# Real Numbers (Chapter 1, Exercise 1.1) – Class 10 Maths Solutions NCERT

The Chapter 1 of Class 10 Mathematics based on NCERT is about Real Numbers. This page contains the Solution of Exercise 1.1, Chapter 1 of Class 10 Mathematics. The Exercise 1.1 of Class 10 Mathematics consists of 5 questions which are based on Euclid’s Division Algorithm or Lemma.

The solution of all the 5 questions of Exercise 1.1 of Class 10 is given below.

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

### i. 135 and 225

Solution: Here,

$225 > 135$

$\therefore c=225$ and $d=135$

Using Euclid’s Division Lemma,

$c = dq +r$

$\Rightarrow 225=135\times 1+90$

Since the remainder is not equal to zero $(0)$, using Euclid’s Division Lemma,

$\Rightarrow 135=90\times 1 +45$

Again, the remainder is not equal to zero $(0)$. Using Euclid’s Division Lemma,

$\Rightarrow 90=45\times 2+0$

The remainder becomes zero $(0)$.

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

### ii. 196 and 38220

Solution: Here,

$38220 > 196$

$\therefore c=38220$ and $d=196$

Using Euclid’s Division Lemma,

$c=dq+r$

$\Rightarrow 38220=196\times 195+0$

The remainder becomes zero $(0)$.

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

### iii. 867 and 255

Solution: Here,

$867 > 255$

$\therefore c=867$ and $d=255$

Using Euclid’s Division Lemma,

$c = dq +r$

$\Rightarrow 867=255\times 3+102$

Since the remainder is not equal to zero $(0)$, using Euclid’s Division Lemma,

$\Rightarrow 255=102\times 2 +51$

Again, the remainder is not equal to zero $(0)$. Using Euclid’s Division Lemma,

$\Rightarrow 102=51\times 2+0$

The remainder becomes zero $(0)$.

Therefore, the HCF of $867$ and $255$ is $51$.

## Question 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 integer and $b=6$

From Euclid’s Division Algorithm,

$a=bq+r$

$\Rightarrow a=6q+r$

Case I: When $r=1$, $a=6q+1$

If we put, $q=1, 2, 3, 4, 5, 6,…$

$a=7, 13, 19, 25, 31,…$

Case II: When $r=3$, $a=6q+3$

If we put, $q=1, 2, 3, 4, 5, 6,…$

$a=9, 15, 21, 27, 33,…$

Case III: Similarly, When $r=5$, $a=6q+5$

If we put, $q=1, 2, 3, 4, 5, 6,…$

$a=11, 17, 23, 29, 35,…$

In all the above three cases, the value of $a$ is a positive odd integer.

Therefore, any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

## Question 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:

Number of members in the army contingent $=616$

Number of members in the army band $=32$

The maximum number of columns in which they can march is the HCF of 616 and 32.

Here, $616 > 32$

Using Euclid’s Division Algorithm,

$c=dq+r$

$\therefore 616=32\times 19+8$

Since the remainder is not equal to zero $(0)$, using Euclid’s Division Algorithm,

$\Rightarrow 32=8\times 4+0$

The remainder becomes zero $(0)$.

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

The maximum number of columns in which they can march is $8$.

## Question 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.

Solution:

Let’s consider a positive integer $a$.

From Euclid’s Division Lemma, for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$, such that $a = bq + r$, $0 ≤ r < b$

Here, $b = 3$ and $r = 0, 1, 2$

$\therefore\, a=3q\,or\,3q + 1\,or\,3q + 2$

$\Rightarrow (a)^2 = (3q)^2 \,or\,(3q + 1)^2\,or\,(3q + 2)^2$

$\Rightarrow a^2 = 3(3q^2)\,or\,(9q^2 + 6q + 1)\,or\,(9q2 + 12q + 4)$

Now,

Case I: $a^2 = 3(3q^2)$  here, $m = 3q^2$

Case II: $a^2 = 3(3q^2 + 2q) + 1$ here, $m = (3q^2 + 2q)$

Case III: $a^2 = 3(3q^2 + 4q +1) + 1$ here, $m = (3q^2 + 4q +1)$

From the above three cases, $a^2$ is of the form $3m$ or $3m + 1$ where, $m$ is any positive integer.

Therefore, the square of any positive integer is either of the form $3m$ or $3m + 1$ for some integer $m$.

Question 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 $a$ be any positive integer and $b=3$.

From Euclid’s Division Lemma we know that

$a = bq + r$, $0 ≤ r < b$

$\therefore a = 3q + r$ and $r=0,\,1,\,2$

Case I: When $r = 0$,

$a = 3q$

$\Rightarrow a^3 = (3q)^3$

$\Rightarrow a^3 = 27q^3$

$\Rightarrow a^3 = 9(3q^3)$

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

Case II: When $r = 1$,

$a = 3q + 1$

$\Rightarrow a^3 = (3q + 1)^3$

$\Rightarrow a^3 = (3q)^3 + 13 + 3 × 3q × 1(3q + 1)$

$\Rightarrow a^3 = 27q^3 + 1 + 9q × (3q + 1)$

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

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

$\Rightarrow a^3 = 9 ( 3q^3 + 3q^2 + q) + 1$

$\Rightarrow a^3 = 9m + 1$, where $m = ( 3q^3 + 3q^2 + q)$

Case III: When $r = 2$,

$a = 3q + 2$

$a^3 = (3q + 2)^3$

$a^3 = (3q)^3 + 23 + 3 × 3q × 2 (3q + 1)$

$a^3 = 27q^3 + 8 + 54q^2 + 36q$

$a^3 = 27q^3 + 54q^2 + 36q + 8$

$a^3 = 9 (3q^3 + 6q^2 + 4q) + 8$

$a^3 = 9m + 8$, where $m = (3q^3 + 6q^2 + 4q)$

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

I hope the solutions of Class 10, Exercise 1.1, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.1, feel free to contact me at [email protected] or [email protected] or fill the form here.