9th Standard Maths Annual Exam Model Question Paper III - 2019 - 2020

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 8

2 x 8 = 16

Draw and locate the centroid of the triangle ABC where right angle at A, AB = 4cm and AC = 3cm

Draw ΔPQR with sides PQ = 7 cm, QR = 8 cm and PR = 5 cm and construct its Orthocentre.

Generic 2

14 x 2 = 28

Find the value of sin 3x. sin 6x. sin 9x when x = 10°

Find the volume of a cube whose surface are is a 96 cm².

Find the six trigo'nometric ratios of the angle 0 using the diagram

Find the coordinates of the point which divides, the line segment joining the point A (3,7) and B (-11, -2) in the ratio 5 : 1.

If A (10, 11) and B (2 ,3) are the coordinates of end points of diameter of circle. Then find the centre of the circle.

A manufacturer tested 7000 LED lights at random and found that 25 of them were defective. If a LED light is selected at random, what is the probability that the selected LED light is a defective one.

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

10.

Factorise the following expressions: pr+qr+pq+p²

11.

The mass of the Earth is 5.97 \(\times\) 10²⁴ kg and that of the Moon is 0.073 \(\times\) 10²⁴ kg. What is their total mass?

12.

Write the following in the form of 5ⁿ
\(\sqrt{5}\)

13.

Without actual division classify the decimal expansion of the following numbers as terminating or non-terminating and recurring.
\(7\over 16\)

14.

If U = {a, b, c, d, e, f, g, h}, A = {b, d, f, h} and B = {a, d, e, h}, find the following sets. A′\(\cup \) B′

15.

Find the distance between the following pairs of points. (3,4) and (– 7, 2)

16.

Which ones are not quadrilaterals?

Multiple Choice Question

14 x 1 = 14

17.

The perimeter of an equilateral triangle is 30 cm. The area is _______.

(a)

\(10\sqrt { 3 } \) cm²

(b)

\(12\sqrt { 3 } \) cm²

(c)

\(15\sqrt { 3 } \) cm²

(d)

\(25\sqrt { 3 } \) cm ²

18.

A number between 0 and 1 that is used to measure uncertainty is called _______.

(a)

Random variable

(b)

Trial

(c)

Simple event

(d)

Probability

19.

The value of 2tan30° tan60° is

(a)

(b)

(c)

\(2\sqrt { 3 } \)

(d)

20.

The linear equation in one variable is __________

(a)

2x + 2 = y

(b)

5x − 7 = 6 − 2x

(c)

2t(5 − t) = 0

(d)

7p − q = 0

21.

In what ratio does the y-axis divides the line joining the points (−5, 1) and (2, 3) internally ______.

(a)

1 :3

(b)

2 :5

(c)

3 :1

(d)

5 :2

22.

Rationalising the denominator \(\cfrac { 1 }{ \sqrt [ 3 ]{ 3 } } \) ___________

(a)

(b)

\(\cfrac { { 3 }^{ \frac { 2 }{ 3 } } }{ 3 } \)

(c)

\(\sqrt { 3 } \)

(d)

\(\sqrt [ 3 ]{ 3 } \)

23.

The perpendicular line from the centre of the circle to the chord divided the chord in the ratio________

(a)

1: 1

(b)

1: 2

(c)

2: 1

(d)

1: 3

24.

The mean of 5, 9, x, 17,and 21 is 13 then find the value of x ___________

(a)

(b)

(c)

(d)

25.

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

(a)

Frequency

(b)

range

(c)

mode

(d)

Median

26.

The length and breadth of a rectangular plot are 5 \(\times\) 10⁵ and 4 \(\times\) 10⁴ metres respectively. Its area is ______.

(a)

9 \(\times\) 10¹ m²

(b)

9 \(\times\) 10⁹ m²

(c)

2 \(\times\) 10¹⁰ m²

(d)

20 \(\times\) 10²⁰ m²

27.

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

(a)

(b)

(c)

(d)

28.

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

(a)

AΔB

(b)

AUB

(c)

A∩B

(d)

A'UB'

29.

