Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 1.
Find the value of the following :
(i) sin 49°
(ii) cos 74° 39′
(iii) tan 54° 26′
(iv) sin 21° 21′
(v) cos 33° 53′
(vi) tan 70° 17′
Solution:
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 1
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 2
Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 2.
Find the value of θ if
(i) sin θ = 0.9975
(ii) cos θ = 0.6763
(iii) tan θ = 0.0720
(iv) cos θ = 0.0410
(v) tan θ = 7.5958
Solution:
(i) From the natural sines table
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 3
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 4

Question 3.
Find the value of the following :
(i) sin 65° 39′ + cos 24° 57′
(ii) tan 70° 58′ + cos 15° 26′ – sin 84° 59′
Solution:
(i) = 0.9111 + 0.9066 + 0.1793 = 1.9970
(ii) = 2.8982 + 0.9639 – 0.9962 = 3.8621 – 0.9962 = 2.8659

Question 4.
Find the area of a right triangle whose hypotenuse is 10cm and one of the acute angle is 24° 24′
Solution:
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 5

Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 5.
Find the angle made by a ladder of length 5m with the ground, if one of its end is
4m away from the wall and the other end is on the wall.
Solution:
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 6
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 50

Question 6.
In the given figure, HT shows the height of a tree standing vertically. From a point P, the angle of elevation of the top of the tree (that is ∠P) measures 42° and the distance to the tree is 60 metres. Find the height of the tree.
Solution:
Samacheer Kalvi 9th Maths Chapter 6 Trigonometry Ex 6.4 7

Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Multiple Choice Questions :

Question 1.
If the y-coordinate of a point is zero, then the point always lies
(1) in the I quadrant
(2) in the II quadrant
(3) on x-axis
(4) on y-axis
Solution:
(3) on x-axis

Question 2.
The points (-5, 2) and (2, -5) lie in the ___.
(1) same quadrant
(2) II and III quadrant respectively
(3) II and IV quadrant respectively
(4) IV and II quadrant respectively
Hint: (-, +) lies IInd quadrant and (+, -) lies in IVth quadrant
Solution:
(3) II and IV quadrant respectively

Question 3.
On plotting the points 0(0, 0), A(3, – 4), B(3, 4) and C(0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(1) Square
(2) Rectangle
(3) Trapezium
(4) Rhombus
Hint: In trapezium one pair of opposite side is parallel.
Solution:
(3) Trapezium

Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Question 4.
If P(-1, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ____.
(1) P and T
(2) Q and R
(3) only S
(4) P and Q
Hint: Points in IVth quadrant are (+, -)
Solution:
(2) Q and R

Question 5.
The point whose ordinate is 4 and which lies on they-axis is
(1) (4, 0)
(2) (0, 4)
(3) (1, 4)
(4) (4, 2)
Hint: Points in y-axis have abscissa a zero
Solution:
(2) (0, 4)

Question 6.
The distance between the two points (2, 3) and (1, 4) is ___ .
(1) 2
(2) \(\sqrt{56}\)
(3) \(\sqrt{10}\)
(4) \(\sqrt{2}\)
Hint: Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
Solution:
(4) \(\sqrt{2}\)

Question 7.
If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is ___.
(1) 0
(2) 2
(3) 3
(4) -6
Hint: Points in y-axis have ordinate zero. ∴ a – 3 = 0 ⇒ a = 3
Solution:
(3) 3

Question 8.
If ( x + 2, 4) = (5, y – 2), then the coordinates (x, y) are ___.
(1) (7, 12)
(2) (6, 3)
(3) (3, 6)
(4) (2, 1)
Hint: x + 2 = 5 ⇒ x = 3; y – 2 = 4 ⇒ y = 6
Solution:
(3) (3, 6)

Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.6

Question 9.
If Q1, Q2, Q3, Q4 are the quadrants in a Cartesian plane then Q2 ∩ Q3 is
(1) Q1 ∪ Q2
(2) Q2 ∪ Q3
(3) Null set
(4) Negative x-axis
Hint: Quadrants do not contain the axis.
Solution:
(3) Null set

Question 10.
The distance between the point (5, -1) and the origin is
(1) \(\sqrt{24}\)
(2) \(\sqrt{37}\)
(3) \(\sqrt{26}\)
(4) \(\sqrt{17}\)
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 1
Solution:
(3) \(\sqrt{26}\)

Question 11.
The coordinates of the point C dividing the line segment joining the points P(2, 4) and Q(5, 7) internally in the ratio 2 : 1 is
(1) \(\left(\frac{7}{2}, \frac{11}{2}\right)\)
(2) (3, 5)
(3) (4, 4)
(4) (4, 6)
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 51

Question 12.
If P\(\left(\frac{a}{3}, \frac{b}{2}\right)\) is the mid-point of the line segment joining A(-4, 3) and B(-2, 4) then (a, b)
(1) (-9, 7)
(2) \(\left(-3, \frac{7}{2}\right)\)
(3) (9, -7)
(4) \(\left(3-\frac{7}{2}\right)\)
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 52
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 53
Solution:
(1) (-9, 7)

