9th Standard Maths Important Questions

Download Tamil Nadu 9th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Generic 2

25 x 2 = 50

Two unbiased coins are tossed simultaneously find the probability of getting
(i) two heads
(ii) one head
(iii) at least one head
(iv) at most one head

Find the value of cot 15°. cot 30°. cot 45°. cot 60°. cot 75°

Using Heron's formula, find the area of a triangle whose sides are 41 m, 15 m, 25 m.

If the centroid of a triangle is at (10, -1) and two vertices are (3, 2) and (5, -11). Find the third vertex of a triangle.

Find the value of s for the following system of equa~on has infinitely many
solutions. 2x - 3y = 7; (s +2) x - (2s + 1)y = 3(2s -1)

Frame two problems in calculating probability, based on the spinner shown here.

The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of Rs. 8.50 per m².

If 3 cot A = 2 , then find the value of = \(\frac { 4\ sin\ A-3\ cos\ A }{ 2\ sin\ A+3\ cos\ A } \)

Find the coordinates of the point which divides the line segment joining A(−5,11) and B(4,−7) in the ratio 7:2.

10.

Express the surds in the simple form \(\sqrt { 27 } \)

11.

Can you reduce the following numbers to surds of same of same order \(\sqrt { 5 } \)

12.

Find the value of X^o

13.

Find the mode of the given data: 3.1, 3.2, 3.3, 2.1,1.3, 3.3, 3.1

14.

A set of numbers consists of five 4’s, four 5’s, nine 6’s,and six 9’s. What is the mode.

15.

Factorise the following: 25a²-10a+1

16.

Simplify the following: \(2\sqrt [ 3 ]{ 40 } +3\sqrt [ 3 ]{ 625 } -4\sqrt [ 3 ]{ 320 } \)

17.

If x = \(\sqrt{5}\) + 2, then find the value of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \)

18.

If A = { x : x = 2ⁿ, n \(\in \) W and n < 4}, B = {x : x = 2n, n \(\in \) N and n ≤ 4} and C = {0, 1, 2, 5, 6} , then verify the associative property of intersection of sets.

19.

In the given diagram PQRS is a parallelogram.
ㄥS = 4x - 60, ㄥQ = 30 - x. Find the angles of P and R.

20.

Let U= {x : -3 : < x < 4} A = {-1,2,3} B = {0,1,2,3} C = {-3;-2,-1,0,1,2}. Find (i) A' UB' (ii) (A ∩ B)' (iii) (A ⋂ C)'

21.

Represent U= {1,2,3,4,5,6,7,8,9,10} A = {1,2,4,6,8,10} and B = {3,6,1,10} in venn-diagram, then find
(i) (A UB)'
(ii) (A ∩ B)'

22.

Procedure:
(I) Make a parallelogram on a chart/graph paper and cut it.
(ii) Draw diagonal of the parallelogram.
(iii) Cut along the diagonal and obtain two triangles.
(iv) Superimpose one triangle onto the other.
What do you conclude?

23.

Find the cardinal number of the following sets.
R = {x : x is an integers, x ∈ Z and –5 ≤ x < 5}

24.

Find the number of zeros of the following polynomials represented by their graphs.

25.

Find the complement of the following angles (1°= 60′ minutes, 1′ = 60′′ seconds)
45°

Generic 3

15 x 3 = 45

26.

Team I and Team II play 10 cricket matches each of 20 overs. Their total scores in each match are tabulated in the table as follows:

Match numbers	1	2	3	4	5	6	7	8	9	10
Team I	200	122	111	88	156	184	99	199	121	156
Team II	143	123	156	92	164	72	100	201	98	157

What is the relative frequency of Team I winning?

27.

Find the value of cos19⁰59'

28.

The lengths of sides of a triangular field are 28 m, 15 m and 41 m. Calculate the area of the field. Find the cost of levelling the field at the rate of Rs. 20 per m²

29.

If A = {2,5,6,7} and B = {3,5,7,8}, then verify the commulative property of intersection of sets

30.

Write in scientific notation: (500000)⁵\(\times\)(3000)³

31.

Find the mean, median and mode of the following distribution:

Weight(in kgs)	25-34	35-44	45-54	55-64	65-74	75-84
Number of students	4	8	10	14	8	6

32.

Find the sum of the deviations from the arithmetic mean for the following observations:
21, 30, 22, 16, 24, 28, 18, 17

33.

Represent the following numbers in scientific notation:
(i) (300000)² \(\times\) (20000)⁴
(ii) (0.000001)¹¹ ÷ (0.005)³
(iii) \( \{ (0.00003 ) ^{ 6 }\times (0.00005)^{ 4 }\} \div \{ (0.009)^{ 3 }\times (0.05)^{ 2 }\} \)

34.

