AP SSC or 10th Class Question Paper Mathematics Paper 2 English Medium 2015 with Solutions – Free Download
Andhra Pradesh SSC (Class 10) maths 2015 question paper 2 with solutions are given here in a downloadable pdf format and also in the text so that the students can easily get them. Along with the solutions, they can also access the maths question paper 2 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 2 can be downloaded easily, and students can practice and verify the answers provided by BYJU’S. Solving 2015 Maths question paper 2 for Class 10 will help the students to predict the type of questions that will appear in the exam.
Download SSC 2015 Question Paper Maths Paper 2
Download SSC 2015 Question Paper Maths Paper 2 With Solutions
Andhra Pradesh SSC Class 10th Maths Question Paper 2 With Solution 2015
QUESTION PAPER CODE 16E(A)
SECTION – I
GROUP – A
(5 * 2 = 10)
Answer ANY 5 Questions choosing two from each of the following groups.
Question 1: What value of x will make DE || AB, in the given figure?
AD = 8x + 9, CD = x + 3, BE = 3x + 4, CE = x.
Solution:
In ∆ABC , DE || BC
CD / DA = CE / BE [ Thales theorem ]
(x + 3) / (8x + 9) = x / (3x + 4)
=> (x + 3) (3x + 4 ) = x(8x + 9)
=> 3x² + 4x + 9x + 12 = 8x² + 9x
=> 3x² + 13x + 12 = 8x² + 9x
=> 5x² – 4x – 12 = 0
=> 5x² – 10x + 6x – 12 = 0
=> 5x(x – 2) + 6 (x – 2) = 0
=> (x – 2)(5x + 6) = 0
x – 2 = 0 or 5x + 6 = 0
Therefore, x = 2 or x = -6 / 5.
Question 2: Prove that “The lengths of tangents drawn from an external point to a circle are equal”.
Solution:
Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA ⟂ PA and OB ⟂ PB
∠OAP = ∠OBP = 90^{o} … (1)
In OPA and OPB,
∠OAP = ∠OBP (Using (1))
OA = OB (Radii of the same circle)
OP = PO (Common side)
Therefore, OPA ⩭ OPB (RHS congruence criterion)
PA = PB (Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
Question 3: The curved surface area of a cone is 470cm^{2}, and its diameter is 70cm. What is its slant height?
Solution:
Curved surface area of cone = 4070sq.cm
Diameter = 70cm
Radius (r) = 70 / 2 = 35
The curved surface of cone =πrl
4070 = 22/7 . 35 . l
4070×7/22×35 = l
4070/22×5 = l
4070/110 = l
37 =l
Therefore, slant height = 37cm
Question 4: Two cubes of each volume 64cm^{3} are joined end to end together. Find the surface area of the resulting cuboid.
Solution:
Given, the volume of each cube is 64 cm³
⇒ s³ = 64
⇒s = ∛64
⇒s = 4 cm
When two cubes are joined, the length of the resulting cuboid(l)
= side + side
= 4 + 4
= 8 cm
Breadth(b) = side = 4cm
Height(h) = side = 4cm
Total surface area of a cuboid = 2(lb + bh + hl)
⇒ = 2[(8) (4) + (4) (4) + (4) (8)]
⇒ = 2 (32 + 16 + 32)
⇒ = 2 (80)
⇒ = 160 cm²
∴ The surface area of the resulting cuboid is 160 cm².
Question 5: Find the mean of the following data: 5, 6, 9, 10, 6, 12, 3, 6, 11, 10.
Solution:
Given,
5, 6, 9, 10, 6, 12, 3, 6, 11, 10
Mean = Sum of observations / No.of observations.
Mean = 5 + 6 + 9 + 10 + 6 + 12 + 3 + 6 + 11 + 10 / 10
=》78 / 10
=》 7.8
Question 6: A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
[a] red
[b] white
[c] not green?
Solution:
Total number of marbles in the box = 5 + 8 + 4 = 17
[a] P(red) = number of red marbles / total number of marblesP(red) = 5 / 17
[b] P(white) = no. of white marbles/total number of marblesP(white) = 8 / 17
[c] P(green) = no. of green marbles/total number of marblesP(green) = 4 / 17
P(not green) = 1 − P(green) = 1 – [4 / 17] = 13 / 17
Question 7: If cos 7A = sin (A – 6^{o}), where 7A is an acute angle. Find the value of A.
Solution:
When the angle is acute,
cosθ = sin (90^{o }− θ)
cos7A = sin(90^{o }− 7A)
⇒ sin (90^{o }− 7A) = sin (A − 6^{o})
⇒ 90^{o }− 7A = A − 6^{o}
⇒ 8A = 96^{o}
⇒ A = 12^{o}
Question 8: Length of the shadow of a 15m high pole is 5√3 meters at 7’o clock in the morning. Then what is the angle of elevation of the sun rays with the ground at that time?
