Solutions: Real Numbers - Test 1

Q1. (b) 2

Smallest prime number = 2. Smallest composite number = 4.
HCF(2, 4) = 2.

Q2. (c) 338

Formula: $\text{HCF} \times \text{LCM} = a \times b$
$13 \times \text{LCM} = 26 \times 169$
$\text{LCM} = \frac{26 \times 169}{13} = 2 \times 169 = 338$.

Q3. (a) always irrational

The product of a non-zero rational number and an irrational number is always irrational.

Q4. (d) A is false but R is true.

Assertion: $6^n = (2 \times 3)^n$. Prime factorization contains 2 and 3. To end in 5, it must have 5 as a factor. Hence, it never ends in 5. (False)
Reason: $9^n = (3^2)^n = 3^{2n}$. No factor 2 or 5. Cannot end in 0. (True)

Q5. $7 \times 11 \times 13 + 13$

Step 1: Take 13 common.
$= 13 \times (7 \times 11 + 1)$ [1]

Step 2: Simplify.
$= 13 \times (77 + 1) = 13 \times 78$.
Since the number has factors 13 and 78 (other than 1 and itself), it is a composite number. [1]

Q6. Prime Factorization:

$6 = 2 \times 3$
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3 \times 5$ [1]

HCF = Product of smallest power of common primes = $2^1 \times 3^1 = 6$.
LCM = Product of highest power of all primes = $2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$. [1]

Q7. Check for $6^n$ ending in 0.

For a number to end with digit 0, its prime factorization must contain at least one pair of 2 and 5. [1]
$6^n = (2 \times 3)^n = 2^n \times 3^n$.
It contains 2 but not 5. By Fundamental Theorem of Arithmetic, this factorization is unique. Hence, $6^n$ can never end with 0. [1]

Q8. Prove $\sqrt{2}$ is irrational.

1. Let $\sqrt{2}$ be rational. $\sqrt{2} = \frac{a}{b}$, where $a, b$ are co-prime integers, $b \neq 0$.
2. $2 = \frac{a^2}{b^2} \Rightarrow a^2 = 2b^2$. Thus $2$ divides $a^2$, so $2$ divides $a$. [1]
3. Let $a = 2c$. Then $(2c)^2 = 2b^2 \Rightarrow 4c^2 = 2b^2 \Rightarrow b^2 = 2c^2$.
4. Thus $2$ divides $b^2$, so $2$ divides $b$. [1]
5. $a$ and $b$ have common factor 2, contradicting that they are co-prime. Hence $\sqrt{2}$ is irrational. [1]

Q9. Find largest number dividing 70 and 125 with remainders 5 and 8.

Numbers are:
$70 - 5 = 65$
$125 - 8 = 117$ [1]

Find HCF(65, 117):
$65 = 5 \times 13$
$117 = 3^2 \times 13$ [1]

HCF = 13.
Answer: 13 [1]

Q10.

Part 1: Prove $\sqrt{5}$ is irrational.
(Similar steps to Q8, replacing 2 with 5).
Conclusion: $\sqrt{5}$ is irrational. [3]

Part 2: Show $3 + 2\sqrt{5}$ is irrational.
Let $3 + 2\sqrt{5} = \frac{p}{q}$ (rational).
$2\sqrt{5} = \frac{p}{q} - 3 = \frac{p-3q}{q}$
$\sqrt{5} = \frac{p-3q}{2q}$ [1]
Here, RHS is rational (since p, q are integers), but LHS ($\sqrt{5}$) is irrational.
This is a contradiction. Hence, $3 + 2\sqrt{5}$ is irrational. [1]

Q11. Participants: Hindi(60), English(84), Maths(108).

(i) Max participants per room = HCF(60, 84, 108).
$60 = 2^2 \times 3 \times 5$
$84 = 2^2 \times 3 \times 7$
$108 = 2^2 \times 3^3$
HCF = $2^2 \times 3 = 12$. 12 participants [1]

(ii) Minimum rooms required = Total Participants / Max per room.
Total = $60 + 84 + 108 = 252$.
Rooms = $252 / 12 = 21$. 21 Rooms [2]

(iii) LCM(60, 84, 108) = $2^2 \times 3^3 \times 5 \times 7$
$= 4 \times 27 \times 35 = 3780$. LCM = 3780 [1]