Andhra Pradesh SSC Board Question Paper for Class 10th Maths Paper 1 2015 In PDF

AP SSC or 10th Class Question Paper Mathematics Paper 1 English Medium 2015 with Solutions – Free Download

Andhra Pradesh SSC (Class 10) Maths 2015 question paper 1 with solutions are provided here in a downloadable pdf format and also in the text so that the students can easily access them. Along with the solutions, they can also get the Maths question paper 1 2015 Class 10 SSC for reference. Students are able to access all the Andhra Pradesh board previous year maths question papers here. AP 10th Class Mathematics Question Paper 2015 Paper 1 can be easily downloaded in PDF format. Students can practice this Andhra Pradesh SSC 10th Class Maths 2015 question paper 1 solutions and verify the answers provided by BYJU’S. Solving 2015 Maths question paper 1 for Class 10 will help the students to predict what type of questions will appear in the exam.

Download AP SSC 2015 Question Paper Maths Paper 1 :-Download PDF

Download AP SSC 2015 Question Paper Maths Paper 1 With Solutions :-Download PDF

Andhra Pradesh SSC Class 10th Maths Question Paper 1 With Solution 2015

QUESTION PAPER CODE 15E(A)

SECTION – I

GROUP – A

(5 * 2 = 10)

Answer ANY 5 Questions choosing two from each of the following groups.

Question 1: Find L.C.M and H.C.F of 72 and 108 by the prime factorisation method.

Solution:

H.C.F of 72 and 108

H.C.F. of 72 and 108 = 2*2*3*3

= 36

L.C.M. of 72 and 108

L.C.M. of 72 and 108 = 2*2*2*3*3*3

= 216

So, H.C.F. of 72 and 108 is 36 and L.C.M of 72 and 108 is 216

Question 2: If V = {a, e, i, o, u} and B = {a, i, k, u}, then find V – B and B – V. Are they equal?

Solution:

V – B = {e, o}

B – V = {k}

No; they are not equal.

Question 3: Find a quadratic polynomial, the sum and product of whose zeroes are 1 / 4, – 1 respectively.

Solution:

Let the roots be a,b.

Then the equation of polynomial is given by

x² – (a + b)x + ab ————–(1)

The sum of roots = 1/4

a + b = 1 / 4

product of roots = -1

ab = -1

Substituting above values in (1)

∴ Equation of the required polynomial is x² – (1 / 4)x – 1 = 0 or 4x² – x – 4 = 0.

Question 4: Find two numbers, whose sum is 27 and product is 182.

Solution:

Let the first number be x and the second number is 27 – x.

Therefore, their product = x (27 – x)

It is given that the product of these numbers is 182.

Therefore, x(27 – x) = 182

⇒ x² – 27x + 182 = 0

⇒ x² – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13) (x -14) = 0

Either x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

If first number = 13, then the other number = 27 – 13 = 14

If first number = 14, then the other number = 27 – 14 = 13

Therefore, the numbers are 13 and 14.

GROUP – B

Question 5: For what value of ‘k’, the pair equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines.

Solution:

The coincident lines are 3x + 4y + 2 = 0 and 9x + 12y + k = 0.

Comparing the equations with the general equations

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Then,

a₁ = 3, b₁ = 4, c₁ = 2, a₂ = 9, b₂ = 12, c₂ = k

Since, the lines are coincident, then,

a₁ / a₂ = b₁ / b₂ = c₁ / c₂

3 / 9 = 4 / 12 = 2 / k

1 / 3 = 1 / 3 = 2 / k

1 / 3 = 2 / k

k = 6

Question 6: In a flower bed, there are 23 rose plants in the first row, 21 in the second row, 19 in the third row, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Solution:

a₁ = 23 , a₂ = 21 , a₃ = 19 and a_n = 5 = l; S_n= ?

a_n = a + (n – 1)d

=> 5 = 23 + (n – 1) (-2)

=> 5 = 23 – 2n + 2

=> 5 = 25 – 2n

=> 2n = 25 – 5

=> n = 20 / 2

=> n = 10

S_n = [n / 2] (2a + [n – 1] d)

S₁₀ = [10 / 2] (2 * 23 + 9 * -2)

S₁₀ = 5 × 28

= 140

So, there are 140 rose plants in the flower bed.

Question 7: Find the coordinates of a point A, where AB is the diameter of a circle, whose centre is (2, -3) and B is (1, 4).

