Class 10 CBSE Mathematics
Real Numbers
CBSE NCERT Solutions of Class 10 Mathematics Chapter 1
EXERCISE 1.1
1. Euclid’s Division Lemma: Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. This result was perhaps known for a long time but was first recorded in Book VII of Euclid’s Elements. Euclid’s division algorithm is based on this lemma
1.Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Step1:Since 225 >135,
Step1:Since 225 >135,
By apply Euclid's division lemma to
a = 225 and b = 135 to find q and r such that
225 = 135q + r, 0 ≤ r < 135
So, dividing 225 by 135 we get 1 as the quotient and 90 as remainder. 225 = (135 × 1) + 90
Step 2:
Remainder r is 90 and is not equal to 0, Again apply Euclid's division lemma
to b = 135 and r = 90 to find whole numbers q and r such that
135 = 90 × q + r, 0 ≤ r < 90
So, dividing 135 by 90 we get 1 as the quotient and 45 as remainder.
135 = (90 × 1) + 45
Step 3:
Again, remainder r is 45 and is not equal to 0, Again apply Euclid's
division lemma to b = 90 and r = 45 to find q and r such that
90 = 45 × q + r, 0 ≤ r < 45
So, dividing 90 by 45 we get 2 as the quotient and 0 as remainder.
90 = (2 × 45) + 0
Step 4:
Since the remainder is zero, the divisor at this stage will be HCF of
(135, 225)
Since the last divisor is 45, therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
38220 >196
By applying Euclid's division lemma
38220 = 196 q + r, 0 ≤ r < 196
So by dividing 38220 by 196, we get 195 as the quotient and 0 as the remainder
38220 = (196 × 195) + 0
As the remainder is zero, divisor, at last, is 196, therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Write the language as before, since 866>255 we will apply Euclid's division lemma
867 = 255q + r, 0 ≤ r < 255. So, by dividing 867 by 255 we get 3 as the quotient and 102 as remainder.
867 = 255 × 3 + 102 , since yet r not 0
255 = 102q + r where 0 ≤ r < 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Now, remainder is zero ,Therefore 51 as a last divisor be the HCF of 867 and 255
Q2.Show that any positive odd integer is of form 6q+1or 6q+3 or 6q+5 where q is some integer.
Let us start with taking a, where a is a positive odd integer. Now, apply the division algorithm with a and b =6
Since the last divisor is 45, therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
38220 >196
By applying Euclid's division lemma
38220 = 196 q + r, 0 ≤ r < 196
So by dividing 38220 by 196, we get 195 as the quotient and 0 as the remainder
38220 = (196 × 195) + 0
As the remainder is zero, divisor, at last, is 196, therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Write the language as before, since 866>255 we will apply Euclid's division lemma
867 = 255q + r, 0 ≤ r < 255. So, by dividing 867 by 255 we get 3 as the quotient and 102 as remainder.
867 = 255 × 3 + 102 , since yet r not 0
255 = 102q + r where 0 ≤ r < 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Now, remainder is zero ,Therefore 51 as a last divisor be the HCF of 867 and 255
Q2.Show that any positive odd integer is of form 6q+1or 6q+3 or 6q+5 where q is some integer.
Let us start with taking a, where a is a positive odd integer. Now, apply the division algorithm with a and b =6
Since 0 ≤ r < 6, the possible remainders are 0, 1, 2,3,4 and5 (r)
That is, a can be 6q, 6q + 1, 6q + 2, 6q + 3,6q + 4, or 6q + 5,where q is the quotient. However, since a is odd, a cannot be 6q or 6q + 2 or 6q+4 (since they are both divisible by 2).
Therefore, any odd integer is of the form 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?
To get the maximum number column to find HCF and for the minimum number find LCM
That is, a can be 6q, 6q + 1, 6q + 2, 6q + 3,6q + 4, or 6q + 5,where q is the quotient. However, since a is odd, a cannot be 6q or 6q + 2 or 6q+4 (since they are both divisible by 2).
Therefore, any odd integer is of the form 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?
To get the maximum number column to find HCF and for the minimum number find LCM
Use Euclid’s algorithm to find the HCF.
Here 616> 32 so divide the greater number with smaller one
When we divide 616 by 32 we get quotient 19 and remainder 8
so, 616 = 32 × 19 + 8
Now again divide 32 by 8 we get quotient 4 and no remainder
32 = 8 × 4 + 0
As there are no remainder our HCF will 8
Here 616> 32 so divide the greater number with smaller one
When we divide 616 by 32 we get quotient 19 and remainder 8
so, 616 = 32 × 19 + 8
Now again divide 32 by 8 we get quotient 4 and no remainder
32 = 8 × 4 + 0
As there are no remainder our HCF will 8
Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
If a is any positive integer and b = 3.
By using Euclid’s algorithm we get a = 3q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 , r can be 0,1, 2 as r < 3
Use the algebraic identity (a+b)² = a² + 2ab +b² to open the square bracket
(3q)² = 9q² if we divide by 3 we get no remainder
Use the algebraic identity (a+b)² = a² + 2ab +b² to open the square bracket
(3q)² = 9q² if we divide by 3 we get no remainder
it can be written as 3×(3q²) so it is in form of 3m here m = 3q²
(3q+1)² = (3q)² + 2×3q×1 + 1² =9q² + 6q +1 now divide by 3 we get 1 as remainder now, it can be written as
(3q+1)² = (3q)² + 2×3q×1 + 1² =9q² + 6q +1 now divide by 3 we get 1 as remainder now, it can be written as
3(3q² + 2q) +1 it is in form of 3m+1 and value of m = 3q² + 2q
(3q+2)² = (3q)² + 2×3q×2 + 2² =9q² + 12q +4 now divide by 3 we get 1 remainder
it can be written as 3(3q² + 4q +1) +1 it is in form of 3m +1 and value of m= 3q² + 4q +1
Hence, Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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.
Let
a be any positive integer and b = 3
a
= 3q + r, where q ≥ 0 and 0 ≤ r < 3
Then,
it is of the form 3q
or, 3q+
1 or, 3q+
2.
Case
1: When a = 3q,
a3
=(3q)3
= 27q3
= 9 (3q3)
= 9m
Where
m is an integer such that m =3q3
Case
2: When a = 3q + 1,
a3
=
(3q +1)3
a3=
27q3
+
27q3
+
9q + 1
a3
=
9(3q3
+
3q2+
q) + 1
a3=
9m + 1
Where
m is an integer such that m = (3q3+
3q2+
q)
Case
3: When a = 3q + 2,
a3=
(3q +2)3
a3
=
27q3+
54q2
+
36q + 8
a3=
9(3q3+
6q2
+
4q) + 8
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, or 9m + 8.
