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.
Answer:-
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.