“Explore the comprehensive NCERT Solutions for Class 10 Mathematics, specifically for Chapter 1: Real Numbers, Exercise 1.1. Our team of expert educators has meticulously crafted these solutions to support Class 10 students in their exam preparations. These solutions have been carefully reviewed and structured to serve as valuable references for students.

Within Real Numbers, Exercise 1.1, students delve into the concept of integer divisibility through the application of Euclid’s Division Algorithm. Our detailed explanations provide a step-by-step breakdown of each question from the NCERT textbook for Class 10. These solutions adhere to the NCERT guidelines, ensuring comprehensive coverage of the syllabus and aiding students in achieving excellent results in their examinations.

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Answers of Maths NCERT Class 10 Chapter 1 – Real Number Exercise 1.1 page number 7

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

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

Solutions:

i. 135 and 225

In the question 225 is greater than 135. By Euclid’s division algorithm, we have
225 = 135 × 1 + 90
Now, the remainder is 90, which is not zero. We continue by applying the division lemma to 90:
135 = 90 × 1 + 45
Again, the remainder is not zero (45), so we repeat the above step:
90 = 45 × 2 + 0
At this stage, the remainder is zero, which tells us our process stops here. The divisor in the last step was 45.
Therefore, the HCF of 225 and 135 is 45.

ii. 196 and 38220
In the question, 38220 is greater than 196, by applying Euclid’s division algorithm we have
38220 = 196 × 195 + 0
We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196
Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 255
867>225, by Euclid’s division algorithm
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
The remainder is zero, and the divisor in the last step was 51.
Hence the HCF of 867 and 255 is 51

2. Show that any positive odd integer is of form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Explanation: 
According to Euclid’s Division Lemma, if a and b are two positive integers, then
a = bq + r
where 0 ≤ r < b
Let’s consider a positive integer ‘a’ in the form 6q + r, where ‘q’ is any integer.
This implies that 0 ≤ r < 6, which means ‘r’ can take on values of 0, 1, 2, 3, 4, or 5, but it cannot be 6 since ‘r’ is always less than 6.
So, we can express ‘a’ as a sum of 6q and the possible values of ‘r’, where ‘q’ is an integer greater than or equal to 0, and ‘r’ can be 0, 1, 2, 3, 4, or 5.
Therefore, the possible forms of ‘a’ are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5.

Now, let’s examine these expressions further:

1. 1. 6q + 1 can be rewritten as 2(3q) + 1, where ‘3q’ is a positive integer ‘k1’.

1. 1. 6q + 3 can be expressed as 6q + 2 + 1, which is equal to 2(3q + 1) + 1, where ‘3q + 1’ is a positive integer ‘k2’.

1. 1. 6q + 5 can be written as 6q + 4 + 1, which simplifies to 2(3q + 2) + 1, where ‘3q + 2’ is a positive integer ‘k3’.

In each case, 6q + 1, 6q + 3, and 6q + 5 are in the form 2k + 1, where ‘k’ is an integer. Consequently, 6q + 1, 6q + 3, and 6q + 5 are not evenly divisible by 2.

Therefore, these expressions represent odd numbers, and it follows that any odd integer can be expressed in one of the forms: 6q + 1, 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:

Explanation:

To determine the maximum number of columns in which both the army contingent and the army band can march together, we need to find their highest common factor (HCF), often referred to as the greatest common divisor (GCD). The HCF of 616 and 32 will give us the answer.

We can use Euclid’s algorithm to calculate the HCF:

1. 1. Since 616 is greater than 32, we start with the following division:616 = 32 × 19 + 8

1. 1. As the remainder is 8 (which is not zero), we continue the process. This time, we take 32 as the new divisor:32 = 8 × 4 + 0

1. 1. Now, the remainder is zero, indicating that our process ends here. The divisor in the last step was 8.

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

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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.

Solution:

Explanation:

Let’s consider a positive integer ‘x’ and set ‘y’ as 3.

According to Euclid’s division algorithm, we can express ‘x’ as follows:

x = 3q + r, where ‘q’ is an integer greater than or equal to 0, and ‘r’ satisfies 0 ≤ r < 3.

This implies that ‘x’ can be expressed as 3q, 3q + 1, or 3q + 2.

Now, in accordance with the problem statement, we square both sides:

x² = (3q)³ = 9q³ = 3 X 3q²

Let’s denote 3q² as ‘m.’

So, we have x^2 = 3m ……………..(1)

Now, consider when x = 3q + 1:

x² = (3q + 1)² = (3q)² + 1² + 2 * 3q * 1 = 9q² + 1 + 6q = 3(3q² + 2q) + 1

Let’s denote 3q² + 2q as ‘m.’

So, we have x^2 = 3m + 1 ……………..(2)

Finally, when x = 3q + 2:

x² = (3q + 2)² = (3q)² + 2² + 2 * 3q * 2 = 9q² + 4 + 12q = 3(3q² + 4q + 1) + 1

Let’s denote 3q² + 4q + 1 as ‘m.’

So, we have x² = 3m + 1 ……………..(3)

From equations (1), (2), and (3), we conclude that the square of any positive integer can be represented in the form 3m or 3m + 1, where ‘m’ is an integer.

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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:

Explanation:

Consider a positive integer ‘a.’ By applying Euclid’s division lemma, we know that for any positive integers ‘a’ and ‘b,’ there exist unique integers ‘q’ and ‘r’ such that a = bq + r, where 0 ≤ r < b.

Now, let’s take b = 3. This implies that 0 ≤ r < 3, meaning r can take the values 0, 1, or 2, but it cannot be 3 since r is always less than 3.

Therefore, the possible expressions for ‘a’ are 3q, 3q + 1, or 3q + 2.

Now, let’s calculate the cube of each of these possible values of ‘a.’

Suppose q is any positive integer; its cube (denoted as “m”) will also be a positive integer.

Upon careful observation, we find that the cube of all positive integers can be represented in the form 9m, 9m + 1, or 9m + 8 for some integer ‘m.’

Now, let’s illustrate this for each case:

Case I: When a = 3q (a)³ = (3q)³ = 27q³
= 9(3q³)
= 9m
where ‘m’ is an integer such that m = 3q³.

Case II: When a = 3q + 1
(a)³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1, where ‘m’ is an integer such that m = 3q³ + 3q² + q.

Case III: When a = 3q + 2
(a)³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8, where ‘m’ is an integer such that m = 3q³ + 6q² + 4q.

Hence, we can conclude that the cube of any positive integer can be expressed in the form 9m, 9m + 1, or 9m + 8.