Regularly solving AP 10th Class Maths Model Papers and AP 10th Class Maths Question Paper April 2022 contributes to the development of problem-solving skills.
AP SSC Maths Question Paper April 2022 with Solutions
Time : 3.15 hours
Max. Marks : 100
Instructions:
- In the duration of 3 hours 15 minutes, 15 minutes of time is allotted to read the question paper.
- All answers shall be written in the answer booklet only.
- Question paper consists of 4 Sections and 33 questions.
- Internal choice is available in section – IV only.
- Answers shall be written neatly and legibly.
Section – I
(12 × 1 = 12 M)
Note:
- Answer all the questions in one word or phrase.
- Each question carries 1 mark.
Question 1.
Express \(\frac{7}{25}\) in decimal form.
Solution:
0.28
Question 2.
Express the set P = {x : x is a prime, x < 5} in roaster form.
Solution:
{2, 3}
Question 3.
Find the number of zeroes of the polynomial p(x), whose graph is given.
Solution:
3
Question 4.
What is the common ratio of the G.P. 25, -5, 1, -1, \(\frac{-1}{5}\), … . ?
Solution:
\(\frac{-1}{5}\)
Question 5.
Match the following:
| Equation |
Solution |
| (a) x + y = 5 |
(p) (3, 3) |
| (b) 2x – y = 9 |
(q) (1, 4) |
| (c) x – y = 0 |
(r) (5, 1) |
Choose the correct answer.
A) a → r, b → p, c → q
B) a → p, b → q, c → r
C) a → q, b → r, c → p
D) a → r, b → q, c → p
Solution:
C) a → q, b → r, c → p
Question 6.
Assertion : (0, 2) is a point on Y-axis.
Reason : Every point on Y-axis is at a distance of zero units from the Y-axis.
Now, choose the correct answer.
A) Both Assertion and Reason are true. Reason is supporting the Assertion.
B) Both Assertion and Reason are true but, Reason is not supporting the Assertion.
C) Assertion is True, but Reason is False.
D) Assertion is False, but Reason is True.
Solution:
A) Both Assertion and Reason are true. Reason is supporting the Assertion.
Question 7.
Statement – I : The lengths 3 cm, 4 cm, 5 cm form a right angled triangle.
Statement – II: If ‘a’ is the side of an equilateral triangle, then its height is \(\sqrt{3}\) a.
Now, choose the correct answer.
A) Statement -1 and statement – II both are true.
B) Statement – I and statement – II both are false.
C) Statement -1 is true. Statement – II is false.
D) Statement – I is false. Statement – II is true.
Solution:
C) Statement -1 is true. Statement – II is false.
Question 8.
The tangents drawn at the end points of a diameter are ……..
Solution:
Parallel
Question 9.
Which of the following is NOT true ?
A) sin (90° – θ) = cosec θ
B) sin2θ + cos2θ = 1
C) cosec θ ∙ A sin θ = 1
D) sin 90° = 1
Solution:
A) sin (90° – θ) = cosec θ
Question 10.
At a particular time, if the angle of elevation of the sun is 45°, then the length of the shadow of a 5 m high tree is ……….
A) 5\(\sqrt{3}\) m
B) 10 m
C) 5 m
D) \(\frac{5}{\sqrt{3}}\) m
Solution:
C) 5 m
Question 11.
If P(E) = 0.3, then P (not E) …………
A) 0.3
B) \(\frac{1}{3}\)
C) 0
D) 0.7
Solution:
D) 0.7
Question 12.
In the classes 35 – 39, 40 – 44, 45 – 49,…. of a frequency distribution, then the upper boundary of the class 40 – 44 is ………..
Solution:
44.5
Section – II
(8 × 2 = 16 M)
Note:
- Answer all the questions.
- Each question carries 2 Marks.
Question 13.
Evaluate: log2 (1 + tan2 45°).
