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