Andhra Pradesh SSC Board Question Paper for Class 10th Maths Paper 2 2018 In PDF

AP SSC or 10th Class Question Paper Mathematics Paper 2 English Medium 2018 with Solutions – Free Download

Andhra Pradesh SSC (Class 10) Maths 2018 question paper 2 with solutions are provided here in a downloadable pdf format and also in the text so that the students can easily get them. Along with the solutions, they can also download the Maths question paper 2 2018 class 10 SSC for reference. Students are open to access all the Andhra Pradesh board previous year Maths question papers here. 10th Class Mathematics Question Paper 2018 Paper 2 can be downloaded easily and students can practice and verify the answers provided by BYJU’S. Solving 2018 Maths question paper 2 for Class 10 will help the students to predict what type of questions will appear in the exam.

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Andhra Pradesh SSC Class 10th Maths Question Paper 2 With Solution 2018

QUESTION PAPER CODE 16E(A)

SECTION – I

(4 * 1 = 4)

Question 1: If A (4, 0), B (o, y) and AB = 5, find the possible values of y.

Solution:

Given :- A (4,0) and B (0,y), and AB = 5

To find:- y

Distance formula = AB = √(x₂ – x₁)² + (y₂ – y₁)²

=> 5 = √(0 – 4)² + (y – 0)²

=> 5 = √(-4)² + (y²)

=> 5 = √16 + y²

Squaring on both sides

(5)² = (√16 + y²)²

=> 25 = 16 + y²

=> y² = 25 – 16

=> y² = 9

=> y = √9

=> y = ± 3

The possible values of y are 3 and -3.

Question 2: A boy observed from the top of an electric pole at an angle of elevation of 30^o when the observation point is 10m away from the foot of the pole. Draw a suitable diagram for the above situation.

Solution:

Let the height of the electric pole AB be H.

Angle of elevation = 30^o

The distance of observer form the base of pole BC = 10 m

Base = 10

Perpendicular = H

tan θ = perpendicular / base

tan 30^o = H / 10

1 / √3 = H / 10

10 / √3 = H

So, Height of pole is 10 / √3 m

The figure is as follows:

Question 3: Find the value of tan² 45^o + cot² 30^o.

Solution:

tan² 45 + cot² 30

From the trigonometric values,

tan 45 = 1 and cot 30 = √3

Substitute,

tan² 45 + cot²30

= (tan 30)² + (cot 45)²

= (1)² + (√3)²

= 1 + 3

= 4

Question 4: If P(E) = 0.546, what is the probability of P(not E)?

Solution:

P(E) = 0.546

P(not E) = 1 – 0.546

= 0.454

P(E’) = 0.454

SECTION – II

(5 * 2 = 10)

Question 5: Find the centroid of the triangle whose vertices are (-4, 4), (-2, 2) and (6, -6).

Solution:

Point 1 (x1, y1) = (-4, 4)

Point 2 (x2, y2) = (-2, 2)

Point 3 (x3, y3) = (6, -6)

Centroid of a Triangle =(x1 + x2 + x3 / 3, y1 + y2 + y3 / 3)

= (-4 – 2 + 6 / 3 , 4 + 2 – 6 / 3)

= (0 / 3 , 0 / 3)

= (0, 0)

Question 6: Triangle ABC ~ triangle DEF and their areas 64 and 121 cm² respectively. If EF = 15.4cm, then find BC.

Solution:

Given that Triangle ABC ~ triangle DEF

The ratio of areas of two similar triangles is equal to the ratio of squares of any two corresponding sides.

Area of triangle ABC / Area of triangle DEF = BC² / EF²

64 / 121 = BC² / (15.4)²

64 / 121 = BC² / 237.16

BC² = 125.44

BC = 11.2 cm

Question 7: Prove that tan² θ – sin² θ = tan² θ * sin² θ.

Solution:

LHS = tan² θ – sin²θ

= sin²θ / cos²θ – sin²θ

= sin²θ [1 / cos²θ – 1] [we know, secx = 1 / cosx, so, 1 / cos²θ = sec²θ]

= sin²θ [ sec²θ – 1]

we also know, sec²x – tan²x = 1

so, sec²θ – tan²θ = 1

or, sec²θ – 1 = tan² θ

then, sin²θ [sec²θ – 1] = sin²θ × tan²θ = RHS

So, (tan²θ – sin²θ) = tan²θ × sin²θ

Question 8: A die is thrown once. Find the probability of getting

[a] an even number

[b] an odd prime number

Solution:

Probability of any one outcome occurring = 1/6

The event that an even number on tossing the die can occur in 3 possible ways.

