# UP Board Solutions for Class 10 Maths Chapter 1 Exercise 1.2 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

Sol.

(i) We use the division method as shown below :

: – 140 = 2 × 2 × 5 x 7
=x5x7

### (ii) We use the division method as shown below:

: – 156 = 2 x 2 × 3 × 13
= x 3 × 13

### (iii) We use the division method as shown below:

=> 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 x 17

## (iv) We use the division method as shown below:

=> 5005 = 5 × 7 x 11 × 13

### (V) We use the division method as shown below:

=> 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) 28 and 91 (ii) 510 and 92 (iii) 336 and 51

Sol.(i) 26 and 91

26 = 2 × 13                     and                          91 = 7 × 13

=>     LCM of 26 and 91 = 2 × 7 × 13 =182

And     HCF of 26 and 91 = 13
Now,                   182 x 13 = 2366 and 26 x 91 = 2366
182 x 13 = 26 × 91
Hence verified.

## (ii) 510 and 92

510 = 2 × 3 × 5 × 17          and               92 = 2 × 2 × 23

=> LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
and HCF of 510 and 92 = 2
Now, 23460 x 2 = 46920 and 510 x 92 = 46920
=>               23460 × 2 = 510 × 92
Hence verified.

### (iii) 336 and 54

336 = 2×2×2×2×3×7
and                                54 = 2 × 3 × 3 × 3
..         LCM of 336 and 54 = 2×2 x2x2x3x3x3x7
= 3024
and HCF of 336 and 54 = 2 × 3 = 6
Now,                    3024 × 6 = 18144
and                      336 x 54 = 18144
3024 × 6 = 336 × 54
Hence verified.

## 3. 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

Sol.

(i) First we write the prime factorisation of each of the given numbers.

12 = 2x2x3 = 3×22x3, 15=3×5
and.           21 = 3 x7
LCM =22x 3 × 5 x 7 = 420
and,        HCF = 3

(ii) First we write the prime factorisation of each of the given numbers.

17 = 17, 23 = 23 and 29 = 29
LCM = 17 x 23 x 29 = 11339
and,       HCF = 1

(iii) First we write the prime factorisation of each of the given numbers.

8 = 2×2 x2 =22 , 9 = 3 × 3 = 32,
25 = 5 × 5 =52
LCM = 23 x 32 x 52 = 8 × 9 × 25 = 1800

and, HCF = 1

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

Sol. We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.

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

Sol. If the number 6n , for any n, ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. This is not possible as the only prime in the factorisation of 6n is 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6n .

### 6. Explain why 7 x 11 x 18 + 18 and 7 × 6 x 5 x 4 x8 × 2 x 1 + 5 are composite numbers.

Sol. Since,                         7 × 11 × 13 + 13 = 13 × (7 × 11 x 1 + 1)
= 13 x (77 + 1)
= 13 × 78
= It is a composite number.
Again, 7×6×5×4×3×1×1+5
=5х (7×6×4×3x1x1+ 1)
It is 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 again at the starting point ?

Sol. To final the LCM of 18 and 12, we have

18 =2 × 3 x 3       and                          12 = 2 x2 x 3
LCM of 18 and 12 = 2 x 2 × 3 × 3 = 36 So, Sonia and Ravi will meet again at the starting point after 36 minutes.