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Real Numbers Exercise 1.1

 Hi Friends and Champs,

Before coming to the exercise 1.1, we need to understand some points, which you have already read in class IX.

Prime Number: Those numbers which could be divide by "1" and itself. i.e. there are only two multiple of the number, example 2,3,5,7...... etc. 2 is the smallest prime number we have. "1" is not a prime number, because for a prime number there should be exactly two multiples of the number, but "1" have only one multiple which is itself.

Prime factors: Prime factors are the product of prime numbers for a given number. ex 2x2x3x5 are the prime factors for the number 60.

 

Exercise 1.1

Q-1 Express each number as a product of its prime factors:

(i) 140        (ii) 156    (ii) 3825    (iv)  5005     (v) 7429

Note: Prime numbers are those numbers, which could be divide by 1 and by itself.

Note: Do not try to skip the steps in prime factorization, start from the lowest prime number.


Solution: 

(1) 140 = 2x2x5x7

(2) 156=2x2x3x13

(3) 3825=3x3x5x5x17

(4) 5005=5x7x11x13

(5) 7429=17x19x23

Q-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


Note: Champs, keep it in mind that the product of LCM and HCF of any two numbers are always equal to the product of both the numbers. 

Note: In earlier class you have already learn to find the LCM and HCF of any numbers.

Reminder: HCF - is the most common factor of the given numbers or GCD (greatest common divisor) and LCM - is the least common multiple mean the number which could be divide by          all the given numbers.

Solution:

(i) 26 and 91

Prime factors of 26 :  2 x 13

Prime factors of 91 :  7 x 13

Here 13 is common in both so :

HCF is 13

Now for LCM we take those factors from the factors of both, which repeat maximum of times  i.e. 2,7 and 13, so 

LCM is 2 x 7 x 13=182 ( hope you understand it )

so, the product of HCF and LCM is

13x182=2,366 Which is equal to the product of both the numbers 

26x91=2,366 Product of the given numbers

Similarly, will do next questions:

(ii) 510 and 92 

Prime factors of 510 = 2 x 3 x 5 x 17

Prime factors of 92 = 2 x 2 x23

Hence, HCF = 2

LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23,460 

Product of HCF and LCM =2*23460=46920

Product of both the numbers =510*92=46,920

(iii) 336 and 54

Prime factors of 336=2x2x2x2x3x7

Prime factors of 54=2x3x3x3

Hence, HCF = 2x3 = 6

LCM = 2 x2 x 2 x 2 x 3 x 3 x 3 x 7 = 3,024

Product of HCF and LCM = 6 x3024=18144  

Product of both the numbers=336 x 54=18,144

Q (3) Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12,15 and 21           (ii) 17,23 and 29         (iii) 8,9 and 25

Solution:

(i)  12,15 and 21 

Prime factorization of 12 = 2 x 2 x 3

Prime factorization of 15 = 3 x 5

Prime factorization of 21 = 3 x 7

Hence,

HCF = 3 ( most common factors )

and

LCM = 2 x 2 x 3 x 5 x 7=420

(ii) 17,23 and 29

Prime factorization of 17 = 1 x 17

Prime factorization of 23 = 1 x 23

Prime factorization of 29 = 1 x 29  ( note : 17,23 and 29 are prime numbers) 

Hence

HCF = 1

LCM = 17 x 23 x 29 = 11,339

(iii) 8,9 and 25

Prime factorization of 8=2 x 2 x 2

Prime factorization of 9=3 x 3

Prime factorization of 25=5 x 5

Hence

HCF = 1 ( because no other factor is common )

LCM = 2 X 2 X 2 X 3 X 3 X 5 X 5 = 1,800

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

Solution: 

You already know that product of any two numbers is equal to the product of their HCF and LCM .

i.e.

HCF X LCM = product of both numbers

=> 9 X LCM = 306 X 657

 ⇒LCM=306×6579=22338

Q (5)  Check whether 6^n  can end with the digit 0 for any natural number n.

Solution:  

Note : Natural numbers are the positive numbers from 1 to infinity.

Since , if a number end with '0' always multiple by 2 and 5 

Example 10,20,30,100 … etc will always a multiple of 2 and 5 ,hence if 6n  can end with the digit 0 then 6n definitely will be a multiple of 2 and 5, but

6n= (2 x 3)n

=   2n x 3n  i.e. factor of 2 and 3 not of 2 and 5 (necessary condition)

Hence, 6n  can not end with the digit 0 for any natural number.

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

Solution:

Note : Prime numbers are the numbers which are divisible by 1 and itself like 2,3,5,7,11,..etc. i.e. a prime number has exactly two factor '1' and itself. In easiest way , a composit number is the number which have more than 2 factors i.e. it is divisible by '1' , itself and by some other number also, like 4,6,8,9….etc.

7 x 11 x 13 + 13=1,014 and

7 x 6 x 5 x 4 x 3 x 2 x 1 +5=5,045

factors of 1014 = 1 x 2 x 3 x 13 x 13

factors of 5045 = 1 x 5 x 1009

clearly seen both have more than 2 factors, and so they are composite numbers.

Q (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 :   

In this question after careful reading you can easily understand that Sonia and Ravi will meet at some particular time where both one round time will divide exactly, in easy way that least particular time which is a multiple of both one round time i.e. we need to know the L.C.M. of both one round time.

Prime factorization of 18 = 2 x 3 x 3

Prime factorization of 12 = 2 x 2 x 3

LCM = 2 x 2 x 3 x 3 = 36

Hence, both will meet exactly after 36 minute.



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