NCERT Solutions Maths Class 10 Chapter 1

# NCERT Solutions Maths Class 10 Chapter 1

Go To:    Exercise 1.4

NCERT Solutions Maths Class 10 Exercise 1.1

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

We have 225 > 135,

So, applying the division lemma to 225 and 135, we get

225 = 135 × 1 + 90

Here remainder 90 ≠ 0, again applying the division lemma to 135 and 90, we get

135 = 90 × 1 + 45

Here remainder 90 ≠ 0, again applying the division lemma to 90 and 45, we get

90 = 2 × 45 + 0

Now, since the remainder is zero, the process gets stop.

The divisor at this stage is 45.

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

We have 38220 > 196,

So, applying the division lemma to 38220 and 196, we get

38220 = 196 × 195 + 0

Now, since the remainder is zero, the process gets stop.

The divisor at this stage is 196.

Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

We have 867 > 255,

So, applying the division lemma to 867 and 255, we get

867 = 255 × 3 + 102

Here remainder 102 ≠ 0, again applying the division lemma to 255 and 102, we get

255 = 102 × 2 + 51

Here remainder 51 ≠ 0, again applying the division lemma to 102 and 51, we get

102 = 51 × 2 + 0

Now, since the remainder is zero, the process stops.

The divisor at this stage is 51.

Therefore, HCF of 867 and 255 is 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 + 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: We have to find the HCF of 616 and 32 to find the maximum number of columns in which they can march.

To find the HCF of 616 and 32, we can use Euclid’s division algorithm.

616 = 32 × 19 + 8

Here remainder 8 ≠ 0, again applying the division algorithm to 32 and 8, we get

32 = 8 × 4 + 0

Hence, the HCF of 616 and 32 is 8.

Therefore, they can march in 8 columns each.

###### [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: Let a be any positive integer and b = 3.

Then, by Euclid’s division lemmaa = 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,

Where k1, k2 and k3 are some positive integers.

Hence, we can say that the 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.

Solution: Let a be any positive integer and b = 3.

Then, by Euclid’s division algorithm, a = 3q + r, for some integer ≥ 0 and r = 0, 1, 2 because 0 ≤ r < 3.

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

Thus, every positive integer can be represented as these three forms.

We have three cases.

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 + 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, 9+ 1 or 9m + 8.

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