Solution:
log2(1 + tan245°)
= log2[1 + (1)2] (∵ tan 45° = 1)
= 1 (∵ \(\log _a^a\) = 1)
Question 14.
Check whether – 3 and 3 are the zeroes of the polynomial x2 -9 or not.
Solution:
Let p(x) = x2 – 9
p(-3) = (-3)2 – 9
= 9 – 9
= 0
∴ -3 is zero of polynomial p(x)
p(x) = x2 – 9
= 32 – 9
= 9 – 9
= 0
∴ 3 is zero of polynomial p(x)
Alternate Method:
To find zeroes of a polynomial,
p(x) = 0
Then x2 – 9 = 0
x2 – 32 = 0
(x + 3)(x – 3) = 0
x + 3 = 0 OR x – 3 = 0
x = -3 OR x = 3
∴ -3, 3 are zeroes of polynomial x – 9.
Question 15.
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. What is its slant height?
Solution:
The curved surface area of a cone
= 4070 cm2
Its diameter (d) = 70 cm
∴ Radius (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{70}{2}\) cm = 35 cm
Now, πrl = 4070 cm2
⇒ \(\frac{22}{7}\) × 35 cm × l = 4070 cm2
l = 4070 × \(\frac{7}{22}\) × \(\frac{1}{35}\) cm = 37 cm
∴ The slant height (l) = 37 cm.
Question 16.
If the slope of the line passing through the points R(2, y) and S(x, 3) is 2, then find the relation between ‘x’ and y’.
Solution:
R = (2, y), S = (x, 3)
Solpe of line joining the points (x1, y1) & (x2, y2) is \(\frac{y_2-y_1}{x_2-x_1}\)
Given, Slope of line RS = 2
∴ \(\frac{3-y}{x-2}\) = 2 ⇒ 3 – y = 2x – 4
2x +y – 4 – 3 = 0 ⇒ 2x + y – 7 = 0
Question 17.
If ABC is an isosceles triangle, right angled at C, then prove that AB2 = 2AC2.
Solution:
In ∆ABC, ∠C = 90° and AC = BC
By Pythagoras theorem,
AB2 = AC2 + BC2
AB2 = AC2 + AC2 (∵ AC = BC)
AB2 = 2AC2
Question 18.
The length of the tangent drawn from an external point R to a circle is 24 cm and the distance of R from the centre of the circle is 25 cm. Find the radius of that circle.
Solution:
Question 19.
A person is flying a kite at an angle of elevation a and the length of the thread from his hand to kite is “l”. Draw a rough diagram for the above situation.
Solution:
For drawing right angled traingle
For representation of value α and representation of value l in the diagram
Question 20.
Median of the observations \(\frac{x}{5}\), x, \(\frac{x}{4}\), \(\frac{x}{2}\), \(\frac{x}{3}\) is 7. Find the value of x.
Solution:
Ascending order of observations
\(\frac{x}{5}\), \(\frac{x}{4}\), \(\frac{x}{3}\), \(\frac{x}{2}\), x
Median of observations is \(\frac{x}{3}\) = 7
x = 21
Section-III
(8 × 4 = 32 M)
Note:
- Answer all the questions.
- Each question carries 4 marks.
Question 21.
Show that the following sets are equal:
i) A = {x : x is a letter in the word ‘FOLLOW’)
ii) B = {x : x is a letter in the word ‘FLOW’)
iii) C : {x : x is a letter in the word ‘WOLF’)
Solution:
Given, A = {x : x is a letter in the word ‘FOLLOW’}
A = {F, O, L, W) → (1)
Given, B = {x : x is a letter in the word ‘FLOW’}
B = {F, L, O, W} → (2)
Given, C = {x : x is a letter in the word ‘WOLF’}
C = {W, O, L, F} → (3)
From, (1), (2) and (3)
Elements of A, B and C are same
∴ A = B = C
Question 22.
Which term of the A.P. : 3, 8, 13, 18,……….. is 78?
