# Telangana SSC Board (TS SSC) Question Paper 1 for Class 10th Maths 2018 In PDF

## Telangana 10th Annual Exam Question Papers 2018 Maths Paper 1 with Solutions – Free Download

Telangana SSC Question Paper Maths Paper 1 with solutions is available at BYJUâ€™S with detailed explanations along with the required formulas and graphs (or images) so that class 10 students can understand them easily.

BYJUâ€™S provided all the previous years question papers of the Telangana SSC board with two papers each having part A and part B. TS SSC Board Question Paper 1 for Class 10th Maths 2018 can be downloaded by the students in a pdf form and practice them at any time. Also, they are provided with solutions to help the students in referring to the accurate solutions.

### Telangana Board SSC Class 10 Maths 2018 Question Paper 1 with Solutions

PART A

SECTION – I

1. Find the distance between the points (1, 5) and (5, 8).

Solution:

Let the given points be:

(x1, y1) = (1, 5)

(x2, y2) = (5, 8)

Distance between two points

= âˆš25

= 5 units

2. Expand log10 385.

Solution:

log10 385

= log10 (5 Ã— 7 Ã— 11)

We know that log abc = log a + log b + log c

= log10 5 + log10 7 + log10 11

3. Give one example each for a finite set and an infinite set.

Solution:

Finite set: A = {3, 5, 7, 9, 11, 13, 15, 17}

n(A) = 8

Infinite set: Z = Set of all integers

= {…..,-4, -3, -2, -1, 0, 1, 2, 3, 4, …..}

n(Z) = âˆž

4. Find the sum and product of roots of the quadratic equation.

x2 – 4âˆš3x + 9 = 0

Solution:

x2 – 4âˆš3x + 9 = 0

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

a = 1, b = -4âˆš3 and c = 9

Sum of the roots = -b/a = -(-4âˆš3)/1 = 4âˆš3

Product of the roots = c/a = 9/1 = 9

5. Is the sequence âˆš3, âˆš6, âˆš9, âˆš12,…. form an Arithmetic Progression? Give reason.

Solution:

Given,

âˆš3, âˆš6, âˆš9, âˆš12,…

First term = a = âˆš3

Second term – First term = âˆš6 – âˆš3

Third term – Second term = âˆš9 – âˆš6 = 3 – âˆš6

Since the common difference is not the same throughout the sequence, it is nor form an AP.

6. If x = a and y = b is the solution for the pair of equations x – y = 2 and x + y = 4, then find the values of a and b.

Solution:

Given pair of equations are:

x – y = 2….(i)

x + y = 4….(ii)

x – y + x + y = 2 + 4

2y = 6

y = 6/2 = 3

Substituting y = 3 in (i),

x – 3 = 2

x = 2 + 3

x = 5

Therefore, a = 5 and b = 3.

7. Verify the relation between zeroes and coefficients of the quadratic polynomial x2 – 4.

Solution:

Let the given quadratic polynomial p(x) = x2 – 4

Let p(x) = 0

x2 – 4 = 0

x2 = 4

x = âˆš4

x = Â±2

Therefore, zeroes of the given polynomial are -2 and 2.

Sum of the zeroes = -2 + 2 = 0 = -0/1 = -Coefficient of x/ Coefficient of x2

Product of the zeroes = (-2)(2) = -4 = -4/1 = Constant term/Coefficient of x2

Hence, verified the relationship between the zeroes and coefficients of the given quadratic polynomial.

SECTION – II

8. Complete the following table for the polynomial y = p(x) = x3 – 2x + 3.

 x -1 0 1 2 x3 -2x 3 y (x, y)

Solution:

Given,

y = p(x) = x3 – 2x + 3

 x -1 0 1 2 x3 -1 0 1 8 -2x 2 0 -2 -4 3 3 3 3 3 y 4 3 2 7 (x, y) (-1, 4) (0, 3) (1, 2) (2, 7)

9. Show that log (162/343) + 2 log(7/9) – log(1/7) = log 2

Solution:

LHS = log (162/343) + 2 log(7/9) – log(1/7)

= log (162/343) + log (7/9)2 – log (1/7)

= log (162/343) + log (49/81) – log (1/7)

= log [(162 Ã— 49)/ (343 Ã— 81)] – log (1/7)

= log (2/7) – log (1/7)

= log [(2/7)/ (1/7)]

= log 2

= RHS

Therefore, log (162/343) + 2 log(7/9) – log(1/7) = log 2

10. If the equation kx2 – 2kx + 6 = 0 has equal roots, then find the value of k.

Solution:

Given,

kx2 – 2kx + 6 = 0

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

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

Given that, the equation has equal roots.

â‡’ b2 – 4ac = 0

â‡’ (-2k)2 – 4(k)(6) = 0

â‡’ 4k2 -24k = 0

â‡’ 4k(k – 6) = 0

â‡’ k(k – 6) = 0

â‡’ k = 0, k = 6

k = 0 is not possible.

