NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

In this article, NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 are given for free. Detailed answers to all the questions in Exercise 1.1 Chapter 1 maths class 10 Real Numbers provided. NCERT Solutions of Class 10 Maths (Real Numbers Ch 1 Ex 1.1) are extremely helpful while doing homework.

QUESTION 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

In this case, 225 > 135, so apply Euclid’s division algorithm to 225 and 135, then we get

225 = 135 × 1 + 90

Since the remainder 90 ≠ 0, so apply the division lemma to 135 and 90, and we get

135 = 90 × 1 + 45

Again the remainder 45 ≠ 0, so apply the division lemma to 90 and 45, and we get

90 = 45 × 2 + 0

Now, the remainder has become zero, so our procedure stops. Since, in this stage, the divisor is 45, the HCF of 225 and 135 is 45.

Here, we notice that HCF (225, 135) = HCF (135, 90) = HCF (90, 45) = 45.

Short Method –

Apply Euclid’s division algorithm to find the HCF of 225 and 135 as follows:

225 = 135 × 1 + 90

135 = 90 × 1 + 45

90 = 45 × 2 + 0

Therefore, HCF (225, 135) = HCF (135, 90) = HCF (90, 45) = 45

(ii) 196 and 38220

In this case, 38220 > 196, so apply Euclid’s division algorithm to 38220 and 196, and we get

38220 = 196 × 195 + 0

Since, the remainder has become zero, stop procedures. Since, in this stage, the divisor is 196, the HCF of 38220 and 196 is 196.

Hence, we notice that HCF (38220, 196) = 196.

Short Method –

Apply Euclid’s division algorithm to find the HCF as follows:

38220 = 196 × 195 + 0

Therefore, HCF (38220, 196) = 196

(iii) 867 and 255

In this case, 867 > 255, so apply Euclid’s division algorithm to 867 and 255, and we get

867 = 255 × 3 + 102

Since the remainder 102 ≠ 0, so apply the division lemma to 255 and 102, and we get

255 = 102 × 2 + 51

Again the remainder 51 ≠ 0, so apply the division lemma to 102 and 51, and we get

102 = 51 × 2 + 0

Now, the remainder has become zero, so our procedure stops. Since, in this stage, the divisor is 51, the HCF of 225 and 135 is 51.

Here, we notice that HCF (867, 255) = HCF (255, 102) = HCF (102, 51) = 51.

Short Method –

Apply Euclid’s division algorithm to find the HCF as follows:

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

Therefore, HCF (867, 255) = HCF (255, 102) = HCF (102, 51) = 51

QUESTION 2 –

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

SOLUTION –

Let $a$ be any positive odd integer. Apply Euclid’s division algorithm to $a$ and $b = 6$ , we get

$a = 6q + r$ , for some integer $q \ge 0$ and $0 \le r < 6$ .

The possible values of $r$ , that is, the possible remainders are, $r = 0,1,2,3,4,5$ , because $0 \le r < 6$ .

Substitute these possible values into $a = 6q + r$ , to find $a$ :

If $r = 0$ , then $a = 6q + 0$ or $a = 6q$

If $r = 1$ , then $a = 6q + 1$

If $r = 2$ , then $a = 6q + 2$

If $r = 3$ , then $a = 6q + 3$

If $r = 4$ , then $a = 6q + 4$

If $r = 5$ , then $a = 6q + 5$

That is, $a$ can be $6q$ , $6q + 1$ , $6q + 2$ , $6q + 3$ , $6q + 4$ or $6q + 5$ .

But $a$ is an odd integer, so $a$ cannot be $6q$ , $6q + 2$ or $6q + 4$ (because these numbers are divisible by 2 and so these are even not odd).

Thus, it is proved that, any positive odd integer is of the form $6q + 1$ , $6q + 3$ or $6q + 5$ , where $q$ is some integer.

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 –

Given Data:

Number of members in an army contingent = 616

Number of members in an army band = 32

Both the groups have to march in the same number of columns.

Our goal is to find the maximum number of columns in which they can march, that is, we need to find the HCF of 616 and 32.

Apply Euclid’s division algorithm to find the HCF of 616 and 32, we get

616 = 32 × 19 + 8

Since the remainder 8 ≠ 0, so again apply the division lemma to 32 and 8, we get

32 = 8 × 4 + 0

Now, the remainder has become zero, so our procedure stops. Since, in this stage, the divisor is 8, the HCF of 616 and 32 is 8.

That is, HCF (616, 32) = HCF (32, 8) = 8.

Hence, the maximum number of columns is 8 in which they can march.