The distance between the points (4, -1) and the origin is___________

(a)

\(\sqrt{24}\)

(b)

\(\sqrt{37}\)

(c)

\(\sqrt{26}\)

(d)

\(\sqrt{17}\)

30.

If U = {x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A′)′ is –––––––––.

(a)

{1, 6, 7, 8, 9}

(b)

{1, 2, 3, 4}

(c)

{2, 3, 4, 5}

(d)

{ }

Generic 3

13 x 3 = 39

31.

The total surface area of a cube is 864 cm². Find its volume

32.

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?

33.

If (x, 3), (6, y), (8, 2) and (9, 4) are the vertices of a parallelogram taken in order, then find the value of x and y.

34.

For the measures in the figure, compute sine, cosine and tangent ratios of the angle \(\theta \)

35.

Show that x+4 is a factor of x³+ 6x²- 7x - 60

36.

Write in scientific notation ;(60000000)³

37.

The median of observation 11,12,14,18, x+12, x+4, 30, 32, 35, 41 arrenged in ascending order is 24. Find the values of x.

38.

(i) Add 3\(\sqrt{7}\) and 5\(\sqrt{7}\) . Check whether the sum is rational or irrational.
(ii) Subtract 4\(\sqrt{5}\) from 7\(\sqrt{5}\) . Is the answer rational or irrational?

39.

If K = {a,b,d e f } L = {b,c,d,g} and M = {a,b,c,d,h}, then find the following:
(i) K \(\cup \)(L \(\cap \) M)
(ii) K \(\cap \)(L \(\cup \) M)
(iii) (K\(\cup \)L)\(\cap \)(K\(\cup \)M)
(iv) (K\(\cap \)L)\(\cup \)(K\(\cap \)M)

40.

Given that A = {1,3,5,7} B = {1,2,4,6,8}. Find
(i) AΔB and
(ii) BΔA

41.

Find quotient and the remainder when f(x) is divided by g(x) x⁴ – 3x³ + 5x² – 7 is divided by x² + x + 1

42.

Express the following decimal expression into rational numbers. \(2.\overline { 327 } \)

43.

Determine whether the given set of points in each case are collinear or not (a, –2), (a, 3), (a, 0)

Generic 5

3 x 5 = 15

44.

Find the quotient and remainder when 5x³ + 7x² + 3x + 2 is divided by 3x + 2

45.

Find the angle of the given cyclic quadrilateral ABCD in the figure.

46.

Factorise the following:
(i) (p - q²) - 6(p - q) -16
(ii) m²+ 2mm - 24n²
(iii) \(\sqrt{5}a^2+2a-3\sqrt5\)
(iv) a⁴- 3a²+ 2
(iv) 8m³- 2m²n -15mn²
(v) \({1\over x^2}+{1\over y^2}+{2\over xy}\)

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Answers Key:

Generic 8

2 x 8 = 16

In \(\triangle \)ABC,
AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°
Step 2: Draw perpendicular bisectors of any two sides (AB and AC) to find the mid points of AB and AC.
Step 3: Draw the medians CD and BE. Let them meet at G.
Step 4: G is the centroid of the given triangle.

Steps for construction:
Step 1: Draw the ΔPQR with the given measurements.
Step 2: Construct altitudes from any two vertices P and Q to their opposite sides QR and PR respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given ΔPQR.

Generic 2

14 x 2 = 28

sin 3 (10°) sin 6 (10°) sin 9 (10°)
= \(\cfrac { 1 }{ 2 } \times \cfrac { \sqrt { 3 } }{ 2 } \times 1=\cfrac { \sqrt { 3 } }{4 } \)

6a² = 96 cm²
a² =\(\frac { 96 }{ 6 } \)
a = 6
Volume = a³ = 6³ = 36 \(\times\) 6 = 216 cm³.

