Class 10 Mathematics Chapter 1 Real Numbers

Key Concept: Prime Factorization Method

b) HCF: 5, LCM: 150
[Solution Description] Prime factors of 10 are: $2 \times 5$
Prime factors of 15 are: $3 \times 5$
Prime factors of 25 are: $5^2$
The common prime factor among them is $5$, hence HCF is $5$.
The greatest powers of all primes are: $2 \times 3 \times 5^2$.
Hence, LCM is $2 \times 3 \times 5^2 = 150$

Click Here to Download Notes

Key Concept: Contradiction Method

b) Assume $\sqrt{3}$ is rational.
[Solution Description] To prove $\sqrt{3}$ is irrational, we start by assuming it is rational. Thus, let $\sqrt{3} = \frac{a}{b}$ where a and b are integers with $\gcd(a, b) = 1$. This implies that $a^2 = 3b^2$, leading to a contradiction as shown in further steps.

Click Here to Download Notes

Key Concept: Complex Proofs, Advanced Applications

a) True

[Solution Description] Assume, for contradiction, that $\sqrt{2} + \sqrt{5}$ is rational. Then it can be expressed as $\frac{a}{b}$ where a and b are integers with no common factors other than 1.

Squaring both sides gives $(\sqrt{2} + \sqrt{5})^2 = \left(\frac{a}{b}\right)^2.$

This simplifies to $2 + 2\sqrt{10} + 5 = \frac{a^2}{b^2}.$

Hence, $7 + 2\sqrt{10} = \frac{a^2}{b^2}.$

Rearranging leads to $2\sqrt{10} = \frac{a^2}{b^2} – 7.$

Since $\frac{a^2}{b^2}$ is rational, the right-hand side is also rational, implying $\sqrt{10}$ is rational, contradicting its known irrationality. Thus, $\sqrt{2} + \sqrt{5}$ is indeed irrational.

Click Here to Download Notes

Key Concept: Factorization Verification

a) Correct
[Solution Description] Recalculate the product from the given prime factors:

$2 \times 3^2 \times 5^2 = 2 \times 9 \times 25$

Calculation: $2 \times 9 = 18$

$18 \times 25 = 450$

Therefore, the given prime factorization is verified as correct.

Key Concept: Basic Division

a) 5
[Solution Description] Use Euclid’s division algorithm to divide 125 by 12. The steps are:

Divide 125 by 12: $125 \div 12 = 10 \text{ remainder } 5$

So, the remainder is 5.

Click Here to Download Notes

Key Concept: Proof by Contradiction, Properties of Irrational Numbers

c) Assume $r + \sqrt{7}$ is a rational number

[Solution Description] Assume, for contradiction, that $s = r + \sqrt{7}$ is rational. Then s can also be written as $c/d$ where c and d are integers, $d \neq 0$.

If $s = c/d$, then rearranging gives: $\sqrt{7} = s – r = \frac{c}{d} – \frac{a}{b} = \frac{cb – ad}{bd}$

This implies $\sqrt{7}$ is a ratio of two integers, contradicting the known irrationality of $\sqrt{7}$. Thus, adding $\sqrt{7}$ to a rational number does not yield a rational result, thereby leading to a contradiction. Therefore, s must be irrational.

Click Here to Download Notes

Key Concept: Prime Factorization

a) 7 divides a
[Solution Description] According to Theorem 1.2, if 7 divides $^2$, then it means that 7 must divide a as well.

Click Here to Download Notes

Key Concept: Operations with Irrational Numbers

d) All of the above

[Solution Description] According to the properties of irrational numbers, adding or subtracting a rational number to/from an irrational number like $\pi$ results in an irrational number, and multiplying or dividing a non-zero rational number with an irrational number also results in an irrational number.

Click Here to Download Notes

Key Concept: Basic Division

b) 1
[Solution Description] To find the remainder when 25 is divided by 4, we use Euclid’s division algorithm: $a = bq + r$ where $0 \leq r < b$. Here, a = 25, b = 4. Dividing 25 by 4 gives quotient q = 6 and remainder r = 1. Therefore, the remainder is $ 25 - 4 \times 6 = 1 $. Click Here to Download Notes

Key Concept: Division Verification

c) $12 \times 8 = 96$
[Solution Description] To verify the division result, multiply the quotient by the divisor:

– Given quotient is 12 and divisor is 8.
– Multiply: $12 \times 8 = 96$.

