Category: Class 9

Students can download Maths Chapter 3 Algebra Ex 3.11 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 3.11

Question 1.
Solve, using the method of substitution.
(i) 2x – 3y – 7; 5x + y = 9
Solution:
2x – 3y = 7 → (1)
5x + y = 9 → (2)
Equation (2) becomes
y = 9 – 5x
Substitute the value of y in (1)
2x – 3 (9 – 5x) = 7
2x – 27 + 15x = 7
17x = 7 + 27
17x = 34
x = \( \frac{34}{17}\)
= 2
Substitute the value of x = 2 (in) (2)
y = 9 – 5 (2) = 9 – 10 = -1
∴ The value of x = 2 and y = -1

(ii) 1.5x + 0.1y = 6.2; 3x – 0.4y = 11.2
Solution:
1.5x + 0.1y = 6.2
Multiply by 10
15x + y = 62
y = 62 – 15x → (1)
3x – 0.4y = 11.2
Multiply by 10
30x – 4y = 112
divided by (2) we get
15x – 2y = 56 → (2)
Substitute the value of y in (2)
15x – 2(62 – 15x) = 56
15x – 124 + 30x = 56
45x = 56 + 124
45x = 180
x = \( \frac{180}{45}\)
= 4
Substitute the value of x = 4 in (1)
y = 62 – 15(4)
= 62 – 60
y = 2
∴ The value of x = 4 and y = 2

(iii) 10% of x + 20% of y = 24; 3x – y = 20
Solution:
\( \frac{10}{100}\) × x + \( \frac{20}{100}\) × y = 24
\( \frac{x}{10}\) + \( \frac{y}{5}\) = 24
Multiply by 10
x + 2y = 240 → (1)
3x – y = 20
– y = 20 – 3x
y = 3x – 20 → (2)
Substitute the value of y in (1)
x + 2 (3x – 20) = 240
x + 6x – 40 = 240
7x – 40 = 240
7x = 240 + 40
7x = 280
x = \( \frac{280}{7}\)
x = 40
Substitute the value of x = 40 in (2)
y = 3 (40) – 20 = 120 – 20
y = 100
∴ The value of x = 40 and y = 100

(iv) √2x – √3y = 1; √3x – √8y = 0
Solution:
√2x – √3y = 1
– √3y = 1 – √2x
√3y = √2x – 1
y = \(\frac{√2x-1}{√3}\) → (1)
√3x – √8y = 0 → (2)
Substitute the value of y in (2)
\(\sqrt{3x}-\frac{√8(√2x-1)}{√3}\)
multiply by √3
⇒ \(\frac{3x-√8(√2x-1)}{√3}\)
3x – √8(√2x – 1) = 0
3x – 4x + √8 = 0
-x = √8
Substitute the value of x in (1)

The value of x = √8 and y = √3

Question 2.
Raman’s age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution:
Let Raman’s age be “x” years and the sum of the ages of two sons be “y” years.
By the given first condition
x = 3y
x – 3y = 0 → (1)
After 5 years Raman’s age is x + 5 years
Sum of sons age is (y + 10) years
(each son age increases by 5 years)
By the given second condition
x + 5 = 2 (y + 10)
x + 5 = 2y + 20
x – 2y = 20 – 5
x – 2y = 15 → (2)
Equation (1) becomes
x = 3y
Substitute the value of x in (2)
3y – 2y = 15
y = 15
Substitute the value of y = 15 in x = 3y
x = 3(15)
x = 45
∴ Raman’s age = 45 years

Question 3.
The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. If the digits are reversed, the number so formed exceeds the original number by 495. Find the number.
Solution:
Let the unit digit be and the 100 is digit be X. The number is XOY (100x + y)
By the given first condition
x + y = 13 ….(1)
If the digits are reversed the number is 100 y + x.
By the given second condition.
100y + x = 100x + y + 495
-99x + 99y = 495
-x + y = 5 …. (2)
x + y = 13 ….(1)
Add (1) and (2)
2y = 18
y = 9
Substitute the value of y = 9 in (1)
x + 9 = 13
x = 13 – 9
x = 4
∴ The number is 409

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

Plot the points (-2, -4) (0, 0) (2, 4) and (3, 6) in the graph sheet we get a straight line.

(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

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.

(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


Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line.

(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


Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line.

Question 2.
Solve graphically (i) x + y = 7, x – y = 3
Solution:
x + y = 7
y = 7 – x

Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet
x – y = 3
-y = -x + 3
y = x – 3

Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet.

The point of intersection is (5, 2) of lines (1) and (2).
The solution set is (5,2).

(ii) 3x + 2y = 4; 9x + 6y – 12 = 0
Solution:
2y = -3x + 4
y = \(\frac{-3x+4}{2}\)
= \(\frac{-3}{2}\)x + 2

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

Plot the points (-2, 5), (0, 2) and (2, -1) the same graph sheet

Here both the equations are identical, but in different form.
Their solution is same.
This equations have an infinite number of solution.

(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

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

Plot the points (-2, 12), (-1, 10), (0, 8) and (2, 4) in the same graph sheet.

The given two lines are parallel.
∴ They do not intersect a point.
∴ There is no solution.

