PYQ Mathematics Class10 Chapter1 Real Numbers (Previous Year Qestions)
Previous Year Questions 2026 (Basic)
Q1. If the HCF of two positive integers $a$ and $b$ is 1, then their LCM is: [1 Mark][cite: 1]
(A) $a+b$
(B) $a$
(C) $b$
(D) $ab$[cite: 1]
Click to View Solution
Answer: (D) $ab$[cite: 1]
Since $HCF(a,b) \times LCM(a,b) = a \times b$ and $HCF=1$, therefore $LCM=ab$.[cite: 1]
Q2. The number $3+\sqrt{2}$ is: [1 Mark][cite: 1]
(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number[cite: 1]
Click to View Solution
Answer: (B) an irrational number[cite: 1]
Since $\sqrt{2}$ is irrational, the sum of a rational number 3 and an irrational number $\sqrt{2}$ is always irrational.[cite: 1]
Q3. 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 $b$. [1 Mark][cite: 1]
Reason (R): HCF of any two natural numbers divides both the numbers.[cite: 1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.[cite: 1]
Click to View Solution
Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).[cite: 1]
Since HCF divides both $a$ and $b$, and $LCM = \frac{a \times b}{HCF}$, the HCF divides LCM. Reason R correctly explains Assertion A.[cite: 1]
Q4(A). Prove that $\sqrt{3}$ is an irrational number. [3 Marks][cite: 1]
Click to View Solution
Answer: Assume $\sqrt{3} = \frac{p}{q}$ where $p, q$ are integers, $q \neq 0$, and $\text{gcd}(p, q) = 1$.[cite: 1]
Squaring: $3 = \frac{p^2}{q^2} \Rightarrow p^2 = 3q^2$. So $3|p^2 \Rightarrow 3|p$.[cite: 1]
Let $p = 3m$. Now $9m^2 = 3q^2 \Rightarrow q^2 = 3m^2 \Rightarrow 3|q^2 \Rightarrow 3|q$.[cite: 1]
Now 3 divides both $p$ and $q$, contradicting $\text{gcd}(p, q) = 1$. Hence $\sqrt{3}$ is irrational.[cite: 1]
Q4(B). The factor tree of a number x is shown below: [3 Marks][cite: 1]
Find the values of $x, y, a$ and $b$. Hence, write the product of the prime factors of the number $x$ so obtained.[cite: 1]
Click to View Solution
Answer: From the tree (bottom up):[cite: 1]
$35 = 5 \times b \Rightarrow b = 7$[cite: 1]
$70 = 2 \times 35$[cite: 1]
$a \times 70 = 210 \Rightarrow a = 3$[cite: 1]
$y = 2 \times 210 = 420$[cite: 1]
$x = 2 \times 420 = 840$[cite: 1]
$840 = 2^3 \times 3 \times 5 \times 7$. Product of prime factors = $2 \times 3 \times 5 \times 7 = 210$.[cite: 1]
Q5. The HCF of the smallest prime number and the smallest 3-digit number is $2^m \cdot 5^n$. The respective values of $m$ and $n$ are: [1 Mark][cite: 1]
(A) 0, 0
(B) 1, 0
(C) 0, 1
(D) 1, 1[cite: 1]
Click to View Solution
Answer: (B) 1, 0[cite: 1]
The smallest prime number is 2. The smallest 3-digit number is 100. $HCF(2, 100) = 2 = 2^1 \cdot 5^0$. So $m=1, n=0$.[cite: 1]
Q6. Which of the following statements is true for HCF and LCM of two distinct natural numbers $a$ and $b$? [1 Mark][cite: 1]
(i) HCF is always greater than LCM.
(ii) HCF is a factor of LCM.
(iii) LCM is a factor of HCF.[cite: 1]
(A) (i) only
(B) (i) and (iii)
(C) (i) and (ii)
(D) (ii) only[cite: 1]
Click to View Solution
Answer: (D) (ii) only[cite: 1]
Q7. Assertion (A): $(2+\sqrt{3})$ is an irrational number. [1 Mark][cite: 1]
Reason (R): The sum of two irrational numbers is always an irrational number.[cite: 1]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.[cite: 1]
Click to View Solution
Answer: (C) Assertion (A) is true, Reason (R) is false.[cite: 1]
Q8. Neha claimed that there does not exist any irrational number between 1 and 2. Raunak claimed that $\sqrt{2}$ lies between 1 and 2 and $\sqrt{2}$ is an irrational number. Who do you think is correct? Justify by proving either $\sqrt{2}$ as an irrational number or otherwise. [3 Marks][cite: 1]
Click to View Solution
Answer: Raunak is correct.[cite: 1]
Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are coprime and $b \neq 0$.[cite: 1]
$2b^2 = a^2$. 2 divides $a^2$, so 2 divides $a$ as well. Let $a = 2p$ (for some integer $p$).[cite: 1]
$a^2 = 4p^2 \Rightarrow 2b^2 = 4p^2 \Rightarrow b^2 = 2p^2$. 2 divides $b^2$ so 2 divides $b$ as well.[cite: 1]
2 is a common factor of $a$ and $b$ which is a contradiction as $a$ and $b$ are coprime. Our assumption is wrong. Hence, $\sqrt{2}$ is an irrational number.[cite: 1]
Q9. The HCF of $2^3$ and $3^2$ is: [1 Mark][cite: 1]
(A) $2^0 \cdot 3^0$
(B) $2^1 \cdot 3^1$
(C) $2^3 \cdot 3^2$
(D) $2^2 \cdot 3^3$[cite: 1]
Click to View Solution
Answer: (A) $2^0 \cdot 3^0$[cite: 1]
Q10. The LCM of two consecutive natural numbers $p$ and $p+1$ is: [1 Mark][cite: 1]
(A) $p$
(B) $p^2+p$
(C) 1
(D) $2p+1$[cite: 1]
Click to View Solution
Answer: (B) $p^2+p$[cite: 1]
Q11. Assertion (A): $(\sqrt{3}+1)^2$ is a rational number.
