# Exercise 1.1 real number class 10 ncert maths solutions

# Exercise 1.1(Real numbers)

**Exercise 1.1 real number class 10 ncert maths solutions**

**Question 1: Express each number as a product of its prime factors:**

**(i) 140**

**(ii) 156**

**(iii) 3825**

**(iv) 5005**

**(v) 7429**

**Solutions: (i) 140**

we will get the product of its prime factor.

Therefore, 140 = 2 Ã— 2 Ã— 5 Ã— 7 Ã— 1 = 2^{2}Ã—5Ã—7

**(ii) 156**

we will get the product of its prime factor.

Hence, 156 = 2 Ã— 2 Ã— 13 Ã— 3 Ã— 1 = 2^{2}Ã— 13 Ã— 3

**(iii) 3825**

we will get the product of its prime factor.

Hence, 3825 = 3 Ã— 3 Ã— 5 Ã— 5 Ã— 17 Ã— 1 = 3^{2}Ã—5^{2}Ã—17

**(iv) 5005**

we will get the product of its prime factor.

Hence, 5005 = 5 Ã— 7 Ã— 11 Ã— 13 Ã— 1 = 5 Ã— 7 Ã— 11 Ã— 13

**(v) 7429**

we will get the product of its prime factor.

Hence, 7429 = 17 Ã— 19 Ã— 23 Ã— 1 = 17 Ã— 19 Ã— 23

**Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM Ã— 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 product of its prime factors, we get,

26 = 2 Ã— 13 Ã— 1

91 = 7 Ã— 13 Ã— 1

Therefore, LCM (26, 91) = 2 Ã— 7 Ã— 13 Ã— 1 = 182

And HCF (26, 91) = 13

**Verification**

Now, product of 26 and 91 = 26 Ã— 91 = 2366

And product of LCM and HCF = 182 Ã— 13 = 2366

**Hence, LCM Ã— HCF = product of the 26 and 91.**

**(ii) 510 and 92**

Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 Ã— 3 Ã— 17 Ã— 5 Ã— 1

92 = 2 Ã— 2 Ã— 23 Ã— 1

Therefore, LCM(510, 92) = 2 Ã— 2 Ã— 3 Ã— 5 Ã— 17 Ã— 23 = 23460

And HCF (510, 92) = 2

**Verification**

Now, product of 510 and 92 = 510 Ã— 92 = 46920

And Product of LCM and HCF = 23460 Ã— 2 = 46920

Hence, LCM Ã— HCF = product of the 510 and 92.

**(iii) 336 and 54**

Expressing 336 and 54 as product of its prime factors, we get,

336 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 7 Ã— 3 Ã— 1

54 = 2 Ã— 3 Ã— 3 Ã— 3 Ã— 1

Therefore, LCM(336, 54) = = 3024

And HCF(336, 54) = 2Ã—3 = 6

**Verification**

Now, product of 336 and 54 = 336 Ã— 54 = 18,144

And product of LCM and HCF = 3024 Ã— 6 = 18,144

Hence, LCM Ã— HCF = product of the 336 and 54.

**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**

**Solutions: **(i) 12, 15 and 21

Writing the product of prime factors for all the three numbers, we get,

12=2Ã—2Ã—3

15=5Ã—3

21=7Ã—3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 Ã— 2 Ã— 3 Ã— 5 Ã— 7 = 420

**(ii) 17, 23 and 29**

Writing the product of prime factors for all the three numbers, we get,

17=17Ã—1

23=23Ã—1

29=29Ã—1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 Ã— 23 Ã— 29 = 11339

**(iii) 8, 9 and 25**

Writing the product of prime factors for all the three numbers, we get,

8=2Ã—2Ã—2Ã—1

9=3Ã—3Ã—1

25=5Ã—5Ã—1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2Ã—2Ã—2Ã—3Ã—3Ã—5Ã—5 = 1800

**Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).**

**Solution:Â **As we know that,

HCFÃ—LCM=Product of the two given numbers

Therefore,

9 Ã— LCM = 306 Ã— 657

LCM = (306Ã—657)/9 = 22338

Hence, LCM(306,657) = 22338

**Question 5: Check whether 6 ^{n}Â can end with the digit 0 for any natural number n.**

**Solution:Â **If the numberÂ 6^{n}Â ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.

Prime factorization of 6^{n}Â = (2Ã—3)^{n}

Therefore, the prime factorization ofÂ 6^{n}Â doesnâ€™t contain prime number 5.

Hence, it is clear that for any natural number n, 6^{nÂ }is not divisible by 5, and thus it proves that 6^{n}Â cannot end with the digit 0 for any natural number n.

**Question 6: Explain why 7 Ã— 11 Ã— 13 + 13 and 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5 are composite numbers.**

**Solution:Â **By the definition of 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 Ã— 11 Ã— 13 + 13**

Taking 13 as common factor, we get,

=13(7Ã—11Ã—1+1) = 13(77+1) = 13Ã—78 = 13Ã—3Ã—2Ã—13

Hence, 7 Ã— 11 Ã— 13 + 13 is a composite number.

Now letâ€™s take the other number,

**7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5**

Taking 5 as a common factor, we get,

=5(7Ã—6Ã—4Ã—3Ã—2Ã—1+1) = 5(1008+1) = 5Ã—1009

Hence, 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5 is a composite number.

**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:Â **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) = 2Ã—3Ã—3Ã—2Ã—1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.