Super 20 Math

TERM - 1 10
CBSE
2021-22
MATHEMATICS
(Standard)
Strictly Based on CBSE Sample Question Paper
SAMPLE PAPERS
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Note from the Publishers
Mathematics (Standard) Sample Papers-X (Term-1) is a combined book

for all subjects, based on the reduced CBSE syllabus and the latest Sample Papers
issued by CBSE on 2nd September 2021 for 2021-22 (Term-1) Board Examination.
Each Sample Paper is developed by highly experienced subject experts as per the
level and the pattern followed by the CBSE. These Sample Papers will familiarise
the students with the questioning pattern of each subject.
The book contains 6 Sample Papers (1 issued by CBSE with Marking Scheme and
5 Sample Papers with Answers). Undoubtedly, students will get some parameter
to evaluate their preparation for better performance
Features of Sample Papers

• Designed exclusively to test the knowledge and preparation level of students.
• Latest CBSE Sample Question Paper has been given with Marking Scheme.
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syllabus.
• 2 OMR Sheets for Practice
Any suggestions for further improvement of the book will be thankfully received
and incorporated in the next edition.
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(iii)
CONTENTS
1. Sample Paper 1 (CBSE Sample issued by CBSE on 2nd September, 2021) ................. D–1
2. Sample Paper 2.............................................................................................................D–16
3. Sample Paper 3.............................................................................................................D–22
4. Sample Paper 4.............................................................................................................D–27
5. Sample Paper 5.............................................................................................................D–32
6. Sample Paper 6.............................................................................................................D–38
l Answers to Sample Papers 2 to 6.................................................................................. D–44
l 2 OMR Sheets
(iv)
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\ 29-Sep-2021 Ved_Goswami Proof-4 Reader’s Sign _______________________ Date __________
MATHEMATICS
(Standard)
MATHEMATICS – STANDARD
COURSE STRUCTURE
CLASS-X (2021-22)
(Term-I)
Times : 90 Minutes Marks : 40
S.No. Unit Name Total marks
I. Number Systems 6
II. Algebra 10
III. Coordinate Geometry 6
IV. Geometry 6
V. Trigonometry 5
VI. Mensuration 4
VII. Statistics & Probability 3
Total 40
Internal Assessment 10
Total 50
Sample Paper–
[Issued by CBSE on 2nd September, 2021]
1
Time Allowed: 90 Minutes Maximum Marks: 40
General Instructions:
1. The question paper contains three parts A, B and C.
2. Section-A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.
3. Section-B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.
4. Section-C consists of 10 questions based on two Case Studies. Attempt any 8 questions.
5. There is no negative marking.
SECTION-A
Section-A consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.
1. The ratio of LCM and HCF of the least composite and the least prime numbers is 1
(a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 3
2. The value of k for which the lines 5x + 7y = 3 and 15x + 21y = k coincide is 1
(a) 9 (b) 5 (c) 7 (d) 18
3. A girl walks 200 m towards East and then 150m towards North. The distance of the girl from
the starting point is 1
(a) 350 m (b) 250 m (c) 300 m (d) 225 m
4. The lengths of the diagonals of a rhombus are 24 cm and 32 cm, then the length of the altitude
of the rhombus is 1
(a) 12 cm (b) 12.8 cm (c) 19 cm (d) 19.2 cm
5. Two fair coins are tossed. What is the probability of getting at the most one head? 1
3 1 1 3
(a) (b) 4 (c) (d)
4 2 8
6. ΔABC ~ ΔPQR. If AM and PN are altitudes of ΔABC and ΔPQR respectively and AB2 :
PQ2 = 4 : 9, then AM : PN = 1
(a) 16 : 81 (b) 4 : 9 (c) 3 : 2 (d) 2 : 3
7. If 2 sin2 β – cos2 β = 2, then β is 1
(a) 0° (b) 90° (c) 45° (d) 30°
8. Prime factors of the denominator of a rational number with the decimal expansion 44.123
are 1
(a) 2, 3 (b) 2, 3, 5 (c) 2, 5 (d) 3, 5
9. The lines x = a and y = b, are 1
(a) intersecting (b) parallel (c) overlapping (d) None of these
D–1
10. The distance of point A (–5, 6) from the origin is 1