Let E denote this event.

E = {2, 4, 6}

P(E) = 3 × (1 / 6) = 1 / 2

Since there are 6 possible outcomes, the probability of an odd prime is 2 out of 6

= 2 / 6

= 1 / 3

Question 9: Write less than cumulative frequency and greater than cumulative frequency table for the following data.

Solution:

SECTION – III

(4 * 4 = 16)

Question 10:

[a] If cosec A + cot A = p, show that [p² + 1] / [p² – 1] = sec A.

OR

[b] Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus.

Solution:

XY = √(x₂ – x₁)² + (y₂ – y₁)² )

AB = √((-1 – (-4))² + (2 – (-7))²

= √(3² + 9²)

= √(90)

AB = 3√10

BC = √(8 – (-1))² + (5 – 2)²

= √9² + 3²

BC = 3√10

CD= √(5 – 8)² + (-4 – 5)²

CD= 3√10

AD= √(5 – (-4))² + (-4 – (-7))²

AD = 3√10

∴ AB = BC = CD = AD

AC = √(8 – (-4))² + (5 – (-7))²

= √144 + 144

AC = 12√2

BD = √(5 – (-1))² + (-4 – 2)²

= √36 + 36

BD = 6√2

∴AC ≠ BD

Since all the sides are equal and diagonals are not equal, it shows that the following vertices are of a rhombus.

Question 11:

[a] Find the mode of the following data.

OR

[b] A chord of a circle of radius 14cm subtends 120^o angle at the centre. Find the area of the corresponding major segment of the circle.

Solution:

49.5 – 52.5

52.5 – 55.5

55.5 – 58.5

58.5 – 61.5

61.5 – 64.5

Maximum frequency = 135

Model class = 55.5 – 58.5

l = 55.5

f₁ = 135

f₂ = 115

f₀ = 110

h = 3

mode = l + (f₁ – f₀ / 2f₁ – f₀ -f₂) × h

= 55.5 + (135 – 110 / 2 × 135 – 110 – 115) × 3

= 55.5 + (25 / 270 – 225) × 3

= 55.5 + (25 / 45) × 3

= 55.5 + 25 / 15

= 55.5 + 5 / 3

= 55.5 + 1.67

Mode = 57.17

𝛉 = 120^o

In radian, 𝛉 = (120 / 180) 𝛑 = 2𝛑 / 3

r = 14 cm

Then, the area of the corresponding major segment of the circle

The area of the corresponding major segment of the circle is 120.54 sq. cm.

Question 12:

[a] A bag contains 20 discs, which are numbered from 1 to 20. If one disc is drawn at random from the bag, find the probability that it bears:

[i] an even number

[ii] prime number

[iii] multiple of 5

[iv] 2 digits odd number

OR

[b] The angle of elevation of the top of the building from the foot of the tower is 30^o and the angle of elevation of the top of the tower from the foot of the building is 60^o. If the tower is 30m high, find the height of the building.

Solution:

Total number of discs = 20

∴ Total number of all possible outcomes = 20

(i) Favourable outcomes are: 2, 4, 6, 8,….., 20

Number of favourable outcomes = 10

∴ P(getting an even number) = 10 / 20 = 1 / 2

(ii) Favourable outcomes are: 2, 3, 5, 7, 11, 13, 17 and 19

Number of favourable outcomes = 8

∴ P(getting a prime number) = 8 / 20 = 2 / 5

(iii) Favourable outcomes are: 5, 10, 15 and 20

Number of favourable outcomes = 4

∴ P(getting a multiple of 5) = 4 / 20 = 1 / 5

(iv) Favourable outcomes are 11, 13, 15, 17 and 19.

