NCERT Solutions Class 10 Chapter 1 Real Numbers Exercise 1.1 (original) (raw)

Last Updated : 11 Sep, 2024

Chapter 1 of Class 10 Mathematics in the NCERT textbook deals with the Real Numbers an essential concept in algebra. This chapter introduces students to the various types of numbers that form the foundation of mathematical operations and calculations. Exercise 1.1 focuses on helping the students understand and apply these concepts through a series of problems.

Real Numbers

The Real Numbers encompass all the numbers that can be found on the number line. This includes:

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

****(i) 140**

****(ii) 156**

****(iii) 3825**

****(iv) 5005**

****(v) 7429**

**Solutions:

****(i) 140**

Taking the LCM of 140,

140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7

Therefore, the product of the prime factors is 22 × 5 × 7

****(ii)** **156

Taking the LCM of 156,

156 = 2 × 2 × 3 × 13 × 1 = 22 × 3 × 13

Therefore, the product of the prime factors is 22 × 3 × 13

****(iii) 3825**

Taking the LCM of 3825,

3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 × 17

Therefore, the product of the prime factors is 32 × 52 × 17

****(iv)** **5005

Taking the LCM of 5005,

5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

Therefore, the product of the prime factors is 5 × 7 × 11 × 13

****(v) 7429**

Taking the LCM of 7429,

7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

Therefore, the product of the prime factors is 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.

****(i) 26 and 91**

****(ii) 510 and 92**

****(iii) 336 and 54**

**Solutions:

****(i)** **26 and 91

Taking the LCM of 26,

26 = 2 × 13 × 1

Taking the LCM of 91,

91 = 7 × 13 × 1

Therefore, LCM of 26 and 91 together = 2 × 7 × 13 × 1 = 182

HCF of 26 and 91 = 13

Now, the product of 26 and 91 = 26 × 91 = 2366

And the product of LCM and HCF = 182 × 13 = 2366

Therefore, LCM × HCF = product of the 26 and 91.

****(ii)** **510 and 92

Taking the LCM of 510,

510 = 2 × 3 × 17 × 5 × 1

Taking the LCM of 92,

92 = 2 × 2 × 23 × 1

Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

HCF of 510 and 92 = 2

Now, the product of 510 and 92 = 510 × 92 = 46920

And the product of LCM and HCF = 23460 × 2 = 46920

Therefore, LCM × HCF = product of the 510 and 92.

****(iii)** **336 and 54

Taking the LCM of 336,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

Taking the LCM of 54,

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM of 336 and 54 = 3024

HCF of 336 and 54 = 2×3 = 6

Now, the product of 336 and 54 = 336 × 54 = 18144

And the product of LCM and HCF = 3024 × 6 = 18144

Therefore, LCM × HCF = product of the 336 and 54.

**Question 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**

**Solutions:

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

Taking the LCM of 12,

12=2×2×3

Taking the LCM of 15,

15=5×3

Taking the LCM of 21,

21=7×3

Therefore,

HCF of 12, 15 and 21 = 3

LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420

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

Taking the LCM of 17,

17=17×1

Taking the LCM of 23,

23=23×1

Taking the LCM of 29,

29=29×1

Therefore,

HCF of 17, 23 and 29 = 1

LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339

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

Taking the LCM of 8,

8=2×2×2×1

Taking the LCM of 9,

9=3×3×1

Taking the LCM of 25,

25=5×5×1

Therefore,

HCF of 8, 9 and 25 =1

LCM of 8, 9 and 25 = 2×2×2×3×3×5×5 = 1800

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

**Solution:

Given: HCF (306, 657) = 9

We know that,

HCF×LCM=Product of the two given numbers

Therefore, by substituting the value we get,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

LCM (306,657) = 22338

Therefore, the LCM (306,657) = 22338

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

**Solution:

If the given number 6n ends with the digit 0, then it should be divisible by 5.

We know that if any number with the unit place as 0 or 5 is divisible by 5.

Therefore,

By prime factorization of 6n = (2×3)n

Since the prime factorization of 6n doesn’t contain prime number 5.

Therefore, 6n cannot end with the digit 0 for any natural number.

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

**Solution:

We know by the definition of a composite number, that, the composite number has factors other than 1 and itself.

Therefore, in the given expression

7 × 11 × 13 + 13

By taking 13 as a common factor, we get,

=13(7×11×1+1)

= 13(77+1)

= 13×78 [taking prime factors of 78]

= 13×3×2×13

Therefore, 7 × 11 × 13 + 13 is a composite number.

Now, for the 2nd number

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

By taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1)

= 5(1008+1)

= 5×1009

Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

**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?

**Solution:

Given: Both Sonia and Ravi move in the same direction and at the same time.

Now, the time when they will be meeting again at the starting point can be calculated by finding the LCM of 18 and 12

Therefore,

LCM (18, 12) = 2×3×3×2×1 = 36

Finally, they both will meet again at the starting point after 36 minutes.

Conclusion

Exercise 1.1 in Chapter 1 of the NCERT Class 10 Mathematics textbook provides the foundational understanding of Real Numbers. By solving these exercises, students will gain a clear grasp of the different types of numbers and their properties preparing them for the more advanced mathematical concepts. Mastery of these fundamental concepts is crucial for the further studies in the mathematics and related fields.