(a) 11 units (b) 61 units (c) 11 units (d) 61 units
23
11. If a2 = , then a is 1
25
(a) rational (b) irrational (c) whole number (d) integer
12. If LCM (x, 18) = 36 and HCF (x, 18) = 2, then x is 1
(a) 2 (b) 3 (c) 4 (d) 5
13. In ΔABC right angled at B, if tan A= 3 , then cos A cos C – sin A sin C = 1
3
(a) – 1 (b) 0 (c) 1 (d)
2
14. If the angles of ΔABC are in ratio 1 : 1 : 2, respectively (the largest angle being angle C),
sec A tan A
then the value of − is 1
cosec B cot B
1 3
(a) 0 (b) (c) 1 (d)
2 2
15. The number of revolutions made by a circular wheel of radius 0.7 m in rolling a distance of
176 m is 1
(a) 22 (b) 24 (c) 75 (d) 40
16. ΔABC is such that AB = 3 cm, BC = 2cm, CA = 2.5 cm. If ΔABC ~ ΔDEF and EF = 4 cm,
then perimeter of ΔDEF is 1
(a) 7.5 cm (b) 15 cm (c) 22.5 cm (d) 30 cm
17. In the figure, if DE || BC, AD = 3 cm, BD = 4 cm and BC = 14 cm, then DE equals 1
(a) 7 cm (b) 6 cm (c) 4 cm (d) 3 cm

4 sin β − 3 cos β
18. If 4 tan β = 3, then = 1
4 sin β + 3 cos β
1 2 3
(a) 0 (b)(c) (d)
3 3 4
19. One equation of a pair of dependent linear equations is – 5x + 7y = 2. The second equation
can be 1
(a) 10x +14y +4 = 0 (b) –10x –14y + 4 = 0 (c) –10x + 14y + 4 = 0(d) 10x – 14y = – 4
D–2 n Mathematics (Standard) – X

20. A letter of English alphabets is chosen at random. What is the probability that it is a letter of
the word ‘MATHEMATICS’? 1
4 9 5 11
(a) (b) (c) (d)
13 26 13 26
SECTION-B
Section-B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted.
21. If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such
numbers are 1
(a) 2 (b) 3 (c) 4 (d) 5
22. Given below is the graph representing two linear
equations by lines AB and CD respectively. What is the
area of the triangle formed by these two lines and the
line x = 0? 1
(a) 3 sq. units
(b) 4 sq. units
(c) 6 sq. units
(d) 8 sq. units
23. If tan α + cot α = 2, then tan20 α + cot20 α = 1
(a) 0 (b) 2
(c) 20 (d) 220
24. If 217x + 131y = 913, 131x + 217y = 827, then x + y is 1
(a) 5 (b) 6 (c) 7 (d) 8
25. The LCM of two prime numbers p and q (p > q) is 221. The value of 3p – q is equal to 1
(a) 4 (b) 28 (c) 38 (d) 48
26. A card is drawn from a well shuffled deck of cards. What is the probability that the card
drawn is neither a king nor a queen? 1
11 12 11 11
(a) (b) (c) (d)
13 13 26 52
27. Two fair dice are rolled simultaneously. The probability that 5 will come up at least once is
1
5 11 12 23
(a) (b) (c) (d)
36 36 36 36
2
28. If 1 + sin α = 3 sin α cos α, then values of cot α are 1
(a) –1, 1 (b) 0, 1 (c) 1, 2 (d) –1, – 1
29. The vertices of a parallelogram in order are A (1, 2), B (4, y), C (x, 6) and D (3, 5). Then
(x, y) is 1
(a) (6, 3) (b) (3, 6) (c) (5, 6) (d) (1, 4)
Sample Paper–1 n D–3

30. In the given figure, ∠ACB = ∠CDA, AC = 8 cm, AD = 3 cm, then BD is 1
22 26 55 64
(a) cm (b) cm (c) cm (d) cm
3 3 3 3
31. The equation of the perpendicular bisector of line segment joining points A (4, 5) and B (–2, 3)
is 1
(a) 2x – y + 7 = 0 (b) 3x +2y – 7 = 0 (c) 3x – y – 7 = 0 (d) 3x + y – 7 = 0
cot y°
32. In the given figure, D is the mid-point of BC, then the value of is 1
cot x°
1 1 1
(a) 2 (b) (c) (d)
2 3 4
1
33. The smallest number by which should be multiplied so that its decimal expansion terminates
13
after two decimal places is 1
13 13 10 100
(a) (b) (c) (d)
100 10 13 13
34. Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm
respectively. The length of the side of largest square FDGB that can be inscribed in the
triangle ABE is 1

