Solutions: Real Numbers - Test 3

Q1. (d) 2520

LCM of numbers from 1 to 10 is $2^3 \times 3^2 \times 5 \times 7 = 2520$.

Q2. (c) 0

$12^n = (2^2 \times 3)^n$. It does not contain the prime factor 5. Hence, it cannot end with 0.

Q3. (a) 70

Let numbers be $3x$ and $4x$. LCM = $12x$.
$12x = 120 \Rightarrow x = 10$. Numbers are 30 and 40. Sum = 70.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Square root of any prime is irrational.

Q5. Required number = LCM(28, 32) - Common Difference.
Diff: $28-8=20$, $32-12=20$.
LCM(28, 32) = 224.
Number = $224 - 20 = 204$. [2]

Q6. HCF must always divide LCM.
$380 \div 18 = 21.11$ (Not divisible).
No, two numbers cannot have 18 as HCF and 380 as LCM. [2]

Q7. Let $2 - 3\sqrt{5} = r$ (rational).
$3\sqrt{5} = 2 - r \Rightarrow \sqrt{5} = \frac{2-r}{3}$.
RHS is rational, LHS is irrational. Contradiction. [2]

Q8. Let $\sqrt{2} + \sqrt{3} = r$ (rational).
$\sqrt{3} = r - \sqrt{2}$. Squaring both sides:
$3 = r^2 + 2 - 2r\sqrt{2} \Rightarrow 2r\sqrt{2} = r^2 - 1$.
$\sqrt{2} = \frac{r^2 - 1}{2r}$.
RHS is rational, LHS is irrational. Contradiction. [3]

Q9. Find LCM(9, 12, 15).
$9 = 3^2$, $12 = 2^2 \times 3$, $15 = 3 \times 5$.
LCM = $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$ minutes.
3 hours [3]

Q10. Part 1: Prove $\sqrt{3}$ is irrational (Standard proof). [3]
Part 2: Rationalize $\frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$.
Since 2 is rational and $\sqrt{3}$ is irrational, their sum is irrational. [2]

(i) Size of tile = HCF(10, 8).
HCF = 2. 2 ft [1]

(ii) Area of bathroom = $10 \times 8 = 80$ sq ft.
Area of 1 tile = $2 \times 2 = 4$ sq ft.
Number of tiles = $80 / 4 = 20$. [2]

(iii) HCF(10, 15) = 5.
Largest tile side = 5 ft. [1]