Hypotenuse = \(\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } \)
= \(\sqrt { 144+25\quad } =\sqrt { 69 } \)
= 13
\(sin\theta =\cfrac { 5 }{ 13 } ;cos\theta =\cfrac { 12 }{ 13 } ;tan\theta =\cfrac { 5 }{ 12 } ; \)
\(cosec\theta =\cfrac { 13 }{ 5 } ;sec\theta =\cfrac { 13 }{ 12 } ;cot\theta =\cfrac { 12 }{ 3 } \)

\((x,y)=\left( \frac { 3(1)+5(-11) }{ 5+1 } ,\frac { 1(7)+5(-2) }{ 6 } \right) \)
\(=\left( \frac { 3-55 }{ 6 } ,\frac { 7-10 }{ 6 } \right) =\left( \frac { -52 }{ 6 } ,\frac { -3 }{ 6 } \right) =\left( \frac { -26 }{ 3 } ,\frac { -1 }{ 2 } \right) \)

Centre of the circle \(=\left( \frac { 10+2 }{ 2 } ,\frac { 11+3 }{ 2 } \right) =(6,7)\)

n(S) = 7000
S - Total no. of lights.
n(A) = 25
A - Defective ones.
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 25 }{ 7000 } =\frac { 1 }{ 280 } \).

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

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

10.

pr+qr+pq+p²= p(p+r)+q(p+r)
= (p+r)(p+q)

11.

Total mass = 5.97\(\times\) 10²⁴ kg + 0.073 \(\times\) 10²⁴ kg
= (5.97 + 0.073) \(\times\) 10²⁴ kg
= 6.043 \(\times\) 10²⁴ kg

12.

\({ 5 }^{ \frac { 1 }{ 2 } }\)

13.

\({7\over 16}={7\over 2^4}={7\over 2^4\times 5^6}\)
\(\therefore {7\over 16}\) has a terminating decimal expansion.

14.

A′\(\cup \) B′ = {a, c, e, g} \(\cup \) {b, c, f, g}
{a, b, c, e, f, g}

15.

Distance between the points (3,4) and (- 7, 2)

= \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } } \)
= \(\sqrt { (-7-3)^{ 2 }+(2-4)^{ 2 } } =\sqrt { (-10)^{ 2 }+(-2)^{ 2 } } \)
= \(\sqrt { 100+4 } =\sqrt { 104 } =\sqrt { { 2 }^{ 2 }\times 2\times 13 } \)
= \(2\sqrt { 26 } \) units

16.

It is a quadrilateral

Multiple Choice Question

14 x 1 = 14

17.

(d)
\(25\sqrt { 3 } \) cm ²

18.

(d)
Probability

19.

(b)
2

20.

(b)
5x − 7 = 6 − 2x

21.

(d)
5 :2

22.

(b)
\(\cfrac { { 3 }^{ \frac { 2 }{ 3 } } }{ 3 } \)

23.

(a)
1: 1

24.

(b)
13

25.

26.

27.

(a)
0

28.

(a)
AΔB

29.

(d)
\(\sqrt{17}\)

30.

Generic 3

13 x 3 = 39

31.

Let ‘a’ be the side of the cube.
Given that, total surface area = 864 cm²
6a²= 864
a²= \(\frac{864}{6}\)
a²= 144
Therefore, side (a) = 12 cm
Now, volume of the cube = a³
= 123 = 12 ×12 ×12 = 1728 cm³

32.

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

33.

Let A(x, 3), B(6, y), C(8, 2) and D(9, 4) be the vertices of the parallelogram ABCD. By definition, diagonals AC and BD bisect each other.

Mid-point of AC = Mid-point of BD
\(\left( \frac { x+8 }{ 2 } ,\frac { 3+2 }{ 2 } \right) =\left( \frac { 6+9 }{ 2 } ,\frac { y+4 }{ 2 } \right) \)
equating the coordinates on both sides, we get
\(\frac { x+8 }{ 2 } =\frac { 15 }{ 2 } \)
x + 8 = 15
x = 7
\(\frac { 5 }{ 2 } =\frac { y+4 }{ 2 } \)
5 = y + 4
y = 1
Hence, x = 7 and y = 1.

34.

