10th Maths 1.1

Exercise 1.1

Q1. Use Euclid's division algorithm to find the HCF of:

i) $135$ $135$ and $225$ $225$ ii) $196$ $196$ and $38220$ $38220$ iii) $867$ $867$ and $255$ $255$

Sol.

By Euclid's division algorithm

a = bq + r

$text{a = bq + r}$

where

a = 225,

$text{a = 225,}$

b =135,

$text{b =135,}$

q = quotient and r = remainder

$text{q = quotient and r = remainder}$

225 = 135 \times 1 + 90

$225=135times1+90$

135 = 90 \times 1 + 45

$135=90times1+45$

90 = 45 \times 2 + 0

$90=45times2+0$

Answer:

The

HCF

$text{HCF}$ of

135

$135$ and

225

$225$ is

45

$45$ .

ii)

By Euclid's division algorithm

a = bq + r

$text{a = bq + r}$

where

a = 38220,

$text{a = 38220,}$

b = 196,

$text{b = 196,}$

q = quotient and r = remainder

$text{q = quotient and r = remainder}$

38220 = 196 \times 195 + 0

$38220 = 196times195 + 0$

Answer:

The

HCF

$text{HCF}$ of

196

$196$ and

38220

$38220$ is

196

$196$ .

iii)

By Euclid's division algorithm

a = bq + r

$text{a = bq + r}$

where

a = 867,

$text{a = 867,}$

b = 225,

$text{b = 225,}$

q = quotient and r = remainder

$text{q = quotient and r = remainder}$

867 = 225 \times 3 + 102

$867=225times3+102$

225 = 102 \times 2 + 51

$225=102times2+51$

102 = 51 \times 2 + 0

$102=51times2+0$

Answer:

The

HCF

$text{HCF}$ of

867

$867$ and

225

$225$ is

51

$51$ .

☝

Like, Share, and Subscribe.

Q2. Show that any positive odd integer is of the form $6 q + 1$ $6q+1$ , or $6 q + 3$ $6q+3$ , or $6 q + 5$ $6q+5$ , where q is some integer.

Sol.

Let

a

$a$ and

b

$b$ be positive integers.

By Euclid's division algorithm

a = bq + r

$text{a = bq + r}$

where

a = positive integer,

$text{a = positive integer,}$

b = 6 and r is remainder (0 \leq r < 6)

$text{b = 6 and r is remainder }(0lerlt6)$

Therefore possible

r = 0, 1, 2, 3, 4, and 5

$r=0, 1, 2, 3, 4, and 5$

6 \times q + 0 = 6 q

$6timesq+0=6q$

6 \times q + 1 = 6 q + 1

$6timesq+1=6q+1$

6 \times q + 2 = 6 q + 2

$6timesq+2=6q+2$

6 \times q + 3 = 6 q + 3

$6timesq+3=6q+3$

6 \times q + 4 = 6 q + 4

$6timesq+4=6q+4$

6 \times q + 5 = 6 q + 5

$6timesq+5=6q+5$

However since

a

$a$ is odd

a

$a$ cannot be

6 q

$6q$ ,

6 q + 2

$6q+2$ and

6 q + 4

$6q+4$ because these numbers are divisible by

2

$2$ , these are even numbers.

Therefore, any positive odd integer is of the form

6 q + 1, 6 q + 3, 6 q + 5

$6q+1, 6q+3, 6q+5$

Proved.

Q3. An army contingent of $616$ $616$ members is to march behind an army band of $32$ $32$ 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?

Sol.

To find maximum number of columns means we find

HCF (616,32)

$text{HCF (616,32)}$ by applying Euclid's algorithm,

a = bq + r

$text{a = bq + r}$

616 = 32 \times 19 + 8

$616=32times19+8$

32 = 8 \times 4 + 0

$32=8times4+0$

Answer:

Thus, the required maximum number of column is

8

$8$ .

Q4. Use Euclid's divisions lemma to show that the square of any positive integer is either of the form $3 m$ $3m$ or $3 m + 1$ $3m+1$ for some integer $m$ $m$ .

Sol.

Let

x

$x$ be any positive integer then it is of the form

3 q, 3 q + 1, 3 q + 2

$3q, 3q+1, 3q+2$ .

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

$(3q)^2=9q^2$

{(3 q)}^{2} = 3 {(3 q)}^{2}

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

{(3 q)}^{2} = 3 m

$(3q)^2=3m$ , where

$m$ is any positive integer.

$(3q+1)^2=9q^2+6q+1$

$(3q+1)^2=3q(3q+2)+1$

$(3q+1)^2=3m+1$ , where

$m=q(3q+2)$ is a positive integer.

$(3q+2)^2=9q^2+12q+4$

$(3q+2)^2=9q^2+12q+3+1$

$(3q+2)^2=3(3q^2+4q+1)+1$

$(3q+2)^2=3(3q+1)(q+1)+1$

$(3q+2)^2=3m+1$ , where

$m=(q+1)(3q+1)$ is a positive integer.

Thus, the square of any positive integer is either of the form

$text{3m}$ or

$text{3m+1}$ .

Proved.

Q5. Use Euclid's division lemma to show that the cubes of any positive integer is of the form $9m, 9m+1$ or $9m+8$ .

Sol.

Let

$x$ be any positive integer then it is the form

$3q, 3q+1, 3q+2$ , where

$q$ is a positive integer.

$(3q)^3=27q^3$

$(3q)^3=9(3q^3)$

$(3q)^3=9m$ , where

$m=3q^3$ is a positive integer.

$(3q+1)^3=27q^3+27q^2+9q+1$

$(3q+1)^3=9q(3q^2+3q+1)+1$

$(3q+1)^3=9m+1$ , where

$m=q(3q^2+3q+1)$ is a positive integer.

$(3q+2)^3=27q^3+54q^2+36q+8$

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

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

$m=q(3q^2+6q+4)$ is a positive integer.

Thus, the cube of any positive integer is of the form

$text{9m, 9m+1, 9m+8}$ .

Proved.

Search This Blog

10th Maths 1.1

NCERT Class 10th solution of Exercise 1.2

NCERT Class 10th solution of Exercise 1.3

NCERT Class 10th solution of Exercise 1.4

NCERT Class 10th Maths Projects

Exercise 1.1

Comments

Popular posts from this blog

10th Maths Chapter 1

CBSE 10th and 12th Result

RSKMP 5th & 8th Result