Solution:
Here a = 3
d = 8 – 3 = 5
and an = 78
an = a + (n – 1) d
78 = 3 + (n – 1) 5
78 – 3 = (n – 1) 5
75 = (n – 1) 5
n – 1 = 15 ⇒ n = 16
∴ 78 is the 16th term of A.P.
Question 23.
The sum of a two digit number and the number obtained by reversing the digits is 66. If the digits
of the number differ by 2, find the number. How many such numbers are there?
Solution:
Let the digit in the units place be x tens place be y
∴ The number is 10y + x
On reversing the digits 10x + y
Given the sum of a two digit number and the number obtained by reversing the digits is 66.
∴ (10y + x) + (10x + y) = 66
10y + x + 10x + y = 66
11x + 11y = 66
x + y = 6 → (1)
Difference of the digits = 2
Substitute the value of x is equation (1)
4 + y = 6
y = 6 – 4
y = 2
∴ The number is 10 × 4 + 2 = 42
There are only two numbers to satisfy the given condition is 42 and 24.
Question 24.
A ladder 25 m long reaches a window of building 24 m above the ground. Determine the distance of the foot of the ladder from the building.
Solution:
In a ∆ABC, ∠C = 90°
AB = length of ladder = 25 m,
AC = height of window = 24 m.
By Pythagoras theorem,
AB2 = AC2 + BC2
252 = 242 + BC2
BC2 = 252 – 242
= (25 + 24) (25 – 24)
= 49 × 1 = 72
∴ BC = 7 m
Distance of the foot of the ladder from the building = 7 m
Question 25.
Find the roots of 2x2 + x – 4 = 0, if they exist, by the method of completing the square
Solution:
2x2 + x – 4 = 0
Given 2x2 + x – 4 = 0
Question 26.
A tower stands vertically on the ground. From a point which is 15 meters away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?
Solution:
From the figure, in triangle ABC
Given, BC = Distance from the tower = 15 m,
Angle of elevation = ∠C = 45°
Let, height of the tower AB = h meters
∴ tan45° = \(\frac{h}{15}\) ⇒ 1 = \(\frac{h}{15}\) ⇒ h = 15
Height of the tower 15 m.
Question 27.
A die is thrown once. Find the probability of getting
(i) â prime number
(ii) a number lying between 1 and 5.
Solution:
When a die is thrown once,
sample space, S = {1, 2, 3, 4, 5, 6}
Total no. of outcomes, n(S) = 6
i) Let ‘E’ be an event of getting a prime number
Favourable outcomes to E arc
E = {2, 3, 5) ⇒ n(E) = 3,
∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
ii) Let ‘F’ be an event of getting a number lying between 1 and 5.
Favourable outcomes to F are,
F = {2, 3, 4}
n(F) = 3
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Question 28.
A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution:
From the figure,
AB height of the lamp = 3.6 m
DE = height of the girl = 90 cm = 0.9 m
speed of the girl = 1.2 m/sec
AD = distance covered in 4 sec
= 4 × 1.2 m = 4.8 m
let DC = length of her shadow = x m
cearly ∆ABC ~ ∆DEC (by AA criterion similarity)
Section-IV
(5 × 8 = 40 M)
Note:
- Answer all the questions.
- Each question carries 8 marks.
- Each question has internal choice.
Question 29.
a) Prove that is irrational.
(OR)
b) If sec θ + tan θ = p, then prove that
sinθ = \(\frac{p^2-1}{p^2+1}\)
Solution:
a) Let us assume \(\sqrt{3}\) is rational.
If it is rational, then there must exist two integers ‘r’ and ‘s’
(S ≠ 0) such that \(\sqrt{3}\) = \(\frac{\mathrm{r}}{\mathrm{s}}\)
If ‘r’ and ‘s have a common factor other than 1. Then we divide by the common factor to get \(\sqrt{3}\) = \(\frac{\mathrm{r}}{\mathrm{s}}\),
where a’ and ‘b’ are co-prime.