If A = {b,e,f,g} and B = {c,e,g,h}, then verify the commutative property of
(i) union of sets
(ii) intersection of sets.

35.

Identify monomials, binomials and trinomial from the following expression.
(i) -8 abc
(ii) 3a²bc+8-9a²
(iii) -9
(iv) a²+b²+c²-k²
(v) a+b
(vi) 7ab³
(vii) -z+\(\sqrt{3}\)z³

36.

Study the following pattern and write the algebraic expression
(i)

Shape Number	1	2	3	4	5
Number of matchsticks	4	7	10	13	16

(ii)

Shape Number	1	2	3	4	5
Number of Square Boxes	1	4	7	10	13

37.

In the figure find x⁰ and y⁰.

38.

Read the coordinates of the vertices of the triangle ABC with the following figure.

39.

The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3) is 10 units, What are the coordinates of A?

40.

Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

Multiple Choice Question

40 x 1 = 40

41.

The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is _______.

(a)

280 cm²

(b)

300 cm²

(c)

360 cm²

(d)

600 cm²

42.

A letter is chosen at random from the word “STATISTICS”. The probability of getting a vowel is

(a)

\(\frac { 1 }{ 10 } \)

(b)

\(\frac { 2 }{ 10 } \)

(c)

\(\frac { 3 }{ 10 } \)

(d)

\(\frac { 4 }{ 10 } \)

43.

A particular result of an experiment is called _______.

(a)

Trial

(b)

Simple event

(c)

Compound event

(d)

Outcome

44.

If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is _______.

(a)

3 cm²

(b)

6 cm²

(c)

9 cm²

(d)

12 cm²

45.

The value of \(\frac { 1-{ tan }^{ 2 }{ 45 }^{ 0 } }{ 1+{ tan }^{ 2 }{ 45 }^{ 0 } } \) is ________.

(a)

(b)

(c)

(d)

\(\frac { 1 }{ 2 } \)

46.

If 2 sin 2\(\theta\) = \(\sqrt { 3 } \) , them the value of \(\theta\) is ________.

(a)

90⁰

(b)

30⁰

(c)

45⁰

(d)

60⁰

47.

Which condition does not satisfy the linear equation ax + by + c = 0.

(a)

a \(\neq \) 0 , b = 0

(b)

a = 0 , b \(\neq \) 0

(c)

a = 0 , b = 0 , c \(\neq \) 0

(d)

a \(\neq \)0, b \(\neq \) 0

48.

The centroid of the triangle with vertices (−1, −6), (−2, 12) and (9, 3) is

(a)

(3, 2)

(b)

(2, 3)

(c)

(4, 3)

(d)

(3, 4)

49.

If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (−3, 2), then the coordinate of the other end of the diameter is ______.

(a)

(0, −3)

(b)

(0, 9)

(c)

(3, 0)

(d)

(−9, 0)

50.

Zero of (7+4x) is_______

(a)

\(\cfrac { 4 }{ 7 } \)

(b)

\(\cfrac { -7 }{ 4 } \)

(c)

(d)

51.

\(\sqrt [ 4 ]{ 405 } =h\sqrt [ 4 ]{ 5 } \), then h = ____________

(a)

(b)

(c)

(d)

52.

The angle subtend by a semicircle at the remaining part of the circumference is___________

(a)

60^o

(b)

90^o

(c)

120^o

(d)

180°

53.

The angle subtend by a semicircle at the centre is_________

(a)

60^o

(b)

90°

(c)

120^o

(d)

180^o

54.

If the length of a chord decreases, then its distance from the centre__________

(a)

Increases

(b)

decreases

(c)

same

(d)

cannot say

55.

The mean of set of numbers is \(\bar{x}\) If each number is multiplied by z, the mean is

(a)

\(\bar{X}+z\)

(b)

\(\bar{X}-z\)

(c)

\(z\bar{X}\)

(d)

\(\bar{X}\)

56.

The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is _______________

(a)

(b)

(c)

(d)

57.

The mean of set of seven number is 81. If one of the nimbers is discarded,the mean of remaining number is 78. The value of discarded number is

(a)

101

(b)

100

(c)

(d)

58.

Which one of the following is not a measure of central tendency?

(a)

Mean

(b)

Range

(c)

Median

(d)

Mode

59.

A particular observation which occurs maximum number of times in a given data is called its _______.

(a)

Frequency

(b)

range

(c)

mode

(d)

Median

60.

Data available in an unorganized form is called __________ data

(a)

Grouped data

(b)

class interval

(c)

mode

(d)

raw data

61.

In the figure, PQRS and PTVS are two cyclic quadrilaterals, If ㄥQRS = 100°, then ㄥTVS = ______.

(a)

80°

(b)

100°

(c)

70°

(d)

90°

62.

