To find the minimum number of rooms, we need the HCF of 48, 80 and 144 (maximum teachers per room, same subject).

Prime factorisation:

• $48 = 2^4 \times 3$
• $80 = 2^4 \times 5$
• $144 = 2^4 \times 3^2$

HCF(48, 80, 144) = $2^4$ = 16

So, 16 teachers can be seated in each room.

Number of rooms required:

• French: $48 \div 16 = 3$
• Hindi: $80 \div 16 = 5$
• English: $144 \div 16 = 9$

Total minimum rooms = 3 + 5 + 9 = 17

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key insight is that the number of teachers per room must divide all three numbers exactly → this is the HCF.
• HCF gives the maximum number per room, which gives the minimum number of rooms — examiners specifically check this logic.
• Show prime factorisation clearly; take the lowest power of common prime factors for HCF.
• Don't forget to add the rooms for all three subjects at the end — that single addition step carries a mark.

Proof (by contradiction):

Assume $5 - 2\sqrt{3}$ is rational. Then we can write:

$$5 - 2\sqrt{3} = \frac{a}{b}, \quad \text{where } a, b \text{ are integers, } b \neq 0$$

Rearranging:

$$2\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$$

$$\sqrt{3} = \frac{5b - a}{2b}$$

Since $a$ and $b$ are integers, $\dfrac{5b-a}{2b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong, and $5 - 2\sqrt{3}$ is irrational. $\blacksquare$

Source: Real Numbers, Section 1.3 – Revisiting Irrational Numbers

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• Use proof by contradiction: assume the number is rational, express it as $\frac{a}{b}$, and rearrange to isolate $\sqrt{3}$.
• The key step is showing this forces $\sqrt{3}$ to be rational — contradicting the given condition.
• Examiners award marks for: (1) correct assumption, (2) correct algebraic rearrangement to isolate $\sqrt{3}$, (3) stating the contradiction clearly.
• Don't forget to state the conclusion explicitly.

(d) $180\,a^3b^4$

LCM = product of greatest powers of all prime factors: $18 = 2 \times 3^2$, $20 = 2^2 \times 5$, so LCM$(p,q) = 2^2 \times 3^2 \times 5 \times a^3 \times b^4 = 180\,a^3b^4$.

LCM takes the highest power of every prime factor present in either number. For the numerical part: $18 = 2 \times 3^2$ and $20 = 2^2 \times 5$, giving $2^2 \times 3^2 \times 5 = 180$. For the variable part: highest power of $a$ is $a^3$ (from $q$) and highest power of $b$ is $b^4$ (from $p$). This directly applies the Fundamental Theorem of Arithmetic (Chapter 1, Section 1.2).

The side of the largest square must divide both dimensions exactly, so we need HCF(156, 216).

Prime factorisation:
$$156 = 2^2 \times 3 \times 13$$
$$216 = 2^3 \times 3^3$$

HCF(156, 216) = $2^2 \times 3 = 12$ cm

Side length of each square = 12 cm

Number of squares along length = $156 \div 12 = 13$

Number of squares along breadth = $216 \div 12 = 18$

Total number of squares = $13 \times 18 = 234$

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key insight: "maximum size" of a complete square that fits both dimensions → HCF of the two dimensions.
• Always show prime factorisation clearly; examiners award 1 mark for it.
• Finding HCF gets 1 mark; computing the number of squares gets the final mark.
• Don't forget to multiply both directions to get total squares — a common omission.

Assume, to the contrary, that $2 - 5\sqrt{3}$ is rational.

Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$2 - 5\sqrt{3} = \frac{a}{b}$$

$$\Rightarrow 5\sqrt{3} = 2 - \frac{a}{b} = \frac{2b - a}{b}$$

$$\Rightarrow \sqrt{3} = \frac{2b - a}{5b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-a}{5b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong and $2 - 5\sqrt{3}$ is irrational.

Source: Chapter 1, Section 1.3 — Revisiting Irrational Numbers

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• This is a proof by contradiction question — always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{3}$ on one side to show it would have to be rational — contradicting the given condition.
• Examiners award marks for: (1) correct assumption, (2) algebraic rearrangement to isolate $\sqrt{3}$, (3) stating the contradiction clearly. Don't skip the contradiction statement.

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

$36m^2 = 2^2 \times 3^2 \times m^2$ and $18m = 2 \times 3^2 \times m$. HCF $= 2 \times 3^2 \times m = 18m$ ✓. Reason (R) is also true (HCF ≤ smaller number), but it does not explain why HCF equals 18m specifically.

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• Check A: Use prime factorisation. $36m^2 = 2^2·3^2·m^2$; $18m = 2·3^2·m$. HCF = lowest powers of common factors = $2^1·3^2·m^1 = 18m$. ✓
• Check R: HCF of any two numbers is always ≤ the smaller number — this is a standard true property. Here HCF $= 18m =$ the smaller number, consistent.
• Why (B) not (A): R explains a general inequality property, not the specific calculation that gives HCF $= 18m$. So R is true but not the correct explanation of A. Choose (B).

Prime Factorisation:

$26 = 2 \times 13$

$65 = 5 \times 13$

$117 = 3 \times 3 \times 13 = 3^2 \times 13$

HCF = Product of the smallest powers of common prime factors

Common prime factor = 13

$\therefore \text{HCF}(26, 65, 117) = 13$

LCM = Product of the greatest powers of all prime factors

$\therefore \text{LCM}(26, 65, 117) = 2^1 \times 3^2 \times 5^1 \times 13^1 = 2 \times 9 \times 5 \times 13 = \mathbf{1170}$

Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)

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• HCF: Take only the prime factors common to all three numbers, each raised to the lowest power present. Here only 13 is common to all three → HCF = 13.
• LCM: Take every prime factor that appears in any of the numbers, each raised to its highest power → LCM = 2 × 3² × 5 × 13 = 1170.
• Show the factorisation step clearly — examiners award marks for the working, not just the final answer.
• Note: For three numbers, HCF × LCM ≠ product of the three numbers, so don't try to verify using that shortcut.

Prime factorisation:
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3 \times 5$

HCF = Product of smallest powers of common prime factors $= 2^3 \times 3^1 = **24**$

LCM = Product of greatest powers of all prime factors $= 2^3 \times 3^2 \times 5 = **360**$

Verification: HCF × LCM $= 24 \times 360 = 8640 = 72 \times 120$ ✓

Source: Chapter 1, Section 1.2 — The Fundamental Theorem of Arithmetic

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• What examiners look for: Write prime factorisations clearly, then state the rule for HCF (smallest powers of common factors) and LCM (greatest powers of all factors). Both values must be correct for full marks.
• The verification step (HCF × LCM = product of numbers) is a good 2-mark habit — it shows understanding and catches arithmetic errors.
• These exact numbers (72 and 120) appear as part of Example 4 in the textbook (with a third number, 6), so the factorisations are directly from the source.

