SEBA Class 10 Mathematics Chapter 1.1 Solutions (Real Numbers) for New Book 2026

Here you will get all the solutions of SEBA Class 10 Mathematics Chapter 1.1 Real Numbers from the New SCERT Textbooks 2026.

1. Express each number as a product of its prime factors:

(i) $140$

By prime factorisation, we get, 

$140 = 2 \times 2 \times 5 \times 7 $

$= 2^{2} \times 5 \times 7$

(ii) $156$

By prime factorisation, we get, 

$ 156 = 2 \times 2 \times 3 \times 13 $

$ = 2^{2} \times 3 \times 13$

(iii) $3825$

By prime factorisation, we get, 

$ 3825 = 3 \times 3 \times 5 \times 5 \times 17 $

$ = 3^{2} \times 5^{2} \times 17$

(iv) $5005$

By prime factorisation, we get, 

$ 5005 = 5 \times 7 \times 11 \times 13$

(v) $7429$

By prime factorisation, we get, 

$ 7429 = 17 \times 19 \times 23$

2. Find the LCM and HCF of the following pairs of integers and verify that $\text{LCM} \times \text{HCF} = \text{product of the two numbers}$.

Solution:

(i) $26$ and $91$

Prime factors of $26 = 2 \times 13$

Prime factors of $91 = 7 \times 13$

$\text{HCF of } 26 \text{ and } 91 = 13$

$\text{LCM of } 26 \text{ and } 91 = 2 \times 7 \times 13 = 182$

$\text{Product of two numbers} = 26 \times 91 = 2366$

$\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$

So, $\text{Product of two numbers} = \text{LCM} \times \text{HCF}$

(ii) $510$ and $92$

Prime factors of $510 = 2 \times 3 \times 5 \times 17$

Prime factors of $92 = 2 \times 2 \times 23 = 2^{2} \times 23$

$\text{HCF of two numbers} = 2$

$\text{LCM of two numbers} = 2^{2} \times 3 \times 5 \times 17 \times 23 = 23460$

$\text{Product of two numbers} = 510 \times 92 = 46920$

$\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920$

So, $\text{Product of two numbers} = \text{LCM} \times \text{HCF}$

(iii) $336$ and $54$

Prime factors of $336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7 = 2^{4} \times 3 \times 7$

Prime factors of $54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^{3}$

$\text{HCF of two numbers} = 2 \times 3 = 6$

$\text{LCM of two numbers} = 2^{4} \times 3^{3} \times 7 = 16 \times 27 \times 7 = 3024$

$\text{Product of two numbers} = 336 \times 54 = 18144$

$\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144$

So, $\text{Product of two numbers} = \text{LCM} \times \text{HCF}$

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

Solution:

(i) $12$, $15$ and $21$

By prime factorisation, we get:

$12 = 2 \times 2 \times 3 $
$ = 2^{2} \times 3$

$15 = 3 \times 5$

$21 = 3 \times 7$

Therefore, $\text{HCF}(12, 15, 21) = 3$

$\text{LCM}(12, 15, 21) = 2^{2} \times 3 \times 5 \times 7 = 420$

(ii) $17$, $23$ and $29$

By prime factorisation, we get:

$17 = 17 \times 1$

$23 = 23 \times 1$

$29 = 29 \times 1$

Therefore, $\text{HCF}(17, 23, 29) = 1$

$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$

(iii) $8$, $9$ and $25$

By prime factorisation, we get:

$8 = 2 \times 2 \times 2 = 2^{3}$

$9 = 3 \times 3 = 3^{2}$

$25 = 5 \times 5 = 5^{2}$

Therefore, $\text{HCF}(8, 9, 25) = 1$

$\text{LCM}(8, 9, 25) = 2^{3} \times 3^{2} \times 5^{2} $
$= 8 \times 9 \times 25 = 1800$

4. Given that $\text{HCF}(306, 657) = 9$, find $\text{LCM}(306, 657)$.

Solution:

Given, $\text{HCF}(306, 657) = 9$

We know that: $\text{LCM} \times \text{HCF} = \text{Product of two numbers}$

$\Rightarrow \text{LCM} \times 9 = 306 \times 657$

$\Rightarrow \text{LCM} = \frac{306 \times 657}{9}$

$\Rightarrow \text{LCM} = 34 \times 657$

$\therefore \text{LCM} = 22338$

5. Check whether $6^{n}$ can end with the digit $0$ for any natural number $n$.

Solution:

If the number $6^{n}$ ends with the digit zero ($0$), then it should be divisible by $5$, as any number ending with $0$ has prime factors $2$ and $5$.

