Goa Board Class 10 Sample Paper 2025 Maths Standard

2025

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GOA BOARD OF SECONDARY AND HIGHER SECONDARY EDUCATION
ALTO-BETIM GOA 403521
SSC FINAL EXAM -PRACTICE QUESTION PAPER (2024 – 2025)
Subject : MATHEMATICS (E) – LEVEL 1 ( Standard Mathematics )
Time : 3 hrs Std : X Max. Marks : 80
INSTRUCTIONS:
i) This question paper consists of 42 questions . All questions are compulsory.
ii) This question paper is divided into four Sections-A, B, C and D
iii) In Section A, Question Nos.1 to 16 are multiple choice questions (MCQs) and Question Nos.
17 to 20 are very short answer type questions (VSA) of 1 mark each.
iv) In Section B, Question Nos. 21 to 29 are short answer type I (SA-I ) questions carrying 2
marks each.

v) In Section C, Question Nos. 30 to 39 are short answer type II (SA-II) questions carrying
3marks each.
vi) In Section D, Question Nos.40 to 42 are long answer (LA ) questions carrying 4marks each.

vii) There is no overall choice. However an internal choice has been provided in two questions
of 2marks each in Section B and two questions of 3marks each in Section C.
viii) In questions on Constructions, the drawing should be clear and exactly as per given
measurements. The construction lines and arcs should also be maintained.
ix) Graph page is provided on the answer booklet.
x) Logarithm and Antilogarithm tables are printed on the last page of the question paper.

xi) Use of calculators is not permitted.
Section A (1 mark each)
Select and write the correct alternative from those given below each statement for question 1to16 :
1 The quadratic polynomial in x , whose zeroes are –7 and 4, is:
• x2 + 3x – 28 • x2 - 3x – 28 • x2 + 3x + 28 • x2 - 3x - 28

2 If one of the zeroes of the quadratic polynomial (k–1)x² + kx+1 is –3, then the value of k is:
• (-4)/3 • (-2)/3 • 2/3 • 4/3

3 The pair of linear equations 5x – 15y = 8 and 3x – 9y = 24/5 have
•No solution •Unique solution •Exactly two solutions • Infinitely many solutions

4 Father’s age is six times his son’s age. Four years hence, the age of the father will be four
times his son’s age. The present ages, in years, of the son and the father are respectively
• 4 and 24 • 5 and 30 • 6 and 36 • 7 and 42

5 The 11th term of the AP : 25, 50, 75, 100…….is
• 225 • 250 • 275 • 300
6 If △ABC ～△ DEF and AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm, then the perimeter
of △ ABC is:
• 18 cm • 20 cm • 21 cm • 22 cm
7 If 1 + tan2 36º= sec2 3A where 3A is an acute angle, then the value of A is:
• 12° • 18° • 36° • 54°

8 The simplified form of (cosecA- cotA) (1+cosA) is :
• sinA • cosA • cosecA • secA

9 If sinA – cosA = 0 then the value of sin4A - sin2A is :
• -3/ 4 • - 1/4 • 1/4 • 3/4

10 If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length
of each tangent is equal to :
• √3 cm •6 cm •3 cm •3√3 cm

11 The circumference of a circle that can be inscribed in a square of side 10cm is:
• 10π cm • 20π cm • 25π cm • 100π cm

12 A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a
22
square, then its area will be: (Take π = 7 )
• 3520 cm2 • 6400 cm2 • 7744 cm2 • 8800 cm2

13 A piece of paper is in the shape of a semicircular region of radius 10 cm. It is rolled to
form a right circular cone. The slant height of the cone is:
• 5 cm • 10 cm •15 cm •20 cm

14 When two same solid hemispheres of equal base radius r cm are attached together along
with their bases, the total surface area of the combination is:
• 6πr2 • 4 πr2 • 3 πr2 • 2 πr2

15 3 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. A card is
drawn at random from the remaining pack. The probability of getting a black card is:
• 23/52 • 22/52 • 22/45 • 23/45

16 The value of log 800 – log 8 is :
• 1 • 2 • 100 • 792
17 Find whether the following pair of linear equations is consistent or inconsistent:
4x + 3y – 1 = 5 and 12x + 9y = -18

18 In the given figure, a circle touches the side DF of △EDF at H
and touches ED and EF produced at K & M respectively.
If EK = 9 cm, calculate the perimeter of △EDF

19 Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.
( Do not substitute for π )
20 Amita calculates that the probability of her winning a lottery is 0.08 . If 6000 tickets
were sold find the number of tickets she bought.
Section B (2 marks each)
21 Find the HCF of 145 and 325 using Euclid’s Division Algorithm.

22 A race track is in the form of a circular ring whose outer and inner circumference are
396m and 352m respectively. Find the width of the track.
23 In ∆ABC,DE∥BC. If AD=4 , AE=8 , A
DB = x-2 and EC = 3x-19, then find the value of x
D E

B C
24 If the distance between the points A(2, -2) and B(-1, x) is equal to 5,then find the value of x

25 If P(-3, 2), Q(7, 6) and R(-1, 4) are the vertices of △PQR and QS is its median find the
area of △PQS
OR
Find the value of k for which the area of the triangle formed by the points A (1, k),
B (4, -3) and C (-9, 7) is 15 square units.
26 In ∆APQ , ∠ APQ = 90° A
15
and cos Q = 17
Find the length of AP and value of cot A
Q P

OR
Evaluate the following trigonometric expression using known trigonometric values:
4 cos 2 60°+ cosec245° - 6 cot230°

27 Prove the trigonometric identity:
sin 𝐴 . 𝑡𝑎𝑛𝐴
= secA+1
1−𝑐𝑜𝑠𝐴

28 In the given figure, a circle is inscribed in a
quadrilateral ABCD in which ∠B = 90°.
If AD = 23 cm, AB = 29 cm and DS = 5 cm,
find the radius of the circle.