In the figure, O is the centre of the circle and ㄥACB = 40° then ㄥAOB = ________.

(a)

80°

(b)

85°

(c)

70°

(d)

65°

63.

When (2\(\sqrt{5}\)-\(\sqrt{2}\))² is simplified, we get ________.

(a)

4\(\sqrt{5}\)+2\(\sqrt{2}\)

(b)

22-4\(\sqrt{10}\)

(c)

8-4\(\sqrt{10}\)

(d)

2\(\sqrt{10}\)-2

64.

For any three sets P, Q and R, P-(Q\(\\ \cap \)R) is ________.

(a)

P-(Q\(\cup \)R)

(b)

(P\(\\ \cap \)Q)-R

(c)

(P-Q)\(\cup \)(P-R)

(d)

(P-Q)\(\\ \cap \)(P-R)

65.

The value of the polynomial f(x) = 6x - 3x²+9 when x = -1 is _____________________

(a)

(b)

(c)

(d)

66.

The roots of the polynominal equation \({ x }^{ 2 }+2x=0\) are_______________

(a)

x = 0, 2

(b)

x = 1, 2

(c)

x = 1, -2

(d)

x = 0, -2

67.

The sum of \({ 5x }^{ 2 };-7x^{ 2 };8x^{ 2 };11x^{ 2 }\ and\ -9x^{ 2 }\) is ___________

(a)

\(2{ x }^{ 2 }\)

(b)

\(4{ x }^{ 2 }\)

(c)

\(6{ x }^{ 2 }\)

(d)

\(8{ x }^{ 2 }\)

68.

The point of concurrency of the medians of a triangle is known as __________

(a)

circumcentre

(b)

incentre

(c)

orthocentre

(d)

centroid

69.

The shade region with adjoint diagram represents ___________

(a)

A - B

(b)

B - A

(c)

(d)

70.

The set (A - B) U(B - A) is ___________

(a)

AΔB

(b)

AUB

(c)

A∩B

(d)

A'UB'

71.

A = {set of odd natural numbers}, B = {set of even natural numbers}, then A and B are ___________

(a)

equal set

(b)

equivalent sets

(c)

overlapping sets

(d)

disjoint sets

72.

Which one of the following has terminating decimal expansion?

(a)

\(\frac { 7 }{ 9 } \)

(b)

\(\frac { 8 }{ 15 } \)

(c)

\(\frac { 1 }{ 2 } \)

(d)

\(\frac { 5 }{ 32 } \)

73.

The decimal form of -\(\frac { 3 }{ 4 } \) is_______________

(a)

- 0.75

(b)

- 0.50

(c)

-0.25

(d)

- 0.125

74.

The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, - 1) is_______________

(a)

(b)

(c)

\(\sqrt{2}\)

(d)

75.

The point which is on y-axis with ordinate - 5 is _____________

(a)

(0, - 5)

(b)

(-5,0)

(c)

(5,0)

(d)

(0,5)

76.

The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then ______________

(a)

\(\sqrt{24}\)

(b)

\(\sqrt{37}\)

(c)

\(\sqrt{26}\)

(d)

\(\sqrt{17}\)

77.

The distance between the points (a, 0) and (0, b) is____________

(a)

a unit

(b)

b unit

(c)

\(\sqrt{a^2+{b^2}}\ unit\)

(d)

\(\sqrt{a^2-{b^2}}\ unit\)

78.

\(0.\overline { 34 } +0.3\bar { 4 } \) = ________.

(a)

\(0.6\overline { 87 } \)

(b)

\(0.\overline { 68 } \)

(c)

\(0.6\bar { 8 } \)

(d)

\(0.68\bar { 7 } \)

79.

The product of the polynomials p(x) = 4x –3 q(x) = 4x + 3 ____________

(a)

1 – x – 8

(b)

16x² – 9

(c)

18x³+ 12x² – 12x – 8

(d)

18x³ – 12x² + 12x + 8

80.

Let A = {∅} and B = P(A), then A∩B is ________.

(a)

{ ∅, {∅} }

(b)

{∅}

(c)

∅

(d)

{0}

Generic 5

8 x 5 = 40

81.

A farmer has a field in the shape of a rhombus. The perimeter of the field is 400 m and one of its diagonal is 120 m. He wants to divide the field into two equal parts to grow two different types of vegetables. Find the area of the field.

82.

In a school, 80 students like Maths,90 students like Science 82 students like History, 21 like both Maths and Science 19 like both science and History 20 like both Maths and History and 8 liked all the three subjects. If each student like atleast one subject, then find
(i) the number of students in the school
(ii)the number of students who like only one subject.

83.

Prove that x -1 is a factor x⁵ - 45x⁴ + 36x³ + 45x²- 36x^-1

84.

