SEBA Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 Textual Question Answer Solutions (New Syllabus 2026–27)

Looking for accurate SEBA Class 10 Maths Exercise 1.1 (Real Numbers) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering the Fundamental Theorem of Arithmetic, prime factorization methods, finding HCF and LCM, and verifying the relation HCF × LCM = Product of Two Numbers to score top marks in your HSLC board exams.

Exercise: 1.1

Q1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(i) 140:

We divide 140 by its smallest prime factors sequentially:

140 / 2 = 70

70 / 2 = 35

35 / 5 = 7

7 / 7 = 1

The prime factors of 140 are: 2, 2, 5, and 7.

Answer: 140 = 2 x 2 x 5 x 7 = 2² x 5 x 7

(ii) 156:

We divide 156 by its smallest prime factors sequentially:

156 / 2 = 78

78 / 2 = 39

39 / 3 = 13

13 / 13 = 1

The prime factors of 156 are: 2, 2, 3, and 13.

Answer: 156 = 2 x 2 x 3 x 13 = 2² x 3 x 13

(iii) 3825:

We divide 3825 by its smallest prime factors sequentially:

3825 / 3 = 1275

1275 / 3 = 425

425 / 5 = 85

85 / 5 = 17

17 / 17 = 1

The prime factors of 3825 are: 3, 3, 5, 5, and 17.

Answer: 3825 = 3 x 3 x 5 x 5 x 17 = 3² x 5² x 17

(iv) 5005:

We divide 5005 by its smallest prime factors sequentially:

5005 / 5 = 1001

1001 / 7 = 143

143 / 11 = 13

13 / 13 = 1

The prime factors of 5005 are: 5, 7, 11, and 13.

Answer: 5005 = 5 x 7 x 11 x 13

(v)7429:

We divide 7429 by its smallest prime factors sequentially:

7429 / 17 = 437

437 / 19 = 23

23 / 23 = 1

The prime factors of 7429 are: 17, 19, and 23.

Answer: 7429 = 17 x 19 x 23

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = product of the twnumbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

(i) 26 and 91
Step 1: Find the prime factorization of both numbers.

· 26 = 2 x 13

· 91 = 7 x 13

Step 2: Find HCF and LCM.

· HCF (Product of the smallest power of each common prime factor) = 13

· LCM (Product of the highest power of each prime factor involved) = 2 x 7 x 13 = 182

Step 3: Verification

· LCM x HCF = 182 x 13 = 2366

· Product of the twnumbers = 26 x 91 = 2366

Since LCM x HCF = Product of the twnumbers (2366 = 2366), the relationship is verified.

(ii) 510 and 92
Step 1: Find the prime factorization of both numbers.

· 510 = 2 x 3 x 5 x 17

· 92 = 2 x 2 x 23 = 2² x 23

Step 2: Find HCF and LCM.

· HCF = 2

· LCM = 2² x 3 x 5 x 17 x 23 = 4 x 3 x 5 x 17 x 23 = 23460

Step 3: Verification

· LCM x HCF = 23460 x 2 = 46920

· Product of the twnumbers = 510 x 92 = 46920

Since LCM x HCF = Product of the twnumbers (46920 = 46920), the relationship is verified.

(iii) 336 and 54
Step 1: Find the prime factorization of both numbers.

· 336 = 2 x 2 x 2 x 2 x 3 x 7 = 2⁴ x 3 x 7

· 54 = 2 x 3 x 3 x 3 = 2 x 3³

Step 2: Find HCF and LCM.

· HCF = 2 x 3 = 6

· LCM = 2⁴ x 3³ x 7 = 16 x 27 x 7 = 3024

Step 3: Verification

· LCM x HCF = 3024 x 6 = 18144

· Product of the twnumbers = 336 x 54 = 18144

Since LCM x HCF = Product of the twnumbers (18144 = 18144), the relationship is verified.

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

(i) 12, 15 and 21
Step 1: Find the prime factorization of each number.

