Ncert Class 10 math chapter-1 exercise-1.1.

                                                               

Full Ncert Class 10 math chapter-1 Solution exercise-1.1 

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                                                 Exercise-1.1

Some important things to know:-

• Euclid’s division Algorithm: For given positive integers ‘a’ and ‘b’ there exist distinctive whole numbers ‘q’ and ‘r’ satisfying the relation a = bq + r, o ≤ r < b.
• Euclid’s division algorithm: HCF of any 2 positive integers a and b with a > b is obtained as follows:
• Step one: Apply Euclid’s division algorithm to a and b to seek out letter of the alphabet      and r such a = bq + r, zero ≤ r < b.
• Step a pair of : If r = zero then HCF (a, b) = b ; if r ≥ zero then again apply Euclid’s algorithm to b and r.
• Repeat the steps until we tend to get r = zero

                                      

Question 1: 

     Use Euclid’s division algorithm to find the HCF of:

    (i) 135 and 225

Here we Have 225 > 135, we apply the division algorithm to 225 and 

135 to obtain

225 = 135 × 1 + 90

Here we Have remainder 90 ≠ 0, we apply the division algorithm to 135

and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division algorithm to obtain

90 = 2 × 45 + 0

Here we Have the remainder is zero, the process stops.

Here we Have the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Here we Have 38220 > 196, we apply the division algorithm to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Here we Have the remainder is zero, the process stops.

Here we Have the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii)  867 and 255

Here we Have 867 > 255, we apply the division algorithm to 867 and 

255 to obtain

867 = 255 × 3 + 102

Here we Have remainder 102 ≠ 0, we apply the division algorithm to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the 

division algorithm to obtain

102 = 51 × 2 + 0

Here we Have the remainder is zero, the process stops.

Here we Have 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 + 2 '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

Similarly, 6q + 1 = 2 × 3q + 1 = 2m₁ + 1, where m₁ is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2m₂ + 1, where m₂ is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2m₃ + 1, where m₃ is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2m + 1, where m 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?

Solution:-

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].

Solution –

            Suppose, a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0, Where, r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

a² = (3q) ²         Or  (3q + 1)²                    Or  (3q + 2)²

a2=9q2             Or  9q2+6q+1               Or  9q2+12q+4 =9q2+12q+3+1 

a2=3*3q2         Or  3(3q2+2q)+1          Or  3(3q2+4q+1)+1 

a2=3m₁,           Or  3m₂+1                     Or  3m₃+1

Where m₁, m₂, and m₃ 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.

Solution –

            Suppose, a be any positive integer and b = 3 so, a = 3q + r, where q ≥ 0 
            and 0 ≤ r < 3

Therefore, every number can be represented as these three forms.

Case 1: When a = (3q)³ = (27q³) =9(3q³) =9m ,Where  m = 3q³

Case 2: When a = 3q + 1,

a³ = (3q +1)³                                       [Formula (a + b)³ = a³ + b³ + 3ab(a + b)]

a³ = 27q³ + 27q² + 9q + 1

           

a³= 9(3q³ + 3q² + q) + 1

a³= 9m + 1                                       Where m = (3q3 + 3q2 + q)

     Case 3: When a = 3q + 2,

a3 = (3q +2)3                                  [Formula (a + b)3 = a3 + b3 + 3ab(a + b)]

a3 = 27q3 + 54q2 + 36q + 8

a3 = 9(3q3 + 6q2 + 4q) + 8

a3 = 9m + 8                                     Where m = (3q3 + 6q2 + 4q)

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