NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.1

Important Topics under NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.1) Exercise 1.1

The first chapter of the Class 10 Maths syllabus is on real numbers. It is a significant chapter in maths that deals with some of the fundamental topics. This chapter has 6 major parts that need to be covered and understood properly to get a good grasp on real numbers. It is recommended that students go through these individual topics carefully for a precise understanding and for better internalization of the same.

Importance of Real Numbers

Real numbers comprise both rational and irrational numbers. Rational numbers include integers, decimals, and fractions, while irrational numbers are numbers like root overs, pi (22/7), and so on. In short, real numbers are all numbers excluding imaginary numbers. 

The importance of real numbers lies in their use in almost every sphere of mathematics and in real life. We encourage students to gather as much knowledge as they can on real numbers so that they are able to solve all sums that have the application of real numbers.

Exercise 1.1 

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

(i) $135$ and $225$ 

Ans: We have to find the HCF of $135$ and $225$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a>b$ is found as :

(i) First find the values of $q$ and $r$, where $a=bq+r$, $0\le r<b$.

(ii) If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

(iii) Continue steps till the remainder is zero. When we get the remainder zero, the divisor will be the HCF.

Let $a=225$ and $b=135$.

Since, $a>b$ 

Using division algorithm, we get

$a=bq+r$

$\Rightarrow 225=135\times 1+90$ 

Here,

$\Rightarrow b=135$ 

$\Rightarrow q=1$ 

$\Rightarrow r=90$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new reminder). We get

$\Rightarrow 135=90\times 1+45$ 

Here, 

$\Rightarrow b=90$ 

$\Rightarrow q=1$ 

$\Rightarrow r=45$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 90=2\times 45+0$ 

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, the divisor will be the HCF.

Therefore, the HCF of $135$ and $225$ is $45$.

(ii) $196$ and $38220$

Ans: We have to find the HCF of $196$ and $38220$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a>b$ is found as :

(i) First find the values of $q$ and $r$, where $a=bq+r$, $0\le r<b$.

(ii) If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

(iii) Continue steps till the remainder is zero. When we get the remainder zero, the divisor will be the HCF.

Let $a=38220$ and $b=196$.

Since, $a>b$ 

Using division algorithm, we get

$a=bq+r$

$\Rightarrow 38220=196\times 195+0$ 

Here, 

$\Rightarrow b=196$ 

$\Rightarrow q=195$ 

$\Rightarrow r=0$ 

Since, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, the divisor will be the HCF.

Therefore, the HCF of $196$ and $38220$ is $196$.

(iii) $867$ and $255$ 

Ans: We have to find the HCF of $867$ and $255$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a>b$ is found as :

(i) First find the values of $q$ and $r$, where $a=bq+r$, $0\le r<b$.

(ii) If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

(iii) Continue steps till the remainder is zero. When we get the remainder zero, the divisor will be the HCF.

Let $a=867$ and $b=255$.

Since, $a>b$ 

Using division algorithm, we get

$a=bq+r$

\[\Rightarrow 867=255\times 3+102\] 

Here,

$\Rightarrow b=255$ 

$\Rightarrow q=3$ 

$\Rightarrow r=102$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new reminder). We get

$\Rightarrow 255=102\times 2+51$ 

Here, 

$\Rightarrow b=102$ 

$\Rightarrow q=2$ 

$\Rightarrow r=51$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 102=51\times 2+0$ 

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, the divisor will be the HCF.

Therefore, the HCF of $867$ and $255$ 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.

Ans: Let $a$ be any positive integer and $b=6$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r<b$.

Here, $0\le r<6$.

Substitute the values, we get

$\Rightarrow a=6q+r$

If $r=0$, we get

$\Rightarrow a=6q+0$

$\Rightarrow a=6q$

If $r=1$, we get

$\Rightarrow a=6q+1$

If $r=2$, we get

$\Rightarrow a=6q+2$ and so on

Therefore, $a=6q$ or $6q+1$ or $6q+2$ or $6q+3$ or $6q+4$ or $6q+5$.