· 12 = 2 x 2 x 3 = 2² x 3

· 15 = 3 x 5

· 21 = 3 x 7

Step 2: Find HCF and LCM.

· HCF (Product of the smallest power of each common prime factor) The only common prime factor is 3. HCF = 3

· LCM (Product of the highest power of each prime factor involved) LCM = 2² x 3 x 5 x 7 = 4 x 3 x 5 x 7 = 420 LCM = 420

(ii) 17, 23 and 29
Step 1: Find the prime factorization of each number.

· 17 = 1 x 17 (Since 17 is a prime number)

· 23 = 1 x 23 (Since 23 is a prime number)

· 29 = 1 x 29 (Since 29 is a prime number)

Step 2: Find HCF and LCM.

· HCF Since there are ncommon prime factors, the only common factor is 1. HCF = 1

· LCM LCM = 17 x 23 x 29 = 11339 LCM = 11339

(iii) 8, 9 and 25
Step 1: Find the prime factorization of each number.

· 8 = 2 x 2 x 2 = 2³

· 9 = 3 x 3 = 3²

· 25 = 5 x 5 = 5²

Step 2: Find HCF and LCM.

· HCF There are ncommon prime factors among 8, 9, and 25. Therefore, the common factor is 1. HCF = 1

· LCM LCM = 2³ x 3² x 5² = 8 x 9 x 25 = 1800 LCM = 1800

Q4. Given that HCF(306, 657) = 9, find LCM(306, 657).

Solution: We know the formula relating HCF and LCM of twnumbers:

· HCF x LCM = Product of the twnumbers

Given:

· First number = 306

· Second number = 657

· HCF = 9

Substituting the values intthe formula:

· 9 x LCM = 306 x 657

· LCM = (306 x 657) / 9

Dividing 306 by 9 gives 34:

· LCM = 34 x 657

· LCM = 22338

Answer: LCM(306, 657) = 22338

SEBA Class 10 Maths Exercise 1.1 Solutions

Q5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution: If any number ends with the digit 0, it must be divisible by 10. This means its prime factorization must contain both 2 and 5 as prime factors.

Let’s look at the prime factorization of 6ⁿ:

· 6 = 2 x 3

· Therefore, 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ

The prime factors of 6ⁿ are only 2 and 3.

By the Fundamental Theorem of Arithmetic, this prime factorization is unique. This means that 5 cannot be a prime factor of 6ⁿ for any natural number n.

Since 5 is not a prime factor, 6ⁿ is not divisible by 10.

Answer: Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Q6. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Definition Note for Students: A composite number is a positive integer that has factors other than 1 and itself. According tthe Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of prime numbers.

Solution:
Part 1: For the expression 7 x 11 x 13 + 13
Step 1: Take out the common factor. We can see that 13 is common tboth terms. Let’s take 13 as a common factor:

· 7 x 11 x 13 + 13 = 13 x (7 x 11 + 1)

Step 2: Simplify the expression inside the bracket.

· = 13 x (77 + 1)

· = 13 x 78

Step 3: Find the prime factors of 78.

· 78 = 2 x 3 x 13

· So, the expression becomes: 13 x 2 x 3 x 13 = 2 x 3 x 13²

Conclusion: Since the given number can be expressed as a product of prime factors (2, 3, and 13), it has factors other than 1 and itself.

Answer: Therefore, 7 x 11 x 13 + 13 is a composite number.

Part 2: For the expression 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5
Step 1: Take out the common factor. We can see that 5 is common tboth terms. Let’s take 5 as a common factor:

· 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)

Step 2: Simplify the expression inside the bracket.

· = 5 x (1008 + 1)

· = 5 x 1009

(Note: 1009 is a prime number and cannot be factored further).

Conclusion: The given number can be expressed as a product of its prime factors (5 and 1009). This means it has factors other than 1 and itself.

Answer: Therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 is a composite number.