In the given right angled triangle, note that for the given angle \(\theta \), PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { PR }{ QR } =\frac { 35 }{ 37 } \)
\(cos\theta =\frac { adjacent\ side }{ hypotenuse } =\frac { PQ }{ QR } =\frac { 12 }{ 37 } \)
\(tan\theta =\frac { opposite\ side }{ adjacent\ side } =\frac { PR }{ PQ } =\frac { 35 }{ 12 } \)
It is enough to leave the ratios as fractions. In case, if you want to simplify each ratio neatly in a terminating decimal form, you may opt for it, but that is not obligatory.

35.

Let p (x) = x³+ 6x²- 7x - 60
By factor theorem (x+4) is a factor of p(-4) = 0
p(-4) = (-4)³+ 6(-4)²-7(-4) -60 = -64 + 96 + 28 - 60 = 0
Therefore,(x+4) is a factor of x³+ 6x²+ (-7x) -60

36.

(60000000)³ = \(\left(6.0 \times 10^{7}\right)^{4}=(6.04)^{4} \times\left(10^{7}\right)^{4}\)
\( =1296 \times 10^{28}=1.296 \times 10^{3} \times 10^{28} \\ =1.296 \times 10^{31} \)

37.

11, 12,14,18, x+12, x+4, 30, 32, 35, 41
The number of values = 10,which is an even number
Median = Average of,\(\left( \cfrac { 10 }{ 2 } \right) ^{ th }\) and \(\left( \cfrac { 10 }{ 2 } +1 \right) ^{ th }\)
\(24=\cfrac{x+2+x+4}{2}\)
\(24=\cfrac { 2x+6 }{ 2 } =\cfrac { 2\left( x+3 \right) }{ 2 } \)
\(\therefore\) x+3 = 24
x = 24-3 = 21

38.

(i) 3\(\sqrt{7}\) + 5\(\sqrt{7}\) = (3 + 5)\(\sqrt{7}\) = 8\(\sqrt{7}\). The answer is irrational.
(ii) 7\(\sqrt{5}\) - 4\(\sqrt{5}\) = (7 - 4)\(\sqrt{5}\) = 3\(\sqrt{5}\). The answer is irrational.

39.

K = {a,b,d,e,f}, L = {b,c,d,g} and M = {a,b,c,d,h}
(i) K\(\cup\)(L\(\cap \)M)
L\(\cap \) M = {b,c,d,g} \(\cap \) {a,b,c,d,h} = {b,c,d}
K\(\cup\)(L\(\cap \)M) = {a,b,d,e,f} \(\cup\) {b,c,d} = {a,b,c,d,e,f}
(ii) K∩(L\(\cup\)M)
L\(\cup\)M = {a,b,c,d,g,h}
K⋂(L \(\cup\) M) = {a,b,d,e,f) \(\cap \) {a,b,c,d,g,h} = {a,b,d}
(iii) (K \(\cup\) L) \(\cap \) (K\(\cup\)M)
K \(\cup\) L = {a,b,c,d,e,f,g}
K \(\cup\) M = {a,b,c,d,e,f,h}
(K \(\cup\) L) \(\cap \) (K \(\cup\) M) = {a,b,c,d,e,f}
(iv) (K \(\cap \) L) \(\cup\) (K \(\cap \) M)
(K \(\cap \) L) = {b,d}
(K \(\cap \) M) = {a,b,d}
(K \(\cap \) L) \(\cup\) (K \(\cap \) M) = {b,d} \(\cup\) {a,b,d} = {a,b,d}

40.

(i) A = {1,3,5,7} ; B = {1,2,4,6,8}
A-B = {1,3,:5,7} - {1,2,4,6,8}
= {3,5,7}
B -A = {1,2,4,6,S} - {1,3,5,7}
= {2,4,6,8}
AΔB = (A - B)U(B-A)
= {3,5,7} U {2,4,6,8}
= {2,3,4,5,6,7,8}.
(ii) BM = (B-A) U (A-B)
= {2,4,6,8} u {3,5,7}
= {2,3,4,516,7,S}

41.

