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

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Questions:

Generic 3

14 x 3 = 42

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?

Find the value of cos19⁰59'

The sides of a triangular park are in the ratio 9:10:11 and its perimeter is 300 m. Find the area of the triangular park.

Expland the following using identities (4a + 3b) (4a - 3b)

Write in scientific notation ;(60000000)³

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.

In the given figure, ㄥCAB = 25°, find ㄥBDC, ㄥDBA and ㄥCOB

Can you reduce the following numbers to surds of same order :
(i) \(\sqrt{3}\)
(ii) \(\sqrt [ 4 ]{ 3 } \)
(iii) \(\sqrt [ 3 ]{ 3 } \)

In the adjacent diagram, if n(U) = 125, y is two times of x and z is 10 more than x, then find the value of x,y and z.

10.

Write the following polynomials in standard form.

S.No	Polynomial	Standard form
1	5m⁴ -3m + 7m² + 8
2	\(\frac { 2 }{ 3 } y+{ 8y }^{ 3 }-12+\sqrt { 5 } { y }^{ 2 }\)
3	12p² -8p⁵ -10p⁴ -7

11.

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

12.

Show that the following points taken in order form an equilateral triangle in each case \(A\left( 2,2 \right) ,B\left( -2,-2 \right) ,C\left( -2\sqrt { 3 } ,2\sqrt { 3 } \right) \)

13.

Express the following decimal expression into rational numbers \(3.1\overline { 7 } \)

14.

Find the distance between the points (–4, 3), (2, –3).

Generic 2

13 x 2 = 26

15.

Find the value of \(\frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}}\)

16.

If 3 (tan \(\theta\)) + 4 (sec \(\theta\) \(\times\) sin 6) = 24. Then find all the trigonometric ratios of the angle \(\theta\)

17.

Find the TSA and LSA of a cuboid whose length, breadth and height are 10 cm, 12 cm and 14 cm respectively.

18.

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.

19.

Find the centroid of the triangle whose vertices are (2, -5), (5, 11) and (9, 9)

20.

In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.

21.

Find any 3 irrational numbers between 0.12 and 0.13.

22.

Use a fractional index to write \(\sqrt [ 3 ]{ 7 } \)

23.

In a rice mill, seven labours are receiving the daily wages of Rs. 500, Rs. 600, Rs. 600, Rs. 800, Rs. 800, Rs. 800 and Rs. 1000, find the modal wage

24.

Expand the following using identities: (2x - 3b)²

25.

Express the following in the form 2ⁿ:
\(\frac{1}{4}\)

26.

Identify the following sets as null set or singleton set.
B = The set of all even natural numbers which are not divisible by 2

27.

Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(0, –4) (0, –2) (0, 4) (0, 5)

Multiple Choice Question

14 x 1 = 14

28.

The lateral surface area of a cube of side 12 cm is _______.

(a)

144 cm²

(b)

196 cm²

(c)

576 cm²

(d)

664 cm²

29.

The probability of an event cannot be _______.

(a)

Equal to zero

(b)

Greater than zero

(c)

Equal to one

(d)

Less than zero

30.

if sin 30⁰ = x and cos 60⁰ = y, then x² + y² is________.

(a)

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

(b)

(c)

sin90°

(d)

cos90°

31.

If U = {x : x\(\in \)W and x < 20}, A = {2,4,6,8}, B = {6,8,12,14} then\(\left[ n\left( A\cup B^{ ' } \right) \right] \)

(a)

15%

(b)

20%

(c)

10%

(d)

32.

Which of the following is not an irrational number?

(a)

\(\sqrt { 2 } \)

(b)

\(\sqrt { 5 } \)

(c)

\(\sqrt { 3 } \)

(d)

\(\sqrt { 25 } \)

33.

If the mean of five observations x, x+2, x+4, x+6, x+8, is 11, then the mean of first three observations is __________

(a)

(b)

(c)

(d)

34.

The mean of the first 10 prime numbers is ___________

(a)

12.6

(b)

12.7

(c)

12.8

(d)

12.9

35.

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

(a)

(b)

(c)

(d)

36.

What is the name of a regular polygon of six sides?

(a)

Square

(b)

Equilateral triangle

(c)

Regular hexagon

(d)

Regular octagon

37.

A point on the y-axis is ________________

(a)

(1, 1)

(b)

(6,0)

(c)

(0,6)

(d)

(-1, -1)

38.

Which one of the following is an irrational number.

(a)

\(\sqrt { 25 } \)

(b)

\(\sqrt { \frac { 9 }{ 4 } } \)

(c)

\(\frac { 7 }{ 11 } \)

(d)

\(\pi\)

39.

The type of the polynomial 4–3x³is ________.

(a)

constant polynomial

(b)

linear polynomial

(c)

quadratic polynomial

(d)

cubic polynomial.