Question 13.
In what ratio does the point Q(1, 6) divide the line segment joining the points P(2, 7) and R(-2, 3)
(1) 1 : 2
(2) 2 : 1
(3) 1 : 3
(4) 3 : 1.
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 54

Question 14.
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its center is (-3, 2), then the coordinate of the other end of the diameter is
(1) (0, -3)
(2) (0, 9)
(3) (3, 0)
(4) (-9, 0)
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 55
Solution:
(4) (-9, 0)

Question 15.
The ratio in which the x-axis divides the line segment joining the points A(a1, b1) and B(a2, b2) is
(1) b1 : b2
(2 ) -b1 : b2
(3 ) a1 : a2
(4 ) -a1 : a2
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 56
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 57
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 58
Solution:
(2 ) -b1 : b2

Question 16.
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is
(1) 2 : 3
(2) 3 : 4
(3) 4 : 7
(4) 4 : 3
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 59
Solution:
(3) 4 : 7

Question 17.
If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are
(1) (3, 2), (2, 4)
(2) (4, 0), (2, 8)
(3) (3, 4), (2, 0)
(4) (4, 3), (2, 4)
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 60
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 61
0 + y2 = 8 ⇒ y2 = 8
(x1, y1) = (4, 0) ; (x2, y2) = (2, 8)
Solution:
(2) (4, 0), (2, 8)

Question 18.
The mid-point of the line joining (-a, 2b) and (-3a, -4b) is
(1) (2a, 3b)
(2) (-2a, -b)
(3) (2a, b)
(4) (-2a, -3b)
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 62
Solution:
(2) (-2a, -b)

Question 19.
In what ratio does the j-axis divides the line joining the points (-5, 1) and (2, 3) internally
(1) 1 : 3
(2) 2 : 5
(3) 3 : 1
(4) 5 : 2
Hint:
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 64
Solution:
(4) 5 : 2

Question 20.
If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is
(1) 6
(2) 5
(3) 4
(4) 3
Hint:
In parallelogram diagonals bisect each other.
Mid point of AC = Mid point of BD
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 66
Samacheer Kalvi 9th Maths Chapter 5 Coordinate Geometry Ex 5.6 67
Solution:
(2) 5

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

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Exercise 2.1

Question 1.
Find only two rational numbers between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 4 }\).
Solution:
A rational number between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 4 }\) = \(\frac { 1 }{ 2 }\) ( \(\frac { 1 }{ 4 }\) + \(\frac { 3 }{ 4 }\)) = \(\frac { 1 }{ 2 }\) (1) = \(\frac { 1 }{ 2 }\)
Another rational number between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) = \(\frac { 1 }{ 2 }\) ( \(\frac { 1 }{ 2 }\) + \(\frac { 3 }{ 4 }\)) = \(\frac { 1 }{ 2 }\) ( \(\frac { 2+3 }{ 4 }\) = \(\frac { 31}{ 2 }\) × \(\frac { 5 }{ 4 }\)) = \(\frac { 5 }{ 8 }\)
The rational numbers \(\frac { 1 }{ 2 }\) and \(\frac { 5 }{ 8 }\) lies between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 2 }\) .

Question 2.
Is zero a rational numbers? Give reasons for your answer.
Solution:
Yes, since \(\frac { 0 }{ 2 }\) = 0, (i.e) it can be written in the form \(\frac { p }{ q }\) where q ≠ 0

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

Exercise 2.2

Question 1.
Express the following decimal expansion is the form \(\frac { p }{ q }\) , where p and q are integers and q ≠ 0.
(i) 0.75
(ii) 0.625
(iii) 0.5625
(iv) 0.28
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 1

Question 2.
Convert \(\overline { 0.9 }\) to a rational number.
Solution:
(i) Let x = \(0.\overline { 9 }\). Then x = 0.99999….
Multiplying by 10 on both sides, we get
10x = 9.99999….. = 9 + 0.9999….. = 9 + x
9x = 9
x = 1. That is, \(0.\overline { 9 }\) = 1 (∵ 1 is rational number).

Exercise 2.3

Question 1.
Classify the following number as rational or irrational.
(i) \(\sqrt { 11 }\)
(ii) \(\sqrt { 81 }\)
(iii) 0.0625
(iv) \(0.8\overline { 3 }\)
Solution:
(i) \(\sqrt { 11 }\) is an irrational number. (11 is not a perfect square number)
(ii) \(\sqrt { 81 }\) = 9 = \(\frac { 9 }{ 1 }\) , a rational number.
(iii) 0.0625 is a terminating decimal
∴ 0. 0625 is a rational number.
(iv) \(0.8\overline { 3 }\) = 0.8333
The decimal expansion is non-terminating and recurring.
∴ \(0.8\overline { 3 }\) is a rational number.

