# Solved AP 10th Class Model Papers for Maths

11:18 amAP Board of Education organize SSC/OSCC public examination twice in a year in which examination schedule for the classes of 10th and 12^{th } 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.

**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 5 * 2 = 10**

**1.**** Find the [HCF X LCM] for the numbers 100 and 190.**

**Solution: **

100 = 5^{2} X 2^{2}

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) = ax ^{2} – 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 sec ^{2}θ (1 + sin θ)(1 – sin θ) = k, then find k.**

**Solution:**

sec

^{2}θ (1 + sin θ) (1 – sin θ) = k

sec

^{2}θ (1 – sin

^{2}) = k

sec

^{2}(cos

^{2}θ) = 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. 4 * 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.** I**f 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 17 ^{th} term of an AP exceeds the 10^{th} 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 x ^{3} + 3x^{2} – 2x – 6.**

**Solution:**

Let P(x) = x

^{3}+ 3x

^{2}– 2x – 6

Given two zeroes of the given polynomial P(x) = x

^{3}+ 3x

^{2}– 2x – 6 are and

→ x

^{2}– 2 is a factor.

Now, we apply the division algorithm to P(x) and x

^{2}– 2

*→ P(x) = x*

^{3}+ 3x^{2}– 2x – 6 = (x^{2}– 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 = t_{2} – t_{1} = 15 – 3 = 12

n^{th} term of A.P. = t_{n} = a + (n – 1)d

Put n = 21

→ t_{21} = 3 + (21 – 1)12

→ t_{21} = 3 + (20)12

→ t_{21} = 243

Let t_{m} be the term which is 120 more than 21^{st} term

→ t_{m} = t_{21} + 120

→ t_{m }= 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 31^{st} 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 AB ^{2} = 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]

→ BA

^{2}= BC . BD

→ AB

^{2}= 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, AD^{2} = AB^{2} – BD^{2}

**SECTION – IV **

**Note: This question carries FIVE marks. 1 × 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)

→ y^{2} – 8y + 15 = 0

→ y^{2} – 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 ) B_{1}, B_{2}, B_{3}, B_{4} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}

4 . Join B_{4} and C

5. Through B_{3}, draw a line parallel to B_{4}C 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: I**n 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 2x ^{2} + px – 15 = 0 and the equation p(x^{2} +x) + k = 0 has discriminant zero, find p and k.**

**Solution:**

Given that -5 is a root of 2x

^{2}+ px – 15 = 0

∴2(-5)

^{2}+ p(-5) – 15 = 0

50 – 5p – 15 = 0

5p = 35

p = 7

Putting p = 7,we get

7x

^{2}+ 7x + k = 0

Since above equation has discriminant zero,

*b*

^{2}– 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, A_{1} be the 1^{st} aero plane A_{2} be the 2^{nd} plane A_{1}A_{2} is the vertical distance between these plane at that instant.

Let, BC = x m and A_{1} A_{2} = h m

From the ΔA_{1}BC we have,

From ΔA_{2}BC 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:**

**For more AP 10th Class Model Papers, follow:**

**Chemistry Sample Papers for Class 10****Physics Sample Paper for Class X****English Sample Paper of Class X****Physical Science Sample paper for Class 10****Sample Paper Social Science for Class 10***State Board Previous years English Papers*

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* Best of luck for your exam preparations..!!
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