(iv) x – y = 0; y + 3 = 0
Solution:
x – y = 0
-y = -x
y = x

Plot the points (-2, -2), (0, 0), (1, 1) and (3, 3) in the same graph sheet.
y + 3 = 0
y = -3

Plot the points (-2, -3), (0, -3), (1, -3) and (2, -3) in the same graph sheet.

The two lines l₁ and l₂ intersect at (-3, -3). The solution set is (-3, -3).

(v) y = 2x + 1; 3x – 6 = 0
Solution:
y = 2x + 1

Plot the points (-3, -5), (-1, -1), (0, 1) and (2, 5) in the graph sheet
3x – 6 = 0
y = -3x + 6

Plot the points (-2, 12), (-1, 9), (0, 6) and (2, 0) in the same graph sheet

The two lines l₁ and l₂ intersect at (1, 3).
∴ The solution set is (1, 3).

(vi) x = -3; y = 3
Solution:
x = -3

Plot the points (-3, -3), (-3, -2), (-3, 2) and (-3, 3) in the graph sheet
y = 3

Plot the points (-3, 3), (-1, 3), (0, 3) and (2, 3) in the same graph sheet

The two lines l₁ and l₂ intersect at (-3, 3)
∴ The solution set is (-3, 3)

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

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

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

Plot the points (40, -10), (50, 0), (60, 10) and (70, 20) in the same graph sheet

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

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

(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

240x⁶ = 2⁴ × 3 × 5 × x⁶
G.C.D = 2⁴ × x⁶
= 16x⁶

(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

(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

Question 2.
Find the GCD for the following:
(i) (2x + 5), (5x + 2)
(ii) a^m+1, a^m+2, a^m+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) a^m+1 = a^m × a¹
a^m+2 = a^m × a²
a^m+3 = a^m × a³
G.C.D.= a^m × a
= a^m+1

(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 )

(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

Students can download Maths Chapter 3 Algebra Ex 3.8 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 3 Algebra Ex 3.8

Question 1.
Factorise each of the following polynomials using synthetic division:
(i) x³ – 3x² – 10x + 24
Solution:
p(x) – x³ – 3x² – 10x + 24
p(1) = 1³ – 3(1)² – 10(1) + 24
= 1 – 3 – 10 + 24
= 25 – 13
≠ 0
x – 1 is not a factor

p(-1) = (-1)³ – 3(-1)² – 10(-1) + 24
= – 1 – 3(1) + 10 + 24
= -1 – 3 + 10 + 24
= 34 – 4
= 30
≠ 0
x + 1 is not a factor

p(2) = 2³ – 3(2)² – 10(2) + 24
= 8 – 3(4) – 20 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
∴ x – 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
∴ The factors of x³ – 3x² – 10x + 24 = (x – 2) (x – 4) (x + 3)

(ii) 2x³ – 3x² – 3x + 2
Solution:
p(x) = 2x³ – 3x² – 3x + 2
P(1) = 2(1)³ – 3(1)² – 3(1) + 2
= 2 – 3 – 3 + 2
= 2 – 6
= -4
≠ 0
x – 1 is not a factor

P(-1) = 2(-1)³ – 3(-1)² – 3(-1) + 2
= -2 – 3 + 3 + 2
= 5 – 5
= 0
∴ x + 1 is a factor

2x² – 5x + 2 = 2x² – 4x – x + 2

= 2x(x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ The factors of 2x³ – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) – 7x + 3 + 4x³
Solution:
p(x) = – 7x + 3 + 4x³
= 4x³ – 7x + 3
P(1) = 4(1)³ – 7(1) + 3
4 – 7 + 3
= 7 – 7
= 0
∴ x – 1 is a factor

4x² + 4x – 3 = 4x² + 6x – 2x – 3

= 2x(2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors of – 7x + 3 + 4x³ = (x – 1) (2x + 3) (2x – 1)

(iv) x³ + x² – 14x – 24
Solution:
p(x) = x³ + x² – 14x – 24
p(1) = (1)³ + (1)² – 14 (1) – 24
= 1 + 1 – 14 – 24
= -36
≠ 0
x + 1 is not a factor.

p(-1) = (-1)³ + (-1)² – 14(-1) – 24
= -1 + 1 + 14 – 24
= 15 – 25
≠ 0
x – 1 is not a factor.

p(2) = (-2)³ + (-2)² – 14 (-2) – 24
= -8 + 4 + 28 – 24
= 32 – 32
= 0
∴ x + 2 is a factor

x² – x – 12 = x² – 4x + 3x – 12

= x(x – 4) + 3 (x – 4)
= (x – 4) (x + 3)
This (x + 2) (x + 3) (x – 4) are the factors.
x³ + x² – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x³ – 7x + 6
Solution:
p(x) = x³ – 7x + 6
P( 1) = 1³ – 7(1) + 6
= 1 – 7 + 6
= 7 – 7
= 0
∴ x – 1 is a factor

x² + x – 6 = x² + 3x – 2x – 6

= x(x + 3) – 2 (x + 3)
= (x + 3) (x – 2)
This (x – 1) (x – 2) (x + 3) are factors.
∴ x³ – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x³ – 10x² – x + 10
p(x) = x³ – 10x² – x + 10
= 1 – 10 – 1 + 10
= 11 – 11
= 0
∴ x – 1 is a factor

x² – 9x – 10 = x² – 10x + x – 10

= x(x – 10) + 1 (x – 10)
= (x – 10) (x + 1)
This (x – 1) (x + 1) (x – 10) are the factors.
∴ x³ – 10x² – x + 10 = (x – 1) (x – 10) (x + 1)

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

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)

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

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

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

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.

