SEBA Class 10 Maths Chapter 1 Exercise 1.1 Solutions – Real Numbers

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

(i) 140
Answer: Given: 140
\(140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7\)

(ii) 156
Answer: Given: 156
\(156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13\)

(iii) 3825
Answer: Given: 3825
\(3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17\)

(iv) 5005
Answer: Given: 5005
\(5005 = 5 \times 7 \times 11 \times 13\)

(v) 7429
Answer: Given: 7429
\(7429 = 17 \times 19 \times 23\)

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers:

(i) 26 and 91
Solution:
We have:
\[ 26 = 2 \times 13,\quad 91 = 7 \times 13 \] Here, 13 is the common prime factor.
So,
\[ \text{HCF}(26, 91) = 13 \] Taking highest powers of all prime factors:
\[ \text{LCM}(26, 91) = 2 \times 7 \times 13 = 182 \] Verification:
\[ LCM \times HCF = 182 \times 13 = 2366 \] \[ 26 \times 91 = 2366 \] Hence verified.

(ii) 510 and 92
Solution:
We have:
\[ 510 = 2 \times 3 \times 5 \times 17,\quad 92 = 2^2 \times 23 \] Common factor = 2
\[ \text{HCF}(510, 92) = 2 \] Taking greatest powers:
\[ \text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \] Verification:
\[ 23460 \times 2 = 510 \times 92 \] Hence verified.

(iii) 336 and 54
Solution:
We have:
\[ 336 = 2^4 \times 3 \times 7,\quad 54 = 2 \times 3^3 \] Common factors with smallest powers:
\[ \text{HCF}(336, 54) = 2 \times 3 = 6 \] Taking greatest powers:
\[ \text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 3024 \] Verification:
\[ 3024 \times 6 = 336 \times 54 \] Hence verified.

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method:

(i) 12, 15 and 21
Solution:
We have:
\[ 12 = 2^2 \times 3,\quad 15 = 3 \times 5,\quad 21 = 3 \times 7 \] Common factor:
\[ \text{HCF} = 3 \] Taking greatest powers:
\[ \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420 \]

(ii) 17, 23 and 29
Solution:
We have:
\[ 17, 23, 29 \text{ are prime numbers} \] So,
\[ \text{HCF} = 1 \] \[ \text{LCM} = 17 \times 23 \times 29 = 11339 \]

(iii) 8, 9 and 25
Solution:
We have:
\[ 8 = 2^3,\quad 9 = 3^2,\quad 25 = 5^2 \] No common factor:
\[ \text{HCF} = 1 \] Taking greatest powers:
\[ \text{LCM} = 2^3 \times 3^2 \times 5^2 = 1800 \]

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

Answer: Given: HCF = 9, numbers = 306, 657
We know, \[ LCM \times HCF = \text{Product of numbers} \] \[ LCM = \frac{306 \times 657}{9} \] First find the product: \[ 306 \times 657 = 201042 \] Now divide: \[ LCM = \frac{201042}{9} = 22338 \]

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

Solution:
We know that a number ending with 0 must be divisible by 10.
That means it must contain factors 2 and 5.

Now,
\[ 6^n = (2 \times 3)^n = 2^n \times 3^n \] It contains factors 2 and 3 only, but no factor 5.

So, it cannot be divisible by 10.

Final Answer: \(6^n\) cannot end with digit 0 for any natural number \(n\).

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

Answer:
(i) \(7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1)\)
So, it is divisible by 13 → composite

(ii) \(7! + 5 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)
\(= 5(remaining terms + 1)\)
So divisible by 5 → composite

Q7. There is a circular path around a sports field. Sonu takes 18 minutes, Ravi takes 12 minutes, and Sona takes 18 minutes to complete one round. If they start together, after how many minutes will they meet again?

Answer:
Time = LCM of 18, 12, 18
\[ LCM = 36 \]
Final Answer: They will meet again after 36 minutes.

Q8. Solve the following:

(i) The soldiers in a regiment can be stood in some rows consisting of 15, 20 or 25 soldiers in each row. Find the least number of soldiers in the regiment.
Answer:
Given: Number of soldiers per row = 15, 20, 25

We need to find the smallest number which is divisible by all three numbers.
So, we find LCM of 15, 20 and 25.

Prime factorisation:
\[ 15 = 3 \times 5,\quad 20 = 2^2 \times 5,\quad 25 = 5^2 \] Taking highest powers:
\[ LCM = 2^2 \times 3 \times 5^2 = 300 \] Final Answer: The least number of soldiers = 300

(ii) All bells ring 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 they will ring simultaneously again.
Answer:
Given: Time intervals = 18 seconds and 60 seconds

To find when they ring together again, we calculate LCM of 18 and 60.

Prime factorisation:
\[ 18 = 2 \times 3^2,\quad 60 = 2^2 \times 3 \times 5 \] Taking highest powers:
\[ LCM = 2^2 \times 3^2 \times 5 = 180 \] Final Answer: They will ring together again after 180 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 stations play the same song on the same day?
Answer:
Given: First station = every 2 days, Second station = every 3 days

To find when both play together, we find LCM of 2 and 3.
\[ LCM = 6 \] So, both stations play the song together every 6 days.

