Learn about smartphone sensors in the next Xcel masterclass with Tinkerbee CEO, Anupam! Learn about smartphone sensors in the next Xcel masterclass with Tinkerbee CEO, Anupam!

Andhra Pradesh SSC Board Question Paper for Class 10th Maths Paper 1 2017 In PDF

AP SSC or 10th Class Question Paper Mathematics Paper 1 English Medium 2017 with Solutions – Free Download

Andhra Pradesh SSC (Class 10) Maths 2017 question paper 1 with solutions are furnished 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 2017 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 2017 Paper 1 can be downloaded easily and students can avail them and verify the answers provided by BYJU’S. Solving 2017 Maths question paper 1 for Class 10 will help the students to predict the pattern and type of questions that will appear in the exam.

Download SSC 2017 Question Paper Maths Paper 1

Download SSC 2017 Question Paper Maths Paper 1 With Solutions

Andhra Pradesh SSC Class 10th Maths Question Paper 1 With Solution 2017

QUESTION PAPER CODE 15E(A)

 

ap class 10 maths question paper 1 sol march 2017 01
ap class 10 maths question paper 1 sol march 2017 02
ap class 10 maths question paper 1 sol march 2017 03
ap class 10 maths question paper 1 sol march 2017 04
ap class 10 maths question paper 1 sol march 2017 05
ap class 10 maths question paper 1 sol march 2017 06
ap class 10 maths question paper 1 sol march 2017 07
ap class 10 maths question paper 1 sol march 2017 08
ap class 10 maths question paper 1 sol march 2017 09
ap class 10 maths question paper 1 sol march 2017 10
ap class 10 maths question paper 1 sol march 2017 11
ap class 10 maths question paper 1 sol march 2017 12
ap class 10 maths question paper 1 sol march 2017 13
ap class 10 maths question paper 1 sol march 2017 14

 

SECTION – I

(4 * 1 = 4)

Question 1: Find the value of log2 512.

Solution:

512 is 29.

⇒ log2 (512) = log2 (29)

By the power Rule, bring the 9 to the front of the log.

= 9 log2 (2)

The logarithm of a to the base a is always 1.

log2(2) = 1

= 9

Question 2: Write A = {1, 4, 9, 16, 25} in set builder form.

Solution:

A = {x : x is the square of a natural number}

A = {x : x = n2, where n ∈ N}

Question 3: Two angles are complementary and one angle is 18o more than the other, then find the angles.

Solution:

Let one angle be x and another angle be x + 18.

Since the angles are complementary,

∠A + ∠B = 90°

x + x + 18° = 90°

2x = 90° – 18°

2x = 72°

x = 36°

The two angles are

∠A = 36°

∠B = 36 + 18 =54°

Question 4: Find the total surface area of a hemisphere of radius 7cm.

Solution:

Total surface area of a hemisphere = 3𝛑r2

Radius = 7 cm

= 3 * (22 / 7) * 7 * 7

= 3 * 22 * 7

= 21 * 22

= 462 sq.cm

SECTION – II

(5 * 2 = 10)

Question 5: Find the zeroes of the quadratic polynomial x2 – 2x – 8 and verify the relationship between zeroes and coefficients.

Solution:

P(x) => x² – 2x – 8

=> x² – 4x + 2x – 8

=> x (x – 4) + 2 (x – 4)

=> (x – 4) (x + 2) = 0

=> (x – 4) = 0 or (x + 2) = 0

=> x = 4 or x = -2

4 and -2 are the two zeroes of the polynomial x² – 2x – 8.

The relation between the zeroes and coefficients:

Sum of zeroes = ɑ + β

= 4 + (-2)

= 4 – 2

= 2 / 1

= coefficient of x / coefficient of x².

Product of zeroes = ɑ × β

= 4 × -2

= -8 / 1

= constant term / coefficient of x²

Question 6: Which term of AP 21, 18, 15, …… is -81?

Solution:

Let an = -81

d = 18 – 21 = -3

a = 21

The general formula of AP is an = a + (n – 1)d

-81 = 21 + (n – 1) -3

-81 – 21 = (n – 1) -3

-102 / -3 = n – 1

34 = n – 1

n = 35

Hence -81 is 35th term of an AP.

Question 7: The curved surface area of a cone is 4070 cm2 and its diameter is 70cm. What will be its slant height?

Solution:

Given that curved surface area of the cone is 4070cm²

Diameter of cone = 70cm

Radius of cone = d / 2 = 70 / 2 = 35cm

Curved surface area of cone = πrl

πrl = 4070

(22 / 7) * 35 * l = 4070

22 * 5 * l = 4070

110 l = 4070

l = 4070 / 110

l = 37 cm

Question 8: Find the discriminant of 2x2 – 4x + 3 = 0 and discuss the nature of its roots.

