# TS SSC BOARD QUESTIONS PAPER WITH SOLUTIONS 2019

Telangana Board SSC Class 10 Maths 2019 QuestionPaper 1 with Solutions

**PART A**

**SECTION - I**

**1. If A = {x: x is a factor of 24}, then find n(A).**

**Solution: **A = {1, 2, 3, 4, 6, 8, 12, 24} n(A) = 8

**2. Find the HCF of 24 and 33 by using a division algorithm.**

**Solution: **24 < 33

**3. Radha says "1, 1, 1,.... are in AP and also in GP". Do you agree with Radha? Give reason.**

**Solution:****Given,**

**1, 1, 1,....**

**I**

**f a, b, c are in AP, then a + c = 2b**

**Thus, 1 + 1 = 2(1)**

**2 = 2**

**T**

**herefore, 1, 1, 1 are in AP.**

**I**

**f a, b, a are in GP, then, ac = b2**

**(1)(1) = (1)²**

**1 = 1**

**Therefore, 1, 1, 1 are in GP.**

**Hence, we can say that 1,1,1 are in AP and also in GP.**

**4. If P(x) = x4 + 1, then find P(2) - P(-2).**

**Solution**:

P(2) - P(-2) = 17 - 17 = 0

**5. Find the roots of the quadratic equation x²+ 2x - 3 = 0.**

**Solution:**Given quadratic equation is: x² + 2x - 3 = 0

x² + 3x - x - 3 = 0

x(x + 3) - 1(x + 3) = 0

(x - 1)(x + 3) = 0

x - 1 = 0, x + 3 = 0

x = 1, x = -3

Hence, the roots of the given quadratic equation are 1 and -3.

**6. Find the centroid of ΔPQR, whose vertices are P(1, 1), Q(2, 2), R(-3, -3).**

**Solution:**

Let the given vertices of a triangle PQR are:

(x1, y1) = (1, 1)

(x2, y2) = (2, 2)

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

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

= [(1 + 2 - 3)/3, (1 + 2 - 3)/3]

= (0/3, 0/3)

= (0, 0)

**7. For what value of 't' the following pair of linear equations has a no solution? ****2x - ty = 5 and 3x + 2y = 11**

**Solution:**Given pair of linear equations are:

2x - ty = 5

3x + 2y = 11

Comparing with the standard form,

a1 = 2, b1 = -t, c1 = -5

a2 = 3, b2 = 2, c2 = -11

Condition for no solution of linear equations is

a1/a2 = b1/b2 ≠ c1/c2

⅔ = -t/2

⇒ t = -4/3

**SECTION - II**

**8. If µ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 8} and B = {0, 3, 5, 7, 10}. Then represent A ⋂ B in the Venn diagram.**

**Solution:**Given,

µ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 5, 8}

B = {0, 3, 5, 7, 10}

A ⋂ B = {2, 3, 5, 8} ⋂ {0, 3, 5, 7, 10} = {3, 5}

**9. Akhila says, "points A(1, 3), B(2, 2), C(5, 1) are collinear". Do you agree with Akhila? Why?**

**Solution:**

Given points are A(1, 3), B(2, 2) and C(5, 1).

If (x1, y1), (x2, y2) and (x3, y3) are collinear, then area of triangle = 0

i.e. ½ [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0

= ½ [1(2 - 1) + 2(1 - 3) + 5(3 - 2)]

= 1 + 2(-2) + 5(1)

= 1 - 4 + 5

= 2 ≠ 0

Thus, the given points are not collinear.

Hence, we do not agree with Akhila.

**10. Write the quadratic equation, whose roots are 2 + √3 and 2 - √3.**

**Solution: **Let α and β be the roots of the quadratic equation.

Given,

2 + √3 and 2 - √3 are the roots of the quadratic equation.

Sum of the roots = α + β = 2 + √3 + 2 - √3 = 4

Product of the roots = αβ = (2 + √3)(2 - √3)

= (2)²

- (√3)²

= 4 - 3

= 1

Hence, the quadratic equation is x²- (α + β)x + αβ = 0

x²- 4x + 1 = 0

**11. Divide x³- 4x²+ 5x - 2 by x - 2.**

**Solution:**

**12. Write the formula of the nth term of GP and explain the terms in it.**

**Solution:**

The formula for nth term of GP is

an = arn - ¹

Here,

an = nth term of the sequence

a = first term

r = common ratio

13. Solve the pair of linear equations 2x + 3y = 8 and x + 2y = 5 by elimination method.

Solution:

Given,

2x + 3y = 8….(i)

x + 2y = 5….(ii)

(ii) × 2 - (i),

2x + 4y - (2x + 3y) = 10 - 8

y = 2

Substituting y = 2 in (ii),

x + 2(2) = 5

x = 5 - 4

x = 1

**SECTION - III**

**14. (a) Draw the graph of the polynomial p(x) = x²- 7x + 12, then find its zeroes from the graph.**

**Solution:**

Given polynomial is p(x) = x²- 7x + 12

**Solution:**

Let P(x, 0) divide the line segment joining the points A(2, -3) and B(5, 6) in the ratio m : n.