∴ Quotient = x² – 4x + 8 and Remainder = –4x – 15

42.

Let \(2.\overline { 327 } \) = 2.327327327 ..... \(\rightarrow\) (1)
Here period of decimal is 3, multiply equation (1) by 1000
1000 x = 2327.327327 ..... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 1000x - x = 999x = 2325
\(x=\frac{2325}{999}\) (or) \(775\over 333\)

43.

Distance AB = \(\sqrt { (a-a)^{ 2 }+(3+2)^{ 2 } } \)
= \(\sqrt { 0+{ 5 }^{ 2 } } =\sqrt { 25 } =5\)
BC = \(\sqrt { (a-a)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { 0+9 } =\sqrt { 9 } =3\)
AC = \(\sqrt { (a-a)^{ 2 }+(0+2)^{ 2 } } \)
= \(\sqrt { 0+2^{ 2 } } =\sqrt { 4 } =2\)
AC + BC = AB ⇒ 2 + 3 = 5
∴ The given three points are collinear

Generic 5

3 x 5 = 15

44.

P (x) = 5x³ + 7x² + 3x + 2
d(x) = 3x + 2
Standard form of p (x) = 5x³+ 7x² + 3x + 2
and d (x) = 3x + 2

5x³ + 7x² + 3x + 2 = \(\left( x+\cfrac { 2 }{ 3 } \right) \left( 5{ x }^{ 2 }-{ \cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } } \right) +\cfrac { 44 }{ 27 } =\left( \cfrac { 3x+2 }{ 3 } \right) \left( 5x2-\cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( \cfrac { 3x+2 }{ 3 } \right) 3\left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( 3x+2 \right) \left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
Hence the quotient \(\cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 21 } \) and remainder IS \(\cfrac { 44 }{ 27 } \)

45.

\(\angle\)In the cyclic quadrilateral \(\angle\) A+\(\angle\)C = 180^o
y + 4^o + 3y^o+ 8^o = 180^o
4y^o + 12^o = 80^o
4y^o = 180^o -12^o = 168
y =\(\cfrac { 168 }{ 4 } \) = 42
\(\angle\)B + \(\angle\)D = 180^o
8x + 12 = 180^o
8x = 180^o- 12^o = 168
x = \(\cfrac { 168 }{ 8 } \) = 21^o
\(\therefore\) \(\angle\)A = y + 4^o = 42^o + 4^o = 46^o
\(\angle\)C = 3y + 8 = 3 \(\times\) 42 + 8 = 126 + 8 = 134^o
\(\angle\)B = 3x + 6 = 3 \(\times\) 21 + 6 = 63 + 6 = 69^o
\(\angle\)D = 5x + 6 = 5 \(\times\) 21 + 6
= 105 + 6 = 111^o

46.

(i) (p-q)²- 6(p - q) -16
= (p - q)²- 8(p - q) +2(p - q) -16
= (p - q)((p - q) -8)+2((p - q)-8)
= (p - q - 8)(p - q + 2)

(ii) m²+ 2mn - 24n²
= m²+ 6mn - 4mn - 24n²
= m(m + 6n) - 4n(m + 6n)
= (m + 6n)(m - 4n)

(iii) √5a²+2a-3√5
=√5a²+2a-3√5
=√5a(a+√5)-3(a+√5)
=(a+√5)(√5a-3)

(iv) a⁴- 3a²+2
= a⁴-2a²-1a²+2
= a²(a²-2)-1(a²-2)
= (a²-2)(a²-1)=(a²-2)(a+1)(a-1)

(v) 8m³-2m²n-15mn²
= m(8m²-2mn-15n²)
= m(8m²-12mn+10mn-15n²)
= m(4m(2m-3n)+5n(2m-3n))
= m(4m+5n)(2m-3n)

(vi) \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\)
=\((\frac{1}{x})^2+2(\frac{1}{x})(\frac{1}{y})+(\frac{1}{y})^2\)
=\(\left({1\over x}+{1\over y}\right)^2\)