### NCERT Solutions Class 10 Maths Chapter – 1 Exercise – 1.1

NCERT Solutions For Class 10 Math Chapter – 1 Real Numbers Exercise – 1.1

Q1. Use Euclid’s division algorithm to find HCF of:

• 135 and 225
• 196 and 38220
• 867 and 255

Solution:-

• 225 > 135 we always divide greater number with smaller one.

Divide  225 by 135 we get 1 quotient and 90 as remainder so that 225 = 135 x 1 + 90

Divide 135 by 90 we get  1 quotient and 45 as remainder so that 135 = 90 x 1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as 90 = 2 x 45 +

0

As  there are no remainder so divisor 45 is our HCF.

• 38220 > 196 we  always divide  greater number with smaller one.

Divide 38220 by 196 then we get quotient 185 and no remainder so we can write it as 38229 = 196 x 195 + 0

As there is no remainder so divisor 196 is our HCF.

• 867 > 255 we always divide greater number with smaller one.

Divide 867 by 255 then we get quotient 2 and remainder is 102 so we can write it as 867 = 225 x 3 + 102

Divide 255   by 102 then we get quotient 2 and remainder is 51 so we can write it as 255 = 102 x 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder  so we can wrote it as 102 = 51 x 2 + 0

As there is no remainder so divisor 51 is our HCF.

Q2. Show that any positive odd integer is of the form 6q + 1 , or 6q + 3 , or 6q + 5 where q is some integer.

Let take a as any positive integer and b = 6.

Then using Euclid algorithm we get a = 6q + r  here r is remainder and value of q is more than or equal to 0 and r = 0 , 1 , 2 , 3 , 4 , 5 because 0 < r < b and the value of b is 6 So total possible forms will 6q + 0 . 6q + 1 , 6q + 2 , 6q + 3 , 6q + 4 , 6q + 5.

6q + 0

6 is divisible by 2 so it is a even number

6q + 1

6 is divisible by 2 but 1 is not divisible by 2 so it a odd number

6q + 2

6 is divisible by 2 and 2 is also divisible by 2 so it is a even number.

6q + 3

6 is divisible by 2 but 3 is not divisible by 2 so it is odd number.

6q + 4

6 is divisible by 2 and 4 is also divisible by  2 it is an even number.

6q + 5

6 is divisible by 2 and 5 is not divisible by 5 it is odd number.

So , odd numbers will in form of 6q + 1 , or 6q + 3 , or 6q + 5.

Q3. 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 ?

Solution:-

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

We can use Euclid algorithm to find the HCF.

616 = 32 x 19 + 8

32 = 8 x 4 + 0

The HCF (616, 32) is 8.

Therefore , they can march in 8 columns each.

Q4. Use Euclid division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Solution:-

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,

A2 = (3q)2 or (3q + 1)2 or (3q + 2 )2

A2 =(9q)2 or 9q2 + 6q + 1 or 9q2 + 12q + 4

= 3 x (3q2) or 3(3q2 + 2q) + 1 or 3(3q2  + 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 integers is either of the form 3m or 3m + 1.

Q5. Use Euclid division lemma to show that the cube of any positive integer is of the form 9m , 9m + 1 or 9m + 8.

Solution:-

Let a be any positive integer and b = 3

A = 3q + r where q > 0 and 0 < r < 3

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

A3 = (3q)3

= 27q3

= 9m

Case 2: When a = 3q + 1

A3 = (3q + 1)3

A3 = 9m + 1

Case 3: When a = 3q + 2

A3 =  (3q + 2)3

A3 = 9m + 8

Where m is an integer such that m = (3q3 + 6q2 + 4q)

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

NCERT Solutions For Class 10 Math Chapter – 1 Real Numbers Exercise – 1.1   Q1. Use Euclid’s division algorithm to…