Telangana SSC Board (TS SSC) Question Paper 1 for Class 10th Maths 2017 In PDF

Telangana 10th Annual Exam Question Papers 2017 Maths Paper 1 with Solutions – Free Download

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Telangana Board SSC Class 10 Maths 2017 Question Paper 1 with Solutions

PART A

SECTION – I

1. Write the nature of roots of the quadratic equation 2x² – 5x + 6 = 0.

Solution:

Given,

2x² – 5x + 6 = 0

Comparing with the standard form ax² + bx + c = 0,

a = 2, b = -5, c = 6

Discriminant = b² – 4ac

= (-5)² – 4(2)(6)

= 25 – 48

= -23 < 0

Hence, the roots of the given quadratic equation are not real, i.e. the roots are imaginary.

2. Find the value of log_√2 256.

Solution:

Let log_√2 256 = x

⇒ (√2)^x = 256

⇒ (√2)^x = 2⁸

⇒ (√2)^x = (√2)¹⁶

⇒ x = 16

Therefore, log_√2 256 = 16

3. In a GP, t_n = (-1)ⁿ 2017. Find the common ratio.

Solution:

Given,

nth term of GP is t_n = (-1)ⁿ 2017

t₁ = (-1)¹ 2017 = -2017

t₂ = (-1)² 2017 = 2017

Common ratio = t2/t1

= -2017/2017

= -1

4. Srikar says that the order of the polynomial (x² – 5)(x³ + 1) is 6. Do you agree with him? How?

Solution:

Given polynomial is

p(x) = (x² – 5)(x³ + 1)

= x²(x³ + 1) – 5(x³ + 1)

= x⁵ + x² – 5x³ – 5

= x⁵ – 5x³ + x² – 5

The highest degree of the polynomial is 5.

Hence, the order of the polynomial is not 6 and not agreeing with Srikar.

5. A(0, 3), B(k, 0) and AB = 5. Find the positive value of k.

Solution:

Given,

A(0, 3), B(k, 0) and AB = 5

Using the distance formula,

AB = √[(k – 0)² + (0 – 3)²]

5 = √(k² + 3²)

Squaring on both sides,

25 = k² + 9

⇒ k² = 25 – 9

⇒ k² = 16

⇒ k = √16

⇒ k = ±4

Therefore, the positive value of k = 4.

6. Show that the pair of linear equations 7x + y = 10 and x + 7y = 10 are consistent.

Solution:

Given,

7x + y = 10

x + 7y = 10

Comparing with the standard form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a₁ = 7, b₁ = 1, c₁ = -10

a₂ = 1, b₂ = 7, c₂ = -10

a₁/a₂ = 7/1

b₁/b₂ = 1/7

a₁/a₂ ≠ b₁/b₂

Hence, the given pair of linear equations are consistent and have a unique solution.

7. Represent A ⋂ B through Venn diagram, where A = {1, 4, 6, 9, 10} and B = {perfect squares less than 25}.

Solution:

Given,

A = {1, 4, 6, 9, 10}

B = {perfect squares less than 25} = {1, 4, 9, 16}

A ⋂ B = {1, 4, 6, 9, 10} ⋂ {1, 4, 9, 16} = {1, 4, 9}

SECTION – II

8. Write any two three digit numbers. Find their LCM and GCD by prime factorisation method.

Solution:

Let 126 and 340 are the two three digit numbers.

Prime factorisation of 126:

126 = 2 × 3 × 3 × 7

Prime factorisation of 340:

340 = 2 × 2 × 5 × 17

LCM = 2 × 2 × 3 × 3 × 7 × 5 × 17 = 21420

GCD = 2

9. Find the sum of first 10 terms of an AP

3, 15, 27, 39,….

Solution:

Given AP:

3, 15, 27, 39,…

First term = a = 3

Common difference = d = 12

Sum of the first n terms

S_n = n/2[2a + (n – 1)d

S₁₀ = (10/2) [2(3) + (10 – 1)12]

= 5(6 + 108)

= 5 × 114

= 570

Hence, the sum of the first 10 terms of the given AP is 570.

