Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

Students can download Maths Chapter 3 Algebra Ex 3.10 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.
Draw the graph for the following:
(i) y = 2x
Solution:
When x = -2, y = 2 (-2) = -4
When x = 0, y = 2 (0) = 0
When x = 2, y = 2 (2) = 4
When x = 3, y = 2 (3) = 6
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 1
Plot the points (-2, -4) (0, 0) (2, 4) and (3, 6) in the graph sheet we get a straight line.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 2

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(ii) y = 4x – 1
Solution:
When x = – 1; y = 4 (-1) -1 ⇒ y = -5
When x = 0; y = 4 (0) – 1 = 0 – 1 ⇒ y = -1
When x = 2; y = 4 (2) -1 = 8 – 1 ⇒ y = l
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 3
Plot the points (-1, -5) (0, -1) and (2, 7) in the graph sheet we get a straight line. At the time of printing change the direction.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 4

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(iii) y = (\(\frac{3}{2}\))x + 3
Solution:
When x = -2;
y = \(\frac{3}{2}\)(-2) + 3
y = -3 + 3 = 0
when x = 0;
y = \(\frac{3}{2}\)(0) + 3
y = 3
when x = 2;
y = \(\frac{3}{2}\)(2) + 3
y = 3 + 3
= 6
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 5
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 6
Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(iv) 3x + 2y = 14
Solution:
y = \(\frac{-3x+14}{2}\)
y = – \(\frac{3}{2}\)x + 7
when x = -2
y = –\(\frac{3}{2}\)(-2) + 7 = 10
when x = 0
y = –\(\frac{3}{2}\)(0) + 7 = 7
when x = 2
y = –\(\frac{3}{2}\)(2) + 7 = 4
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 7
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 8
Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

Question 2.
Solve graphically (i) x + y = 7, x – y = 3
Solution:
x + y = 7
y = 7 – x
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 9
Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet
x – y = 3
-y = -x + 3
y = x – 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 10
Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 11
The point of intersection is (5, 2) of lines (1) and (2).
The solution set is (5,2).

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(ii) 3x + 2y = 4; 9x + 6y – 12 = 0
Solution:
2y = -3x + 4
y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 12
Plot the points (-2, 5), (0, 2) and (2, -1) in the graph sheet
9x + 6y= 12 (÷3)
3x + 2y = 4
2y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 13
Plot the points (-2, 5), (0, 2) and (2, -1) the same graph sheet
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 14
Here both the equations are identical, but in different form.
Their solution is same.
This equations have an infinite number of solution.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(iii) \(\frac{x}{2}\) + \(\frac{y}{4}\) = 1: \(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
Solution:
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 1
multiply by 4
2x + y = 4
y = -2x + 4
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 15
Plot the points (-3, 10), (-1, 6), (0, 4) and (2, 0) in the graph sheet
\(\frac{x}{2}\) + \(\frac{y}{4}\) = 2
multiply by 4
2x + y = 8
y = -2x + 8
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 16
Plot the points (-2, 12), (-1, 10), (0, 8) and (2, 4) in the same graph sheet.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 17
The given two lines are parallel.
∴ They do not intersect a point.
∴ There is no solution.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(iv) x – y = 0; y + 3 = 0
Solution:
x – y = 0
-y = -x
y = x
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 18
Plot the points (-2, -2), (0, 0), (1, 1) and (3, 3) in the same graph sheet.
y + 3 = 0
y = -3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 19
Plot the points (-2, -3), (0, -3), (1, -3) and (2, -3) in the same graph sheet.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 20
The two lines l1 and l2 intersect at (-3, -3). The solution set is (-3, -3).

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

(v) y = 2x + 1; 3x – 6 = 0
Solution:
y = 2x + 1
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 21
Plot the points (-3, -5), (-1, -1), (0, 1) and (2, 5) in the graph sheet
3x – 6 = 0
y = -3x + 6
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 22
Plot the points (-2, 12), (-1, 9), (0, 6) and (2, 0) in the same graph sheet
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 23
The two lines l1 and l2 intersect at (1, 3).
∴ The solution set is (1, 3).

(vi) x = -3; y = 3
Solution:
x = -3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 24
Plot the points (-3, -3), (-3, -2), (-3, 2) and (-3, 3) in the graph sheet
y = 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 25
Plot the points (-3, 3), (-1, 3), (0, 3) and (2, 3) in the same graph sheet
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 26
The two lines l1 and l2 intersect at (-3, 3)
∴ The solution set is (-3, 3)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10

Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let the speed of the two cars be “x” and “y”.
By the given first condition
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 27
x+ y = 100 → (1)
(They travel in opposite direction)
By the given second condition.
\(\)\frac{100}{x-y}= 2 [time taken in 2 hours in the same direction]
2x – 2y = 100
x – y = 50 → (2)
x + y = 100
y = 100 – x
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 28
Plot the points (30, 70), (50, 50), (60, 40) and (70,30) in the graph sheet
x – y = 50
-y = -x + 50
y = x – 50
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 29
Plot the points (40, -10), (50, 0), (60, 10) and (70, 20) in the same graph sheet
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.10 30
The two cars intersect at (75, 25)
The speed of the first car 75 km/hr
The speed of the second car 25 km/hr

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Find the GCD for the following:
(i) P5, P11, P3
Solution:
p5 = p5
p11 = p11
P9 = P9
G.C.D. is p5 (Highest common power is 5)

(ii) 4x3, y3, z3
Solution:
4x3 = 2 × 2 × x3
y3 = y3
z3 = z3
G.C.D. of 4x3, y3 and z3 = 1

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

(iii) 9a²b²c3, 15a3b2c4
Solution:
9a²b²c3 = 3 × 3 × a² × b² × c3
15a3b²c3 = 3 × 5 × a3 × b2 × c4
G.C.D = 3 × a2 × b2 × c3
= 3a2b2c3

(iv) 64x8, 240x6
Solution:
64x8 = 2 × 2 × 2 × 2 × 2 × 2 × x8
= 26 × x8
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9 1
240x6 = 24 × 3 × 5 × x6
G.C.D = 24 × x6
= 16x6

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

(v) ab²c3, a²b3c, a3ac²
Solution:
ab²c3 = a × b² × c3
a²b3c = a² × b3 × c
a3bc² = a3 × b × c²
G.C.D. = abc

(vi) 35x5y3z4, 49x2yz3, 14xy2z2
Solution:
35x5y3z4 = 5 × 7 × x5 × y3 × z4
49x²yz3 = 7 × 7 × x2 × z3
14xy²z² = 2 × 7 × x × y² × z²
G.C.D. = 7 × x × y × z²
= 7xyz²

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

(vii) 25ab3c, 100a²bc, 125 ab
Solution:
25ab3c = 5 × 5 × a × b3 × c
100a²be = 2 × 2 × 5 × 5 × a² × b × c
125ab = 5 × 5 × 5 × a × b
G.C.D. = 5 × 5 × a × b
= 25ab