Hence, the value of k is 6.

11. Find the 7th term from the end of the arithmetic progression 7, 10, 13,…, 184.

Solution:

Given AP is:

7, 10, 13, â€¦., 184

First term = a = 7

Common difference = d = 3

nth term of AP

an = a + (n – 1)d

a7 = 7 + (7 – 1)3

= 7 + (6) (3)

= 7 + 18

= 25

Therefore, the 7th term of the AP is 25.

12. In the diagram on a Lunar eclipse, if the positions of Sun, Earth and Moon are shown by (-4, 6), (k, -2) and (5, -6) respectively, then find the value of k.

Solution:

Given that the diagram on a Lunar eclipse, if the positions of Sun, Earth and Moon are shown by (-4, 6), (k, -2) and (5, -6) respectively.

That means the points lie on a straight line.

Hence, the area of the triangle formed by these points will be 0.

â‡’ 1/2 [-4(-2 + 6) + k(-6 – 6) + 5(6 + 2)] = 0

â‡’ -4(4) + k(-12) + 5(8) = 0

â‡’ -16 – 12k + 40 = 0

â‡’ 12k = 24

â‡’ k = 24/12

â‡’ k = 2

13. Given the linear equation 3x + 4y = 11, write linear equations in two variables such that their geometrical representations form parallel lines and intersecting lines.

Solution:

Given,

Equation of first line is 3x + 4y = 11

Let ax + by + c be the equation of the second line.

Condition for parallel lines:

a1/a2 = b1/b2 â‰  c1/c2

3/a = 4/b â‰  11/c

Let us consider some values of a, b, c which satisfies the above condition.

a = 6, b = 8, c = 14

Therefore, the equation of line which is parallel to the given line is 6x + 8y + 14 = 0

Condition for intersecting lines:

a1/a2 â‰  b1/b2

3/a â‰  4/b

Let us consider some values of a, b, c which satisfies the above condition.

a = 8, b = 6, c = 11

8x + 6y + 11 = 0

Therefore, the equation of line which is perpendicular to the given line is 8x + 6y + 11 = 0

SECTION – III

14. Find the points of trisection of the line segment joining the points (-2, 1) and (7, 4).

Solution:

Let the given points be:

A = (x1, y1) = (-2, 1)

B = (x2, y2) = (7, 4)

P divides AB in the ratio 1 : 2.

Here, m : n = 1 : 2

Using the section formula,

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

= [(7 – 4)/ (1 + 2), (4 + 2)/ (1 + 2)]

= (3/3, 6/3)

= (1, 2)

Q is the midpoint of PB.

Q = [(1 + 7)/2, (2 + 4)/3]

= (8/2, 6/2)

= (4, 3)

Hence, the required points are (1, 2) and (4, 3).

OR

Sum of squares of two consecutive even numbers is 580. Find the numbers by writing a suitable quadratic equation.

Solution:

Let x and (x + 2) be the two consecutive even numbers.

According to the given,

x2 + (x + 2)2 = 580

x2 + x2 + 4x + 4 – 580 = 0

2x2 + 4x – 576 = 0

2(x2 + 2x – 288) = 0

x2 + 2x – 288 = 0

x2 + 18x – 16x – 288 = 0

x(x + 18) – 16(x + 18) = 0

(x – 16)(x + 18) = 0

x = 16, x = -18

If x = 16, x + 2 = 18

If x = -18, x + 2 = -18 + 2 = -16

Hence, the two consecutive even numbers are 16, 18 or -18, 16.

15. Prove that âˆš3 + âˆš5 is an irrational number.

Solution:

Let âˆš3 + âˆš5 be a rational number.

âˆš3 + âˆš5 = a, where a is an integer.

Squaring on both sides,

(âˆš3 + âˆš5)2 = a2

(âˆš3)2 + (âˆš5)2 + 2(âˆš3)(âˆš5) = a2

3 + 5 + 2âˆš15 = a2

8 + 2âˆš15 = a2

2âˆš15 = a2 – 8

âˆš15 = (a2 – 8)/2

(a2 – 8)/2 is a rational number since a is an integer.

Therefore, âˆš15 is also an integer.

We know that integers are not rational numbers.

Thus, our assumption that âˆš3 + âˆš5 is a rational number is wrong.

Hence, âˆš3 + âˆš5 is an irrational number.

OR

Show that cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number.

Solution:

Let a be the positive integer.