Now the rational numbers between

We can take any seven of them.

Question 2.
Find any three rational numbers between \(\frac{1}{2}\) and \(\frac{1}{5}\)
Solution:

Thus the three rational numbers are \(\frac{7}{20}\), \(\frac{17}{40}\) and \(\frac{37}{80}\)

Question 3.
Represent \(-\frac{2}{11}\), \(-\frac{5}{11}\) and \(-\frac{9}{11}\) on the number lines.
Solution:

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.\(\overline { 245 }\) = \(\frac{27}{110}\)

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.

Question 6.
Find the value of \(\sqrt{27}\) + \(\sqrt{75}\) – \(\sqrt{108}\) + \(\sqrt{48}\)
Solution:

= 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:

= 2√2 + 3
= 2 × 1.414 + 3
= 2.828 + 3
= 5.828

Question 8.

Solution:

= 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

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

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

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 × 10^10-16
= 144 × 10²⁶
= 1.44 × 10²⁸

(ii) (0.000001)¹¹ ÷ (0.005)³

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}

= 2.5 × 10^-49+13
= 2.5 × 10^-36

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

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

= 150500000
= 1.505 × 10⁸

(ii) (1.598 × 10¹⁷) – (4.58 × 10¹⁵) = 1552,20000000000000

= 1.5522 × 10¹⁷

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

Students can download Maths Chapter 2 Real Numbers Ex 2.7 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 2.7

Question 1.
Rationalise the denominator:
(i) \( \frac{1}{\sqrt{50}}\)
(ii) \( \frac{5}{3\sqrt{5}}\)
(iii) \( \frac{\sqrt{75}}{\sqrt{18}}\)
(iv) \( \frac{3\sqrt{5}}{\sqrt{6}}\)
Solution:
(i) \( \frac{1}{\sqrt{50}}\) = \(\frac{1}{\sqrt{25 \times 2}}=\frac{1}{5 \sqrt{2}}=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{5 \times 2}=\frac{\sqrt{2}}{10}\)

(ii) \( \frac{5}{3\sqrt{5}}\) = \(\frac{5}{3 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{5 \sqrt{5}}{3 \times 5}=\frac{\sqrt{5}}{3}\)

(iii) \( \frac{\sqrt{75}}{\sqrt{18}}\) = \(\frac{\sqrt{3 \times 25}}{\sqrt{2 \times 9}}=\frac{5 \sqrt{3}}{3 \sqrt{2}}=\frac{5 \sqrt{3}}{3 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{5 \sqrt{6}}{3 \times 2}=\frac{5 \sqrt{6}}{6}\)

(iv) \( \frac{3\sqrt{5}}{\sqrt{6}}\) = \( \frac{3 \sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=\frac{3 \sqrt{30}}{6}=\frac{\sqrt{30}}{2} \)

Question 2.
Rationalise the denominator and simplify:
(i) \(\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}\)
Solution:

(ii) \(\frac{5\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
Solution:

(iii) \(\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\)
Solution:

(iv) \(\frac{\sqrt{5}}{\sqrt{6}+2} – \frac{\sqrt{5}}{\sqrt{6}-2}\)
Solution:

Question 3.
Find the value of a and b if \(\frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b\).
Solution:

Question 4.
If x = \(\sqrt{7}\) + 2, then find the value of x² + \(\frac{1}{x^2}\)
Solution:
\(\sqrt{7}\) + 2 ⇒ x² = \((\sqrt{5}+2)^{2}\)
= \((\sqrt{5})^{2}\) + 2 × 2 × \(\sqrt{5}\) + 2² = 5 + 4 \(\sqrt{5}\) + 4 = 9 + 4\(\sqrt{5}\)
\(\frac{1}{x}=\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}=\frac{\sqrt{5}-2}{(\sqrt{5})^{2}-2^{2}}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\)
\(\frac{1}{x^{2}}\) = (\(\sqrt{5} – 2)^{2}\)
= \((\sqrt{5})^{2}\) – 2 × \(\sqrt{5}\) × 2 + 2² = 5 – 4 \(\sqrt{5}\) + 4 = 9 – 4 \(\sqrt{5}\)
∴ x² + \(\frac{1}{x^{2}}\) = 9 + \(4\sqrt{5}\) + 9 – \(4\sqrt{5}\) = 18
The value of x² + \(\frac{1}{x^{2}}\) = 18

Question 5.
Given \(\sqrt{2}\) = 1.414, find the value of \(\frac{8 – 5\sqrt{2}}{3 – 2\sqrt{2}}\) (to 3 places of decimals).
Solution:

= 4 + \(\sqrt{2}\) = 4 + 1.414 = 5.414

Students can download Maths Chapter 2 Real Numbers Ex 2.6 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 2.6

Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\)
(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\)
(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\)
(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)
Solution:
(i) 5\(\sqrt{3}\) + 18\(\sqrt{3}\) – 2\(\sqrt{3}\) = (5 + 18 – 2)\(\sqrt{3}\)
= (23 – 2) \(\sqrt{3}\) = 21\(\sqrt{3}\)

