--- title: "Real Numbers: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/real-numbers --- # Real Numbers: CBSE Class 10 Maths previous year questions 27 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 1 *1 mark · MCQ* The HCF of 960 and 432 is : - 48 - 54 - 72 - 36 **Solution** 1. Find the prime factorisation of 960 and 432: $960 = 2^6 \times 3 \times 5$ and $432 = 2^4 \times 3^3$. 2. The HCF is the product of the smallest power of each common prime factor in the numbers: $\text{HCF}(960, 432) = 2^4 \times 3^1 = 16 \times 3 = 48$. **Answer:** (A) 48 > Common mistake: Taking the highest powers instead of the lowest powers for HCF. ### Question 2 *1 mark · MCQ* The natural number $2$ is : - a prime number - a composite number - prime as well as composite - neither prime nor composite **Solution** 1. A natural number greater than 1 is called a prime number if it has exactly two distinct factors, 1 and itself. 2. The number 2 has only two factors, 1 and 2, and hence it is a prime number. **Answer:** (a) a prime number > Common mistake: Students sometimes think 2 is a composite number or neither. ### Question 3 *1 mark · MCQ* For any natural number $n$, $6^n$ ends with the digit : - 0 - 6 - 3 - 2 **Solution** 1. If any number $6^n$ were to end with the digit 0, it would be divisible by 5, meaning its prime factorisation must contain the prime 5. 2. The prime factorisation of $6^n$ is $(2 \times 3)^n = 2^n \times 3^n$, which shows that 2 and 3 are the only prime factors, so by the uniqueness of the Fundamental Theorem of Arithmetic, 5 is not present. **Answer:** (B) 6 > Common mistake: Assuming $6^n$ ends in 0 for some large $n$ because $6 \times 6 = 36$ ends in 6. ### Question 26 *3 marks · Proof* Prove that $\sqrt{3}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $\sqrt{3}$ is a rational number. 2. So, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \frac{a}{b}$. 3. Squaring both sides, we get $3 = \frac{a^2}{b^2}$, which means $3b^2 = a^2$. 4. Therefore, $3$ divides $a^2$, and consequently $3$ divides $a$ by Theorem 1.3. 5. So, we can write $a = 3c$ for some integer $c$. 6. Substituting $a = 3c$ in $3b^2 = a^2$, we get $3b^2 = 9c^2$, which gives $b^2 = 3c^2$. 7. This means $3$ divides $b^2$, and so $3$ divides $b$. 8. Thus, $a$ and $b$ have at least $3$ as a common factor, which contradicts the fact that $a$ and $b$ are co-prime. 9. This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. 10. Hence, $\sqrt{3}$ is an irrational number. **Answer:** Hence proved. > Common mistake: Stating that $a$ and $b$ are integers instead of co-prime integers. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 1 *1 mark · MCQ* If the HCF of two positive integers a and b is 1, then their LCM is: - a + b - a - b - ab **Solution** 1. We know that for any two positive integers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$. 2. Given that $\text{HCF}(a, b) = 1$, substituting this gives $1 \times \text{LCM}(a, b) = ab$, which means $\text{LCM}(a, b) = ab$. **Answer:** (d) ab > Common mistake: Students sometimes confuse HCF with LCM and choose $1$ or $a+b$. ### Question 2 *1 mark · MCQ* The number $3+\sqrt{2}$ is: - a rational number - an irrational number - an integer - a natural number **Solution** 1. We know that the sum of a rational number and an irrational number is always an irrational number. 2. Since $3$ is rational and $\sqrt{2}$ is irrational, $3+\sqrt{2}$ is an irrational number. **Answer:** (b) an irrational number > Common mistake: Students may incorrectly assume adding a rational and irrational yields a rational number. ### Question 19 *1 mark · Assertion and reason* Assertion (A): For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b. Reason (R): HCF of any two natural numbers divides both the numbers. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). - Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. For any two positive integers a and b, the HCF is always a factor of their LCM. 2. The reason states that the HCF of two numbers divides both numbers, which is true by definition of HCF, but it does not explain why the HCF is a factor of the LCM. **Answer:** Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). > Common mistake: Confusing the properties of HCF dividing numbers with the relation between HCF and LCM. ### Question 26 (a) *3 marks · Proof* Prove that $\sqrt{3}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $\sqrt{3}$ is a rational number. 2. So, we can find coprime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \frac{a}{b}$. 3. Squaring both sides, we get $3 = \frac{a^2}{b^2}$, which means $3b^2 = a^2$. 4. This implies that $3$ divides $a^2$, and therefore $3$ divides $a$ by the theorem. 5. So, we can write $a = 3c$ for some integer $c$. Substituting this, $3b^2 = (3c)^2 = 9c^2$, which gives $b^2 = 3c^2$. 6. This means $3$ divides $b^2$, and therefore $3$ divides $b$. 7. Thus, $a$ and $b$ have at least $3$ as a common factor, which contradicts the fact that $a$ and $b$ are coprime. 8. This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence proved. **Answer:** Hence proved that $\sqrt{3}$ is irrational. > Common mistake: Failing to state that $a$ and $b$ are coprime integers. ### Question 26 (OR) *3 marks · Short answer* The factor tree of a number x is shown below : Find the values of x, y, a and b. Hence, write the product of the prime factors of the number x so obtained. **Solution** 1. From the given factor tree, working from the bottom: $b = 35 / 5 = 7$. 2. Next, $a = 70 / 7 = 10$. 3. Then, $y = 2 \times 210 = 420$. 4. Finally, $x = 2 \times y = 2 \times 420 = 840$. 5. The values are $x = 840$, $y = 420$, $a = 10$, and $b = 7$. 6. The prime factorisation of $x$ is $840 = 2^3 \times 3 \times 5 \times 7$. **Answer:** x = 840, y = 420, a = 10, b = 7; Prime factorisation = $2^3 \times 3 \times 5 \times 7$ > Common mistake: Making calculation errors while working bottom-up through the factor tree. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 6 *1 mark · MCQ* If $\text{HCF}(98, 28) = m$ and $\text{LCM}(98, 28) = n$, then the value of $n - 7m$ is : - 0 - 28 - 98 - 198 **Solution** 1. Find the prime factorisation of $98$ and $28$: $98 = 2 \times 7^2$ and $28 = 2^2 \times 7$. 2. Calculate $\text{HCF}(98, 28) = 2 \times 7 = 14$ (so $m = 14$) and $\text{LCM}(98, 28) = 2^2 \times 7^2 = 196$ (so $n = 196$). 3. Evaluate $n - 7m$: $196 - 7(14) = 196 - 98 = 98$. **Answer:** (c) 98 > Common mistake: Arithmetic error while multiplying $7$ and $14$. ### Question 9 *1 mark · MCQ* If $(-1)^n + (-1)^8 = 0$, then $n$ is : - any positive integer - any negative integer - any odd number - any even number **Solution** 1. We are given $(-1)^n + (-1)^8 = 0$, which means $(-1)^n + 1 = 0$, so $(-1)^n = -1$. 2. A power of $-1$ is $-1$ only when the exponent is an odd number. **Answer:** (c) any odd number > Common mistake: Thinking that $(-1)^n = -1$ for even numbers. ### Question 14 *1 mark · MCQ* Which of the following is a rational number between $\sqrt{3}$ and $\sqrt{5}$ ? - 1.4142387954012.... - 2.326 (bar over 26) - π - 1.857142 **Solution** 1. We know $\sqrt{3} \approx 1.732$ and $\sqrt{5} \approx 2.236$. 2. A rational number must have a terminating or repeating decimal expansion. 