# Real Numbers Chapter 1 Class 10 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 10 Maths Chapter 1‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 10 Maths Chapter 1 Exercise Solution’ using the download button.

### Chapter 1: Real Numbers

#### 1.1 Introduction

In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers. We continue our discussion on real numbers in this chapter.

We begin with two very important properties of positive integers in Sections 1.2 and 1.3, namely Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.

Euclid’s division algorithm, as the name suggests, has to do with the divisibility of integers.

Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b.

Many of you probably recognize this as the usual long division process.

Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers. We touch upon a few of them and use them mainly to compute the HCF of two positive integers.

### 1.2 Euclid’s Division Lemma

Consider the following folk puzzle*. A trader was moving along a road selling eggs. An idler who didn’t have much work to do started to get the trader into a wordy duel.

This grew into a fight, he pulled the basket with eggs and dashed it to the floor. The eggs broke.

The trader requested the Panchayat to ask the idler to pay for the broken eggs. The Panchayat asked the trader how many eggs were broken. He gave the following response:

If counted in pairs, one will remain;
If counted in threes, two will remain;
If counted in fours, three will remain;
If counted in fives, four will remain;
If counted in sixes, five will remain;
If counted in sevens, nothing will remain;
My basket cannot accommodate more than 150 eggs.
So, how many eggs were there? Let us try and solve the puzzle. Let the number of eggs be a. Then working backward, we see that a is less than or equal to 150

### NCERT Solutions Class 10 Maths Chapter 1 Real Numbers

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, 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 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 apply now 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 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 (867,255) = 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.