Solution:

The given equation is of the form ax2 + bx + c = 0, where

a = 2

b = – 4

c = 3

Therefore, the discriminant is

D = b² – 4ac

= (– 4)2 – (4 × 2 × 3)

= 16 – 24

= – 8 < 0

Since, D < 0, the equation has no real roots.

Question 9: Express as algebraic expressions of the following:

[a] Five times of a number, when increased by 10 gives 20.

[b] The digits in ones and tens place of a two-digit number are x and y, then find the number.

Solution:

[a] 5x + 10 = 20

[b] Since, a number is written as the sum of all the place value of all digits in the number,

So, a two-digit number can be written as,

10 × ( tens place digit ) + ones place digit,

Here, ones place digit = X and the tens place digit = Y,

Thus, the required two-digit number is,

10 × Y + X = 10Y + X

SECTION – III

(4 * 4 = 16)

Question 10:

[a] Solve the following pair of equations by reducing them to a pair of linear equations.

(5 / [x – 1]) + (1 / [y – 2]) = 2, (6 / [x – 1]) – (3 / [y – 2]) = 1.

OR

[b] A well of diameter 14m is dug 15m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7m to form an embankment. Find the height of the embankment.

Solution:

[a] (5 / [x – 1]) + (1 / [y – 2]) = 2 —- (1)

(6 / [x – 1]) – (3 / [y – 2]) = 1 —– (2)

Let 1 / [x – 1] = u [from (1)]

1 / [y – 2] = v [from (2)]

The equations become,

5u + v = 2 —- (3)

6u – 3v = 1 —– (4)

From (3), 5u + v = 2

v = 2 – 5u

Putting the value of v in (4),

6u – 3v = 1

6u – 3 * (2 – 5u) = 1

6u + 15u = 7

21u = 7

u = 1 / 3

On putting u = (1 / 3) in (3),

5 * (1 / 3) + v = 2

v = 1 / 3

To find the values of x and y, substitute u and v in (1) and (2),

x = 4, y = 5 is the solution.

[b] Inner Diameter of the well= 14 m

Inner Radius of the well (r) = 14 / 2 m = 7 m

Height of the well (h) = 15 m

The volume of the earth taken out of the well = πr²h

= (22 / 7) × (7)² × 15

= 22 × 7 × 15

= 2310 m³

Width= 7m

Outer radius of the embankment R = inner radius + width

Outer radius (R)= 7 + 7 = 14m

The embankment is in the form of cylindrical shell, so the area of embankment

Area of embankment = outer area – inner area

= πR² – πr² = π (R² – r²)

= (22 / 7) ( 14² – 7²)

= 22 / 7 (196 – 49)

= 22/7 × 147

= 22 × 21

= 462 m²

The volume of embankment= volume of earth taken out on digging the well

Area of embankment × height of embankment= volume of earth dugout

Height of embankment= volume of earth dug out/area of the embankment

Height of the embankment = 2310 / 462

Height of embankment= 5 m

Hence, the height of the embankment so formed is 5 m.

Question 11:

[a] Show that the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8, where m is an integer.

OR

[b] If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}; then check whether A U B = B U A and A – B = B – A.

Solution:

[a] Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0, 1, 2

Therefore, every number can be represented as these three forms.

There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9 (3q ³ + 3q ² + q) + 1

a = 9m + 1 [ where m = 3q³ + 3q² + q ).

Case 3: When a = 3q + 2,

a = (3q +2)³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

[b] AUB = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21}

B U A = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21}

B U A = A U B is the same.

A – B = {3, 6, 9, 15, 18, 21}

B – A = {4, 8, 16, 20}

A – B is not the same as B – A.

Question 12:

[a] A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find

[i] The production in the 1st year.

[ii] Total production in the 10th year.

[iii] Total production in the first seven years.

OR

[b] There is a motorboat, whose speed in still water is 18km/h. It takes 1 hour more to go 24km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

[a] Let the number of sets produced in 1st year be ‘a’ and ‘d’ be the increase in the production every year.

a + 2d = 600 —– (1)

a + 6d = 700 —– (2)

Subtracting equation (1) from (2),

4d= 100 or d= 25

Substituting d = 25 in equation (1),

a = 550

(a) Production in the first year = a = 550

(b) Production in 10th year = a + 9d = 550 + 9 x 25 = 775

(c) Total production in first 7 years = a + (a + d) + (a + 2d) +….. + (a + 6d)

= (7 / 2) (2 * 550 + (7 – 1) 25)

= 4375

Question 13:

[a] Solve the quadratic polynomial x2 – 3x – 4 by graphical method.

OR

[b] Half of the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.