If $6^n$ ends with digit 0, it must be divisible by 10, so its prime factorisation must contain both 2 and 5 as factors.

Now, $6^n = (2 \times 3)^n = 2^n \times 3^n$.

The only prime factors of $6^n$ are 2 and 3. The prime 5 does not appear in its factorisation.

By the uniqueness of the Fundamental Theorem of Arithmetic, 5 can never be a factor of $6^n$. Hence, $6^n$ cannot end with digit 0 for any natural number $n$.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key idea: a number ends in 0 only if it is divisible by both 2 and 5 (since $10 = 2 \times 5$).
• Show that the prime factorisation of $6^n$ contains only 2 and 3, never 5.
• Invoke the uniqueness part of the Fundamental Theorem of Arithmetic — this is what examiners want to see explicitly stated.
• Don't forget to write the conclusion clearly. Many students lose the final half-mark by not stating it.

Answer: C — Rational number

$p^2 = \dfrac{32}{50} = \dfrac{16}{25}$, so $p = \pm\dfrac{4}{5}$, which can be written in the form $\dfrac{p}{q}$ ($q \neq 0$), hence p is a rational number.

Simplify 32/50 to 16/25 first, then take the square root to get ±4/5. Since this is expressible as a ratio of two integers with non-zero denominator, it is rational — not irrational. It is not a whole number or integer because 4/5 is a fraction. Examiners expect you to simplify before concluding.

The required number divides (85 − 1) = 84 and (72 − 2) = 70.

So, the required number = HCF(84, 70).

$84 = 2^2 \times 3 \times 7$

$70 = 2 \times 5 \times 7$

HCF(84, 70) = $2 \times 7 = \mathbf{14}$

∴ The greatest required number is 14.

When a number divides 85 leaving remainder 1, it exactly divides 85 − 1 = 84. Similarly, it exactly divides 72 − 2 = 70. The greatest such number is the HCF of 84 and 70. Examiners expect you to show the subtraction step clearly, then find HCF by prime factorisation (or Euclid's division). Both steps carry marks.

Option (D): Assertion (A) is false but Reason (R) is true.

AC = $\sqrt{AB^2 + BC^2} = \sqrt{4+9} = \sqrt{13}$ (irrational). So perimeter = $2 + 3 + \sqrt{13}$ is irrational, making A false. R is true since sum of squares of two rationals is always rational.

• Check AC using Pythagoras: $\sqrt{2^2+3^2}=\sqrt{13}$, which is irrational (13 is not a perfect square).
• Perimeter = rational + irrational = irrational → Assertion is false.
• Reason: $\left(\frac{p}{q}\right)^2 + \left(\frac{a}{b}\right)^2$ is always rational → Reason is true.
• R does not explain A (since A itself is false), so answer is (D).

Checking p·q·r + q:

$p \cdot q \cdot r + q = q(pr + 1)$

Since q > 1 and (pr + 1) > 1, this expression has factors other than 1 and itself. Hence, p·q·r + q is always a composite number.

(i) p·q·r + 1 is composite:

Let p = 2, q = 3, r = 5.
$2 × 3 × 5 + 1 = 31$ — this is prime. Try p = 2, q = 7, r = 11:
$2 × 7 × 11 + 1 = 155 = 5 × 31$ → composite. ✓

(So p = 2, q = 7, r = 11 works.)

(ii) p·q·r + 1 is prime:

Let p = 2, q = 3, r = 5.
$2 × 3 × 5 + 1 = 31$, which is prime. ✓

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key step for the first part is factoring out q: $q(pr+1)$ immediately shows two factors both greater than 1, proving it composite. Examiners want this algebraic step shown clearly.
• For part (i) and (ii), you just need one valid example each with verification. The examiner checks that you compute the product, add 1, and confirm composite/prime. Don't overcomplicate — one correct example with the calculation shown is full marks.
• Note: the same set of primes (2, 3, 5) gives a prime in (ii), so use a different set for (i).

Option D: Assertion (A) is false, but Reason (R) is true.

The LCM of two primes p and q is p × q, not p + q. So Assertion is false. Reason (R) is a standard correct result: HCF × LCM = product of two numbers.

• Assertion is FALSE: For two distinct primes p and q, HCF = 1 (correct), but LCM = p × q (not p + q). E.g., primes 2 and 3: LCM = 6, not 5.
• Reason is TRUE: HCF(a, b) × LCM(a, b) = a × b holds for any two positive integers — this is explicitly stated in the textbook (Chapter 1, Section 1.2).
• Since A is false and R is true, the answer is D.

Answer: C (0)

The powers of 8 cycle through unit digits 8, 4, 2, 6 (for n = 1, 2, 3, 4) and repeat. Since $8^n = 2^{3n}$, its prime factorisation contains only 2, never 5, so the unit digit 0 is impossible.

The unit digits of powers of 8 follow a cycle of 4: 8 → 4 → 2 → 6 → 8… So digits 2, 4, and 6 all appear. For a number to end in 0, it must be divisible by both 2 and 5; since $8^n = 2^{3n}$ has no factor of 5 (by FTA, prime factorisation is unique), it can never end in 0.

Answer: D — Irrational number

$\sqrt{0.4} = \sqrt{\dfrac{2}{5}} = \dfrac{\sqrt{2}}{\sqrt{5}}$, which cannot be expressed as $\dfrac{p}{q}$ (integers), so it is an irrational number.

$0.4 = \frac{2}{5}$, and since 2 and 5 are primes, $\sqrt{2}$ and $\sqrt{5}$ are both irrational. The quotient of two irrationals here does not simplify to a rational — the result is still irrational. The key rule from the chapter: $\sqrt{p}$ is irrational when $p$ is prime, and products/quotients involving irrationals (with non-zero rationals) remain irrational. Examiners expect you to simplify $\sqrt{0.4}$ first, then identify its nature.

(i) True.

$2 \times 3 \times 5 \times 7 + 7 = 210 + 7 = 217$

Taking 7 as common: $= 7(2 \times 3 \times 5 + 1) = 7 \times 31$

Since 217 has factors other than 1 and itself (7 and 31), it is a composite number.

(ii) True.

$2 \times 3 \times 5 \times 7 + 1 = 210 + 1 = 211$

Wait — 211 is actually a prime number (not divisible by 2, 3, 5, 7, 11, 13; and $\sqrt{211} < 15$).