29 Find the median of the following data:
CI 0 - 10 10 -20 20 -30 30 -40 40- 50
Frequency 5 15 30 8 2

(Write your answer correct up to one place of decimal)
Section C (3 mark each)
30 On dividing the polynomial 6x3 +8x2 -3x + 8 by the polynomial g(x), the quotient and the
remainder are 3x + 4 and 6x + 20 respectively . Find g(x)

31 Find the solution of the pair of linear equations:
11x - 7y = 57 and 14x + 5y = 3 by elimination method
OR
Find the solution of the pair of linear equations:
7x – 3y = 15 and 5x + 11y - 37 = 0 by cross multiplication method

32 Find the roots of the quadratic equation:
15x2 - 4x - 3 = 0 by factorisation method.
OR
Find the roots of the quadratic equation:
2x2 - 3x - 5 = 0 by completing the square method.

33 From 15th January, Sheena decided to save her pocket money daily. The amount of
saving is equal to the date of the month. On 31st January, she decided to give a gift to
her mother worth ` 200 from her savings. What is the amount left with Sheena at the end
of the month?
34 In the given figure, □DEFG is a square and C
∠BAC = 90°. Show that FG2 = BD × EC. E

F
┐
A G B

35 The angles of depression of the top and bottom of a
50 m high building DC from the top of a tower AB
are 45° and 60° respectively. Find the height of the
tower and the horizontal distance between the
tower and the building, (take √3 = 1.73)

36 Draw a circle with centre O and radius 3.5cm Take a point R at a distance of 7.7cm from
O. Construct tangents RS and RP from external point R. Measure and state the length of
the tangent segments.

37 Using a pair of compasses and ruler construct △PQR with PQ =8.1cm ,QR = 7.5cm and
∠PQR= 75°. Then construct △P’Q’R whose sides are 5/2 of the corresponding sides of
△PQR.

38 A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper
and lower circular ends of the frustum be 14 m and 26 m respectively, the height of the
frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the
area of canvas required to make the tent. (Assume that the radii of the upper circular end
22
of the frustum and the base of surmounted conical portion are equal). (take 𝜋 = )
7

39 Evaluate the following expression by using the logarithm method:

(6.782)3 × √0.0043
35.29

Section D (4 mark each)
40 Find the solution of the following pair of linear equations graphically.
3x + 2y = - 4 and 2x – 3y = 19
Rewrite and complete the following tables:
3x + 2y = - 4 2x – 3y = 19
x x
y y

41 Abdul takes 6 days less than the time taken by Ramesh to finish a piece of work. If both
Abdul and Ramesh together can finish that work in 4 days, find the time taken by
Ramesh to finish the work independently.
42 The following table shows the data regarding the height in cms of students of class X .
Find the mean height of students of the class taking ‘a’ as assumed mean of the interval
162-168 using assumed mean method.

Heights fi xi di fidi
150-156 4
156-162 7
162-168 12
168-174 8
174-180 6
180-186 3
Σ fi= Σ fidi=

SSC PRACTICE PAPER
Mathematics (E) Level 1 – Standard Mathematics
Model Answers and Marking Scheme
Note: Any alternative method unless otherwise specified should be considered for full credit

Section A (1 mark each)
1 x2 + 3x – 28 1
2 4/3 1
3 Infinitely many solutions 1
4 6 and 36 1
5 40 1
6 18 1
7 12° 1
8 Sin A 1
9 -1/4 1
10 3√3 1
11 10π cm 1
12 7744cm2 1
13 10cm 1
14 4 π r2 1
15 22/45 1
16 2 1
17 a1 4 1 b1 3 1 c1 −6 −1
= 12 = 3 , 𝑏2 = 9 = 3 , 𝑐2 = 18 = 3 ½
𝑎2
a1 b1 c1
= 𝑏2 ≠ 𝑐2 ∴ No solution hence inconsistent ½ 1
𝑎2

18 Perimeter of △EDF=ED+DF+EF
=ED+DH+HF+EF
=ED+DK+FM+EF ½
= EK+EM=2(EK) =2 x 9 =18cm ½ 1

19 𝜃 60
Area of the sector = 360 x πr2 = 360 x π x 62 ½
1
= 6π cm2 ½

20 Let the number of tickets she bought be x
𝑥
P(she wins the lottery ) = 6000 =0.08 ½
x = 0.08 x 6000 = 480 ½ 1

21 2 ½
145 325
290 4 ½
35 145
140 7 ½
5 35
35
00 2
HCF (145, 325) = 5 ½

22 Let the radius of the outer track and inner track be R and r respectively
2πR= 396 and 2 π r = 352 ½
396 x 7 352 x 7
R= r = 2 x 22 ½
2 x 22
R = 63 = 56 ½
2
Width of the track = 63 – 56 = 7cm ½

