NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers Exercise 1.1: Detailed Guide for Assam State Board Syllabus 2025-2026 (English Medium)

NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers Exercise 1.1: Detailed Guide for Assam State Board Syllabus 2025-2026 (English Medium)

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 – Digital Pipal Academy

Digital Pipal Academy offers concise and easy-to-understand NCERT Solutions for Class 10 Maths Chapter 1, Exercise 1.1, covered under the Assam State Board Syllabus 2025-2026. Our expert faculty provides step-by-step solutions, focusing on Euclid’s Division Algorithm to help students grasp key concepts and solve problems efficiently. Aligned with NCERT guidelines, our solutions ensure thorough exam preparation and help students score better in their exams.

Real Number	Solution Link
Exercise 1.1 (Real Number)	Click Here
Exercise 1.2 (Real Number)	Click Here
Exercise 1.3 (Real Number)	Click Here
Exercise 1.4 (Real Number)	Click Here

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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
(iv) 272 and 1032
(v) 405 and 2520
(vi) 155 and 1385
(vii) 384 and 1296
(viii) 1848 and 3058

Solution:

(i) Applying Euclid’s division algorithm on 135 and 225:
225 = 1 × 135 + 90
135 = 1 × 90 + 45
90 = 2 × 45 + 0
HCF (225, 135) = 45

(ii) Applying Euclid’s division algorithm on 196 and 38220:
38220 = 195 × 196 + 0
HCF (38220, 196) = 196

(iii) Applying Euclid’s division algorithm on 867 and 255:
867 = 3 × 255 + 102
255 = 2 × 102 + 51
102 = 2 × 51 + 0
HCF (867, 255) = 51

(iv) Applying Euclid’s division algorithm on 272 and 1032:
1032 = 272 × 3 + 216
272 = 216 × 1 + 56
216 = 56 × 3 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
HCF (272, 1032) = 8

(v) Applying Euclid’s division algorithm on 405 and 2520:
2520 = 405 × 6 + 90
405 = 90 × 4 + 45
90 = 45 × 2 + 0
HCF (405, 2520) = 45

(vi) Applying Euclid’s division algorithm on 155 and 1385:
1385 = 155 × 8 + 145
155 = 145 × 1 + 10
145 = 10 × 14 + 5
10 = 5 × 2 + 0
HCF (155, 1385) = 5

(vii) Applying Euclid’s division algorithm on 384 and 1296:
1296 = 384 × 3 + 144
384 = 144 × 2 + 96
144 = 96 × 1 + 48
96 = 48 × 2 + 0
HCF (384, 1296) = 48

(viii) Applying Euclid’s division algorithm on 1848 and 3058:
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 5 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
HCF (1848, 3058) = 22

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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 a positive integer. Applying the division algorithm with divisor $b = 6$, we get:
$a = 6q + r$ where $0 \leq r < 6$, and $r$ can take values: $0, 1, 2, 3, 4, 5$.

Now:

• If $r = 0$, $a = 6q$ (divisible by 6, even).
• If $r = 1$, $a = 6q + 1$ (odd).
• If $r = 2$, $a = 6q + 2$ (divisible by 2, even).
• If $r = 3$, $a = 6q + 3$ (odd).
• If $r = 4$, $a = 6q + 4$ (divisible by 2, even).
• If $r = 5$, $a = 6q + 5$ (odd).

Thus, positive odd integers are of the form $6q + 1$, $6q + 3$, or $6q + 5$.

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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:
To find the maximum number of columns, we need to find the HCF of $616$ and $32$ using Euclid's division algorithm:

1. $616 = 32 \times 19 + 8$
2. $32 = 8 \times 4 + 0$

Thus, $\text{HCF}(616, 32) = 8$.

Answer: The maximum number of columns is 8.