(ii) 4\(\sqrt[3]{5}\) + 2\(\sqrt[3]{5}\) – 3\(\sqrt[3]{5}\) = (4 + 2 – 3) \(\sqrt[3]{5}\)
= (6 – 3) \(\sqrt[3]{5}\) = 3\(\sqrt[3]{5}\)

(iii) 3\(\sqrt{75}\) + 5\(\sqrt{48}\) – \(\sqrt{243}\) = \(3\sqrt{5^{2}×3} + 5\sqrt{2^{4}×3} – \sqrt{3^{5}}\)

= 3 × 5\(\sqrt{3}\) + 5 × 2²\(\sqrt{3}\) – 3²\(\sqrt{3}\) = 15\(\sqrt{3}\) + 20\(\sqrt{3}\) – 9\(\sqrt{3}\)
= (15 + 20 – 9)\(\sqrt{3}\)
= (35 – 9)\(\sqrt{3}\)
= 26 \(\sqrt{3}\)

(iv) 5\(\sqrt[3]{40}\) + 2\(\sqrt[3]{625}\) – 3\(\sqrt[3]{320}\)

= 5\(\sqrt[3]{2^{3}×5} + 2\sqrt[3]{5^{3}×5} – 3\sqrt[3]{2^{3}×2^{3}×5}\)
5 × 2\(\sqrt[3]{5} + 2 × 5\sqrt[3]{5} – 3 × 2 × 2\sqrt[3]{5} \)
= 10\(\sqrt[3]{5} + 10\sqrt[3]{5} – 12\sqrt[3]{5} \)
= 20\(\sqrt[3]{5} – 12\sqrt[3]{5}\)
= 8\(\sqrt[3]{5}\)

Question 2.
Simplify the following using multiplication and division properties of surds:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\)
(ii) \(\sqrt{35}\) ÷ \(\sqrt{7}\)
(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\)
(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)
Solution:
(i) \(\sqrt{3}\) × \(\sqrt{5}\) × \(\sqrt{2}\) = \(\sqrt{3×5×2} = \sqrt{30}\)

(ii) \(\sqrt{37} ÷ \sqrt{7} = \frac{\sqrt{35}}{\sqrt{7}} = \sqrt{\frac{35}{7}} = \sqrt{5}\)

(iii) \(\sqrt[3]{27}\) × \(\sqrt[3]{8}\) × \(\sqrt[3]{125}\) = \(\sqrt[3]{27×8×125}\)
= \(\sqrt[3]{3^{3}×2^{3}×5^{3}}\) = 3 × 2 × 5 = 30

(iv) (7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\))
[using a2 – b2 = (a + b) (a – b)]
(7\(\sqrt{a}\) – 5\(\sqrt{b}\)) (7\(\sqrt{a}\) + 5\(\sqrt{b}\)) = \((7\sqrt{a})^{2} – (5\sqrt{b})^{2}\) = 49a – 25b

(v) (\(\sqrt{\frac{225}{729}} – \sqrt{\frac{25}{144}}\)) ÷ \(\sqrt{\frac{16}{81}}\)


= \(\frac{5}{36}\) × \(\frac{9}{4}\)
= \(\frac{5×1}{4×4}\)
= \(\frac{5}{16}\)

Question 3.
If \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732, \(\sqrt{5}\) = 2.236, \(\sqrt{10}\) = 3.162, then find the values of the following correct to 3 places of decimals.
(i) \(\sqrt{40}\) – \(\sqrt{20}\)
(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\)
Solution:
(i) \(\sqrt{40}\) – \(\sqrt{20}\) = \(\sqrt{4×10} – \sqrt{4×5} = 2\sqrt{10} – 2\sqrt{5}\)
= 2 × 3.162 – 2 × 2.236 = 6.324 – 4.472 = 1.852

(ii) \(\sqrt{300}\) + \(\sqrt{90}\) – \(\sqrt{8}\) = \(\sqrt{3×100} + \sqrt{9×10} – \sqrt{4×2}\)
= 10\(\sqrt{3}\) + 3\(\sqrt{10}\) – 2\(\sqrt{2}\)
= 10 × 1.732 + 3 × 3.162 – 2 × 1.414
= 17.32 + 9.486 – 2.828
= 26.806 – 2.828
= 23.978

Question 4.
Arrange surds in descending order
(i) \(\sqrt[3]{5}\), \(\sqrt[9]{4}\), \(\sqrt[6]{3}\)
Solution:
LCM of 3, 9 and 6 is 18

\(\sqrt[3]{5}\) = \(\sqrt[3×6]{5^{6}}\) = \(\sqrt[18]{15625}\)
\(\sqrt[9]{4}\) = \(\sqrt[2×9]{4^{2}}\) = \(\sqrt[18]{16}\)
\(\sqrt[6]{3}\) = \(\sqrt[3×6]{3^{3}}\) = \(\sqrt[18]{27}\)
\(\sqrt[18]{15625}\) > \(\sqrt[18]{27}\) > \(\sqrt[18]{16}\)
\(\sqrt[3]{5}\) > \(\sqrt[6]{3}\) > \(\sqrt[9]{4}\)

(ii) \(\sqrt[2]{\sqrt[3]{5}}\), \(\sqrt[3]{\sqrt[4]{7}}\), \(\sqrt{\sqrt{3}}\)
Solution:
\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\); \(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\); \(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\)
LCM of 6, 12 and 4 is 12

