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NCERT Class 10 Mathematics Chapter 1 Real Number
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 10 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 10 Mathematics Chapter 1 Real Number Solutions for All Subject, You can practice these here.
Real Number
Chapter – 1
| Exercise 1.1 |
1. Express each number as a product of its prime factors:
(i) 140
Ans:
∴ 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7
(ii) 156
Ans:
∴ 156 = 2 x 2 x 3 x 13
(iii) 3825
Ans:
∴ 3825 = 3 x 3 x 5 x 17
(iv) 5005
Ans:
∴ 5005 = 5 x 7 x 11 x 13
(v) 7429
Ans:
∴ 7429 = 17 x 19 x 23
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
Ans: We have 26 and 91
Thus LCM (26, 96) = 2 × 7 × 13 = 182
HCF (26,91) = 13
Now LCM (26,91) × HCF (26,91) = 182 × 13 = 2366
And product of the numbers = 26 × 91 = 2366.
Hence the LCM X HCF = product of two numbers.
(ii) 510 and 92
Ans: We have 510 and 92
92 = 22 X 23
LCM (510, 92 ) = 22 X 33 X 23 = 23460
HCH (510, 92) = 2
Now LCM (510, 92) x HCF (510, 92) = 23460 x 2 = 46920
Product of two numbers = 510 x 92 = 46920
Hence LCM x HCF = Product of two numbers.
(iii) 336 and 54
Ans: We have 336 = 2⁴ x 3 × 7 54 = 2× 3³
HCF (336,92) = 2 × 3 = 6
Now LCM (336,54) HCF (336,54) 3024 × 6 = 18,144 and product of two numbers = 336 × 54
= 18,144.
Hence LCM × HCF = product of two numbers.
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Ans: (i) 12 = 22 x 315 = 3 x 5 and 21 = 3 x 7
The HCF of these integers is 3 2, 3, 5 and are the greatest powers involved in the prime factors of 12, 15 and 21.
(ii) 17, 23 and 28
Ans: 17, 23 and 29
There is no common factor as 17,23, 29 they are primes. Hence HCF is 1.
LCM is the product of all prime factors 17, 23, 29 = 17 x 23 x 29=11339
HCF (17, 23, 29) = 1.
(iii) 8, 9 and 25
Ans: 8, 9 and 25
8 = 2 x 2 x 2 = 2¹
HCF (8,9,25)=1 [∵ There is no common factor of 8,9,25]
LCM = 23 x 32 x 52 = 1800
4. Given that HCF (306, 657) = 9, find LCM(306, 657)
Ans: We have
HCF (306, 657) = 9
We know that, LCM x HCF = Product of two numbers.
306 × 657
LCM × 9 = 306 × 657 = LCM = 22338
5. Check whether 6″ can end with the digit 0 for many natural numbers.
Ans: If the number 6″ ends with the digit zero, then it must have 5
as a factor.
But, we know that the only prime factors of 6″ are 2 and 3.
Also, we know from the fundamental theorem of arithmetic, that the
prime factorisation of each number is unique.
So, there is no value of n natural numbers for which 6″ ends with the digit zero.
6. Explain why 7 x 11 x 13 +13 and 7 x 6 x composite numbers.
Ans: We have 7 x 1×13+13-1001+13=1014
1014 2 x 3 x 13 x 13
So, it is the product of prime factors, 2 x 3 x 13 x 13
Hence, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1+ 5 = 5 x 17 x 6 x 4 × 3 × 2 × 1 + 1)
-5 x 1009.
5 × 1009.
= 5045 Thus 5045 can be expressed as products of primes.
It is the of prime factor 5 × 5 101
Hence 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 meet again 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?
Ans: Required number of minutes is the LCM of 18 and 12
We have, 18 = 12 × 2 × 3²
LCM of 18 and 12 is 2 x 3 = 36
Hence, Ravish and Sonia will meet again at the starting point after 36 Minutes.
| Exercise 1.2 |
1. Prove that √5 is rational number.
Ans: Let √5 represents a rational number.
Then √5 can be expressed have no common factor, q ≠ 0. in the form, where p,q are integers and
2. Prove that 3+2√5 is irrational.
Ans: Let us assume on the contrary that 3 + 2 sqrt(5) is rational. Then there exist coprime positive integers a and b (b ≠ 0) such that
3 + 2 √5 = a/b ⇒ 2 √5 = a/b -3 ⇒ √ 5 = a – 3b/2b
⇒ √ 5 is rational {∴ a,b are integers ∵ a – 3b /2b is rational}
This contradicts the fact that √ 5 is irrational. So our supposition is incorrect.
Hence 3 + 2 √ 5 is an irrational number.
3. Prove that the following irrationals
(i) 1/ √2 (ii) 7/ √5 (iii) 6 / √2
(ii) 7/ √5
(iii) 6 / √2
Ans: Let 6 + √2 be a rational number equal to a/b
where a, b are positive co-primes. Then,
This is a contradiction.
Hence, 6 + √2 is irrational.