Prime factorisation of $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$

Therefore, the prime factorisation of $6^{n}$ does not contain the prime number $5$.

Hence, $6^{n}$ cannot end with the digit $0$ for any natural number $n$.

6. 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.

Solution:

A composite number is a number that has factors other than $1$ and number itself.

For the first expression:

$7 \times 11 \times 13 + 13$
$= 13 \times (7 \times 11 + 1)$ [Taking $13$ as a common factor]
$= 13 \times (77 + 1)$ $= 13 \times 78$
$= 13 \times 2 \times 3 \times 13$

Hence, $7 \times 11 \times 13 + 13$ is a composite number.

For the second expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

$= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$ [Taking $5$ as a common factor] $= 5 \times (1008 + 1)$
$= 5 \times 1009$

Since it has factors other than $1$ and the number itself, $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ is a composite number.

7. 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?

Solution:

The time after which Sonia and Ravi meet again at the starting point will be the LCM of $18$ and $12$.

$18 = 2 \times 3 \times 3 = 2 \times 3^{2}$

$12 = 2 \times 2 \times 3 = 2^{2} \times 3$

$\text{LCM}(18, 12) = 2^{2} \times 3^{2} = 4 \times 9 = 36$

Hence, Sonia and Ravi will meet again at the starting point after $36\text{ minutes}$.

8. (i) The soldiers in a regiment can be stood in some rows consisting of 15, 20 or 25 numbers of soldiers. Find the least number of soldiers in the regiment.

Solution:

The least number of soldiers in the regiment will be the LCM of $15$, $20$ and $25$.

$15 = 3 \times 5$

$20 = 2^{2} \times 5$

$25 = 5^{2}$

$\text{LCM} = 2^{2} \times 3 \times 5^{2} $

$ = 4 \times 3 \times 25 $

$= 300$

So, the least number of soldiers in the regiment is $300$.

(ii) A bell rings at every 18 seconds, another bell rings at every 60 seconds. If these two bells ring simultaneously at an instant, then find after how many seconds will the bells ring simultaneously again.

Solution:

The time after which the bells will ring simultaneously again will be the LCM of $18$ and $60$.

So,

$18 = 2 \times 3^{2}$
$60 = 2^{2} \times 3 \times 5$

Now,

$\text{LCM} = 2^{2} \times 3^{2} \times 5 $
$= 4 \times 9 \times 5$
$= 180$

So, the bells will ring together again after $180\text{ seconds}$.

(iii) A radio station plays ‘Assam Sangeet’ once every two days. Another radio station plays the same song once every three days. How many times in 30 days will both the radio stations play the same song on the same day.

Solution:

The first station plays the song once every 2 days.
The second station plays the song once every 3 days.

Now, $\text{LCM}(2, 3) = 6$

So, both stations play the song together once every $6\text{ days}$.

So, the number of times both play together in $30\text{ days} = 30 \div 6 = 5\text{ times}$

(iv) An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

The maximum number of columns in which they can march will be the HCF of $ 616$ and $32$.

$616 = 2^{3} \times 7 \times 11$

$32 = 2^{5}$

$\therefore \text{HCF} = 2^{3} = 8$

So, the maximum number of columns in which they can march is $8$.

(v) Himadri has a collection of 625 Indian postal stamps and 325 International postal stamps. She wants to display them in identical groups of Indian and International stamps with no stamp left out. What is greatest number of groups Himadri can display the stamps?

Solution:

The greatest number of groups Himadri can display the stamps will be the HCF of $625$ and $325$.

$625 = 5^{4}$

$325 = 5^{2} \times 13$

$\text{HCF} = 5^{2} = 25$

So, the greatest number of groups is $25$.

