Solved AP 10th Class Model Papers for Maths

Anshita Shandilya 11:18 am

Jul 16 • Board Sample Papers • 13741 Views • 53 Comments on Solved AP 10th Class Model Papers for Maths

AP Board of Education organize SSC/OSCC public examination twice in a year in which examination schedule for the classes of 10th and 12th  is mentioned. It conducts exam every year in the month of March and April and famous for its quality education. In the following Solved AP 10th Class Model Papers for Maths, I have tried to cover all important topics of Maths for your quick revision. As an Engineer, I am setting this Solved AP 10th Class Model Papers for Maths  students by taking help from text book based on syllabus and previous year sample papers. You can take help from Class 10th formulae of Maths for your better preparations.

AP 10th Class Model Papers for Maths

10th Solved Maths Sample Paper

General Instruction:
(A) This paper consists of two parts namely Part-A and Part-B
(B) Part -A include four sections for time duration of 2 hours and 35 marks as total.
(C) Part-B is for time duration 1/2 hour and contain 15 marks as total.
(D) Marks are allotted to respective questions.

PART – A.                                               Time: 2 Hours PART- A Marks: 35

Note: Answer any  FIVE questions choosing at least TWO from each group. Each question carries TWO marks.

GROUP – A      2 = 10

1. Find the [HCF X LCM] for the numbers 100 and 190.
Solution: 
100 = 52 X 22
190 = 2 X 5 X 19
∴ H.C.F. (100, 190) = 2 X 5 = 10
L.C.M(100, 190) = 1900
∴ H.C.F. X L.C.M. = 10 X 1900= 19000

2. If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1)x – 1, then find the value of a.
Solution:
Given that 1 is a zero of a polynomial
Therefore,
p(1) = 0
a(1)2 – 3 (a – 1) (1) – 1 = 0
a – 3(a – 1) – 1 = 0
a – 3a + 3 – 1 = 0
-2a + 2 = 0
2a = 2
a = 1
Hence value of a is 1

3. In ΔLMN, ∠L = 50° and ∠N = 60°. If Δ LMN ~ ΔPQR, then find ∠Q.
Solution:
In ΔLMN,  ∠L + ∠M+ ∠N = 180° ( sum of 3 angles in a Δ is 180°)
50° + ∠M + 60° = 180°
∠M = 70°
Since ΔLMN ~ΔPQR
∠M = ∠Q = 70°
Hence ∠Q is 70°

GROUP – B

 4. If sec2θ (1 + sin θ)(1 – sin θ) = k, then find k.
Solution:
sec2θ (1 + sin θ) (1 – sin θ) = k
sec2θ (1 – sin2) = k
sec2(cos2θ) = k
1 = k
∴ k = 1

5. If the diameter of a semicircular protractor is 14  cm, what is its perimeter.
Solution:
Diameter = 14 cm
Radius
Perimeter is = pr+ 2r = 36 cm

SECTION – II           

Note: Answer any FOUR. Each question carries ONE mark.    1 = 4

6. Find the number of solutions of the following pair of linear equations:x + 2y – 8 =0 and 2x + 4y =1
Solution: Infinitely many solutions

7. Find the discriminant of the equation .
Solution:  Discriminant =  100 – 36 = 6

8. If one of the roots of the quadratic equation  is 4, then the value of p is:
A. 7
B. -7
C. 5
D. 8

9. If the 17th term of an AP exceeds the 10th term by 7, then the common difference is:
A. 3
B. 2
C. 1
D. 4

SECTION – III  

Note: Answer any FOUR questions. Each question carries FOUR marks.   4 *4 = 16

 10. Find all the zeroes of the polynomial x3 + 3x2 – 2x – 6.
Solution:
Let P(x) = x3 + 3x2 – 2x – 6
Given two zeroes of the given polynomial P(x) = x3 + 3x2 – 2x – 6  are  and 
→ x2 – 2  is a factor.
Now, we apply the division algorithm to P(x) and x2 – 2
 → P(x) = x3 + 3x2 – 2x – 6 = (x2 – 2) (x + 3)
→ x + 3 is a factor of P(x)
∴  -3 is the other zero of the polynomial.

11. In the A.P. 3, 15, 27, 39, … which term will be 120 more than the twenty first term?
Solution:
Given A.P. is  3, 15, 27, 39, …
Here a = 3, d = t2 – t1 = 15 – 3 = 12
nth term of A.P. = tn = a + (n – 1)d
Put n = 21
→ t21 = 3 + (21 – 1)12
→ t21 = 3 + (20)12
→ t21 = 243
Let tm be the term which is 120 more than 21st term
→ tm = t21 + 120
→ tm­ = 243 + 120
→ a + (m – 1)d = 363
→ 3+ (m – 1)12 = 363
→  (m – 1)12 = 363 – 3= 360
→ m – 1 = 30
→ m = 31
Hence, the 31st term of the A.P. is 120 more than its twenty first term.     