(viii) 3abc, 5xyz, 7pqr
Solution:
3abc = 3 × a × b × c
5xyz = 5 × x × y × z
7pqr = 7 × p × q × r
G.C.D. = 1

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

Question 2.
Find the GCD for the following:
(i) (2x + 5), (5x + 2)
(ii) am+1, am+2, am+3
(iii) 2a² + a, 4a² – 1
(iv) 3a², 5b3, 7c4
(v) x4 – 1, x² – 1
(vi) a3 – 9ax², (a – 3x)²
Solution:
(i) (2x + 5) = 2x + 5
5x + 2 = 5x + 2
G.C.D. = 1

(ii) am+1 = am × a1
am+2 = am × a2
am+3 = am × a3
G.C.D.= am × a
= am+1

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

(iii) 2a² + a = a(2a + 1)
4a² – 1 = (2a)2 – 1
(Using a² – b² = (a + b)(a – b)
= (2a + 1)(2a – 1)
G.C.D. = 2a + 1

(iv) 3a² = 3 × a²
5b3 = 5 × b3
7c4 = 7 × c4
G.C.D. = 1

(v) x4 – 1 = (x²)² – 1
= (x² + 1 ) (x² – 1)
= (x² + 1 ) (x + 1 ) (x – 1 )
x² – 1 = (x + 1 ) (x – 1 )
G.C.D. = (x + 1 ) (x – 1 )

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.9

(vi) a3 – 9ax2 = a(a2 – 9x2)
= a[a2 – (3x)2]
= a(a + 3x)(a – 3x)
(a – 3x)2 = (a – 3x)2
G.C.D. = a – 3x

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Students can download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1.
Expand the following:
(i) (2x + 3y + 4z)2
(ii) (-p + 2q + 3r)2
(iii) (2p + 3) (2p – 4) (2p – 5)
(iv) (3a + 1) (3a – 2) (3a + 4)
Solution:
We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
(i) (2x + 3y + 4z)2 = (2x)2 + (3y)2 + (4z)2 + 2(2x) (3y) + 2(3y) (4z) + 2(4z) (2x)
= 4x2 + 9y2 + 16z2 + 12xy + 24yz + 16xz

(ii) (-p + 2q + 3r)2 = (-p)2 + (2q)2 + (3r)2 + 2(-p) (2q) + 2(2q)(3r) + 2(3r) (- p)
= p2+ 4q2 + 9r2 – 4pq + 12qr – 6pr

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

(iii) (2p + 3) (2p – 4) (2p – 5)
[Here x = 2p, a = 3, b = -4 and c = -5]
= (2p)3 + (3 – 4 – 5) (2p)2 + [(3)(-4) + (-4)(-5) + (3) (-5)] 2p + (3) (-4) (-5)
= 8p3 + (-6)(4p2) + (-12 + 20 – 15) 2p + 60
= 8p3 – 24p2 – 14p + 60

(iv) (3a + 1) (3a – 2) (3a + 4)
[Here x = 3a, a = 1, b = -2 and c = 4]
= (3a)3 + (1 – 2 + 4) (3a)2 + [(1)(-2) + (-2) (4) + (4) (1)] (3a) + (1) (-2) (4)
= 27a3 + 3(9a2) + (-2 – 8 + 4) (3a) – 8
= 27a3 + 27a2 – 18a – 8

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 2.
Using algebraic identity, find the coefficients of x2, x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
Solution:
[Here x = x, a = 5, b = 6, c = 7]
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc
coefficient of x2 = 5 + 6 + 7
= 18
coefficient of x = 30 + 42 + 35
= 107
constant term = (5) (6) (7)
= 210

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

(ii) (2x + 3)(2x – 5) (2x – 6)
Solution:
[Here x = 2x, a = 3, b = -5, c = -6]
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ac)x + abc
coefficient of x2 = (3 – 5 – 6)4 [(2x)2 = 4x2]
= (-8) (4)
= -32
coefficient of x = [(3)(-5) + (-5)(-6) + (-6)(3)](2)
= (-15 + 30-18) (2)
= (-3) (2)
= -6
constant term = (3) (-5) (-6)
= 90

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 3.
If (x + a)(x + b)(x + c) = x3 + 14x2 + 59x + 70, find the value of
(i) a + b + c
(ii) \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)
(iii) a2 + b2 + c2
(iv) \(\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}\)
Solution:
(x + a) (x + b) (x + c) = x3 + 14x2 + 59x + 70
x3 + (a + b + c)x2 + (ab + bc + ac)x + abc = x3 + 14x2 + 59x + 70
a + b + c = 14, ab + bc + ac = 59, abc = 70
(i) a + b + c = 14

(ii) \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) = \(\frac{bc+ac+ab}{abc}\)
= \(\frac{59}{70}\)

(iii) a2 + b2 + c2 = (a + b + c)2 – 2 (ab + bc + ac)
= (14)2 – 2(59)
= 196 – 118
= 78

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 1

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 4.
Expand:
(i) (3a – 4b)3
Solution:
(a – b)3 = a3 – b3 – 3ab (a – b)
(3a – 4b)3 = (3a)3 – (4b)3 – 3(3a)(4b)(3a – 4b)
= 27a3 – 64b3 – 36ab (3a – 4b)
= 27a3 – 64b3 – 108a2b + 144ab2

(ii) [x + \(\frac{1}{y}]^{3}\)
Solution:
(a + b)3 = a3 + b3 + 3ab (a + b)
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 2

Question 5.
Evaluate the following by using identities:
(i) 983
Solution:
983 = (100 – 2)3 [(a – b)3 = a3 – b3 – 3ab (a – b)]
= 1003 – (2)3 – 3(100) (2) (100 – 2)
= 1000000 – 8 – 600(98)
= 1000000 – 8 – 58800
= 1000000 – 58808
= 941192

(ii) 10013
Solution:
(1001)3 = (1000 + 1)3
[(a + b)3 = a3 + b3 + 3ab (a + b)]
= (1000)3 + 13 + 3(1000) (1) (1000 + 1)
= 1000000000 + 1 + 3000 (1001)
= 1000000001 + 3003000
= 1003003001

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 6.
If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x2 + y2 + z2.
Solution:
x + y + z = 9; xy + yz + zx = 26
x2 + y2 + z2 = (x + y + z)2 – 2xy – 2yz – 2xz
= (x + y + z)2 – 2 (xy + yz + zx)
= 92 – 2(26)
= 81 – 52
= 29

Question 7.
Find 27a3 + 64b3, If 3a + 4b = 10 and ab = 2
Solution:
3a + Ab = 10, ab = 2
27a3 + 64b3 = (3a)3 + (4b)3
[a3 + b3 = (a + b)3 – 3 ab (a + b)]
= (3a + 4b)3 – 3 × 3a × 4b (3a + 4b)
= 103 – 36ab (10)
= 1000 – 36(2)(10)
= 1000 – 720
= 280