Now, in 30 days:
\[ 30 \div 6 = 5 \] Final Answer: They will play together 5 times in 30 days.

Q8 (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?

Answer:
Given: Army = 616 members, Band = 32 members

They must march in same number of columns.
So, we find HCF of 616 and 32.

Prime factorisation:
\[ 616 = 2^3 \times 7 \times 11,\quad 32 = 2^5 \] Common factor:
\[ HCF = 2^3 = 8 \] Final Answer: Maximum number of columns = 8

Q8 (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 stamps left out. What is the greatest number of groups Himadri can display the stamps?

Answer:
Given: Indian stamps = 625, International stamps = 325

We need maximum equal groups → find HCF of 625 and 325.

Prime factorisation:
\[ 625 = 5^4,\quad 325 = 5^2 \times 13 \] Common factor:
\[ HCF = 5^2 = 25 \] Final Answer: Maximum number of groups = 25

Q8 (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 each piece?

Answer:
Given: Lengths = 64 cm and 80 cm

We need largest possible equal length → find HCF.

Prime factorisation:
\[ 64 = 2^6,\quad 80 = 2^4 \times 5 \] Common factor:
\[ HCF = 2^4 = 16 \] Final Answer: Maximum length of each piece = 16 cm

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

Answer:
Given: Numbers = 18, 24, 36

First find LCM:
\[ 18 = 2 \times 3^2,\quad 24 = 2^3 \times 3,\quad 36 = 2^2 \times 3^2 \] \[ LCM = 2^3 \times 3^2 = 72 \] Largest 4-digit number = 9999

Now divide:
\[ 9999 \div 72 = 138 \text{ remainder } 63 \] So required number:
\[ 9999 - 63 = 9936 \] Final Answer: 9936

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 × q
(iv) p ÷ q

Choose the correct answer:

(A) only (iii)
(B) only (ii) and (ii)
(C) only (i), (ii) and (iii)
(D) All (i), (ii), (iii) and (iv)

Answer: (B) only (ii) and (iii)

Q11. Match the columns

Choose the correct option:

A) P-4, Q-3, R-2, S-1
B) P-3, Q-2, R-1, S-4
C) P-2, Q-4, R-3, S-1
D) P-1, Q-2, R-3, S-4

Answer: (D) P-1, Q-2, R-3, S-4

Q12. Which of the following statement is true or false

Statement (P): HCF of two consecutive natural numbers is 1
Statement (Q): HCF of two co-prime numbers 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

Answer: (C) Both P & Q are true

Q13. Two positive integers P and Q can be expressed as \(P = ab^2\) and \(Q = a^2b\), where a and b are prime numbers. The LCM of P and Q is

A) \(a^2b^2\)
B) \(a^2b^2\)
C) \(ab\)
D) \(ab^2\)

Answer: (A) \(a^2b^2\)

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

A) P, Q, R, S
B) P, R, Q, S
C) Q, P, S, R
D) R, S, P, Q

Answer: (B) P, R, Q, S

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

Choose the correct option:

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 correct explanation
C) Assertion (A) is true but Reason (R) is false
D) Assertion (A) is false but Reason (R) is true

Answer: (A)

Q16. Observe the factor tree and answer the questions:

(i) The value of x is
Answer: (d) 3825

(ii) The value of y is
Answer: (b) 25

(iii) The value of z is
Answer: (b) 17

(iv) The value of x + y + z is
Answer: (b) 3847

Q17. A inter school seminar 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.
Answer:
Given: Science = 60, English = 84, Mathematics = 108

We need maximum equal number → find HCF of 60, 84 and 108.

Prime factorisation:
\[ 60 = 2^2 \times 3 \times 5 \] \[ 84 = 2^2 \times 3 \times 7 \] \[ 108 = 2^2 \times 3^3 \] Common factors with smallest powers:
\[ HCF = 2^2 \times 3 = 12 \] Final Answer: Maximum participants in each room = 12

(ii) What is the minimum number of rooms required for the event?
Answer:
Each room has 12 participants.

Rooms required:
\[ \frac{60}{12} = 5,\quad \frac{84}{12} = 7,\quad \frac{108}{12} = 9 \] Total rooms:
\[ 5 + 7 + 9 = 21 \] Final Answer: Minimum number of rooms = 21

(iii) Find the LCM of 60, 84 and 108.
Answer:
Using prime factors:
\[ 60 = 2^2 \times 3 \times 5 \] \[ 84 = 2^2 \times 3 \times 7 \] \[ 108 = 2^2 \times 3^3 \] Taking greatest powers:
\[ LCM = 2^2 \times 3^3 \times 5 \times 7 \] \[ = 4 \times 27 \times 5 \times 7 = 3780 \] Final Answer: LCM = 3780

(iv) Find the product of HCF and LCM of 60, 84 and 108.
Answer:
\[ HCF = 12,\quad LCM = 3780 \] \[ \text{Product} = 12 \times 3780 = 45360 \] Final Answer: Product = 45360