Find the Arithmetic Mean of the following data using Step Deviation Method

Age	15-19	20-24	25-29	30-34	35-39	40-44
No.of persons	4	20	38	24	10	9

85.

Represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line.

86.

Find any seven rational numbers between \(\frac { 5 }{ 8 } \) and \(\frac { 5 }{ 6 } \)

87.

Diagonal AC of a parallelogram ABCD bisects ㄥA. Show that
(i) it bisects ㄥC also
(ii) ABCD is a rhombus.

88.

Find the type of triangle formed by (-1, -1), (1, 1) and (\(-\sqrt{13},\sqrt{13}\))

Generic 8

1 x 8 = 8

89.

Construct the centroid of \(\triangle\)PQR such that PQ = 9 cm, PQ = 7cm, RP = 8 cm.

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Answers Key:

Generic 2

25 x 2 = 50

S = {HH, HT, TH, TT}
(i) probability of two heads = \(\frac { 1 }{ 4 } \)
(ii) probability of one head = \(\frac { 1 }{ 2 } \)
(iii) probability of at least one head = \(\frac { 3 }{ 4 } \)
(iv) probability of at most one head = \(\frac { 3 }{ 4 } \).

cot (90° - 75) cot (90° - 60°) cot 45° cot 60° cot 75°
= tan 75° tan 6.0° (1) cot 60° cot 75° = 1

s = \(\frac { 41+15+25 }{ 2 } \) = 40.5
Area = \(\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 40.5\times 0.5\times 25.5\times 15.5 } \)
=\(\sqrt { \frac { 405 }{ 10 } \times \frac { 5 }{ 10 } \times \frac { 255 }{ 10 } \times \frac { 155 }{ 10 } } =\sqrt { \frac { 5\times 9\times 9\times 5\times 5\times 3\times 17\times 5\times 31 }{ 10\times 10\times 10\times 10 } } \)
=\(\frac { 1 }{ 100 } \times 5\times 9\times 5\times \sqrt { 3\times 17\times 31 } \)

\((10,-1)=\left( \frac { 3+5+x }{ 3 } ,\frac { 2-11+y }{ 3 } \right) \)
\(10=\frac { 3+5+x }{ 3 } ,\)
30 = 3 + 5 + x
22 = x
\(-1=\frac { 2-11+y }{ 3 } \)
-3 = 2 - 11 + y
-3 + 9 = y
y = 6
(22, 6)

2x- 3y = 7
(s + 2) x - (2s + 1)y == 3 (2s - 1)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } =\cfrac { 7 }{ 3(2s-1) } \)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } \)
2(2s + 1 ) = 3 (s + 2)
4s + 2 = 3s + 6
4s - 3s = 6- 2
s = 4

(i) What is the probability that the spinner will not land on a multiple of 2?
(ii) What is the probability that the spinner will land on an odd number?

Dimensions of a hall 10 m \(\times\) 9 m \(\times\) 8 m
l = 10 m
b = 9 m
h = 8 m
White washing to be done for the area of the surface
= 2 (lh + bh) + lb
= 2 (10 \(\times\) 8 + 9 \(\times\) 8) + 10 \(\times\) 9
= 2 (80 + 72) + 90 =2 \(\times\) 152 + 90
= 304 + 90 = 394 m²
Cost of white washing per m²= Rs. 8.50
Cost of white washing 394 m² = 394 \(\times\) 8.50
Total cost = Rs. 3349

\(cot\ A=\cfrac { 2 }{ 3 } \)
\(cot\ A=\cfrac { Adjacent\ side }{ Opposite\ side } \)
\(tan\ A=\ \cfrac { Opp.side }{ Adj.side } =\cfrac { 3 }{ 2 } \)
\(sin\ A=\cfrac { 3 }{ \sqrt { 13 } } \)
\(cotA=\cfrac { 2 }{ \sqrt { 13 } } \)
\(\therefore \cfrac { 4sinA-3cosA }{ 2ainA+3cosA } =\cfrac { 4\times \cfrac { 3 }{ \sqrt { 13 } } -3\times \cfrac { 2 }{ \sqrt { 13 } } }{ 2\times \cfrac { 3 }{ \sqrt { 13 } } +3\cfrac { 2 }{ \sqrt { 13 } } } \)
= \(\cfrac { \frac { 12 }{ \sqrt { 13 } } -\frac { 6 }{ \sqrt { 13 } } }{ \frac { 6 }{ \sqrt { 13 } } +\frac { 6 }{ \sqrt { 13 } } } =\cfrac { \frac { 6 }{ \sqrt { 13 } } }{ \frac { 2 }{ \sqrt { 13 } } } =\cfrac { 6 }{ \sqrt { 13 } } \times \cfrac { \sqrt { 13 } }{ 12 } \)
= \(\cfrac { 6 }{ 12 } =\cfrac { 1 }{ 2 } \)