23 𝐴𝐷 𝐴𝐸
= 𝐸𝐶 ( DE ‖BC)
𝐷𝐵
4 8
= 3𝑥−19 ½
𝑥−2
4(3x-19) = 8(x-2) ½
12x-76 = 8x – 16 ½
4x = 60
x = 15 ½
2

24 AB =√(−1 − 2)2 + (x + 2)2 = 5 ½
9 + x2 + 4x + 4 = 25 ½
x2 + 4x -12 = 0 ½
(x + 6)(x-2) = 0
x = -6 or x =2 ½
2
25 −3−1 2+4
S is the midpoint of PR ∴ S( 2 , 2 ) = S (-2, 3) ½
1
ar (△PQS) = 2 [-3(6-3) +7(3-2) -2( 2 -6)] ½
1
= 2 [ -9 +7 +8 ] ½
1
= 2 [6] = 3 sq units ½
OR
1
ar (△ABC) = 2 [1(-3-7) +4(7-k) -9( k +3)] = 15 ½
[ -10 +28 -4k -9k -27 ] = 30 ½
-9 -13k = 30 ½
- 13k = 39
2
k = -3 ½

26 cos Q = 15/17 , Let QP =15k , QA =17k
( 8, 15 , 17 ) is a Pythagorean triplet ∴ AP = 8k ½ +½
8𝑘 8
Cot A = = 15 ½ +½
15𝑘
OR
4 cos 2 60°+ cosec245° - 6 cot230°
1
4( 2 )2 + ( √2 )2 - 6 (√3 )2 ½ +½+½
1 + 2 - 18 = -15 ½ 2

27 Sin A.tan A
= secA+1
(1−cosA)
Sin A.tan A 𝑠𝑖𝑛2 A
LHS = (1−cosA) = cosA(1−cosA) ½
1−𝑐𝑜𝑠 2 A (1−𝑐𝑜𝑠A)(1+cosA)
= cosA(1−cosA) = cosA(1−cosA)
½
(1+cosA) 1 𝑐𝑜𝑠𝐴
= = 𝑐𝑜𝑠𝐴 + 𝑐𝑜𝑠𝐴 ½
cosA
2
= secA +1 =RHS ½

28 DS =DR = 5cm ½
AR= AD-DR = 23-5= 18 ½
AQ=AR= 18

BQ= AB-AQ = 29- 18 = 11 ½ 2
radius = OQ= BQ = 11cm ½
29
CI Frequency Cumulative Freq n = 60, n/2 = 30
0 - 10 5 5 Median class 20 – 30 l =20
𝑛
10 -20 15 5+15=20 Median = l + [ 2 𝑓
− 𝑐𝑓
] ×ℎ ½
20 -30 30 20 + 30 = 50 30 − 20
= 20 + [ ]× 10 ½+½
30 -40 8 50 + 8 = 58 30
10
40- 50 2 58+2 =60 = 20 + 3
= 23. 3 ½ 2

30 p (x) = g(x) x q(x) + r(x)
6x3 +8x2 -3x + 8 = g(x) x (3x + 4) + (6x + 20) ½
6x3 +8x2 −3x + 8 − 6x – 20
= g(x) ½
3x + 4

2 x2 – 3 ½
3x +4 6x3 + 8x2 – 9x - 12
6x3+ 8x2 ½
- -
- 9x – 12
- 9x - 12 ½
+ +
3
g(x) = 2 x2 – 3 ½

31 11x - 7y = 57 --------------(i) x 5
14x + 5y = 3 --------------(ii) x 7
55x - 35y = 285 ½
98x + 35y = 21
153x = 306 ½
306
x = 153 = 2 ½
substitute x = 2 in 11x - 7y = 57
11(2) – 7y = 57 ½
22 -7y = 57
-7y = 57 -22
35
y=
−7
y=-5 ½
The solution is x = 2 and y = - 5 ½
OR
7x - 3 y – 15 = 0 and 5x + 11y - 37 = 0
b c a b
-3 -15 7 -3 ½
11 -37 5 11
𝑥 𝑦 1
= −75−(−259) ½+½
111−(−165) 77−(−15)
𝑥 𝑦 1
276
= 184 = 92 ½

276 184
x = 92 = 3 and y = 92 = 2 ½

3
The solution is x=3 and y = 2 ½

32 15x2 - 4x - 3 = 0
15x2 +5x - 9x - 3 = 0 ½
5x(3x + 1) -3(3x + 1)=0 ½
(3x + 1) (5x -3 )=0 ½
(3x + 1)=0 or (5x -3 )=0 ½
x = -1/3 or x = 3/5 ½
The roots of the equation are -1/3 and 3/5 ½
OR
2x2 - 3x - 5 = 0
2x2 - 3x = 5
3 5
x2 - 2 x = 2 ½
3 9 5 9
x2 - 2 x + 16 = 2 + 16 ½
3 49
(x- 4 )2 = 16 ½

3 7
x-4=∓ 4 ½

3 7
x= ∓
4 4
3 7 3 7
x= + or x = −
4 4 4 4
10 5 −4
x= 4 = 2 or x= 4 = −1 ½ 3
The roots of the equation are 5/2 and -1 ½