10. Which of √2 and 2 is a zero of the polynomial p(x) = x³ – 2x? Why?

Solution:

Given,

p(x) = x³ – 2x

p(√2) = (√2)³ – 2(√2) = 2√2 – 2√2 = 0

p(2) = (2)³ – 2(2) = 8 – 4 = 4 ≠ 0

Therefore, √2 is the zero of the given polynomial.

11. The sum of a number and its reciprocal is 10/3. Find the number.

Solution:

Let x be any number.

According to the given,

x + 1/x = 10/3

(x² + 1)/x = 10/3

3(x² + 1) = 10x

3x² + 3 – 10x = 0

3x² – 9x – x + 3 = 0

3x(x – 3) – 1(x – 3) = 0

(3x – 1)(x – 3) = 0

3x – 1 = 0, x – 3 = 0

x = 1/3, x = 3

Hence, the required number is 3 or 1/3.

12. Two vertices of a triangle are (3, 2), (-2, 1) and its centroid is (5/3, -1/3). Find the third vertex of the triangle.

Solution:

Given,

Two vertices of a triangle are (3, 2), (-2, 1).

Centroid = (5/3, -⅓)

Let (x, y) be the third vertex.

Using centroid formula,

(5/3, -1/3) = [(x₁ + x₂ + x₃)/2, (y₁ + y₂ + y₃)/3]

(5/3, -1/3) = [(3 – 2 + x)/3, (2 + 1 + y)/3]

(5/3, -1/3) = [(1 + x)/3, (3 + y)/3]

⇒ (1 + x)/3 = 5/3, (3 + y)3 = -1/3

⇒ 1 + x = 5, 3 + y = -1

⇒ x = 4, y = -4

Hence, the third vertex of the triangle is (4, -4).

13. Find the angle made by the line joining (5, 3) and (-1, -3) with the positive direction of X-axis.

Solution:

Let the given points be:

(x₁, y₁) = (5, 3)

(x₂, y₂) = (-1, -3)

Slope = m = (y₂ – y₁)/(x₂ – x₁)

= (-3 – 3)/(-1 – 5)

= -6/-6

= 1

Let θ be the angle made by the line with x-axis.

We know that,

m = tan θ

1 = tan θ

tan 45° = tan θ

θ = 45°

Hence, the required angle is 45°.

SECTION – III

14. From the following Venn diagram, write the elements of the sets of A and B. And verify n(A ⋃ B) + n(A ⋂ B) = n(A) + n(B).

Solution:

A = {a, c, d, f, h}

B = {a, b, d, e, g, h}

A ⋃ B = {a, c, d, f, h} ⋃ {a, b, d, e, g, h} = {a, b, c, d, e, f, g, h}

A ⋂ B = {a, c, d, f, h} ⋂ {a, b, d, e, g, h} = {a, d, h}

n(A) = 5

n(B) = 6

n(A ⋃ B} = 8

n(A ⋂ B) = 3

n(A ⋃ B) + n(A ⋂ B) = 8 + 3 = 11

n(A) + n(B) = 5 + 6 = 11

Therefore, n(A ⋃ B) + n(A ⋂ B) = n(A) + n(B)

OR

Use Euclid’s division lemma to show that the square of any positive integer is of the form 5n or 5n + 1 or 5n + 4, where n is a whole number.

Solution:

Let a be any positive integer and b = 5.

By Euclid division lemma,

a = bq + r, where 0 ≤ r < b

⇒ a = 5q + r; r = 0, 1, 2, 3, 4

When r = 0,

a = 5q

a² = (5q)²

a² = 25q²

a² = 5(5q²)

a² = 5n

where n = 5q²

When r = 1,

a = 5q + 1

a² = (5q + 1)²

a² = 25q² + 10q + 1

a² = 5(5q² + 2q) + 1

a² = 5n + 1

where n = 5q² + 2q

When r = 2,

a = 5q + 2

a² = (5q + 2)²

a² = 25q² + 20q + 4

a² = 5(5q² + 4q) + 4

a² = 5n + 4

where n = 5q² + 4q

When r = 3,

a = 5q + 3

a² = (5q + 3)²

a² = 25q² + 30q + 5 + 4

a² = 5(5q² + 6q + 1) + 4

a² = 5n + 4

where n = 5q² + 6q + 1

When r = 4,

a = 5q + 4

a² = (5q + 4)²

a² = 25q² + 40q + 15 + 1

a² = 5(5q² + 8q + 3) + 1

a² = 5n + 1

where n = 5q² + 8q + 3

Hence, the square of any positive integer is of the form 5n or 5n + 1 or 5n + 4, where n is a whole number.

