In Class 10 Math Ch 1 Real Numbers Ex 1.1 begins with the fundamental concept of Real Numbers, forming the foundation for many topics ahead. Exercise 1.1 mainly deals with the Euclid’s Division Lemma, a powerful method used to find the highest common factor (HCF) of two positive integers. This exercise helps students understand how any two numbers can be divided in a systematic way using Euclid’s algorithm. Learning this concept not only strengthens your number system basics but also prepares you for advanced number theory problems.
Class 10 Math Ch 1 Real Numbers Ex 1.1 – Textbook
Question 1.Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Apply Euclid’s Division Algorithm:
225 ÷ 135 = 1, remainder = 90
135 ÷ 90 = 1, remainder = 45
90 ÷ 45 = 2, remainder = 0
Since the remainder is 0,
HCF = 45
(ii) 196 and 38220
Apply Euclid’s Division Algorithm:
38220 ÷ 196 = 195, remainder = 0
Since the remainder is 0,
HCF = 196
(iii) 867 and 255
Apply Euclid’s Division Algorithm:
867 ÷ 255 = 3, remainder = 102
255 ÷ 102 = 2, remainder = 51
102 ÷ 51 = 2, remainder = 0
Since the remainder is 0,
HCF = 51
Ex 1.1 Class 10 Maths Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution
Any positive odd integer, when divided by 6, will have a remainder of 0, 1, 2, 3, 4, or 5. So, any integer can be written as 6 times some integer q plus a remainder r, where r is one of those six numbers.
Now, check which remainders give odd numbers:
- If the remainder is 0, the number is 6q, which is even.
- If the remainder is 1, the number is 6q + 1, which is odd.
- If the remainder is 2, the number is 6q + 2, which is even.
- If the remainder is 3, the number is 6q + 3, which is odd.
- If the remainder is 4, the number is 6q + 4, which is even.
- If the remainder is 5, the number is 6q + 5, which is odd.
Therefore, any positive odd integer must be of the form 6q + 1, 6q + 3, or 6q + 5, where q is an integer.
Ex 1.1 Class 10 Maths Question 3.
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:
The army contingent has 616 members, and the army band has 32 members.
Both groups must march in the same number of columns.
To find the maximum number of columns, we need to find the greatest common divisor (GCD) of 616 and 32.
Find GCD of 616 and 32:
- Divide 616 by 32:
616 ÷ 32 = 19 remainder 8 - Divide 32 by 8:
32 ÷ 8 = 4 remainder 0
Since the remainder is now 0, the GCD is 8.
Ex 1.1 Class 10 Maths Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
Any positive integer when divided by 3 can leave a remainder 0, 1, or 2. So, any number n can be written as 3q, 3q + 1, or 3q + 2, where q is an integer.
Now, square each form:
- If n = 3q, then n² = (3q)² = 9q² = 3 × (3q²), which is of the form 3m.
- If n = 3q + 1, then n² = (3q + 1)² = 9q² + 6q + 1 = 3 × (3q² + 2q) + 1, which is of the form 3m + 1.
- If n = 3q + 2, then n² = (3q + 2)² = 9q² + 12q + 4 = 3 × (3q² + 4q + 1) + 1, which is also of the form 3m + 1.
So, the square of any positive integer is either of the form 3m or 3m + 1, where m is an integer.
Ex 1.1 Class 10 Maths Question 5.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:
Any positive integer when divided by 3 can be written as 3q, 3q + 1, or 3q + 2, where q is an integer.
Now, cube each form:
- If n = 3q, then n³ = (3q)³ = 27q³ = 9 × (3q³), which is of the form 9m.
- If n = 3q + 1, then n³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9 × (3q³ + 3q² + q) + 1, which is of the form 9m + 1.
- If n = 3q + 2, then n³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9 × (3q³ + 6q² + 4q) + 8, which is of the form 9m + 8.
- So, the cube of any positive integer is either of the form 9m, 9m + 1, or 9m + 8, where m is an integer.