Q7. There is a circular path around a sports field. Sonia takes 18 minutes tdrive 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 gin the same direction. After how many minutes will they meet again at the starting point?

Solution:
Given: * Time taken by Sonia tcomplete 1 round = 18 minutes

Time taken by Ravi tcomplete 1 round = 12 minutes

They both start at the same point, at the same time, and move in the same direction.

TFind: * The number of minutes after which they will meet again at the starting point.

Solution: Tfind the time when they meet again at the starting point, we need tcalculate the Lowest Common Multiple (LCM) of 18 and 12.

First, find the prime factorization of both numbers:

18 = 2 x 3 x 3 = 2 x 3²

12 = 2 x 2 x 3 = 2² x 3

Now, take the product of the highest power of each prime factor involved tfind the LCM:

LCM(18, 12) = 2² x 3²

LCM(18, 12) = 4 x 9

LCM(18, 12) = 36

Answer: Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.

Q8 . (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.

Given: Number of soldiers in each row can be 15, 20, or 25.

TFind: The least number of soldiers in the regiment.

Solution: Tfind the least number of soldiers, we need tfind the Lowest Common Multiple (LCM) of 15, 20, and 25.

First, find the prime factorization of each number:

15 = 3 x 5

20 = 2 x 2 x 5 = 2² x 5

25 = 5 x 5 = 5²

Now, take the product of the highest power of each prime factor involved tfind the LCM:

LCM(15, 20, 25) = 2² x 3 x 5²

LCM(15, 20, 25) = 4 x 3 x 25

LCM(15, 20, 25) = 300

Answer: Therefore, 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 twbells ring simultaneously at an instant, then find after how many seconds will the bells ring simultaneously again.

Given: Time interval for the first bell = 18 seconds

Time interval for the second bell = 60 seconds

TFind: The number of seconds after which the bells will ring simultaneously again.

Solution: Tfind the time when they will ring together again, we need tfind the Lowest Common Multiple (LCM) of 18 and 60.

First, find the prime factorization of both numbers:

18 = 2 x 3 x 3 = 2 x 3²

60 = 2 x 2 x 3 x 5 = 2² x 3 x 5

Now, take the product of the highest power of each prime factor involved tfind the LCM:

LCM(18, 60) = 2² x 3² x 5

LCM(18, 60) = 4 x 9 x 5

LCM(18, 60) = 180

Answer: Therefore, the bells will ring simultaneously again after 180 seconds (or 3 minutes).

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

Given: First radistation plays the song every 2 days.

Second radistation plays the song every 3 days.

Total period of observation = 30 days.

TFind: The number of times both stations play the song on the same day within 30 days.

Solution: First, we need tfind out how often they play the song on the same day by finding the LCM of 2 and 3.

LCM(2, 3) = 2 x 3 = 6 days. So, both stations play the song together once every 6 days.

Now, tfind how many times this happens in 30 days, we divide the total days by the LCM:

Number of times = 30 / 6 = 5

Answer: Therefore, both radistations will play the same song on the same day 5 times in 30 days.

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

Given: Total number of army contingent members = 616

Total number of army band members = 32

TFind: The maximum number of columns in which they can march.

Solution: Tfind the maximum number of columns, we need tcalculate the Highest Common Factor (HCF) of 616 and 32.

Using prime factorization:

616 = 2 x 2 x 2 x 7 x 11 = 2³ x 7 x 11

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

Now, take the product of the lowest power of the common prime factor tfind the HCF:

Common prime factor is 2, and its lowest power is 2³.

HCF(616, 32) = 2³ = 8

Answer: Therefore, 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 tdisplay them in identical groups of Indian and International stamps with nstamp left out. What is greatest number of groups Himadri can display the stamps?

Given: Number of Indian postal stamps = 625

Number of International postal stamps = 325

TFind: The greatest number of identical groups Himadri can display.

Solution: Tfind the greatest number of identical groups, we need tfind the Highest Common Factor (HCF) of 625 and 325.

