Mathematics solution NCERT
Class 10 - Chapter 1: Real Numbers
Exercise 1.1 Solution
Q1. Express each number as a product of its prime factors.
(i) 140
Expressing each number as a product of prime means we have to
take LCM of that number by using prime numbers only.
Note: Prime numbers are: 2,3,5,7,11,13,17.....etc
140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7
Answer: 140 = 22 × 5 × 7
(ii) 156
156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
Answer: 156 = 22 × 3 × 13
(iii) 3825
3825 = 5 × 765
= 5 × 5 × 153
= 5 × 5 × 3 × 51
= 5 × 5 × 3 × 3 × 17
Answer: 3825 = 32 × 52 × 17
(iv) 5005
5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13
Answer: 5005 = 5 × 7 × 11 × 13
(v) 7429
7429 = 17 × 437
= 17 × 19 × 23
Answer: 7429 = 17 × 19 × 23
IMP NOTE- Learn divisibility rules of 2,3,5,6,7,8,9,11 etc for fast calculation of LCM.
Q2. Find the LCM and HCF of the following pairs and verify that: LCM × HCF = Product of the two numbers.
(Note- Important Question for board exam)
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
LCM × HCF = 182 × 13 = 2366
26 × 91 = 2366
Hence verified.
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 22 × 3 × 5 × 17 × 23 = 23460
LCM × HCF = 23460 × 2 = 46920
510 × 92 = 46920
Hence verified.
(iii) 336 and 54
336 = 24 × 3 × 7
54 = 2 × 33
HCF = 2 × 3 = 6
LCM = 24 × 33 × 7 = 3024
LCM × HCF = 3024 × 6 = 18144
336 × 54 = 18144
Hence verified.
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
LCM = 22 × 3 × 5 × 7 = 420
Answer: HCF = 3, LCM = 420
(ii) 17, 23 and 29
All are prime numbers.
HCF = 1
LCM = 17 × 23 × 29 = 11339
Answer: HCF = 1, LCM = 11339
(iii) 8, 9 and 25
8 = 23
9 = 32
25 = 52
HCF = 1
LCM = 23 × 32 × 52
= 1800
Answer: HCF = 1, LCM = 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
We know that: LCM × HCF = Product of numbers
LCM × 9 = 306 × 657
LCM =
LCM = 22338
Answer: LCM = 22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
(Note- Important Question for board exams)
6 = 2 × 3
6n = 2n × 3n
Here note one thing students, a number ends with 0 only if it contains factors 2 and 5 both.
Since 6n has no factor 5, it can never end with 0.
Answer: No, 6n can never end with the digit 0.
Q6. Explain why the following numbers are composite.
(i) 7 × 11 × 13 + 13
In this question, First we will take common, here we can take 13 as common:
we get,
= 13(7 × 11 + 1)
= 13(77 + 1)
= 13 × 78
Since it has factors other than 1 and itself, it is composite.
(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5(1008 + 1)
= 5 × 1009
Hence it is composite.
(Note- Every number which has more than two two factors, is called composite number.)
you must be thinking that this number has only two factors but not more than two,
then why it is composite
- so here we have not included 1.
Q7. 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?
Sonia completes one round in 18 minutes.
Ravi completes one round in 12 minutes.
They will meet again at the starting point after the LCM of 18 and 12.
18 = 2 × 32
12 = 22 × 3
LCM = 22 × 32
= 36 minutes
Answer: Sonia and Ravi will meet again at the starting point after 36 minutes.
Note: Use HCF when the question contains words like:
✅ Divide
✅ Distribute equally
✅ Arrange into largest possible groups
✅ Maximum size
✅ Greatest length
✅ Biggest possible measure
✅ Largest square
Breaking a big thing into equal pieces → HCF
Use LCM when the question contains words like:
✅ Together again
✅ Simultaneously
✅ Repeated events
✅ Minimum number
✅ First time together
✅ Next common time
✅ Least number divisible by all
Think: When separate cycles meet again → LCM
Jab bhi bada number chahiye to LCM lena hai
aur jab chota number chahiye to HCF lena hai.