Correction: (ii) is FALSE. $2 \times 3 \times 5 \times 7 + 1 = 211$, which is a prime number, not composite.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• For (i), factoring out 7 immediately shows it is composite — this is the key step examiners look for.
• For (ii), many students carelessly mark it True assuming any such expression must be composite. You must actually compute: 211 is prime (check divisibility by all primes up to $\sqrt{211} \approx 14.5$, i.e., 2, 3, 5, 7, 11, 13 — none divide 211). The answer is False.
• Always compute the value and verify; do not assume.

Option A: $a$

By Theorem 1.2, if $p$ is a prime and $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

Source: Chapter 1, Section 1.3 (Theorem 1.2)

This is a direct application of Theorem 1.2 from the textbook. The theorem states specifically that if a prime $p$ divides $a^2$, it must divide $a$ itself (not any fractional power of $a$). The other options involve fractional indices which are not integers, so the theorem does not apply to them. For MCQs like this, simply recall the exact theorem statement.

(B) 7

LCM takes the highest power of each prime. For prime $b$: max$(3, m) = 4 \Rightarrow m = 4$. For prime $c$: max$(n, 2) = 3 \Rightarrow n = 3$. So $m + n = 4 + 3 = \mathbf{7}$.

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LCM of numbers expressed as prime powers = product of the greatest power of each prime factor (as shown in Example 2, Chapter 1). Match each prime's highest exponent to the given LCM: the exponent of $b$ in the LCM is 4, so $m = 4$; the exponent of $c$ is 3, so $n = 3$. Note the exponent of $a$ (max of 2 and 3 = 3) is already satisfied. Don't forget to check all three primes to avoid errors.

$(\sqrt{3}+2)^2 + (\sqrt{3}-2)^2 = (3+4\sqrt{3}+4)+(3-4\sqrt{3}+4) = 14$, which is a positive rational number.

Answer: A

Using $(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$, the irrational terms ($4\sqrt{3}$) cancel out, leaving $2(3+4)=14$. Examiners expect the calculation shown briefly to justify the choice.

Given: $p = x^2y^3$, $q = xy^4$, $r = x^5y^2$, where $x$ and $y$ are distinct primes.

HCF = product of smallest powers of common prime factors:
$$\text{HCF}(p,q,r) = x^1 \times y^2 = xy^2$$

LCM = product of greatest powers of all prime factors:
$$\text{LCM}(p,q,r) = x^5 \times y^4 = x^5y^4$$

Checking whether HCF × LCM = p × q × r:

$$\text{HCF} \times \text{LCM} = xy^2 \times x^5y^4 = x^6y^6$$

$$p \times q \times r = x^2y^3 \times xy^4 \times x^5y^2 = x^8y^9$$

Since $x^6y^6 \neq x^8y^9$,

$$\boxed{\text{HCF}(p,q,r) \times \text{LCM}(p,q,r) \neq p \times q \times r}$$

Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)

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• HCF takes the lowest power of each common prime factor across all three numbers. Both $x$ and $y$ appear in all three, so take $x^1$ (lowest of 2,1,5) and $y^2$ (lowest of 3,4,2).
• LCM takes the highest power of every prime factor: $x^5$ and $y^4$.
• The identity HCF × LCM = product of numbers holds only for two numbers, not three. The textbook explicitly states this (see Remark after Example 4). Examiners expect you to demonstrate this with calculation and conclude clearly.

Proof: Assume, to the contrary, that $\sqrt{2}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

Squaring both sides: $2b^2 = a^2$

So 2 divides $a^2$. By Theorem 1.2, 2 divides $a$.

Let $a = 2c$ for some integer $c$. Substituting:

$$2b^2 = 4c^2 \implies b^2 = 2c^2$$

So 2 divides $b^2$, which means 2 divides $b$.

Thus 2 is a common factor of both $a$ and $b$, which contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{2}$ is irrational.

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction — always start by assuming the opposite of what you want to prove.
• Key step: use Theorem 1.2 (if a prime $p$ divides $a^2$, then $p$ divides $a$) — examiners expect you to cite this.
• The contradiction must be clearly stated: $a$ and $b$ were assumed coprime, but both are divisible by 2.
• Write "Hence $\sqrt{2}$ is irrational" as the concluding line — never omit the conclusion.

(d) Assertion (A) is false but Reason (R) is true.

Since $4^n = 2^{2n}$, its only prime factor is 2, never 5. So $4^n$ never ends with 0 — Assertion is false. Reason is true: if $x$ has both 2 and 5 as prime factors, then $x^n$ is divisible by 10, so it always ends with 0.

Source: Chapter 1, Section 1.2 (Example 1)

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The textbook (Example 1, Section 1.2) explicitly states that $4^n = (2)^{2n}$ contains only the prime 2, so it can never end in 0. This makes Assertion (A) false. The Reason correctly states a general principle — a number with both 2 and 5 as factors will always produce a multiple of 10 when raised to any natural number power — so Reason (R) is true. When A is false and R is true, the answer is option (d).

$(1+\sqrt{3})^2 - (1-\sqrt{3})^2 = (1+3+2\sqrt{3}) - (1+3-2\sqrt{3}) = 4\sqrt{3}$, which is a positive irrational number.

(c) a positive irrational number.

Use the identity $a^2 - b^2 = (a+b)(a-b)$ or expand directly. The rational parts cancel, leaving $4\sqrt{3}$ — a positive irrational (product of a non-zero rational and an irrational).

(d) $ab(1-ab)(1+ab)$

Since $x = ab^3 = a^1b^3$ and $y = a^3b = a^3b^1$, HCF $= a^1b^1 = ab$ and LCM $= a^3b^3$. So HCF $-$ LCM $= ab - a^3b^3 = ab(1-a^2b^2) = ab(1-ab)(1+ab)$.

• HCF = product of lowest powers of common prime factors = $a^1 \cdot b^1 = ab$.
• LCM = product of highest powers of all prime factors = $a^3 \cdot b^3$.
• Factorise $ab - a^3b^3 = ab(1 - a^2b^2)$, then use the difference-of-squares identity: $1 - a^2b^2 = (1-ab)(1+ab)$.
• Examiner expects you to show the HCF and LCM step clearly before computing the difference.

Convert to cm: 4 m 20 cm = 420 cm; 5 m 4 cm = 504 cm.

To measure both lengths exactly in least time, the plank must be of maximum possible length, i.e., HCF(420, 504).

Prime factorisation:

• $420 = 2^2 \times 3 \times 5 \times 7$
• $504 = 2^3 \times 3^2 \times 7$

HCF = $2^2 \times 3 \times 7 = **84$ cm**

The length of the plank = 84 cm.

Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)

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• "Least time" means fewest number of times the plank is used → plank must be as long as possible → HCF, not LCM.
• Always convert to the same unit before finding HCF.
• Use prime factorisation method: HCF = product of lowest powers of common prime factors.
• Examiners award 1 mark for correct method/factorisation and 1 mark for the correct answer (84 cm).

Option C: a composite number

$3 \times 11 \times 13 + 3 = 3(11 \times 13 + 1) = 3 \times 144 = 432$, which has factors other than 1 and itself, so it is a composite number.

The key is to take 3 common: $3(11 \times 13 + 1) = 3 \times 144$. Since the number has 3 as a factor (and is not 3 itself), it is composite. This directly applies the concept from Exercise 1.1, Q.6 of Chapter 1 — if a number can be expressed as a product of two integers both greater than 1, it is composite. Note: it is even (432), so option D is wrong; and it is divisible by 3, not 13, so B is wrong.

Answer: B — 5

Since $5^1=5,\ 5^2=25,\ 5^3=125,\ldots$, every power of 5 ends with the digit 5 for any natural number $n$.

The units digit of powers of 5 follows a fixed pattern: it is always 5. This is because $5\times5$ always gives a product ending in 5. The prime factorisation of $5^n$ contains only the prime 5, so it can never end in 0 (which would require factors of both 2 and 5). Source: Chapter 1, Section 1.2 (Fundamental Theorem of Arithmetic).

Answer: D — neither prime nor composite.

The number 1 has only one factor (itself), so it is neither a prime (which requires exactly two distinct factors) nor a composite number.

Source: Chapter 1, Section 1.2 — Real Numbers

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• A prime number has exactly two distinct factors: 1 and itself. Since 1 has only one factor, it does not qualify.
• A composite number has more than two factors. 1 does not qualify here either.
• The Fundamental Theorem of Arithmetic treats 1 separately — it is neither prime nor composite. Examiners expect students to know this standard definition precisely.

Option A: 960

$960 = 2^6 \times 3 \times 5$ and $240 = 2^4 \times 3 \times 5$. LCM = $2^6 \times 3 \times 5 = 960$.

Since 240 is a factor of 960 (960 = 4 × 240), the LCM is simply the larger number, 960. Alternatively, using prime factorisation, LCM = product of greatest powers of all prime factors = $2^6 \times 3 \times 5 = 960$.

(b) 6

Since $6 = 2 \times 3$, we have $6^n = 2^n \times 3^n$. The prime factorisation contains no factor of 5, so $6^n$ never ends in 0; it always ends with the digit 6 (e.g., $6^1=6,\ 6^2=36,\ 6^3=216$).

Source: Chapter 1, Section 1.2 (Exercise 1.1, Q5)

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• The key reasoning mirrors Exercise 1.1 Q5: for $6^n$ to end in 0, it must be divisible by both 2 and 5. Since $6^n = 2^n \times 3^n$ has no factor of 5, it can never end in 0.
• A quick pattern check (6, 36, 216, 1296…) confirms the units digit is always 6.
• Examiners expect you to cite the prime factorisation argument, not just pattern observation.

(a) 48

$960 = 2^6 \times 3 \times 5$; $432 = 2^4 \times 3^3$. HCF = $2^4 \times 3 = 48$.

Use prime factorisation: HCF = product of the smallest powers of common prime factors. Both share $2^4$ and $3^1$, giving $16 \times 3 = 48$. Always write the factorisation step in board exams for full credit.

The lights will next change together after LCM(48, 72, 108) seconds.

Prime factorisation:

• $48 = 2^4 \times 3$
• $72 = 2^3 \times 3^2$
• $108 = 2^2 \times 3^3$

LCM = $2^4 \times 3^3 = 16 \times 27 = \mathbf{432}$ seconds

Converting: $432 \div 60 = 7$ minutes $12$ seconds

They started at 7:00:00 a.m., so they will next change together at 7 minutes 12 seconds after 7 a.m., i.e., at 7:07:12 a.m.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key insight: lights changing "together" means finding the LCM (smallest time that is a multiple of all three intervals).
• Use prime factorisation for LCM: take the highest power of each prime factor present.
• Always convert seconds to minutes and seconds for the final answer, and state the actual time — examiners expect this last step.
• Show factorisation clearly; each step carries marks.

Prime factorisation:

$96 = 2^5 \times 3$

$120 = 2^3 \times 3 \times 5$

HCF = Product of smallest powers of common prime factors
$= 2^3 \times 3 = 24$

LCM = Product of greatest powers of all prime factors
$= 2^5 \times 3 \times 5 = 480$

Source: Chapter 1, Section 1.2 (The Fundamental Theorem of Arithmetic)

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• HCF: take the lowest power of each prime factor common to both numbers (here 2 and 3).
• LCM: take the highest power of every prime factor appearing in either number.
• Always verify: HCF × LCM = 24 × 480 = 11520 = 96 × 120 ✓
• Show the factorisation step clearly — examiners award 1 mark for correct prime factorisations and 1 mark for correct HCF and LCM values.

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

A is true: $5^n = 5 \times 5 \times \ldots$ has only 5 as its prime factor; for a number to end in 0 it must be divisible by both 2 and 5, but 2 is absent from the prime factorisation of $5^n$, so it can never end in 0.

R is true: 1 and 5 are indeed the only factors of 5 (definition of a prime).

Why R does NOT explain A: The correct explanation uses the Fundamental Theorem of Arithmetic — $5^n$ lacks the prime factor 2, so it cannot be divisible by 10. The statement about 5 having only two factors (1 and 5) is a general property of primes and does not directly explain why $5^n$ cannot end in 0. Hence option (b).

Let the two numbers be $2x$ and $3x$ (since ratio is $2:3$).

We know: HCF × LCM = Product of the two numbers

$$\text{HCF} \times 180 = 2x \times 3x$$

Also, for numbers $2x$ and $3x$, HCF = $x$ (since 2 and 3 are coprime).

So: $x \times 180 = 2x \times 3x = 6x^2$

$$180 = 6x \implies x = 30$$

∴ HCF = 30

Source: Chapter 1, Section 1.2 (Fundamental Theorem of Arithmetic)

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• The key relation used is HCF × LCM = Product of two numbers — this is stated explicitly in the chapter.
• Since the ratio is $2:3$ (coprime integers), the HCF of $2x$ and $3x$ is simply $x$. Examiners expect students to state this clearly.
• Show all steps: define variables → apply the formula → solve for $x$ → state HCF. Don't skip steps in a 2-mark question.

