NCERT Solutions Class 10 Maths
Chapter – 1 (Real Numbers)
The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 1 Real Numbers Exercise 1.1 has been provided here to help the students in solving the questions from this exercise.
Chapter : 1 Real Numbers
- NCERT Class 10 Maths Solution Ex – 1.2
- NCERT Class 10 Maths Solution Ex – 1.3
- NCERT Class 10 Maths Solution Ex – 1.4
Exercise – 1.1
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
Solutions – Since 225 > 135, we apply the division lemma to 225 and 135 to obtain.
Step 1 : By Euclid division lemma
225 = 135 × 1 + 90
Step 2 : Here remainder is not zero.
So, for divisor 135 and remainder 90.
135 = 90 × 1 + 45
Step 3 : Here remainder is not zero.
So for divisor 18 and remainder 3.
90 = 45 × 2 + 0
Remainder = 0
Hence HCF of 135 and 225 = 45
(ii) 196 and 38220
Solutions – Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
Step 1 : By Euclid division lemma
38220 = 196 × 195 + 0
Remainder = 0
Step 2 : Hence HCF of 196 and 38220 = 196.
(iii) 867 and 255
Solutions – Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
Step 1 : By Euclid division lemma
867 = 255 × 3 + 102
Step 2 : Here remainder is not zero.
So, for divisor 255 and remainder 102.
255 = 102 × 2 + 51
Step 3 : Here remainder is not zero.
So for divisor 102 and remainder 51.
102 = 51 × 2 + 0
Step 4 : Here remainder is zero.
Since the divisor at this stage is 51,
Hence HCF of 867 and 255 = 51.
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution – Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5
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?
Solution – Given, Number of army contingent members = 616
Number of army band members = 32
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.
4. 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.
Solutions – 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,
a2 = (3q)2 = 9q2 = 3 × 3q2 = 3k1
Where k1 is an integer such that k1 = 3q2
Case 2: When a = 3q + 1,
a2 = (3q + 1)2 = (3q)2 + 12 + 2 × 3q × 1
= 9q2 + 1 +6q = 3(3q2+2q) + 1 = 3k2 + 1
Where k2 is an integer such that k2 = 3q2 + 2q
Case 3: When a = 3q + 2,
a2 = (3q + 2)2 = (3q)2 + 22 + 2 × 3q × 2
= 9q2 + 4 + 12q = 3 (3q2 + 4q + 1) + 1 = 3k3 + 1
Where k3 is an integer such that k3 = (3q2 + 4q + 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.
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.
Solution – Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q + r, where q ≥ 0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, putting the value of r, we get,
x = 3q
or
x = 3q + 1
or
x = 3q + 2
Now, by taking the cube of all the three above expressions, we get,
Case (i): When r = 0, then,
x2= (3q)3 = 27q3= 9(3q3) = 9m; where m = 3q3
Case (ii): When r = 1, then,
x3 = (3q + 1)3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1) = 27q3 + 1 + 27q2 + 9q
Taking 9 as common factor, we get,
x3 = 9(3q3+3q2+q)+1
Putting = m, we get,
Putting (3q3 + 3q2 + q) = m, we get ,
x3 = 9m+1
Case (iii): When r = 2, then,
x3 = (3q+2)3 = (3q)3 + 23 + 3 × 3q × 2(3q + 2) = 27q3 + 54q2 + 36q + 8
Taking 9 as common factor, we get,
x3 = 9(3q3 + 6q2 + 4q) + 8
Putting (3q3 + 6q2 + 4q) = m, we get,
x3 = 9m + 8
Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