By Pythagoras theorem In\(\triangle\)OAB,
AB² = OA²+ OB²
= 3² + 2²
= 9+4
= 13
\(AB=\sqrt { 13 } \)

x₁ y₁
Here (-5,11)
x₂ y₂
(4,-7)
m n
7 : 2
\(=P(x,y)\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) \)
\(=\left( \frac { 7\times 4+2\times -5 }{ 7+2 } ,\frac { 7\times -7+2\times 11) }{ 7+2 } \right) \)
\(=\left( \frac { 28-10 }{ 9 } ,\frac { -49+22 }{ 9 } \right) \)
\(=\left( \frac { 18 }{ 9 } ,\frac { -27 }{ 9 } \right) =(2,-3)\)

10.

\(\sqrt { 27 } \) =\(\sqrt { 3\times 3\times 3 } =\sqrt [ 3 ]{ 3 } \)

11.

\(\sqrt { 5 } \) = \({ 5 }^{ \frac { 1 }{ 2 } }={ 5 }^{ \frac { 6 }{ 12 } }=\sqrt [ 12 ]{ { 5 }^{ 6 } } =\sqrt [ 12 ]{ 15625 } \)

12.

MPN = 90^o(Angle subtended by the diameter is 90°)
\(\angle\)OMP +\(\angle\) OPM = 180^o-120^o= 60^o
\(\angle\)MPO = \(\cfrac { 60^{ o } }{ 2 } \) = 30^o
\(\therefore\)x = \(\angle \)OPN = 90^o- 30^o= 60^o

13.

3.1, 3.2, 3.3, 2.1,1.3, 3.3, 3.1
In this given data 3.1, 3.3 occurs twice
\(\therefore\) mode = 3.1 and 3.3(bimodal)

14.

Size of item	4	5	6	9
Frequency	5	4	9	6

6 has the maximum frequency 9. Therefore 6 is the mode.

15.

25a²-10a+1 = (5a)²-2(5a)(1)+1²
= (5a-1)² [∵ a²-2ab+b²=(a-b)²]

16.

\(2\sqrt [ 3 ]{ 40 } +3\sqrt [ 3 ]{ 625 } -4\sqrt [ 3 ]{ 320 } =2\sqrt [ 3 ]{ 8\times 5 } +3\sqrt [ 3 ]{ 125\times 5 } -4\sqrt [ 3 ]{ 64\times 5 } \)
= \(2\sqrt [ 3 ]{ 2^{ 3 }\times 5 } +3\sqrt [ 3 ]{ { 5 }^{ 3 }\times 5 } -4\sqrt [ 3 ]{ { 4 }^{ 3 }\times 5 } \)
= \(2\times 2\sqrt [ 3 ]{ 5 } +3\times 5\sqrt [ 3 ]{ 5 } -4\times 4\sqrt [ 3 ]{ 5 } \)
= \(4\sqrt [ 3 ]{ 5 } +15\sqrt [ 3 ]{ 5 } -16\sqrt [ 3 ]{ 5 } \)
= \((4+15-16)\sqrt [ 3 ]{ 5 } =3\sqrt [ 3 ]{ 5 } \).

17.

\( x^{2} =(\sqrt{5}+2)^{2}=5+4+4 \sqrt{5}=9+4 \sqrt{5} \)
\(\frac{1}{x^{2}} =\frac{1}{9+4 \sqrt{5}} \)
\( =\frac{1}{9+4 \sqrt{5}} \times \frac{9-4 \sqrt{5}}{9-4 \sqrt{5}}=\frac{9-4 \sqrt{5}}{81-80} \)
\( ^{2}+\frac{1}{x^{2}} =9+4 \sqrt{5}+9-4 \sqrt{5} \)
\( =9+9+4 \sqrt{5}-4 \sqrt{5} \)
= 18

18.

A = {x : x = 2ⁿ, n ∈ W, n < 4}
x = 2⁰= 1
x = 2¹= 2
x = 2²= 4
x = 2³= 8
ஃ A = {1,2,4,8}
B = {x : x = 2n, n∈N and n ≤ 4}
x = 2 \(\times\) 1 = 2
x = 2 \(\times\) 2 = 4
x = 2 \(\times\) 3 = 6
x = 2 \(\times\)4 = 8
ஃ B = {2,4,6,8}
C = {0,1,2,5,6}
Associative property of intersection of sets
A∩(B∩C) = (A∩B)∩C
B∩C = {2,6}
A∩(B∩C) = {1,2,4,6,8}∩{2,6}
= {2} .............(1)
A∩B = {1,2,4,8} ∩ {2,4,6,8} = {2,4,8}
(A∩B) ∩C = {2,4,8} ∩ {0,1,2,5,6}
= {2} ............(2)
From (1) and (2), It is verified that A∩(B∩C) = (A∩B)∩C

19.