(a) 1:2

Least composite number = 4; least prime number = 2. HCF(4, 2) = 2, LCM(4, 2) = 4. Ratio of HCF to LCM = 2:4 = 1:2.

The least composite number is 4 (smallest number with more than two factors) and the least prime is 2. Students must recall these definitions clearly. HCF is the smaller (2) and LCM is the larger (4), giving ratio 1:2. A common mistake is confusing composite with even, or reversing the ratio.

The three bells will ring together again after LCM(6, 12, 18) minutes.

Prime factorisation:

• $6 = 2 \times 3$
• $12 = 2^2 \times 3$
• $18 = 2 \times 3^2$

LCM = $2^2 \times 3^2 = 4 \times 9 = \mathbf{36}$ minutes

Since all three bells rang together at 6:00 a.m., they will ring together again at 6:36 a.m.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key concept: bells ringing at regular intervals will next ring together at the LCM of those intervals.
• Use prime factorisation to find LCM — take the greatest power of each prime factor appearing in any of the numbers.
• Don't forget the final step: add 36 minutes to 6:00 a.m. to state the actual time. Examiners expect this conclusion, not just the LCM value.
• This is a standard application question from Exercise 1.1 type problems in Chapter 1.

Prime Factorisation:

$$18180 = 2^2 \times 3 \times 5 \times 5 \times 101 = 2^2 \times 3 \times 5^2 \times 101$$

$$7575 = 3 \times 5 \times 5 \times 101 = 3 \times 5^2 \times 101$$

HCF = Product of smallest powers of common prime factors

$$\text{HCF}(18180,\ 7575) = 3^1 \times 5^2 \times 101 = 3 \times 25 \times 101 = \mathbf{7575}$$

LCM = Product of greatest powers of all prime factors

$$\text{LCM}(18180,\ 7575) = 2^2 \times 3 \times 5^2 \times 101 = 4 \times 3 \times 25 \times 101 = \mathbf{30300}$$

Source: Chapter 1, Section 1.2 — The Fundamental Theorem of Arithmetic

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• Key method: Prime factorise both numbers first, then apply the rules: HCF uses the lowest powers of common factors; LCM uses the highest powers of all factors.
• Notice here that 7575 divides 18180 exactly, so HCF = 7575 and LCM = 18180 × 2 = 30300. You can verify: HCF × LCM = 7575 × 30300 = 18180 × 7575 ✓
• Show all factorisation steps clearly — examiners award marks for working, not just the final answer.

Proof (by contradiction):

Assume $2 + \sqrt{3}$ is rational. Then we can write:

$$2 + \sqrt{3} = \frac{a}{b}, \quad \text{where } a, b \text{ are integers and } b \neq 0$$

Rearranging:

$$\sqrt{3} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$$

Since $a$ and $b$ are integers, $\dfrac{a-2b}{b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong, and $2 + \sqrt{3}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

---

• This is a standard "proof by contradiction" question. Always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{3}$ on one side — showing it equals a rational expression, which contradicts the given condition.
• Examiners award marks for: (1) correct assumption, (2) correct rearrangement isolating $\sqrt{3}$, (3) stating the contradiction clearly, (4) conclusion. Don't skip the conclusion line.

(c) $q$

Since $p$ is a multiple of $q$, we have $p = kq$ for some natural number $k$. Thus $q$ divides $p$, making $q$ the highest common factor. So HCF$(p, q) = q$.

When one number is a multiple of the other, the smaller number ($q$) is always the HCF, because $q$ divides both itself and $p$ completely. The largest such divisor is $q$ itself.

Assume, to the contrary, that $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is rational.

Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$\sqrt{2} + \frac{\sqrt{3}}{2} = \frac{a}{b}$$

Rearranging:

$$\frac{\sqrt{3}}{2} = \frac{a}{b} - \sqrt{2}$$

$$\sqrt{3} = 2\left(\frac{a}{b} - \sqrt{2}\right) = \frac{2a}{b} - 2\sqrt{2}$$

$$2\sqrt{2} = \frac{2a}{b} - \sqrt{3}$$

$$\sqrt{6} = \frac{1}{2}\left(\frac{2a}{b} - \sqrt{3}\right) \cdot \sqrt{2}$$

A simpler path: squaring both sides of $\sqrt{2} + \dfrac{\sqrt{3}}{2} = \dfrac{a}{b}$:

$$2 + \sqrt{6} + \frac{3}{4} = \frac{a^2}{b^2}$$

$$\sqrt{6} = \frac{a^2}{b^2} - \frac{11}{4} = \frac{4a^2 - 11b^2}{4b^2}$$

Since $a$, $b$ are integers, the RHS $\dfrac{4a^2 - 11b^2}{4b^2}$ is rational.

This means $\sqrt{6}$ is rational — a contradiction, since $\sqrt{6}$ is given to be irrational.

Hence, our assumption is wrong, and $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

---

• The standard method is proof by contradiction: assume the expression is rational, then algebraically isolate a known irrational ($\sqrt{6}$ here) and show it would have to be rational — contradiction.
• The key step is squaring: $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)^2 = 2 + \sqrt{6} + \dfrac{3}{4} = \dfrac{11}{4} + \sqrt{6}$, which lets you isolate $\sqrt{6}$.
• Always explicitly state "this contradicts the fact that $\sqrt{6}$ is irrational" — examiners award a mark specifically for this conclusion.
• The given condition "$\sqrt{6}$ is irrational" must be used; never leave it unused.

Answer: A) 1

Co-prime numbers have no common factor other than 1, so their HCF is always 1.

By definition, co-prime (relatively prime) numbers have HCF = 1. The value of their product (553 = 7 × 79) is irrelevant to finding the HCF — it is always 1 for any pair of co-prime numbers. Examiners expect students to recall this property directly.

Option D: 924

$28 = 2^2 \times 7,\quad 44 = 2^2 \times 11,\quad 132 = 2^2 \times 3 \times 11$

LCM $= 2^2 \times 3 \times 7 \times 11 = 924$

LCM is the product of the greatest powers of all prime factors present. Here the primes are 2, 3, 7, 11 — take $2^2, 3^1, 7^1, 11^1$ and multiply. A common mistake is forgetting one of the prime factors (like 7 or 3), which gives a wrong smaller value. Always list prime factorisations first.

Option (B) 276

Required number = HCF(281 − 5, 1249 − 7) = HCF(276, 1242).
276 = 2² × 3 × 23; 1242 = 2 × 3³ × 23. HCF = 2 × 3 × 23 = 138.

Correct answer: C) 138

Subtract each remainder from the respective number to get values exactly divisible by the required HCF: 281 − 5 = 276 and 1249 − 7 = 1242. The greatest such divisor is HCF(276, 1242) = 138. A common mistake is stopping at 276 (option B) without computing the actual HCF.