So b\(\sqrt{3}\) = a
On squaring both sides and rearranging, we get
3b2 = a2
∴ 3 divides a2
⇒ 3 divides a
So, we can write a = 3c, for some integer ‘c’
Substituting for a, we get 3b2 = (3)2
⇒ 3b2 = 9c2
⇒ b2 = 3cb2
This means 3 divides b2
⇒ 3 divides b
∴ Both ‘a’ and ‘b’ have 3 as a common factor.
But this contradicts the fact that ‘a’ and ‘b’ are co-prime.
This contradiction has arisen because of our assumption that \(\sqrt{3}\) is rational.
Thus our assumption is false.
\(\sqrt{3}\) is irrational.
b) Given sec θ + tan θ = p ………. (1)
We have sec<2θ – tan2θ = 1
⇒ (secθ + tanθ) (secθ – tanθ)
⇒ secθ – tanθ = \(\frac{1}{\mathrm{p}}\) ……… (2)
By adding equations (1) and (2), we get
secθ – tanθ = p
secθ – tanθ = \(\frac{1}{\mathrm{p}}\)
2secθ = p + \(\frac{1}{\mathrm{p}}\) = \(\frac{p^2+1}{p}\)
secθ = \(\frac{p^2+1}{2 p}\) ………. (3)
By substracting equation (2) from (1), we get
Question 30.
a) If A = {2, 3, 4, 5, 6}, B = {1, 3, 5, 7} C = {2, 4, 8}, D = {2, 3, 5, 7}, then find
i) A∪B
ii) B∩D
iii) C∩D
iv) D – A
(OR)
b) A right circular cylinder has base radius 14 cm and height 21 cm. Find:
(i) Area of base or area of each end
(ii) Curved surface area
(iii) Total surface area and
(iv) Volumé of the right circular cylinder.
Solution:
a) Given A = {2, 3, 4, 5, 6}
B = {1, 3, 5, 7}
C = {2, 4, 6, 8}
D = {2, 3, 5, 7}
(i) A ∪ B = {2, 3, 4, 5, 6} ∪ {1, 3, 5, 7}
= {2, 3, 4, 5, 6, 1, 7}
= {1, 2, 3, 4, 5, 6, 7}
(ii) B ∩ D = {1, 3, 5, 7} – {2, 3, 5, 7}
= {3, 5, 7}
(iii) C ∩ D = {2, 4, 6, 8) ∩ {2, 3, 5, 7}
= {2}
(iv) D – A = {2, 3, 5, 7} – {2, 3, 4, 5, 6}
= {7}
(OR)
b)
i) Area of base = πr2 = \(\frac{22}{7}\)(14)2
= 316 cm2
ii) Curved surface are = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 21 = 1848 cm2
iii) Total surface area = 2 × Base area + curved surface area
= 2 × 616 + 1848
= 3080 cm2
iv) Volume of cylinder = πr2h
= Base area × height
= 616 × 21 = 12936 cm3
Question 31.
a) Find the value of ‘k’ for which the points (7, -2), (5, 1), (3, k) are collinear.
(OR)
b) The table below shows the daily expenditure on food of 30 households in a locality:
| Daily Expenditure (in Rupees) |
100 -150 |
150 – 200 |
200-250 |
250-300 |
300 – 350 |
| Number of households |
4 |
5 |
12 |
6 |
3 |
Find the mean daily expenditure on food by a suitable method.