32 16 8 4
(a)cm (b) cm (c) cm (d) cm
3 3 3 3
35. Point P divides the line segment joining R (– 1, 3) and S (9, 8) in ratio k :1. If P lies on the line
x – y + 2 = 0, then value of k is 1
2 1 1 1
(a) (b) (c) (d) 4
3 2 3
36. In the figure given below, ABCD is a square of side 14 cm with E, F, G and H as the mid
points of sides AB, BC, CD and DA respectively. The area of the shaded portion is 1
49π
(a) 44 cm² (b) 49 cm² (c) 98 cm²
cm² (d)
2
37. Given below is the picture of the Olympic rings made by taking five congruent circles of
radius 1cm each, intersecting in such a way that the chord formed by joining the point of
intersection of two circles is also of length 1cm. Total area of all the dotted regions assuming
the thickness of the rings to be negligible is 1
π 3 π 3 π 3 π 3
(a) 4  − cm² (b)  − cm² (c) 4  − cm² (d) 8  6 − 4  cm²

 12 4  6 4  6

4   
1
38. If 2 and are the zeroes of px2 + 5x + r, then 1
2
(a) p = r = 2 (b) p = r = – 2 (c) p = 2, r = – 2 (d) p = – 2, r = 2
39. The circumference of a circle is 100 cm. The side of a square inscribed in the circle is 1
100 50 2 100 2
(a) 50 2 cm (b) cm (c) cm (d) cm
π π π
40. The number of solutions of 3x + y = 243 and 243x – y = 3 is 1
(a) 0 (b) 1 (c) 2 (d) infinite

SECTION-C (Case Study Based Questions)

Section-C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted.
Q41 – Q45 are based on Case Study-1
Case Study-1
The figure given here shows the path of a diver, when she takes a jump from the diving board.
Clearly it is a parabola.
Annie was standing on a diving board, 48 feet above the water level. She took a dive into the pool.
Her height (in feet) above the water level at any time ‘t’ in seconds is given by the polynomial
h(t) such that h(t) = – 16t² + 8t + k
41. What is the value of k? 48 1

(a) 0 (b) – 48 (c) 48 (d)
−16
42. At what time will she touch the water in the pool? 1
(a) 30 seconds (b) 2 seconds (c) 1.5 seconds (d) 0.5 seconds
43. Rita’s height (in feet) above the water level is given by another polynomial p(t) with zeroes
– 1 and 2. Then p(t) is given by 1
(a) t² + t – 2 (b) t² + 2t – 1 (c) 24t² – 24t + 48 (d) – 24t² + 24t + 48
44. A polynomial q(t) with sum of zeroes as 1 and the product as – 6 is modelling Anu’s height
in feet above the water at any time t (in seconds). Then q(t) is given by 1
(a) t² + t + 6 (b) t² + t – 6 (c) – 8t² + 8t + 48 (d) 8t² – 8t + 48
45. The zeroes of the polynomial r(t) = – 12t² + (k – 3)t + 48 are negative of each other. Then k
is 1
(a) 3 (b) 0 (c) – 1.5 (d) – 3
Case Study-2
A hockey field is the playing surface for the game of hockey. Historically, the game was played
on natural turf (grass) but nowadays it is predominantly played on an artificial turf. It is rectangular
in shape – 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the
centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts
must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7
feet) above the ground.

Each team plays with 11 players on the field during the game including the goalie. Positions you
might play include:
• Forward: As shown by players A, B, C and D.
• Midfielders: As shown by players E, F and G.
• Fullbacks: As shown by players H, I and J.
• Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
46. The coordinates of the centroid of ΔEHJ are 1

 2   −2  2   2 
(a)  − , 1 (b) 1,  (c)  , 1 (d)  − , − 1
 3   3  3   3 
47. If a player P needs to be at equal distances from A and G, such that A, P and G are in straight
line, then position of P will be given by 1
 3   3  3
(a)  − 2 , 2
 (b)  2, − 2  (c)  2,  (d) (– 2, – 3)
 2
48. The point on x-axis equidistant from I and E is 1
1   1   1   1 
(a)  , 0 (b)  0, −  (c)  − , 0 (d)  0, 
2  2 2 2
49. What are the coordinates of the position of a player Q such that his distance from K is twice
his distance from E and K, Q and E are collinear? 1
(a) (1, 0) (b) (0, 1) (c) (– 2, 1) (d) (– 1, 0)
50. The point on y-axis equidistant from B and C is 1
(a) (– 1, 0) (b) (0, – 1) (c) (1, 0) (d) (0, 1)