Using prime factorization:

625 = 5 x 5 x 5 x 5 = 5⁴

325 = 5 x 5 x 13 = 5² x 13

Now, take the product of the lowest power of the common prime factor tfind the HCF:

Common prime factor is 5, and its lowest power is 5².

HCF(625, 325) = 5² = 25

Answer: Therefore, the greatest number of groups Himadri can display the stamps is 25.

(vi) Twropes are of length 64 cm and 80 cm. Both are tbe cut intpieces of equal length. What should be the maximum length of the pieces?

Given: Length of the first rope = 64 cm

Length of the second rope = 80 cm

TFind: The maximum length of the pieces when both ropes are cut equally.

Solution: Tfind the maximum equal length of the pieces, we need tcalculate the Highest Common Factor (HCF) of 64 and 80.

Using prime factorization:

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

80 = 2 x 2 x 2 x 2 x 5 = 2⁴ x 5

Now, take the product of the lowest power of the common prime factor tfind the HCF:

Common prime factor is 2, and its lowest power is 2⁴.

HCF(64, 80) = 2⁴ = 16

Answer: Therefore, the maximum length of each piece should be 16 cm.

Q9 Find the greatest number of 4 digits which is exactly divisible by 18, 24 and 36.

Solution:
Given: The divisor numbers = 18, 24, and 36.

The target number must be the greatest 4-digit number.

TFind: The greatest 4-digit number that is exactly divisible by 18, 24, and 36.

Solution: Step 1: Find the LCM of 18, 24, and 36. Any number exactly divisible by 18, 24, and 36 must alsbe a multiple of their Lowest Common Multiple (LCM).

Using prime factorization:

18 = 2 x 3 x 3 = 2 x 3²

24 = 2 x 2 x 2 x 3 = 2³ x 3

36 = 2 x 2 x 3 x 3 = 2² x 3²

Taking the highest power of each prime factor involved:

LCM(18, 24, 36) = 2³ x 3²

LCM(18, 24, 36) = 8 x 9

LCM(18, 24, 36) = 72

Step 2: Find the greatest 4-digit number. The greatest 4-digit number is 9999.

Step 3: Divide the greatest 4-digit number by the LCM tfind the remainder. Let’s divide 9999 by 72:

9999 / 72 = 138 with a remainder.

Calculation: 72 x 138 = 9936

Remainder: 9999 – 9936 = 63

Step 4: Subtract the remainder from 9999 tget the exact multiple.

Required Number = 9999 – 63 = 9936

Answer: Therefore, the greatest 4-digit number exactly divisible by 18, 24, and 36 is 9936.

Class 10 Maths Chapter 1 Exercise 1.1 Question Answer SEBA 2026-27

Q10. 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 x q

(iv) p + 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:
Given: 1245 is a factor of both p and q.

This means we can write:

§ p = 1245x (where x is an integer)

§ q = 1245y (where y is an integer)

TFind: Which of the given expressions—(i) p + q, (ii) p – q, (iii) p x q, or (iv) p + q—will alshave 1245 as a factor.

Solution: Let’s test each expression by substituting the values of p and q:

For (i) and (iv) p + q:

p + q = 1245x + 1245y

p + q = 1245(x + y) Since 1245 is multiplied by an integer (x + y), 1245 is a factor of p + q.

For (ii) p – q:

p – q = 1245x – 1245y

p – q = 1245(x – y) Since 1245 is multiplied by an integer (x – y), 1245 is a factor of p – q.

For (iii) p x q:

p x q = (1245x) x (1245y)

p x q = 1245 x (1245xy) Since 1245 is multiplied by an integer (1245xy), 1245 is a factor of p x q.

Since all options (i), (ii), (iii), and (iv) have 1245 as a factor, the correct choice is (D).