3. The number $1.857142$ lies between $1.732$ and $2.236$ and is a terminating decimal, hence it is rational. **Answer:** (d) 1.857142 > Common mistake: Choosing $\pi$ or the non-terminating non-recurring decimal. ### Question 29 *3 marks · Proof* Prove that $\frac{1}{\sqrt{5}}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $\frac{1}{\sqrt{5}}$ is a rational number. 2. Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\frac{1}{\sqrt{5}} = \frac{a}{b}$. 3. Rearranging the equation gives $\sqrt{5} = \frac{b}{a}$. 4. Since $a$ and $b$ are integers, $\frac{b}{a}$ is a rational number, which means $\sqrt{5}$ must be a rational number. 5. This contradicts the established fact that $\sqrt{5}$ is an irrational number. 6. Therefore, our assumption is false and $\frac{1}{\sqrt{5}}$ is an irrational number. Hence proved. **Answer:** Hence proved. > Common mistake: Forgetting to mention that $a$ and $b$ are co-prime integers. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 5 *1 mark · MCQ* $\text{HCF}(132, 77)$ is : - $11$ - $77$ - $22$ - $44$ **Solution** 1. Prime factorization of $132$ is $2^2 \times 3 \times 11$. 2. Prime factorization of $77$ is $7 \times 11$. 3. The common prime factor with the lowest power is $11$, so $\text{HCF}(132, 77) = 11$. **Answer:** (a) $11$ > Common mistake: Calculating LCM instead of HCF. ### Question 23 *2 marks · Proof* Prove that $6 - 4\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $6 - 4\sqrt{5}$ is a rational number. 2. Then, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $6 - 4\sqrt{5} = \frac{a}{b}$. 3. Rearranging the equation, we get $6 - \frac{a}{b} = 4\sqrt{5}$, which means $\sqrt{5} = \frac{1}{4}\left(6 - \frac{a}{b}\right)$. 4. Since $a$ and $b$ are integers, $\frac{1}{4}\left(6 - \frac{a}{b}\right)$ is a rational number. 5. This implies that $\sqrt{5}$ is rational, which contradicts the given fact that $\sqrt{5}$ is irrational. 6. Our assumption is false. Hence, $6 - 4\sqrt{5}$ is an irrational number. **Answer:** Hence proved. > Common mistake: Not stating the contradiction with the given fact that $\sqrt{5}$ is irrational. ### Question 23 (OR) *2 marks · Proof* Show that $11 \times 19 \times 23 + 3 \times 11$ is not a prime number. **Solution** 1. The given number is $11 \times 19 \times 23 + 3 \times 11$. 2. Take 11 common from both terms: $11 \times (19 \times 23 + 3)$. 3. Calculate the product inside the bracket: $19 \times 23 = 437$. 4. Add 3 to the product: $437 + 3 = 440$. 5. Thus, the expression can be written as $11 \times 440 = 11 \times 11 \times 40 = 11^2 \times 2^3 \times 5$. 6. Since the number can be expressed as a product of prime factors other than 1 and itself, it is a composite number and therefore not a prime number. **Answer:** Hence shown that it is not a prime number. > Common mistake: Multiplying out the entire large number instead of taking out common factors. ### Question 26 *3 marks · Short answer* Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time ? **Solution** 1. Find the LCM of the time intervals of the two alarm clocks, which are $20$ minutes and $25$ minutes. 2. Prime factorization of $20 = 2^2 \times 5$ and $25 = 5^2$. 3. The LCM of $20$ and $25$ is $2^2 \times 5^2 = 4 \times 25 = 100$ minutes. 4. Convert $100$ minutes into hours and minutes, which is $1$ hour and $40$ minutes. 5. Add $1$ hour $40$ minutes to the initial time of $12:00$ noon to get the next beep time at $1:40$ p.m. **Answer:** 1:40 p.m. > Common mistake: Calculating the HCF instead of the LCM of the two time intervals. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 5 *1 mark · MCQ* If two positive integers $p$ and $q$ can be expressed as $p = 18a^2b^4$ and $q = 20a^3b^2$, where $a$ and $b$ are prime numbers, then $\text{LCM}(p, q)$ is : - $2a^2b^2$ - $180a^2b^2$ - $12a^3b^2$ - $180a^3b^4$ **Solution** 1. Given $p = 18a^2b^4 = 2 \times 3^2 \times a^2 \times b^4$ and $q = 20a^3b^2 = 2^2 \times 5 \times a^3 \times b^2$, where $a$ and $b$ are prime numbers. 