Question 2.
Find the decimal expansion of \(\sqrt { 3 }\).
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 2

Question 3.
Find any 4 irrational numbers between \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 3 }\).
Solution:
\(\frac { 1 }{ 4 }\) = 0.25 and \(\frac { 1 }{ 3 }\) = 0.3333 = \(0.\overline { 3 }\)
In between 0.25 and \(0.\overline { 3 }\) there are infinitely many irrational numbers .
Four irrational numbers between 0.25 and \(0.\overline { 3 }\) are
0.2601001000100001 ……
0.2701001000100001 ……
0.2801001000100001 …..
0.3101001000100001 ……

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

Exercise 2.4

Question 1.
Visualise \(6.7\overline { 3 }\) on the number line, upto 4 decimal places.
Solution:
We locate 6.73 on the number line, by the process of successive magnification.
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 3
Step 1 : First we note that \(6.7\overline { 3 }\) lies between 6 and 7.
Step 2 : Divide the portion between 6 and 7 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.7 and 6.8.
Step 3 : Divide the portion between 6.7 and 6.8 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.73 and 6.74.
Step 4 : Divide the portion between 6.73 and 6.74 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.733 and 6.734.
Step 5 : Divide the portion between 6.733 and 6.734 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.7332 and 6.7334.
We note that \(6.7\overline { 3 }\) is visualised closed to 6.7332 than to 6.7334.

Question 2.
Find whether x and y are rational or irrational in the following:
(i) a = 2 + \(\sqrt{3}\), b = 2 – \(\sqrt{3}\); x = a + b, y = a – b
(ii) a = \(\sqrt{2}\) + 7, b = x = a + b, y = a – b
Solution:
(i) Given that a = 2 + \(\sqrt{3}\), b = 2 – \(\sqrt{3}\)
x = a + b = (2+ \(\sqrt{3}\)) +(2 – \(\sqrt{3}\)) = 4, a rational number.
y = a – b = {2 + \(\sqrt{3}\)) – (2 – \(\sqrt{3}\)) = 2\(\sqrt{3}\) , an irrational number.

(ii) Given that a = \(\sqrt{2}\) + 7,b = \(\sqrt{2}\) – 7
x = a + b = (\(\sqrt{2}\) + 7)+ (\(\sqrt{2}\) – 7) = 2\(\sqrt{2}\), an irrational number.
y = a – b = (\(\sqrt{2}\) + 7 ) – (\(\sqrt{2}\) – 7) = 14, a rational number.

Exercise 2.5

Question 1.
Evaluate :
(i) 10-4
(ii) (\(\frac { 1 }{ 9 }\))-3
(iii) (0.01)-2
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 4

Question 2.
Find the value of 625\(\frac { 3 }{ 4 }\) :
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 5

Question 3.
Find the value of 729\(\frac { -5 }{ 6 }\) :
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 6

Question 4.
Use a fractional index to write :
(i) (5\(\sqrt { 125 }\))7
(ii) \(\sqrt [ 3 ]{ 7 }\)
Solution:
(i) (5\(\sqrt { 125 }\))7 = 125\(\frac { 7 }{ 5 }\)
(ii) \(\sqrt [ 3 ]{ 7 }\) = 7\(\frac { 1 }{ 3 }\)

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

Exercise 2.6

Question 1.
Can you reduce the following numbers to surds of same order.
(i) \(\sqrt{ 5 }\)
(ii) \(\sqrt [ 3 ]{ 5 }\)
(iii) \(\sqrt [ 4 ]{ 5 }\)
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 7
Now the surds have same order

Question 2.
Express the following surds in its simplest form
(i) \(\sqrt { 27 }\)
(ii) \(\sqrt [ 3 ]{ 128 }\)
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 8

Question 3.
Show that \(\sqrt [ 3 ]{ 2 }\) > \(\sqrt [ 5 ]{ 3 }\).
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 9

Question 4.
Express the following surds in its simplest form \(\sqrt [ 4 ]{ 324 }\).
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 10
order = 4 ; radicand = 4; Coefficient = 3

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

Question 5.
Simplify \(\sqrt { 63 }\) – \(\sqrt { 175 }\) + \(\sqrt { 28 }\)
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 11

Question 6.
Arrange in ascending order: \(\sqrt [ 3 ]{ 2 }\), \(\sqrt [ 2 ]{ 4 }\), \(\sqrt [ 4 ]{ 3 }\)
Solution:
The order of the surds \(\sqrt [ 3 ]{ 2 }\), \(\sqrt [ 2 ]{ 4 }\), \(\sqrt [ 4 ]{ 3 }\) are 3, 2, 4
L.CM. of 3, 2, 4 = 12.
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 12

Exercise 2.7

Question 1.
Subtract 6\(\sqrt { 7 }\) from 9\(\sqrt { 7 }\). Is the answer rational or irrational?
Solution:
9\(\sqrt { 7 }\) – 6\(\sqrt { 7 }\) = (9 – 6) \(\sqrt { 7 }\) = 3\(\sqrt { 7 }\)
The answer is irrational.

