Class 10 Maths — NCERT Exercise Solutions

Concept Statement:

By the Fundamental Theorem of Arithmetic, every composite number can be expressed uniquely as a product of prime powers, irrespective of the order of the prime factors.

Step-by-step Prime Factorisations:

1. (i) 140:
   $$140 = 2 \times 70 = 2 \times 2 \times 35 = 2^2 \times 5 \times 7$$ Prime Factorisation: $140 = 2^2 \times 5^1 \times 7^1$.
2. (ii) 156:
   $$156 = 2 \times 78 = 2 \times 2 \times 39 = 2^2 \times 3 \times 13$$ Prime Factorisation: $156 = 2^2 \times 3^1 \times 13^1$.
3. (iii) 3825:
   Sum of digits = $3 + 8 + 2 + 5 = 18$ (divisible by 3).
   $$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 5 \times 85 = 3^2 \times 5^2 \times 17$$ Prime Factorisation: $3825 = 3^2 \times 5^2 \times 17^1$.
4. (iv) 5005:
   Ends in 5: $5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$.
   Prime Factorisation: $5005 = 5^1 \times 7^1 \times 11^1 \times 13^1$.
5. (v) 7429:
   Testing primes 2, 3, 5, 7, 11, 13 (fails), test 17: $7429 = 17 \times 437 = 17 \times 19 \times 23$.
   Prime Factorisation: $7429 = 17^1 \times 19^1 \times 23^1$.

Fundamental Relations:

• $\text{HCF}(a, b) = \text{Product of smallest powers of each common prime factor.}$
• $\text{LCM}(a, b) = \text{Product of greatest powers of each prime factor involved.}$
• Identity: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.

(i) 26 and 91:

• $26 = 2 \times 13$, $91 = 7 \times 13$.
• $\text{HCF} = 13$, $\text{LCM} = 2 \times 7 \times 13 = 182$.
• $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$. Product of numbers = $26 \times 91 = 2366$. Verified: $2366 = 2366$.

(ii) 510 and 92:

• $510 = 2 \times 3 \times 5 \times 17$, $92 = 2^2 \times 23$.
• $\text{HCF} = 2$, $\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$.
• $\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920$. Product = $510 \times 92 = 46920$. Verified.

(iii) 336 and 54:

• $336 = 2^4 \times 3 \times 7$, $54 = 2 \times 3^3$.
• $\text{HCF} = 2 \times 3 = 6$, $\text{LCM} = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$.
• $\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144$. Product = $336 \times 54 = 18144$. Verified.

Applying Prime Factorisation:

1. (i) 12, 15 and 21:
   $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$.
   $\text{HCF}(12, 15, 21) = 3^1 = 3$.
   $\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 35 = 420$.
2. (ii) 17, 23 and 29:
   17, 23, and 29 are all prime numbers.
   $\text{HCF}(17, 23, 29) = 1$.
   $\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 391 \times 29 = 11339$.
3. (iii) 8, 9 and 25:
   $8 = 2^3$, $9 = 3^2$, $25 = 5^2$. (Co-prime numbers)
   $\text{HCF}(8, 9, 25) = 1$.
   $\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$.

Mathematical Theorem:

For any two positive integers $a$ and $b$:

Given: $a = 306, b = 657$, and $\text{HCF}(306, 657) = 9$.

Step-by-step Evaluation:

Final Answer: The LCM of 306 and 657 is 22,338.

Step 1: Necessary Condition for a Number to End with Digit 0

If any number ends with the digit 0, it must be divisible by 10. That is, its prime factorisation must contain both the prime factors 2 and 5, as $10 = 2 \times 5$.

Step 2: Prime Factorisation of $6^n$

The only prime factors of $6^n$ are 2 and 3.

Step 3: Uniqueness Theorem

By the Uniqueness of the Fundamental Theorem of Arithmetic, there are no other prime factors in the factorisation of $6^n$. Since 5 is not a prime factor of $6^n$, $6^n$ cannot be divisible by 5.

Conclusion: $6^n$ cannot end with the digit 0 for any natural number $n \in \mathbb{N}$.

Definition of Prime & Composite Numbers:

• A prime number has only two distinct factors: 1 and the number itself.
• A composite number has more than two factors (it can be written as a product of two or more prime factors).

Case 1: Analyze $7 \times 11 \times 13 + 13$:

Since this number has factors other than 1 and itself (namely 2, 3, and 13), it is a composite number.

