## Telangana 10th Annual Exam Question Papers 2016 Maths Paper 2 with Solutions – Free Download

Telangana SSC Class 10 maths 2016 Question Paper 2 with solutions can be accessed at BYJUâ€™S, which are prepared by the expert faculty. Different patterns of questions appear in Telangana SSC exams which test the studentsâ€™ ability to understand the concepts of maths. For more knowledge on class 10 Maths concepts, students can visit BYJUâ€™S. Also, they can get all TS SSC class 10 maths previous year papers along with the solutions. Students can also get other year question papers of TS SSC class 10 maths so that they can practice them before going to attempt the main exam. This will help the students in overcoming stress and boost confidence over the exam.

## Download TS SSC 2016 Maths Question Paper 2

## Download TS SSC 2016 Maths Question Paper 2 With Solutions

### Telangana Board SSC Class 10 Maths 2016 Question Paper 2 with Solutions

**PART A**

**SECTION – I**

**1. **In the given figure, Î”ABC ~ Î”ADE, then find the value of x.

**Solution:**

In Î”ADE and Î”ABC,

âˆ A = âˆ A (common)

âˆ ADE = âˆ ABC (corresponding angles)

By SS similarity criterion,

Î”ADE ~ Î”ABC

AC/AE = BC/DE

(3 + 6)/3 = x/5

x/5 = 9/3

x = 15 cm

**2. **Find the probability of getting a sum of the numbers on them is 7, when two dice are rolled at a time.

**Solution:**

Total number of outcomes = 6^{2} = 36

n(S) = 36

Let E be the event of getting a sum of numbers on the dice is 7.

E = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

Number of outcomes favourable to E = n(E) = 7

P(E) = n(E)/n(S) = 7/36

**3. **If tan Î¸ = âˆš3 (where Î¸ is acute), then find the value of 1 + cos Î¸.

**Solution:**

Given,

tan Î¸ = âˆš3

tan Î¸ = tan 60Â°

Î¸ = 60Â°

1 + cos Î¸ = 1 + cos 60Â°

= 1 + (1/2)

= 3/2

**4.** “A conical solid block is exactly fitted inside the cubical box of side ‘a’, then the volume of conical solid block is 4/3 Ï€a^{3}“. Is this statement true? Justify your answer.

**Solution:**

Given,

Side of a cube = a

Volume of cubical box = a3

Volume of conical solid block = 4/3 Ï€a3

4/3 Ï€3 > a3

Hence, the given statement is false since the volume of the conical block should not be greater than the volume of the cube if it is fitted exactly inside it.

**5.** If the surface area of a hemisphere is ‘S’, then express ‘r’ in terms of ‘S’.

**Solution:**

Surface area of hemisphere S = 2Ï€r^{2}

r^{2} = S/2Ï€

r = âˆš(S/2Ï€)

**6. **Write the formula to find the median for grouped data and explain each term.

**Solution:**

Median = l + {[(n/2) – cf]/ f} Ã— h

Here,

l = Lower limit of the median class

n = Sum of frequencies

cf = Cumulative frequency of the class preceding the median class

f = Frequency of the median class

h = Class height

**7. **“If the angle of elevation of the Sun increases from 0Â° to 90Â°, then the length of shadow of a tower decreases.” Is this statement true? Justify your answer.

**Solution:**

Let h be the height of the tower and s be the length of its shadow.

Initially, the angle of elevation = 0Â°

tan 0Â° = h/s

h/s = 0

Thus, the length of shadow is infinity.

Finally, the angle of elevation = 90Â°

tan 90Â° = h/s

h/s = âˆž

h/s = 1/0

s = 0

Thus, the length of shadow is minimum.

Hence, the given statement is true.