Solution:

2 = [x +1] / 2

4 – x = 1

x = 3

-3 = [y + 4] / 2

-6 – y = 4

y = -10

The coordinates of A are (3, -10).

Question 8: Find the area of the triangle, whose vertices are (-5, -1), (3, -5), (5, 2).

Solution:

Area = [1 / 2] [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= [1 / 2] [-5 * (-5 -2) + 3 (2 + 1) + 5 (-1 + 5)]

= [1 / 2] [35 + 9 + 20]

= [1 / 2] * 64

= 32 cm²

SECTION – II

Answer ANY 4 of the following six questions.

Question 9: Expand log 15.

Solution:

log 15 = log5 3

= log5 log3

Question 10: Write roster and set builder form of “The set of all-natural numbers, which divide 42”.

Solution:

(i) x = {1, 2, 3, 6, 7}

(ii) x = {x : x = 1, 2, 3, 6, 7}

Question 11: If p(t) = t³ – 1, find the values of p (1), p (–2).

Solution:

p(1) = 1³ – 1 = 0

p(-2) = -2³ – 1 = -8 – 1 = -9

Question 12: Formulate a pair of linear equations in two variables “5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46”.

Solution:

Let pencil = x

pen = y

5x + 7y = 50 * 5

7x + 5y = 46 * 7

25x + 35y = 250

49x + 35y = 322

______________

24x = 72

x = 42 / 24

x = 3

Pencil = 3

y = [46 – 7x] / 5

= [46 – 7 * 3] / 5

= 25 / 5

= 5

y = 5

Pen = 5

Question 13: If b² – 4ac ≥ 0, then write the roots of a quadratic equation ax² + bx + c = 0.

Solution:

Real, equal and distinct.

Question 14: Write the formula for the sum of first ‘n’ positive integers.

Solution:

n [n + 1] / 2

SECTION – III

[4 * 4 = 16]

Answer ANY 4 questions, choosing atleast two from each of the following groups.

GROUP – A

Question 15: Prove that √5 is irrational by the method of Contradiction.

Solution:

Say, √5 is a rational number.

∴ It can be expressed in the form p / q where p,q are co-prime integers.

⇒√5 = p / q

⇒ 5 = p² / q² {Squaring both the sides}

⇒ 5q² = p² — (1)

⇒ p² is a multiple of 5. {Euclid’s Division Lemma}

⇒ p is also a multiple of 5. {Fundamental Theorem of arithmetic}

⇒ p = 5m

⇒ p² = 25m² — (2)

From equations (1) and (2), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5. {Euclid’s Division Lemma}

⇒ q is a multiple of 5.{Fundamental Theorem of Arithmetic}

Hence, p,q have a common factor 5.

This contradicts that they are co-primes.

Therefore, p / q is not a rational number.

So, √5 is an irrational number.

Question 16: If A = {3, 6, 9, 12, 15, 18, 21}; B = {4, 8, 12, 16, 20}; C = {2, 4, 6, 8, 10, 12, 14, 16}; D = {5, 10, 15, 20}; find

(i) A – B

(ii) B – A

(iii) C – A

(iv) D – A

(v) B – C

(vi) B – D

(vii) C – B

(viii) D – B

Solution:

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) B – A = {4, 8, 16, 20}

(iii) C – A = {2, 4, 8, 10, 14, 16}

(iv) D – A = {5, 10, 20}

(v) B – C = {20}

(vi) B – D = {4, 8, 12, 16}

(vii) C – B = {2, 6, 10, 14}

(viii) D – B = {5, 10, 15}

Question 17: Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficient.

Solution:

To verify roots, put root inside equations x³ + 3x² – x – 3

Root 1 ⟶ 1³ + 3(1)² – 1 – 3 = 1 + 3 – 1 – 3 = 0

Root (-1) = -1³ + 3 (-1)² + 1 – 3 = – 1 + 3 + 1 – 3 = 0

Root (-3) = (-3)³ + 3 (-3)² – (-3) – 3 = – 27 + 27 + 3 – 3 = 0

Sum of root = -b / a = -3 / 1 = -3

From roots = (1 – 1 – 3) = –3

Product of roots = -d / a = 3 / 1 = 3

From roots = 1 * -1 * -3 = 3

Product of roots = pq + qr + rp = c / a = -1 / 1 = -1

From roots = 1 * –1 + –1 * –3 + –3 * 1 = – 1 + 3 – 3 = –1

Hence verified.