Solution:

[a]

If x = 0,

y= 02 – 3(0) – 4 = – 4

If x = 1

y= 12 – 3(1) – 4 = – 4 = 1 – 3 – 4 = – 6

If x = -1

y = y= (-1)2 – 3(-1) – 4 = 1+3 – 4 = 0

If x = 4

y = ​​​​​​​(-1)2 – 3(4) – 4 = 16 – 12 – 4 = 0

The coordinate points of the given equation are

x 0 1 – 1 4
y – 4 – 6 0 0

AP Class 10 Maths QP solutions 2017 Paper 1 Question Number 13b

[b] Perimeter = 2 (l+b)

half of the perimeter = 36m = (l+b)

Let breadth = x, length= 4x

Perimeter = 2 × 36 = 72m

Perimeter = 2 (l+b)

72 = 2 (4 + x + x)

36 = 4 + 2x

36 – 4 = 2x

32 = 2x

32 / 2 = x

16 = x

Breadth = 16

Length = x + 4 = 16 + 4 = 20

SECTION – IV

(20 * 0.5 = 10)

Question 14: A rational number that equals to (2.6[bar]) is

(A) 7 / 3 (B) 8 / 3 (C) 16 / 7 (D) 17 / 7

Answer: B

Question 15: The value of log25 5 is

(A) 1 / 2 (B) 2 (C) 5 (D) 25

Answer: A

Question 16: If ‘4’ is one of the zeroes of p(x) = x2 + kx – 8, then the value of k is

(A) 1 (B) -1 (C) 2 (D) -2

Answer: C

Question 17: If the pair of equations 2x + 3y + k = 0, 6x + 9y + 3 = 0 having infinite solutions, the value of k is ____

(A) 2 (B) 3 (C) 0 (D) 1

Answer: D

Question 18: If the roots of x2 + 6x + 5 = 0 are a and b, then a + b is

(A) 5 (B) -6 (C) 6 (D) -1

Answer: B

Question 19: Which term of GP 3, 3√3, 9 ……… equals to 243?

(A) 6 (B) 7 (C) 8 (D) 9

Answer: D

Question 20: If n(A) = 12 and n(A ∩ B) = 5, then n(A – B) =

(A) 4 (B) 7 (C) 17 (D) 0

Answer: B

Question 21: If x, x + 2, x + 6 are three consecutive terms in GP, find the value of x.

(A) 3 (B) 4 (C) 2 (D) 1

Answer: C

Question 22: A quadratic equation, whose roots are 2 + √3 and 2 – √3 =

(A) x2 – x – 4 = 0

(B) x2 – 4x + 1 = 0

(C) x2 + 4x + 3 = 0

(D) x2 + x – 3 = 0

Answer: B

Question 23: If an = n (n + 3) / (n + 2), then find a17.

(A) 340 / 20 (B) 341 / 19 (C) 340 / 19 (D) 341 / 20

Answer: C

Question 24: The curved surface area of a sphere will be ____ whose radius is 10cm.

(A) 239𝛑 (B) 400𝛑 (C) 221𝛑 (D) 129𝛑

Answer: B

Question 25: The volume of a cube will be ___ if the total surface area is 216cm2.

(A) 216 (B) 196 (C) 212 (D) 144

Answer: A

Question 26: A famous book written by ancient mathematician Aryabhata is

(A) Arya tharkam

(B) Aryabhatteeyam

(C) Siddhanta Shiromani

(D) Karana Kuthuhalam

Answer: B

Question 27: The degree of the polynomial √2x2 – 3x + 1 =

(A) √2 (B) 3 (C) 1 (D) 2

Answer: C

Question 28: Which of the following equations has the solution of (2, -3)?

(A) 2x – 3y = 10 (B) 2x + 3y = 13 (C) 2x – 3y = 13 (D) 2x + 3y = -13

Answer: The set of equations should be two.

Question 29: If A = {x : x is the letter of the word HEADMASTER, then its roster form is

(A) A = {h, e, a, d, m, a, s, t, e, r}

(B) A = {h, e, a, d, m, s, t, r}

(C) A = {h, e, a, d, m, s, t, e, r}

(D) A = {h, e, a, d, m, a, s, t, r}

Answer: B

Question 30: The following Venn diagram indicates

AP Class 10 Maths QP solutions 2017 Paper 1 Question Number 30

(A) A ⊂ B

(B) B ⊂ A

(C) A, B are disjoint sets

(D) ᶬ ⊂ B

Solution: B

Question 31:

AP Class 10 Maths QP solutions 2017 Paper 1 Question Number 31

The above diagram shows

(A) b² – 4ac > 0

(B) b² – 4ac = 0

(C) b² – 4ac < 0

(D) None of the above

Solution: A

Question 32: Which of the following vessels can be filled with more water (A, B are in cylindrical shape)?

AP Class 10 Maths QP solutions 2017 Paper 1 Question Number 32

(A) A

(B) B

(C) Both

(D) cannot be determined

Answer: B

Question 33:

AP Class 10 Maths QP solutions 2017 Paper 1 Question Number 33

The number of zeroes can be identified by the adjacent figure.

(A) 0 (B) 1 (C) 2 (D) 3

Answer: D

 

 

Leave a Comment

Your Mobile number and Email id will not be published.

*

*

close
close

DOWNLOAD

App Now