Question 2.
Simplify,: \(\sqrt { 44 }\) + \(\sqrt { 99 }\) – \(\sqrt { 275 }\).
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 13

Question 3.
Compute and give the answer in the simplest form : 3 \(\sqrt { 162 }\) x 7 \(\sqrt { 50 }\) x 6 \(\sqrt { 98 }\)
Solution:
3 \(\sqrt { 162 }\) × 7 \(\sqrt { 50 }\) × 6 \(\sqrt { 98 }\) = \((3 \times 9 \sqrt{2} \times 7 \times 5 \sqrt{2} \times 6 \times 7 \sqrt{2})\)
= \(3 \times 7 \times 6 \times 9 \times 5 \times 7 \times \sqrt{2} \times \sqrt{2} \times \sqrt{2}=79380 \sqrt{2}\)

Exercise 2.8

Question 1.
Write in scientific notation : (60000000)3
Solution:
(60000000)3 = (6.0 × 107)4 = (6.0)4 × (107)4
= 1296 × 1028
= 1.296 × 103 × 1028 = 1.296 × 1031

Question 2.
Write in scientific notation : (0.00000004)3
Solution:
(0.00000004)3 = (4.0 × 10-8)3 = (4.0)3 × (10-8)3
= 64 × 10-24 = 6.4 × 10 × 10-24 = 6.4 × 10-23

Question 3.
Write in scientific notation : (500000)5 × (3000)3
Solution:
(500000)5 × (3000)3 = (5.0 × 105)3 × (3.0 × 103)3
= (5.0)2 × (105)2 × (3.0)3 × (103)3
= 25 × 1010 × 27 × 109 = 675 × 1019
= 675.0 × 1019 = 6.75 × 102 × 1019= 6.75 × 1021

Question 4.
Write in scientific notation : (6000000)3 ÷ (0.00003)2
Solution:
(6000000)3 + (0.00003)2 = (6.0 × 106)3 + (3.0 × 10-5)2
= (6.0 × 106)3 ÷ (3.0 × 10-5)2 = 216 × 1018 ÷ 9 × 10-10
= \(\frac{216 \times 10^{9}}{9 \times 10^{-10}}\)
= 24 × 1018 × 1010 = 24 × 1028
= 24.0 × 1028 = 2.4 × 10 × 1028 = 2.4 × 1029

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

Exercise 2.9

Multiple Choice Questions :
Question 1.
A number having non-terminating and recurring decimal expansion is
(1) an integer
(2) a rational number
(3) an irrational number
(4) a whole number
Solution:
(2) a rational number
Hint:
Irrational number have nonterminating and non recurring decimal expansion.

Question 2.
If a number has a non-terminating and non-recurring decimal expansion, then it is
(1) a rational number
(2) a natural number
(3) an irrational number
(4) an integer
Solution:
(3) an irrational number
Hint: Rational number gave terminating or recurring and non-terminating decimal expansion.

Question 3.
Decimal form of \(\frac { -3 }{ 4 }\) is
(1) -0.75
(2) -0.50
(3) -0.25
(4) -0.125
Solution:
(1) -0.75
Hint:
\(\frac { 1 }{ 4 }\) = 0.25; \(\frac {1 }{ 2 }\) = 0.5; \(\frac { 3 }{ 4 }\) = 0.75

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

Question 4.
Which one of the following has a terminating decimal expansion?
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 14
Solution:
(1) \(\frac { 5 }{ 32 }\)
Hint:
32 = 25 ⇒ \(\frac { 5 }{ 32 }\) has terminating decimal expansion

Question 5.
Which one of the following is an irrational number?
(1) π
(2) √9
(3) \(\frac { 1 }{ 4 }\)
(4) \(\frac { 1 }{ 5 }\)
Solution:
(1) π

Question 6.
Which one of the following are irrational numbers?
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 15
(a) (ii), (iii) and (iv)
(b) (i), (ii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
Hint:
\(\sqrt{4+\sqrt{25}}=\sqrt{9}=3 ; \sqrt{8-\sqrt[3]{8}}=\sqrt{8-2}=\sqrt{6}\)

Question 7.
Which of the following is not an irrational number?
(1) \(\sqrt {2}\)
(2) \(\sqrt {5}\)
(3) \(\sqrt {3}\)
(4) \(\sqrt {25}\)
Solution:
(4) \(\sqrt {25}\)

Question 8.
In simple form, \(\sqrt [ 3 ]{ 54 }\) is?
(1) 3 \(\sqrt [ 3 ]{ 2 }\)
(2) 3 \(\sqrt [ 3 ]{ 27 }\)
(3) 3 \(\sqrt [ 3 ]{ 2 }\)
(4) \(\sqrt { 3 }\)
Solution:
(1) 3 \(\sqrt [ 3 ]{ 2 }\)

Question 9.
\(\sqrt [ 3 ]{ 192 }\) + \(\sqrt [ 3 ]{ 24 }\)
(1) 3\(\sqrt [ 3 ]{ 6 }\)
(2) 6\(\sqrt [ 3 ]{ 3 }\)
(3) 3\(\sqrt [ 3 ]{ 216 }\)
(4) 3\(\sqrt [ 6 ]{ 216 }\)
Solution:
(2) 6\(\sqrt [ 3 ]{ 3 }\)

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

Question 10.
5\(\sqrt { 21 }\) × 6\(\sqrt { 10 }\)
(1) 30\(\sqrt { 210 }\)
(2) 30
(3) \(\sqrt { 210 }\)
(4) 210\(\sqrt { 30 }\)
Solution:
(1) 30\(\sqrt { 210 }\)

Text Book Activities

Activity – 1
Is it interesting to see this pattern? \(\sqrt{4 \frac{4}{15}}=4 \sqrt{\frac{4}{15}} \text { and } \sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}}\) Verify it. Can you frame 4 such new surds?
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 16

