Class 10 Maths · Chapter 1

Real Numbers

82 Class 10 Maths questions on Real Numbers from the NCERT textbook, NCERT Exemplar and CBSE board papers 2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 1 of 5

1. Q1 NCERT · Exercise 1.1
   Express each ​number as a product of ​its ​prime factors: ​

   1. (i)$140$ ​
   2. (ii)$156$
   3. (iii)$3825$
   4. (iv)$5005$ ​
   5. (v)$7429$
   Show answer

   $140=2^2\times 5\times 7$; $156=2^2\times 3\times 13$; $3825=3^2\times 5^2\times 17$; $5005=5\times 7\times 11\times 13$; $7429=17\times 19\times 23$.

   Step-by-step solution →
2. Q2 NCERT · Exercise 1.1
   Find the LCM ​and HCF of ​the following pairs of integers and verify that $\text{LCM}\times \text{HCF}=$ ​product ​of ​the two numbers.

   1. (i)$26$ and $91$ ​
   2. (ii)$510$ and $92$
   3. (iii)$336$ and $54$
   Show answer

   (i) HCF $=13$, LCM $=182$; (ii) HCF $=2$, LCM $=23460$; (iii) HCF $=6$, LCM $=3024$. In each case $\text{HCF}\times \text{LCM}$ equals the product of the two numbers.

   Step-by-step solution →
3. Q3 NCERT · Exercise 1.1
   Find the ​LCM and HCF of the following ​integers ​by applying ​the prime factorisation method.

   1. (i)$12$, $15$ ​and $21$ ​
   2. (ii)$17$, $23$ and $29$ ​
   3. (iii)$8$, $9$ and $25$
   Show answer

   (i) HCF $=3$, LCM $=420$; (ii) HCF $=1$, LCM $=11339$; (iii) HCF $=1$, LCM $=1800$.

   Step-by-step solution →
4. Q4 NCERT · Exercise 1.1
   Given ​that ​HCF (306, ​657) = ​9, find ​LCM ​(306, ​657).
   Show answer

   $\text{LCM}(306,657)=22338$.

   Step-by-step solution →
5. Q5 NCERT · Exercise 1.1
   Check whether $6^n$ can end with ​the digit $0$ ​for ​any natural ​number $n$.
   Show answer

   No. $6^n=2^n\times 3^n$ has no factor of $5$, so it is never a multiple of $10$ and cannot end in $0$ for any natural number $n$. Its unit digit is always $6$.

   Step-by-step solution →
6. Q6 NCERT · Exercise 1.1
   Explain ​why $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are ​composite ​numbers.
   Show answer

   $7\times 11\times 13+13=13\times 78=1014$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times 1009=5045$. Each is a product of two whole numbers greater than $1$, so both are composite.

   Step-by-step solution →
7. Q7 NCERT · Exercise 1.1
   There ​is a circular path around a sports ​field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes ​for the same. Suppose ​they both start at the same point and at the same time, and go in ​the same direction. After how many minutes will they meet again at ​the starting point?
   Show answer

   They will meet again at the starting point after $36$ minutes.

   Step-by-step solution →
8. Q8 NCERT · Exercise 1.2
   Prove ​that $\sqrt{5}$ is ​irrational.
   Show answer

   The assumption that $\sqrt{5}=\frac{p}{q}$ (in lowest terms) forces $5$ to divide both $p$ and $q$, contradicting that they are coprime. Hence $\sqrt{5}$ is irrational.

   Step-by-step solution →
9. Q9 NCERT · Exercise 1.2
   Prove ​that $3+2\sqrt{5}$ ​is irrational.
   Show answer

   If $3+2\sqrt{5}$ were rational, then $\sqrt{5}=\frac{p-3q}{2q}$ would also be rational, contradicting the known irrationality of $\sqrt{5}$. Hence $3+2\sqrt{5}$ is irrational.

