.
\(6^n\) contains factor \(3^n\).
There is no factor 5 in \(6^n\).

Step 4: Conclusion

Since factor 5 is missing, \(6^n\) is not divisible by 10.

\[\therefore \, 6^n \text{ can never end with digit } 0\]

Why this question is important for Class 10 Boards?

Tests understanding of divisibility rules and prime factors.
Common reasoning-based question in board exams.
Strengthens concept of factor-based digit properties.
Helps avoid common mistake: assuming powers increase factors.

← Q4

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Q6 →

Theory: Composite Numbers

A composite number has more than two factors.
If a number can be written as a product of two integers greater than 1, it is composite.
Strategy: Try to factorise the given expression.

Solution Roadmap

Rewrite expression to identify common factors.
Factorise using distributive property.
Express number as product of two integers > 1.
Conclude composite nature.

First Expression:

\[7 \times 11 \times 13 + 13\]

Step 1: Take common factor 13

\[= 13 \times (7 \times 11 + 1)\]

Step 2: Simplify inside bracket

\[7 \times 11 = 77\] \[77 + 1 = 78\]

Step 3: Multiply

\[= 13 \times 78\]

Conclusion:

Since it is written as product of two numbers greater than 1,

\[\therefore 13 \times 78 = 1014 \text{ is a composite number}\]

Second Expression:

\[7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\]

Step 1: Recognize factorial

\[= 7! + 5\] \[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\]

Step 2: Add 5

\[= 5040 + 5 = 5045\]

Step 3: Factorise

\[5045 = 5 \times 1009\]

Conclusion:

Since it can be written as product of two numbers greater than 1,

\[\therefore 5045 \text{ is a composite number}\]

Why this question is important for Class 10 Boards?

Tests ability to identify hidden factorisation.
Common pattern: \(a \times b + b = b(a+1)\).
Encourages algebraic manipulation instead of direct calculation.
Frequently asked as reasoning-based question.

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Q7 →

Theory: Repetition and LCM

When events repeat at fixed intervals, their common repetition time is given by LCM.
Meeting again at the starting point means both complete whole number of rounds simultaneously.
This occurs at the LCM of their individual times.

Solution Roadmap

Write the time taken by both persons.
Find prime factorisation.
Compute LCM using highest powers.
Interpret result in context.

Given:

Time taken by Sonia = 18 minutes
Time taken by Ravi = 12 minutes

Step 1: Prime Factorisation

\[\begin{aligned}12 \div 2 &= 6,\\ 6 \div 2 &= 3,\\ 3 \div 3 &= 1\end{aligned}\] \[12 = 2^2 \times 3\] \[\begin{aligned}18 \div 2 &= 9,\\ 9 \div 3 &= 3,\\ 3 \div 3 &= 1\end{aligned}\] \[18 = 2 \times 3^2\]

Step 2: Find LCM

\[\text{LCM}(12, 18) = 2^2 \times 3^2\] \[= 4 \times 9 = 36\]

Step 3: Interpretation

After 36 minutes, both Sonia and Ravi will complete whole number of rounds and reach the starting point together.

Final Answer:

\[\therefore \text{They will meet again after } 36 \text{ minutes}\]

Why this question is important for Class 10 Boards?