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 8.
Find x3 – y3, if x – y = 5 and xy = 14.
Solution:
x – y = 5, xy = 14
x3 – y= (x – y)3 + 3xy (x – y)
= 53 + 3(14) (5)
= 125 + 210
= 335

Question 9.
If a + \(\frac{1}{a}\) = 6, then find the value of a3 +\(\frac{1}{a^3}\)
Solution:
a + \(\frac{1}{a}\) = 6 [a3 + b3 = (a + b)3 – 3ab (a + b)]
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 3
= 63 – 3(6)
= 216 – 18
= 198

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 10.
If x2 + \(\frac{1}{x^2}\) = 23, then find the value of x + \(\frac{1}{x}\) and x3 + \(\frac{1}{x^3}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 4
When x = 5 [a3 + b3 = (a + b)3 – 3ab (a + b)]
= (5)3 – 3(5)
= 125 – 15
= 110
when x = -5
x3 + \(\frac{1}{x^3}\) = (-5)3 – 3(-5)
= -125 + 15
= -110
∴ x3 + \(\frac{1}{x^3}\) = ±110

Question 11.
If (y – \(\frac{1}{y})^{3}\) = 27 then find the value of y3 – \(\frac{1}{y^3}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 5
= 33 + 3(3)
= 27 + 9
= 36

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 12.
Simplify:
(i) (2a + 3b + 4c) (4a2 + 9b2 + 16c2 – 6ab – 12bc – 8ca)
(ii) (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
Solution:
x3 + y3 + z3 – 3xyz ≡ (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
(i) (2a + 3b + 4c) (4a2 + 9b2 + 16c2 – 6ab – 12bc – 8ea)
= (2a)3 + (3b)3 + (4c)3 – 3 (2a) (3b) (4c)
= 8a3 + 27b3 + 64c3 – 72abc

(ii) (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
= x3 + (-2y)3 + (3z)3 – 3(x) (-2y) (3z)
= x3 – 8y3 + 27z3 + 18xyz

Question 13.
By using identity evaluate the following:
(i) 73 – 103 + 33
Solution:
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
We know that a + b + c = 0 then a3 + b3 + c3 = 3ab
a + b + c = 7 + (-10) + 3
= 10 – 10
= 0
∴ 73 – 103 + 33 = 3(7) (-10) (3)
= -630

(ii) 1 + \(\frac{1}{8}\) – \(\frac{27}{8}\)
Solution:
We know that a3 + b3 + c3 = 0 then a + b + c = 3abc
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4 6

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4

Question 14.
If 2x -3y – 4z = 0, then find 8x3 – 27y3 – 64z3.
Solution:
We know x3 +y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 = (x + y + z) (x2 +y2 + z2 – xy – yz – zx) + 3xyz
8x3 – 27y3 – 64z3 = (2x)3 + (-3y)3 + (-4z)3
= (2x – 3y- 4z) [(2x)2 + (-3y)2 + (-4z)2 – (2x)(-3y) – (-3y) (-4z) -(-4z)(2x)] + 3(2x)(-3y)(-4z)
= 0 (4x2 + 9y2 + 16z2 + 6xy – 12yz + 8xz) + 72xyz
= 72xyz
8x3 – 27y3 – 64z3 = 72xyz

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Students can download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Check whether p(x) is a multiple of g(x) or not.
(i) p(x) = x3 – 5x2 + 4x – 3; g(x) = x – 2
Solution:
p(x) = x3 – 5x2 + 4x – 3
P(2) = (2)3 – 5(2)2 + 4(2) – 3
= 8 – 5(4) + 8 – 3
= 8 – 20 + 8 – 3
= 16 – 23
= -7
p{2) ≠ 0
∴ p(x) is not a multiple of g(x)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 2.
By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(i) p(x) = x3 – 2x2 – 4x – 1; g(x) = x + 1
Solution:
p(x) = x3 – 2x2 – 4x – 1
p(-1) = (-1)3 – 2(-1)2 – 4(-1) – 1
= 1 – 2 + 4 – 1
= 4 – 4 = 0
∴ The remainder = 0

(ii) p(x) = 4x3 – 12x2 + 14x – 3; g(x) = 2x – 1
Solution:
p(x) = 4x3 – 12x2 + 14x – 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3 1
= 4 × \(\frac{1}{8}\) – 12 × \(\frac{1}{4}\) + 14 × \(\frac{1}{2}\) – 3
= \(\frac{1}{2}\) – 3 + 7 – 3
= \(\frac{1}{2}\) – 6 + 7
= \(\frac{1}{2}\) + 1
= \(\frac{3}{2}\)
∴ The reminder is \(\frac{3}{2}\)

(iii) p(x) = x3 – 3x2 + 4x + 50; g(x) = x – 3
Solution:
p(x) = x3 – 3x2 + 4x + 50
p(3) = 33 – 3(3)2 + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
The remainder is 62.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 3.
Find the remainder when 3x3 – 4x2 + 7x – 5 is divided by (x + 3)
Solution:
p(x) = 3x3 – 4x2 + 7x – 5
When it is divided by x +3,
p(-3) = 3(-3)3 – 4(-3)2 + 7(-3) – 5
= 3(-27) – 4(9) – 21 – 5
= -81 – 36 – 21 – 5
= -143
The remainder is -143.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 4.
What is the remainder when x2018 + 2018 is divided by x – 1.
Solution:
p(x) = x2018 + 2018
When it is divided by x – 1,
p(1) = 12018 + 2018
= 1 + 2018
= 2019
The remainder is 2019.

Question 5.
For what value of k is the polynomial
p(x) = 2x3 – kx2 + 3x + 10 exactly divisible by x – 2
Solution:
p(x) = 2x3 – kx2 + 3x + 10
When it is exactly divided by x – 2,
P(2) = 0
2(2)3 – k(2)2 + 3(2) + 10 = 0
2(8) – k(4) + 6 + 10 = 0
16 – k(4) + 6 + 10 = 0
16 – 4k + 6 + 10 = 0
32 – 4k = 0
32 = 4k
∴ k = \(\frac{32}{4}\)
= 8
The value of k = 8

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 6.
If two polynomials 2x3 + ax2 + 4x – 12 and x3 + x2 – 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.
Solution:
p(x1) = 2x3 + ax2 + 4x – 12
When it is divided by x – 3,
p(3) = 2(3)3 + a(3)2 + 4(3) – 12
= 54 + 9a + 12 – 12
= 54 + 9a ……….(R1)
p(x2) = x3 + x2 – 2x + a
When it is divided by x – 3,
p(3) = 33 + 32 – 2(3) + a
= 27 + 9 – 6 + a
= 30 + a ………(R2)
The given remainders are same (R1 = R2)
∴ 54 + 9a = 30 + a
9a – a = 30 – 54
8a = -24
∴ a = -24/8
= -3
Consider R2,
Remainder = 30 – 3
= 27