   Step-by-step solution →
10. Q10 NCERT · Exercise 1.2
    Prove ​that the ​following ​are ​irrationals:

    1. (i)$\frac{1}{\sqrt{2}}$
    2. (ii)$7\sqrt{5}$ ​
    3. (iii)$6+\sqrt{2}$
    Show answer

    Each assumption that the number is rational leads to $\sqrt{2}$ or $\sqrt{5}$ being rational, which is false. Hence all three are irrational.

    Step-by-step solution →
11. Q11 Exemplar · Exercise 1.1
    For some integer $m$, every ​even integer ​is ​of ​the ​form

    1. (a)$m$
    2. (b)$m+1$
    3. (c)$2m$
    4. (d)$2m+1$
    Show answer

    (c) $2m$

    Step-by-step solution →
12. Q12 Exemplar · Exercise 1.1
    For ​some integer $q$, ​every ​odd integer ​is ​of the ​form

    1. (a)$q$
    2. (b)$q+1$
    3. (c)$2q$
    4. (d)$2q+1$
    Show answer

    (d) $2q+1$

    Step-by-step solution →
13. Q13 Exemplar · Exercise 1.1
    $n^2-1$ ​is ​divisible by ​8, ​if $n$ is

    1. (a)an integer
    2. (b)a natural number
    3. (c)an odd integer
    4. (d)an even integer
    Show answer

    (c) an odd integer

    Step-by-step solution →
14. Q14 Exemplar · Exercise 1.1
    If the HCF ​of $65$ and $117$ is ​expressible ​in ​the form $65m-117$, ​then ​the value ​of $m$ ​is

    1. (a)$4$
    2. (b)$2$
    3. (c)$1$
    4. (d)$3$
    Show answer

    (b) $2$

    Step-by-step solution →
15. Q15 Exemplar · Exercise 1.1
    The ​largest ​number which divides $70$ and $125$, ​leaving remainders $5$ and $8$, ​respectively, is

    1. (a)$13$
    2. (b)$65$
    3. (c)$875$
    4. (d)$1750$
    Show answer

    (a) $13$

    Step-by-step solution →
16. Q16 Exemplar · Exercise 1.1
    If two positive ​integers $a$ and $b$ are ​written as $a=x^3{y}^2$ ​and $b=x{y}^3$; $x$, $y$ ​are ​prime numbers, then ​HCF $(a,b)$ is

    1. (a)$xy$
    2. (b)$x{y}^2$
    3. (c)$x^3{y}^3$
    4. (d)$x^2{y}^2$
    Show answer

    (b) $x{y}^2$

    Step-by-step solution →
17. Q17 Exemplar · Exercise 1.1
    If ​two positive ​integers $p$ ​and $q$ can be expressed ​as $p=a{b}^2$ and $q=a^{3}b$; $a,b$ being ​prime numbers, then ​LCM $(p,q)$ is

    1. (a)$ab$
    2. (b)$a^2{b}^2$
    3. (c)$a^3{b}^2$
    4. (d)$a^3{b}^3$
    Show answer

    (c) $a^3{b}^2$

    Step-by-step solution →
18. Q18 Exemplar · Exercise 1.1
    The product of ​a non-zero rational ​and an ​irrational ​number ​is

    1. (a)always irrational
    2. (b)always rational
    3. (c)rational or irrational
    4. (d)one
    Show answer

    (a) always irrational

    Step-by-step solution →
19. Q19 Exemplar · Exercise 1.1
    The ​least number that ​is divisible by all the numbers from ​1 ​to 10 (both inclusive)

    1. (a)$10$
    2. (b)$100$
    3. (c)$504$
    4. (d)$2520$
    Show answer

    (d) $2520$

    Step-by-step solution →
20. Q20 Exemplar · Exercise 1.1
    The decimal ​expansion ​of ​the rational ​number $\frac{14587}{1250}$ ​will terminate ​after:

    1. (a)one decimal place
    2. (b)two decimal places
    3. (c)three decimal places
    4. (d)four decimal places
    Show answer

    (d) four decimal places

    Step-by-step solution →