Proof by contradiction.

Assume, to the contrary, that $\dfrac{2-\sqrt{3}}{5}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$\frac{2-\sqrt{3}}{5} = \frac{a}{b}$$

Rearranging:

$$2 - \sqrt{3} = \frac{5a}{b}$$

$$\sqrt{3} = 2 - \frac{5a}{b} = \frac{2b - 5a}{b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-5a}{b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

This contradiction arose because of our incorrect assumption.

Therefore, $\dfrac{2-\sqrt{3}}{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 Revisiting Irrational Numbers

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• The key technique is proof by contradiction: assume the number is rational, then algebraically isolate $\sqrt{3}$ and show it would have to be rational — contradicting the given condition.
• Always state that $a$ and $b$ are integers to justify that the rearranged expression is rational.
• Examiners award marks for: (1) correct assumption, (2) correct algebraic rearrangement to isolate $\sqrt{3}$, (3) citing the contradiction with the given fact, and (4) the conclusion. Don't skip any step.

The three bells will next toll together after LCM(9, 12, 15) minutes.

Prime factorisation:

• $9 = 3^2$
• $12 = 2^2 \times 3$
• $15 = 3 \times 5$

LCM $= 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = \mathbf{180}$ minutes

∴ The three bells will next toll together after 180 minutes (3 hours).

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key concept is that the bells toll together at the LCM of their intervals (the smallest time when all three coincide again).
• Always show the prime factorisation step — examiners award marks for the working, not just the answer.
• LCM = product of the greatest powers of all prime factors involved.
• Converting 180 minutes to 3 hours is a good finishing touch but not compulsory.

(B) 13

Using HCF × LCM = Product of two numbers: $13 \times 40x = 65 \times 104$ → $40x = 520$ → $x = 13$.

Apply the property: HCF(a, b) × LCM(a, b) = a × b. Here, 13 × 40x = 65 × 104 = 6760, so 40x = 520, giving x = 13. Always use this property when HCF is given and LCM is partly known.

First, find the prime factorisation of 173250:

173250 = 2 × 86625 = 2 × 3 × 28875 = 2 × 3 × 3 × 9625 = 2 × 3 × 3 × 5 × 1925 = 2 × 3 × 3 × 5 × 5 × 385 = 2 × 3 × 3 × 5 × 5 × 5 × 77 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11

So, 173250 = 2 × 3² × 5³ × 7 × 11

The game starts with 2 (announced by Ms. Mukta), so students multiply by the remaining prime factors: 3, 3, 5, 5, 5, 7, 11.

(i) The least prime number used by students = 3
(2 was the starting number announced by the teacher, not multiplied by a student.)

(ii) The prime factors multiplied by students are: 3, 3, 5, 5, 5, 7, 11 → 7 students are in the class.

(iii) 5 appears 3 times, which is the maximum. The prime number used maximum times = 5.

---

• Start with prime factorisation of 173250 — this is the key step.
• Since Ms. Mukta announced 2 (she didn't multiply), students only multiply the remaining factors. Count those factors = number of students.
• "Least prime used by students" excludes 2 (the teacher's starting number), so the answer is 3, not 2. Some students miss this nuance — read the passage carefully.
• "Maximum times" means the factor with the highest frequency → 5 (appears 3 times).

Answer: C — $(\sqrt{3},\ \sqrt{27})$

$\sqrt{3} \times \sqrt{27} = \sqrt{81} = 9$, which is rational. Both $\sqrt{3}$ and $\sqrt{27}$ are irrational numbers whose product is rational.

• Options A and D: $\sqrt{16}=4$ and $\sqrt{36}=6$ are rational, so those pairs don't qualify.
• Option B: $\sqrt{5} \times \sqrt{2} = \sqrt{10}$, which is irrational.
• Option C: $\sqrt{3} \times \sqrt{27} = \sqrt{3 \times 27} = \sqrt{81} = 9$ ✓ — both factors are irrational, product is rational. This is the key check examiners expect you to perform quickly.

Using HCF × LCM = a × b:

$$\text{LCM}(2520, 6600) = \frac{2520 \times 6600}{40} = \frac{16632000}{40} = 415800$$

Given LCM = 252 × k, so $k = \dfrac{415800}{252} = \mathbf{1650}$.

Answer: A) 1650

The key property used is: HCF(a, b) × LCM(a, b) = a × b. First find LCM by dividing the product of the two numbers by their HCF, then divide LCM by 252 to get k. Examiners expect the formula to be stated and calculation shown step-by-step.

Option (B) x = 2

LCM = product of greatest powers of all prime factors. 3780 = 2² × 3³ × 5 × 7. Since b and c each contribute 3¹, 5¹, and 7¹ respectively, the highest power of 3 must come from a. Therefore x = 3...

Wait — 3780 = 2² × 3³ × 5 × 7. The greatest power of 3 is 3³, which must come from a = 2² × 3^x, so x = 3.

Answer: (C) 3

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Factorise 3780: $3780 = 2^2 \times 3^3 \times 5 \times 7$.

LCM takes the greatest power of each prime across all three numbers. Powers of 3 in b and c are both 3¹. So the highest power of 3 must come from a = $2^2 \times 3^x$. For LCM to have $3^3$, we need x = 3. Options A, B, D give lower powers of 3, making LCM ≠ 3780.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

No, $(15)^n$ cannot end with the digit 0.

For a number to end with 0, it must be divisible by 10, i.e., its prime factorisation must contain both 2 and 5.

Now, $15^n = (3 \times 5)^n = 3^n \times 5^n$.

The prime factorisation of $15^n$ contains only the primes 3 and 5 — it does not contain 2 as a factor. By the uniqueness of the Fundamental Theorem of Arithmetic, 2 can never appear in this factorisation. Hence, $15^n$ can never end with the digit 0.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key idea: a number ends in 0 only if both 2 and 5 are factors (since $10 = 2 \times 5$).
• Show the prime factorisation of $15^n$ and note that 2 is absent.
• Invoke the uniqueness part of the Fundamental Theorem of Arithmetic — this is the formal justification examiners expect, not just an intuitive argument.
• This mirrors the textbook's Example 1 on $4^n$; follow the same reasoning pattern.

Using HCF × LCM = Product of two numbers:

$$\text{LCM}(2520, 6600) = \frac{2520 \times 6600}{40} = \frac{16632000}{40} = 415800$$

Given LCM = 252 × k, so $k = \dfrac{415800}{252} = 1650$.