33 AP : 15,16,17,18,………..31 ½
a =15, d = 1, l = 31 n = 31-14 = 17 ½
𝑛
Sn = 2 [a + l ] ½
17
S17 = 2 [15 + 31 ] ½

17
= [46 ] = 391 ½
2
Amount left with Sheena at the end of the month is ` (391 – 200) = ` 191 ½
3
34 In △CEF and △BGD, ∠CEF = ∠BDG = 90--------------------1 ½
In △CEF , ∠CFE + ∠FCE =90-------------------------2
In △ABC, ∠ABC + ∠ACB =90
∠GBD + ∠FCE =90--------------------------3
From 2 & 3, ∠CFE = ∠GBD--------------------------------4 ½
From 1 & 4, △CEF ~ △GDB ½

𝐶𝐸 𝐸𝐹 𝐶𝐹
= 𝐷𝐵 = 𝐺𝐵 ½
𝐺𝐷

CE X DB = GD X EF
CE X DB = FG X FG ( □ DEFG is a square) ½

FG2 = CE X DB 3

½

𝑥
35 In △AB)C, ℎ = cot 60°
1
x = h x cot 60°= h x -----------------1 ½
√3
𝑥
In △AED , = cot 45°
ℎ−50
x = (h-50) x cot 45°= (h-50) x 1 = h-50-------------------------2 ½
1
From 1 & 2 , hx 3= h-50 ½
√
1
h-hx = 50
√3
1
h(1 - 3 )= 50
√
√3 − 1
h( )= 50
√3
√3 √3 +1
h = 50 x x 3+1 ½
√3−1 √
3+ √3
h = 50 x 2
h = 25 x (3 + √3 )
= 25 x 4.73
Height of tower = 118.25m ½
Distance between tower and building = 118.25m -50 = 68.25m ½ 3

36 To draw a circle with centre O and radius 3.5 cm ½
To draw a line segment OR = 7.7 cm ½
To construct the perpendicular bisector of line segment OR ½
To mark the points P and Q on the circle ½
To draw the tangent segments RP and RQ ½
To measure the tangent segments RP = RQ = (6.8±0.1)cm ½ 3

37 To construct Δ PQR with the given data 1
To draw a ray making an acute angle with side QR. ½
To locate 5 points on the ray using a pair of compasses ½
To join R5 Q’ ∥ R2 Q and Q’P’ ∥ QP ½+½
ΔP’Q’R is the required triangle.
3
38 Frustum of the cone
d =14m r = 7 ; D=26m , R=13m h=8m
Slant height of Frustum of the cone L = √(𝑅 − 𝑟)2 + ℎ2
√(13 − 7)2 + 82
= √36 + 64
= √100 =10m ½
Cone l = 12m
Area of canvas required = CSA(frustum of the cone) + CSA (cone) ½
= π (R+r)L + π r l
= π { (13 + 7)x10 + 7x12 } ½+½
= π(200 + 84)
22
= x 284 ½
7
6248
= 7
= 892.57m2 ½ 3

3
(6.782) x √0.0043
Let x =
39 35.29
1
Log x = 3 log 6.782 + 2 log 0.0043 – log 35.29 ½
1
= 3 x 0. 8313 + x 3.6335 – 1.5476 ½+½+½
2
=2.4939 + 2. 8168 – 1.5476
= 1.3107 -1.5476
=1.7631 ½
x = antilog 1.7631
= 0. 5795 ½ 3

40 For completing the table1 ½
For completing the table2 ½
Plotting the points and drawing the line for equation 1 1
Plotting the points and drawing the line for equation 2 1
The solution is x = 2 and y = -5 1 4

41 Let Ramesh alone take x days to finish the work
Abdul alone will take x-6 days to finish the work ½
1
Ramesh’s one days work = 𝑥
1
Abdul’s one days work = 𝑥−6
1
Together they finish the work in 4 days ∴ Their one days work = 4 ½
1 1 1
+ 𝑥−6 = 4 ½
𝑥
4(x-6) + 4x = x(x-6)
4x -24 +4x = x2 -6x ½
x2 -14x +24 = 0 ½
x = 2 or x = 12 ½
x=2 is discarded since x has to be greater than 6 ½
∴ x = 12
4
Ramesh alone take 12 days to finish the work. ½

42
Heights fi xi di fidi
150-156 4 153 -12 -48 ( ½ mk for 3
156-162 7 159 -6 -42 correct entries
162-168 12 165 0 0 in each of the
168-174 8 171 6 48 3 columns)
174-180 6 177 12 72
180-186 3 183 18 54
Σfi=40 Σfidi=84

Σfidi
Mean = a + Σfi
84
= 165 + 40 ½
=165 + 2.1
Mean height is = 167.1cm ½ 4

*************************************************************************************************************************************************************************

GOA BOARD OF SECONDARY AND HIGHER SECONDARY EDUCATION
ALTO-BETIM GOA 403521
SSC FINAL EXAM -PRACTICE QUESTION PAPER (2024 – 2025)
Subject : MATHEMATICS (E) – LEVEL 2 ( Basic Mathematics )
Time : 3 hrs Std : X Max. Marks : 80
INSTRUCTIONS:
i) This question paper consists of 42 questions. All questions are compulsory.
ii) This question paper is divided into four Sections.-A , B , C and D.
iii) In Section A, Question Nos. 1 to 16 are multiple choice questions (MCQs) and
Question Nos. 17 to 20 are very short answer type questions (VSA) carrying 1 mark
each.
iv) In Section B, Question Nos. 21 to 28 are short answer type I (SA-I) questions
carrying 2 marks each.
v) In Section C, Question Nos. 29 to 40 are short answer type II (SA-II) questions
carrying 3 marks each.
vi) In Section D, Question Nos. 41 and 42 are long answer (LA) questions carrying
4 marks each.
vii) There is no overall choice. However, an internal choice has been provided in two
questions of 2 marks each in Section B and two questions of 3 marks each in Section C.
viii) In questions on constructions, the drawing should be clear and exactly as per given
measurements. The construction lines and arcs should also be maintained.
ix) Graph page is provided on the answer booklet.
x) Logarithm and Antilogarithm tables are printed on the last pages of the question paper.
xi) Use of calculator is not permitted.