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 7.
Determine whether (x – 1) is a factor of the following polynomials:
(i) x3 + 5x2 – 10x + 4
Solution:
p(x) = x3 + 5x2 – 10x + 4
p(1) = 13 + 5(1) – 10(1) + 4
= 1 + 5 – 10 + 4
= 10 – 10
= 0
∴ x – 1 is a factor of p(x)

(ii) x4 + 5x2 – 5x + 1
Solution:
p(1) = 14 + 5(1)2 – 5(1) + 1
= 1 + 5 – 5 + 1
= 7 – 5
= 2
= 0
∴ x – 1 is not a factor of p(x)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 8.
Using factor theorem, show that (x – 5) is a factor of the polynomial
2x3 – 5x2 – 28x + 15
Solution:
p(x) = 2x3 – 5x2 – 28x + 15
x – 5 is a factor
p(5) = 2(5)3 – 5(5)2 – 28(5) + 15
= 250 – 125 – 140 + 15
= 265 – 265
= 0
∴ x – 5 is a factor of p(x)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 9.
Determine the value of m, if (x + 3) is a factor of x3 – 3x2 – mx + 24.
Solution:
p(x) = x3 – 3x2 – mx + 24
when x + 3 is a factor
P(-3) = 0
(-3)3 – 3(-3)2 – m(-3) + 24 = 0
-27 – 27 + 3m + 24 = 0
-54 + 24 + 3m = 0
-30 + 3m = 0
3m = 30
m = \(\frac{30}{3}\)
= 10
The value of m = 10

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 10.
If both (x-2) and (x – \(\frac{1}{2}\)) are the factors of ax2 + 5x + b, then show that a = b.
Solution:
p(x) = ax2 + 5x + b
when (x-2) is a factor
P(2) = 0
a(2)2 + 5(2) + b = 0
4a + 10 + b = 0
4a + b = -10 …….(1)
when (x – \(\frac{1}{2}\)) is a factor
p(\(\frac{1}{2}\)) = 0
a\((\frac{1}{2})^2\) + 5(\(\frac{1}{2}\)) + b = 0
Multiply by 4
a + 10 + 4b = 0
a + 46 = -10 …….(2)
From (1) and (2) we get
4a + b = a + 4b
4a – a = 4b – b
3a = 3b
a = b
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 11.
If (x – 1) divides the polynomial kx3 – 2x2 + 25x – 26 without remainder, then find the value of k.
Solution:
p(x) = kx3 – 2x2 + 25x – 26
When it is divided by x – 1
P(1) = 0
k(1)3 – 2(1)2 + 25(1) – 26 = 0
k – 2 + 25 – 26 = 0
k + 25 – 28 = 0
k – 3 = 0
k = 3
The value of k = 3

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.3

Question 12.
Check if (x + 2) and (x – 4) are the sides of a rectangle whose area is x2 – 2x – 8 by using factor theorem.
Solution:
Let the area of a rectangle be p(x)
p(x) = x2 – 2x – 8
When x + 2 is the side of the rectangle
p(-2) = (-2)2 – 2(-2) – 8
= 4 + 4 – 8
= 8 – 8
= 0
When x – 4 is the side of the rectangle.
P(4) = (4)2 – 2(4) – 8
= 16 – 8 – 8
= 16 – 16
= 0
(x + 2) and (x – 4) are the sides of a rectangle

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

Students can download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the value of the polynomial f(y) = 6y – 3y2 + 3 at
(i) y = 1
(ii) y = -1
(iii) y = 0
Solution:
(i) When y = 1
f(y) = 6y – 3y2 + 3
f(1) = 6(1) – 3(1)2 + 3
= 6 – 3 + 3 = 6

(ii) When y = – 1
f(y) = 6y – 3y2 + 3
f(-1) = 6(-1) – 3(-1)2 + 3
= – 6 – 3 + 3
= – 6

(iii) When y = 0
f(y) = 6y – 3y2 + 3
f(0) = 6(0) – 3(0)2 + 3
= 0 – 0 + 3
= 3

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

Question 2.
If p(x) = x2 – 2√2x + 1, find p(2√2).
Solution:
p(x) = x2 – 2√2x + 1
p(2√2) = (2√2)2 – 2√2 (2√2) + 1
= 8 – 8 + 1
= 0 + 1
= 1

Question 3.
Find the zeros of the polynomial in each of the following.
(i) P(x) = x – 3
Solution:
p( 3) = 3 – 3
= 0
p(3) is the zero of p(x)

(ii) p(x) = 2x + 5
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2 1
= -5 + 5
= 2(0)
= 0
Hence –\(\frac{5}{2}\) is the zero of p(x).

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

(iii) q(y) = 2y – 3
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2 2
= 2 × 0
= 0
Hence \(\frac{3}{2}\) is the zero of q(y).

(iv) f(z) = 8z
Solution:
f(0) = 8 × 0
= 0
Hence 0 is the zero of f(z)

(v) p(x) = ax when a ≠ 0
Solution:
p(0) = a(0)
= 0
Hence, 0 is the zero of p(x)

(vi) h(x) = ax + b, a ≠ 0, a, b∈R
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2 3
Hence –\(\frac{b}{a}\) is the zero of h(x).

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

Question 4.
Find the roots of the polynomial equations.
(i) 5x – 6 = 0
Solution:
5x = 6
x = \(\frac{6}{5}\)
\(\frac{6}{5}\) is the root of the polynomial.

(ii) x + 3 = 0
Solution:
x = -3
-3 is the root of the polynomial.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

(iii) 10x + 9 = 0
Solution:
10x = -9
x = –\(\frac{9}{10}\)
–\(\frac{9}{10}\) is the root of the polynomial.

(iv) 9x – 4 = 0
Solution:
9x = 4
x = \(\frac{4}{9}\)
\(\frac{4}{9}\) is the root of the polynomial.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

Question 5.
Verify whether the following are zeros of the polynomial, indicated against them,or not.
(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)
Solution:
p (\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) – 1
= 1 – 1
= 0
∴ \(\frac{1}{2}\) is the zero of the polynomial.

(ii) p(x) = x3 – 1, x = 1
Solution:
p(1) = 13 – 1
= 1 – 1
= 0
∴ 1 is the zero of the polynomial

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)
Solution:
p(\(\frac{-b}{a}\)) = a(\(\frac{-b}{a}\)) + b
= -b + b
= 0
∴ \(\frac{-b}{a}\) is the zero of the polynomial. a

(iv) p(x) = (x + 3) (x – 4); x = -3, x = 4
Solution:
P(-3) = (-3 + 3) (-3 – 4)
= (0) (-7)
= 0
P( 4) = (4 + 3) (4 – 4)
= (7) (0)
= 0
∴ -3 and 4 are the zeros of the polynomial.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2

Question 6.
Find the number of zeros of the following polynomials represented by their graphs.
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.2 4
Solution:
(i) Number of zeros = 2 (The curve is intersecting the x-axis at 2 points)
(ii) Number of zeros = 3 (The curve is intersecting the x-axis at 3 points)
(iii) Number of zeros = 0 (The curve is not intersecting the x-axis)
(iv) Number of zeros = 1 (The curve is intersecting at the origin)
(v) Number of zeros = 1 (The curve is intersecting the x-axis at one point)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Students can download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Which of the following expressions are polynomials. If not give reason:
(i) \(\frac{1}{x^2}\) + 3x – 4
Solution:
(i) \(\frac{1}{x^2}\) + 3x – 4 is not a polynomial. Since the exponent of x2 is not a whole number, but it is (\(\frac{1}{x^2}\) = x-2) negative number.