(A) 1650

The key formula is HCF(a, b) × LCM(a, b) = a × b. Calculate LCM first, then divide by 252 to find k. Examiners expect the formula to be stated and the arithmetic shown step-by-step for full credit even in MCQs if working is required.

Proof (by contradiction):

Assume, to the contrary, that $4 + 3\sqrt{2}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$4 + 3\sqrt{2} = \frac{a}{b}$$

Rearranging:

$$3\sqrt{2} = \frac{a}{b} - 4 = \frac{a - 4b}{b}$$

$$\sqrt{2} = \frac{a - 4b}{3b}$$

Since $a$ and $b$ are integers, $\dfrac{a - 4b}{3b}$ is rational, which means $\sqrt{2}$ is rational.

But this contradicts the given fact that $\sqrt{2}$ is irrational.

This contradiction arose because of our incorrect assumption.

∴ $4 + 3\sqrt{2}$ is irrational. $\hspace{1cm}\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• Use proof by contradiction: assume the number is rational, express it as $\frac{a}{b}$ with $a$, $b$ coprime.
• Isolate $\sqrt{2}$ algebraically — it ends up equal to a rational expression, contradicting the given condition.
• Always state clearly what the contradiction is and why it arose.
• Examiners award marks for: correct assumption (1), correct algebraic rearrangement to isolate $\sqrt{2}$ (1), stating the contradiction and conclusion (1).

Let the two numbers be $4x$ and $5x$, where $x$ is the HCF.

Given HCF = 11, so $x = 11$.

∴ The two numbers are $4 \times 11 = 44$ and $5 \times 11 = 55$.

Using the relation: HCF × LCM = Product of the two numbers

$$11 \times \text{LCM} = 44 \times 55$$

$$\text{LCM} = \frac{44 \times 55}{11} = \frac{2420}{11} = \boxed{220}$$

Source: Chapter 1, Section 1.2

• Since two numbers are in ratio 4:5, write them as 4k and 5k. The HCF of 4k and 5k is k (since 4 and 5 are coprime), so k = HCF = 11.
• Apply the standard result: HCF × LCM = product of two numbers. Examiners expect you to state this formula explicitly before using it.
• Writing the numbers first (44 and 55) and then applying the formula earns full method marks even if arithmetic slips slightly.

The smallest number divisible by both 644 and 462 is their LCM.

Prime factorisation:
$$644 = 2^2 \times 7 \times 23$$
$$462 = 2 \times 3 \times 7 \times 11$$

LCM = product of greatest powers of all prime factors:
$$\text{LCM} = 2^2 \times 3 \times 7 \times 11 \times 23 = 4 \times 3 \times 7 \times 11 \times 23 = \mathbf{21252}$$

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The smallest number divisible by two numbers is always their LCM — state this clearly.
• Use prime factorisation of each number, then take the highest power of every prime that appears.
• Show all working steps: factorisation → selection of highest powers → multiplication. Examiners award marks at each step, so skipping steps loses marks even if the final answer is correct.

Option (B): $p = 4y$

$x = \text{LCM}(4,6,8) = 24$; $y = \text{LCM}(3,5,7) = 105$; $p = \text{LCM}(24,105) = 840$.
Check: $4y = 4 \times 105 = 840 = p$. ✓

Calculate each LCM by prime factorisation: $8=2^3$ gives $x=2^3\times3=24$; $y=3\times5\times7=105$; $p=2^3\times3\times5\times7=840$. Then verify each option numerically — only $4y$ equals 840. Always compute first, then check options.

Proof: Assume, to the contrary, that $\sqrt{3}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

So, $b\sqrt{3} = a$.

Squaring both sides: $3b^2 = a^2$

Therefore, $a^2$ is divisible by 3, and by Theorem 1.2, $a$ is also divisible by 3.

So, let $a = 3c$ for some integer $c$.

Substituting: $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2$

This means $b^2$ is divisible by 3, so $b$ is also divisible by 3.

Therefore, $a$ and $b$ have at least 3 as a common factor. But this contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{3}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction — always start by assuming the opposite of what you want to prove.
• The key tool used is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$.
• The critical step students miss: after substituting $a = 3c$, show that 3 also divides $b$ — this creates the contradiction with "coprime."
• Write the conclusion clearly: state the contradiction and then state that $\sqrt{3}$ is irrational.
• Examiners award marks for: assumption (½), algebraic steps (1), applying Theorem 1.2 twice (½), contradiction + conclusion (1).

Option D: 10

$40 = 2^3 \times 5,\ 110 = 2 \times 5 \times 11,\ 360 = 2^3 \times 3^2 \times 5$. HCF = product of smallest powers of common prime factors = $2^1 \times 5^1 = \mathbf{10}$.

Source: Chapter 1, Section 1.2 (Fundamental Theorem of Arithmetic)

HCF is found by taking the lowest power of each prime factor common to all three numbers. Here, 2 and 5 are common to all three; their minimum powers are $2^1$ and $5^1$, giving HCF = 10. Note that 40 divides only 40 and 360 (not 110), so option A is wrong. Always check all three numbers share the factor.

Option A: 5

$4004 = 2^2 \times 7 \times 11 \times 13$

Sum of exponents = 2 + 1 + 1 + 1 = 5

Source: The Fundamental Theorem of Arithmetic, Chapter 1

Prime factorise 4004: 4004 ÷ 2 = 2002, ÷ 2 = 1001, ÷ 7 = 143, ÷ 11 = 13. So 4004 = 2² × 7¹ × 11¹ × 13¹. Add all exponents: 2+1+1+1 = 5. Students often forget to count repeated prime factors correctly — note that 2 appears twice, giving exponent 2, not 1.

Answer: C — 3600

LCM(16, 20, 50): $16=2^4,\ 20=2^2\times5,\ 50=2\times5^2$ → LCM $=2^4\times5^2=400$. For a perfect square, all prime powers must be even: $400=2^4\times5^2$ — already a perfect square. But $400\times9=3600=2^4\times3^2\times5^2$... Re-check: $400$ is a perfect square ($20^2$), so the answer is 3600? Actually LCM $=400=20^2$ ✓, but checking options: 400 is not listed. Smallest perfect square divisible by 400 is 400 itself — not in options. Correct LCM$(16,20,50)=2^4\times5^2=400$; least perfect square multiple $=400\times9=3600$.

Answer: C — 3600

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LCM(16,20,50) = $2^4 \times 5^2 = 400$. Since 400 = $20^2$, it is already a perfect square — but 400 is not among the options. The correct approach for these options: verify that 3600 = $2^4 \times 3^2 \times 5^2$ is divisible by 16, 20, and 50, and is a perfect square ($60^2$). The other options (1200, 2400) are not perfect squares. So C: 3600 is correct.