40.

Point (0, –7) lies ________

(a)

on the x-axis

(b)

in the II quadrant

(c)

on the y-axis

(d)

in the IV quadrant

41.

From the adjacent diagram n[P(AΔB)] is ________.

(a)

(b)

(c)

(d)

Generic 5

4 x 5 = 20

42.

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

43.

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

44.

Verify x³+y³+z³-3xyz = \(\frac{1}{2}\)[x+y+z][(x-y)²+(y-z)²+(z-x)²]

45.

In the given Fig, ∠A = 64° , ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°.

Generic 8

1 x 8 = 8

46.

Draw a right triangle whose hypotenuse is 10 cm and one of the legs is 8 cm. Locate its incentre and also draw the incircle.

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Answers Key:

Generic 3

14 x 3 = 42

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

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

Given the sides are in the ratio 9:10:11, let the sides be 9k, 10k, 11k
The perimeter of the triangular park = 300 m
9k +10k +11k = 300m
30k = 300
k = 10m
Therefore, the sides are a = 90 m, b = 100 m, c = 110 m
s=\(\frac{1+b+c}{2}=\frac{90+100+110}{2}=\frac{300}{2}\)= 150 m
Hence, Area of triangular park = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 150\times (150-90)(150-100)(150-110) } \)
= \(\sqrt { 150\times 60\times 50\times 40 } \)
= \(\sqrt { 3\times 50\times 20\times 3\times 50\times 2\times 20 } \)
= 50 \(\times\) 20 \(\times\) 3\(\sqrt{2}\)
= 3000 \(\times\)1.414 = 4242 m²

(4a+3b) (4a-3b) = (4a)²- (3b)²= 16a²- 9b²[We have (a+b)(a-b) = a²- b²] Put [a = 4a, b = 3b]

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

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

(i)\(\angle\)CAB = 25⁰
\(\therefore\) \(\angle\)BDC = 25⁰
(ii) \(\angle\)DBA = \(\angle\)DCA =180-(90+25⁰)
= 180⁰- 115⁰
= 65
(iii)\(\angle\)COB = 2\(\angle\)CAB = 2\(\times\)25⁰= 50⁰

(i) \( \sqrt{3} =3^{\frac{1}{2}} =3^{\frac{6}{12}} =\sqrt[12]{3^{6}} =\sqrt[12]{729} \)
(ii) \( \sqrt[4]{3} =3^{\frac{1}{4}} =3^{\frac{3}{12}} =\sqrt[12]{3^{3}} =\sqrt[12]{27} \)
(iii) \( \sqrt[3]{3} =3^{\frac{1}{3}} =3^{\frac{4}{12}} =\sqrt[12]{3^{4}} =\sqrt[12]{81} \)
The last row has surds of same order.

n(U) = 125
y = 2x
z = x + 10
ஃ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 - 45
4x = 80
x = 20
ஃ y = 2x = 2 \(\times\) 20 = 40
z = x +10 = 20 + 10 = 30
Hence x = 20; y = 40; z = 30

10.

S.No	Polynomial	Standard form
1	5m⁴ -3m + 7m² + 8	5m⁴+7m²-3m +8
2	\(\frac { 2 }{ 3 } y+{ 8y }^{ 3 }-12+\sqrt { 5 } { y }^{ 2 }\)	\({ 8y }^{ 3 }+\sqrt { 5 } { y }^{ 2 }+\frac { 2 }{ 3 } y-12\)
3	12p² -8p⁵ -10p⁴ -7	-8p⁵-10p⁴ + 12p²- 7

11.

(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}

12.

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB= \(\sqrt { (-2-2)^{ 2 }+(-2-2)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(-4)^{ 2 } } =\sqrt { 16+16 } =\sqrt { 32 } \)
BC = \(\sqrt { (-2\sqrt { 3 } +2)^{ 2 }+(2\sqrt { 3 } +2)^{ 2 } } \)
= \(\sqrt { 12+4-8\sqrt { 3 } +12+4+8\sqrt { 3 } } \)
= \(\sqrt { 16+16 } =\sqrt { 32 } \)
AC = \(\sqrt { (-2\sqrt { 3 } -2)^{ 2 }+(2\sqrt { 3 } -2)^{ 2 } } \)
= \(\sqrt { 2(\sqrt { 3 } +2)^{ 2 }+2(\sqrt { 3 } -2)^{ 2 } } \)
= \(\sqrt { 12+4+8\sqrt { 3 } +12+4-8\sqrt { 3 } } \)
= \(\sqrt { 16+16 } =\sqrt { 32 } \)
AB = BC = AC (Three sides are equal)
∴ ABC is an equilateral triangle.

13.

Let x = \(3.1\overline { 7 } \) = 3.1777 ....... (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 ......... (2)
(2)-(1)

\( =\frac{143}{45} \)

14.