Answer: (D) All (i), (ii), (iii) and (iv)

Q11. Match the columns:

Column I Column II
P) Number which is neither Prime nor Composite is 1) 18
Q) Only even prime number is 2) 3
R) HCF of 12, 15, 21 is 3) 2
S) LCM of 2 and 9 is 4) 1
Choose the correct option:

A) 4 3 2 1

B) 3 2 1 4

C) 2 4 3 1

D) 1 2 3 4

Solution:

Given: Twcolumns containing mathematical statements (Column I) and their corresponding numerical values (Column II).

Column I items: P, Q, R, S
Column II items: 1, 2, 3, 4
TFind: The correct matching sequence for P, Q, R, and S from the given multiple-choice options.

Let’s solve and match each item from Column I one by one:

For P: “Number which is neither Prime nor Composite is”
The number 1 fits this definition because prime numbers must be greater than 1.

Therefore, P matches with 4.

For Q: “Only even prime number is”
The number 2 is the smallest prime number and the only one that is even.

Therefore, Q matches with 3.

For R: “HCF of 12, 15, 21 is”
Let’s find the prime factorization:

12 = 2² x 3

15 = 3 x 5

21 = 3 x 7

The highest common factor is 3.

Therefore, R matches with 2.

For S: “LCM of 2 and 9 is”
Since 2 and 9 share ncommon factors other than 1, their LCM is their product: 2 x 9 = 18.

Therefore, S matches with 1.

Combining the matching numbers in order of P, Q, R, S gives the sequence: 4, 3, 2, 1. This corresponds toption A.

Answer: A) 4 3 2 1

Q12. Which of the following statement is true or false Statement

(P) HCF of twconsecutive natural number is 1 Statement

(Q) HCF of twcprime number is 1

Choose the correct option:

A) P is true, Q is false

B) P is false, Q is true

C) Both P & Q are true

D) Both P & Q are false

Solution:

Given: Two mathematical statements regarding the Highest Common Factor (HCF):

To Find: Whether Statement (P) and Statement (Q) are true or false, and choose the correct option.

Analysis of Statement (P): Consecutive natural numbers are numbers that follow each other in order without gaps (for example: 3 and 4, or 14 and 15). Since two consecutive natural numbers never share any common factors other than 1, their Highest Common Factor (HCF) is always 1.

Analysis of Statement (Q): By definition, co-prime numbers are a pair of numbers that have no common positive factor other than 1. Because they share no other common factors, their Highest Common Factor (HCF) is always 1.

Conclusion: Since both Statement (P) and Statement (Q) are correct, Option C is the right choice.

Answer: C) Both P & Q are true

Real Numbers Class 10 SEBA Ex 1.1 Solutions

Q13. Two positive integers P and Q can be expressed as P = ab² and Q = a²b, where a and b are prime numbers. The LCM of P and Q is A) a²b B) a²b² C) ab D) ab²

Solution:
Solution:

Given: Two positive integers expressed in terms of their prime factors: P = ab² Q = a²b Here, a and b are prime numbers.

To Find: The Lowest Common Multiple (LCM) of P and Q.

Analysis: To find the LCM of two expressions given in terms of their prime factors, take the highest power of each prime factor present across both expressions.

Powers of prime factor ‘a’:

Powers of prime factor ‘b’:

Multiplying the highest powers together gives: LCM(P, Q) = a² * b² = a²b²

Conclusion: This matches with Option B.

Answer: B) a²b²

Q14. 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 P Q R S A) P Q R S B) P R Q S C) Q P S R D) R S P Q

Solution:
Given: Four positive integers:

To Find: The correct increasing sequence based on the total number of prime factors each number has.

Solution: Find the prime factorization and count the number of distinct prime factors for each number:

For P = 119:

For Q = 462:

For R = 105:

For S = 2310:

Arranging the numbers in an increasing sequence based on their count of prime factors: 2 < 3 < 4 < 5 P < R < Q < S

Conclusion: The correct sequence is P R Q S, which matches Option B.

Answer: B) P R Q S

Q15. Assertion (A): 3 and 10 are co-prime numbers. Reason (R): Two numbers a and b are co-prime if they have no common factors other than 1.

