Solutions: Real Numbers - Test 2

Q1. (a) an even number

Square of an odd number is odd. Difference of two odd numbers is even.
Example: $5^2 - 3^2 = 25 - 9 = 16$ (Even).

Q2. (b) $xy^2$

$a = x^3y^2$, $b = xy^3$.
HCF is the product of the smallest power of each common prime factor.
HCF = $x^1 \times y^2 = xy^2$.

Q3. (c) 20

Smallest 2-digit composite number = 10.
Smallest composite number = 4.
LCM(10, 4) = 20.

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

Check Assertion: $HCF \times LCM = 18 \times 169 = 3042$.
Given Product = 3072. Since $3042 \neq 3072$, Assertion is False.

Q5. $3 \times 5 \times 7 + 7 = 7(3 \times 5 + 1)$ [1]
$= 7(15 + 1) = 7 \times 16$.
Since it has factors other than 1 and itself (7 and 16), it is a composite number. [1]

Q6. $6 = 2 \times 3$ and $20 = 2^2 \times 5$. [1]
HCF = $2^1 = 2$.
LCM = $2^2 \times 3 \times 5 = 60$. [1]

Q7. Let $7\sqrt{5}$ be rational. $7\sqrt{5} = \frac{a}{b}$ ($b \neq 0$).
$\sqrt{5} = \frac{a}{7b}$. [1]
Since $a, b$ are integers, $\frac{a}{7b}$ is rational. But $\sqrt{5}$ is irrational. Contradiction.
Hence $7\sqrt{5}$ is irrational. [1]

Q8. Let $5 - \sqrt{3}$ be rational, say $r$.
$5 - \sqrt{3} = r \Rightarrow \sqrt{3} = 5 - r$. [1]
Since 5 and $r$ are rational, $5 - r$ is rational.
But $\sqrt{3}$ is irrational. Rational $\neq$ Irrational. [1]
Contradiction. Hence $5 - \sqrt{3}$ is irrational. [1]

Q9. Convert to cm: 825 cm, 675 cm, 450 cm.
Find HCF(825, 675, 450). [1]
$825 = 3 \times 5^2 \times 11$
$675 = 3^3 \times 5^2$
$450 = 2 \times 3^2 \times 5^2$
HCF = $3^1 \times 5^2 = 3 \times 25 = 75$. [1]
Length of rod = 75 cm [1]

Q10. Part 1: Prove $\sqrt{2}$ is irrational (Standard proof using $p/q$ form). [3]
Part 2: Let $3 - \sqrt{2} = r$ (rational).
$\sqrt{2} = 3 - r$.
RHS ($3-r$) is rational, LHS ($\sqrt{2}$) is irrational.
Contradiction. Hence $3 - \sqrt{2}$ is irrational. [2]

(i) $36 = 2^2 \times 3^2$. [1]

(ii) $32 = 2^5$, $36 = 2^2 \times 3^2$.
HCF = $2^2 = 4$. [1]

(iii) Minimum books = LCM(32, 36).
LCM = $2^5 \times 3^2 = 32 \times 9 = 288$.
288 Books [2]