The distance between the points (-4, 3), (2, -3) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2+4 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( { 6 }^{ 2 }+\left( -6 \right) ^{ 2 } \right) } =\sqrt { \left( 36+36 \right) } \)
\(=\sqrt { 36\times 2 } \)
\(=6\sqrt { 2 } \)

Generic 2

13 x 2 = 26

15.

\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)

16.

3 tan \(\theta\) + 4 (sec \(\theta\) \(\times\) sin\(\theta\)) = 24
\(3tan\theta +4\left( \cfrac { 1 }{ cos\theta } \times sin\theta \right) =24\)
7 tan \(\theta\) = 24
\(tan\theta =\cfrac { 24 }{ 7 } \)
hypetenuse = \(\sqrt { { 24 }^{ 2 }+49 } =\sqrt { 576+49 } =\sqrt { 625 } =25\)
\(sin\theta =\cfrac { 24 }{ 25 } ;cos\theta =\cfrac { 7 }{ 25 } ;cosec\theta =\cfrac { 25 }{ 24 } ;sec\theta =\cfrac { 25 }{ 27 } ;cot\theta =\cfrac { 7 }{ 24 } \)

17.

TSA = 2 (lb + bh + lh)
= 2 (10 \(\times\)12 + 12 \(\times\) 14 +10 \(\times\) 14)
= 2 (120 + 168 + 140)
= 856 cm²
LSA = 2 (bh + lh)
= 2 (12 \(\times\) 14 + 10 \(\times\) 14)
= 2 (168 + 140)
= 2 (308) = 616 cm².

18.

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

19.

\(G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =\left( \frac { 2+5+9 }{ 3 } ,\frac { -5+11+9 }{ 9 } \right) =\left( \frac { 16 }{ 3 } ,5 \right) \)

20.

No. of youngsters = 400
n(S)
No. of youngsters having voter id = 191
n(A)
No. of youngsters do not have their voter id
n(B) = 400 - 191 = 209
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 209 }{ 400 } \).

21.

Three irrational numbers between 0.12 and 0.13 are 0.12010010001…, 0.12040040004…, 0.12070070007…

22.

\(\sqrt [ 3 ]{ 7 } \) = \({ 7 }^{ \cfrac { 1 }{ 3 } }\)

23.

In the given data Rs. 800 occurs thrice. Hence the mode is Rs. 800

24.

(2x - 3b)² [we have (a - b)²= a²- 2ab + b²]
(2a - 3b)²= (2a)²-2(2a)(3b)+(3b)² put [a = 2a, b = 3b]
= 4a²-12ab + 9b²

25.

\(\frac { 1 }{ 4 } =\frac { 1 }{ 2\times 2 } =\frac { 1 }{ { 2 }^{ 2 } } \) = 2^-2

26.

The set B is a Null set

27.

The line is on the y - axis

Multiple Choice Question

14 x 1 = 14

28.

29.

(d)
Less than zero

30.

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

31.

(b)
20%

32.

(d)
\(\sqrt { 25 } \)

33.

(a)
9

34.

(d)
12.9

35.

(a)
0

36.

37.

38.

(d)
\(\pi\)

39.

(d)
cubic polynomial.

40.

41.

Generic 5

4 x 5 = 20

42.

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)

43.

\(\angle\)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

44.

R.H.S = \(\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]\)
= \(\frac{1}{2}(x+y+z)[x^2-2xy+y^2+y^2-2yz+z^2-2zx+x^2]\)
= \(\frac{1}{2}[(x+y+z)[2x^2+2y^2+2z^2-2xy-2yz-2zx^2]]\)

= x³+y³+z³-3xyz = L.H.S
Hence it is verified.

45.

In the given \(\triangle ABC\)
\(\angle A=64^0\ and\ \angle B=58^0\)
\(\angle C=180^0-(64^0+58^0)\)
= 180° - 122°
= 58°
Since OC is the bisector of \(\angle C\)
\(y=\frac{58^0}{2}=29^0\)
Given \(\triangle OBC\)
\(\angle OBC=\frac{58^0}{2}=29^0\)
\(\angle OCB=29^0\)
\(\therefore\angle BOC=180^0-(29^0+29^0)\)
x = 180° - 58°
x = 122°
\(\angle x=122^0\ and\ \angle y=29^0\).

Generic 8

1 x 8 = 8

46.

Construction :
Step 1: Draw \(\triangle \)ABC with BC = 8 cm. AC = 10 cm with right angle at B.
Step 2: Construct angle bisectors of any two angles (B and C) and let them meet at I. I is the incentre.
Step 3: Draw perpendicular from I to any side of the triangle to meet BC at D.
Step 4: With I as centre, ID as radius draw the incircle, which touches all the three sides of the triangle internally. In radius = 1.9 cm.