Real Numbers 3-Mark Questions Class 10: CBSE PYQs with Answers

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Given: 3 is a prime number.

To prove: √3 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √3 is rational. Write √3 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 3b². Since 3 is prime and divides a², it divides a.

Let a = 3c. Then 9c² = 3b², so b² = 3c².

Thus 3 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √3 is irrational.

Given: 2 is a prime number.

To prove: √2 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √2 is rational. Write √2 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 2b². Since 2 is prime and divides a², it divides a.

Let a = 2c. Then 4c² = 2b², so b² = 2c².

Thus 2 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √2 is irrational.

Given: Expression: 3 + 7√2. √2 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 3 + 7√2 = r, where r is rational.

Rearranging gives √2 = $\frac{\text{r − 3}}{\text{7}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: 3 + 7√2 is irrational.

Given: Expression: 2 − 3√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 2 − 3√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{2 − r}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 2 − 3√5 is irrational.

Given: Expression: 5 + 2√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 5 + 2√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 5}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 5 + 2√3 is irrational.

Given: Expression: 2 + 3√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 2 + 3√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{r − 2}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 2 + 3√5 is irrational.

Given: Expression: 2 + 3√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 2 + 3√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 2}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 2 + 3√3 is irrational.

Given: Expression: 5√3 + $\frac{\text{2}}{\text{3}}$. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 5√3 + $\frac{\text{2}}{\text{3}}$ = r, where r is rational.

Rearranging gives √3 = $\frac{\text{3r − 2}}{\text{15}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 5√3 + $\frac{\text{2}}{\text{3}}$ is irrational.

Given: Expression: $\frac{\text{√3 + 5}}{\text{2}}$. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume $\frac{\text{√3 + 5}}{\text{2}}$ = r, where r is rational.

Rearranging gives √3 = 2r − 5.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: $\frac{\text{√3 + 5}}{\text{2}}$ is irrational.

Given: Expression: $\frac{\text{3 + 2√5}}{\text{7}}$. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume $\frac{\text{3 + 2√5}}{\text{7}}$ = r, where r is rational.

Rearranging gives √5 = $\frac{\text{7r − 3}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: $\frac{\text{3 + 2√5}}{\text{7}}$ is irrational.

Given: Expression: $\frac{\text{2 + 3√5}}{\text{7}}$. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume $\frac{\text{2 + 3√5}}{\text{7}}$ = r, where r is rational.

Rearranging gives √5 = $\frac{\text{7r − 2}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: $\frac{\text{2 + 3√5}}{\text{7}}$ is irrational.

Given: √2 and √5 are irrational.

To prove: √2 − √5 is irrational.

Principle: A nonzero rational number has a rational reciprocal. Rational numbers are closed under addition and division.

Assume √2 − √5 = r for a rational number r. Since √2 ≠ √5, r ≠ 0.

(√2 − √5)(√2 + √5) = 2 − 5 = −3.

Therefore √2 + √5 = $\frac{\text{−3}}{\text{r}}$, which is rational.

Adding the two equations gives 2√2 = r − $\frac{\text{3}}{\text{r}}$. Dividing by 2 would make √2 rational, a contradiction.

Conclusion: √2 − √5 is irrational.

Given: Expression: 5 − √3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 5 − √3 = r, where r is rational.

Rearranging gives √3 = 5 − r.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 5 − √3 is irrational.

Given: Expression: 4 + 2√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 4 + 2√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 4}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 4 + 2√3 is irrational.

Given: Expression: 10 + 2√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 10 + 2√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 10}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 10 + 2√3 is irrational.

Given: Expression: 7 − 2√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 7 − 2√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{7 − r}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 7 − 2√3 is irrational.

Given: Expression: 8 + 5√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 8 + 5√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{r − 8}}{\text{5}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 8 + 5√5 is irrational.

Given: Expression: 11 + 3√2. √2 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 11 + 3√2 = r, where r is rational.

Rearranging gives √2 = $\frac{\text{r − 11}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: 11 + 3√2 is irrational.

Given: Expression: 7 + 4√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 7 + 4√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{r − 7}}{\text{4}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 7 + 4√5 is irrational.

