Class 10 Mathematics Chapter 1 - Real Numbers Exercise 1.2

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

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

Solution:

(i) 140:

To find the prime factorization of 140, we can use either trial division or prime decomposition (factor tree). Let’s use prime decomposition here: Factor tree for 140:


      140
     /    \
    2      70
         /    \
        2      35
             /    \
            5      7

Prime factorization of 140:  2 × 2 × 5 × 7 .

(ii) 156:

Factor tree for 156:

      156
     /    \
    2      78
         /    \
        2      39
             /    \
            3      13

Prime factorization of 156:
2 × 2 × 3 × 13.

(iii) 3825:

Factor tree for 3825:

      3825
     /    \
    3      1275
         /    \
        3      425
             /    \
            5      85
                 /    \
                5      17

Prime factorization of 3825:  3 × 3 × 5 × 5 × 17

(iv) 5005:

Factor tree for 5005:

      5005
     /    \
   5      1001
         /    \
        7      143
             /    \
            11     13

Prime factorization of 5005:  5 × 7 × 11 × 13

(v) 7429:
To find the prime factorization of 7429, we can directly check if any smaller prime numbers divide it. In this case, we find that 7429 is a prime number, as it is not divisible by any other prime numbers up to its square root.

Prime factorization of 7429: 7429 itself (since it is a prime number).

2). Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Solution:

(i) 26 and 91

Step 1: Finding LCM and HCF

Prime Factorization Method:

26 = 2 × 13

91 = 7 × 13

LCM: The LCM is the product of all the unique prime factors, each taken with the highest power it appears in any of the numbers.

LCM(26, 91) = 2 × 7 × 13 = 182

HCF: The HCF is the product of all the common prime factors, each taken with the lowest power it appears in any of the numbers.

HCF(26, 91) = 13

Step 2: Verification Now, let’s check if LCM × HCF equals the product of the two numbers.

Product of the two numbers: 26 × 91 = 2366

LCM × HCF = 182 × 13 = 2366

Which verifies the property: LCM × HCF = Product of the two numbers.

(ii) 510 and 92

Step 1: Finding LCM and HCF

510 = 2 × 3 × 5 × 17

92 = 22 × 23

LCM: (510, 92) = 22 × 3 × 5 × 17 × 23 = 23460

HCF: (510, 92) = 2 (since 2 is the only common prime factor)

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

Again, we can observe that LCM × HCF is equal to the product of the two numbers, verifying the property.

(iii) 336 and 54

Step 1: Finding LCM and HCF

336 = 24 × 3 × 7

54 = 2 × 33

LCM: (336, 54) = 24 × 33 × 7 = 3024

HCF: (336, 54) = 2 × 3 = 6

Step 2: Verification Product of the two numbers: 336 × 54 = 18144 LCM × HCF = 3024 × 6 = 18144

Once again, we find that LCM × HCF equals the product of the two numbers, confirming the property.

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

Solution:

(i) 12, 15, and 21:
Prime factorization of each number:

12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3 (since 3 is the only common prime factor)

LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23, and 29:
As all the given numbers are prime numbers, they do not have any common factors other than 1. Therefore, the HCF will be 1.

LCM is simply the product of all three numbers since they are all prime and not divisible by each other:
LCM = 17 × 23 × 29 = 11339

(iii) 8, 9, and 25:
Prime factorization of each number:

8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
HCF = 1 (no common prime factors)

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 900

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

Solution:

We know :  LCM(a, b) × HCF(a, b) = a × b
So,  HCF(306, 657) = 9
Therefore, LCM(306, 657) × 9 = (306 × 657)
LCM(306, 657) = (306 × 657) / 9
LCM(306, 657) = 201042 / 9
LCM(306, 657) = 22338

Therefore, the LCM of 306 and 657 is 22338.

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

Solution:

We know any number with power n can end with the digit zero if it has atleast 2 and 5 as its prime factors.
Prime factors of 6 = 2 and 3
Since 5 is not a prime factor of 6, thus 6n cannot end with the digit 0.

6). 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

Solution:

We know a number is composite when it has factors other than 1 and itself.

For the first expression,

 7 x 11 x 13 + 13

⇒ 13(7 x 11 + 1)

Since 7 x 11 + 1  is not a prime number, we can conclude that 7 x 11 x 13 + 13 is a composite number.

For the second expression,

 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

⇒ 5(7 x 6 x 4 x 3 x 2 x 1 + 1)

Since 7 x 6 x 4 x 3 x 2 x 1 + 1 is not a prime number, we can conclude that 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 is also a composite number.

Therefore, both 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

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:

Solution:

To find out after minutes after which Sonia and Ravi will meet again at the starting point, we need to find LCM of the times it takes for them to complete one round of the field.

Time taken by Sonia to complete one round = 18 minutes 

Time taken by Ravi to complete one round = 12 minutes 

Prime factorization of 18:

18 = 2 × 3 × 3

Prime factorization of 12:

12 = 2 × 2 × 3

So, LCM = 22 × 32 = 36

Therefore, the LCM of 18 and 12 is 36 minutes.

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

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