Choose the correct option: A) Both (A) and (R) are true and Reason (R) is the 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

Solution:

Given:

To Find: Determine the correctness of Assertion (A) and Reason (R), and check if the reason correctly explains the assertion.

Checking Assertion (A):

Checking Reason (R):

Conclusion: Since 3 and 10 are co-prime precisely because they satisfy the definition provided in Reason (R), the reason is the correct explanation for the assertion. This matches Option A.

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

SEBA Class 10 Maths Chapter 1 Exercise 1.1 Textbook Answers

Q16. Observe the factor tree and answer the questions:

[ x ]
/ \
[3] [1275]
/ \
[3] [425]
/ \
[ y ] [85]
/ \
[5] [ z ]

(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

Solution:

Given: A factor tree showing the prime factor decomposition of a number x.

To Find: The individual values of missing variables x, y, and z, and the total value of the expression x + y + z.

Analysis: In a factor tree, a parent node is equal to the product of its two child branches.

  1. Finding the value of x: The top node x splits into 3 and 1275.
  1. Finding the value of y: The node 425 splits into y and 85.
  1. Finding the value of z: The node 85 splits into 5 and z.
  1. Finding the value of x + y + z: Substitute the calculated values:

Answer: (i) (b) 3825 (ii) (a) 5 (iii) (b) 17 (iv) (b) 3847

Class 10 Maths Exercise 1.1 Prime Factorization HCF LCM

Q17. An inter school seminar being conducted by an NGrelated teducation, 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 tbe seated and all of them being from the same subject. Find the maximum number of participants that can be accommodated in each room.

Given: Number of Science participants = 60

Number of English participants = 84

Number of Mathematics participants = 108

To Find: The maximum number of participants per room.

The minimum number of rooms required.

The LCM of 60, 84, and 108.

The product of HCF and LCM of these three numbers.

Solution: First, let’s find the prime factorization of the three numbers:

60 = 2 x 2 x 3 x 5 = 2² x 3 x 5

84 = 2 x 2 x 3 x 7 = 2² x 3 x 7

108 = 2 x 2 x 3 x 3 x 3 = 2² x 3³

Maximum number of participants in each room Tfind the maximum number of participants that can sit in each room under the given conditions, we must find the Highest Common Factor (HCF) of 60, 84, and 108.

Take the product of the lowest power of each common prime factor:

Common factors are 2 and 3. The lowest powers are 2² and 3¹.

HCF(60, 84, 108) = 2² x 3 = 4 x 3 = 12

Answer: The maximum number of participants in each room is 12.

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

Minimum number of rooms required Tuse the minimum number of rooms, each room must hold the maximum possible number of participants (which is 12).

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

Minimum rooms needed = Total participants / Participants per room

Minimum rooms needed = 252 / 12 = 21

Answer The minimum number of rooms required is 21.

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

LCM of 60, 84 and 108 Tfind the LCM, we take the product of the highest power of each prime factor involved:

Highest power of 2 = 2²

Highest power of 3 = 3³

Highest power of 5 = 5¹

Highest power of 7 = 7¹

LCM(60, 84, 108) = 2² x 3³ x 5 x 7

LCM(60, 84, 108) = 4 x 27 x 5 x 7

LCM(60, 84, 108) = 3780

AnswerThe LCM of 60, 84, and 108 is 3780.

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

Product of HCF and LCM

We have HCF = 12 and LCM = 3780.

Product = HCF x LCM

Product = 12 x 3780

Product = 45360

Answer: The product of the HCF and LCM is 45360.

📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 10 Mathematics Chapter 1 (Real Numbers) Exercise 1.1 Textual Solutions updated for the 2026–27 academic session. All prime factorization steps, Fundamental Theorem of Arithmetic applications, and HCF-LCM verification formulas strictly follow the latest revised Assam Board (SEBA/ASSEB) curriculum.

💡 Master Class 10 Maths Concepts:

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