Section A (1 mark each)
Select and write the correct alternative from those given below each statement for question 1 to 16:
1. The product of the zeroes of the quadratic polynomial 2𝑥 2 – 6𝓍 – 9 is:
−9
• 2
• -3
• 3
9
• 2

2. A pair of linear equations a₁x+ b₁y +c₁=0 ; a₂x+ b₂y +c₂=0 is said to be inconsistent, if:
𝑎₁ 𝑏₁ 𝑐₁
• 𝑎₂ = 𝑏₂ = 𝑐₂
𝑎₁ 𝑏₁ 𝑐₁
• = 𝑏₂ ≠ 𝑐₂
𝑎₂
𝑎₁ 𝑏₁
• ≠ 𝑏₂
𝑎₂
𝑎₁ 𝑐₁
• ≠ 𝑐₂
𝑎₂

3.The solution of a pair of linear equations 2𝑥 + 3y = 1 and 𝑥 − 3y = 5 is:
• 𝓍 = 2, y = -1
• 𝓍 = -2, y = 1
• 𝓍 = 1, y = -2
• 𝓍 = -1, y = 2

4. The roots of the quadratic equation 3𝑥 2 − 2𝑥 + 6 = 0 are :
• rational and unequal
• irrational and unequal
• rational and equal
• non real

5. In an AP, if a = 17, d = -5 and an = -13 then the value of n is:
• 5
• 6
• 7
• 8

6. A 15m high tower casts a shadow 24m long at a certain time and at the same time , a telephone pole
casts a shadow 16m long. Therefore the height of the telephone pole is:
• 10m
• 12m
• 22.5m
• 25.6m
√3
7.If sin 2𝜃 = 2 , then the value of 𝜃 is :
• 30°
• 45°
• 60°
• 90°
𝑠𝑖𝑛2 72°
8. The value of is:
𝑐𝑜𝑠 2 18
• 0
• 1
• 4
• 90

9. The simplified form of √1 − 𝑐𝑜𝑠 2 𝐵 is:
• sin B
• cos B
• 𝑠𝑖𝑛2 𝐵
• 𝑐𝑜𝑠 2 𝐵

10. PA and PB are tangents from an external point P to a circle with centre O. If ∠AOB = 110°,
then ∠PAB is: A
• 35°
• 55° O 110⁰ P
• 70°
• 110° B
11. The area of the sector of a circle of radius 9 cm that subtends an angle of 120° at the centre is:
• 3𝜋 cm²
• 9𝜋 cm²
• 27𝜋 cm²
• 54𝜋 cm²
12. The area of the circle that can be inscribed in a square of side 8 cm is:
• 4𝜋 cm²
• 8𝜋 cm²
• 16𝜋 cm²
• 64𝜋 cm²

13. If the lateral surface area of a cube is 324 cm², then its total surface area is:
• 9 cm2
• 81 cm2
• 405 cm2
• 486 cm2
22
14. The curved surface area of a hemispherical bowl of radius 3.5 cm is: (Take π = 7 )
• 22 𝑐𝑚2
• 33 𝑐𝑚2
• 44 𝑐𝑚2
• 77 𝑐𝑚2
15. If a dice is thrown once then probability of getting an odd number on it’s top face is:
1
• 3
1
• 2
2
• 3
• 1
16. The value of log 3 729 is:
• 3
• 6
• 9
• 27

17. Find the zeroes of the quadratic polynomial 𝓍² - 9

18. If 3𝓍 + 2y = 16 and 2𝓍 + 3y = 19, then find the value of 𝓍 + y.

19. In the figure given below, XPY is an arc of a circle with centre O and radius 6 cm. If ∠ XOY = 40° and
point P lies on arc XY, then Find the length of arc XPY.
(Do not substitute for 𝜋 )
O
40°
X Y
P

20. Cards numbered from 1 to 30 are put in a box and mixed thoroughly. A card is drawn at
random from the box. Find the probability that the number on the card drawn is a multiple
of 3 or 5.
Section B (2 marks each)
23
21. Without actually performing the long division, show that the rational number 250 has
terminating decimal expansion. Also find its decimal expansion.
OR
Find the LCM of 294 and 420 by prime factorisation method. 7 cm

22. In the trapezium ABCD, AB ∥ CD, AB = 7 cm, CD = 12 cm,
DE = 5 cm, ∠C = 90°. A quadrant BCE is cut from it.
Find the area of the remaining shaded part.
22
(Take 𝜋 = 7 )

23. Find the mode of the following distribution table:

Class interval Frequency
10 - 14 3
14 - 18 6
18 - 22 8
22 - 26 14
26 - 30 4
A
° 7
24. In ∆APQ, ∠ APQ = 90 and tan Q = 24
Find the length of AQ and the value of cosec A.