**SECTION – II**

**8. **Prove that âˆš[(1 – sin Î¸)/(1 + sin Î¸)] = sec Î¸ – tan Î¸, where Î¸ is acute.

**Solution:**

LHS = âˆš[(1 – sin Î¸)/(1 + sin Î¸)]

= âˆš{[(1 – sin Î¸)/(1 + sin Î¸)] [(1 – sin Î¸)/(1 – sin Î¸)]}

= âˆš[(1 – sin Î¸)^{2}/ (1 – sin^{2}Î¸)]

= âˆš[(1 – sin Î¸)^{2}/cos^{2}Î¸]

= (1 – sin Î¸)/cos Î¸

= (1/cos Î¸) – (sin Î¸/cos Î¸)

= sec Î¸ – tan Î¸

= RHS

Hence proved.

**9. **ABC is an isosceles triangle and âˆ B = 90Â°, then show that AC^{2} = 2AB^{2}.

**Solution:**

Given,

AB = BC

By Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = AB^{2} + AB^{2}

AC^{2} = 2AB^{2}

Hence proved.

**10. **Find the volume and surface area of a sphere of radius 42 cm. (Ï€ = 22/7)

**Solution:**

Given,

Radius of sphere = r = 42 cm

Surface area of sphere = 4Ï€r^{2}

= 4 Ã— (22/7) Ã— 42 Ã— 42

= 22176 cm^{2}

Volume of sphere = (4/3)Ï€r^{3}

= (4/3) Ã— (22/7) Ã— 42 Ã— 42 Ã— 42

= 310464 cm^{3}

**11. **If tan (A + B) = 1 and cos (A – B) = âˆš3/2, 0Â° < A + B < 90Â° and A > B; find A and B.

**Solution:**

Given,

tan (A + B) = 1

tan (A + B) = tan 45Â°

A + B = 45Â°….(i)

And

cos (A – B) = âˆš3/2

cos (A – B) = cos 30Â°

A – B = 30Â°….(ii)

Adding (i) and (ii),

A + B + A – B = 45Â° + 30Â°

2A = 75Â°

A = 75Â°/2

A = 37.5Â°

Substituting A = 37.5Â° in (i)

37.5Â° + B = 45Â°

B = 45Â° – 37.5Â°

B = 7.5Â°

**12. **A solid metallic ball of volume 64 cm^{3} melted and made into a solid cube. Find the side of the solid cube.

**Solution:**

Let a be the side of a solid cube.

According to the given,

Volume of metallic ball = Volume of solid cube

64 = a^{3}

â‡’ a^{3} = 4^{3}

â‡’ a = 4

Hence, the side of the solid cube = 4 cm

**13. **A boat has to cross a river. It crosses the river by making an angle of 60Â° with the bank of the river due to the stream of the river and travels a distance of 450 m to reach the other side of the river. Draw the diagram for this data.

**Solution:**

Let AB be the width of the river and C be the position of boat.

**SECTION – III**

**14.** A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.

**Solution:**

Given,

Number of red balls = 5

Let x be the number of blue balls.

Total number of balls = x + 5

P(getting a red ball) = 5/(x + 5)

P(getting a blue ball) = x/(x + 5)

According to the given,

2[5/(x + 5)] = x/(x + 5)

â‡’ x = 10

Hence, the number of blue balls in the bag = 10

**OR**

Evaluate: (tan^{2}60Â° + 4 cos^{2}45Â° + 3 sec^{2}30Â° + 5 cos^{2}90Â°)/ (cosec 30Â° + sec 60Â° – cot^{2}30Â°)

**Solution:**

(tan^{2}60Â° + 4 cos^{2}45Â° + 3 sec^{2}30Â° + 5 cos^{2}90Â°)/ (cosec 30Â° + sec 60Â° – cot^{2}30Â°)

= [(âˆš3)^{2} + 4 (1/âˆš2)^{2} + 3 (2/âˆš3)^{2} + 5 (0)^{2}]/ [2 + 2 – (âˆš3)^{2}]

= [3 + 2 + 4 + 0]/ (4 – 3)

= 9/1

= 9

**15. **Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages in Rupees | 200 – 250 | 250 – 300 | 300 – 350 | 350 – 400 | 400 – 450 |

Number of workers | 6 | 8 | 14 | 10 | 12 |

Find the mean daily wages of the workers in the factory by using step-deviation method.