Since the product of the quotient and divisor equals the original number 96, the division is verified.

Click Here to Download Notes

Key Concept: Ending Digit Check

b) No, never.
[Solution Description] A number ending with digit 3 must have a unit place which matches when raised to certain powers. We consider powers of 5:

– $5^1 = 5$ ends in 5
– $5^2 = 25$ ends in 5
– $5^3 = 125$ ends in 5

It can be observed that $5^n$ always ends with 5 regardless of the value of n since $5 \cdot 5 = 25$, and any power higher than this will also continue this pattern due to multiplication properties, hence it cannot end in 3.

Click Here to Download Notes

Key Concept: LCM and HCF for Multiple Numbers

c) 504
[Solution Description] The prime factorization of 8 is $2^3$, for 18 is $2 \cdot 3^2$, and for 28 is $2^2 \cdot 7$. To find the LCM, take the highest power of each prime: $2^3, 3^2, 7^1$. Therefore, the LCM is $2^3 \times 3^2 \times 7 = 504$.

Click Here to Download Notes

Key Concept: Prime Factorization Verification

c) $2^2 \times 3^2 \times 5$
[Solution Description] Start by dividing 180 by the smallest prime number, which is 2:
$180 \div 2 = 90$
$90 \div 2 = 45$
$45 \div 3 = 15$
$15 \div 3 = 5$
Finally, divide by 5: $5 \div 5 = 1$
So the prime factorization of 180 is $2^2 \times 3^2 \times 5$.

Click Here to Download Notes

Key Concept: Complex Factorization, Advanced HCF/LCM

c) 2640
[Solution Description] The prime factorization of 264 is $2^3 \times 3 \times 11$, and for 440 it is $2^3 \times 5 \times 11$.
To find the LCM, take the highest power of each prime:

– For 2: $\max(2^3, 2^3) = 2^3$
– For 3: $\max(3^1, 3^0) = 3^1$
– For 5: $\max(5^0, 5^1) = 5^1$
– For 11: $\max(11^1, 11^1) = 11^1$

Thus, the LCM is $2^3 \times 3 \times 5 \times 11$.

Calculate as follows: $2^3 \times 3 \times 5 \times 11 = 8 \times 3 \times 5 \times 11 = 2640$
So, the LCM is 2640.

Click Here to Download Notes

Key Concept: Simple Divisibility

b) No, never possible
[Solution Description] For a number to end in 0, it must be divisible by 10. Any power of 9 results in only the factor of 9 which consists of only the factor 3. Since 10 requires at least a factor of 2 and 5, and these are not present in $9^n$, it cannot end in 0.

Click Here to Download Notes

Key Concept: Prime Factorization Application

b) Non-terminating repeating
[Solution Description] The prime factorization includes the prime number 7, which is neither 2 nor 5. Therefore, the decimal expansion is non-terminating repeating.

Click Here to Download Notes

Key Concept: Complex Proof Application, Advanced Divisibility

d) $\sqrt{6}$
[Solution Description] Assume $\sqrt{6}$ is rational, meaning it can be expressed as $\frac{p}{q}$ where p and q are coprime integers. Then $6 = \frac{p^2}{q^2}$implies $p^2 = 6q^2$. Therefore, 2 divides $p^2$, implying 2 divides p. Let p = 2k, so $4k^2 = 6q^2$. This simplifies to $2k^2 = 3q^2$, hence 3 divides $2k^2$, implying 3 divides k. However, this invalidates the coprime ness of p and q, proving $\sqrt{6}$ is indeed irrational.

Click Here to Download Notes

Key Concept: Factorization Verification

c) $2^2 \times 3^2 \times 5$
[Solution Description] Let’s verify the prime factorization:

Divide by 2: $180 \div 2 = 90$
Divide by 2 again: $90 \div 2 = 45$
Divide by 3: $45 \div 3 = 15$
Divide by 3 again: $15 \div 3 = 5$

So, the correct prime factorization of 180 is $2^2 \times 3^2 \times 5$.

Click Here to Download Notes