15. Find the sum of all three digit natural numbers, which are divisible by 3 and not divisible by 6.

Solution:

Three digit natural numbers that are divisible by 3:

102, 105, 108, 111, 114,…,993, 996, 999

Three digit natural numbers that are divisible by 3 but not by 6:

105, 111, 117,…, 993, 999

This is an AP with a = 105, d = 6 and a_n = 999

nth term of AP

a_n = a + (n – 1)d

999 = 105 + (n – 1)6

⇒ (n – 1)6 = 999 – 105

⇒ n – 1 = 894/6

⇒ n – 1 = 149

⇒ n = 149 + 1

⇒ n = 150

Sum of the first n terms

S_n = n/2(a + an)

S₁₅₀ = (150/2) × (105 + 999)

= 75 × 1104

= 82800

Hence, the required sum is 82800.

OR

Divide 3x⁴ – 5x³ + 4x² + 3x – 5 by x² – 3 and verify the division algorithm.

Solution:

Let p(x) = 3x⁴ – 5x³ + 4x² + 3x – 5

g(x) = x² – 3

Quotient = q(x) = 3x² – 5x + 13

Remainder = r(x) = -12x + 34

By division algorithm,

p(x) = [g(x) q(x)] + r(x)

g(x) q(x) + r(x) = (x² – 3)(3x² – 5x + 13) -12x + 34

= 3x⁴ – 5x³ + 13x² – 9x² + 15x – 39 – 12x + 34

= 3x⁴ – 5x³ + 4x² + 3x – 5

= p(x)

Hence verified.

16. The perimeter of a right-angles triangle is 60 cm and its hypotenuse is 25 cm. Find the remaining two sides.

Solution:

Let ABC be the right triangle in which B is the right angle.

Given,

Hypotenuse = AC = 25 cm

Perimeter = 60 cm

AB + BC + AC = 60

AB + BC + 25 = 60

AB + BC = 35 cm

By Pythagoras theorem,

AC² = AB² + BC²

(25)² = AB² + (35 – AB)²

625 = AB² + 1225 + AB² – 70AB

2AB² – 70AB + 1225 – 625 = 0

2AB² – 70AB + 600 = 0

AB² – 35AB + 300 = 0

Let AB = x

x² – 35x + 300 = 0

x² – 20x – 15x + 300 = 0

x(x – 20) – 15(x – 20) = 0

(x – 15)(x – 20) = 0

x = 15, x = 20

Therefore, the remaining sides of the triangle are 15 cm and 20 cm.

OR

The points C and D are on the line segment joining A(-4, 7) and B(5, 13) such that AC = CD = DB. Then find coordinates of points C and D.

Solution:

Given,

The points C and D are on the line segment joining A(-4, 7) and B(5, 13) such that AC = CD = DB.

C divided AB in the ratio 1 : 2.

m : n = 1 : 2

Using section formula,

C = [(mx₂ + nx₁)/ (m + n), (my₂ + ny₁)/ (m + n)]

= [(5 – 8)/ (1 + 2), (13 + 14)/(1 + 2)]

= (-3/3, 27/3)

= (-1, 9)

D is the midpoint of BC.

D = [(-1 + 5)/2, (9 + 13)/2]

= (4/2, 22/2)

= (2, 11)

Therefore, C = (-1, 9) and D = (2, 11)

17. Draw the graph for the polynomial p(x) = x² – 5x + 6 and find the zeroes from the graph.

Solution:

Given,

p(x) = x² – 5x + 6

Parabola intersecting the x-axis at (2, 0) and (3, 0).

Hence, the zeroes of the given polynomial are 2 and 3.

OR

Draw the graph of 2x + y = 6 and 2x – y + 2 = 0 and find the solution from the graph.

Solution:

Given,

2x + y = 6

2x – y + 2 = 0

Consider the first equation:

2x + y = 6

y = -2x + 6

Now, consider another equation:

2x – y + 2 = 0

y = 2x + 2

Graph:

The lines representing the given pair of equations intersecting each other at (1, 4).

Hence, the solution is x = 1 and y = 4.