 12.  In the figure below, Δ ABD is a right triangle, 90 degree at A and AC ^ BD. Prove that   AB2 = BC .
Solution:
In ΔABC,
∠BAC + ∠CBA = 90° … (1)
In ΔABD
∠BDA + ∠CBA = 90° … (2)
From (1) and (2)
∠BAC = ∠BDA
In ΔBAC and  ΔBDA
∠BAC = ∠BDA
∠ACB = ∠DAB [Each is equal to 90°]
∴ ΔBAC ~ ΔBDA [A.A similarity criterion]
[ sides are proportional]
→ BA2 = BC . BD
→ AB2 = BC . BD

GROUP – B
 
13. Calculate tan 60 practically.
Solution:
ABC is an equilateral triangle.
So each angle is 60°,
therefore ∠A = ∠B = ∠C = 60°
A perpendicular AD from A to BC is drawn.
Now in triangles ABD and ACD, AB = AC
[ ΔABC is equilateral]
AD = AD = common
∠ADB = ∠ADC = 90°
∴ ΔADB @ ΔADC [RHS rule]
∴BD = DC [CPCT]

Now, AD2 = AB2 – BD2

SECTION – IV 

Note: This question carries FIVE marks. × 5 = 5

14. The sum of two numbers is 8. if the sum of their reciprocals is What are the nos.
Solution:
Given sum of two numbers = 8
Sum of their reciprocals =
Let the numbers be x and y
∴ x + y = 8 ……..(1)
And
∴  xy = 15
Substitute x value in equation (1)
→ y2 – 8y + 15 = 0
→ y2 – 5y – 3y + 15 = 0
→ y(y – 5) – 3(y – 5) = 0
→  (y – 3) (y – 5) = 0
→ y = 3 or 5
Case (i) If y = 3,
∴ x = 5 and y = 3
Case (ii): If y = 5,
∴ x = 3 and y = 5

PART – B                                                                                                Time: 30 mins Marks: 15

Note:  Each question carries 1/2 mark.

15.  Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Thereafter draw another triangle where sides are  times the corresponding sides of the first triangle.
Solution
 Steps of construction
1. Draw a right triangle using given conditions. Consider the triangle as ΔABC in which BC = 8 cm and AB = 6 cm, and ∠B = 90°
2. Make any acute angle ∠CBX below the line BC
3.Locate 4 points (greater of 3 and 4 in ) B1, B2, B3, B4 on BX such that BB1 = B1B2 = B2B3 = B3B4
4 . Join B4 and C
5. Through B3, draw a line parallel to B4C meeting BC in C’.
6. Again draw a line through C’ parallel to CA meeting BA in A’ on being. ΔA’BC’ is the    required triangle whole sides are ¾ times the sides of ΔABC.

17. In the figure below, M is mid-point of CD. The line BM intersects AC at L and AD produced at E.  Prove that EL = 2 BL.
Solution: In Δ’s DME and BMC
DM = MC[ M is midpoint of DC]
∠DME = ∠BMC (vertically opposite angles)
∠MDE = ∠MCB [ alternate angles]
ΔDME @ ΔBMC [ASA congruence rule]
∴ BC = DE (CPCT)
But BC = AD [ ABCD is a parallogerm]
AE = AD + DE
= BC + BC
AE = 2 BC — (1)
In Δ’s BLC and ALE
∠BLC = ∠ALE [vertically opposite angles]
∠CBL = ∠AEL [alternate angles]
ΔBLC ~ ΔALE [AA similarity criterion]
∴ EL = 2 BL

18. Find the area of the quadrilateral ABCD whose vertices are A(-4, -2), B(-3, -5),  C(3, -2) and  D(2,3)
Solution:
A(-4, -2) ,B(-3, -5), C(3, -2) and D(2, 3).
Area of the quadrilateral
ABCD = Area of ΔABD + area of ΔBDC
∴ Area of Δ BDC =
∴ Area of quadrilateral ABCD =

19. If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the equation p(x2 +x) + k = 0 has discriminant zero, find  p and k.
Solution:
Given that -5 is a root of 2x2 + px – 15 = 0
∴2(-5)2 + p(-5) – 15 = 0
50 – 5p – 15 = 0
5p = 35
p = 7
Putting p = 7,we get
7x2 + 7x + k = 0
Since above equation has discriminant zero, b2 – 4ac = 0
∴ 49 – 4(7)k = 0  ( a = 7, b = 7, c = k)
28k = 49

20. Prove that the lengths of the tangents drawn from an external point to a circle are equal.Using the above theorem, prove than If quadrilateral ABCD is circumscribing a circle, then AB + CD = AD + BC
Solution: Ist part:
CB = CD (Radii of the same circle)
AC = AC (common)
∠CBA=∠CDA=90°
ΔCBA @ ΔCDA (RHS)
→ AB = AD (CPCT)
Hence the tangents are equal.
IInd part:
The tangents drawn from an external point to a circle are equal and
A is a point outside the circle and AP, AS are the tangents to the circle
→ AP = AS …..(1)
BP = BQ ……….(2)
CR = CQ …………..(3)
And DR = DS ……..(4)
Adding,  (AP + BP) + (CR + DR) = AS + BQ + CQ + DS
→ AB + CD = (AS + DS) + (BQ + CQ)
→AB + CD = AD + BC

 21.A plane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the length of ground between the two planes at that instant.
Solution:
Let at some instant, A1 be the 1st aero plane A2 be the 2nd plane A1A2 is the vertical distance between these plane at that instant.
Let, BC = x m and A1 A2 = h m
From the ΔA1BC we have,
From ΔA2BC we have,
→ h = 3 X 3125 – 3125
→ h = 2 X 3125 = 6250
Hence the length of ground between the two planes at that instant is 6250 m.

For more Maths AP 10th Class Model Papers in pdf,follow:

  1. Maths Sample Paper for Class 10 AP Board 2012-13
  2. AP 10th Class Model Papers for Maths 2011-12

For more AP 10th Class Model Papers, follow:

  1. Chemistry Sample Papers for Class 10
  2. Physics Sample Paper for Class X 
  3.  English Sample Paper of Class X
  4. Physical Science Sample paper for Class 10
  5. Sample Paper Social Science for Class 10
  6.  State Board Previous years English Papers

If you have any suggestion about the content of this paper, than feel free to give below in the commenting section.
Best of luck for your exam preparations..!!
 

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53 Responses to Solved AP 10th Class Model Papers for Maths

  1. Madhuri says:

    Maths

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