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 2 of 5

1. Q21 Exemplar · Exercise 1.2
   Write ​whether ​every positive integer can be of the form 4q + ​2, ​where q ​is an integer. Justify ​your answer.
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   No - not every positive integer is of the form $4q+2$. By Euclid's division lemma an integer can be $4q$, $4q+1$, $4q+2$ or $4q+3$; the form $4q+2$ covers only those leaving remainder 2 (e.g. it misses 1, 3, 4, 5).

   Step-by-step solution →
2. Q22 Exemplar · Exercise 1.2
   "The ​product of ​two consecutive ​positive integers ​is ​divisible by 2". Is ​this statement true ​or false? Give ​reasons.
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   True. Of any two consecutive integers one is even, so the product $n(n+1)$ always has an even factor and is divisible by $2$.

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3. Q23 Exemplar · Exercise 1.2
   "The product of three ​consecutive positive integers ​is ​divisible ​by 6." Is this statement true or ​false? ​Justify your ​answer.
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   True. Among three consecutive integers one is a multiple of 3 and at least one is even, so the product is divisible by both 2 and 3, hence by 6.

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4. Q24 Exemplar · Exercise 1.2
   Write whether ​the square ​of any ​positive ​integer can be of ​the form $3m+2$, where $m$ is a natural ​number. Justify your ​answer.
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   No. The square of any positive integer is always of the form $3m$ or $3m+1$, never $3m+2$.

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5. Q25 Exemplar · Exercise 1.2
   A positive integer is of the form 3q + 1, q being a ​natural number. ​Can ​you write ​its square in any form other than 3m + ​1, i.e., 3m or 3m ​+ 2 for some integer m? Justify ​your ​answer.
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   No - the square of a number of the form $3q+1$ is always of the form $3m+1$. It can never be $3m$ or $3m+2$, because $(3q+1{)}^2=3(3q^2+2q)+1$.

   Step-by-step solution →
6. Q26 Exemplar · Exercise 1.2
   The ​numbers ​525 and 3000 are both divisible only by 3, ​5, 15, 25 ​and 75. ​What ​is HCF ​(525, 3000)? ​Justify your answer.
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   HCF$(525,3000)=75$, since 75 is the greatest of the common divisors $3,5,15,25,75$.

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7. Q27 Exemplar · Exercise 1.2
   Explain ​why $3\times 5\times 7+7$ is ​a ​composite ​number.
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   $3\times 5\times 7+7=7(3\times 5+1)=7\times 16=112$, a product of two integers greater than 1, so it is composite.

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8. Q28 Exemplar · Exercise 1.2
   Can two numbers have 18 ​as their ​HCF and 380 as ​their ​LCM? Give ​reasons.
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   No. The HCF must divide the LCM, but $18$ does not divide $380$ ($380=18\times 21+2$). So no two numbers can have HCF 18 and LCM 380.

   Step-by-step solution →
9. Q29 Exemplar · Exercise 1.2
   Without actually ​performing the ​long division, find if $\frac{987}{10500}$ will have terminating ​or non-terminating (repeating) ​decimal ​expansion. Give reasons for your ​answer.
   Show answer

   $\frac{987}{10500}=\frac{47}{500}$ and $500=2^2\times 5^3$, so the decimal expansion is terminating.

   Step-by-step solution →
10. Q30 Exemplar · Exercise 1.2
    A ​rational ​number in ​its decimal ​expansion ​is $327.7081$. What can you say about the prime factors ​of $q$, when this number ​is expressed in the form $\frac{p}{q}$? ​Give reasons.
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    Since 327.7081 has a terminating decimal expansion, $q$ in $\frac{p}{q}$ (lowest terms) has prime factors only of the form $2^m\times 5^n$ -- that is, $q$ is made up only of 2s and 5s.

    Step-by-step solution →
11. Q31 Exemplar · Exercise 1.3
    Show that ​the square of ​any positive ​integer ​is ​either of the ​form $4q$ ​or $4q+1$ for some integer $q$.
    Show answer

    If $n$ is even, $n^2=4m^2=4q$; if $n$ is odd, $n^2=4(m^2+m)+1=4q+1$. So every square is of the form $4q$ or $4q+1$. Proved.

    Step-by-step solution →
12. Q32 Exemplar · Exercise 1.3
    Show that ​cube of any positive ​integer ​is ​of ​the form $4m$, $4m+1$ or $4m+3$, for some integer $m$.
    Show answer

    For every positive integer $n$, the cube $n^3$ is of the form $4m$, $4m+1$ or $4m+3$. Hence proved.

    Step-by-step solution →
13. Q33 Exemplar · Exercise 1.3
    Show that the ​square ​of any positive ​integer cannot ​be of ​the ​form $5q+2$ or $5q+3$ ​for any ​integer $q$.
    Show answer

    Every square of a positive integer is of the form $5q$, $5q+1$ or $5q+4$, hence never $5q+2$ or $5q+3$. Proved.

    Step-by-step solution →
14. Q34 Exemplar · Exercise 1.3
    Show ​that ​the square ​of any positive integer cannot be ​of the form $6m+2$ ​or $6m+5$ ​for ​any ​integer $m$.
    Show answer

    A perfect square mod 6 can only be $0,1,3$ or $4$, never $2$ or $5$; hence no square has the form $6m+2$ or $6m+5$.

    Step-by-step solution →
15. Q35 Exemplar · Exercise 1.3
    Show ​that the square of ​any ​odd ​integer is of ​the ​form $4q+1$, ​for some integer $q$.
    Show answer

    For any odd $n=2m+1$, $n^2=4(m^2+m)+1=4q+1$ with $q=m^2+m$ an integer.

    Step-by-step solution →
16. Q36 Exemplar · Exercise 1.3
    If $n$ ​is an ​odd integer, ​then ​show that $n^2-1$ is ​divisible by ​8.
    Show answer

    With $n=2k+1$, $n^2-1=4k(k+1)$; since $k(k+1)$ is even, $n^2-1=8m$, so it is divisible by 8.

    Step-by-step solution →
17. Q37 Exemplar · Exercise 1.3
    Prove that ​if $x$ and $y$ ​are both odd ​positive integers, ​then $x^2+y^2$ ​is even but not ​divisible ​by ​4.
    Show answer

    Proved: $x^2+y^2=4(m^2+m+n^2+n)+2$, which is even (a multiple of 2) but leaves remainder 2 on division by 4.

    Step-by-step solution →
18. Q38 Exemplar · Exercise 1.3
    Use Euclid's ​division ​algorithm ​to ​find ​the ​HCF of $441,567,693$.
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    The HCF of $441,567,693$ is $63$.

    Step-by-step solution →
19. Q39 Exemplar · Exercise 1.3
    Using Euclid's division algorithm, find the ​largest ​number that divides 1251, 9377 and 15628 ​leaving remainders 1, 2 and ​3, ​respectively.
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    The largest such number is $625$.

    Step-by-step solution →
20. Q40 Exemplar · Exercise 1.3
    Prove ​that $\sqrt{3}+\sqrt{5}$ is ​irrational.
    Show answer

    By contradiction, assuming $\sqrt{3}+\sqrt{5}=r$ rational forces $\sqrt{3}=\frac{r^2-2}{2r}$ to be rational, which is false. Hence $\sqrt{3}+\sqrt{5}$ is irrational.

    Step-by-step solution →