Reason (R): $(\sqrt{3})^2=3$ is a rational number. [1 Mark][cite: 1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.[cite: 1]
Click to View Solution
Answer: (D) Assertion (A) is false but Reason (R) is true.[cite: 1]
Q12 & Q16. Prove that $\sqrt{5}$ is an irrational number. [3 Marks][cite: 1]
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Answer: Let $\sqrt{5}$ be a rational number such that $\sqrt{5} = \frac{p}{q}$ where $p$ and $q$ are coprime, $q \neq 0$.[cite: 1]
$5q^2 = p^2 \Rightarrow 5$ divides $p^2 \Rightarrow 5$ divides $p$ as well.[cite: 1]
Let $p = 5m$ (for some integer $m$). $5q^2 = 25m^2 \Rightarrow q^2 = 5m^2$.[cite: 1]
$5$ divides $q^2 \Rightarrow 5$ divides $q$ as well.[cite: 1]
$p$ and $q$ have a common factor 5, which is a contradiction as $p$ and $q$ are co-prime integers. Our assumption is wrong. Hence, $\sqrt{5}$ is an irrational number.[cite: 1]
Q13. If HCF (850, 325) is 25, then LCM (850, 325) is: [1 Mark][cite: 1]
(A) 442
(B) 11050
(C) 8450
(D) 2210[cite: 1]
Click to View Solution
Answer: (B) 11050[cite: 1]
Q14. $7 \times 29 \times 23 + 1$ is: [1 Mark][cite: 1]
(A) a prime number
(B) divisible by 23
(C) an odd number
(D) a composite number[cite: 1]
Click to View Solution
Answer: (D) a composite number[cite: 1]
Q15. Assertion (A): $4^n$ can not end with the digit zero. [1 Mark]
Reason (R): Prime factorisation of $4^n$ is unique.[cite: 1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.[cite: 1]
Click to View Solution
Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).[cite: 1]
Q17. $7 \times 11 \times 13 + 5$ is [1 Mark][cite: 1]
(A) a prime number.
(B) an odd number.
(C) a composite number.
(D) a multiple of 5.[cite: 1]
Click to View Solution
Answer: (C) a composite number.[cite: 1]
Q18. Find the H.C.F. and L.C.M. of 1530 and 2040. [2 Marks][cite: 1]
Click to View Solution
Answer:
$1530 = 2 \times 3^2 \times 5 \times 17$
$2040 = 2^3 \times 3 \times 5 \times 17$[cite: 1]
$HCF(1530, 2040) = 2 \times 3 \times 5 \times 17 = 510$[cite: 1]
$LCM(1530, 2040) = 2^3 \times 3^2 \times 5 \times 17 = 6120$[cite: 1]
Q19. Prove that $\sqrt{2}$ is an irrational number. [3 Marks][cite: 1]
Click to View Solution
Answer: Let $\sqrt{2}$ be a rational number such that $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ are coprime numbers and $b \neq 0$.[cite: 1]
$2b^2 = a^2 \Rightarrow 2$ divides $a^2 \Rightarrow 2$ divides $a$ as well. So $a = 2p$ for some integer $p$.[cite: 1]
$a^2 = 4p^2 \Rightarrow 2b^2 = 4p^2 \Rightarrow b^2 = 2p^2 \Rightarrow 2$ divides $b^2 \Rightarrow 2$ divides $b$ as well.[cite: 1]
2 is a common factor of $a$ and $b$ which is a contradiction as $a$ and $b$ are coprime numbers. Our assumption is wrong. Hence $\sqrt{2}$ is an irrational number.[cite: 1]
Previous Year Questions 2026 (Standard)
Q1. The HCF of 960 and 432 is: [1 Mark][cite: 1]
(a) 48
(b) 54
(c) 72
(d) 36[cite: 1]
Click to View Solution
Answer: (a) 48[cite: 1]
Q2. The natural number 2 is: [1 Mark][cite: 1]
(a) a prime number
(b) a composite number
(c) prime as well as composite
(d) neither prime nor composite[cite: 1]
Click to View Solution
Answer: (a) a prime number[cite: 1]
Q3. For any natural number $n$, $6^n$ ends with the digit: [1 Mark][cite: 1]
(a) 0
(b) 6
(c) 3
(d) 2[cite: 1]
Click to View Solution
Answer: (b) 6[cite: 1]
Q5. The LCM of 960 and 240 is: [1 Mark][cite: 1]
(A) 960
(B) 240
(C) 60
(D) 15[cite: 1]
Click to View Solution
Answer: (A) 960[cite: 1]
Q6. The natural number 1 is: [1 Mark][cite: 1]
(A) a prime number.