Solution:
The problem is in the form of a right triangle, where the height of the pole is 15m, the shadow of the pole is 5√3 m.
Let Ф be the angle of elevation.
tan Ф = opposite / adjacent
=> tan Ф = 15 / 5√3
=> tan Ф = √3
=> tan 60° = √3
Ф = 60°
Hence the angle of elevation at that time was 60°.
SECTION – II
(4 * 1 = 4)
Answer ANY 4 of the following six questions.
Question 9: Write the formula for the median of a grouped data. Explain the symbols along with their usual meanings.
Solution:
Median = L + (n / 2 – cf)* h / f
where, L = lower limit of median class
n = number of observations
cf = cumulative frequency of class preceding the median class
f = frequency of median class
h = class size
Question 10: Find the total surface area of a hemisphere of radius 3.5cm.
Solution:
The volume of a hemisphere = 2πr³ / 3
= [2 / 3] * [22 / 7] * r³
= (2 * 22 * 3.5 * 3.5 * 3.5) / [7 * 3]
= (44 * 0.5 * 3.5 * 3.5) / 3
= 269.5 / 3
= 89.83 cm^{3}
Question 11: If P(E) = 0.05, what is the probability of “not E’?
Solution:
P(E) = 0.05
P(not E) = 1 – P(E)
= 1 – 0.05
= 0.95
Question 12: Calculate the length of the tangent from a point 15cm away from the centre of a circle of radius 9cm.
Solution:
Length of the tangent will be 12cm.
The radius of the circle = 9 cm (Given)
Distance to the tangent from the centre = 15 cm (Given)
Let the length of the tangent be = x
(Distance from the centre)² = (radius)² + (length of the tangent)²
Therefore,
Length of tangent^{2} = (15)² – (9)²
= 225 – 81
= 144
= √144
= 12
Therefore, the length of the tangent is 12cm with the endpoints of diameter being parallel.
Question 13: Can mode be calculated for grouped data with unequal class sizes?
Solution:
Yes, a mode can be calculated for grouped data with unequal class sizes.
Question 14: If sin A = cos A, then find the value of A in degrees.
Solution:
sin A = cos A
sin A = cos A
sin A = sin (90 – A)
A = 90 – A
A + A = 90
2A = 90
A = 90 / 2
A = 45^{o}
SECTION – III
(4 * 4 = 16)
Answer ANY 4 of the following questions choosing atleast 2 from each group.
Question 15: In the adjoining figure, MP, LR and NQ are perpendicular to the straight line MLN. If MP = x, LR = z, NQ = y, prove that [1 / x] + [1 / y] = [1 / z].
Solution:
In triangle NMP and NLR,
PM || RL
So △NMP ~ △NLR
z / x = LN / MN —- (1)
So △MNP ~ △MLR
QN || RL
z / y = ML / MN —- (2)
Adding equation (1) and (2) we get,
[z / x] + [z / y] = [LN / MN] + [ML / MN] [z / x] + [z / y] = [MN / MN] [z / x] + [z / y] = 1 [1 / x] + [1 / y] = [1 / z]Question 16: A hemispherical bowl of internal radius 15cm contains a liquid. The liquid is to be filled into cylindrical bottles of diameter 5cm and height 6cm. How many bottles are necessary to empty the bowl?
Solution:
Let the number of cylindrical bottles be ‘n’. Then,
n × Volume of 1 bottle = Volume of hemisphere bowl
⇒ n × πr^{2}h = [2 / 3] πR^{3}
⇒ n × (2.5)^{2 }× 6 = [2 / 3] × (15)^{3}
⇒ n = [2 × 15 × 15 × 15] / {[2.5^{2} * 6 * 3]}
= 60
Question 17: “O” is any point inside a rectangle ABCD. Prove that OB^{2} + OD^{2} = OA^{2} + OC^{2}.
Solution:
We draw PQ ║AB ║CD as shown in the figure,
ABCD is a rectangle; it means ABPQ and PQDC are also rectangles.
For ABPQ
AP = BQ [opposite sides are equal ]
For, PQDC
PD = QC [ opposite sides are equal ]
now, for ∆OPD,
OD² = OP² + PD² ——(1)
For, ∆OQB,
OB² = OQ² + BQ² ——–(2)
Add equations (1) and (2),
OB² + OD²
= (OP² + PD²)+ (OQ² + BQ²)
= (OP² + CQ²) + (OQ² + AP²)
As you can see in the figure,
∆OPA and ∆OQC are right-angled triangles.
For, ∆OCQ ⇒ OQ² + CQ² = OC²
For, ∆OPA ⇒ OP² + AP² = OA²,
Now,
OB² + OD² = OC² + OA²
Question 18: A heap of rice is in the form of a cone of a diameter 12m and height 8m. Find its volume. How much canvas cloth is required to cover the heap.