New Syllabus – Class 10 Math Ch 1 Real Numbers Ex 1.1
Q1. Express each number as a product of its prime factors:
(i) 140
Solution:
140 = 2 × 70
= 2 × 2 × 35
= 2² × 5 × 7
Answer: 2² × 5 × 7
(ii) 156
Solution:
156 = 2 × 78
= 2 × 2 × 39
= 2² × 3 × 13
Answer: 2² × 3 × 13
(iii) 3825
Solution:
3825 = 3 × 1275
= 3 × 3 × 425
= 3² × 5 × 85
= 3² × 5 × 5 × 17
Answer: 3² × 5² × 17
(iv) 5005
Solution:
5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13
Answer: 5 × 7 × 11 × 13
(v) 7429
Solution:
7429 = 17 × 19 × 23 (Direct prime factorization)
Answer: 17 × 19 × 23
Q2. Find the LCM and HCF of the following pairs and verify: LCM × HCF = Product of the numbers
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Verification: 26 × 91 = 2366 = 13 × 182
✔️ Verified
Answer: HCF = 13, LCM = 182
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 11730
Verification: 510 × 92 = 46920 = 2 × 23460
✔️ Verified
Answer: HCF = 2, LCM = 11730
(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Verification: 336 × 54 = 18144 = 6 × 3024
✔️ Verified
Answer: HCF = 6, LCM = 3024
Q3. Find the LCM and HCF by prime factorisation method:
(i) 12, 15 and 21
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3 (common)
LCM = 2² × 3 × 5 × 7 = 420
Answer: HCF = 3, LCM = 420
(ii) 17, 23 and 29
All are primes and have no common factor.
HCF = 1
LCM = 17 × 23 × 29 = 11339
Answer: HCF = 1, LCM = 11339
(iii) 8, 9 and 25
8 = 2³
9 = 3²
25 = 5²
No common factors.
HCF = 1
LCM = 2³ × 3² × 5² = 1800
Answer: HCF = 1, LCM = 1800
Q4. Given HCF(306, 657) = 9, find LCM(306, 657).
LCM = (306 × 657) ÷ 9
= 201042 ÷ 9 = 22338
Answer: LCM = 22338
Q5. Check whether 6ⁿ can end with digit 0 for any natural number n.
6ⁿ = (2 × 3)ⁿ
It has no factor 5 → cannot end in 0.
Answer: No, 6ⁿ cannot end with digit 0 for any natural number n.
Q6. Explain why these numbers are composite:
(i) 7 × 11 × 13 + 13 = 1001 + 13 = 1014
1014 = 13 × 78
⇒ More than two factors → Composite
(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045
5045 = 5 × 1009
⇒ Composite
Answer: Both are composite because they have more than two factors.
Q7. Circular path: Sonia takes 18 min, Ravi takes 12 min. When will they meet again?
Find LCM(18, 12)
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 36
Answer: They will meet again at the starting point after 36 minutes.
Refer to NCERT for Class 10 Maths Official Book
For additional reference and to access the official NCERT Class 10 Maths textbook, visit the NCERT website. This will help you understand the concepts covered in Class 10 Math Ch 1 Real Numbers Ex 1.1 more effectively.
🔗 Visit NCERT Official Website: NCERT Class 10 Maths Book
This resource will provide you with the complete Class 10 Maths syllabus and ensure that you’re following the latest NCERT guidelines. Happy learning!
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Understanding the foundation of real numbers is essential for mastering higher-level mathematics. On this page, we have provided Class 10 Math Ch 1 Real Numbers Ex 1.1 all solution to help students grasp the concepts clearly and thoroughly. These exercises cover important topics like Euclid’s Division Lemma, the Fundamental Theorem of Arithmetic, and rational and irrational numbers.
By going through Class 10 Math Ch 1 Real Numbers Ex 1.1 all solution, students can strengthen their problem-solving skills and gain confidence in tackling board exam questions. Each solution is crafted to explain the steps in a student-friendly manner, ensuring clarity in understanding.
Whether you’re revising for exams or learning these concepts for the first time, our comprehensive guide to Class 10 Math Ch 1 Real Numbers Ex 1.1 offers the support you need. With regular practice and the right guidance, success in mathematics becomes much more achievable.