Activity – 2
Take a graph sheet and mark O, A, B, C as follows:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 17
Consider the following graphs:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 18
Are they equal? Discuss. Can you verify the same by taking different squares of different lengths?
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 19

Activity – 3
The following list shows the mean distance of the planets of the solar system from the Sun. Complete the following table. Then arrange in order of magnitude starting with the distance of the planet closest to the Sun.
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Additional Questions 20
Arrange the planets in order of distance from the sun.
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

Students can Download Maths Chapter 5 Information Processing Ex 5.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

I. Choose the best answer.

Question 1.
How many outcomes can you get when you toss three coins once?
(A) 6
(B) 8
(C) 3
(D) 2
Solution:
(B) 8

Question 2.
In how many ways can you answer 3 multiple choice questions, with the choices A, B, C and D ?
(1) 4
(2) 3
(3) 12
(4) 64
Solution:
(4) 64

Question 3.
How many 2 digit numbers contain the number 7 ?
(A) 10
(B) 18
(C) 19
(D) 20
Solution:
(B) 18

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

II. Answer the following.

Question 1.
You are going to have an ice cream or a cake. There are three flavours (chocolate, strawberry , vanilla) in ice creams, and two flavours (orange or red velvet) in the cake. In how many possible ways can you choose an ice cream or the cake?
Samacheer Kalvi 8th Maths Term 1 Chapter 5 Information Processing Ex 5.1 1
Solution:
We are going to have either a ice cream or a cake.
Ice cream can be selected from 3 flavors and cake from two flavors. Both the events cannot occur simultaneously selecting ice cream and cake.
∴ Number of possible ways = 3 + 2
= 5 ways

Question 2.
In how many ways, can the teacher choose 3 students, in all one each from 10 students in VI std, 15 students in VII std and 20 students in VIII std to go to an excursion?
Solution:
The teacher is going to selected one student from class VI out of 10 students in 10 ways from
class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
∴ Number of ways 3 students can be selected
= 10 + 15 + 20 = 45 ways

Question 3.
If you have 2 school bags and 3 water bottles then, in how many different ways can you carry a school bag and a water bottle, while going to school?
Solution:
We can select one school bag from 2 and one bottle from 3 as follows.
Samacheer Kalvi 8th Maths Term 1 Chapter 5 Information Processing Ex 5.1 3
∴ A bag and a water bottle can be selected in 2 × 3 = 6 ways.

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

Question 4.
Roll numbers are created with a letter followed by 3 digits in it. From the letter A, B, C, D, E and any 3 digits from 0 to 9. Then, in how many possible ways can the roll numbers be generated?
Solution:
We have a letter followed by 3 digits in the roll number.
The letter is selected from A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits 0 to 9.
Once place can be formed using any one of the 10 number 0 to 9 in 10 ways.
Tens place can be formed in 10 ways.
∴ A two digit number can be formed in 10 × 10 = 100 ways.
Thousands place can be formed in 10 ways
∴ A 3 digit number can be formed in 10 × 10 × 10 = 1000 ways.
∴ 5 letters can be attached in 5 × 1000 = 5000 ways.
∴ The roll number can be formed in 5000 ways.

Question 5.
A safety locker in a jewel shop requires a 4 digit unique code. The code has the digits from 0 to 9. How many unique codes are possible ?
Samacheer Kalvi 8th Maths Term 1 Chapter 5 Information Processing Ex 5.1 4
Solution:
The unique code has 4 digits.
Each digit is formed using any of the 10 numbers from 0 to 9.
∴ Single digit number can be formed in 10 ways.
A double digit number can be formed in 10 × 10 ways.
A three digit number can be formed in 10 × 10 × 10 ways.
A four digit number can be formed in 10 × 10 × 10 × 10 ways. = 10,000 ways

Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.3

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.3

Question 1.
Find the volume of a cuboid whose dimensions are
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length = 60 m, breadth = 25 m, height = 1.5 m
Solution:
(i) l = 12 cm
b = 8 cm
h = 6 cm
Volume of the cuboid = lbh
= 12 × 8 × 6 cm3 = 576 cm3

(ii) l = 60 m
b = 25 m
h = 1.5 m
Volume of the cuboid = lbh = 60 × 25 × 1.5m3 = 2250 m3
Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.3
Question 2.
The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such match boxes.
Solution:
Dimensions of a match box = 6 cm × 3.5 cm × 2.5 cm
V = (6 × 3.5 × 2.5) cm3 = 52.5 cm3
Volume of 12 such boxes = 12 × 52.5 cm3 = 630 cm3

Question 3.
The length, breadth and height of a chocolate box are in the ratio 5 : 4 : 3. If its volume is 7500 cm3, then find its dimensions.
Solution:
Let 1 ratio = x then 5 : 4 : 3
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Ex 7.3 1
x = 5 cm
∴ 5x = 5 × 5 = 25 cm
4x = 4 × 5 = 20 cm
3x = 3 × 5 = 15 cm
∴ Dimensions of a chocolate box are 25 cm × 20 cm × 15 cm