(ii) x2 (x – 1)
Solution:
x2 (x – 1) is a polynomial.

(iii) \(\frac{1}{x}\) (x + 5)
Solution:
\(\frac{1}{x}\) (x + 5) is not a polynomial. Since the exponent of x is not a whole number, but it is (\(\frac{1}{x}\) = x-1) negative number.

(iv) \(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7
Solution:
\(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7 is a polynomial. (\(\frac{1}{x^{-2}}\) = x2 and \(\frac{1}{x^{-1}}\) = x)

(v) √5x2 + √3x + √2
Solution:
√5x2 + √3x + √2 is a polynomial.

(vi) m2 – \(\sqrt[3]{m}\) + 7m – 10
m2 –\(\sqrt[3]{m}\) + 7m – 10 is not a polynomial. Since the exponent of m is not a whole number.
(\(\sqrt[3]{m}\) = m1/3)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 2.
Write the coefficient of x2 and x in each of the following polynomials.
(i) 4 + \(\frac{2}{5}\) x2 – 3x
Solution:
Coefficient of x2 is \(\frac{2}{5}\) and coefficient of x is -3.

(ii) 6 – 2x2 + 3x3 – √7x
Solution:
Coefficient of x2 is -2 and coefficient of x is -√7

(iii) π x2 – x + 2
Solution:
Coefficient of x2 is π and coefficient of x is -1.

(iv) √3x2 + √2x + 0.5
Solution:
Coefficient of x2 is √3 and coefficient of x is √2

(v) x2 – \(\frac{7}{2}\) x + 8
Solution:
Coefficient of x2 is 1 and coefficient of x is –\(\frac{7}{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 3.
Find the degree of the following polynomials.
(i) 1 – √2 y2 + y7
(ii) \(\frac{x^{3}-x^{4}+6 x^{6}}{x^{2}}\)
(iii) x3 (x2 + x)
(iv) 3x4 + 9x2 + 27x6
(v) 2√5p4 \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
Solution:
(i) 1 – √2 y2 + y7
The degree of the polynomial is 7.

(ii) Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1 1
= x – x2 + 6x4
The degree of the polynomial is 4.

(iii) x3 (x2 + x) = x5 + x4
The degree of the polynomial is 5.

(iv) 3x4 + 9x2 + 27x6
The degree of the polynomial is 6.

(v) 2√5p4 \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
The degree of the polynomial is 4.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 4.
Rewrite the following polynomial in standard form.
(i) x – 9 + √7x3 + 6x2
Solution:
The standard form is √7x3 + 6x2 – x – 9
(or) – 9 + x + 6x2 + √7x3

(ii) √2x2 – \(\frac{7}{2}\) x4 + x – 5x3
Solution:
The standard form is – \(\frac{7}{2}\) x4 – 5x3 + √2x2 + x
(or) x + √2x2 – 5x3 – \(\frac{7}{2}\) x4

(iii) 7x3 – \(\frac{6}{5}\) x2 + 4x – 1
Solution:
The given polynomial is in standard form (or) – 1 + 4x – \(\frac{6}{5}\) x2 + 7x3

(iv) y2 + √5y3 – 11 – \(\frac{7}{3}\) y + 9y4
Solution:
The standard form is 9y4 + √5y3 + y2 – \(\frac{7}{3}\) y – 11
(or) – 11 – \(\frac{7}{3}\) y + y2 + √5y3 + 9y4

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 5.
Add the following polynomials and find the degree of the resultant polynomial
(i) p(x) = 6x2 – 7x + 2; q(x) = 6x3 – 7x + 15
Solution:
p(x) + q(x) = 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x3 + 6x2 – 7x – 7x + 2 + 15
= 6x3 + 6x2 – 14x + 17
The degree of the polynomial is 3.

(ii) h(x) = 7x3 – 6x + 1; f(x) = 7x2 + 17x – 9
Solution:
h(x) + f(x) = 7x3 – 6x + 1 + 7x2 + 17x – 9
= 7x3 + 7x2 + 11x – 8
The degree of the polynomial is 3.

(iii) f(x) = 16x4 – 5x2 + 9; g(x) = -6x3 + 7x – 15
Solution:
f(x) + g(x) = 16x4 – 5x2 + 9 – 6x3 + 7x – 15
= 16x4 – 6x3 – 5x2 + 7x + 9 – 15
= 16x4 – 6x3 – 5x2 + 7x – 6
The degree of the polynomial is 4.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 6.
Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x2 + 6x – 1; q(x) = 6x – 9
Solution:
p(x) – q(x) = 7x2 + 6x – 1 – (6x – 9)
= 7x2 + 6x – 1 – 6x + 9
= 7x2 + 6x – 6x – 1 + 9
= 7x2 + 8
The degree of the polynomial is 2.

(ii) f(y) = 6y2 – 7y + 2; g(y) = 7y + y3
Solution:
f(y) – g(y) = 6y2 – 7y + 2 – (7y + y3)
= 6y2 – 7y + 2 – 7y – y3
= -y3 + 6y2 – 7y – 7y + 2
= -y3 + 6y2 – 14y + 2
The degree of the polynomial is 3.

(iii) h(z) = z5 – 6z4 + z; f(z) = 6z2 + 10z – 7
Solution:
h(z) – f(z) = z5 – 6z4 + z – (6z2 + 10z – 7)
= z5 – 6z4 + z – 6z2 – 10z + 7
= z5 – 6z4 – 6z2 + z – 10z + 7
= z5 – 6z4 – 6z2 – 9z + 7
The degree of the polynomial is 5.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 7.
What should be added to 2x3 + 6x2 – 5x + 8 to get 3x3 – 2x2 + 6x + 15?
Solution:
3x³ – 2x2 + 6x + 15 – (2x³ + 6x2 – 5x + 8)
= 3x³ – 2x2 + 6x + 15 – 2x³ – 6x2 + 5x – 8
= 3x³ – 2x³- 2x2 – 6x2 + 6x + 5x + 15 – 8
= x³ – 8x2 + 11x + 7
x³ – 8x2 + 11x + 7 must be added to get 3x³ – 2x2 + 6x + 15.