\(\sqrt[2]{\sqrt[3]{5}}\) = \(\sqrt[6]{5}\) = \(\sqrt[12]{5^{2}}\) = \(\sqrt[12]{25}\)
\(\sqrt[3]{\sqrt[4]{7}}\) = \(\sqrt[12]{7}\) = \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) = \(\sqrt[4]{3}\) = \(\sqrt[12]{3^{3}}\) = \(\sqrt[12]{27}\)
\(\sqrt[12]{27}\) > \(\sqrt[12]{25}\) > \(\sqrt[12]{7}\)
\(\sqrt{\sqrt{3}}\) > \(\sqrt[2]{\sqrt[3]{5}}\) > \(\sqrt[3]{\sqrt[4]{7}}\)

Question 5.
Can you get a pure surd when you find:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes we can get a surd.
Example:
(a) 3\(\sqrt{2}\) + 5\(\sqrt{2}\) = (3 + 5)\(\sqrt{2}\) = 8\(\sqrt{2}\)
(b) 3\(\sqrt{6}\) + 2\(\sqrt{6}\) = (3 + 2)\(\sqrt{6}\) = 5\(\sqrt{6}\)

(ii) Yes we can get a surd.
Example:
(a) \(\sqrt{75}\) – \(\sqrt{48}\) = \(\sqrt{25×3}\) – \(\sqrt{16×3}\) = (5 – 4) \(\sqrt{3}\) = \(\sqrt{3}\)
(b) \(\sqrt{98}\) – \(\sqrt{72}\) = \(\sqrt{49×2}\) – \(\sqrt{36×2}\) = (7 – 6) \(\sqrt{2}\) = \(\sqrt{2}\)

(iii) Yes we can get a surd.
Example:
(a) \(\sqrt{8}\) × \(\sqrt{6}\) = \(\sqrt{8×6}\) = \(\sqrt{48}\)
(b) \(\sqrt{11}\) × \(\sqrt{3}\) = \(\sqrt{11×3}\) = \(\sqrt{33}\)

(iv) Yes we can get a surd.
Example:
(a) \(\sqrt{55}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{11×5}}{\sqrt{5}} = \sqrt{11}\)
(b) \(\sqrt{65}\) ÷ \(\sqrt{5}\) = \(\frac{\sqrt{13×5}}{\sqrt{13}} = \sqrt{5}\)

Question 6.
Can you get a rational number when you compute:
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes, the sum of two surds will give a rational number.
Example:
(a) (2 + \(\sqrt{3}\)) + (2 – \(\sqrt{3}\)) = 4
(b) (\(\sqrt{5}\) + 4) + (7 – \(\sqrt{5}\)) = 11

(ii) Yes, the difference of two surds will give a rational number.
Example:
(a) (5 + \(\sqrt{7}\)) – (- 5 + \(\sqrt{7}\)) = 10
(b) (\(\sqrt{11}\) + 5) – (-3 + \(\sqrt{11}\)) = 8

(iii) Yes, the product of two surds will give a rational number.
Example:
(a) \(\sqrt{125}\) × \(\sqrt{45}\) = \(\sqrt{25×5}\) × \(\sqrt{9×5}\) = 5\(\sqrt{5}\) × 3\(\sqrt{5}\) = 15 × 5 = 75
(b) \(\sqrt{150}\) × \(\sqrt{6}\) = \(\sqrt{25×6}\) × \(\sqrt{6}\) = 5\(\sqrt{6}\) × \(\sqrt{6}\) = 5 × 6 = 30

(iv) Yes. The quotient of two surds will give a rational number.
Example:
(a) \(\sqrt{32}\) ÷ \(\sqrt{8}\) = \(\frac{\sqrt{8×4}}{\sqrt{8}} = \sqrt{4}\) = 2
(b) \(\sqrt{50}\) ÷ \(\sqrt{2}\) = \(\frac{\sqrt{25×2}}{\sqrt{2}} = \sqrt{25}\) = 5

Students can download Maths Chapter 2 Real Numbers Ex 2.9 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 2 Real Numbers Ex 2.9

Question 1.
If n is a natural number then √n is……….
(a) always a natural number
(b) always an irrational number
(c) always a rational number
(d) may be rational or irrational
Solution:
(d) may be rational or irrational

Question 2.
Which of the following is not true?
(a) Every rational number is a real number
(b) Every integer is a rational number
(c) Every real number is an irrational number
(d) Every natural number is a whole number
Solution:
(c) Every real number is an irrational number

Question 3.
Which one of the following, regarding sum of two irrational numbers, is true?
(a) always an irrational number
(b) may be a rational or irrational number
(c) always a rational number
(d) always an integer
Solution:
(b) may be a rational or irrational number

Question 4.
Which one of the following has a terminating decimal expansion?
(a) \(\frac{5}{64}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{14}{15}\)
(d) \(\frac{1}{12}\)
Solution:
(a) \(\frac{5}{64}\)
Hint:
\(\frac{5}{64}\) = \(\frac{5}{2^{6}}\)

Question 5.
Which one of the following is an irrational number?
(a) \(\sqrt{25}\)
(b) \(\sqrt{\frac{9}{4}}\)
(c) \(\frac{7}{11}\)
(d) π
Solution:
(d) π
Hint:
We take frequently π as \(\frac{22}{7}\) (which gives the value of 3.1428571428571…….) to be its correct value, but in reality these are only approximations