2. The LCM of two numbers is the product of the greatest power of each prime factor involved in the numbers: $\text{LCM}(p, q) = 2^2 \times 3^2 \times 5 \times a^3 \times b^4 = 180a^3b^4$. **Answer:** (d) $180a^3b^4$ > Common mistake: Students often find HCF instead of LCM by taking the lowest powers of the prime factors. ### Question 25 *2 marks · Proof* Prove that $5 - 2\sqrt{3}$ is an irrational number. It is given that $\sqrt{3}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $5 - 2\sqrt{3}$ is a rational number. 2. Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $5 - 2\sqrt{3} = \frac{a}{b}$. 3. Rearranging the terms, we get $5 - \frac{a}{b} = 2\sqrt{3}$, which means $\sqrt{3} = \frac{1}{2}\left(5 - \frac{a}{b}\right)$. 4. Since $a$ and $b$ are integers, $\frac{1}{2}\left(5 - \frac{a}{b}\right)$ is a rational number, so $\sqrt{3}$ must be rational. 5. This contradicts the given fact that $\sqrt{3}$ is irrational. 6. Hence, our assumption is false and $5 - 2\sqrt{3}$ is an irrational number. Hence proved. **Answer:** $5 - 2\sqrt{3}$ is irrational. > Common mistake: Not explicitly stating the contradiction with the given fact that root 3 is irrational. ### Question 25 (OR) *2 marks · Proof* Show that the number $5 \times 11 \times 17 + 3 \times 11$ is a composite number. **Solution** 1. Consider the given number: $5 \times 11 \times 17 + 3 \times 11$. 2. Take out the common factor $11$ from both terms: $11 \times (5 \times 17 + 3)$. 3. Simplify the expression inside the bracket: $11 \times (85 + 3) = 11 \times 88$. 4. Express $88$ as product of primes: $11 \times (11 \times 8) = 11^2 \times 2^3$. 5. Since the number can be expressed as a product of prime factors greater than 1, it is a composite number. **Answer:** Hence proved that the given number is a composite number. > Common mistake: Multiplying out the entire large number instead of taking out common factors to show prime factorization. ### Question 27 *3 marks · Short answer* In a teachers' workshop, the number of teachers teaching French, Hindi and English are $48, 80$ and $144$ respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject. **Solution** 1. Given number of teachers teaching French, Hindi and English are $48, 80$ and $144$ respectively. 2. To find the minimum number of rooms, the number of teachers in each room must be the maximum possible, which is the HCF of $48, 80$ and $144$. 3. Prime factorisation of $48 = 2^4 \times 3$ 4. Prime factorisation of $80 = 2^4 \times 5$ 5. Prime factorisation of $144 = 2^4 \times 3^2$ 6. HCF$(48, 80, 144) = 2^4 = 16$. 7. Maximum number of teachers in each room = $16$. 8. Total number of teachers = $48 + 80 + 144 = 272$. 9. Minimum number of rooms required = $\frac{272}{16} = 17$. **Answer:** 17 rooms > Common mistake: Dividing individual counts incorrectly or finding LCM instead of HCF. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 1 *1 mark · MCQ* The prime factorisation of natural number $288$ is - $2^4 \times 3^3$ - $2^4 \times 3^2$ - $2^5 \times 3^2$ - $2^5 \times 3^1$ **Solution** 1. Divide $288$ by prime factors: $288 = 2 \times 144 = 2 \times 2 \times 72 = 2^3 \times 36 = 2^3 \times 4 \times 9 = 2^5 \times 3^2$. 2. Thus, the prime factorisation of $288$ is $2^5 \times 3^2$. **Answer:** (c) $2^5 \times 3^2$ > Common mistake: Making errors in counting the number of times 2 divides 288. ### Question 23 *2 marks · Short answer* Find the LCM and HCF of $92$ and $510$, using prime factorisation. **Solution** 1. Find the prime factorisation of $92$: $92 = 2^2 \times 23$. 