P Q

25. Evaluate the following expression using known numerical values of trigonometric ratios:
2
3sin²30° +
1+𝑠𝑒𝑐²45°

26. If ΔABC ~ ΔPQR, ar (ABC) = 144 cm², ar (PQR) = 81 cm² and QR = 27 cm, then find the length of BC.

27. Find the distance between the points A (-2, 5) and B (7, -1).

OR

Find the coordinates of the midpoint of the join of the points P (7, -3) and Q (-5, -1)
28. If the area of a ΔPNG formed by points P (6, 2), N (k, 4) and G (4, -3) is 24.5 sq. units, then
find the value of k.

. Section C (3 marks each)
29. Divide the polynomial 2𝑥 3 − 7𝑥 2 + 7𝑥 - 9 by the polynomial 2𝑥 − 1 and find the quotient and remainder.
Hence write the result in the form: Dividend = Divisor × Quotient + Remainder.
30. Find the solution of the pair of linear equations 7𝑥 - 2y = 20 and 4𝑥 + 3y = -1 by elimination method.
OR
Find the solution of the pair of linear equations 5𝑥 + y = 14 and 3𝑥 − 7y = 16 by substitution method.

31. Find the roots of the quadratic equation 4𝑥 2 - 17𝑥 + 15 = 0 by factorisation Method.

32. Find the roots of the quadratic equation 6𝑥 2 - 11𝑥 - 7 = 0 by using the quadratic Formula.

33. Find the 15𝑡ℎ term and the sum of first 51 terms of the AP : -30, -23, -16, -9, -2, …
34. Evaluate the following expression by using the logarithm method.
(17.64)2 × 0.0486
0.9642
35. Draw line segment RS of length 6.7cm. Taking R as centre and radius 2.5 cm draw a circle.
Using a pair of compasses and ruler, construct two tangents SP and SQ to the circle.
Measure and state the length of the tangent segments.

36. Using a pair of compasses and ruler, construct ∆XYZ with sides YZ = 6.8cm, XY = 7cm
4
and XZ = 5.5 cm. Then construct ∆X′YZ′ whose sides are 3 of the corresponding sides
of ∆XYZ.

37. With reference to the given figure and the given conditions, write only the proof with reasons of the
following theorem :
P

Q R

Given: In ∆𝑃𝑄𝑅 , DE ∥ QR where the points D and E lie on PQ and PR respectively and
DF ⊥ PR.
𝑃𝐷 𝑃𝐸
Prove that : =
𝐷𝑄 𝐸𝑅

OR

With reference to the given figure and the given conditions, write only the proof with
reasons of the following theorem :

Given : In ΔDEF, DE² + EF² = DF² and ΔXYZ is constructed such that DE = XY and EF = YZ,
∠Y = 90°
Prove that : ΔDEF is a right triangle.
38. The angle of elevation of the top ‘X’ of a building ‘XY’ from a point Z at a distance of 50 m
from its foot on a horizontal plane is 30°. Find the height of the building. (Take √3 = 1.73)

39. ABC is a right triangle, right angled at B. A circle is inscribed in it. The lengths of the two sides
containing the right angle are 6 cm and 8 cm. find the radius of the circle

40. A solid metallic right circular cone of height 50 cm and radius of the base 21 cm is melted and
recast into solid cylinders each of height 10 cm and radius 3.5 cm. Find the number of such
cylinders formed.
Section D (4 marks each)
41. The following table shows the weekly pocket money of 80 students of a class.

Weekly pocket Number of
money (in ₹) workers 𝑥𝑖 𝑓𝑖 𝑥𝑖
(C.I) (𝑓𝑖 )
15 - 25 12
25 - 35 14
35 - 45 21
45 - 55 16
55 - 65 9
65 - 75 8
Total Σ𝑓𝑖 = 80 Σ𝑓𝑖 𝑥𝑖 = ___

Rewrite and complete the table and find the mean weekly pocket money by the Direct Method.
42. Find the solution of the following pair of linear equations graphically:
2𝑥 − 𝑦 = 10 and 𝑥 + 3𝑦 = - 2
2𝑥 − 𝑦 = 10 𝑥 + 3𝑦 = −2
𝑥
𝑥
𝑦
𝑦

(Plot at least 3 points for each line on a graph paper.)

********************************************************************************

SSC PRACTICE PAPER
Mathematics (E) Level 2 – Basic Mathematics
Model Answers and Marking Scheme
Note: Any alternative method unless otherwise specified should be considered for full credit
Section A
1 −9 1
2
2 𝑎₁ 𝑏₁ 𝑐₁
= ≠ 1
𝑎₂ 𝑏₂ 𝑐₂
3 x = 2, y = - 1 1
4 Non real 1
5 7 1
6 10m 1
7 30° 1
8 1 1
9 sin B 1
10 55° 1
11 27𝜋 cm² 1
12 16𝜋 cm² 1
13 486 cm² 1
14 77 cm2 1
15 1 1
2
16 6 1
17 𝓍²-9
= (𝓍 – 3) (𝓍 + 3) ½
The value of the polynomial 𝓍 ² - 9 is zero when
(𝓍 – 3) = 0 or (𝓍 + 3) = 0
∴ 𝓍 = 3 or 𝓍 = − 3
∴ the zeroes are 3 and -3 ½
1
18 3𝓍 + 2𝑦 = 16
2𝓍 + 3𝑦 = 19
5𝓍 + 5𝑦 = 35 ½
∴ 𝓍+𝑦=7 ½
1
19 𝜃
Length of arc = 360 × 2𝜋r
40
= 360 × 2 𝜋 ×6 ½
4𝜋 ½
= cm
3
1
20 P(E) = 30
14 ½
7
= 15 ½