**Solution:**

Daily wages (in Rs) | Number of workers (f_{i}) |
Mid values (x_{i}) |
d_{i} = x_{i} – 325 |
u_{i} = (x_{i} – 325)/50 |
f_{i}u_{i} |

200 – 250 | 6 | 225 | -100 | -2 | -12 |

250 – 300 | 8 | 275 | -50 | -1 | -8 |

300 – 350 | 14 | 325 = A | 0 | 0 | 0 |

350 – 400 | 10 | 375 | 50 | 1 | 10 |

400 – 450 | 12 | 425 | 100 | 2 | 24 |

âˆ‘f_{i} = 50 |
âˆ‘f_{i}u_{i} = 14 |

Mean = a + (âˆ‘f_{i}u_{i}/ âˆ‘f_{i}) Ã— h

= 325 + (14/50) Ã— 50

= 325 + 14

= 339

**OR**

Draw a circle of radius 5 cm. From a point 8 cm away from its centre, construct a pair of tangents to the circle. Find the lengths of tangents.

**Solution:**

PA and PB are the required tangents to the circle.

Length of tangents = PA = PB = 6 cm

**16. **The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (Quintal/ HEc.) | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 | 75 – 80 |

Number of farmers | 2 | 24 | 16 | 8 | 38 | 12 |

Draw both Ogives for the above data. Hence obtain the median production yield.

**Solution:**

CI | Less than cumulative frequency | CI | More than cumulative frequency |

Less than 55 | 2 | More than 50 | 100 |

Less than 60 | 26 | More than 55 | 98 |

Less than 65 | 42 | More than 60 | 74 |

Less than 70 | 50 | More than 65 | 58 |

Less than 75 | 88 | More than 70 | 50 |

Less than 80 | 100 | More than 75 | 12 |

**OR**

Construct a triangle of sides 5 cm, 6 cm and 7 cm, then construct a triangle similar to it, whose sides are 2/3 of the corresponding sides of the first triangle.

**Solution:**

Hence, Î”AB’C’ be the required triangle similar to the Î”ABC.

**17. **DWACRA is supplied with a cuboidal shaped wax block with measurements 88 cm Ã— 42 cm Ã— 35 cm. From this how many cylindrical candles of 2.8 cm diameter and 8 cm of height can be prepared?

**Solution:**

Given,

Length of block = 88 cm

Width of block = 42 cm

Height of block = 35 cm

Volume of block = Length Ã— Height Ã— width

= 88 Ã— 35 Ã— 42

= 129360 cm^{3}

Also,

Diameter of candle = 2.8 cm

Radius of candle = r = 2.8/2 = 1.4 cm

Height of candle = h = 8 cm

Volume of candle = Ï€r^{2}h

= (22/7) Ã— (1.4)^{2} Ã— 8

= 49.28 cm^{3}

Number of candles formed = Voume of block/Voluem of one candle

= 129360/49.28

= 2625

Hence, the required number of candles is 2625.

**OR**

Two poles of equal heights are standing opposite to each other, on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of top of the poles are 60Â° and 30Â° respectively. Find the height of the poles.

**Solution:**

Let AB and CD be the two poles.

BC is the width of the road and P be the point of observation.

In right triangle PCD,

tan 30Â° = CD/PC

1/âˆš3 = CD/PC

CD = PC/âˆš3 ….(i)

In right triangle PBA,

tan 60Â° = AB/PB

âˆš3 = AB/PB

AB = âˆš3 BP

CD = âˆš3 BP….(ii) (given that heights of two poles are equal)

From (i) and (ii),

PC/âˆš3 = âˆš3 BP

PC = âˆš3 Ã— âˆš3 Ã— BP

PC = 3BP

Now,

BP + PC = BC

BP + 3BP = 80

4BP = 80

BP = 80/4

BP = 20 m

Again, PC = BC – BP

PC = 80 – 20 = 60 m

From (ii),

CD = âˆš3 (20) = 20âˆš3 ,

Hence, the height of the poles = 20âˆš3 m