Question 8.
What must be subtracted from 2x4 + 4x2 – 3x + 7 to get 3x3 – x2 + 2x + 1?
Solution:
2x4 + 4x2 – 3x + 7 – (3x3 – x2 + 2x + 1)
= 2x4 + 4x2 – 3x + 7 – 3x3 + x2 – 2x – 1
= 2x4 – 3x3 + 4x2 + x2 – 3x – 2x + 7 – 1
= 2x4 – 3x3 + 5x2 – 5x + 6
2x4 – 3x3 + 5x2 – 5x + 6 must be subtracted to get 3x3 – x2 + 2x + 1.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x2 – 9, q(x) = 6x2 + 7x – 2
Solution:
p(x) × q(x) = (x2 – 9) (6x2 + 7x – 2)
= 6x4 + 7x3 – 2x2 – 54x2 – 63x + 18
= 6x4 + 7x3 – 56x2 – 63x + 18
The degree of the polynomial is 4.

(ii) f(x) = 7x + 2, g(x) = 15x – 9
Solution:
f(x) × g(x) = (7x + 2) (15x – 9)
= 105x2 – 63x + 30x – 18
= 105x2 – 33x – 18
The degree of the polynomial is 2.

(iii) h(x) = 6x2 – 7x + 1, f(x) = 5x – 7
Solution:
h(x) × f(x) = (6x2 – 7x + 1) (5x – 7)
= 30x3 – 42x2 – 35x2 + 49x + 5x – 7
= 30x3 – 77x2 + 54x – 7
The degree of the polynomial is 3.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 10.
The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x and y. If x = 10, y = 5 find the amount paid by him.
Solution:
The cost of a chocolate = (x + y)
Number of chocolates bought by Amir = x + y
Total amount paid by him = (x + y) (x + y)
= x2 + xy + xy + y2
= x2 + 2xy + y2
When x = 10 and y = 5
The total amount paid by him = (10)2 + 2(10)(5) + (5)2
= 100 + 100 + 25 = 225

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 11.
The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution:
Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x – 2 units
Area of the rectangle = (3x + 2) (3x – 2)
= 9x2 – 6x + 6x – 4
= 9x2 – 4
When x = 20
Area of the rectangle = 9(20)2 – 4
= 9(400) – 4
= 3600 – 4
= 3596 sq.units.

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.1

Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial is p(x) × q(x)?
Solution:
Degree of the polynomial p(x) = 1
Degree of the polynomial q(x) = 2
Degree of p(x) × q(x) = 3
The polynomial is a cubic polynomial (or) Polynomial of degree 3.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Students can download Maths Chapter 2 Real Numbers Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Additional Questions

I. Multiple choice question

Question 1.
The decimal form of –\(\frac{3}{4}\) is ………
(a) – 0.75
(b) – 0.50
(c) – 0.25
(d) – 0.125
Solution:
(a) – 0.75

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 2.
If a number has a non-terminating and non-recurring decimal expansion, then it is……….
(a) a rational number
(b) a natural number
(c) an irrational number
(d) an integer
Solution:
(c) an irrational number

Question 3.
Which one of the following has terminating decimal expansion?
(a) \(\frac{7}{9}\)
(b) \(\frac{8}{15}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{5}{32}\)
Solution:
(d) \(\frac{5}{32}\)

Question 4.
Which of the following are irrational numbers?
(i) \(\sqrt{2+\sqrt3}\)
(ii) \(\sqrt{4+\sqrt25}\)
(iii) \(\sqrt[3]{5+\sqrt7}\)
(iv) \(\sqrt{8-\sqrt[3]8}\)
(a) (ii), (iii) and (iv)
(b) (i), (iii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 5.
Irrational number has a
(a) terminating decimal
(b) no decimal part
(c) non-terminating and recurring decimal
(d) non-terminating and non-recurring decimal
Solution:
(d) non-terminating and non-recurring decimal

Question 6.
If \(\frac{1}{7}\) = 0.142857, then the value of \(\frac{3}{7}\) is……..
(a) 0.285741
(b) 0.428571
(c) 0.285714
(d) 0.574128
Solution:
(b) 0.428571

Question 7.
Which of the following are not rational numbers?
(a) 7√5
(b) \(\frac{7}{\sqrt{5}}\)
(c) \(\sqrt{36}\) – 9
(d) π + 2
Solution:
(c) \(\sqrt{36}\) – 9

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 8.
The product of 2√5 and 6√5 is……….
(a) 12√5
(b) 60
(c) 40
(d) 8√5
Solution:
(b) 60

Question 9.
The rational number lying between \(\frac{1}{5}\) and \(\frac{1}{2}\)
(a) \(\frac{7}{20}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{3}{10}\)
Solution:
(a) \(\frac{7}{20}\)

Question 10.
The value of 0.03 + 0.03 is ……….
(a) 0.\(\overline { 09 }\)
(b) 0.\(\overline { 0303 }\)
(c) 0.\(\overline { 06 }\)
(d) 0
Solution:
(c) 0.06

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 11.
The sum of \(\sqrt{343}\) + \(\sqrt{567}\) is
(a) 18√3
(b) 16√7
(c) 15√3
(d) 14√7
Solution:
(b) 16√7

Question 12.
If \(\sqrt{363}\) = x√3 then x = ………
(a) 8
(b) 9
(c) 10
(d) 11
Solution:
(d) 11

Question 13.
The rationalising factor of \(\frac{1}{\sqrt{7}}\) is ……….
(i) 7
(b) √7
(c) \(\frac{1}{7}\)
(d) \(\frac{1}{\sqrt{7}}\)
Solution:
(b) √7

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 14.
The value of \((\frac{1}{3^5})^4\) is ……..
(a) 320
(b) 3-20
(c) \(\frac{1}{3^{-20}}\)
(d) \(\frac{1}{3^{9}}\)
Solution:
(b) 3-20

Question 15.
What is 3.976 × 10-4 written in decimal form?
(a) 0.003976
(b) 0.0003976
(c) 39760
(d) 0.03976
Solution:
(b) 0.0003976

II. Answer the following Questions.

Question 1.
Find any seven rational numbers between \(\frac{5}{8}\) and –\(\frac{5}{6}\)
Solution:
Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 1
Now the rational numbers between
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 2
We can take any seven of them.
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 3

Question 2.
Find any three rational numbers between \(\frac{1}{2}\) and \(\frac{1}{5}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 4
Thus the three rational numbers are \(\frac{7}{20}\), \(\frac{17}{40}\) and \(\frac{37}{80}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 3.
Represent \(-\frac{2}{11}\), \(-\frac{5}{11}\) and \(-\frac{9}{11}\) on the number lines.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 5
To Represent \(-\frac{2}{11}\), \(-\frac{5}{11}\) and \(-\frac{9}{11}\) on the number line we make 11 markings each being equal distence \(\frac{1}{11}\) on the left of 0.
The point A represent \((-\frac{2}{11})\), the point B represents \((-\frac{5}{11})\) and the point C represents \((-\frac{9}{11})\)