(vi) Two ropes are of length 64 cm and 80 cm. Both are to be cut into pieces of equal length. What should be the maximum length of the pieces?

Solution:

The maximum length of each piece will be the HCF of $64$ and $80$.

$64 = 2^{6}$

$80 = 2^{4} \times 5$

$\text{HCF} = 2^{4} = 16$

So, the maximum length of the pieces should be $16\text{ cm}$.

9. Find the greatest number of 4 digits which is exactly divisible by 18, 24 and 36.

Solution:

The number exactly divisible by $18$, $24$, and $36$, must also be a multiple of their LCM.

$18 = 2 \times 3^{2}$

$24 = 2^{3} \times 3$

$36 = 2^{2} \times 3^{2}$

So,

$\text{LCM} = 2^{3} \times 3^{2} $

$= 8 \times 9 $

$= 72$

Now, we know that the greatest 4-digit number is $9999$.

Dividing $9999$ by $72$, we get

$9999 = 72 \times 138 + 63$

∴ Required number $= 9999 – 63 = 9936$

So, the greatest 4-digit number exactly divisible by $18$, $24$ and $36$ is $9936$.

10. 1245 is a factor of the numbers p and q. Which of the following will have 1245 as a factor?

(i) $p + q$
(ii) $p – q$
(iii) $p \times q$
(iv) $p \div q$

Choose the correct options:
(A) only (iii)
(B) only (i) and (ii)
(C) only (i), (ii) and (iii)
(D) All (i), (ii), (iii) and (iv)

Solution: (C) only (i), (ii) and (iii)

Explanation:-

Let $p = 1245a$ and $q = 1245b$ (where $a$ and $b$ are integers).

(i) $p + q $

$ = 1245a + 1245b $

$ = 1245(a + b) $

$\therefore1245\text{ is a factor of  p  + q }$

(ii) $p – q $

$= 1245a – 1245b $

$= 1245(a – b) $

$\therefore1245\text{ is a factor of p – q}$

(iii) $p \times q $

$ = (1245a)(1245b) $

$= 1245(1245ab) $

$\therefore1245\text{ is a factor of } p \times q$ 

(iv) $p \div q $

$ = \frac{1245a}{1245b} $

$= \frac{a}{b}$, which is not a multiple of $1245$.

$\therefore1245\text{ is not a factor of } p \div q$ 

Therefore, only (i), (ii), and (iii) have $1245$ as a factor.

Correct option is (C) only (i), (ii) and (iii)

11. Match the columns:

Ans: A) P-4, Q-3, R-2, S-1

P) Number which is neither prime nor composite is $1 \rightarrow 4$

Q) Only even prime number is $2 \rightarrow 3$

R) HCF of $12$, $15$, $21$:

$12 = 2^{2} \times 3$,
$15 = 3 \times 5$,
$21 = 3 \times 7 $

$\therefore \text{HCF} = 3 \rightarrow 2$

S) LCM of $2$ and $9$:

$2 = 2$,
$9 = 3^{2} $

$\therefore \text{LCM} = 2 \times 9 = 18 \rightarrow 1$

So, the correct option is option (A) P-4, Q-3, R-2, S-1

12. Which of the following statement is true or false:

Statement (P): HCF of two consecutive natural numbers is 1
Statement (Q): HCF of two coprime numbers is 1

Choose the correct option
A) Pis true, Q is false
B) Pis false, Q is true
C) Both P & Q are true
D) Both P & Q are false

Ans:- C) Both P & Q are true

• Any two consecutive natural numbers (like $n$ and $n+1$) share no common factor other than $1$, so $\text{HCF} = 1$. Hence, Statement (P) is true.

• By definition, coprime numbers have no common factor other than $1$, so their $\text{HCF} = 1$. Hence, Statement (Q) is true.

