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# Exercise 1.2 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

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## Exercise 1.2 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

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### You can Find Mathematics solution Class 10 Chapter 1. Real Numbers

- All Chapter review quick revision notes for chapter 1. Real Numbers Class 10
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- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 1.2 class 10 Mathematics Chapter 1. Real Numbers

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## 1. Real Numbers

### | Exercise 1.2 |

## Exercise 1.2 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

******EXERCISE 1.2 Chapter ****1. Real Numbers class 10 Maths **

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

**(i) 140 **

**Solution:**

= 2^{2} × 5 × 7

**(ii) 156**

**Solution:**

= 2^{2} × 3 × 13

**(iii) 3825**

**Solution:**

= 3^{2} × 5^{2} × 17

**(iv) 5005**

**Solution:**

= 5 × 7 × 11 × 13

**(v) 7429**

**Solution:**

**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:**

26 = 2 × 13

91 = 7 × 13

Common factors = 13

∴ HCF = 13

LCM = 2 × 7 × 13 = 182

Now varification,

product of the two numbers = LCM × HCF

N_{1} × N_{2} = LCM × HCF

26 × 91 = 13 × 182

2366 = 2366

Hence Varified,

**(ii) 510 and 92**

**Solution:**

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

Common factors = 2

∴ HCF = 2

LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Now varification,

product of the two numbers = LCM × HCF

N_{1} × N_{2} = LCM × HCF

510 × 92 = 2 × 23460

46920 = 46920

Hence Varified,

**(iii) 336 and 54**

**Solution:**

336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

Common factors = 2 × 3

∴ HCF = 6

LCM = 2 × 2 × 2× 2 × 3 × 3 × 3 × 7 = 3024

Now varification,

product of the two numbers = LCM × HCF

N_{1} × N_{2} = LCM × HCF

336 × 54 = 6 × 3024

18144 = 18144

Hence Varified,

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

**(i) 12, 15 and 21 **

**Solution:**

12 = 2 × 2 × 3

15 = 5 × 3

21 = 7 × 3

Common Factors = 3

HCF = 3

LCM = 3 × 2 × 2 × 5 × 7 = 420

**(ii) 17, 23 and 29 **

**Solution:**

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

**(iii) 8, 9 and 25**

**Solution:**

8 = 2 × 2 × 2

9 = 3 × 3

25 = 5 × 5

There is no common factor except 1.

∴ HCF = 1

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5

= 8 × 9 × 25

= 1800

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

**Solution:**

HCF (306, 657) = 9

LCM × HCF = N_{1} × N_{2}

LCM = 22338

**Q5. Check whether 6 ^{n} can end with the digit 0 for any natural number n.**

**Solution:**

Prime factorisation of 6^{n} = (2 × 3 )^{n}

While, Any natural number which end with digit 0 has

the prime factorisation as form of (2 × 5 )^{n}

Therefore, 6^{n} will not end with digit 0.

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

**Solution:**

Let A = 7 × 11 × 13 + 13

= 13 (7 × 11 + 1)

= 13 (77 + 1)

= 13 × 78

Hence this is composite number because It has at least one positive divisor other than one.

Similarily,

Let B = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 × 1009

Hence this is also a composite number because It has at least one positive divisor other than one.

**Q7. 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: **

Sonia takes 18 minutes in one round.

Ravi takes 12 minutes in one round

they will meet again at the starting point after LCM(18, 12) minutes

18 = 2 × 3 × 3

12 = 2 × 2 × 3

HCF = 2 × 3 = 6

= 36 minutes

##### Other Pages of this Chapter: 1. Real Numbers

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