We can write the obtained expressions as 

\[6q+1=2\times 3q+1\]

$\Rightarrow 6q+1=2{{k}_{1}}+1$ 

Where, ${{k}_{1}}$ is an integer.

\[6q+3=6q+2+1\]

$\Rightarrow 6q+3=2\left( 3q+1 \right)+1$ 

$\Rightarrow 6q+3=2{{k}_{2}}+1$ 

Where, ${{k}_{2}}$ is an integer.

\[6q+5=6q+4+1\]

$\Rightarrow 6q+5=2\left( 3q+2 \right)+1$ 

$\Rightarrow 6q+5=2{{k}_{3}}+1$  

Where, ${{k}_{3}}$ is an integer.

Thus, $6q+1,6q+3,6q+5$ are of the form $2k+1$ and are not exactly divisible by $2$.

Also, all these expressions are of odd numbers.

Therefore, any positive odd integer can be expressed in the form, $6q+1$ or $6q+3$, or $6q+5$, 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?  

Ans: Given that 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.

We have to find the maximum number of columns in which they can march.

We need to find the HCF of $616$ and $32$ to find the maximum number of columns.

We will use Euclid's division algorithm to find the HCF.

Let $a=616$ and $b=32$.

Since, $a>b$ 

Using division algorithm, we get

$a=bq+r$

\[\Rightarrow 616=32\times 19+8\] 

Here,

$\Rightarrow b=32$ 

$\Rightarrow q=19$ 

$\Rightarrow r=8$ 

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor)  and $r$ (new reminder). We get

$\Rightarrow 32=8\times 4+0$ 

Since, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, the divisor will be the HCF.

Therefore, the HCF of $616$ and $32$ is $8$.

Therefore, in $8$ columns members can march.

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 $m$.

(Hint: Let $x$ be any positive integer then it is of the form $3q,3q+1$ or $3q+2$. Now, square each of these and show that they can be rewritten in the form $3m$ or $3m+1$.)

Ans: Let $a$ be any positive integer and $b=3$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r<b$.

Here, $0\le r<3$.

Substitute the values, we get

$\Rightarrow a=3q+r$

If $r=0$, we get

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

If $r=1$, we get

$\Rightarrow a=3q+1$

If $r=2$, we get

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, squaring each of these, we get

$\Rightarrow {{a}^{2}}={{\left( 3q \right)}^{2}}$ or ${{\left( 3q+1 \right)}^{2}}$ or ${{\left( 3q+2 \right)}^{2}}$

Now, applying the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$

Thus, we get

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}$

$\Rightarrow {{a}^{2}}=3m$, where, $m=3{{q}^{2}}$ 

And, 

${{a}^{2}}=3\times 3{{q}^{2}}+3\times 2q+1$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+2q \right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+2q$ 

And, 

${{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+4$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q \right)+3+1$ 

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q +1\right)+1$ 

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+4q+1$

Therefore, 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 $9m+8$.

Ans: Let $a$ be any positive integer and $b=3$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r<b$.

Here, $0\le r<3$.

Substitute the values, we get

$\Rightarrow a=3q+r$

If $r=0$, we get

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

If $r=1$, we get

$\Rightarrow a=3q+1$

If $r=2$, we get

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, consider $a=3q$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q \right)}^{3}}$ 

$\Rightarrow {{a}^{3}}=27{{q}^{3}}$ 

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}} \right)$

$\Rightarrow {{a}^{3}}=9m$, where, $m=3{{q}^{3}}$ 

Now, consider $a=3q+1$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+1 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$

$\Rightarrow {{a}^{3}}=9m+1$, where, \[m=3{{q}^{3}}+3{{q}^{2}}+q\]  

Now, consider $a=3q+2$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+2 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+54{{q}^{2}}+36q+8$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$

$\Rightarrow {{a}^{3}}=9m+8$, where, \[m=3{{q}^{3}}+6{{q}^{2}}+4q\]  

Therefore, we can say that a cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Opting for the NCERT solutions for Ex 1.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 1.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Real Numbers textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 1 Exercise 1.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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