ㄥP = ㄥR= 168⁰

20.

(i) {-3,-2,-1,0,1,4} (ii) {-3,-2,-1,0,1,4} (iii) {-3,-2,0,1,3,4}

21.

(i) {5,9}
(ii) {1,2,3,4,5,7,8,9}

22.

23.

n(R) = 10

24.

Number of zeros = 1 (The curve is intersecting the x-axis at one point)

25.

Complement of 45° = 90⁰-45°
= 45°

Generic 3

15 x 3 = 45

26.

In this experiment, each trial is a match where Team I faces Team II.
We are concerned about the winning status of Team I.
There are 10 trials in total; out of which Team I wins in the 1^st, 6^th and 9^th matches.
The relative frequency of Team I winning the matches = \(\frac { 3 }{ 10 } \)or 0.3

27.

12'

18'

24'

30'

36'

42'

48'

54'

Mean Difference

0.0⁰

0.1⁰

0.2⁰

0.3⁰

0.4⁰

0.5⁰

0.6⁰

0.7⁰

0.8⁰

0.9⁰

19⁰

0.9403

write 19⁰59' = 19⁰54' + 5'
From the table we have, cos19⁰54' = 0.9403

28.

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.

29.

\(A\cap B\) = {5,7}
\(B\cap A\)= {5,7}
From (3) and (4) we get, \(A\cap B=B\cap C\)
It is verifield that insersection of sets is commutative.

30.

(500000)⁵\(\times\) (3000)³
\( =\left(5.0 \times 10^{5}\right)^{3} \times\left(3.0 \times 10^{3}\right)^{3} \\ =(5.0)^{2} \times\left(10^{5}\right)^{2} \times(3.0)^{3} \times\left(10^{3}\right)^{3} \\ =25 \times 10^{10} \times 27 \times 10^{9}=675 \times 10^{19} \\ =675.0 \times 1019=6.75 \times 102 \times 1019=6.75 \times 10^{21} \)

31.

Weight in kgs	Number of Students f	Mid x	fx	cf
24.5-34.5	4	29.5	118	4
34.5-44.5	8	39.5	316	12
44.5-54.5	10	49.5	495	22
54.5-64.5	14	59.5	833	36
64.5-74.5	8	69.5	556	44
74.5-84.5	6	79.5	477	50
	50		2975

Mean\(\bar{x}=\cfrac{\Sigma fx}{\Sigma f}=\cfrac{2795}{50}=55.9\)
Median = \(\left(\cfrac{N}{2}\right)^{th} \) term is called the median class
N = 50
\(\cfrac{N}{2}=\cfrac{50}{2}=25\)
\(\therefore\) Median class = 25^th value = 54.5 - 64.5
l = 54.5, m = 22, f = 14, c = 10
Median = \( l+\cfrac { \left( \cfrac { N }{ 2 } -m \right) }{ f } \times c\)
= 54.5 + \(\cfrac{25-22}{14}\times10\)
= 54.5 + \(\cfrac{3}{14}\times10\)
= 54.5 + \(\cfrac{30}{14}\)
= 54.5 + 2.14
= 56.64
Model class is 54.5 - 64.5 since it has the maximum frequency.
l = 54.5, f = 14, f₁= 10, f₂= 8, c = 10
\(\therefore\) Mode =\(l+\left[\cfrac{f-{f}_{1}}{2f-{f}_{1}{f}_{2}}\right]\times c\)
= 54.5+\(\left[\cfrac{14-10}{2\times 14-10-8}\right]\times 8\)
= 54.5+ \(\left(\cfrac{4}{28-18}\right)\times 10\)
= 54.5+ \(\cfrac{4}{10}\times 10=58.5\)
\(\therefore \) Mean = 55.9
Median = 56.64
Mode = 58.5

32.

\(\bar{X}=\frac{\sum_{i=1}^{8} x_{i}}{n}=\frac{21+30+22+16+24+28+18+17}{8}=\frac{176}{8}=22\)
Deviation of an entry x_i from the arithmetic mean \(\bar X\) is \(x_i-\bar X\), i = 1, 2,........8
Sum of the deviations = (21-22)+(30-22)+(22-22)+(16-22)+(24-22)+(28-22)+(18-22)+(17-22)
= 16–16 = 0. or equivalently, \(\sum_{i=1}^{8}\left(x_{i}-\bar{X}\right)=0\)
Hence, we conclude that sum of the deviations from the Arithmetic Mean is zero.

33.

