NCERT solution class 10 chapter 1 Real numbers exercise 1.1

Exercise 1.1

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Introduction of Real Numbers, Maths Class 10th

Q 1, Ex 1.1 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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Question 1
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
Sol :
(i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220
Sol :
(ii)196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255
Sol :
(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

Q 2, Ex 1.1 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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Question 2
Show that any positive odd integer is of the form 6q+1 , or 6q+3 or 6q+5 where q is some integer.
Sol :
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, $6q + 1 = 2 × 3q + 1 = 2k_1 + 1$, where k₁ is a positive integer
$6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k_2+ 1$, where k₂ is an integer
$6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k_3+ 1$, 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 $6q + 5$

Q 3, Ex 1.1 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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Question 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?
Sol :
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 4, Ex 1.1 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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Question 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.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Sol :
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,
$a^2=(3q)^2~or~(3q+1)^2~or~(3q+2)^2$
$a^2=(9q^2)~or~(9q^2+6q+1)~or~9q^2+12q+4$
$=3\times(3q^2)~or~3(3q^2+2q)+1~or~3(3q^2+4q+1)+1$
$=3k_1~or~3k_2+1~or~3k_3+1$
Where $ k_1 , k_2~and ~k_3  $are some positive integers
Hence, it can be said that the square of any positive integer is either of the form $3m~or~3m + 1.$

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Q 5, Ex 1.1 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT

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Question 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
Sol :
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=9(3q³)
Where m is an integer such that m = 3q³
Case 2: When a = 3q + 1,
a³ = (3q +1)³
a³ = 27q³ + 27q² + 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.