Question 4.
The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres.
Solution:
l = 20.5 m
b = 16 m
h = 8 m
∴ Capacity = Volume = lb h = (20.5 × 16 × 8) m3 = 2624 m3
1 m3 = 1000 litres
∴ 2624 m3 = 2624000 litres
Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.3
Question 5.
The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?
Solution:
l = 24 cm
b = 12 cm
h = 8 cm
Volume of the brick = lbh = (24 × 12 × 8) cm3 = 2304 cm3
Wall dimensions are :
L = 20 m = 2000 cm
B = 48 cm
H = 6 m = 600 cm
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Ex 7.3 20
No. of bricks required = 25000

Question 6.
The volume of container is 1440 m3. The length and breadth of the container are 15 m and 8 m respectively. Find its height.
Solution:
V = l × b × h = 1440 m3
15 × 8 × h = 1440
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Ex 7.3 30

Question 7.
Find the volume of a cube each of whose side is
(i) 5 cm
(ii) 3.5 m
(iii) 21 cm
Solution:
(i) side of a cube (a) = 5 cm
Volume of a cube V = a3 = 5 × 5 × 5 = 125 cm3

(ii) a = 3.5 m
V = a3 = 3.5 × 3.5 × 3.5 m3 = 42.875 m3

(iii) a = 21 cm 23 = 21 × 21 × 21 cm3 = 9261 cm3

Question 8.
A cubical milk tank holds 125000 litres of milk. Find the length of its side in metres.
Solution:
V = a3 = 125000 litres = 125 m3
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Ex 7.3 31
∴ length of its side = 5 m.

Question 9.
A metallic cube with side 15 cm is melted and formed into a cuboid. If the length and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the cuboid.
Solution:
The volume of the cuboid formed = The volume of the cube melted.
Samacheer Kalvi 9th Maths Chapter 7 Mensuration Ex 7.3 32
height of the cuboid = 15 cm

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Multiple Choice Questions :
Question 1.
If n is a natural number then \(\sqrt{n}\) is
(1) always a natural number
(2) always an irrational number
(3) always a rational number
(4) may be rational or irrational
Solution:
(4) may be rational or irrational

Question 2.
Which of the following is not true?
(1) Every rational number is a real number.
(2) Every integer is a rational number.
(3) Every real number is an irrational number.
(4) Every natural number is a whole number.
Solution:
(3) Every real number is an irrational number
Hint:
Real numbers contain rationals and irrationals.

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

Question 4.
Which one of the following has a terminating decimal expansion?
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 1
Solution:
(1) \(\frac { 5 }{ 64 }\)
Hint:
\(\frac { 5 }{ 64 }\) = \(\frac{5}{2^{6}}\)

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 5.
Which one of the following is an irrational number?
(1) \(\sqrt { 25 }\)
(2) \(\sqrt { \frac { 9 }{ 4 } }\)
(3) \(\frac { 7 }{ 11 }\)
(4) π
Solution:
(4) π
Hint:
π represents a irrational number

Question 6.
An irrational number between 2 and 2.5 is
(1) \(\sqrt { 11 }\)
(2) \(\sqrt { 5 }\)
(3) \(\sqrt { 2.5 }\)
(4) \(\sqrt { 8 }\)
Solution:
(2) \(\sqrt { 5 }\)
Hint:
22 = 4 and 2.52 = 6.25

Question 7.
The smallest rational number by which – should be multiplied so that its decimal expansion terminates after one place of decimal is
(1) \(\frac { 1 }{ 10 }\)
(2) \(\frac { 3 }{ 10 }\)
(3) 3
(4) 30
Solution:
(2) \(\frac { 3 }{ 10 }\)
Hint:
\(\frac { 3 }{ 10 }\) is the small number.

Question 8.
If \(\frac { 1 }{ 7 }\) = \(0.\overline { 142857 }\) then the value of \(\frac { 5 }{ 7 }\) is
(1) \(0.\overline { 142857 }\)
(2) \(0.\overline { 714285 }\)
(3) \(1.\overline { 571428 }\)
(4) 0.714285
Solution:
(2) \(0.\overline { 714285 }\)
Hint:
5 × \(\frac { 1 }{ 7 }\) = 5 × \(0.\overline { 142857 }\) = \(0.\overline { 714285 }\)

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 9.
Find the odd one out of the following.
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 2
Solution:
(4) \(\frac{\sqrt{54}}{\sqrt{18}}\)
Hint:
\(\sqrt { 72 }\) × \(\sqrt { 8 }\) = \(\sqrt { 9\times8 }\) × \(\sqrt { 8 }\) = 3 × 8 = 24

Question 10.
\(0.\overline { 34 }\) + \(0.3\overline { 4 }\) =
(1) \(0.6\overline { 87 }\)
(2) \(0.\overline { 68 }\)
(3) \(0.6\overline { 8 }\)
(4) \(0.68\overline { 7 }\)
Solution:
(1) \(0.6\overline { 87 }\)
Hint:
0.343434 … + 0.344444 … = \(0.6\overline { 87 }\)