(i) \( (300000)^{2} \times(20000)^{4}\)
\(=\left(3.0 \times 10^{5}\right)^{2} \times\left(2.0 \times 10^{4}\right)^{4} \)
\(=3^{2} \times 10^{10} \times 2^{4} \times 10^{16} \)
\(=9 \times 16 \times 10^{10+16} \)
\(=144 \times 10^{26} \)
\(=1.44 \times 10^{28} \)
(ii) (0.000001)¹¹ ÷ (0.005)³
\( =\left(1.0 \times 10^{-6}\right)^{11} \div\left(5.0 \times 10^{-3}\right)^{3} \)
\( =\frac{1.0 \times 10^{-66}}{125.0 \times 10^{9}} \)
\( =\frac{1000.0 \times 10^{-69}}{125.0 \times 10^{-9}} \)
\( =8.0 \times 10^{-69} \times 10^{9} \)
\( =8.0 \times 10^{-60} \)
(iii) \(\left\{(0.00003)^{6} \times(0.00005)^{4}\right\} \div\left\{(0.009)^{3} \times(0.05)^{2}\right\}\)
=\(\frac{(3.0\times10^{-5})^6\times(5.0\times10^{-5})^4}{(9.0\times10^{-3})^3\times(5.0\times10^{-2})^2}\)
=\(\frac{3^6\times10^{-30}\times5^4\times10^{20}}{9^3\times10^{-9}\times5^{2}\times10^{-4}}=\frac{3^6\times5^4\times10^{-30-20}}{(3^{2})^3\times10^{-9-4}\times5^{2}}=\frac{ ̶3̶^6̶\times5^4\times10^{-50}}{ ̶3̶^6̶\times5^2\times10^{-13}}\)
\( =5^{4-2} \times 10^{-50+13} \)
\(=5^{2} \times 10^{-37} \)
\(=25 \times 10^{-37} \)
\(=2.5 \times 10^{1} \times 10^{-37}\)
\(=2.5 \times 10^{-36} \)

34.

Given, A = {b,e,f,g} and B = {c,e,g,h}
(i) \(A\cup B\) = {b,c,e,f,g,h} .....(1)
\(B\cup A\) = {b,c,e,f,g,h} ......(2)
From (1) and (2) we have \(A\cup B\) = \(B\cup A\)
It is verified that union of sets is commutative.
(ii) \(A\cap B\) = {e,g} .....(3)
\(B\cap A\) = {e,g} .....(4)
From (3) and (4) we get, \(A\cap B\) = \(B\cap A\)
It is verified that intersection of sets is commutative.

35.

In the given expressin (i), (iii) and (vi) contain only one term.
\(\therefore\) (i), (iii) and (vi) are monomials.
The expression (v) contains two terms. So it is a binomial.
The expression (ii) contains three terms. So it is a trinomial.
The expression (iv) contains four terms. So it is none of the given type. It is a polynomial

36.

(i) The algebraic expression is 3n + 1
(ii) The Algebraic expression is 3n - 2

37.

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60° (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

38.

A (- 6, 4), B (- 3, -3) and C (2, 2)

39.

Let the point A be (a, a), B is (1, 3)
Distance AB = 10 (Given)
By distance formula \(\sqrt { (a-1)^{ 2 }+(a-3)^{ 2 } } =10\)
Simplifying
2a²- 8a + 10 = 100
a²- 4a - 45 = 0
(a - 9)(a + 5) = 0
⇒ a = -5 ; A = (-5,-5)
a = 9 ; A = (9,9)

40.

In the given diagram
AB = CD (Given)
BD is common.
\(\angle ABD=\angle BDC\) (alternate angles)
(Since AC and CD are parallel and BD is the transversal)
By SAS congruency
\(\therefore\triangle ABD\cong\triangle CDB\)

Multiple Choice Question

40 x 1 = 40

41.

(a)
280 cm²

42.

43.

(d)
Outcome

44.

(b)
6 cm²

45.

46.

(b)
30⁰

47.

48.

(b)
(2, 3)

49.

(d)
(−9, 0)

50.

(b)
\(\cfrac { -7 }{ 4 } \)

51.

(d)
3

52.

(b)
90^o

53.

(d)
180^o

54.

(a)
Increases

55.

56.

(d)
34

57.

(b)
100

58.

(b)
Range

59.

60.

(d)
raw data

61.

(a)
80°

62.

(a)
80°

63.

(b)
22-4\(\sqrt{10}\)

64.

65.

(a)
0

66.

(d)
x = 0, -2

67.

(d)
\(8{ x }^{ 2 }\)

68.

(d)
centroid

69.

70.

(a)
AΔB

71.

(d)
disjoint sets

72.

(d)
\(\frac { 5 }{ 32 } \)

73.

(a)
- 0.75

74.

(a)
2

75.