Number of favourable outcomes = 5

∴ P(getting a two-digit odd number) = 5 / 20 = 1 / 4

Height of tower = CD = 50 m

The angle of elevation of the top of a building from the foot of the tower is 30° i.e. ∠ACB =30°

The angle of elevation of the top of the tower from the foot of the building is 60° i.e. ∠DBC =60°

In ΔDBC

tan 𝛉 = perpendicular / base

tan 𝛉 = DC / BC

tan 60^o = 30 / BC

√3 = 30 / BC

BC = 30 / √3

In triangle ABC,

tan 𝛉 = perpendicular / base

tan 𝛉 = AB / BC

tan 30^o = AB / [30 / √3]

1 / √3 = √3AB / 30

AB = 30 / 3

AB = 10m

Thus the height of the building is 10 m.

Question 13:

[a] Construct a triangle similar to the given triangle ABC, with its sides equal to 3 / 4 of the corresponding sides of triangle ABC.

OR

[b] Draw a circle of radius 4cm. From a point 7.5cm away from the centre, construct a pair of tangents to the circle.

Solution:

SECTION – IV

(20 * 0.5 = 10)

Question 14: Slope of the line passing through the points (0, sin 60^o) and (cos 30^o, 0) is

(A) 0 (B) 1 (C) -1 (D) √3

Answer: C

Question 15: △ABC ~ △PQR and ∠A + ∠B = 115^o, then ∠R =

(A) 55 (B) 65 (C) 75 (D) 45

Answer: B

Question 16: The area of a sector whose radius is 7cm and the angle is 120^o, is

(A) 51.3 (B) 51.4 (C) 51.5 (D) 51.6

Answer: A

Question 17: If sec a – tan a = 3, then sec a + tan a is

(A) 1 (B) 1 / 2 (C) 1 / 3 (D) √2

Answer: C

Question 18: In the given figure, BC = ____ units.

(A) 7√3 (B) 7√2 (C) 7 (D) 5

Answer: A

Question 19: From a deck of cards, a card is drawn at random, then the probability of getting a red king is

(A) 1 / 13 (B) 3 / 14 (C) 3 / 26 (D) 1 / 26

Answer: D

Question 20: The mean of the first four odd prime numbers is

(A) 6.5 (B) 7.5 (C) 8.5 (D) 9.5

Answer: A

Question 21: The distance of a point (3, 4) from the origin is ___ units.

(A) 5 (B) 6 (C) 7 (D) 8

Answer: A

Question 22: In triangle ABC, AB = ____ cm.

(A) 5 (B) 6 (C) 7 (D) 8

Answer: D

Question 23: In the given figure, ∠AOB = 120^o, then ∠APO = _____.

(A) 30 (B) 45 (C) 60 (D) 90

Answer: A

Question 24: Arithmetic mean of x – 5, x, x + 5 is _____.

(A) x / 2 (B) x (C) 2x (D) 5x

Answer: B

Question 25: The area of the given triangle is 60 square units, then x = ___ units.

(A) 6 (B) 8 (C) 10 (D) 12

Answer: C

Question 26: If sin 2θ = cos 3θ, then θ =

(A) 15 (B) 18 (C) 21 (D) 24

Answer: B

Question 27: A boy observed 20m away from the base of a 20m high pole, the angle of the elevation of the top is _____.

(A) 15 (B) 30 (C) 45 (D) 60

Answer: C

Question 28: If P(E) = 1, then P(E bar) = ____.

(A) 0 (B) 1 (C) 2 / 3 (D) 3 / 2

Answer: A

Question 29: In triangle ABC, DE || BC, AD = 2cm, DE = 3cm and AB = 6cm, then BC = ____ cm.

(A) 3 (B) 6 (C) 9 (D) 12

Answer: C

Question 30: The length of the tangent drawn from a point 6cm away from the centre of the circle with radius 3cm is ___ cm.

(A) 2√3 (B) 3√3 (C) 3 (D) 4

Answer: B

Question 31: When a dice is rolled, the probability of getting an odd prime number is ____.

(A) 1 /3 (B) 2 / 3 (C) 1 / 6 (D) 3

Answer: A

Question 32: If cos θ = 3 / 5, then sin θ = ____.

(A) 3 /4 (B) 4 / 5 (C) 5 / 12 (D) 5 / 13

Answer: B

Question 33: Mode of 3, 4, 5 and x is 5, then x = ____.

(A) 3 (B) 5 (C) 4 (D) 8

Answer: B