QUESTION 4 –

Use Euclid’s division lemma to show that the square of any positive integer is either of the form $3m$ or $3m + 1$ for some integer $m$ .

SOLUTION –

Let $a$ be any positive odd integer. Apply the Euclid’s division algorithm to $a$ and $b = 3$ , we get

$a = 3q + r$ , for some integer $q \ge 0$ and $0 \le r < 3$ .

The possible values of $r$ , that is, the possible remainders are, $r = 0,1,2$ , because $0 \le r < 3$ .

Substitute these possible values into $a = 3q + r$ , to find $a$ :

If $r = 0$ , then $a = 3q + 0$ or $a = 3q$

If $r = 1$ , then $a = 3q + 1$

If $r = 2$ , then $a = 3q + 2$

That is, $a$ can be $3q$ , $3q + 1$ or $3q + 2$ .

Now, find the square of $a$ , as per the question given,

If $a = 3q$ , then,

${a^2} = {\left( {3q} \right)^2} = 9{q^2} = 3 \times 3{q^2}$

Assume that, $m = 3{q^2}$ , then

${a^2} = 3m$ – – – Equation (1)

If $a = 3q + 1$ , then,

${a^2} = {\left( {3q + 1} \right)^2} = 9{q^2} + 6q + 1 = 3 \times \left( {3{q^2} + 2q} \right) + 1$

Assume that, $m = 3{q^2} + 2q$ , then

${a^2} = 3m + 1$ – – – Equation (2)

If $a = 3q + 2$ , then,

${a^2} = {\left( {3q + 2} \right)^2} = 9{q^2} + 12q + 4 = 9{q^2} + 12q + 3 + 1 = 3\left( {3{q^2} + 4q + 1} \right) + 1$

Assume that, $m = 3{q^2} + 4q + 1$ , then

${a^2} = 3m + 1$ – – – Equation (3)

Hence, from equations (1), (2) and (3), it is verified that the square of any positive integer $a$ is either of the form $3m$ or $3m + 1$ for some integer $m$ .

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 –

Let $a$ be any positive odd integer. Apply the Euclid’s division algorithm to $a$ and $b = 3$ , we get

$a = 3q + r$ , for some integer $q \ge 0$ and $0 \le r < 3$ .

The possible values of $r$ , that is, the possible remainders are, $r = 0,1,2$ , because $0 \le r < 3$ .

Substitute these possible values into $a = 3q + r$ , to find $a$ :

If $r = 0$ , then $a = 3q + 0$ or $a = 3q$

If $r = 1$ , then $a = 3q + 1$

If $r = 2$ , then $a = 3q + 2$

That is, $a$ can be $3q$ , $3q + 1$ or $3q + 2$ .

Now, find the cube of $a$ , as per the question given,

If $a = 3q$ , then,

${a^3} = {\left( {3q} \right)^3} = 27{q^3} = 9 \times 3{q^3}$

Assume that, $m = 3{q^3}$ , then

${a^3} = 9m$ – – – Equation (1)

If $a = 3q + 1$ , then,

${a^3} = {\left( {3q + 1} \right)^3} = 27{q^3} + 27{q^2} + 9q + 1 = 9\left( {3{q^3} + 3{q^2} + q} \right) + 1$

Assume that, $m = 3{q^3} + 3{q^2} + q$ , then

${a^3} = 9m + 1$ – – – Equation (2)

If $a = 3q + 2$ , then,

${a^3} = {\left( {3q + 2} \right)^3} = 27{q^3} + 54{q^2} + 36q + 8 = 9\left( {3{q^3} + 6{q^2} + 4q} \right) + 8$

Assume that, $m = 3{q^3} + 6{q^2} + 4q$ , then

${a^3} = 9m + 8$ – – – Equation (3)

Hence, from equation (1), (2) and (3), it is verified that, the cube of any positive integer $a$ is of the form $9m$ , $9m + 1$ or $9m + 8$ for some integer $m$ .

Tags: The first chapter of class 10th maths is Real Numbers. How do you find the HCF of Class 10? NCERT class 10 maths chapter 1 ex 1.1, CBSE ncert class 10 maths chapter 1 exercise 1.1 solutions, ncert class 10 maths ch 1 ex 1.1 solutions, Ex 1.1 Class 10 Maths Question 1, Ex 1.1 Class 10 Maths Question 2, Ex 1.1 Class 10 Maths Question 3, Ex 1.1 Class 10 Maths Question 4, Ex 1.1 Class 10 Maths Question 5

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NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

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NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

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