Real Number ( NCERT )

 Here, we have providing chapter 1 real numbers NCERT solutions for class 10th math which is beneficial for students. This solutions are updated according to 2020-21. Syllabus.

                    Exercise : 1.1
  
Use Euclid's division algorithm to find the HCF of 
( 1 )   135 and 225.      ( 2 )   196 and 38220. 
( 3 )   867 and 255.

Answer :

( 1 )  135 and 225

since 225 > 135 . We apply the division lemma to obtain 
 225 = 135* 1 + 90
135 =  90* 1 +45
 90 = 45*2 + 0 
Scince the reminder is  zero. The process stop.
Lastly our answers is 45.  

 ( 2 ) 196 and 38220

Here , 38220 > 196 .we apply the division lemma ,to obtain 

38220 =  196*195 +0 

Since the remainder is zero, The process stop. 
So, lastly our answer is. 196

( 3 )  867 and 225.

 Here, 867 >225.  We apply the division lemma , to obtain.
 867 = 225*3 + 102.
 225 = 102*2 + 51.
 102 = 51*2 + 0 
 Scince, the remaider is zero.,the process stop.
So, lastly our answer is , 51

Q,No.2..Show that any positive odd integer is the form 6q+1,or 6q+3,or 6q+5, where q is some integer.

    Answer :

Let a, be any positive integer and b=6. Then,by Euclid's algorithm  a= 6q+r where r= 0,1,2,3,4,5
Therefore
a=6q+0  or 6q ,or 6q+1'or 6q+2,or 6q+3,or 6q+4,or 6q+5.
also,
6q+1= 2(3q) +1.  =2k+1

6q+3= 2(3q+1)+1.  =2k+1

6q+5= 2(3q+2)+1.  =2k+1

Here all the integers have 1 remainder.( if they are even number then it is exactly divisible by 2 and no remainder leave.)

Therefore any odd positive integers are is the form of 6q+1,or 6q+3,or 6q+5.   Proved.

Q. No.3.. An army contingent of 616 members is to march behind an army band of 32 members in a parade.The two groups are to march in the same number of columns.What is the maximum number of columns in which they can march.

Answer :

Hcf( 616,32) will give the maximum number of columns in which they can march.

We can use Euclid's algorithm to find the HCF.

616= 32*19+8.

32=8*4+0

The HCF ( 616,32 ) is 8.

Therefore, they can march in 8 columns each.

Q. No. 4. Use Euclid's division lemma to show that the square of any positive integer is either of form 3m or 3m+1 for some integer m. 

  Answer :

 Let, a be any positive integer and b=3.
Then a =3q+r for some integer q >_0
and r =0,1,2 because 0<_ r<3
therefore, a =3q or 3q+1 or 3q+2
Or
 a^2= (3q)^2 or (3q +1)^2 or (3q+2)^2
 a^2= (9q^2) or 9q^2 +6q+1 or 9q^2 +12q +4
 = 3(3q^2) or 3(3q^2 +2q) +1 or 3(3q^2. +4q+1) +1
=3k1 or 3k2+1 or 3k3 +1
where k1, k2 and k3 are some positive integers

Hence,  it can be said that the square of any positive integer is either of the form 3m or 3m+1.

Q. No 5. 

Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

Answer :

Let a be any positive integer and b= 3q+r, where q>_0 and 0 <_ r< 3.
Therefore
 a=3q or 3q+1 or 3q+2

Therefore, every number can be represented as these three forms.  There are three cases.

Case 1: When a=3q,

a^3 = (3q)^3 =27q^3 = 9(3q^3) = 9m,
where m is an integer such that m=3q^3.

Case 2 :  When a = 3q+1,

a^3= (3q+1)^3 

a^3 = 27q^3 +27q^2 + 9q+1

a^3 = 9(3q^3+ 3q^2 + q) +1.

a^3 = 9m+1

Where,  m is an integer such that m= (3q^3 + 3q^2 +q).

Case 3 : when a=3q+2 ,

a^3 = (3q+2)^3

a^3 = 27q^3 + 54q^2 + 36q + 8

a^3 = 9(3q^3 + 6q^2 + 4q) + 8

a^3 = 9m+8

Where m is integer such that m= (3q^3 +6q^2 + 4q)

Therefore,  the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

       


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March 11, 2020 at 1:18 AM ×

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