Solved Andhra Pradesh Board Maths Sample Paper for Class 9

Anshita Shandilya 5:15 pm

Jul 16 • Board Sample Papers • 17715 Views • 8 Comments on Solved Andhra Pradesh Board Maths Sample Paper for Class 9

The Andhra Pradesh Board of Secondary Education was established in 1953 and is known for its quality education standards not just in India but also abroad. It work as an autonomous body under the department of education.  It executes and supervise various activities like conducts exams, provides direction, support and leadership for the educational system. The  exams are conducted in the month of March and April. Here in this post, we have tried to made available Solved Andhra Pradesh Board Maths Sample Paper for Class 9. Solved papers gives better idea of Paper pattern and the right way to answer the question asked. While preparing this paper some help has been taken from few Evergreen sample papers. The questions are based on the syllabus of ANDHRA PRADESH BOARD.  You can revise Physics and Chemistry Formulae to save your time.

Maths Sample Paper for Class 9 Andhra Pradesh Board

Instructions to be followed: 

AP 9th Class Model Papers

Sample Papers for Class 9 Maths

(a) This Sample Consists of four sections as

  • Section A:1 to 10 Questions with 1 Mark allotment.
  • Section B:11 to 18 Questions with 3 Mark allotment.
  • Section C: 19 to 27 Questions with 4 Mark allotment.
  • Section D: 28 to 32 Questions with 5 Mark allotment.

(b) All questions are compulsory. 

(c) Marks are allotted to each question for your convenience.

SECTION A [ ONE MARK QUESTIONS ]

1. The coordinates of the point of intersection of the lines x = −8 and y = 5 are:
A. (8, 5)
B. (-8, 5)
C. (8, -5)
D. (-8, -5)

ANS:- D

2. x = 2 and y = 5 is a solution of:
A. x + 2y = 12
B. x + y = 12
C. 2x + y = 10
D. x – 2y = 10

ANS:- A

3. If the angles of a quadrilateral are in the ratio of 2 : 4 : 5 : 7, then their angle measure is:
A. 36 °, 58 °, 125 °, 141 °
B. 40 °, 80 °, 100 °, 140 °
C. 48 °, 72 °, 96 °, 144 °
D. 48 °, 52 °, 174 °, 86 °

ANS:- B

4. PQRS is a parallelogram with PR = 7.8 cm and QS = 6 cm. If PQ and QS intersect each other at O, then find the value of  OP and OQ?
A. 3.7 cm, 3 cm
B. 4.9 cm, 4 cm
C. 3.9 cm, 3 cm
D. 3 cm, 3 cm

ANS:- C

5. The region occupied by a simple closed figure in a plane is called:
A. Length
B. Volume
C. Perimeter
D. Area

ANS:- D

6. One-fourth of a circular disk is called a:
A. Semi-circle
B. Quadrant
C. Sector
D. Arc

ANS:- B

7. In the figure given here, if O is the centre of the circle, find the value of x
A. 65 degree
B. 50 degree
C. 60 degree
D. 70 degree

ANS:- A

8. ABCD is a trapezium in which AB || DC. Then ar (∆AOD)= ____.
A. ar(BOC)
B.ar (COD)
C. ar (BDC)
D. ar (ABD)

ANS:- A

9. If 5.5 cm is the edge of a cube  then its surface area is:
A. 172 m2
B. 181.5 m2
C. 160 m2
D. 150.2 m

ANS:- D

10. The probability of getting a number less than 5 in a single throw of a die is:
A. 1/6
B. 3/4
C. 2/3
D. 1/4

ANS:- C

SECTION B [ THREE MARK QUESTIONS ]

11. Write four solutions of the equation 3x + y = 4.

 SOL:- Given: 3x + y − 4 = 0, y = 4 − 3x
Putting x = 0, we get y = 4 − 3(0) = 4− 0 = 4
Putting x = 1, we get y = 4 −3(1) = 4− 3 = 1
Putting x = 2, we get y = 4−3(2) = 4− 6 = −2
Putting x = 3, we get y = 4 −3(3) = 4−9 = −5
Solutions of the given equation are (0, 4), (1, 1), (2, −2) and (3, −5).