(B) a composite number.
(C) prime as well as composite.
(D) neither prime nor composite.[cite: 1]
Click to View Solution
Answer: (D) neither prime nor composite[cite: 1]
Q9. $(3 \times 11 \times 13 + 3)$ is: [1 Mark][cite: 1]
(A) a prime number
(B) divisible by 13
(C) a composite number
(D) an odd number[cite: 1]
Click to View Solution
Answer: (C) a composite number[cite: 1]
Q10. Find the length of the plank that can be used to measure the lengths 4 m 20 cm and 5 m 4 cm exactly, in the least time. [2 Marks][cite: 1]
Click to View Solution
Answer: 4 m 20 cm = 420 cm and 5 m 4 cm = 504 cm.[cite: 1]
Size of plank should be maximum, so we will find $HCF(420, 504)$.[cite: 1]
$420 = 2^2 \times 3 \times 5 \times 7$ and $504 = 2^3 \times 3^2 \times 7$.[cite: 1]
$HCF(420, 504) = 84$. Therefore, the required length of the plank is 84 cm.[cite: 1]
Q12. Assertion (A): $(3+\sqrt{5})$ is an irrational number. [1 Mark]
Reason (R): Sum of any two irrational numbers is always irrational.[cite: 1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.[cite: 1]
Click to View Solution
Answer: (C) Assertion (A) is true, but Reason (R) is false.[cite: 1]
Q13(a). Prove that $2+3\sqrt{5}$ is an irrational number given that $\sqrt{5}$ is irrational. [2 Marks][cite: 1]
Click to View Solution
Answer: Let $2+3\sqrt{5}$ be a rational number. $2+3\sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and $p, q$ are integers.[cite: 1]
$\Rightarrow \sqrt{5} = \frac{p-2q}{3q}$. As $\frac{p-2q}{3q}$ is a rational number, so $\sqrt{5}$ is rational.[cite: 1]
But we know that $\sqrt{5}$ is irrational. Therefore our assumption is wrong. Hence, $2+3\sqrt{5}$ is an irrational number.[cite: 1]
Q13(b). If the HCF of 210 and 55 is expressed as $210 \times 5 + 55m$ then find the value of $m$. [2 Marks][cite: 1]
Click to View Solution
Answer: $210 = 2 \times 3 \times 5 \times 7$
$55 = 5 \times 11$[cite: 1]
$H.C.F. (210, 55) = 5$[cite: 1]
$\therefore 5 = 210 \times 5 + 55m$[cite: 1]
Q14. Find the greatest number less than 10,000 which is exactly divisible by 48, 60 and 65. [3 Marks][cite: 1]
Click to View Solution
Answer: $48 = 2^4 \times 3$, $60 = 2^2 \times 3 \times 5$, $65 = 5 \times 13$[cite: 1]
$L.C.M.(48, 60, 65) = 2^4 \times 3 \times 5 \times 13 = 3120$[cite: 1]
Highest multiple of 3120, less than $10,000 = 3120 \times 3 = 9360$[cite: 1]
Q15. Assertion (A): $H.C.F. (36m^2, 18m) = 18m$, where $m$ is a prime number. [1 Mark]
Reason (R): H.C.F. of two numbers is always less than or equal to the smaller number.[cite: 1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.[cite: 1]
Click to View Solution
Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).[cite: 1]
Q16. Prove that $2-5\sqrt{3}$ is an irrational number given that $\sqrt{3}$ is irrational. [2 Marks][cite: 1]
Click to View Solution
Answer: Let $2-5\sqrt{3}$ be a rational number.[cite: 1]
$\therefore 2-5\sqrt{3} = \frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$[cite: 1]
$\sqrt{3} = \frac{2b-a}{5b}$[cite: 1]
RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence $2-5\sqrt{3}$ is an irrational number.[cite: 1]
Q17. The dimensions of a window are $156 \text{ cm} \times 216 \text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed. [3 Marks][cite: 1]
Click to View Solution
Answer: $156 = 2^2 \times 3 \times 13$
$216 = 2^3 \times 3^3$[cite: 1]
Required side length of the square $= HCF(156, 216) = 12 \text{ cm}$[cite: 1]
Number of squares formed $= \frac{156 \times 216}{12 \times 12} = 234$[cite: 1]