Solution:
a) 2nd Method:
Let the given points are A (7, -2), B(5, 1), C€(3, k)
If the points are collinear,
Slope of \(\overleftrightarrow{\mathrm{AB}}\) = Slope of \(\overleftrightarrow{\mathrm{BC}}\) (or slope of \(\overleftrightarrow{\mathrm{AC}}\))
Slope of the line joining the points
(x1, y1) & (x2, y2) = \(\frac{y_2-y_1}{x_2-x_1}\)
Slope of the line joining A(7, -2), 6(5, 1) is \(\frac{1-(-2)}{5-7}\) = \(\frac{1+2}{-2}\) = \(\frac{-3}{2}\)
Slope of the line joining B(5, 1), C(3, k) is
\(\frac{k-1}{3-5}\) = \(\frac{k-1}{-2}\) = \(\frac{1-\mathrm{k}}{2}\)
Since slopes are equal, we have \(\frac{-3}{2}\) = \(\frac{1-\mathrm{k}}{2}\)
⇒ -3 = 1 – k ⇒ k – 3 = 1
⇒ k = 1 + 3 = 4
k = 4
(OR)
b)
Mean \(\bar{x}\) = A + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
Here A = Assumed mean = 225
Σfiui = -1; Σfi = 30; h = 50
∴ \(\overline{\mathrm{X}}\) = 225 + \(\frac{(-1)}{30}\) × 50 = 225 – \(\frac{1 \times 5}{3}\)
= 225 – 1.67
= 223.33 (approximately)
Mean daily expenditure = ₹ 223.33
Note : Deviation / Direct Method also may be considered.
Question 32.
a) 5 pencils and 7 pens together cost ₹ 95. Whereas 7 pencils and 5 pens together cost ₹ 85. Find cost one pencil and that of one pen.
(OR)
b) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of black colour
ii) a face card
iii) a spade
iv) a card not a heart
Solution:
a) Let Cost of a pencil = ₹ x
Cost of a pen = ₹ y
Total cost of 5 pencils and 7 pens = ₹ 95
5x + 7y = 95 …………. (1)
Total cost of 7 pencils and 5 pens = ₹ 85
7x + 5y = 85 …….. (2)
Substitute y = 10 in equation (1)
5x + 7(10) = 95
5x + 70 = 95
5x = 25
Cost of a pencil = ₹ 5
Cost of a pen = ₹ 10
b)
When one card is drawn from a well shuffled deck of 52 cards
The total number of outcomes in sample space is n(S) = 52
Probability \(=\frac{\text { No.of favourable outcomes }}{\text { Total No.of outcomes }}\)
Event (i) : Consider E1 be the Event of getting a king of black colour.
Number of favourable outcomes for the event E1 is n(E1) = 2
P(E1) = \(\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
Event (ii) : Consider E2 be the Event of getting a face card.
Number of favourable outcomes for the event E2 is n(E2) = 12
P(E2) = \(\frac{\mathrm{n}\left(\mathrm{E}_2\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\).
Event (iii): Consider E3 be the Event of getting a spade.
Number of favourable outcomes for the event E3 is n(E3) = 13
P(E3) = \(\frac{\mathrm{n}\left(\mathrm{E}_3\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\)
Event (iv): Consider E4 be the Event of getting a card not to be a heart.
Number of favourable outcomes for the event E4 is n(E4) = 39
P(E4) = \(\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}\) = \(\frac{39}{52}\) = \(\frac{3}{4}\)
Question 33.
a) Draw the graph of the polynomial p(x) = x2 – 3x – 4 and find the zeroes.
(OR)
b) Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle 60°.
Solution:
a)
| x |
-2 |
-1 |
0 |
1 |
2 |
3 |
4 |
5 |
| y = x2 – 3x – 4 |
6 |
0 |
-4 |
-6 |
-6 |
-4 |
0 |
6 |
| (x, y) |
(-2, 6) |
(-1,0) |
(0,-4) |
(P-6) |
(2,-6) |
(3,-4) |
(4, 0) |
(5,6) |
b)
From
∆OAP, sin 30° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\frac{1}{2}\) = \(\frac{4}{\text { OP }}\) ⇒ OP = 8 cm
i) Draw a circle of radius 4 cm with centre ‘O’.
ii) Plot a point P, such that OP = 8 cm
iii) Bisect OP at M and draw circle with radius OM or MP
iv) Draw tangents from P to the intersecting points of two circles.
CH2O + H2O + O2
