NCERT Solutions Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Here are the Class 10 Math Chapter 1 Real Numbers NCERT Solutions. Our knowledgeable faculty created these solutions to aid students in getting ready for their board exams. To assist students in finding simple solutions to difficulties, they offer NCERT Solutions for Maths. They also put a lot of effort into making the solutions simple enough for their students to understand. Each and every question in the exercises from the NCERT Books is explained in great detail and step-by-step.

Class 10 Maths Chapter 1 Exercise 1.1

1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) 140 = 2 × 2 × 5 × 7 = 2²×5×7
(ii) 156 = 2 × 2 × 13 × 3 = 2²× 13 × 3
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3²×5²×17
(iv) 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23 = 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) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:

(i) By prime factorisation, we get:
26 = 2 x 13
91 = 7 x 13
∴ HCF of 26 and 91 = 13
and LCM of 26 and 91 = 2 x 7 x 13 = 182
Now, HCF x LCM = 182 x 13 = 2366 … (i)
Product of numbers = 26 x 91 = 2366 … (ii)
From (i) and (ii), we get:
HCF x LCM = Product of number
Hence, verified.

(ii) By prime factorisation, we get:
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23
∴ HCF of 510 and 92 = 2
and LCM of 510 and 92
= 2² x 3 x 5 x 17 x 23 = 23460
Now, HCF x LCM = 2 x 23460 = 46920 … (i)
Product of numbers
= 510 x 92 = 46920 … (ii)
From (i) and (ii), we get:
LCM x HCF = Product of numbers
Hence, verified.

(iii) By prime factorisation, we get:
336 = 2 x 2 x 2 x 2 x 3 x 7
54 = 2 x 3 x 3 x 3
∴ HCF of 336 and 54 = 2 x 3 = 6
and LCM of 336 and 54 = 2⁴ x 3³ x 7 = 3024
Now, LCM x HCF = 3024 x 6 = 18144… (i)
Product of numbers
= 336 x 54 = 18144 … (ii)
From (i) and (ii), we get:
LCM x HCF = Product of number
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
Solution:

(i) 12, 15 and 21
Writing the product of 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
Writing the product of 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
Writing the product of 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).
Solution:
As we know, HCF×LCM=Product of the two given numbers
Therefore,
9 × LCM = 306 × 657
LCM = (306×657)/9 = 22338
Hence, LCM(306,657) = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with the unit place as 0 or 5 is divisible by 5.
Prime factorization of 6ⁿ = (2×3)ⁿ
Therefore, the prime factorization of 6ⁿ doesn’t contain the prime number 5.
Hence, it is clear that for any natural number n, 6ⁿis not divisible by 5, and thus it proves that 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
7 × 11 × 13 + 13
Taking 13 as a common factor, we get,
=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13
Hence, 7 × 11 × 13 + 13 is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 as a common factor, we get,
=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009
Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 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 starting point?
Solution:
Since Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.
Therefore, LCM(18,12) = 2×3×3×2×1=36
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Class 10 Maths Chapter 1 Exercise 1.2

1. Prove that √5 is irrational.
Solutions:
Let us assume, that √5 is a rational number.
i.e. √5 = x/y (where, x and y are co-primes)
y√5= x
Squaring both sides, we get,
(y√5)² = x²
⇒5y² = x²……………………………….. (1)
Thus, x² is divisible by 5, so x is also divisible by 5.
Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,
5y² = (5k)²
⇒y² = 5k²
is divisible by 5 it means y is divisible by 5.
Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.
Hence, √5 is an irrational number.

2. Prove that 3 + 2√5 + is irrational.
Solutions:
Let us assume 3 + 2√5 is rational.
Then we can find co-prime a and b (b ≠ 0) such that 3 + 2√5 = a/b
Rearranging, we get,
b(3 + 2√5) = a
3b + 2√5b = a
2√5b = a – 3b
√5 = (a – 3b)/2b
Since (a – 3b)/2b is a rational number, then √5 is also a rational number.
But, we know that √5 is irrational.
Therefore, our assumption was wrong that 3 + 2√5 is rational. Hence, 3 + 2√5 is irrational.

3. Prove that the following are irrational:
(i) 1/√2
(ii) 7√5
(iii) 6 + √2
Solutions:

i) 1/√2
Let us assume 1/√2 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y
Rearranging, we get,
√2 = y/x
Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.
Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5
Let us assume 7√5 is a rational number.
Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y
Rearranging, we get,
√5 = x/7y
Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.
Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2
Let us assume 6 +√2 is a rational number.
Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅
Rearranging, we get,
√2 = (x/y) – 6
Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.
Hence, we can conclude that 6 +√2 is irrational.