NCERT solution class 10 chapter 1 Real numbers

Exercise 1.1    Exercise 1.2   Exercise 1.3   Exercise 1.4

Exercise 1.1

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.

 

 

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 \ge  0, and r = 0, 1, 2, 3, 4, 5 because 0 \le 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

 

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.

 

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 \ge  0

And r = 0, 1, 2 because 0 \le 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.

 

 

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 \ge  0 and 0 \le 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,

\\a^3=(3q)^3

=27q^3

=9(3q^3)

Where m is an integer such that m=3q^3

Case 2: When a = 3q + 1,

a^3= (3q +1)^3

a^3 = 27q^3 + 27q^2+ 9q + 1

a^3 = 9(3q^3 + 3q^2+ q) + 1

a^3= 9m + 1

Where m is an integer such that m = (3q^3+ 3q^2+ q)

Case 3: When a = 3q + 2,

a^3 = (3q +2)^3

a^3 = 27q^3 + 54q^2 + 36q + 8

a^3 = 9(3q^3 + 6q^2 + 4q) + 8

a^3 = 9m + 8

Where m is an integer such that m = (3q^3 + 6q^2 + 4q)

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

Leave a Reply

Your email address will not be published. Required fields are marked *