Question 4.
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0.
(i) 0.\(\overline { 47 }\)
Solution:
Let x = 0.474747…….. →(1)
100 x = 47.4747…….. →(2)
(2) – (1) ⇒ 100x – x = 47.4747……..
(-) 0.4747……..
99 x = 47.0000
x = \(\frac{47}{99}\)
∴ 0.\(\overline { 47 }\) = \(\frac{47}{99}\)

(ii) 0.\(\overline { 57 }\)
Solution:
Let x = 0.57777…….. →(1)
10 x = 5.77777…….. →(2)
100 x = 57.7777…….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 57.7777……..
(-) 5.7777……..
99 x = 52.0000
x = \(\frac{52}{90}\) = \(\frac{26}{45}\)
∴ 0.\(\overline { 57 }\) = \(\frac{26}{45}\)

(iii) 0.\(\overline { 245 }\)
Solution:
Let x = 0.2454545…….. →(1)
10 x = 2.454545…….. →(2)
1000 x = 245.4545…….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 245.4545
(-) 2.4545………
990 x = 243.00000
x = \(\frac{243}{990}\) (or) \(\frac{27}{110}\)
∴ 0.\(\overline { 245 }\) = \(\frac{27}{110}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 5.
Without actual division classify the decimal expansion of the following numbers as terminating or non-terminating and recurring.
(i) \(\frac{7}{16}\)
(ii) \(\frac{13}{150}\)
(ii) –\(\frac{11}{75}\)
(iv) \(\frac{17}{200}\)
Solution:
(i) \(\frac{7}{16}\) = \(\frac{7}{2^4}\) = \(\frac{7}{2^{4} \times 5^{0}}\)
∴ \(\frac{7}{16}\) has a terminating decimal expansion.

(ii) \(\frac{13}{150}=\frac{13}{2 \times 3 \times 5^{2}}\)
Since it is not in the form of \(\frac{P}{2^{m} \times 5^{n}}\)
∴ \(\frac{13}{150}\) as non-terminating and recurring decimal expansion.

(iii) \(-\frac{11}{75}=-\frac{11}{3 \times 5^{2}}\)
Since it is not in the form of \(\frac{P}{2^{m} \times 5^{n}}\)
∴ –\(\frac{11}{75}\) as non-terminating and recurring decimal expansion.

(iv) \(\frac{17}{200}=\frac{17}{2^{3} \times 5^{2}}\)
∴ \(\frac{17}{200}\) has a terminating decimal expansion.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 6.
Find the value of \(\sqrt{27}\) + \(\sqrt{75}\) – \(\sqrt{108}\) + \(\sqrt{48}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 6
= 3√3 + 5√3 – 6√3 + 4√3
= 12√3 – 6√3
= 6√3
= 6 × 1.732
= 10.392

Question 7.
Evaluate \(\frac{\sqrt{2}+1}{\sqrt{2-1}}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 7
= 2√2 + 3
= 2 × 1.414 + 3
= 2.828 + 3
= 5.828

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Question 8.
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 8
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions 9
= 69984 × 1021-21-20+9
= 69984 × 10-32
= 6.9984 × 104 × 10-32
= 6.9984 × 10-32+4
= 6.9984 × 10-28

Question 9.
Write
(a) 9.87 × 109
(b) 4.134 × 10-4 and
(c) 1.432 × 10-9 in decimal form.
Solution:
(a) 9.87 × 109 = 9870000000
(b) 4.134 × 10-4 = 0.0004134
(c) 1.432 × 10-9 = 0.000000001432

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Additional Questions

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

Students can download Maths Chapter 2 Real Numbers Ex 2.8 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.8

Question 1.
Represent the following numbers in the scientific notation:
(i) 569430000000
(ii) 2000.57
(iii) 0.0000006000
(iv) 0.0009000002
Solution:
(i) 569430000000 = 5.6943 × 1011
(ii) 2000.57 = 2.00057 × 103
(iii) 0.0000006000 = 6.0 × 10-7
(iv) 0.0009000002 = 9.000002 × 10-4

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

Question 2.
Write the following numbers in decimal form:
(i) 3.459 × 106
(ii) 5.678 × 104
(iii) 1.00005 × 10-5
(iv) 2.530009 × 10-7
Solution:
(i) 3.459 × 106
= 3459000
(ii) 5.678 × 104
= 56780
(iii) 1.00005 × 10-5
= 0.0000100005
(iv) 2.530009 × 10-7
= 0.0000002530009

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

Question 3.
Represent the following numbers in scientific notation:
(i) (300000)2 × (20000)4
(ii) (0.000001)11 ÷ (0.005)3
(iii) {(0.00003)6 × (0.00005)4} ÷ {(0.009)3 × (0.05)2}
Solution:
(i) (300000)2 × (20000)4 = (3 × 105)2 × (2 × 104)4
= 32 × (105)2 × 24 × (104)4
= 9 × 1010 × 16 × 1016
= 9 × 16 × 1010-16
= 144 × 1026
= 1.44 × 1028

(ii) (0.000001)11 ÷ (0.005)3
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8 1
0.008 × 10-66+9
= 8.0 × 10-3 × 10-57
= 8.0 × 10-3-57
= 8.0 × 10-60

(iii) {(0.00003)6 × (0.00005)4} ÷ {(0.009)3 × (0.05)2}
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8 2
= 2.5 × 10-49+13
= 2.5 × 10-36

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

Question 4.
Represent the following information in scientific notation:
(i) The world population is nearly 7000,000,000.
(ii) One light year means the distance 9460528400000000 km.
(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg.
Solution:
(i) World population = 7.0 × 109
(ii) Distance = 9.4605 × 1015 km.
(iii) Mass of an electron = 9.1093822 × 10-31 kg

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

Question 5.
Simplify:
(2.75 × 107) + (1.23 × 108)
(ii) (1.598 × 1017) – (4.58 × 1015)
(iii) (1.02 × 1010) × (1.20 × 10-3)
(iv) (8.41 × 104) ÷ (4.3 × 105)
Solution:
(i) (2.75 × 107) + (1.23 × 108) = 27500000 + 123000000
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8 3
= 150500000
= 1.505 × 108

(ii) (1.598 × 1017) – (4.58 × 1015) = 1552,20000000000000
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8 4
= 1.5522 × 1017

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.8

(iii) (1.02 × 1010) × (1.20 × 10-3) = 1.02 × 1.20 × 1010 × 10-3
=1.224 × 107

(iv) (8.41 × 104) ÷ (4.3 × 105) = \(\frac{8.41×10^{4}}{4.3×10^{5}}\)
= \(\frac{8.41}{4.3}\) × 104-5
= \(\frac{8.41}{4.3}\) × 10-1
= 1.9558139 × 10-1

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Students can download Maths Chapter 2 Real Numbers Ex 2.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.5