Given: Expression: 3 − 2√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 3 − 2√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{3 − r}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 3 − 2√5 is irrational.

Given: Expression: 3 + 2√2. √2 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 3 + 2√2 = r, where r is rational.

Rearranging gives √2 = $\frac{\text{r − 3}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: 3 + 2√2 is irrational.

Given: Expression: 3 + 2√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 3 + 2√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{r − 3}}{\text{2}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 3 + 2√5 is irrational.

Given: Expression: 7 − 3√5. √5 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 7 − 3√5 = r, where r is rational.

Rearranging gives √5 = $\frac{\text{7 − r}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √5.

Conclusion: 7 − 3√5 is irrational.

Given: Expression: 5√2 − 3. √2 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 5√2 − 3 = r, where r is rational.

Rearranging gives √2 = $\frac{\text{r + 3}}{\text{5}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: 5√2 − 3 is irrational.

Given: Expression: 2 + 5√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 2 + 5√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 2}}{\text{5}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 2 + 5√3 is irrational.

Given: Expression: 6 + 3√2. √2 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 6 + 3√2 = r, where r is rational.

Rearranging gives √2 = $\frac{\text{r − 6}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: 6 + 3√2 is irrational.

Given: Expression: $\frac{\text{2 − √3}}{\text{5}}$. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume $\frac{\text{2 − √3}}{\text{5}}$ = r, where r is rational.

Rearranging gives √3 = 2 − 5r.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: $\frac{\text{2 − √3}}{\text{5}}$ is irrational.

Given: 5 is a prime number.

To prove: $\frac{\text{1}}{\text{√5}}$ is irrational.

Principle: If a prime divides the square of an integer, it divides the integer. A nonzero rational number has a rational reciprocal.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

If $\frac{\text{1}}{\text{√5}}$ were rational, its reciprocal √5 would be rational, which contradicts the proof above.

Conclusion: $\frac{\text{1}}{\text{√5}}$ is irrational.

Given: n is a natural number.

To prove: 4ⁿ cannot have units digit 0.

Principle: A positive integer ending in 0 is divisible by 10 = 2 × 5.

4ⁿ = (2²)ⁿ = 2²ⁿ. Its only prime factor is 2.

It has no factor 5, so it cannot be divisible by 10.

Conclusion: 4ⁿ never ends in 0.

Given: 336, 54 are positive integers.

To find: The HCF and LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

336 = 2⁴ × 3 × 7.

54 = 2 × 3³.

HCF = 2 × 3 = 6.

LCM = 2⁴ × 3³ × 7 = 3024.

Answer: HCF = 6. LCM = 3024.

Related method: Question 30.

Given: 96, 404 are positive integers.

To find: The HCF and LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

96 = 2⁵ × 3.

404 = 2² × 101.

HCF = 2² = 4.

LCM = 2⁵ × 3 × 101 = 9696.

Answer: HCF = 4. LCM = 9696.

Related method: Question 30.

Given: 260, 910 are positive integers.

To find: The HCF and LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

260 = 2² × 5 × 13.

910 = 2 × 5 × 7 × 13.

HCF = 2 × 5 × 13 = 130.

LCM = 2² × 5 × 7 × 13 = 1820.

Answer: HCF = 130. LCM = 1820.

Related method: Question 30.

Given: Group sizes are 48, 80, 144.

To find: Minimum number of rooms.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

48 = 2⁴ × 3.

80 = 2⁴ × 5.

144 = 2⁴ × 3².

HCF = 2⁴ = 16.

Each room holds 16. The three groups need 48 ÷ 16 = 3, 80 ÷ 16 = 5, 144 ÷ 16 = 9.

Total = 3 + 5 + 9 = 17.

Answer: 17 rooms.

Given: Group sizes are 42, 56, 70.

To find: Minimum number of tables.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

42 = 2 × 3 × 7.

56 = 2³ × 7.

70 = 2 × 5 × 7.

HCF = 2 × 7 = 14.

Each table holds 14. The three groups need 42 ÷ 14 = 3, 56 ÷ 14 = 4, 70 ÷ 14 = 5.

Total = 3 + 4 + 5 = 12.

Answer: 12 tables.