Here,

(x1, y1) = (2, -3)

(x2, y2) = (5, 6)

Using the section formula,

P(x, 0) = [(mx2 + nx1)/ (m + n), (my2 + ny1)/ (m + n)]

(x, 0) = [ (5m + 2n)/ (m + n), (6m - 3n)/ (m + n)]

⇒ (6m - 3n)/ (m + n) = 0

⇒ 6m = 3n

⇒ m/n = 3/6

⇒ m/n = 1/2

Therefore, the required ratio is 1 : 2.

Now,

x = (5m + 2n)/ (m + n)

x = [5 + 2(2)]/ (1 + 2)

x = 9/3

x= 3

Hence, the required point on the x-axis is (3, 0).

**OR**

**(b) Find the sum of all two digit odd multiples of 3.**

**Solution:**

Two digit odd multiples of 3 are 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99.

This is an AP with a = 15

Common difference = d = 6

n = 15

Sum of first n terms

Sn = n/2[2a + (n - 1)d]

S15 = (15/2) [2(15) + (15 - 1)6]

= (15/2) [30 + 14(6)]

= (15/2) [30 + 84]

= (15/2) × 114

= 855

Hence, the required sum is 855.

**16. (a) If A = {x : 2x + 1, x ∈ N, x ≤ 5}, B = {x : x is a composite number, x ≤ 12}, then show that (A ⋃ B)-(A ⋂ B) = (A - B) ⋃ (B - A).**

**Solution:**

A = {x : 2x + 1, x ∈ lN, x ≤ 5}

= {3, 5, 7, 9, 11}

B = {x : x is a composite number, x ≤ 12}

= {4, 6, 8, 9, 10, 12}

A ⋃ B = {3, 5, 7, 9, 11} ⋃ {4, 6, 8, 9, 10, 12}

= {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

A ⋂ B = {3, 5, 7, 9, 11} ⋂ {4, 6, 8, 9, 10, 12}

= {9}

(A ⋃ B) - (A ⋂ B) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12} - {9}

= {3, 4, 5, 6, 7, 8, 10, 11, 12}

A - B = {3, 5, 7, 9, 11} - {4, 6, 8, 9, 10, 12} = {3, 5, 7, 11}

B - A = {4, 6, 8, 9, 10, 12} - {3, 5, 7, 9, 11} = {4, 6, 8, 10, 12}

(A - B) ⋃ (B - A) = {3, 5, 7, 11} ⋃ {4, 6, 8, 10, 12} = {3, 4, 5, 6, 7, 8, 10, 11, 12}

Therefore, (A ⋃ B) - (A ⋂ B) = (A - B) ⋃ (B - A)

**OR**

**(b) Prove that √2 + √7 is an irrational number.**

**Solution:**

Let √2 + √7 be a rational number.

√2 + √7 = a, where a is an integer.

Squaring on both sides,

(√2 + √7)² = a²

(√2)² + (√7)² + 2(√2)(√7) = a²

2 + 7 + 2√14 = a²

9 + 2√14 = a²

2√14 = a²

- 9

√14 = (a²- 9)/2

(a²- 9)/2 is a rational number since a is an integer.

Therefore, √14 is also an integer.

We know that integers are not rational numbers.

Thus, our assumption that √2 + √7 is a rational number is wrong.

Hence, √2 + √7 is an irrational number.

**17. (a) Sum of the areas of two squares is 850 m2****. If the difference of their perimeters is 40 m. Find the sides of ****t****he two squares.**

**Solution**:

Let x and y be the sides of two squares.

According to the given,

Sum of the areas of two squares = 850 m2

x²+ y²= 850….(i)

Difference of their perimeters = 40 m

4x - 4y = 40

4(x - y) = 40

x - y = 10

x = y + 10….(ii)

Substituting (ii) in (i),

(y + 10)²+ y² = 850

y² + 100 + 20y + y²

- 850 = 0

2y²+ 20y - 750 = 0

2(y² + 10y - 375) = 0

y² + 10y - 375 = 0

y²+ 25y - 15y - 375 = 0

y(y + 25) - 15(y + 25) = 0

(y - 15)(y + 25) = 0

y - 15 = 0, y + 25 = 0

y = 15, y = -25

Measure cannot be negative.

Therefore, y = 15

Substitute y = 15 in (ii),

x = 15 + 10 = 25

Hence, the sides of two squares are 25 cm and 15 cm.

OR

**(b) Sum of the present ages of two friends are 23 years, five years ago the product of their ages was 42. Find their ****a****ges 5 years hence.**

**Solution:**

Let x and (23 - x) be the present ages (in years) of two friends.

According to the the given

(x - 5)(23 - 5 - x) = 42

(x - 5)(18 - x) = 42

18x - x²

- 90 + 5x = 42

⇒ x²- 5x - 18x + 90 + 42 = 0

⇒ x²- 23x + 132 = 0

⇒ x²- 11x - 12x + 132 = 0

⇒ x(x - 11) - 12(x - 11) = 0

⇒ (x - 11)(x - 12) = 0

⇒ x = 12, x = 11

If x = 12, then 23 - x = 23 - 12 = 11

If x = 11, then 23 - x = 23 - 11 = 12

Therefore, the present ages of two friends is 11 and 12 years.

Hence, their ages after 5 years will be 16 and 17 years.