Question 6.
An irrational number between 2 and 2.5 is………
(a) \(\sqrt{11}\)
(b) √5
(c) \(\sqrt{2.5}\)
(d) √8
Solution:
(b) √5
Hint:
√5 = 2.236, it lies between 2 and 2.5

Question 7.
The smallest rational number by which \(\frac{1}{3}\) should be multiplied so that its decimal expansion terminates with one place of decimal is ………
(a) \(\frac{1}{10}\)
(b) \(\frac{3}{10}\)
(c) 3
(d) 30
Solution:
(b) \(\frac{3}{10}\)
Hint:
\(\frac{1}{3}\) × \(\frac{3}{10}\) = \(\frac{1}{10}\) = \(\frac{1}{2×5}\) it has terminating decimal expansion.

Question 8.
If \(\frac{1}{7}\) = 0.\(\overline { 142857 }\) then the value of \(\frac{5}{7}\) is ………..
(a) 0.\(\overline { 142857 }\)
(b) 0.\(\overline { 714285 }\)
(c) 0.\(\overline { 571428 }\)
(d) 0.714285
Solution:
(b) 0.\(\overline { 714285 }\)

Question 9.
Find the odd one out of the following.
(a) \(\sqrt{32}×\sqrt{2}\)
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
(c) \(\sqrt{72}×\sqrt{8}\)
(d) \(\frac{\sqrt{54}}{\sqrt{18}}\)
Solution:
(b) \(\frac{\sqrt{27}}{\sqrt{3}}\)
Hint:
\(\frac{\sqrt{27}}{\sqrt{3}}\) = \(\frac{\sqrt{27}}{\sqrt{3}}\) = √9 = 3. It is an odd number

Question 10.
0.\(\overline { 34 }\) + 0.3\(\overline { 4 }\) = ……….
(a) 0.6\(\overline { 87 }\)
(b) 0.\(\overline { 68 }\)
(c) 0.6\(\overline { 8 }\)
(d) 0.68\(\overline { 7 }\)
Solution:
(a) 0.6\(\overline { 87 }\)
Hint:
0.34343434
0.34444444
0.68787878

Question 11.
Which of the following statement is false?
(a) The square root of 25 is 5 or -5
(b) \(-\sqrt{25}\) = -5
(c) \(\sqrt{25}\) = 5
(d) \(\sqrt{25}\) = ±5
Solution:
(d) \(\sqrt{25}\) = ±5

Question 12.
Which one of the following is not a rational number?
(a) \(\sqrt{\frac{8}{18}}\)
(b) \(\frac{7}{3}\)
(c) \(\sqrt{0.01}\)
(d) \(\sqrt{13}\)
Solution:
(d) \(\sqrt{13}\)

Question 13.
\(\sqrt{27}\) + \(\sqrt{12}\) = ……….
(a) \(\sqrt{39}\)
(b) 5√6
(c) 5√3
(d) 3√5
Solution:
(d) 3√5
Hint:
\(\sqrt{27}\) + \(\sqrt{12}\) = \(\sqrt{9×3}\) + \(\sqrt{3×4}\) = 3√3 + 2√3 = 5√3

Question 14.
If \(\sqrt{80}\) = k√5, then k = ………
(a) 2
(b) 4
(c) 8
(d) 16
Solution:
(b) 4
Hint:
\(\sqrt{80}\) = k√5
\(\sqrt{16×5}\) = k√5
4√5 = k√5
∴ k = 4

Question 15.
4√7 × 2√3 = ……….
(a) 6\(\sqrt{10}\)
(b) 8\(\sqrt{21}\)
(c) 8\(\sqrt{10}\)
(d) 6\(\sqrt{21}\)
Solution:
(b) 8\(\sqrt{21}\)
Hint:
4√7 × 2√3 = 4 × 2\(\sqrt{7×3}\) = 8\(\sqrt{21}\)

Question 16.
When written with a rational denominator, the expression \(\frac {2\sqrt{3}}{3\sqrt{2}}\) can be simplified as……..
(a) \(\frac {\sqrt{2}}{3}\)
(b) \(\frac {\sqrt{3}}{2}\)
(c) \(\frac {\sqrt{6}}{3}\)
(d) \(\frac {2}{3}\)
Solution:
(c) \(\frac {\sqrt{6}}{3}\)

Question 17.
When (2√5 – √2)² is simplified, we get………
(a) 4√5 + 2√2
(b) 22 – 4\(\sqrt{10}\)
(c) 8 – 4\(\sqrt{10}\)
(d) 2\(\sqrt{10}\) – 2
Solution:
(b) 22 – 4\(\sqrt{10}\)
Hint:
(2√5 – √2)² = (2√5)² + (√2)² – 2 × 2√5 × √2
= 20 – 4\(\sqrt{10}\) + 2
= 22 – 4\(\sqrt{10}\)

Question 18.
\((0.000729)^\frac{-3}{4}\) × \((0.09)^\frac{-3}{4}\) = ……..
(a) \(\frac {10^3}{3^3}\)
(b) \(\frac {10^5}{3^5}\)
(c) \(\frac {10^2}{3^2}\)
(d) \(\frac {10^6}{3^6}\)
Solution:
(d) \(\frac {10^6}{3^6}\)
Hint:

Question 19.
If √9^x = \(\sqrt[3]{9^2}\), then x = …….
(a) \(\frac {2}{3}\)
(b) \(\frac {4}{3}\)
(c) \(\frac {1}{3}\)
(d) \(\frac {5}{3}\)
Solution:
(b) \(\frac {4}{3}\)
Hint:

Question 20.
The length and breadth of a rectangular plot are 5 × 10⁵ and 4 × 10⁴ metres respectively. Its area is ……….
(a) 9 × 10¹ m²
(b) 9 × 10⁹ m²
(c) 2 × 10¹⁰ m²
(d) 20 × 10²⁰ m²
Solution:
(c) 2 × 10¹⁰ m²
Hint:
Area of a rectangle = l × b = 5 × 10⁵ × 4 × 10⁴
= 5 × 4 × 10⁵⁺⁴
= 20 × 10⁹
= 2.0 × 10 × 10⁹
= 2 × 10¹⁰ m²

Students can download Maths Chapter 4 Geometry 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 4 Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
The angle sum of a convex polygon with number of sides 7 is ………
(a) 900°
(b) 1080°
(c) 1444°
(d) 720°
Solution:
(a) 900°

Question 2.
What is the name of a regular polygon of six sides?
(a) Square
(b) Equilateral triangle
(c) Regular hexagon
(d) Regular octagon
Solution:
(c) Regular hexagon

Question 3.
One angle of a parallelogram is a right angle. The name of the quadrilateral is ………
(a) square
(b) rectangle
(c) rhombus
(d) kite
Solution:
(b) rectangle

Question 4.
If all the four sides of a parallelogram are equal and the adjacent angles are of 120° and 60°, then the name of the quadrilateral is ………
(a) rectangle
(b) square
(c) rhombus
(d) kite
Solution:
(c) rhombus

Question 5.
In a parallelogram ∠A : ∠B = 1 : 2. Then ∠A ………
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
(b) 60°

Question 6.
Which of the following is a formula to find the sum of interior angles of a quadrilateral of w-sides?
(a) \(\frac{n}{2}\) × 180°
(b) (\(\frac{n+1}{2}\)) 180°
(c) (\(\frac{n-1}{2}\)) 180°
(d) (n – 2) 180°
Solution:
(d) (n – 2) 180°

Question 7.
Diagonal of which of the following quadrilaterals do not bisect it into two congruent triangles?
(a) rhombus
(b) trapezium
(c) square
(d) rectangle
Solution:
(b) trapezium

Question 8.
The point of concurrency of the medians of a triangle is known as ………
(a) circumcentre
(b) incentre
(c) orthocentre
(d) centroid
Solution:
(d) centroid

Question 9.
Orthocentre of a triangle is the point of concurrency of ………
(a) medians
(b) altitudes
(c) angle bisectors
(d) perpendicular bisectors of side
Solution:
(b) altitudes

Question 10.
ABCD is a parallelogram as shown. Find x and y.

(a) 1, 7
(b) 2, 6
(c) 3, 5
(d) 4, 4
Solution:
(c) 3, 5

Question 11.
A circle divides the plane into part ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 12.
The longest chord of a circle is a of the circle……….
(a) radius
(b) diameter
(c) chord
(d) secant
Solution:
(b) diameter

Question 13.
Opposite angles of a cyclic quadrilateral are ………
(a) supplementary
(b) complementary
(c) equal
(d) none of these
Solution:
(a) supplementary

Question 14.
The value of x from figure is if ‘O’ is the centre of the circle ………

(a) 20 cm
(b) 15 cm
(c) 12 cm
(d) 5 cm
Solution:
(d) 5 cm

Question 15.
If PQ = x and ‘O’ is the centre of the circle, then x= ………

(a) 7 cm
(b) 14 cm
(c) 8 cm
(d) 13 cm
Solution:
(b) 14 cm

Question 16.
In figure OM = ON = 8cm and AB = 30 cm, then CD = ………

(a) 15 cm
(b) 30 cm
(c) 40 cm
(d) 10 cm
Solution:
(b) 30 cm

Question 17.
O is the centre of a circle, ∠AOB = 100°. Then angle ∠ ACB = ………

(a) 80°
(b) 40°
(c) 50°
(d) 60°
Solution:
(c) 50°

Question 18.
In,a circle, with center O, ∠AOB = 20°, ∠BOC = 40°, arc BC = 4 Then length of arc AB will be ………

(a) 8 cm
(b) 6 cm
(c) 2 cm
(d) 1 cm
Solution:
(c) 2 cm

Question 19.
In the figure , OC = 3cm and radius of circle is 5 cm. Then AB = ………

(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
(d) 8 cm

Question 20.
O is the centre of the circle. The value of x in the given diagram is ………

(a) 100°
(b) 160°
(c) 200°
(d) 80°
Solution:
(d) 80°

II. Answer the following questions.

Question 1.
In the figure find x° and y°.
Solution:
∠ACD = ∠A + ∠B
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x°
x° = 120° – 50°
= 70°
In the triangle ABC

∠A + ∠B + ∠ACB = 180° (Sum of the angles of a Δ)
50° + x + ∠ACB = 180°
50° + 70° + ∠ACB = 180°
∠ACB = 180° – 120°
y = 60°
(OR)
∠ACD + ∠ACB = 180° (Angles of a linear pair)
∠ACB = 180° – 120°
= 60°
The value of x = 70° and y = 60°.

