Based on New NCERT PATTERN [2023-24]

## Exercise 1.1

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

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(i) 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825 = 3 x 3 x 5 x 17 = 32 x 52 x x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM xHCF = product of the two numbers.

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

Answer:

(i) 26 and 91

Let, a = 26, b = 91

26 = 2 × 13

91 = 7 × 13

**L.C.M =** 2 × 7 × 13 = 14 × 13 = 182

**H.C.F =** 13

Verification:

H.C.F × L.C.M = a × b

13 × 182 = 26 × 96

**2366 = 2366**

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

(ii) 510 and 92

Let, a = 510, b = 92

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

**L.C.M =** 2 × 2 × 3 × 5 × 17 × 23 = 23460

**H.C.F =** 2

Verification:

H.C.F × L.C.M = a × b

23460 × 2 = 92 × 510

46920 = 46920

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

(iii) 336 and 54

Let a = 336 , b = 54

336= 2 x 2 x 2 x 2 x 3 x 7

54 = 2 x 3 x 3 x 3 =

LCM = 24 x 33 x 7= 3024

HCF = 2 x 3 = 6

Verification:

H.C.F × L.C.M = a × b

3024 × 6 = 336 × 54

18144 = 18144

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

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**

Answer:

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

Prime factors for all the three numbers, we get,

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

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

Prime factors for all the three numbers, we get,

17=17×1

23=23×1

29=29×1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

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

Prime factors for all the three numbers, we get,

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

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

Answer:

HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

L.C.M x H.C.F = first Number x Second Number

L.C.M x 9 = 306 x 657

LCM = (306×657)/9 = 22338

LCM = 22338

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

Answer:

Prime factorization of 6n = (2×3)n

Therefore, prime factorization of 6ndoes not contain 5 and 2 as a factor together.

OR, 6n cannot end with the digit 0 for any natural number n. In order for a number to end with 0, it must be divisible by 10. However, for any natural number n, 6n will never be divisible by 10.

Hence 6ncan never end with the digit 0 for any natural number n

6. Explain why 7 x 11 x 13 and 7 x 6 x 5 x 3 x 2 x 1 + 5 are composite numbers.

Answer:

A composite number is a whole number greater than 1 that has multiple factors, meaning it can be divided by more than just 1 and itself. If a number has only two factors (1 and itself), it is called a prime number.

**1. 7 × 11 × 13 + 13:**

When we calculate 7 × 11 × 13 and add 13 to it, we get the number 1014.

This number has several divisors other than 1 and itself, like 2, 3, 6, 9, 18, 27, and so on. Because it has more than two divisors, it is a composite number.

**2. 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5:**

When we calculate 7 × 6 × 5 × 4 × 3 × 2 × 1 and add 5 to it, we get the number 5045.

Just like the previous example, this number also has divisors other than 1 and itself, like 5, 1009, and others. Because it has more than two divisors, it is also a composite number.

So, both expressions, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5, are composite numbers because they have more than two different numbers that can divide them evenly.

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?

Answer:

To find the time at which Sonia and Ravi will meet again at the starting point, we need to determine the least common multiple (LCM) of their individual time intervals.

Sonia takes 18 minutes to complete one round of the field, and Ravi takes 12 minutes to do the same. The LCM of 18 and 12 will give us the time at which both of them meet again at the starting point.

To find the LCM of 18 and 12, let's first find their prime factorizations:

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM of 12 and 18= 2 x 2 x 3 x 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

**Exercise 1.2**

1. Prove that √2 is irrational.

**Explanation :**

⇒ (√2)2 = (a/b)2

⇒ 2 = a2/b2

⇒ a2 = 2b2

⇒ a2 is multiple of 2.

So, a is a multiple of 2.

From this equation , we can see that 'a2' is a multiple of 2. This means that 'a' must be a multiple of 2 as well since squaring an integer will result in another integer. Let's represent 'a' as '2k' where 'k' is an integer:

⇒ a = 2k for some integer k.

⇒ a2 = 4k2

⇒ 2b2 = 4k2

⇒ b2 = 2k2

⇒ b2 is a multiple of 2

So, b is a multiple of 2

Now, we can see that 'b2' is also a multiple of 2. This implies that 'b' must be a multiple of 2 as well since squaring an integer will result in another integer.

**false**, and

**√2 is irrational.**

2. Prove that (3 + 2√5) is irrational.

**Explanation :**

**3 + √5 is rational.**

**√5 is irrational.**So, our assumption is incorrect. Hence,

**6 + √5 is irrational.**

3. Prove that the following are irrationals.

(i) ** 1/**√**2**

(ii) **7**√**5**

(iii) **6 +**√**2**

Answer:

(i) Let us assume 1/√2 is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

1/√2 = x/y

⇒ √2 = y/x

Since, x and y are integers, thus, √2 is a rational number, But, this contradicts the fact that **√2 is irrational.** So, our assumption is incorrect.

Hence, we can conclude that 1/√2 is irrational.

(ii) Let us assume 7√5 is a rational number.

It means we have some co-prime integers a and b (b ≠ 0)

7√5 = x/y

⇒ √5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, But, this contradicts the fact that **√5 is irrational.** So, our assumption is incorrect.

Hence, we can conclude that 7√5 is irrational.

(iii) Let us assume, to the contrary, that ** **6 + √2 is rational.

**√2 is irrational.**So, our assumption is incorrect. Hence,

**6 + √2 is irrational.**

**Key Features of NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers:**

**Comprehensive Coverage:**The NCERT Solutions cover all the topics and concepts of Class 10 Maths Chapter 1, ensuring a thorough understanding of real numbers.

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