# Telangana Board SSC Class 10 Maths 2016 Question with Solutions PART A

**PART A**

**SECTION - I**

**1. Find the value of log5 125**.

Solution:

Let log5 125 = x

⇒ 5x = 125

⇒ 5x = 53

⇒ x = 3

Therefore, log5 125 = 3

**2. If A = {1, 1/4, 1/9, 1/16, 1/25}, then write A in set-builder form.**

Solution:

A = {1, 1/4, 1/9, 1/16, 1/25}

= {1/1², 1/2², 1/3², 1/4², 1/5²}

The set-builder form: A = {1/x²: x ≤ 5, x ∈ N}

**3. Write an example for a quadratic polynomial that has no zeroes.**

Solution:

x² + x + 11 is one of the polynomials which do not have zeroes.

**4. If b²- 4ac > 0 in ax²+ bx + c = 0, then what can you say about roots of the equation? (a ≠ 0)**

Solution:

Given,

ax² + bx + c = 0

And

b²- 4ac > 0

Hence, the roots of the equation are real and unequal.

**5. Find the sum of the first 200 natural numbers.**

Solution:

First 200 natural numbers: 1, 2, 3, 4,...,200

n = 200

We know that,

Sum of first n natural numbers =

n(n + 1)/2

Sum of the first 200 natural numbers = 200(200 + 1)/2

= 100 × 201

= 20100

**6. For what values of m, the pair of equations 3x + my = 10 and 9x + 12y = 30 have a unique solution.**

Solution:

Given.

3x + my = 10

9x + 12y = 30

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

a1 = 3, b1 = m, c1 = -10

a2 = 9, b2 = 12, c2 = -30

Condition for unique solution

a1/a2 ≠ b1/b2

3/9 ≠ m/12

⇒ m ≠ 12/3

⇒ m ≠ 4

Hence, m takes all the real values except 4.

**7. Find the midpoint of the line segment joining the points (-5, 5) and (5, -5).**

Solution:

Let the given points be:

(x1, y1) = (-5, 5)

(x2, y2) = (5, -5)

Midpoint = [(x1 + x2)/2, (y1 + y2)/2]

= [(-5 + 5)/2, (5 - 5)/2]

= (0/2, 0/2)

= (0, 0)

**SECTION - II**

**8. If x2 + y2 = 7xy, then show that 2 log(x + y) = log x + log y + 2 log 3.**

Solution:

Given,

x²+ y² = 7xy

Adding 2xy on both the sides,

x²+ y²+ 2xy = 7xy + 2xy

(x + y)² = 9xy

(x + y)²= (3)²xy

Taking log on both sides,

log (x + y)² = log (3)²xy

2 log (x + y) = log 32 + log x + log y

2 log (x + y) = 2 log 3 + log x + log y

**9. Length of a rectangle is 5 units more than its breadth. Express its perimeter in polynomial form.**

Solution:

Let x be the breadth of a rectangle.

Length = (x + 5) units

Perimeter of rectangle = 2 (Length + Breadth)

= 2(x + 5 + x)

= 2(2x + 5)

= 4x + 10

**10. Measures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30 cm and the difference**

**between the longest and shortest side is 4 cm, then find the measures of the sides.**

Solution:

Let a - d, a, a + d be the measures of three sides of a triangle.

According to the given,

Perimeter = 30 cm

a - d + a + a + d = 30

3a = 30

a = 30/3

a = 10

Also,

a + d - (a - d) = 4

2d = 4

d = 4/2

d = 2

Thus, a - d = 10 - 2 = 8

a + d = 10 + 2 = 12

Hence, the measures of the triangle are 8 cm, 10 am and 12 cm.

**11. Show that the points A(-3, 3), B(0, 0), C(3, -3) are collinear.**

Solution:

If (x1, y1), (x2, y2) and (x3, y3) are collinear, then the area of the triangle formed by these vertices is 0.

That means ½ [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0

Given,

A(-3, 3), B(0, 0), C(3, -3)

Area of triangle ABC = ½ [-3(0 + 3) + 0(-3 - 3) + 3(3 - 0)]

= ½ [-9 + 0 + 9]

= ½ (0)

= 0

Hence, the given points are collinear.

**12. Solve the following pair of linear equations by substitution method.**

2x - 3y = 19

3x - 2y = 21

Solution:

Given,

2x - 3y = 19….(i)

3x - 2y = 21….(ii)

From (i),

2x - 3y = 19

2x = 3y + 19

x = (3y + 19)/2….(iii)

Substituting (iii) in (ii),

3[(3y + 19)/2] - 2y = 21

9y + 57 - 4y = 42

5y = 42 - 57

5y = -15

y = -15/5

y = -3

Substituting y = -3 in (iii),

x = [3(-3) + 19)]/2

= (-9 + 19)/2

= 10/2

= 5

Hence, the solution of the given pair of linear equations is x = 5 and y = -3.

**13. If 9x2 + kx + 1 = 0 has equal roots, find the value of k.**

Solution:

Given,

9x2 + kx + 1 = 0

Comparing with the standard form ax

2 + bx + c = 0,

a = 9, b = k, c = 1

Condition for equal roots:

b

2

- 4ac = 0

k

2

- 4(9)(1) = 0

k

2

- 36 = 0

k

2 = 36

k = √36

k = ±6

**SECTION - III**

**14. Use Euclid's division lemma to show that the cube of any positive integer is of the form 7m or 7m + 1 or 7m +**

**6.**

Solution:

Let a be any positive integer and b = 7.

By Euclid's division lemma,

a = bq + r, 0 ≤ r < b

a = 7q + r; r = 0, 1, 2, 3, 4, 5, 6

When r = 0,

a = 7q

a³ = (7q)³

a³ = 343q³

a³= 7(49q³)

a³ = 7m, where m = 49q³

Also,in (7q + r)³

consider r³

and divide by 7. The remainder will give the result in each case.

When r = 1,

1³ = 1 and a³ = 7m + 1

When r = 2,

2³ = 8, divided by 7, the remainder is 1.

Therefore, a³= 7m + 1

When r = 3,

3³ = 27 divided by 7, the remainder is 6.

Therefore, a³ = 7m + 6

When r = 4,

3³ = 64 divided by 7, the remainder is

1.

Therefore, a³ = 7m + 1

When r = 5,

3⁵ = 125 divided by 7, the remainder is 6.

Therefore, a³ = 7m + 6

When r = 6,

3⁶ = 216 divided by 7, the remainder is 6.

Therefore, a³ = 7m + 6

Hence, the cube of any positive integer is of the form 7m or 7m + 1 or 7m + 6

**OR**

**Prove that √2 - 3√5 is an irrational number**

Solution:

Let √2 - 3√5 be a rational number.

√2 - 3√5 = a, where a is an integer.

Squaring on both sides,

(√2 - 3√5)²= a²

(√2)² + (3√5)²- 2(√2)(3√5) = a²

2 + 45 - 6√10 = a²

47 - 6√10 = a²

6√10 = a²- 47

(47 - a²)/6 is a rational number since a is an integer.

Therefore, √10 is also an integer.

We know that integers are not rational numbers.

Thus, our assumption that √2 - 3√5 is a rational number is wrong.

Hence, √2 - 3√5 is an irrational number.

**15. Draw the graph for the polynomial p(x) = x²**

**- 3x + 2 and find the zeroes from the graph.**

Solution:

Given,

p(x) = x²- 3x + 2

Let y = x²- 3x + 2