(a)
(0, - 5)

76.

77.

78.

(a)
\(0.6\overline { 87 } \)

79.

(b)
16x² – 9

80.

(b)
{∅}

Generic 5

8 x 5 = 40

81.

Let ABCD be the rhombus.
Its perimeter = 4 × side = 400 m
Therefore, each side of the rhombus = 100 m
Given the length of the diagonal AC = 120 m
In \(\triangle\)ABC, let a =100 m, b =100 m, c = 120 m

s = \(\frac{a+b+c}{2}=\frac{100+100+120}{2}\) = 160 m
Area of \(\triangle\)ABC =\(\sqrt{160(160-100)(160-100)(160-120)}\)
= \(\sqrt{160 \times 60\times 60 \times40}\)
= \(\sqrt{40 \times 2 \times \times2\times60\times60\times40}\)
= 40 × 2 × 60 = 4800 m²
Therefore, Area of the field ABCD = 2 × Area of \(\triangle\)ABC = 2 × 4800 = 9600 m²

82.

Let M.S and H represent sets students who like Maths,Science and history respectively.
Then, n(M) = 80,n(S) = 90,n(H) = 82,\(n\left( M\cap S \right) \) = 21,\(n\left( S\cap H \right) \) = 19,\(n\left( M\cap H \right) \) = 20,\(n\left( M\cap S\cap H \right) \)= 8
Let us represents the given data in a venn diagram

(i) The number of student in the school = 52 + 59 + 55 + 12 + 11 + 8 + 8 = 205
(ii) The number of students who like only one subject = 52 + 59 + 55 = 166

83.

Letp (x) x⁵ - 45x⁴ + 36x³ + 45x² - 36x- 1
Sum of co-efficients = 1 - 45 + 36 + 45 - 36 - 1 = 0
Thus x -1 is a factor of p (x)

84.

Step Deviation Method
\(\bar { x } =A+\left[ \frac { \Sigma fd }{ \Sigma f } \times c \right] \) Where \(d=\frac { x-A }{ c } \)
Let Assumed mean A = 32
Class Interval c = 45

Age Class Interval	x	Mid x	No.of person f	d=\(\cfrac{x-A}{c}\)	fd
15-19	14.5-19.5	17	4	-3	-12
20-24	19.5-24.5	22	20	-2	-40
25-29	24.5-29.5	27	38	-1	-38
30-34	29.5-34.5	32	24	0	0
35-39	34.5-39.5	37	10	1	10
40-44	39.5-44.5	42	9	2	18
			\(\Sigma f=105\)		\(\Sigma fd=-62\)

Mean \(\bar { x } =A+\left[ \cfrac { \Sigma fd }{ \Sigma f } \times c \right] \)
\(=32+\left[ \cfrac { -62 }{ 105 } \times 5 \right] \)
= 32 + (-2.952)
= 29.05

85.

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

86.

Let us convert the given rational numbers having the same denominators.
L.C.M of 8 and 6 is 24.
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
\(\frac { 5 }{ 8 } \times \frac { 5\times 3 }{ 8\times 3 } =\frac { 15 }{ 24 } \)
Now the rational numbers between \(-\frac { 202 }{ 24 } and\frac { 15 }{ 24 } are-\frac { 19 }{ 24 } ,-\frac { 18 }{ 24 } ,-\frac { 17 }{ 24 } ,....,\frac { 0 }{ 24 } ,\frac { 1 }{ 24 } ,\frac { 22 }{ 24 } ,...,\frac { 14 }{ 24 } \)
we can take any seven of them \(\frac { 1 }{ 24 } ,\frac { 2 }{ 24 } ,\frac { 3 }{ 24 } ,\frac { 4 }{ 24 } ,\frac { 5 }{ 24 } ,\frac { 6 }{ 24 } ,\frac { 7 }{ 24 } \)

87.

We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects LC
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle) .........(1)
.Also BC II AD and AC is a transversal.
∴ ㄥ2 = ㄥ4 (Alternate interior angle) ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ 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.

88.

Let the point A (-1, -1), (1, 1) and (\(-\sqrt{13},\sqrt{13}\))
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(1+1)^2+(1+1)^2}\)
\(\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}\)
BC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}\)
\(\sqrt{3+1+3+1}=\sqrt{8}\)
AC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}+1)^2}\)
\(=\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}}\)
\(=\sqrt{4+4}=\sqrt{8}\)

AB= BC=AC= \(\sqrt{8}\)
\(\therefore\)ABC is an equilateral triangle.

Generic 8

1 x 8 = 8

89.

In \(\triangle\)PQR, PQ = 5 cm, PR = 6 cm, \(\angle\)QPR = 60°

Construction:
Step 1: Draw \(\triangle\) PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2: Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given \(\triangle\)PQR.