1

Section B
21 2 250
5 125
5 25
5 5
1
23 23
= 2 ×5³ ½
250

23
∴ 250 has a terminating decimal expansion ½
23 23
= 2 ×5³
250

23 ×2² ½
= 2³ ×5³

92
= 1000

½
= 0.092
2
OR

294 = 2 × 3 × 7² ½
420 =2² × 3 × 5 × 7 ½
LCM = 2² × 3 × 5 × 7² ½
= 2940 ½

2
22 1
Area of trapezium = 2 × (sum of parallel sides) × height
1
= 2 ×(12 + 7) ×7
1
= 2 × 19 ×7
= 66.5 cm²
½
1
Area of quadrant = 4 𝜋 r²

1 22
=4× 7 ×7×7
½
= 38.5 cm²
Area of the shaded region = 66.5 – 38.5
½
= 28 cm²
½

2

23 ½

14−8
= 22 + (2(14)−8−4) × 4 ½+½
6
= 22 + 16 × 4
= 23.5 ½
2
24 AQ² = (7k)² + (24k)² ½
= 49k² + 576k²
= 625k²
AQ = 25k ½

cosec A = 24𝑘 = 24
25𝑘 25 ½ +½

2
25 3sin²30° +
2
1+𝑠𝑒𝑐²45°
1 2 2
= 3(2) + 1+ (√2)² ½+½
3 2
=4+3 ½

=
17 ½
12

2
26 𝑎𝑟(𝐴𝐵𝐶) 𝐵𝐶 2
= (𝑄𝑅)
𝑎𝑟(𝑃𝑄𝑅)
144 𝐵𝐶 2 ½
= ( 27 )
81
12 𝐵𝐶
= ½
9 27
BC = 9
27 ×12 ½
½
BC = 36 cm
2
27
AB =
2
= √(7 − (−2)) + (−1 − 5)² ½
= √81 + 36 ½
= √117 ½
= 3√13 units ½

OR 2

7−5 −3−1
𝓍= 2 y= 2
½+½
𝓍 =1 y = -2 ½
∴ The coordinates of the midpoint are (1, -2) ½
2
28 ½[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 24.5 ½
∴ [6(4 + 3) + k(-3 – 2) + 4(2 – 4) = 24.5 × 2 ½
∴ 42 – 5k – 8 = 49 ½
∴ - 5k + 34 = 49
∴ - 5k = 15
∴ k = -3 ½
2

Section C
29

𝓍² - 3𝓍 + 2
2𝓍 - 1 2𝓍³ - 7𝓍² + 7𝓍 – 9
2𝓍³ - 𝓍² ½
- +
- 6𝓍² + 7𝓍 – 9
- 6𝓍² + 3𝓍 ½
+ -
4𝓍 – 9
4𝓍 – 2 ½
- +
-7
Q = 𝓍² - 3𝓍 + 2 ½
R=-7 ½

2𝓍³ - 7𝓍² + 7𝓍 – 9 = (2𝓍 – 1) (𝓍² - 3𝓍 + 2) + (-7) ½

3
30 7𝓍 – 2y = 20-----------(1)
4𝓍 + 3y = -1------------(2)

28𝓍 – 8y = 80-----------(3) ½
28𝓍 + 21y = -7----------(4) ½
- - +
- 29y = 87
y = -3 ½
substituting the value of y in equation (1) we get
7𝓍 – 2(-3) = 20 ½
∴ 7𝓍 + 6 = 20
∴ 7𝓍 = 14
∴ 𝓍 =2 ½
The solution is 𝓍 = 2 and y = -3 ½

OR 3

y = 14 - 5𝓍------------(1) ½
Substituting the value of y in equation (2) we get
3𝓍 – 7y = 16-----------(2)
3𝓍 – 7(14 - 5𝓍) = 16 ½
3𝓍 – 98 + 35𝓍 = 16
3𝓍 + 35𝓍 = 16 + 98
38𝓍 = 114
𝓍 = 3 ½

From (1) y = 14 -5(3) ½
y = 14 – 15
y = -1 ½
∴ The solution is 𝓍 = 3, y = -1 ½

3

31 4𝓍² - 17𝓍 + 15 = 0
4𝓍² - 12𝓍 - 5𝓍 + 15 = 0 ½
4𝓍 (𝓍 – 3) – 5 (𝓍 – 3) = 0 ½
(𝓍 – 3) (4𝓍 – 5) = 0 ½
(𝓍 – 3) = 0 or (4𝓍 – 5) = 0 ½
𝓍 = 3 or 𝓍 = 4
5 ½
5
∴ The roots are 3 and 4 ½

3
32 D = b² - 4ac
= (-11)² - 4 × 6 × -7
= 121 + 168
= 289 ½
−𝑏 ± √𝐷
𝓍= 2𝑎 ½
11 ± √289
= 2×6
11 ± 17 ½
= 12
11+ 17 11− 17
x= 12
𝑜𝑟 𝑥= 12
½

