NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 are provided here. These solutions are prepared by our subject expert faculty to help students in their Class 10 exam preparations. These experts review these NCERT Maths Solutions chapter-wise to help students to solve problems easily while using them as a reference. They also focus on preparing the NCERT Solutions for these exercises in such a way that it is easy to understand for the students. The first exercise in Real Numbers- Exercise 1.1 explains the divisibility of integers using Euclid’s Division Algorithm. They provide a detailed and step-wise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 10. The solutions are always prepared by following NCERT guidelines so students can cover the whole syllabus accordingly. These are very helpful in scoring well in examinations.

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### Access 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

As you can see from the question, 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have

225 = 135 × 1 + 90

Now, the remainder 90 ≠ 0; thus, again, using the division lemma for 90, we get

135 = 90 × 1 + 45

Again, 45 ≠ 0; repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45; therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220

In the given question, 38220>196; therefore, by applying Euclid’s division algorithm and taking 38220 as the divisor, we get

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

As we know, 867 is greater than 255. Let us now apply Euclid’s division algorithm on 867 to get

867 = 255 × 3 + 102

Remainder 102 ≠ 0; therefore, taking 255 as the divisor and applying the division lemma method, we get

255 = 102 × 2 + 51

Again, 51 ≠ 0. Now, 102 is the new divisor, so by repeating the same step, we get

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. In the last step, the divisor is 51; therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 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: **

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.

Now, substituting the value of r, we get

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

**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,

Number of army contingent members=616

Number of army band members = 32

If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF (616, 32) gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm to find their HCF, we get

Since 616>32, therefore,

616 = 32 × 19 + 8

Since 8 ≠ 0, therefore, taking 32 as the new divisor, we have

32 = 8 × 4 + 0

Now, we have the remainder as 0; therefore, HCF (616, 32) = 8

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

**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.**

**Solutions: **

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3

Therefore, x = 3q, 3q+1 and 3q+2

Now, as per the question given, by squaring both sides, we get

x^{2} = (3q)^{2} = 9q^{2} = 3 × 3q^{2}

Let 3q^{2} = m

Therefore, x^{2}= 3m ……………………..(1)

x^{2 }= (3q + 1)^{2 }= (3q)^{2}+1^{2}+2×3q×1 = 9q^{2} + 1 +6q = 3(3q^{2}+2q) +1

Substitute 3q^{2}+2q = m to get

x^{2}= 3m + 1 ……………………………. (2)

x^{2}= (3q + 2)^{2 }= (3q)^{2}+2^{2}+2×3q×2 = 9q^{2 }+ 4 + 12q = 3 (3q^{2 } + 4q + 1)+1

Again, substitute 3q^{2}+4q+1 = m to get

x^{2}= 3m + 1…………………………… (3)

Hence, from equations 1, 2 and 3, we can say that the square of any positive integer is either of 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 x be any positive integer and y = 3.

By Euclid’s division algorithm, then

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get

**Case (i):** When r = 0, then,

x^{2}= (3q)^{3} = 27q^{3}= 9(3q^{3})= 9m; where m = 3q^{3}

**Case (ii):** When r = 1, then,

x^{3} = (3q+1)^{3} = (3q)^{3 }+1^{3}+3×3q×1(3q+1) = 27q^{3}+1+27q^{2}+9q

Taking 9 as a common factor, we get

x^{3 }= 9(3q^{3}+3q^{2}+q)+1

Putting = m, we get

Putting (3q^{3}+3q^{2+}q) = m, we get

x^{3} = 9m+1

**Case (iii): When r = 2, then,**

x^{3} = (3q+2)^{3}= (3q)^{3}+2^{3}+3×3q×2(3q+2) = 27q^{3}+54q^{2}+36q+8

Taking 9 as a common factor, we get

x^{3}=9(3q^{3}+6q^{2}+4q)+8

Putting (3q^{3}+6q^{2}+4q) = m, we get

x^{3} = 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Exercise 1.1 of NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers is the first exercise of Chapter 1 of Class 10 Maths. Real Numbers is introduced in Class 9 and is discussed further in detail in Class 10 by studying Euclid’s division Algorithm. The exercise discusses the divisibility of integers. The divisibility of integers using Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b, such that the remainder will be which is smaller than b.

- Euclid’s Division Algorithm – It includes 5 questions based on Theorem 1.1 – Euclid’s Division Lemma.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 1 – Real Number Exercise 1.1 Page number 7

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