By Euclid’s division lemma,

a = bq + r where 0 â‰¤ r < b

Let b = 8 then a = 8q + r

Where, r = 0, 1, 2, 3, 4, 5, 6, 7

When r = 0

a = 8q

aÂ³ = (8q)Â³

= 512qÂ³

= 8(64qÂ³)

= 8m where m = 64qÂ³

When r = 1,

a = 8q + 1

aÂ³ = (8q + 1)Â³

aÂ³ = 512qÂ³ + 1 + 3(8q)(8q + 1)

= 512qÂ³ + 1 + 24q(8q + 1)

= 512qÂ³ + 1 + 192qÂ² + 24q

= 8( 64qÂ³ + 24qÂ² + 3q) + 1

= 8m + 1 where m = 64qÂ³ + 24qÂ² + 3q

When r = 2,

a = 8q + 2

aÂ³ = (8q + 2)Â³

= 512qÂ³ + 8 + 48q(8q + 2)

= 512qÂ³ + 8 + 384qÂ²+ 96q

= 512qÂ³ + 384qÂ²+ 96q + 8

= 8 ( 64qÂ³ + 48qÂ² + 12q + 1 )

= 8m

where m = 64qÂ³ + 48qÂ² + 12q+ 1

When r = 3,

a = 8q + 3

aÂ³ = (8q + 3)Â³

= 512qÂ³ + 27 + 72q(8q + 3)

= 8(64qÂ³+ 3 + 72qÂ²+ 27q ) + 3

= 8m+ 3

where m = 64qÂ³ + 72qÂ² + 27q+3)

When r = 4,

a = 8q + 4

aÂ³ = (8q + 4)Â³

= 512qÂ³ + 64 + 768qÂ² + 384q

aÂ³ = 8( 64qÂ³ + 8 + 96qÂ² + 48q)

aÂ³ = 8m

where m = 64qÂ³ + 8 + 96qÂ² + 48q

when r = 5,

a = 8q + 5

aÂ³ = (8q + 5)Â³

= 512qÂ³ + 960qÂ² + 600q + 125

= 8 ( 64qÂ³ + 120qÂ² + 75q + 15) + 5

= 8m + 5

where m = 64qÂ³ + 120qÂ² + 75q + 15

When r = 6,

a = 8q + 6

aÂ³ = (8q + 6)Â³

= 512qÂ³ + 1152qÂ² + 864q + 216

= 8 ( 64qÂ³ + 144qÂ² + 108q + 27 )

= 8m

where m = 64qÂ³ + 144qÂ² + 108q + 27

When r = 7,

a = 8q + 7

aÂ³ = (8q + 7)Â³

= 512qÂ³ + 343 + 1344qÂ² + 1176 q

= 8( 64qÂ³ + 168qÂ² + 147q + 42) + 7

= 8m + 7

where m = 64qÂ³ + 168qÂ² + 147q + 42

Therefore, the cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number.

16. Find the solution of x + 2y = 10 and 2x + 4y = 8 graphically.

Solution:

Given,

x + 2y = 10

2x + 4y = 8

Consider the first equation,

x + 2y = 10

2y = -x + 10

y = -(1/2)x + 5

 x -2 0 2 4 y 6 5 4 3

Now, consider the another equation,

2x + 4y = 8

4y = -2x + 8

y = -(1/2)x + 2

 x -4 -2 0 4 y 4 3 2 0

The lines representing the given pair of equations are parallel to each other.

Hence, there is no solution for the given pair of equations.

OR

A = {x : x is a perfect square, x < 50, x âˆˆ N}

B = {x : x = 8m + 1, where m âˆˆ W, x < 50, x âˆˆ N}

Find A â‹‚ B and display it with a Venn diagram.

Solution:

Given,

A = {x : x is a perfect square, x < 50, x âˆˆ N}

A = {1, 4, 9, 16, 25, 36, 49}

B = {x : x = 8m + 1, where m âˆˆ W, x < 50, x âˆˆ N}

B = {1, 9, 17, 25, 33, 41, 49}

A â‹‚ B = {1, 4, 9, 16, 25, 36, 49} â‹‚ {1, 9, 17, 25, 33, 41, 49}

= {1, 9, 25, 49}

17. Find the sum of all two digit positive integers which are divisible by 3 but not by 2.

Solution:

Two digit positive numbers which are divisible by 3 but not by 2 are

15, 21, 27, 33, ….., 99

This is an AP with a = 15 and d = 6.

Last term = l = 99

n = 15

Sn = n/2 (a + l)

= (15/2) Ã— (15 + 99)

= (15/2) Ã— 114

= 15 Ã— 57

= 855

Hence, the required sum is 855.

OR

Total number of pencils required are given by 4x4 + 2x3 – 2x2 + 62x – 66. If each box contains x2 + 2x – 3 pencils, then find the number of boxes to be purchased.

Solution:

Given,

Total number of pencils = 4x4 + 2x3 – 2x2 + 62x – 66

Number of pencils in each box = x2 + 2x – 3

Hence, the total number of boxes required to be purchased = 4x2 – 6x + 22