Related method: Question 34.

Given: Group sizes are 36, 60, 84.

To find: Minimum number of rooms.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

36 = 2² × 3².

60 = 2² × 3 × 5.

84 = 2² × 3 × 7.

HCF = 2² × 3 = 12.

Each room holds 12. The three groups need 36 ÷ 12 = 3, 60 ÷ 12 = 5, 84 ÷ 12 = 7.

Total = 3 + 5 + 7 = 15.

Answer: 15 rooms.

Related method: Question 34.

Given: Intervals are 20 min and 25 min. The first simultaneous ring is at noon.

To find: The time of the next simultaneous ring.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

20 = 2² × 5.

25 = 5².

LCM = 2² × 5² = 100.

100 minutes = 1 hour 40 minutes. Add this to 12 noon.

Answer: 1:40 p.m.

Related method: Question 34.

Given: Step lengths are 48 cm, 52 cm and 56 cm. The requested distance is ten times the least common distance.

To find: Ten times the least common distance.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

48 = 2⁴ × 3.

52 = 2² × 13.

56 = 2³ × 7.

LCM = 2⁴ × 3 × 7 × 13 = 4368.

The least common distance is 4,368 cm.

Required distance = 10 × 4,368 = 43,680 cm.

Answer: 43,680 cm or 436.8 m.

Related method: Question 34.

Given: Book counts are 144, 180 and 192. Each stack has the same number of books.

To find: Stacks for each subject.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

144 = 2⁴ × 3².

180 = 2² × 3² × 5.

192 = 2⁶ × 3.

HCF = 2² × 3 = 12.

Use 12 books per stack. Physics: 144 ÷ 12 = 12.

Chemistry: 180 ÷ 12 = 15. Mathematics: 192 ÷ 12 = 16.

Answer: 12 Physics, 15 Chemistry and 16 Mathematics stacks.

Related method: Question 34.

Given: Times per sweater are 15, 18 and 20 days.

To find: First simultaneous finish and completed totals.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

15 = 3 × 5.

18 = 2 × 3².

20 = 2² × 5.

LCM = 2² × 3² × 5 = 180.

Completed sweaters: 180 ÷ 15 = 12, 180 ÷ 18 = 10 and 180 ÷ 20 = 9.

Answer: 180 days. The totals are 12, 10 and 9 sweaters.

Related method: Question 34.

Given: Window dimensions are 156 cm and 216 cm.

To find: The square side and number of panes.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

156 = 2² × 3 × 13.

216 = 2³ × 3³.

HCF = 2² × 3 = 12.

Number of panes along the two sides = 156 ÷ 12 = 13 and 216 ÷ 12 = 18.

Total = 13 × 18 = 234.

Answer: Side 12 cm. Total 234 square panes.

Related method: Question 34.

Given: Divisors are 48, 60 and 65. The result is less than 10,000.

To find: Greatest qualifying integer.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

48 = 2⁴ × 3.

60 = 2² × 3 × 5.

65 = 5 × 13.

LCM = 2⁴ × 3 × 5 × 13 = 3120.

3 × 3,120 = 9,360 and 4 × 3,120 = 12,480. The greatest allowed multiple is the third.

Answer: 9,360.

Related method: Question 34.

Given: Oil quantities are 870 L, 812 L and 638 L.

To find: Maximum container capacity and total containers.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

870 = 2 × 3 × 5 × 29.

812 = 2² × 7 × 29.

638 = 2 × 11 × 29.

HCF = 2 × 29 = 58.

The batches fill 870 ÷ 58 = 15, 812 ÷ 58 = 14 and 638 ÷ 58 = 11 containers.

Total = 15 + 14 + 11 = 40.

Answer: 58 L per container. 40 containers.

Related method: Question 34.

Given: Dividend 764 leaves 8. Dividend 1,198 leaves 10.

To find: Greatest possible divisor.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

Subtract the remainders: 764 − 8 = 756 and 1,198 − 10 = 1,188.

756 = 2² × 3³ × 7.

1188 = 2² × 3³ × 11.

HCF = 2² × 3³ = 108.

108 exceeds both remainders, so it is a valid divisor.

Answer: 108.

