Class 10 Maths NCERT Solutions Chapter 1

EXERCISE 1.1

Question 1:

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

• (i) 135 and 225
• (ii) 196 and 38220
• (iii) 867 and 225

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 × 1 + 90

Since the remainder is 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain 135 = 90 × 1 + 45

We consider the new divisor 90 and the 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

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, the HCF of 196 and 38220 is 196.

(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 the 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, the HCF of 867 and 255 is 51.

Question 2:

Show that any positive odd integer is of form 6q + 1, or 6q + 3, or 6q + 5 where q is some integer.

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, and 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 form 6q + 1, 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? Answer:

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 form 3q, 3q + 1, or 3q + 2. Now  square each of these and show that they can be rewritten in the form of 3m or 3m

+ 1.]

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,

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.

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.

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

so, a =3q 0r 3q +1 or 3q +2

Therefore, every number can be represented in these three forms. There are three cases.

Case 1: When a = 3q

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a3 = (3q +1)

a3 = 27q3 + 27q2 + 9q +

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)

a3 = 27q3 + 54q2 + 36q +

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.

EXERCISE 1.2

Question 1:

Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

(i) 140 = 2×2×5×7

(ii) 156 = 2×2×3×13

(iii) 3825 = 3×3×5×5×17

(iv) 7429 = 17×19×23

Question 2:

Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Question 3:

Find the LCM and HCF of the following integers by applying the prime factorization method:
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25

Question 4:

Given that HCF (306, 657) = 9, find LCM (306, 657).

Question 5:

Check whether 6n can end with the digit 0 for any natural number n.

Prime factors of 6 = 2 and 3.

since 5 is not the prime factor of 6.

so, 6n cannot end with the digit 0.

Question 6:

Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Question 7:

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?