12. In ΔABC, D and E are the mid-points of AC and BC, respectively. If DE = 11.5 cm, then find the length of AB.

SOL:- Given, DE = 11.5cm
By the mid-point theorem,
DE is parallel to AB and half of it
AB=2×DE
= 2 × 11.5
AB = 23 cm
Hence, the length of AB is 23 cm.

 13. ABCD is a parallelogram. P is a point on AD such that  AP = 1/3 AD.  If Q is a point on BC such that CQ = 1/3 BC  , then show that AQCP is a parallelogram.

SOL:- Given:
AP=1/3 AD —– (i)
CQ=1/3 BC —– (ii)
AD = BC ( Opposite sides of a parallelogram are equal)
Multiplying by 1/3 on both the sides
1/3 AD = 1/3BC
AP = CQ —– (iii) (from (i) and (ii))
AD is parallel to  BC
so  AP is parallel to CQ —– (iv)
From (iii) and (iv):
AQCP is a quadrilateral in which one pair of opposite sides AP and CQ are equal and parallel.
Hence, AQCP forms a parallelogram.

 14. In ∆ABC, AD is a median. Prove that ar (ΔABD) = ar (ΔACD).

SOL:- Given: AD is a median of ΔABC.
To prove: ar(ΔABD) = ar(ΔACD)

Proof:
ar(ΔABD) = ½ BD X AN  – – – – (1)
ar(ΔACD) = ½ CD X AN – – – – – (2)
We have BD = DC – – – – -(3) [∵AD is the median on side BC]
From equations (1), (2) and (3) we get
ar(ΔABD) = ar(ΔACD)
Hence proved.

 15. In the figure given here, PQR = 75 and PRQ = 35. Find QSR.

SOL:- Do it by yourself.

 16. The length of the sides of a triangle are 9 cm, 12 cm and 15 then what is its area.

SOL:- Area of the triangle
A = √s(s-a) (s-b) (s-c)
S = (a+b+c) / 2
Here,
=  (9+12+15) / 2
= 36 / 2 = 18 cm
A = √s(s-a) (s-b) (s-c)
= √18 (18-9) (18-12) (18- 15)
= √18 x 9 x 6 x 3
= 9 x 3 x 2
= 54 sq.cm
Hence, the area of the triangle is 54 sq. cm.

17. Find the total surface area of a hemispherical bowl which is 5 cm thick and has inner radius 35 cm.

SOL:- Let R be outer radius and r be the inner radius of the bowl.
Thickness of ring = 5 cm
R = ( r+5) cm
= 35 + 5
= 40 cm

Total surface area = CSA of outer hemisphere + CSA of inner hemisphere + Area of ring
= 2 ΠR² + 2Πr² + Π( R² – r²)
= Π [2R² + 2r² + ( R ² – r²)]
= 22/7 (3R² + r²)
= 22/7 (4800 + 1225)
= 22/7 (6025)
= 18935.71 sq. cm

18. A coin is tossed 500 times with the following frequencies:Head: 245, Tail: 255. Compute the probability of each event.

SOL:- Since the coin is tossed 500 times, the total number of trials is 500.
Let the event of getting a head be E  and  the event of getting a tail be F.
Then the number of trials of getting E  is 245.
And the number of trials of getting F is 255.
P (Getting a head)
P(E) = Number of trials which event E happens / Total number of trails= 245/500
=  49 /100
= 0.49

P (Getting a tail)
P (F) = Number of trials in which event F happens / Total number of trials= 255 / 500
=  0.51

SECTION C [ FOUR MARK QUESTIONS ]

19. Draw the graph of the equation 2x + 3y = 13, and determine whether x = 4, y = 2 is a solution not.

 SOL:- The given equation is 2x + 3y = 13.
y = (13 – 2x) / 3
If x = 2,
Y = ( 13 – 4) / 3 = 3
If x = 5
Y = (13 – 10) / 3 = 1
Thus, we have the following table for 2x + 3y = 13:

x

2

5

y

3

1

Plot the points A (2, 3) and B (5, 1) on a graph paper. Draw a line that passes  through points A and B.

Line AB represents the graph for 2x + 3y = 13. We can see that the point (4, 2) does not lie on the graph of the equation 2x + 3y = 13.
Hence, x = 4, y = 2 is not a solution of the equation 2x + 3y = 13.

20.Two opposite angles of a parallelogram measure (60 – x)° and (3x – 4)°. Find the measure of each angle of the parallelogram.