Related method: Question 34.

Given: Divisors are 72, 90 and 150. The increase is 20.

To find: The smallest qualifying number.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

72 = 2³ × 3².

90 = 2 × 3² × 5.

150 = 2 × 3 × 5².

LCM = 2³ × 3² × 5² = 1800.

The least positive common multiple is 1,800.

Original number = 1,800 − 20 = 1,780.

Answer: 1,780.

Related method: Question 34.

Given: Intervals are 45, 75 and 60 seconds. The initial time is 5 a.m.

To find: Next simultaneous change.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

45 = 3² × 5.

75 = 3 × 5².

60 = 2² × 3 × 5.

LCM = 2² × 3² × 5² = 900.

900 seconds = 15 minutes. Add this to 5 a.m.

Answer: 5:15 a.m.

Related method: Question 34.

Given: Rod lengths are 120, 100 and 150 cm.

To find: The shortest length and number of uses.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

120 = 2³ × 3 × 5.

100 = 2² × 5².

150 = 2 × 3 × 5².

LCM = 2³ × 3 × 5² = 600.

120 cm rod: 600 ÷ 120 = 5 uses.

100 cm rod: 600 ÷ 100 = 6 uses.

150 cm rod: 600 ÷ 150 = 4 uses.

Answer: 600 cm. Use the rods 5, 6 and 4 times respectively.

Related method: Question 34.

Given: Expression: $\frac{\text{3 + √3}}{\text{5}}$. √3 is irrational.

To prove: The expression is irrational. Find a suitable irrational addend.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume $\frac{\text{3 + √3}}{\text{5}}$ = r, where r is rational.

Rearranging gives √3 = 5r − 3.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Choose √3 as the irrational addend. The sum is $\frac{\text{3 + 6√3}}{\text{5}}$.

If that sum equalled a rational number s, then √3 = $\frac{\text{5s − 3}}{\text{6}}$ would be rational. Hence the sum is irrational.

Conclusion: $\frac{\text{3 + √3}}{\text{5}}$ is irrational. One valid addend is √3.

Given: Cloth widths are 144 cm and 96 cm.

To find: Maximum strip width and total number.

Formula or rule: A common size must divide the quantities exactly. Use the HCF and take the smallest exponent of each common prime.

144 = 2⁴ × 3².

96 = 2⁵ × 3.

HCF = 2⁴ × 3 = 48.

There are 144 ÷ 48 = 3 and 96 ÷ 48 = 2 strips.

Total = 3 + 2 = 5.

Answer: 48 cm and 5 strips.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: Expression: $\frac{\text{2 + 3√2}}{\text{5}}$. 2 is prime.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers. A prime that divides an integer square also divides the integer.

Assume √2 is rational. Write √2 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 2b². Since 2 is prime and divides a², it divides a.

Let a = 2c. Then 4c² = 2b², so b² = 2c².

Thus 2 also divides b. This contradicts HCF(a, b) = 1.

Assume $\frac{\text{2 + 3√2}}{\text{5}}$ = r, where r is rational.

Rearranging gives √2 = $\frac{\text{5r − 2}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √2.

Conclusion: $\frac{\text{2 + 3√2}}{\text{5}}$ is irrational.

Given: 26, 65, 117 are positive integers.

To find: The HCF and LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

26 = 2 × 13.

65 = 5 × 13.

117 = 3² × 13.

HCF = 13.

LCM = 2 × 3² × 5 × 13 = 1170.

Answer: HCF = 13. LCM = 1170.

Given: 2 is a prime number.

To prove: √2 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √2 is rational. Write √2 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 2b². Since 2 is prime and divides a², it divides a.

Let a = 2c. Then 4c² = 2b², so b² = 2c².

Thus 2 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √2 is irrational.

Related method: Question 3.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: 36, 54 are positive integers.

To find: The LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

Every composite number is a product of primes, unique apart from their order.

36 = 2² × 3².

54 = 2 × 3³.

LCM = 2² × 3³ = 108.

Answer: Every composite number is a product of primes, unique apart from their order. LCM = 108.

Given: 2 is a prime number.

To prove: √2 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √2 is rational. Write √2 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 2b². Since 2 is prime and divides a², it divides a.