2. Find the prime factorisation of $510$: $510 = 2 \times 3 \times 5 \times 17$. 3. The HCF is the product of the smallest power of each common prime factor in the numbers: $\operatorname{HCF}(92, 510) = 2$. 4. The LCM is the product of the greatest power of each prime factor involved in the numbers: $\operatorname{LCM}(92, 510) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$. **Answer:** HCF = $2$, LCM = $23460$ > Common mistake: Multiplying numbers incorrectly when calculating the LCM. ### Question 30 *3 marks · Proof* Prove that $3 + 7\sqrt{2}$ is an irrational number, given that $\sqrt{2}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $3 + 7\sqrt{2}$ is a rational number. 2. Then, there exist co-prime integers $a$ and $b$ (where $b \neq 0$) such that $3 + 7\sqrt{2} = \frac{a}{b}$. 3. Rearranging the equation, we get $7\sqrt{2} = \frac{a}{b} - 3$ 4. $\sqrt{2} = \frac{a - 3b}{7b}$. 5. Since $a$ and $b$ are integers, $\frac{a - 3b}{7b}$ is a rational number. 6. This implies that $\sqrt{2}$ is a rational number. 7. This contradicts the given fact that $\sqrt{2}$ is an irrational number. 8. This contradiction has arisen because of our incorrect assumption that $3 + 7\sqrt{2}$ is rational. 9. Hence, $3 + 7\sqrt{2}$ is an irrational number. Hence proved. **Answer:** Hence proved. > Common mistake: Forgetting to write the final concluding statement linking it back to the contradiction. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 20 *1 mark · Assertion and reason* Assertion (A) : The perimeter of $\Delta ABC$ is a rational number. Reason (R) : The sum of the squares of two rational numbers is always rational. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). - Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. In right-angled triangle $ABC$ with $AB = 2\text{ cm}$ and $BC = 3\text{ cm}$, the hypotenuse $AC = \sqrt{2^2 + 3^2} = \sqrt{13}\text{ cm}$, which is an irrational number. 2. The perimeter of $\Delta ABC$ is $AB + BC + AC = 2 + 3 + \sqrt{13} = 5 + \sqrt{13}$, which is irrational, making Assertion (A) false. **Answer:** Assertion (A) is false but Reason (R) is true. > Common mistake: Assuming the sum of rational numbers and a square root is rational. ### Question 24 *2 marks · Short answer* Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively. **Solution** 1. Subtract the respective remainders from the given numbers: $85 - 1 = 84$ and $72 - 2 = 70$. 2. The required greatest number is the HCF of 84 and 70. 3. Prime factorisation of $84 = 2^2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$. 4. The HCF is $2 \times 7 = 14$. **Answer:** $14$ > Common mistake: Finding the HCF of 85 and 72 directly without subtracting the remainders first. ### Question 27 *3 marks · Proof* Prove that $\sqrt{5}$ is an irrational number. **Solution** 1. Let us assume, to the contrary, that $\sqrt{5}$ is a rational number. 2. So, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \frac{a}{b}$. 3. Squaring on both sides, we get $5 = \frac{a^2}{b^2}$, which means $5b^2 = a^2$. 4. This implies that 5 divides $a^2$, and by the theorem, 5 also divides $a$. 5. So, we can write $a = 5c$ for some integer $c$. Substituting this in $5b^2 = a^2$ gives $5b^2 = 25c^2$, or $b^2 = 5c^2$. 6. This means 5 divides $b^2$, so 5 also divides $b$. 7. Thus, $a$ and $b$ have at least 5 as a common factor, which contradicts the fact that $a$ and $b$ are co-prime. 8. Hence proved that $\sqrt{5}$ is an irrational number. **Answer:** Hence proved that $\sqrt{5}$ is irrational. > Common mistake: Omitting the mention that $a$ and $b$ are co-prime integers. ## Related pages - [Real Numbers: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student