## 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)

**SECTION – I**

** (4 * 1 = 4)**

**Question 1: Find the value of log _{2} 512.**

**Solution: **

512 is 2^{9}.

⇒ log_{2 }(512) = log_{2 }(2^{9})

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

= 9 log_{2 }(2)

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

log_{2}(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 = n^{2}, where n ∈ N}

**Question 3: Two angles are complementary and one angle is 18 ^{o } 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 = 3𝛑r^{2}

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 x ^{2} – 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 a_{n} = -81

d = 18 – 21 = -3

a = 21

The general formula of AP is a_{n }= 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 35^{th} term of an AP.

**Question 7: The curved surface area of a cone is 4070 cm ^{2} 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 2x ^{2} – 4x + 3 = 0 and discuss the nature of its roots.**

**Solution:**

The given equation is of the form ax^{2} + 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:**

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:**

(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 mInner 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 = 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 7 ^{th} year. Assuming that the production increases uniformly by a fixed number every year, find **

**[i] The production in the 1 ^{st} year.**

**[ii] Total production in the 10 ^{th} 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 + 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 10^{th} 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 x ^{2} – 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:**

If x = 0,

**⇒ ** y= 0^{2} – 3(0) – 4 = – 4

If x = 1

**⇒ **y= 1^{2} – 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 |

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 log _{25} 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) = x ^{2 }+ 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 x ^{2} + 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) x ^{2} – x – 4 = 0 **

**(B) x ^{2} – 4x + 1 = 0 **

**(C) x ^{2} + 4x + 3 = 0 **

**(D) x ^{2} + x – 3 = 0**

**Answer: B**

**Question 23: If a _{n} = n (n + 3) / (n + 2), then find a_{17}.**

**(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 216cm ^{2}.**

**(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 √2x ^{2} – 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**

** **

**(A) A ⊂ B**

**(B) B ⊂ A**

**(C) A, B are disjoint sets**

**(D) ᶬ ⊂ B**

**Solution: B**

**Question 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)?**

**(A) A**

**(B) B**

**(C) Both**

**(D) cannot be determined **

**Answer: B**

**Question 33: **

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

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

**Answer: D**