Case 2: Analyze $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$:

Both 5 and 1009 are positive integers greater than 1. Since the expression is a product of two factors other than 1 and itself, it is a composite number.

Step 1: Problem Analysis

• Sonia takes $18\text{ minutes}$ for one complete round.
• Ravi takes $12\text{ minutes}$ for one complete round.
• They both start at the same point, at the same time, in the same direction.
• The time when they meet again at the starting point must be a common multiple of 18 and 12, and to find the first time they meet, we need the Lowest Common Multiple (LCM) of 18 and 12.

Step 2: Calculate $\text{LCM}(18, 12)$

Step 3: Verification

• In 36 minutes, Sonia completes $36 / 18 = 2$ full rounds.
• In 36 minutes, Ravi completes $36 / 12 = 3$ full rounds.
• Both are exactly at the starting point.

Final Answer: Sonia and Ravi will meet again at the starting point after 36 minutes.

Proof by Contradiction:

Step 1: Assumption: Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.

Then, there exist co-prime positive integers $a$ and $b$ ($b \neq 0$, $\text{HCF}(a, b) = 1$) such that:

Step 2: Squaring both sides:

This implies that $5$ divides $a^2$. By the Fundamental Theorem of Arithmetic (Theorem: if a prime $p$ divides $a^2$, then $p$ divides $a$), it follows that $5$ divides $a$.

Step 3: Expressing $a$ in terms of 5:

Let $a = 5c$ for some integer $c$. Substituting into Equation 1:

This implies that $5$ divides $b^2$. Consequently, $5$ divides $b$.

Step 4: Arriving at Contradiction:

From Steps 2 and 3, $5$ divides both $a$ and $b$. Therefore, $a$ and $b$ have at least $5$ as a common factor.
However, this contradicts our initial assumption that $a$ and $b$ are co-prime ($\text{HCF}(a, b) = 1$).

Conclusion: This contradiction arose because of our incorrect assumption that $\sqrt{5}$ is rational. Hence, $\sqrt{5}$ is an irrational number.

Proof by Contradiction:

Step 1: Assumption: Let us assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.

Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:

Step 2: Rearranging to isolate $\sqrt{5}$:

Step 3: Contradiction analysis:

Since $a$ and $b$ are integers and $b \neq 0$, the numerator $(a - 3b)$ and denominator $2b$ are both integers with $2b \neq 0$. Therefore, $\frac{a - 3b}{2b}$ is a rational number.

This implies that $\sqrt{5}$ is rational.
However, this contradicts the established mathematical fact that $\sqrt{5}$ is irrational.

Conclusion: Our assumption that $3 + 2\sqrt{5}$ is rational was incorrect. Hence, $3 + 2\sqrt{5}$ is irrational.

(i) Prove that $\frac{1}{\sqrt{2}}$ is irrational:

1. Let $\frac{1}{\sqrt{2}}$ be rational, so $\frac{1}{\sqrt{2}} = \frac{a}{b}$ where $a, b \in \mathbb{Z}, a \neq 0, b \neq 0$ and $\text{HCF}(a, b) = 1$.
2. Taking reciprocal: $\sqrt{2} = \frac{b}{a}$.
3. Since $a, b$ are integers, $\frac{b}{a}$ is rational. This implies $\sqrt{2}$ is rational, which contradicts the fact that $\sqrt{2}$ is irrational.
4. Hence, $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Prove that $7\sqrt{5}$ is irrational:

1. Let $7\sqrt{5} = \frac{a}{b}$ where $a, b \in \mathbb{Z}, b \neq 0$.
2. Then $\sqrt{5} = \frac{a}{7b}$.
3. Since $a, b$ are integers, $\frac{a}{7b}$ is rational. This implies $\sqrt{5}$ is rational, contradicting the fact that $\sqrt{5}$ is irrational.
4. Hence, $7\sqrt{5}$ is irrational.

(iii) Prove that $6 + \sqrt{2}$ is irrational:

1. Let $6 + \sqrt{2} = \frac{a}{b}$ where $a, b \in \mathbb{Z}, b \neq 0$.
2. Then $\sqrt{2} = \frac{a}{b} - 6 = \frac{a - 6b}{b}$.
3. Since $a, b$ are integers, $\frac{a - 6b}{b}$ is rational. This implies $\sqrt{2}$ is rational, which contradicts that $\sqrt{2}$ is irrational.
4. Hence, $6 + \sqrt{2}$ is irrational.