NCERT Solutions for class 10 Maths Chapter 1- Real Number Exercise 1.1

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 board exam preparations. These experts review these NCERT Maths solutions chapter wise, so as to help students to solve the problems easily while using it 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 stepwise 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 that it should 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 225

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, remainder 90 ≠ 0, thus again using 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 this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as 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 225

As we know, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,

225 = 102 × 2 + 51

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

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 225 is 51.

2. Show that any positive odd integer is of the 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 new divisor, we have,

32 = 8 × 4 + 0

Now we have got 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 the 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 the sides, we get,

x2 = (3q)2 = 9q2 = 3 × 3q2

Let 3q2 = m

Therefore, x2= 3m ……………………..(1)

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

Substitute, 3q2+2q = m, to get,

x2= 3m + 1 ……………………………. (2)

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

Again, substitute, 3q2+4q+1 = m, to get,

x2= 3m + 1…………………………… (3)

Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the 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,

x2= (3q)3 = 27q3= 9(3q3)= 9m; where m = 3q3

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

x3 = (3q+1)3 = (3q)3 +13+3×3q×1(3q+1) = 27q3+1+27q2+9q

Taking 9 as common factor, we get,

x3 = 9(3q3+3q2+q)+1

Putting = m, we get,

Putting (3q3+3q2+q) = m, we get ,

x3 = 9m+1

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

x3 = (3q+2)3= (3q)3+23+3×3q×2(3q+2) = 27q3+54q2+36q+8

Taking 9 as common factor, we get,

x3=9(3q3+6q2+4q)+8

Putting (3q3+6q2+4q) = m, we get ,

x3 = 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 this 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

  • These NCERT Solutions help you solve and revise all questions of Exercise 1.1.
  • After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
  • It helps in scoring well in maths in exams.
  • It follows NCERT guidelines which help in preparing the students accordingly.
  • It contains all the important questions from the examination point of view.

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