NCERT Solutions Class 10 Maths Chapter 1 Real Numbers are provided here. These solutions are prepared by our expert faculty to help students in their board exam preparations. They solve and provide the NCERT Solutions for Maths to aid the students to solve the problems easily. They also focus on preparing the solutions in such a way that it is easy to understand for the students. A detailed and step-wise explanation is given to each question given in the exercises of NCERT books.
Answers to the questions present in Real Numbers are given in the first chapter of Maths Solutions of NCERT Class 10. Here, students are introduced to several important concepts that will be useful for those who wish to pursue mathematics as a subject in their Class 11. Based on these solutions of NCERT, students can prepare for their upcoming board exam. These solutions are helpful as they are in accordance with the CBSE Syllabus for 2022-23.
Class 10 Maths Chapter 1 Exercises 1.1 Question 1 Solutions
Use Euclid’s division algorithm to find the HCF of:
- (i) 135 and 225
- (ii) 196 and 38220
- (iii) 867 and 255
Solution of Exercise 1.1 Question 1
- (i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45. - (ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196, therefore, HCF of 196 and 38220 is 196. - (iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma - to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.
Class 10 Maths Chapter 1 Exercises 1.1 Question 2 Solutions
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also,
6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers and therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.
Class 10 Maths Chapter 1 Exercises 1.1 Question 3 Solutions
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.
Class 10 Maths Chapter 1 Exercises 1.1 Question 4 Solutions
Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Solution:
Let a be any positive integer and b = 3. Using Euclid’s Division Lemma, a = 3q + r for some integer q ≥ 0 where r = 0, 1, 2 because 0 ≤ r < 3.
Therefore, a = 3q or 3q + 1 or 3q + 2
a² = (3q)² or (3q + 1)² or (3q + 2)²
=(3q)² or 9q² + 6q + 1 or 9q² + 12q + 4
= 3×(3q²) or 3×(3q² + 2q) + 1 or 3×(3q² + 4q + 1) + 1
= 3k₁ or 3k₂ + 1 or 3k₃ + 1
Where k₁, k₂ and k₃ are some positive integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
Class 10 Maths Chapter 1 Exercises 1.1 Question 5 Solutions
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let a be any positive integer and b = 3, using Euclid’s Division Lemma, a = 3q + r, where q ≥ 0 and 0 ≤ r < 3. Therefore, a = 3q or 3q + 1 or 3q + 2.
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a³ = (3q)³ = 27q³ = 9(3q³)= 9m
Where m is an integer such that m = 3q³
Case 2: When a = 3q + 1,
a³ = (3q +1)³
a³ = 27q³ + 27q² + 9q + 1
a³ = 9(3q³ + 3q² + q) + 1 = 9m + 1
Where m is an integer such that m = (3q³ + 3q² + q)
Case 3: When a = 3q + 2,
a³ = (3q +2)³
a³ = 27q³ + 54q² + 36q + 8
a³ = 9(3q³ + 6q² + 4q) + 8
a³ = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Important Questions for Practice on 10th Maths Chapter 1
- Write the general form of an even integer. [Answer: 2m]
- Find the largest integer which divides 245 and 1029 leaving remainder 5 in each case. [Answer: 16]
- Write the form in which every odd integer can be written taking t as variable. [Answer: 2t +1]
- If the HCF of 408 and 1032 is expressible in the form 1032y – 408×5, then find y. [Answer: 2]
- What would be the value of n for n²–1 divisible by 8. [Answer: An odd integer]
- Two tankers contain 650 litres and 1170 litres of petrol respectively. Find the maximum capacity of a tanker which can measure the petrol of either tanker in exact number of times. [Answer: 130 litres]
- State whether 7 × 11 × 13 + 7 is a composite number or a prime number. [Answer: Composite]
- A bookseller purchased 117 books out of which 45 books are of mathematics and the remaining 72 books are of Physics. Each book has the same size. Mathematics and Physics books are to be packed in separate bundles and each bundle must contain the same number of books. Find the least number of bundles which can be made for these 117 books. [Answer: 13]
- Is 5.131131113… a rational number or irrational number? [Answer: Irrational]
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NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.2
Class 10 Maths Chapter 1 Exercise 1.2 Question 1
Express each number as 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
Discuss this question in your opinion.
(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
If you have any other suggestion, please suggest us.
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
More options related to above question.
(iv) 5005 = 5 × 7 × 11 × 13
if you have any other idea, please comment here.
(v) 7429 = 17 × 19 × 23
What do you think about 10th Maths Exercise 1.2 Part v solutions?
Class 10 Maths Chapter 1 Exercise 1.2 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
Solution:
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Product of the two numbers = 26×91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM
Do you have more ideas about this solutions?
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the two numbers = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence, product of two numbers = HCF × LCM
You can discuss this question with us.
(iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Product of the two numbers = 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM
Class 10 Maths Chapter 1 Exercise 1.2 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
Solution:
(i) 12, 15 and 21
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420
For more suggestion, see the solution here.
(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
Prove more ways to solve this question.
(iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
How can we solve this question alternatively?
Class 10 Maths Chapter 1 Exercise 1.2 Question 4
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
We know that,
LCM × HCF = Product of two numbers
Therefore,
LCM= (Product of two numbers)/HCF = (306 × 657)/9 = 22338
Hence, LCM (306, 657) = 22338
The alternative way to solve this question.
Class 10 Maths Chapter 1 Exercise 1.2 Question 5
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.
Put your opinion about question 5 of 10th Maths exercise 1.2
Class 10 Maths Chapter 1 Exercise 1.2 Question 6
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that:
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 × 13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
Now: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.
Class 10 Maths Chapter 1 Exercise 1.2 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:
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 × 3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.3 (Ex. 1.3 Class 10) Real numbers in English Medium.UP Board students are also using the same NCERT Textbooks as the CBSE students.All NCERT Solutions Apps 2022-23 and NCERT Solutions are updated according to latest CBSE / NCERT Curriculum 2022-23 for UP Board, Gujrat Board as well as CBSE Board students.
NCERT Solutions for class 10 Maths Chapter 1 Exercise 1.3
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