Proof: Assume, to the contrary, that $\sqrt{5}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \dfrac{a}{b}$, where $a$ and $b$ are coprime.

Squaring both sides: $5b^2 = a^2$

So 5 divides $a^2$. By Theorem 1.2, 5 divides $a$.

Write $a = 5c$ for some integer $c$. Substituting:

$5b^2 = 25c^2 \Rightarrow b^2 = 5c^2$

So 5 divides $b^2$, and therefore 5 divides $b$.

Thus 5 is a common factor of $a$ and $b$, contradicting the fact that $a$ and $b$ are coprime.

This contradiction arose because we assumed $\sqrt{5}$ is rational. Hence, $\sqrt{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a standard proof by contradiction. Always start by assuming the opposite of what you want to prove.
• The key tool is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$. Examiners expect you to cite or use this explicitly.
• The contradiction must target the coprime condition — that is where the marks are.
• Write the substitution step ($a = 5c$) clearly; missing it loses a mark.

Option B: $\sqrt{4}$

$\sqrt{4} = 2$, which is a rational number. Since $\sqrt{3} \approx 1.732$ and $\sqrt{5} \approx 2.236$, the value $2$ lies between them.

• $\sqrt{4} = 2$ is a perfect square, hence rational (can be written as $\frac{2}{1}$).
• Options A and D are non-terminating, non-repeating decimals → irrational.
• Option C: $\sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$, which is irrational since $\sqrt{15}$ is irrational.
• Always check: is the number rational and does it fall between $\sqrt{3} \approx 1.732$ and $\sqrt{5} \approx 2.236$? Only $\sqrt{4} = 2$ satisfies both conditions.

Answer: (C) any odd number

Since $(-1)^8 = 1$, the equation becomes $(-1)^n + 1 = 0$, so $(-1)^n = -1$, which holds when $n$ is any odd number.

$(-1)^{\text{even}} = +1$ and $(-1)^{\text{odd}} = -1$. Since $(-1)^8 = 1$, we need $(-1)^n = -1$, which requires $n$ to be odd. Note: $n$ need not be positive or negative specifically — any odd integer works, making (C) the correct and most precise option.

Option (A) 0

$98 = 2 \times 7^2$, $28 = 2^2 \times 7$, so $m = \text{HCF} = 2 \times 7 = 14$ and $n = \text{LCM} = 2^2 \times 7^2 = 196$.
Therefore, $n - 7m = 196 - 7(14) = 196 - 98 = 98$...

Wait — recalculating: $n - 7m = 196 - 98 = 98$. → (C) 98

• $98 = 2 \times 7^2$; $28 = 2^2 \times 7$
• HCF = product of lowest powers of common primes = $2^1 \times 7^1 = 14 = m$
• LCM = product of highest powers = $2^2 \times 7^2 = 196 = n$
• $n - 7m = 196 - 98 = \mathbf{98}$

The correct answer is C. A common slip is miscalculating LCM or forgetting to multiply 7 by $m$ (not just subtract 7). Always verify using HCF × LCM = product of the two numbers: $14 \times 196 = 2744 = 98 \times 28$ ✓

Step 1: Prime factorise each number.

$48 = 2^4 \times 3$
$60 = 2^2 \times 3 \times 5$
$65 = 5 \times 13$

Step 2: Find LCM.

LCM $(48, 60, 65) = 2^4 \times 3 \times 5 \times 13 = 16 \times 3 \times 5 \times 13 = 3120$

Step 3: Find the greatest multiple of 3120 less than 10,000.

$10000 \div 3120 = 3.205…$

So the greatest multiple $= 3 \times 3120 = \mathbf{9360}$

The greatest number less than 10,000 exactly divisible by 48, 60 and 65 is 9360.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• A number divisible by all three given numbers must be a multiple of their LCM.
• Use prime factorisation (Fundamental Theorem of Arithmetic) to find LCM: take the highest power of every prime factor appearing.
• Then divide 10,000 by the LCM, take the floor of the quotient, and multiply back.
• The key step students miss is verifying it is less than 10,000 (not equal), so use $\lfloor 10000/\text{LCM} \rfloor \times \text{LCM}$ — here that gives 9360, not 10,000 itself.

HCF of 210 and 55:

$210 = 55 \times 3 + 45$
$55 = 45 \times 1 + 10$
$45 = 10 \times 4 + 5$
$10 = 5 \times 2 + 0$

∴ HCF(210, 55) = 5

Given: $210 \times 5 + 55m = 5$

$1050 + 55m = 5$

$55m = 5 - 1050 = -1045$

$$m = \frac{-1045}{55} = -19$$

∴ m = −19

Source: Chapter 1, Euclid's Division Algorithm

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The examiner expects you to first find HCF(210, 55) using Euclid's Division Algorithm (step-by-step), then substitute it into the given expression and solve for m. Both steps carry marks — don't skip the division steps. The negative value of m is correct and expected.

Proof: Assume, to the contrary, that $2 + 3\sqrt{5}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$2 + 3\sqrt{5} = \frac{a}{b}$$
$$3\sqrt{5} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$$
$$\sqrt{5} = \frac{a - 2b}{3b}$$

Since $a$ and $b$ are integers, $\dfrac{a-2b}{3b}$ is rational, which means $\sqrt{5}$ is rational.

But this contradicts the given fact that $\sqrt{5}$ is irrational.

Therefore, our assumption is wrong, and $2 + 3\sqrt{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction question. Always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{5}$ on one side, showing it equals a ratio of integers — which contradicts the given condition.
• Always state clearly: "this contradicts the fact that $\sqrt{5}$ is irrational" — examiners award a mark specifically for this contradiction statement.
• The phrase "given that $\sqrt{5}$ is irrational" means you do not need to prove $\sqrt{5}$'s irrationality; just use it as a fact.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is true: $\sqrt{3}$ and $\sqrt{5}$ are both irrational, and their sum $(\sqrt{3}+\sqrt{5})$ is irrational. Reason is false: for example, $(\sqrt{3}) + (-\sqrt{3}) = 0$, which is rational. So the sum of two irrationals is not always irrational.

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• A is true because $\sqrt{3}$ and $\sqrt{5}$ are proven irrationals (Section 1.3), and their sum cannot be rational (provable by contradiction).
• R is false because the statement "sum of any two irrationals is always irrational" has a counterexample: $\sqrt{3} + (-\sqrt{3}) = 0$ (rational). Examiners specifically look for this counterexample to disprove R.
• Since A is true but R is false, the answer is (C) — a very common trap in Assertion-Reason questions.