Let a = 2c. Then 4c² = 2b², so b² = 2c².

Thus 2 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √2 is irrational.

Related method: Question 3.

Given: (i) 7 × 11 × 13 + 13. (ii) 6 × 5 × 4 + 4. (iii) 7 × 13 + 6.

To find: Classify each number.

Formula or rule: A composite integer greater than 1 has a factor other than 1 and itself. To test primality, check primes up to the square root.

7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 1,014. It is composite.

6 × 5 × 4 + 4 = 4(6 × 5 + 1) = 4 × 31 = 124. It is composite.

7 × 13 + 6 = 97. Since √97 < 10 and none of 2, 3, 5 and 7 divides 97, it is prime.

Answer: 1,014 and 124 are composite. 97 is prime.

Given: 3 is a prime number.

To prove: √3 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √3 is rational. Write √3 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 3b². Since 3 is prime and divides a², it divides a.

Let a = 3c. Then 9c² = 3b², so b² = 3c².

Thus 3 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √3 is irrational.

Related method: Question 2.

Given: The expressions are 2 × 3 × 5 × 7 + 7 and 2 × 3 × 5 × 7 + 1.

To find: The truth of each claim.

Formula or rule: A composite number has a nontrivial divisor. If a number is composite, a prime divisor is at most its square root.

2 × 3 × 5 × 7 + 7 = 7(2 × 3 × 5 + 1) = 7 × 31 = 217. The first claim is true.

2 × 3 × 5 × 7 + 1 = 211. Since √211 < 15 and none of 2, 3, 5, 7, 11 or 13 divides 211, it is prime.

Answer: (i) True. (ii) False.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: p, q and r are three distinct prime numbers.

To find: Prove the first claim and test the second.

Formula or rule: A product of two integers greater than 1 is composite.

pqr + q = q(pr + 1). Both factors exceed 1, so it is composite.

For p = 3, q = 5 and r = 7, pqr + 1 = 106 = 2 × 53.

For p = 2, q = 3 and r = 5, pqr + 1 = 31. It has no prime divisor among 2, 3 and 5 and is prime.

Answer: pqr + q is always composite. pqr + 1 can be prime or composite.

Given: 2 is a prime number.

To prove: √2 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √2 is rational. Write √2 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 2b². Since 2 is prime and divides a², it divides a.

Let a = 2c. Then 4c² = 2b², so b² = 2c².

Thus 2 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √2 is irrational.

Related method: Question 3.

Given: p = x²y³, q = xy⁴, r = x⁵y². The primes x and y are distinct.

To find: HCF, LCM and the comparison.

Formula or rule: Use the smallest common exponent for HCF and the largest exponent for LCM. Add exponents of like bases when multiplying.

For x, the exponents are 2, 1 and 5. For y, the exponents are 3, 4 and 2.

HCF = x¹y² = xy². LCM = x⁵y⁴.

HCF × LCM = x⁶y⁶, while pqr = x⁸y⁹. Their ratio pqr ÷ (HCF × LCM) = x²y³ > 1.

Answer: HCF = xy². LCM = x⁵y⁴. HCF × LCM < pqr.

Given: 18180, 7575 are positive integers.

To find: The HCF and LCM.

Formula or rule: By the Fundamental Theorem of Arithmetic, prime factorisation is unique apart from the order of the factors. HCF uses the smallest exponent of every common prime. LCM uses the largest exponent of each prime present.

18180 = 2² × 3² × 5 × 101.

7575 = 3 × 5² × 101.

HCF = 3 × 5 × 101 = 1515.

LCM = 2² × 3² × 5² × 101 = 90900.

Answer: HCF = 1515. LCM = 90900.

Given: Intervals are 6, 12 and 18 min. The initial time is 6 a.m.

To find: Next simultaneous ring.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

6 = 2 × 3.

12 = 2² × 3.

18 = 2 × 3².

LCM = 2² × 3² = 36.

Add 36 minutes to 6 a.m.

Answer: 6:36 a.m.

Given: 3 is a prime number.

To prove: √3 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √3 is rational. Write √3 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 3b². Since 3 is prime and divides a², it divides a.