Question 1.
Write the following in the form of \(5^n\):
(i) 625
(ii) \(\frac{1}{5}\)
(iii) \(\sqrt{5}\)
(iv) \(\sqrt{125}\)
Solution:
(i) 625 = 54
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5 1
(ii) \(\frac{1}{5}\) = 5-1
(iii) \(\sqrt{5}\) = \(5^\frac{1}{2}\)
(iv) \(\sqrt{125}\) = \(\sqrt{5^3}\) = \((5^3)^\frac{1}{2} = 5^\frac{3}{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Question 2.
Write the following in the form of \(4^n\):
(i) 16
(ii) 8
(iii) 32
Solution:
(i) 16
= 4 × 4
= 4²

(ii) 8
= 4 × 2
= 4 × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4 \(\times 4^{\frac{1}{2}} \)
= 4\(^{1+\frac{1}{2}} \)
= 4\(^{\frac{2+1}{2}} \)
= 4\(^{3 / 2}\)

(iii) 32
= 4 × 4 × 2
= 4² × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4\(^{2} \times 4^{\frac{1}{2}} \)
= 4\(^{2+\frac{1}{2}} \)
= 4\(^{\frac{4+1}{2}} \)
= 4\(^{\frac{5}{2}} \)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Question 3.
Find the value of
(i) (49)\(^\frac{1}{2}\)
(ii) (243)\(^\frac{2}{5}\)
(iii) (9)\(^\frac{-3}{2}\)
(iv) \((\frac{64}{125})^\frac{-2}{3}\)
Solution:
(i) 49\(^\frac{1}{2}\) = \((7^2)^\frac{1}{2}\) = 7\(^{2 × \frac{1}{2}}\) = 7
(ii) (243)\(^\frac{2}{5}\) = \((3^5)^\frac{2}{5}\) = 3\(^{5 × \frac{2}{5}}\) = 3² = 9
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5 2
(iii) \(9^{\frac{-3}{2}}=\left(3^{2}\right)^{\frac{-3}{2}}=3^{2 \times \frac{-3}{2}}=3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\)
(iv) \(\left(\frac{64}{125}\right)^{\frac{-2}{3}}=\left(\frac{4^{3}}{5^{3}}\right)^{\frac{-2}{3}}=\left[\left(\frac{4}{5}\right)^{3}\right]^{\frac{-2}{3}}=\left(\frac{4}{5}\right)^{3 \times \frac{-2}{3}}=\left(\frac{4}{5}\right)^{-2}=\frac{4^{-2}}{5^{-2}}=\frac{5^{2}}{4^{2}}=\frac{25}{16} \)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Question 4.
Use a fractional index to write:
(i) \(\sqrt{5}\)
(ii) \(\sqrt[2]{7}\)
(iii) (\(\sqrt[3]{49})^{5}\)
(iv) \((\frac{1}{\sqrt[3]{100}})^{7}\)
Solution:
(i) \(\sqrt{5}\) = (5)\(^\frac{1}{2}\)
(ii) \(\sqrt[2]{7}\) = 7\(^\frac{1}{2}\)
(iii) \((\sqrt[3]{49})^{5}=\left[(49)^{\frac{1}{3}}\right]^{5}=\left[\left(7^{2}\right)^{\frac{1}{3}}\right]^{5}=\left(7^{\frac{2}{3}}\right)^{5}=7^{\frac{2}{3} \times 5}=7^{\frac{10}{3}}\)
(iv) \(\left(\frac{1}{\sqrt[3]{100}}\right)^{7}=\left[\frac{1}{\sqrt[3]{10^{2}}}\right]^{7}=\left[\frac{1}{\left(10^{2}\right)^{1 / 3}}\right]^{7}=\left[\frac{1}{10^{2 / 3}}\right]^{7}=\left(10^{\frac{-2}{3}}\right)^{7}=10^{\frac{-2}{3} \times 7}=10^{\frac{-14}{3}}\)

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Question 5.
Find the 5th root of:
(i) 32
(ii) 243
(iii) 100000
(iv) \(\frac{1024}{3125}\)
Solution:
(i) \(\sqrt[5]{32}=(32)^{\frac{1}{5}}=\left(2^{5}\right)^{\frac{1}{5}}=2^{5 \times \frac{1}{5}} \) = 2
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5 3
(ii) \(\sqrt[5]{243}=(243)^{\frac{1}{5}}=\left(3^{5}\right)^{\frac{1}{5}}=3^{5 \times \frac{1}{5}}\) = 3
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5 4
(iii) \(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=\left(10^{5}\right)^{\frac{1}{5}}\)
= \(10^{5}\times{\frac{1}{5}}\)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5 5

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.5

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4

Students can download Maths Chapter 2 Real Numbers Ex 2.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.4

Question 1.
Represent the following numbers on the number line.
(i) 5.348
Solution:
5.348 lies between 5 and 6.
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4 1
Steps of construction:
1. Divide the distance between 5 and 6 into 10 equal intervals.
2. Mark the point 5.3 which is the sixth from the left of 6 and 3 from the right of 5.
3. 5.34 lies between 5.3 and 5.4. Divide the distance into 10 equal intervals.
4. Mark the point 5.34 which is sixth from the left of 5.40
5. 5.348 lies between 5.34 and 5.35. Divide the distance into 10 equal intervals.
6. Mark a point 5.348 which is second from the left of 5.350 and seventh form the right of 5.340

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4

(ii) 6.\(\overline {4}\) upto 3 decimal places.
Solution:
6.\(\overline {4}\) = 6.4444
6.\(\overline {4}\) = 6.444 (correct to 3 decimal places)
The number lies between 6 and 7.
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4 2
Steps of construction:
1. Divide the distance between 6 and 7 into 10 equal intervals.
2. Mark the point 6.4 which is the sixth from the left of 7 and fourth from the right of 6.
3. 6.44 lies between 6.44 and 6.45. Divide the distance into 10 equal intervals.
4. Mark the point 6.44 which is sixth from the left of 6.5 and fourth from the right of 6.40.
5. Mark the point 6.444 which is sixth from the left of 6.450 and fourth from the right of 6.440.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4

(ii) 4.\(\overline {73}\) upto 4 decimal places.
Solution:
4.\(\overline {73}\) = 4.737373……..
= 4.737374 (correct to 4 decimal places 4.7374 lies between 4 and 5)
Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4 3
Steps of construction:
1. Divide the distance between 4 and 5 into 10 equal parts.
2. Mark the point 4.7 which is third from the left of 5 and seventh from the right of 4.
3. 4.73 lies between 4.7 and 4.8. Divide the distance into 10 equal intervals.
4. Mark the point 4.73 which is seventh from the left of 4.80 and third from the left of 4.70.
5. 4.737 lies between 4.73 and 4.74. Divide the distance into 10 equal intervals.
6. Mark the point 4.737 which is third from the left of 4.740 and seventh from the right of 4.730.
7. 4.7374 lies between 4.737 and 4.738. Divide the distance into 10 equal intervals.
8. Mark the point 4.7374 which is sixth from the left of 4.7380 and fourth from the right of 4.7370.

Samacheer Kalvi 9th Maths Guide Chapter 2 Real Numbers Ex 2.4