Real Numbers - Exercise 1.2 | CBSE Class 10th Mathematics | Solutions

Real Numbers - Exercise 1.2 | CBSE Class 10th Mathematics | Solutions

Question 1. Express each number as a product of its prime factors:
(i) 140   (ii) 156    (iii) 3825   (iv) 5005    (v) 7429

Solution (i) To express the given number 140 as product of its prime factors, we employ the division method as shown below :
2|140
2|70
5|35
7|7
|1

∴ 140 = 2×2×5×7
       = 22×5×7




Solution (ii) To express the given number 156 as product of its prime factors, we employ the division method as shown below :
2|156
2|78
3|39
13|13
|1

∴ 156 = 2×2×3×13
       = 22×3×13


Solution (iii) To express the given number 3825 as product of its prime factors, we employ the division method as shown below : 
3|3825
3|1275
5|425
5|85
17|17
|1

∴ 3825 = 3×3×5×5×17
       = 32×52×17


Solution (iv) To express the given number 5005 as product of its prime factors, we employ the division method as shown below : 
5|5005
7|1001
11|143
13|13
|1

∴ 5005 = 5×7×11×13


Solution (v) To express the given number 2429 as product of its prime factors, we employ the division method as shown below : 
17|7429
19|437
23|23
|1

∴ 7429 = 17×19×23


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. 
2|26
13|13
|1
7|91
13|13
|1

26 = 2× 1391 = 7× 13

∴    LCM of 26 and 91=2×7×13 = 182
HCF of 26 and 91=13
Now, LCM×HCF=182×13 = 2366
And 26×91=2366
∵     182×13=26×91
Hence verified that LCM&timesHCF=Product of the Two Numbers


Solution (ii) 510 and 92. 
2|510
3|255
5|85
17|17
|1
2|92
2|46
23|23
|1

510 = 2×3×5×592 = 2× 2×23

∴    LCM of 510 and 92=2×2×3×5×17×23 = 23460
HCF of 510 and 92=2
Now, LCM×HCF=23460×2 = 46920
And 510×92=46920
∵     23460×2=510×92
Hence verified that LCM&timesHCF=Product of the Two Numbers


Solution (iii) 336 and 54. 
2|336
2|168
2|84
2|42
3|21
7|7
|1
2|54
3|27
3|9
3|3
|1

336 = 2×2×2×2×3×754 = 2× 3×3×3

∴    LCM of 336 and 54=2×2×2×2×3×3×3×7 = 3024
HCF of 336 and 54=2×3 = 6
Now, LCM×HCF=3024×6 = 18144
And 336×54=18144
∵     3024×6=336×54
Hence verified that LCM&timesHCF=Product of the Two Numbers


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

Note: 
  • The HCF of given numbers is equal to the Product of the smallest power of each common prime factor in the numbers
  • The LCM of given numbers is equal to the Product of the greatest power of each prime factor, involved in the numbers.
Solution (i) 12, 15 and 21 
Prime factorisation of the given numbers
12=2×2×3 = 22×3
15=3×5
21=3×7
Here, 22, 31, 51 and 71 are the greatest powers of the prime factors 2, 3 , 5 and 7 respectively involved in the three numbers.
So, LCM (12, 15 , 21) = 22× 31× 51 × 71 = 420
Also, 31 is the smallest power of the common factor 3, respectively.
So, HCF (12, 15 , 21) = 31 = 3 

Solution (ii) 17, 23 and 29 
Prime factorisation of the given numbers
17=17
23=23
29=29
Here, 171, 231 and 291 are the greatest powers of the prime factors 17, 23 and 29 respectively involved in the three numbers.
So, LCM (17, 23 , 29) = 171× 231× 291 = 11339
Also, 11 is the smallest power of the common factor 1 , respectively.
So, HCF (17, 23 , 29) = 11 = 1 

Solution (iii) 8, 9 and 25 
Prime factorisation of the given numbers
8=2×2×2 = 23
9=3×3 = 32
25=5×5 = 52
Here, 23, 32 and 52 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So, LCM (8, 9 , 25) = 23× 32 ×52 = 8×9×25= 1800
Also, 11 is the smallest power of the common factor 1 , respectively.
So, HCF (8, 9 , 25) = 11 = 1 

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

Solution : We know that, For any two positive integers a and b, product of HCF of the numbers and LCM of numbers is equal to products of numbers
i.e. HCF (a, b) × LCM (a, b)=a × b
∴ HCF (306, 657) × LCM (306, 657)=306 × 657
9 × LCM (306, 657)=306 × 657
 
306 × 657
LCM (306, 657)=
9
 
=22338


Question 5. Check whether 6n can end with the digit 0 for any natural number n.

Solution : If the number 6n, for any natural number n, were to end with the digit zero, then it must be divisible by 5. That is, the prime factorisation of 6n would contain the prime 5. This is not possible because 6n = (2×3)n; so the only prime in the factorisation of 6n is 2 and 3. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6n. 
So, there is no natural number n for which 6n ends with the digit zero. 

Question 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution : 
&#8757   7 × 11 × 13 + 13=13 × (7 × 11+1)
=13 × (77+1)
=13 × 78
=13 × 78
As given number (7 × 11 × 13 + 13 ) has 2 other factors i.e. 13 and 78, other than 1 and the number itself
∴ it is a composite number.
Similarly
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5=5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
=5 × (1008+1)
=5 × 1009
As given number (7 × 6 × 5 × 4 × 3 × 2 × 1 + 5) has 2 other factors i.e. 5 and 1009, other than 1 and the number itself
∴ it 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 : As the time periods of Sonia and Ravi to complete one round of the sports field are different, therefore, after starting, at the same point and at the same time, they will meet again at the starting point after a time, which must be the least multiple of the time periods of Sonia and Ravi to complete one round each. Clearly, this will be given by the LCM of theirs respective time periods to complete one round 

Time taken by Sonia to complete one round = 18 minutes 
Time taken by Ravi to complete one round = 12 minutes 

To find LCM of 18 and 12, we have to find the Prime factors of both. 
2|18
3|9
3|3
|1
2|12
2|6
3|3
|1

18 = 2×3×3and12 = 2× 2×3

∴    LCM of 18 and 12=2×2×3×3 = 36

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

Points to remember

  • Prime Number : A number, which has no factor other than 1 and the number itself, is called a prime number. For example numbers 1, 2, 3, 5, 7 etc. are all prime numbers as each one of them is either divisible by 1 or by the numbers itself.
  • Composite Number : A number, which has at least one factor other than 1 and the number itself, is called a composite number.
  • Fundamental Theorem of Arithmetic : Every composite number can be expressed ( factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
  • The prime factorisation of a natural number is unique, except for the order of its factors.
  • Highest Common Factor : The HCF of two or more numbers is the largest number that divides all the given numbers complletely
  • Prime Factorisation Method to find HCF :
    • First we find the prime factorisation of each of the given numbers.
    • Then we identify the common prime factors
    • The product of all the common prime factors, using each common prime factor the least number of times it appears in the prime factorisation of any of the given numbers, is the required HCF.
  • Lowest Common Multiple : The LCM of two or more numbers is the smallest number which is a multiple of each of the numbers
  • The HCF of given numbers is equal to the Product of the smallest power of each common prime factor in the numbers
  • The LCM of given numbers is equal to the Product of the greatest power of each prime factor, involved in the numbers.
  • For any two positive integers a and b, product of HCF of the numbers and LCM of numbers is equal to products of numbers i.e. HCF (a, b) × LCM (a, b) = a × b
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