Question 2.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Solution:
Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360°}{30}\)
= 12°
3x = 3 × 12 = 36°
5x = 5 × 12 = 60°
9x = 9 × 12 = 108°
13x = 13 × 12 = 156°
The required angles of quadrilateral are 36°, 60°, 108° and 156°.

Question 3.
Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also
(ii) ABCD is a rhombus.
Solution:
We have a parallelogram ABCD in which diagonals AC bisect ∠A.
∠DAC = ∠BAC
(i) To prove that AC bisects ∠C
∴ ABCD is a parallelogram

∴ AB || DC and AC is a transversal
∴ ∠1 = ∠3 (Alternate interior angle) …….(1)
Also BC || AD and AC is a transversal
∴ ∠2 = ∠4 (Alternate interior angle) …….(2)
But AC bisects ∠A
∴ ∠1 = ∠2 ……… (3)
From (1), (2) and (3) we get
∠3 = ∠4
∴ AC bisects ∠C.

(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ∠1 = ∠4 [∴ ∠ 1 = ∠2 = ∠4]
∴ BC = AB (side opposite to equal angles are equal) ……..(4)
Similarly AD = DC …….. (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram) ………(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertex A and C on diagonal BD.
Show that

(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ.
Solution:
(i) In ΔAPB and ΔCQD we have
∠APB = ∠CQD (90° each)
AB = CD (opposite sides of parallelogram ABCD)
∠ABP = ∠CDQ (AB || CD and AD is a transversal)
Using ASA congruency we have,
ΔAPB ≅ ΔCQD

(ii) Since ΔAPB ≅ ΔCQD
∴ Their corresponding parts are equal.
∴ AP = CQ.

Question 5.
ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:
In rectangle ABCD, P is the mid-point of AB.
Q is the mid-point of BC. R is the mid-point of CD.
S is the mid-point of DA. AC is the diagonal.
Now in ΔABC,
PQ = \(\frac{1}{2}\) AC and PQ || AC ……. (1)
Similarly in ΔACD,
SR = \(\frac{1}{2}\) AC and SR || AC ……. (2)
From (1) and (2) we get,
PQ = SR and PQ || SR
Similarly by joining BD, we have
PS = QR and PS || QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.

Question 6.
In the figure A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:

O is the centre of the circle. ∠AOB = 60° and ∠BOC = 30°
Sum of all the angles of quadrilateral = 360°.
∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 30° + 60°
= 90°
∠ADC = \(\frac{1}{2}\) × 90°
= 45°

Question 7.
In the given figure A, B, C and D are four points on a circle, AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ΔCDE,

Exterior ∠BEC = Sum of interior opposite angle
130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
130° – 20° = ∠EDC
110° = ∠EDC
∴ ∠BAC = ∠BDC = 110° (Both the triangles are standing on the same base)
∠BAC = 110°

Question 8.
In the given figure KLMN is a cyclic quadrilateral. KD is the tangent at K. If ∠N is a diameter ∠NLK = 40° and ∠LNM = 50°. Find ∠MLN and ∠DKL.

Solution:
LN is a diameter
∠LMN = ∠LKN = 90° (Angle in a semi-circle)
∴ ∠MLN = 90° – 50°
= 40°
∠LNK = 90° – 40°
= 50°
∠DKL = ∠LKN = 50° (angles in the alternate segment)
∴ ∠DKL = 50°
∠MLN = 40° and ∠DKL = 50°

Question 9.
In the given figure ∠PQR = 100°, where P, Q and R are points on a circle with centre ‘O’. Find ∠OPR.

Solution:
∠PQR is the angle subtended by the chord PR in the minor segment.
reflex ∠POR = 2∠PQR
= 2 × 100°
= 200°
Now ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
= 160°
From the given diagram, POR is an isosceles triangle (∴ PO = OR = radii)
∴ ∠OPR = ∠ORP (angles opposite to equal sides)
In ΔOPR,
∠OPR + ∠ORP + ∠POR = 180°
∠OPR + ∠OPR + 160° = 180°
2∠OPR = 180° – 160°
2∠OPR = 20°
∠OPR = \(\frac{20°}{2}\)
= 10°
∴ ∠OPR = 10°

Question 10.
AB and CD are two parallel chords of a circle which are on either sides of the centre such that AB = 10 cm and CD = 24 cm. Find the radius if the distance between AB and CD is 17 cm.
Solution:

Given AB = 10 cm, CD = 24 cm and PQ = 17 cm.
PC = PD = \(\frac{24}{2}\)
= 12 cm
AQ = QB = \(\frac{10}{2}\)
= 5 cm
Let OP be x; OQ = (17 – x)
In the ΔOPC,
OC² = OP² + PC²
= x² + 12²
In the ΔOAQ,
OA² = AQ² + QO²
= 5² + (17 – x)²
= 25 + 289 + x² – 34x
= 314 + x² – 34x
But OA² = OC²
314 + x² – 34x = x² + 144
-34x = 144 – 314
-34x = -170
34x = 170
x = \(\frac{170}{34}\)
= \(\frac{10}{2}\)
= 5
We know,
OC² = x²+ 144
= 5² + 144
= 25 + 144
OC² = 169
But OC = \(\sqrt{169}\)
= 13
Radius of the circle = 13 cm
= x² + 144