SOL:-  Let ABCD be a parallelogram, where ∠A = (3x – 4)° and ∠C = (60 – x)°.
Then ∠A = ∠C
= 3x – 4 = 60 – x
= 4x = 64
x = 16
∠A = (3 × 16 – 4)°
= (48 – 4)°
∠A = 44°
∠C = (60 – 16)° = 44°
Now, ∠A + ∠B = 180°
44° + ∠B = 180°
∠B = 180° – 44° = 136°
Also, ∠D = ∠B = 136°
Thus, the angles of the parallelogram are 44°, 136°, 44°, 136°.

 22. PQRS and ABRS are two parallelograms, and X is any point on side BR. Show that:

(i) ar(PQRS) = ar(ABRS)
SOL:=  i) Parallelograms PQRS and ABRS are on the same base RS and between the same parallels RS and BP.
ar (PQRS) = ar (ABRS).
Hence, ar(PQRS) = ar(ABRS).

(ii) ar(AXS) =

SOL:=   ΔAXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
Ar (ΔAXS) = ½ ar (ABRS)
Ar (ΔAXS) – ½ ar ( PQRS)
Hence, ar(AXS) =  ½ ar ( PQRS) .

 23. Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

SOL:- Steps of construction:

1. Draw a line segment BC of length 7 cm.
2. At B, construct ΔXBC = 75°.
3. With B as the centre and radius equal to 13 cm, draw an arc cutting XB at D.
4. Joint DC.
5. Draw PQ, the perpendicular bisector of DC, and let it intersect DB at A.
6. Join AC.
Then, ΔABC is the required triangle.

Justification:

Since A lies on the perpendicular bisector of DC, it is equidistant from points D and C.
AD = AC … (1)
Now, BD = 13 cm [By construction]
BA + AD = 13 cm
BA + AC = 13 cm [Using (1)]
Hence, ΔABC is the required triangle

24. A cylinder is 12 cm high, and the circumference of its base is 44 cm. Find its curved surface area and total surface area.

SOL:- Let r be the base radius and h be the height of the cylinder.
Then h = 12 cm.
Given: Circumference of base = 44 cm
⇒ 2 π r = 44 cm
= 2 X 22/7 X r = 44 cm
r  = 44 X 7/ 2 X 22
r = 7 cm … (1)
Curved surface area of a cylinder = 2Πrh
= 2 X 22/7 X 7 X 12
= 44 × 12
= 528 cm2

Total surface area a cylinder = 2Πrh + 2Πr2
= 2Πr(r + h)
= 44(7 + 12) [r = 7 cm, from (1)]
= 44(19)
= 836 cm2

25. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make ten such caps.

SOL:- Let r be the base radius, h the height, and l the slant height of the cone.
r = 7 cm and h = 24 cm
Slant height (l) = √r² + h²
= √7² + 24²
= √49 + 576
= √625
= 25 cm … (1)

Curved surface area of one cap = Πrl = 22/7 X 7 X 25 cm2 [Using (1)]
Curved surface area of 10 caps = 22/7 X 7 X 25 X 10 = 5500 cm2

Hence, the area of the sheet required for 10 caps is 5500 cm2.

 26. In a cricket match, a batsman hits a boundary 6 times off the 30 balls that he plays. Find the probability that he does not hit a boundary off a delivery.

SOL:- The batsman played 30 balls, so the total number of trials is 30.
He hits a boundary 6 times, so he did not hit a boundary (30 – 6) = 24 times.
Let E be the event that the batsman did not hit a boundary.
Then the number of trials in which event E happened = 24
P(Batsman did not hit a boundary) = P(E) =
Number of trials in which event E happened/ Total number of trials = 24/ 30 = 4 /5

27. Three coins are tossed simultaneously 200 times, with the following frequencies of different outcomes:

Outcome

3 heads

2 heads

1 head

No head

Frequency

23

72

77

28

Result

Sol:-  The three coins are tossed simultaneously 200 times.
Therefore, the total number of trials is 200.
Let E be the event of getting two heads
Then the number of trials in which event E happens = 72
P(Getting two heads) = P(E) = Number of trials in which event E happened/ Total number of trials
= 72 / 200
= 9 /25

SECTION D [ FIVE MARK QUESTIONS ]

28. Use this table to draw a graph.

x

–5

–1

3

b

13

y

–2

a

2

5

7

find the values of a and b.

