Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 focuses on one of the most important concepts in number theory, i.e., the Fundamental Theorem of Arithmetic, as per the CBSE Class 10th syllabus. This exercise helps you understand how to break down composite numbers into their prime factors and use this concept to find HCF and LCM, which are part of the CBSE most important questions.
The solutions are explained in a clear, step-by-step manner so that you can easily follow each method and apply it to similar problems. Practising these questions not only strengthens conceptual understanding but also improves accuracy and problem-solving speed, which is essential for board exams.
NCERT Solutions for 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
Solutions:
(i) 140 By taking the LCM of 140, we will get the product of its prime factors.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2 2 ×5×7
(ii) 156 By taking the LCM of 156, we will get the product of its prime factors.
Hence, 156 = 2 × 2 × 13 × 3 × 1 = 2 2 × 13 × 3
(iii) 3825 By taking the LCM of 3825, we will get the product of its prime factors.
Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1= 3 2 ×5 2 ×17
(iv) 5005 By taking the LCM of 5005, we will get the product of its prime factors.
Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13
(v) 7429. By taking the LCM of 7429, we will get the product of its prime factors.
Hence, 7429 = 17 × 19 × 23 × 1 = 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
Solutions:
(i) 26 and 91
Expressing 26 and 91 as the product of its prime factors, we get
26 = 2 × 13 × 1 91
= 7 × 13 × 1.
Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
And HCF (26, 91) = 13 Verification
Now, product of 26 and 91 = 26 × 91 = 2366
And product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91
(ii) 510 and 92
Expressing 510 and 92 as the product of its prime factors, we get
510 = 2 × 3 × 17 × 5 × 1 92
= 2 × 2 × 23 × 1
Therefore, LCM (510, 92)
= 2 × 2 × 3 × 5 × 17 × 23 = 23460
And HCF (510, 92) = 2 Verification
Now, product of 510 and 92 = 510 × 92 = 46920
And product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92
(iii) 336 and 54
Expressing 336 and 54 as the product of its prime factors, we get
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 54
= 2 × 3 × 3 × 3 × 1 Therefore, LCM (336, 54) = 3024
And HCF (336, 54) = 2×3 = 6 Verification
Now, product of 336 and 54 = 336 × 54 = 18,144
And product of LCM and HCF = 3024 × 6 = 18,144
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
Solutions:
(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 6 n can end with the digit 0 for any natural number n.
Solution:
If the number 6 n 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 factorisation of 6 n = (2×3) n. Therefore, the prime factorisation of 6 n doesn’t contain the prime number 5. Hence, it is clear that for any natural number n, 6 n is not divisible by 5, and thus it proves that 6 n 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:
By the definition of a composite number, we know if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression 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. Now let’s take the other 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 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.
How to Score Better in Class 10 Maths Exam?
Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:
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Build Strong Concepts:
Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.
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Practise Regularly:
Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.
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Revise Formulas Daily:
Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.
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Solve Previous Year Papers:
Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.
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Attempt Sample Papers:
Solving PW Class 10 Maths sample papers improves time management and gives you exam-like practice.