Exercise 1.1 – Real Numbers NCERT Solutions – Class 10 Maths

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

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

 

(i) 140

Answer:

Start with the smallest prime number that divides 140.
140 ÷ 2 = 70
70 ÷ 2 = 35
35 ÷ 5 = 7
7 is a prime number.

Therefore, the prime factorisation of 140 is:
👉 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156

Answer:

Begin dividing 156 by the smallest prime.
156 ÷ 2 = 78
78 ÷ 2 = 39
39 ÷ 3 = 13
13 is a prime number.

Therefore, the prime factorisation of 156 is:
👉 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825

Answer:

Start dividing 3825 by 3 (smallest odd prime):
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 is a prime number.

Therefore, the prime factorisation of 3825 is:
👉 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005

Answer:

Begin with the smallest prime factor:
5005 ÷ 5 = 1001
Now factor 1001:
1001 ÷ 7 = 143
143 ÷ 11 = 13
13 is prime.

Therefore, the prime factorisation of 5005 is:
👉 5005 = 5 × 7 × 11 × 13

(v) 7429

Answer:

7429 is not divisible by 2, 3, 5, or 7.

Let’s test higher primes:
7429 ÷ 17 = 437
437 ÷ 19 = 23
23 is a prime number.

Therefore, the prime factorisation of 7429 is:
👉 7429 = 17 × 19 × 23

2. 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                   (ii) 510 and 92                            (iii) 336 and 54

 

(i) 26 and 91

Answer:

Step 1: Prime factorisation

• 26 = 2 × 13

• 91 = 7 × 13

Step 2: HCF (Highest Common Factor)
Common factor = 13
👉 HCF = 13

Step 3: LCM (Least Common Multiple)
LCM is the product of all prime factors without repetition:
LCM = 2 × 7 × 13 = 182

Step 4: Verification
LCM × HCF = 182 × 13 = 2366
Product of numbers = 26 × 91 = 2366

(ii) 510 and 92

Answer:

Step 1: Prime factorisation

• 510 = 2 × 3 × 5 × 17

• 92 = 2 × 2 × 23 = 2² × 23

Step 2: HCF
Common prime factor = 2
👉 HCF = 2

Step 3: LCM
Take all prime factors involved:
LCM = 2² × 3 × 5 × 17 × 23 = 23460

Step 4: Verification
LCM × HCF = 23460 × 2 = 46920
Product of numbers = 510 × 92 = 46920

(iii) 336 and 54

Answer:

Step 1: Prime factorisation

• 336 = 2⁴ × 3 × 7

• 54 = 2 × 3³

Step 2: HCF
Common prime factors: 2¹ × 3¹ = 6

Step 3: LCM
Take the highest powers of all primes:
LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024

Step 4: Verification
LCM × HCF = 3024 × 6 = 18144
Product = 336 × 54 = 18144

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

(i) 12, 15 and 21            (ii) 17, 23 and 29                         (iii) 8, 9 and 25

 

(i) 12, 15 and 21

Answer:

Step 1: Prime Factorisation

• 12 = 2² × 3

• 15 = 3 × 5

• 21 = 3 × 7

Step 2: HCF (Highest Common Factor)
Look for common prime factors in all three:
✅ Only 3 is common.
👉 HCF = 3

Step 3: LCM (Least Common Multiple)
Take the highest power of each prime number appearing in any number:

• 2² (from 12)

• 3 (common in all)

• 5 (from 15)

• 7 (from 21)

👉 LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Answer:

Step 1: Prime Factorisation

All three numbers are prime:

• 17 = 17

• 23 = 23

• 29 = 29

Step 2: HCF
No common factors other than 1.
👉 HCF = 1

Step 3: LCM
Since all are distinct primes, LCM is the product of all:

👉 LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Answer:

Step 1: Prime Factorisation

• 8 = 2³

• 9 = 3²

• 25 = 5²

Step 2: HCF
No common prime factor.
👉 HCF = 1

Step 3: LCM
Take all prime factors with their highest powers:

• 2³ (from 8)

• 3² (from 9)

• 5² (from 25)

👉 LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800

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

 

Answer:

We are given:

• HCF (306, 657) = 9

• We need to find LCM (306, 657)

Step-by-Step Solution Using the Identity:

               LCM × HCF = Product of the two numbers

Step 1: Write the identity

LCM × HCF = 306 × 657

We are given HCF = 9, so:

​

Step 2: Multiply and divide

First, multiply:

306 × 657 = 201042

Now divide:

 

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

Answer:

Yes, 6n can end with the digit 0,
but only when n is a multiple of 5.

Because:

• 6 = 2 × 3 (has no 5),

• And a number ends in 0 only if it has both 2 and 5 as factors (i.e., divisible by 10),

• So n must contain 5 for 6n to end in 0.

 

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

 

Answer:

(i) Check: 7 × 11 × 13 + 13

Let’s simplify:

=(7 × 11 × 13) + 13

First, multiply:

• 7 × 11 = 77

• 77 × 13 = 1001

So:

1001 + 13 = 1014

Now check:

1014 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78

So, 1014 = 13 × 78, which means it has factors other than 1 and itself.

Therefore, 1014 is a composite number.

(ii) Check: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

That is:

= 7! + 5 = 5040 + 5 = 5045

Now check divisibility:

Let’s try dividing 5045 by 5:

• A number ending in 0 or 5 is divisible by 5.

• 5045 ÷ 5 = 1009

So:

5045 = 5 × 1009

This means 5045 has more than 2 factors.

Therefore, 5045 is also 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?

 

Answer:

This is a Least Common Multiple (LCM) problem.

We are asked after how many minutes both Sonia and Ravi will be at the starting point together again.

So we need to find the LCM of 18 and 12.

Sonia and Ravi will meet again at the starting point after 36 minutes.