28 −6
x= or x=
12 12

7 −1
x= 3 𝑜𝑟 𝑥= 2 ½

7
∴ The roots are 3 𝑎𝑛𝑑
−1 ½
2
3
33 a = -30 d = -23 – (-30) = 7
a15 = -30 + (15 – 1) × 7 ½
= -30 + 98
= 68 ½
𝑛
Sn = 2 [2a + (n – 1) d] ½
51
= 2 [2(-30) + (51 – 1) × 7] ½
51
= 2 [ -60 + 350]
51 ½
= 2 × 290
= 7395 ½
3
34 (17.64)2 ×0.0486
Let 𝓍 =
0.9642

log 𝓍 = 2log 17.64 + log 0.0486 – log 0.9642 ½
= 2 × 1.2465 +2̅. 6866 - 1̅. 9842 ½+½+½
= 2.4930 +2̅. 6866 - 1̅. 9842
= 1.1954 ½

𝓍 = antilog 1.1954
= 15.68 ½

3

35 To draw a line segment RS = 6.7 cm ½
To draw a circle with centre R and radius 2.5 cm ½
To construct the perpendicular bisector of line segment RS ½
To mark the points P and Q on the circle ½
To draw the tangent segments from the point S ½
To measure and write the length of the tangent segments.
SP = SQ = 6.2 ± 0.1 ½

3
36 To construct Δ XYZ with the given data 1
To draw a ray making an acute angle with side YZ. ½
To locate 4 points on the ray using a pair of compasses ½
To join Y4Z’ ∥ Y3Z and Z’X’ ∥ ZX ½+½
ΔX’YZ’ is the required triangle.
3
37 𝐴𝑟(𝛥𝑃𝐷𝐸) ½ ×𝑃𝐸 ×𝐷𝐹
(i) 𝐴𝑟(𝛥𝑅𝐷𝐸) = ½ ×𝐸𝑅 ×𝐷𝐹 Area of Δ = ½ × 𝑏 × ℎ ½
𝐴𝑟(𝛥𝑃𝐷𝐸) 𝑃𝐸
(ii) = 𝐸𝑅 ½
𝐴𝑟(𝛥𝑅𝐷𝐸)

𝐴𝑟(𝛥𝑃𝐷𝐸) 𝑃𝐷
(iii) Similarly 𝐴𝑟(𝛥𝑄𝐷𝐸) = 𝐷𝑄
½

(iv) Ar(ΔRDE) = Ar(ΔQDE) Triangles having same base DE ½+½
and lying between the same
parallels DE and QR
𝐴𝑟(𝛥𝑃𝐷𝐸) 𝐴𝑟(𝛥𝑃𝐷𝐸)
(v) 𝐴𝑟(𝛥𝑅𝐷𝐸) = 𝐴𝑟(𝛥𝑄𝐷𝐸) from (ii), (iii) and (iv)
𝑃𝐸 𝑃𝐷
(vi) = 𝐷𝑄 ½
𝐸𝑅

OR 3

(i) XY² + YZ² = XZ² Pythagoras theorem
(ii) DE² + EF² = DF² Given ½
(iii) XY² + YZ² = DF² DE = XY and EF = YZ ½
(iv) XZ² = DF² from(i) and(iii)
(v) XZ = DF ½
(vi) DE = XY and EF = YZ Given
(vii) XZ = DF Step (v)
(viii) ΔDEF ≅ ΔXYZ SSS congruence rule ½
(ix) ∠E = ∠Y CPCT ½
(x) ∠Y = 90° Given
(xi) ∠E = 90°
(xii) 𝛥DEF is a right triangle. ½

3
38 tan 30° = 𝑌𝑍
𝑋𝑌 ½

1 𝑋𝑌
= 50 ½
√3

√3 XY = 50
XY =
50 ½
√3

50√3
XY = 3

XY =
50 ×1.73 ½
3

½
XY = 28.83
½
∴ the height of the building is 28.83 m
3
39 Let OD = OE = OF = r
Since the tangents to a circle from an external point are equal
AF = AD = (8 – r) cm
and CF = CE = (6 – r) cm ½
∴ AC = AF + CF = (8 – r) + (6 – r)
= 14 – 2r ½
AC² = AB² + BC²
AC² = 8² + 6²
AC²= 64 + 36 ½
AC²= 100
∴ AC = 10 ½
∴ 14 − 2𝑟 = 10 ½
2r = 4
r =2 ½
∴ The radius of the circle is 2 cm
3
40 Let the number of cylindrical pieces be n
n=
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑜𝑛𝑒 ½
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝐶𝑦𝑙𝑖𝑛𝑑𝑒𝑟

1
𝜋𝑅²𝐻
= 3
𝜋𝑟 2 ℎ
½+½
1
×21 ×21 ×50
= 3.5 ×3.5 ×10
3
½+½
= 60
∴ the number of cylinders formed is 60 ½

3
Section D
41
Weekly pocket Number of 𝓍i fi𝓍i
money (in ₹) students
C. I. fi
15 – 25 12 20 240
25 - 35 14 30 420
35 - 45 21 40 840
45 - 55 16 50 800
55 - 65 9 60 540
65 - 75 8 70 560
∑ fi = 80 ∑ fi𝓍i = 3400
3400
Mean = 80 ½

= 42.5 ½
4

42 For completing both tables ½+½

For plotting the points of each line correctly ½+½

Drawing lines for each equation ½+½

Solution is 𝓍 = 4, y = -2 1

4
THE END

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