## 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.

## Download SSC 2018 Question Paper Maths Paper 2

## Download SSC 2018 Question Paper Maths Paper 2 With Solutions

### 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 = √(x_{2 }– x_{1})² + (y_{2} – y_{1})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 tan ^{2} 45^{o} + cot^{2} 30^{o}.**

**Solution:**

tan^{2 }45 + cot^{2 }30

From the trigonometric values,

tan 45 = 1 and cot 30 = √3

Substitute,

tan^{2 }45 + cot^{2}30

= (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 cm ^{2} 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 = BC^{2} / EF^{2}

64 / 121 = BC^{2} / (15.4)^{2}

64 / 121 = BC^{2} / 237.16

BC^{2} = 125.44

BC = 11.2 cm

**Question 7: Prove that tan ^{2} θ – sin^{2} θ = tan^{2} θ * sin^{2} 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

[b] The probable odd prime numbers appearing on a die roll are 3 and 5 (1 is not considered to be prime).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.**

CI |
5 – 10 |
10 – 15 |
15 – 20 |
20 – 25 |
25 – 30 |

f |
4 |
45 |
20 |
13 |
9 |

**Solution:**

CI |
5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 |

f |
4 | 45 | 20 | 13 | 9 |

Less than CF |
4 | 49 | 69 | 82 | 91 |

Greater than CF |
91 | 87 | 42 | 22 | 9 |

**SECTION – III**

** (4 * 4 = 16)**

**Question 10: **

**[a] If cosec A + cot A = p, show that [p ^{2} + 1] / [p^{2} – 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 = √(x_{2} – x_{1})² + (y_{2} – y_{1})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.**

CI |
50 – 52 |
53 – 55 |
56 – 58 |
59 – 61 |
62 – 64 |

f |
15 |
110 |
135 |
115 |
25 |

** 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

f_{1} = 135

f_{2} = 115

f_{0} = 110

h = 3

mode = l + (f_{1} – f_{0} / 2f_{1} – f_{0} -f_{2}) × 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

[b] Area of segment = (1 / 2) * (𝛉 – sin 𝛉) * r^{2}, where 𝛉 = central angle, r = radius

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

[b] Height of building = ABHeight 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**