Question 11.
Which of the following statement is false?
(1) The square root of 25 is 5 or -5
(2) \(\sqrt { 25 }\) = 5
(3) –\(\sqrt { 25 }\) = -5
(4) \(\sqrt { 25 }\)= ±5
Solution:
(4) \(\sqrt { 25 }\) = ±5

Question 12.
Which one of the following is not a rational number?
(1) \(\sqrt { \frac { 8 }{ 18 } }\)
(2) \(\frac { 7 }{ 3 }\)
(3) \(\sqrt { 0.01 }\)
(4) \(\sqrt { 13 }\)
Solution:
(4) \(\sqrt { 13 }\)
Hint:
(1) \(\sqrt { \frac { 8 }{ 18 } }\) = \(\sqrt { \frac { 4 }{ 9 } }\) = \(\frac { 2 }{ 3 }\) is a arational number
(2) \(\frac { 7 }{ 3 }\) is a rational number
(3) \(\sqrt { 0.01 }\) = \(\sqrt { \frac { 1 }{ 100 } }\) = \(\frac { 2 }{ 3 }\) is a rational number
(4) \(\sqrt { 13 }\) is a rational number

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 13.
\(\sqrt { 27 }\) + \(\sqrt { 12 }\) =
(1) \(\sqrt { 39 }\)
(2) \(5\sqrt { 6 }\)
(3) \(5\sqrt { 3 }\)
(4) \(3\sqrt { 5 }\)
Solution:
(3) \(5\sqrt { 3 }\)
Hint:
\(\sqrt { 27 }\) + \(\sqrt { 12 }\) = \(\sqrt{9 \times 3}+\sqrt{4 \times 3}=3 \sqrt{3}+2 \sqrt{3}=5 \sqrt{3}\)

Question 14.
if \(\sqrt { 80 }\) = k\(\sqrt { 5 }\), then k =
(1) 2
(2) 4
(3) 8
(4) 16
Solution:
(2) 4
Hint: \(\sqrt { 80 }\) = \(\sqrt{16 \times 5}=4 \sqrt{5}=k \sqrt{5}\) ⇒ k = 4

Question 15.
\(4 \sqrt{7} \times 2 \sqrt{3}\) =
(1) 6\(\sqrt{10}\)
(2) 8\(\sqrt{21}\)
(3) 8\(\sqrt{10}\)
(4) 6\(\sqrt{21}\)
Solution:
(2) 8\(\sqrt{21}\)
Hint:
\(4 \sqrt{7} \times 2 \sqrt{3}\) = \(8\times\sqrt{7 \times 3}\) = 8\(\sqrt{21}\)

Question 16.
When written with a rational denominator, the expression \(\frac{2 \sqrt{3}}{3 \sqrt{2}}\) can be simplified as
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 3
Solution:
(3) \(\frac{\sqrt{6}}{3}\)
Hint:
\(\frac{2 \sqrt{3}}{3 \sqrt{2}}=\frac{2 \sqrt{3}}{3 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2 \sqrt{6}}{3 \times 2}=\frac{2 \sqrt{6}}{63}\)

Question 17.
When (2\(\sqrt{5}\) – \(\sqrt{2}\))2 is simplified, we get
(1) 4\(\sqrt{5}\) + 2\(\sqrt{2}\)
(2) 22 – 4\(\sqrt{10}\)
(3) 8 – 4\(\sqrt{10}\)
(4) 2\(\sqrt{10}\) – 2
Solution:
(2) 22 – 4\(\sqrt{10}\)
Hint:
(2\(\sqrt{5}\) – \(\sqrt{2}\))2 = (2\(\sqrt{5}\))2 – 2 × 2\(\sqrt{5}\) × \(\sqrt{2}\) + \(\sqrt{2^{2}}\)
= 4 × 5 – 4\(\sqrt{10}\) + 2 = 22 – 4\(\sqrt{10}\)

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.9

Question 18.
(0.000729)\(\frac{-3}{4}\) × (0.09)\(\frac{-3}{4}\) = ____.
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 4
Solution:
(4) \(\frac{10^{6}}{3^{6}}\)
Hint :
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 5

Question 19.
If \( \sqrt{9^{x}}=\sqrt[3]{9^{2}}\) , than x = ___
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 6
Solution:
(2) \(\frac { 4 }{ 3 }\)
Hint:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.9 7

Question 20.
The length and breadth of a rectangular plot are 5 x 105 and 4 x 104 metres respectively. Its area is .
(1) 9 × 101 m2
(2) 9 × 109 m2
(3) 2 × 1010 m2
(4) 20 × 1020 m2
Solution:
(3) 2 × 1010 m2
Hint:
l = 5 × 105 metres; b = 4 × 104 metres
∴ Area = l × b = 5 x 105 × 4 × 104
= 20 × 105+4= 20 × 109= 2.0 × 101 × 109 = 2 × 1010m2

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

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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 × 1013
(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

Samacheer Kalvi 9th Maths Solutions 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 x (20000)4 = (3.0 x 105)2 x (2.0 x 104)4
= 32 × 1010 × 24 × 1016
= 9 × 16 × 1010+16
= 144 × 1026 = 1.44 × 1028