So, the correct option is option (C) Both P & Q are true

13. Two positive integers P and Q can be expressed as $P = ab^{2}$ and $Q = a^{2}b$, where $a$ and $b$ are prime numbers. The LCM of P and Q is:

A) $a^2b$
B) $a^2b^2$
C) $ab$
D) $ab^2$

Ans:-

$P = a^{1} \times b^{2}$

$Q = a^{2} \times b^{1}$

 $\text{LCM}(P, Q) = a^{2} \times b^{2} $

$= a^{2}b^{2}$

So, the correct option is option (B) $a^{2}b^{2}$

14. For the numbers $P = 119$; $Q = 462$; $R = 105$; $S = 2310$, choose the option that represents the correct increasing sequence of number of prime factors:

Ans: B) P, R, Q, S

Prime factorisation of each:

$P = 119 = 7 \times 17 \rightarrow 2\text{ prime factors}$

$R = 105 = 3 \times 5 \times 7 \rightarrow 3\text{ prime factors}$

$Q = 462 = 2 \times 3 \times 7 \times 11 \rightarrow 4\text{ prime factors}$

$S = 2310 = 2 \times 3 \times 5 \times 7 \times 11 \rightarrow 5\text{ prime factors}$

Arranging in increasing order, we get: P < R < Q < S

So, the correct option is option (B) P, R, Q, S

15. Assertion (A): 3 and 10 are coprime numbers
Reason (R): a and b are coprime numbers if they have no common factors other than 1.

A) Both (A) and (R) are true and Reason (R) is correct explanation of Assertion (A)
B) Both (A) and (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.

Ans: A) Both (A) and (R) are true and Reason (R) is correct explanation of Assertion (A)

Explanation:-

Factors of $3 = 1, 3$

Factors of $10 = 1, 2, 5, 10$

The only common factor is $1$, so $3$ and $10$ are coprime numbers. Thus, Assertion (A) is true.

Reason (R) correctly gives the exact definition of coprime numbers and correctly explains (A).

Ans: A) Both (A) and (R) are true and Reason (R) is correct explanation of Assertion (A)

16. Observe the factor tree and answer the questions:

(i) The value of x is
(a) 8325 (b) 3825 (c) 835 (d) 3325

(ii) The value of y is
(a) 5 (b) 25 (c) 17 (d) 3

(iii) The value of z is
(a) 3 (b) 17 (c) 5 (d) 13

(iv) The value of x+y+z is
(a) 3842 (b) 3847 (c) 3825 (d) 3874

Ans:-

From the given factor tree:

• $x = 3 \times 1275 = 3825$

• $1275 = 3 \times 425$

• $425 = y \times 85 $
  $\Rightarrow y = \frac{425}{85} = 5$

• $85 = 5 \times z $
  $\Rightarrow z = \frac{85}{5} = 17$

(i) The value of $x$ is:

Ans: (b) 3825

(ii) The value of $y$ is:

Ans: (a) 5

(iii) The value of $z$ is:

Ans: (b) 17

(iv) The value of $x + y + z$ is:

$x + y + z $
$= 3825 + 5 + 17 $
$= 3847$

Ans: (b) 3847

17. An inter-school seminar is being conducted by an NGO related to education, where the participants will be educators of different subjects. The number of participants in Science, English and Mathematics are 60, 84 and 108 respectively.

(i) In each room the same number of participants are to be seated and all of them being from the same subject. Find the maximum number of participants that can be accommodated in each room.

Solution:- 

By prime factorisation of the numbers, we get

$60 = 2^{2} \times 3 \times 5$

$84 = 2^{2} \times 3 \times 7$

$108 = 2^{2} \times 3^{3}$

The maximum number of participants per room $= \text{HCF}(60, 84, 108)$

∴ $\text{HCF}(60, 84, 108) = 2^{2} \times 3 = 12$

So, the maximum number of participants in each room is $12$.

(ii) What is the minimum number of rooms required for the event?

Total number of participants $= 60 + 84 + 108 = 252$

$\text{Minimum number of rooms} = \frac{\text{Total participants}}{\text{Participants per room}}$

$ = \frac{252}{12} $

$ = 21\text{ rooms}$

(iii) Find the LCM of 60, 84 and 108.

$\text{LCM}(60, 84, 108) = 2^{2} \times 3^{3} \times 5 \times 7 $

$ = 4 \times 27 \times 5 \times 7 $

$ = 3780$

(iv) Find the product of HCF and LCM of 60, 84 and 108.

$\text{Product} = \text{HCF} \times \text{LCM} $

$ = 12 \times 3780 $

$= 45360$