Question 18: Find the root of equation 2x² – x – 4 = 0 by the method of completing the square.

Solution:

2𝑥² − 1𝑥 − 4 = 0

𝑎 ≠ 1 so divide through by 2

𝑥² − [1 / 2] 𝑥 − 2 = 0

Keep 𝑥 terms on the left and move the constant to the right side by adding it on both sides

𝑥² − [1 / 2] 𝑥 = 2

Take half of the 𝑥 term and square it

{[−1 / 2] ⋅ [1 / 2]}² = 1 / 16

Add the result to both sides

𝑥² − [1 / 2] 𝑥 + [1 / 16] = 2 + [1 / 16]

Rewrite the perfect square on the left

(𝑥 − (1 / 4))² = 2 + [1 / 16]

Combine terms on the right

(𝑥 − [1 / 4])² = 33 / 16

Take the square root of both sides

𝑥 − [1 / 4] = ± √33 / 16

Simplify the Radical term (0):

𝑥 − [1 / 4] = ± √33 / 4

Isolate the x on the left side and solve for x (1)

𝑥 = [1 / 4] ± √33 / 4

𝑥 = [1 / 4] + √33 / 4

𝑥 = [1 / 4] − √33 / 4

GROUP – B

Question 19: Solve the equations [5] / [x – 1] + [1] / [y – 2] and [6] / [x – 1] – [3] / [y – 2] = 1.

Solution:

let assume, 1 / [x – 1] = u, 1 / [y – 2] = v

15 * u = 3 * v = 6 —) 3

Add equations 2 and 3

21* u = 7

u = (1 / 3)

x – 1 = 3

x = 4

b = 1 / 3

y – 2 = 3

y = 5

Question 20: A fraction becomes 4 / 5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes 1 / 2. What is the fraction?

Solution:

Let the fraction be p / q.

The fraction will become 4 / 5 if 1 is added to numerator and denominator

So,

4 / 5 = (p + 1) / (q + 1)

4q + 4 = 5p + 5

(p – 5) / (q – 5) = 1 / 2

2p – 10 = q – 5

q = 2p – 5

4(2p – 5) + 4 = 5p + 5

8p – 20 + 4 = 5p + 5

3p = 21

p = 7

Therefore q = 2 * 7 – 5

q = 9

So the fraction is 7/9.

Question 21: The sum of the 4^th and 8^th terms of an A.P is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.

Solution:

Let the first term of an AP = a and the common difference of the given AP = d.

a_n = a + (n − 1) d

a₄ = a + (4 − 1) d

a₄ = a + 3d

Similarly,

a₈ = a + 7d

a₆ = a + 5d

a₁₀ = a + 9d

The sum of 4^th and 8^th term = 24

a₄ + a₈ = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12……………….. (i)

The sum of 6^th and 10t^h term = 44

a₆ + a₁₀ = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 ………………….(ii)

Solving (i) and (ii),

a + 7d = 22

a + 5d = 12

2d = 10

d = 10 / 2

d = 5

From equation (i), we get,

a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a₂ = a + d = − 13 + 5 = −8

a₃ = a₂ + d = − 8 + 5 = −3

Question 22: Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the vertices of a parallelogram.

Solution:

Let say ABCD be the points of the parallelogram.

To prove: Correct order of vertices, diagonals bisect each other in the parallelogram. Hence point O must be some from A and C.

C = [-7 + 25] / 2, [-3 + 8] / 2

= [4, (5 / 2)]

From B and D

C = [-5 + 3] / 2, [10 – 5] / 2

= [4, (5 / 2)]

Hence, both are correct.

SECTION – IV

(1 * 5 = 5)

Answer ANY ONE question from the following.

Question 23: Draw the graph of y = x² – x – 6 and find zeroes. Justify the answer.

Solution:

Clearly the graph of y = x² − x − 6 cut the x-axis at x = −2, 3.

⇒x = −2, 3 are zeros of y = x² − x − 6.

Justification:

when x = −2 ; y = 4 + 2 − 6 = 0

when x = 3 ; y = 9 − 3 − 6 = 0

Question 24: Solve the pair linear equations graphically.

2x + y – 6 = 0

4x – 2y – 4 = 0

Solution:

Hence, the graph solution is (2,2).