Solution:
Diameter = 12 m
Radius = 12 / 2 = 6 m
Height = 8 m
The volume of rice in the heap = (1 / 3) πr²h
= (1 / 3) * (22 / 7) * (6)² * 8
= 301.71 m³
l = √r² + h²
= √6² + 8²
= 10 m
∴ Canvas cloth required to cover the heap = πrl
= (22 / 7) * 6 * 10
= 188.57 cm²
Question 19: If 4 sin^{2} θ – 1 = 0 and θ is less than 90^{o}, find the value of θ and the value of cos^{2} θ + tan^{2} θ.
Solution:
Given that, 4sin^{2 }θ − 1 = 0 or, sin^{2 }θ = 1 / 4
sin θ = √1 / 4 = 1 / 2 (θ < 90^{o})
θ = 30^{o}
cos^{2 }θ + tan^{2 }θ
= cos^{2 }30^{o }+ tan^{2} 30^{o}
= (√3 / 2)^{2 }+ (1 / √3)^{2}
= [3 / 4] + [1 / 3]
= 13 / 12
Question 20: Suppose you are shooting an arrow from the top of a building at the height of 6m to a target on the ground at an angle of depression of 60^{o}. What is the distance between you and the object?
Solution:
Let AB be the building you are standing.
AB = 6 m and let the angle of depression be ∠C.
=> ∠C = 60°
According to the diagram, the distance between you and the object is AC, which is the hypotenuse.
sin Ф = Opposite / Hypotenuse
sin 60° = 6 m / Hypotenuse
=> √3 / 2 = 6 / Hypotenuse
=> Hypotenuse = 6 * 2 / √3
=> Hypotenuse = 12 / √3
=> Hypotenuse = 12 √3 / 3
=> Hypotenuse = 4 √3 m
Question 21: Archana wants to buy a tv that costs Rs. 19,000 but she has only Rs. 15,000. So she decides to invest her money at 8% simple interest per year. After how many years, will she be able to buy the TV?
Solution:
Amount of money she needs extra = 19000 −15000 = Rs. 4000
So, she needs Rs. 4000 extra as simple interest.
To find time T when SI = 4000
SI = [P × R × T] / 100
⇒ 4000 = [15000 × 8 × T] / 100
⇒ 400000 = [15000 × 8 × T]
⇒T = 3.33 years
Question 22: The following frequency distribution gives the monthly consumption of 68 consumers of a locality. Find the median.
Solution:
Monthly consumption |
65 – 85 |
85 – 105 |
105 – 125 |
125 – 145 |
145 – 165 |
165 – 185 |
185 – 205 |
Number of consumers |
4 |
5 |
13 |
20 |
14 |
8 |
4 |
Cumulative frequency |
4 |
9 |
22 |
42 |
56 |
64 |
68 |
In the above distribution, n = 68
n / 2 = 34
125 – 145 is the class whose cumulative frequency 42 is greater than (and nearest to) n / 2 , i.e 34.
Therefore ,
125 – 145 is the median class.
l = 125 , cf = 22 ; f = 20 , h = 20 , n = 68
Median ( M ) = l + [(n / 2 – cf ) / f ] × h
M = 125 + [(34 – 22) / 20 ] × 20
M = 137 units
SECTION – IV
(5 * 1 = 5)
Answer ANY ONE question from the following.
Question 23: Construct a triangle of sides 4.2cm, 5.1cm and 6cm. Then construct a triangle similar to it, whose sides are [2 / 3] of corresponding sides of the first triangle.
Solution:
Steps of construction:
- Draw a line segment AB = 4.2 cm.
- Draw an arc at 5.1 cm from point A.
- Draw an arc at 6 cm from point B so that it intersects the previous arc.
- Joint the point of intersection from A and B.
This gives the required △ABC.
- Divide the base in 2:3 ratio.
- Draw a ray AX at an acute angle from AB.
- Plot three points on AX so that; AA_{1} = A_{1}A_{2} = A_{2}A_{3}
- Join A_{3} to B.
- Draw a line from point A_{2} so that this line is parallel to A_{3}.
- B and intersects AB at point B’.
- Draw a line from point B’ parallel to BC so that this line intersects AC at point C’.
Question 24: Teja and Murali on either side of a temple of 30m height, observe its top at the angles of elevation 30^{o} and 60^{o} respectively. Find the distance between the two boys.
Solution:
Height of the temple AB = 30m
In right angle triangle ABC
BC / AB = cot 30˚
= √3
= BC / 30 = √3
= BC = 30√3 ….(i)
Also in right-angled triangle ABD,
BD / AB = cot 60˚
= BD / 30 = 1 / √3
= BD = 30 / √3 × √3/√3
= 10√3 ……(ii)
Thus, the distance between the two men = CD
= BC + BD
= 30√3 + 10√3
= 40√3
Hence, the distance between two men will be 40√3 m.