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

Solution:
Let $a$ be any positive integer. Using Euclid’s division algorithm, for $b = 3$, we have:
$a = 3q + r, \; \text{where } r = 0, 1, 2.$

Case 1: When $r = 0$:
$a = 3q + 0$
$⇒ a = 3q$
$⇒ a^2 = (3q)^2$
$⇒ a^2 = 9q^2$
$⇒ a^2 = 3(3q^2)$
$⇒ a^2 = 3m$
$\text{[where \( m = 3q^2 \)]}$

Case 2: When $r = 1$:
$a = 3q + 1$
$⇒ a^2 = (3q + 1)^2$
$⇒ a^2 = (3q)^2 + 2 \cdot (3q) \cdot 1 + 1^2$
$⇒ a^2 = 9q^2 + 6q + 1$
$⇒ a^2 = 3(3q^2 + 2q) + 1$
$⇒ a^2 = 3m + 1$
$\text{[where \( m = 3q^2 + 2q \)]}$

Case 3: When $r = 2$:
$a = 3q + 2$
$⇒ a^2 = (3q + 2)^2$
$⇒ a^2 = (3q)^2 + 2 \cdot (3q) \cdot 2 + 2^2$
$⇒ a^2 = 9q^2 + 12q + 4$
$⇒ a^2 = 3(3q^2 + 4q + 1) + 1$
$⇒ a^2 = 3m + 1$
$\text{[where \( m = 3q^2 + 4q + 1 \)]}$

Thus, in all cases, the square of any positive integer is either of the form $3m$ or $3m + 1$.

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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 $9m + 8$.

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Solution:
Let $a$ be any positive integer, and $b = 3$.
Using Euclid's division lemma, we have:

Using Euclid’s division lemma,
$a = 3q + r$,
where $r = 0, 1, 2$.

Case 1: When $r = 0$:
$a = 3q + 0$
$⇒ a = 3q$
$⇒ a^3 = (3q)^3$
$⇒ a^3 = 27q^3$
$⇒ a^3 = 9 \cdot 3q^3$
$⇒ a^3 = 9m$
$\text{[where \( m = 3q^3 \)]}$

Case 2: When $r = 1$:
$a = 3q + 1$
$⇒ a^3 = (3q + 1)^3$
$⇒ a^3 = (3q)^3 + 1^3 + 3 \cdot (3q) \cdot 1 \cdot (3q + 1)$
$⇒ a^3 = 27q^3 + 1 + 9q (3q + 1)$
$⇒ a^3 = 27q^3 + 1 + 27q^2 + 9q$
$⇒ a^3 = 27q^3 + 27q^2 + 9q + 1$
$⇒ a^3 = 9 (3q^3 + 3q^2 + q) + 1$
$⇒ a^3 = 9m + 1$
$\text{[where \( m = 3q^3 + 3q^2 + q \)]}$

Case 3: When $r = 2$:
$a = 3q + 2$
$⇒ a^3 = (3q + 2)^3$
$⇒ a^3 = (3q)^3 + 2^3 + 3 \cdot (3q) \cdot 2 \cdot (3q + 2)$
$⇒ a^3 = 27q^3 + 8 + 18q (3q + 2)$
$⇒ a^3 = 27q^3 + 8 + 54q^2 + 36q$
$⇒ a^3 = 27q^3 + 54q^2 + 36q + 8$
$⇒ a^3 = 9 (3q^3 + 6q^2 + 4q) + 8$
$⇒ a^3 = 9m + 8$
$\text{[where \( m = 3q^3 + 6q^2 + 4q \)]}$

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

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6. Himadri has a collection of 625 Indian postal stamps and 325 International postal stamps. She wants to display them in identical groups of Indian and International stamps with no stamps left out. What is the greatest number of groups Himadri can display?

Solution:
To find the greatest number of groups, we need to find the HCF of $625$ and $325$:

1. $625 = 325 \times 1 + 300$
2. $325 = 300 \times 1 + 25$
3. $300 = 25 \times 12 + 0$

Thus, $\text{HCF}(625, 325) = 25$.

Answer: The greatest number of groups is 25.

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7. Two ropes are of length 64 cm and 80 cm. Both are to be cut into pieces of equal length. What should be the maximum length of the pieces?

Solution:
To find the maximum length of the pieces, we need to find the HCF of $64$ and $80$:

1. $80 = 64 \times 1 + 16$
2. $64 = 16 \times 4 + 0$

Thus, $\text{HCF}(80, 64) = 16$.

Answer: The maximum length of the pieces is 16 cm.

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