Let a = 3c. Then 9c² = 3b², so b² = 3c².

Thus 3 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √3 is irrational.

Related method: Question 2.

Given: Intervals are 48, 72 and 108 s. The starting time is 7 a.m.

To find: Next simultaneous change.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

48 = 2⁴ × 3.

72 = 2³ × 3².

108 = 2² × 3³.

LCM = 2⁴ × 3³ = 432.

432 seconds = 7 minutes 12 seconds. Add this to 7 a.m.

Answer: 7:07:12 a.m.

Given: Divisors are 520 and 468. The increase is 17.

To find: The original number.

Formula or rule: The target is a common multiple of the given quantities. Use the LCM and take the greatest exponent of each prime present.

520 = 2³ × 5 × 13.

468 = 2² × 3² × 13.

LCM = 2³ × 3² × 5 × 13 = 4680.

The smallest adjusted number is 4,680.

Original number = 4,680 − 17 = 4,663.

Answer: 4,663.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: The branches are x → 2, y; y → 2, 819; 819 → 3, z; z → 3, w; w → 13, 7.

To find: Missing node values and the prime factorisation of x.

Formula or rule: Each parent is the product of its two children.

w = 13 × 7 = 91.

z = 3 × 91 = 273.

Check: 3 × 273 = 819.

y = 2 × 819 = 1,638.

x = 2 × 1,638 = 3,276.

The prime leaves give 2² × 3² × 7 × 13.

Answer: x = 3,276, y = 1,638, z = 273, w = 91. Prime factorisation: 2² × 3² × 7 × 13.

Given: 3 is a prime number.

To prove: √3 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √3 is rational. Write √3 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 3b². Since 3 is prime and divides a², it divides a.

Let a = 3c. Then 9c² = 3b², so b² = 3c².

Thus 3 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √3 is irrational.

Related method: Question 2.

Given: The branches are x → 2, y; y → 2, 210; 210 → a, 70; 70 → 2, 35; 35 → 5, b.

To find: Missing node values and the prime factorisation of x.

Formula or rule: Each parent is the product of its two children.

b = 35 ÷ 5 = 7.

a = 210 ÷ 70 = 3.

y = 2 × 210 = 420.

x = 2 × 420 = 840.

The prime leaves give 2³ × 3 × 5 × 7.

Answer: x = 840, y = 420, a = 3, b = 7. Prime factorisation: 2³ × 3 × 5 × 7.

Given: Expression: 2 + 3√3. √3 is irrational.

To prove: The expression is irrational.

Principle: Rational numbers are closed under addition, subtraction, multiplication and division by nonzero rational numbers.

Assume 2 + 3√3 = r, where r is rational.

Rearranging gives √3 = $\frac{\text{r − 2}}{\text{3}}$.

The right-hand side is rational by closure under rational arithmetic. This contradicts the irrationality of √3.

Conclusion: 2 + 3√3 is irrational.

Given: 5 is a prime number.

To prove: √5 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √5 is rational. Write √5 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 5b². Since 5 is prime and divides a², it divides a.

Let a = 5c. Then 25c² = 5b², so b² = 5c².

Thus 5 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √5 is irrational.

Related method: Question 1.

Given: 3 is a prime number.

To prove: √3 is irrational.

Principle: If a prime divides the square of an integer, it divides the integer.

Assume √3 is rational. Write √3 = $\frac{\text{a}}{\text{b}}$, where a and b are positive coprime integers.

Squaring gives a² = 3b². Since 3 is prime and divides a², it divides a.

Let a = 3c. Then 9c² = 3b², so b² = 3c².

Thus 3 also divides b. This contradicts HCF(a, b) = 1.

Conclusion: √3 is irrational.

Related method: Question 2.

Given: √6 is irrational.

To prove: (√2 + √3)² is irrational.

Principle: Use (a + b)² = a² + 2ab + b² and closure of rational numbers.

(√2 + √3)² = 2 + 2√6 + 3 = 5 + 2√6.

If this equals a rational number r, then √6 = $\frac{\text{r − 5}}{\text{2}}$, a rational number. This contradicts the given fact.

Conclusion: (√2 + √3)² is irrational.