10th Maths Ex 1.1 NCERT Solution

Question 1:

Express each number as product of its prime factors:

1. 140
2. 156
3. 3825
4. 5005
5. 7429

(i) 140140 = 2 × 2 × 5 × 7 140 = 2² × 5 × 7

(ii) 156156 = 2 × 2 × 3 × 13 156 = 2² × 3 × 13

(iii) 38253825 = 3² × 5² × 17 3825 = 3² × 5² × 17

(iv) 50055005 = 5 × 7 × 11 × 13 5005 = 5 × 7 × 11 × 13

(v) 74297429 = 17 × 19 × 23 7429 = 17 × 19 × 23

Question 2:

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

1. 26 and 91
2. 510 and 92
3. 336 and 54

(i) Numbers: 26 and 91 26 = 2 × 13,
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182

LCM × HCF = 182 × 13 = 2366
26 × 91 = 2366 ✅

(ii) Numbers: 510 and 92 510 = 2 × 3 × 5 × 17,
92 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460

LCM × HCF = 23460 × 2 = 46920
510 × 92 = 46920 ✅

(iii) Numbers: 336 and 54 336 = 2⁴ × 3 × 7,
54 = 2 × 3³
HCF = 6
LCM = 2⁴ × 3³ × 7 = 3024

LCM × HCF = 3024 × 6 = 18144
336 × 54 = 18144 ✅

Question 3:

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

1. 12, 15 and 21
2. 17, 23 and 29
3. 8, 9 and 25

(i) Numbers: 12, 15, 21 12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420

(ii) Numbers: 17, 23, 29 All are primes. No common factors
HCF = 1
LCM = 17 × 23 × 29 = 11339

(iii) Numbers: 8, 9, 25 8 = 2³
9 = 3²
25 = 5²
HCF = 1
LCM = 8 × 9 × 25 = 1800

Question 4:

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

Given: HCF = 9
Product = 306 × 657 = 201042
LCM = 201042 ÷ 9 = 22338
Therefore, LCM(306, 657) = 22338

Question 5:

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

To check: Can 6ⁿ end in 0 for any natural number n?

Observation: A number ending in 0 must be divisible by 10.
That means it must be divisible by both 2 and 5.

Step 1: Analyze 6ⁿ
6 = 2 × 3
So, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
This means 6ⁿ is divisible by 2 and 3 for any n.

Step 2: Is 6ⁿ divisible by 5?
There is no factor of 5 in 6.
So 6ⁿ is not divisible by 5 for any natural number n.

Conclusion:
Since 6ⁿ is not divisible by 5, it cannot be divisible by 10.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Question 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?

We are given:
Sonia’s time for one round = 18 minutes
Ravi’s time for one round = 12 minutes

To find: The time after which both will be at the starting point together again.
This is the LCM (Least Common Multiple) of 18 and 12.

Step 1: Prime factorization
18 = 2 × 3²
12 = 2² × 3

Step 2: Take the highest powers of all prime factors
LCM = 2² × 3² = 4 × 9 = 36

Answer: They will meet again at the starting point after 36 minutes.