(ii) (0.000001)11 ÷ (0. 005)3
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 1
= 1 x× 10-66+9 × 53
= 125 × 10-57
= 1.25 × 102 × 10-57
= 1.25 × 10(-57+2)
= 1.25 × 10-55

(iii) {(0.00003)6 x (0.00005)4} ÷ {(0. 009)3x (0.05)2}
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 2
= 54-2 × 10-50+13
= 52 × 10-37 = 25 × 10-37 = 2.5 × 101 × 10-37 = 2.5 × 10-36

Samacheer Kalvi 9th Maths Solutions 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) The world population is nearly 7000,000,000 = 7.0 × 109
(ii) One light year means the distance 9460528400000000 km = 9.4605284 × 1015 km
(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg. = 9.1093822 × 10-31 kg

Question 5.
Simplify:
(i) (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 x 107 = 27500000
1.23 x 108 – 123000000
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 3

(ii) (1.598 × 1017) – (4.58 × 1015)
1.598 × 1017 = 159800000000000000
4.58 × 1015= 4580000000000000
(1.598 × 1017) – (4.58 × 1015)
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 4

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

(iv) (8.41 x 104) ÷ (4.3 x 105)
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 5
= 1.95581395 × 101 × 102-4
= 1.95581395 × 101 × 10-2
= 1.95581395 × 10-1
= 0. 195581395

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.7

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.7

Question 1.
Rationalise the denominator
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 1
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 2

Question 2.
Rationalise the denominator and simplify
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 3
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 4
Multiply the numerator and denominator by the rationatising factor (\(\sqrt{27}\) + \(\sqrt{18}\))
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 5
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 6
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 7

Question 3.
Find the value of a and b if \(\frac{\sqrt{7}-2}{\sqrt{7}+2}=a \sqrt{7}+b\).
Solution:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 8

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.7

Question 4.
If x = \(\sqrt{5}\) + 2 , then find the value of \(x^{2}+\frac{1}{x^{2}}\) .
Solution:
If x = \(\sqrt{5}\) + 2
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 9

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:
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.7 10

Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

Exercise 8.1

Question 1.
The following data gives the number of residents in an area based on their age. Find the average age of the residents.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 1
Solution:
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 2
Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions
Question 2.
Find the mean for the following frequency table :
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 3
Solution:
Let Assumed mean A = 170
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 4

Question 3.
Find the mean for the following distribution using step Deviation Method.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 5
Solution:
Let Assumed mean A = 28, Class width C = 8
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 6

Exercise 8.2

Question 1.
For the following up grouped data 8, 15, 14, 19, 11, 16, 10, 8, 17, 20. Find the median.
Solution:
Arrange the values in ascending order 8, 8, 10, 11, 14, 15, 16, 17, 19, 20
The number of values = 10
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 7

Question 2.
The following table gives the weekly expenditure of 200 families. Find the median of the weekly expenditure.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 8
Solution:
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 9
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 10
Question 3.
The median of the following data is 24. Find the value of x.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 11
Solution:
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 12
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 13

Question 4.
The following are the scores obtained by 11 players in a cricket match 7, 21, 45, 12, 56, 35, 25, 0, 58, 66, 29. Find the median score.
Solution:
Let us arrange the values in ascending order 0, 7, 12, 21, 25, 29, 35, 45, 56, 58, 66
The number of values = 11 which is odd.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 60

Exercise 8.3
Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions
Question 1.
Find the mode of the given data : 65, 65, 71, 71, 72, 75, 82, 72, 47, 72.
Solution:
In the given data 72 occurs thrice. Hence the mode is 72.

Question 2.
Find the mode:
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 61
Solution:
7 has the maximum frequency 21. Therefore 7 is the mode.

Question 3.
Find the mode for the following data.
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 62
Solution:
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 63

Question 4.
In a distribution, the mean and mode are 46 and 40 respectively. Calculate the median.
Solution:
Given, Mean = 46 and mode = 40
Using mode ≈ 3 median – 2 mean ,
40 ≈ 3 Median – 2 (46)
3 Median ≈ 40 + 92
Therefore, Median ≈ \(\frac{132}{3}\) = 44

Exercise 8.4
Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions
Multiple Choice Questions :

Question 1.
The mean of first 10 natural numbers.
(1) 25
(2) 55
(3) 5.5
(4) 2.5
Samacheer Kalvi 9th Maths Chapter 8 Statistics Additional Questions 71
Solution:
(3) 5.5

Question 2.
The mean of a distribution is 23, the median is 24 and the mode is 25.5. It is most likely that this distribution is :
(1) Positively skewed
(2) Symmetrical
(3) Asymptotic
(4) Negatively skewed
Hint: For Negatively skewed means is likely to be less than mode and median
Solution:
(4) Negatively skewed

Question 3.
The middle value of an ordered array of numbers is the
(1) Mode
(2) Mean
(3) Median
(4) Mid point
Solution:
(3) Median

Question 4.
The weights of students in a school is a :
(1) Discrete variable
(2) Continuous variable
(3) Qualitative variable
(4) None of these
Solution:
(2) Continuous variable
Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions
Question 5.
The first hand and unorganized form data is called
(1) Secondary data
(2) Organised data
(3) Primary data
(4) None of these
Solution:
(3) Primary data