SOL:- Plot the points A (–5, –2), B (3, 2) and C (13, 7).
Draw a straight line passing through A, B and C.
The y coordinate that corresponds  to the x coordinate – 1 is 0:a = 0
Through y = 5, draw a horizontal line that meets the graph at a point, say P. Through P, draw a vertical line that meets the X-axis. It will meet the X-axis at x = 9:b = 9

 23. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

SOL:-   Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. PR and SQ are joined. Join AC, PQ, QR, RS and SP.
In ΔABC, P is the mid-point of AB, and Q is the mid-point of BC.
PQ  AC and  [By mid-point Theorem] …. (1)
In Δ ADC, S is the mid-point of DA and R is the mid-point of DC. [Given]
SRAC and  [By mid-point Theorem] … (2)
From equations (1) and (2), we get
PQ  SR and PQ = SR
Thus, in quadrilateral PQRS, one pair of opposite sides is parallel and equal. PQRS is a parallelogram.
We know that the diagonals of a parallelogram bisect each other so PR and SQ bisect each other.
Hence, the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

30.In the figure given here, CDis parallel to AB and ΔCBA = 25°. Find x.

SOL:- Join AC and BE.
ΔACB = 90° [Angle inscribed in a semi-circle] …. (1)
In ΔACB, we have
∠ACB + ∠CBA + ∠CAB = 180° [Sum of angles of a triangle]
⇒ 90° + 25° + ∠CAB = 180° [Using (1) and ∠CBA = 25°, given]
∠CAB = 180° – 115 °
∠CAB = 65° … (2)
∠CEB = ∠CAB [Angles in the same segment] …. (3)
∠CEB = 65° [Using (2) and (3)] … (4)
CD is parallel AB (given) and CB is the transversa
∠DCB = ∠CBA [Alternate angles] …. (5)
∠CBA = 25° [Given] … (6)
∠DCB = 25° [Using (5) and (6)] …. (7)
∠DEB = ∠DCB [Angles in the same segment] … (8)
∠DEB = 25° [Using (7) and (8) …. (9)
Now x = ∠CED = ∠CEB – ∠DEB
x = 65° – 25° [Using (4) and (9)]
x = 40°
Hence, x = 40°.

 31. If each diagonal of a quadrilateral separates it into two triangles of equal areas, then show that the quadrilateral is a parallelogram.

Sol:- Let ABCD be the given quadrilateral. Diagonal AC separates it into two equal areas such that ar(ΔACD) = ar(CAB), while diagonal BD separates it into two equal areas such that ar(DAB) = ar(BCD).
Draw CP ⊥ AB produced, and DQ ⊥AB.
ar(quad ABCD) = ar(ΔCAB) + ar(ΔACD)]
ar(quad ABCD) = 2ar (ΔCAB) [ ar(ΔACD) = ar(ΔCAB)] ….. (1)
Also, ar (quad ABCD) = ar(ΔDAB) + ar(ΔBCD)
ar(quad ABCD) = 2ar (ΔDAB) [ ar(ΔDAB) = ar(ΔBCD) ….. (2)

From equations (1) and (2), we get
ar(ΔCAB) = ar(ΔDAB)
= 1/ 2 AB X CP = ½ ABX DQ
CP = DQ

Altitude from C of ΔABC = Altitude from D of ΔABD AND DC is parallel AB
Similarly, AD  is parallel BC. Hence, ABCD is a parallelogram.

 32. The difference between the semi-perimeter and the sides of triangle ABC are 8 cm, 7 cm and 5 cm, respectively. Find its semi-perimeter.

SOL:-  Let the semi-perimeter of the triangle be s. Sides of the triangle is  a, b and c.

Given:
s – a = 8
s – b = 7
s – c = 5

Adding all the equations, we get

(s – a) + (s – b) + (s – c) = 8 + 7 + 5
⇒ 3s – (a + b + c) = 20
⇒ 3s – 2s = 20 AND s = 20 cm

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8 Responses to Solved Andhra Pradesh Board Maths Sample Paper for Class 9

  1. Gadi Sureshu says:

    M Sc&B.Ed maths

  2. abhiram says:

    9th class

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