Marking Scheme
1. (b) Least composite number is 4 and the least prime number is 2.
LCM (4, 2) : HCF (4, 2) = 4 : 2 = 2 : 1
a b c
2. (a) For lines to coincide: 1 = 1 = 1
a2 b2 c2
5 7 −3
So, = =
15 21 − k
i.e. k= 9
3. (b) By Pythagoras theorem
The required distance = (200 2
+ 1502 )
= ( 40000 + 22500) = (62500) = 250 m.
So the distance of the girl from the starting point is 250 m.
1 1
4. (d) Area of the Rhombus = × d1 × d2 = 2 × 24 × 32 = 384 cm²
2
Using Pythagoras theorem
2 2
1  1 
Side² =  d1  +  d 2  = 12² +16² = 144 + 256 = 400
2  2 
Side = 20 cm
Area of the Rhombus = base × altitude
384 = 20 × altitude
384
So altitude = = 19.2 cm
20
5. (a) Possible outcomes are (HH), (HT), (TH), (TT).
Favorable outcomes (at the most one head) are (HT), (TH), (TT).
3
So probability of getting at the most one head = .
4
6. (d) Ratio of altitudes = Ratio of sides for similar triangles
So, AM : PN = AB : PQ = 2 : 3
7. (b) 2 sin β – cos2 β = 2
2
Then,2 sin2 β – (1 – sin2 β) = 2

3 sin2 β = 3 or sin2 β =1
β is 90°
8. (c) Since it has a terminating decimal expansion, so prime factors of the denominator will
be 2, 5.

9. (a) Lines x = a is a line parallel to y axis and y = b is a line parallel to x axis. So they will
intersect.
10. (d) Distance of point A (– 5, 6) from the origin (0, 0) is
( 0 + 5 )2 + ( 0 − 6 )2 = 25 + 36 = 61 units
23
11. (b) a² = 23/25, then a = , which is irrational
5
12. (c) LCM × HCF = Product of two numbers
36 × 2 = 18 × x
x= 4
13. (b) tan A = 3 = tan 60°
So, ∠A = 60°,
Hence, ∠C = 30°.
 1  3  3  1
So, cos A cos C – sin A sin C =   ×   –   ×   = 0
2  2   2   2
14. (a) 1x + 1x + 2x = 180°, x = 45°
∠A , ∠B and ∠C are 45°, 45° and 90° respectively
sec A tan A sec 45° tan 45° 2 1
− = − = − =1–1=0
cosec B cot B cosec 45° cot 45° 2 1
total distance 176
15. (d) Number of revolutions = = = 40
circumference 2 × 22 × 0.7
7
perimeter of ∆ABC BC
16. (b) =
perimeter of ∆DEF EF
7.5 2
=
perimeter of ∆DEF 4
So perimeter of ΔDEF = 15 cm
17. (b) Since DE || BC, ΔABC ~ ΔADE (By AA rule of similarity)
AD DE 3 DE
So = i.e. = . So DE = 6 cm
AB BC 7 14
18. (a) Dividing both numerator and denominator by cos β,
4 sin β − 3 cos β 4 tan β − 3 3 − 3
= =
4 sin β + 3 cos β 4 tan β + 3 3 + 3 = 0
19. (d) – 2(– 5x + 7y = 2) gives 10x – 14y = – 4.
a1 b1 c1
Now = = =–2
a2 b2 c2
20. (a) Number of Possible outcomes are 26
Favorable outcomes are M, A, T, H, E, I, C, S

8 4
=
Probability =
26 13
21. (c) Since HCF = 81, two numbers can be taken as 81x and 81y,
ATQ 81x + 81y = 1215
or x + y = 15
which gives four co-prime pairs
1, 14
2, 13
4, 11
7, 8
1
22. (c) Required Area is area of triangle ACD = × 6 × 2 = 6 sq units
2
23. (b) tan α + cot α = 2 gives α = 45°. So tan α = cot α = 1
tan20 α + cot20 α = 120 + 120 = 1 + 1 = 2
24. (a) Adding the two given equations we get: 348x + 348y = 1740.
So, x+y= 5
25. (c) LCM of two prime numbers = product of the numbers
221 = 13 × 17.
So, p = 17 and q = 13
∴ 3p – q = 51 – 13 =38
26. (a) Probability that the card drawn is neither a king nor a queen
52 − 8
=
52
44 11
= =
52 13
27. (b) Outcomes when 5 will come up at least once are
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4) and (5, 6)
11
Probability that 5 will come up at least > 36
28. (c) 1+ sin2 α = 3 sin α cos α
sin2 α + cos2 α + sin2 α = 3 sin α cos α
2 sin2 α – 3 sin α cos α + cos2 α = 0
(2 sin α – cos α) (sin α – cos α) = 0
∴ cot α = 2 or cot α = 1
29. (a) Since ABCD is a parallelogram, diagonals AC and BD bisect each other.
∴ mid point of AC = mid point of BD
 x + 1 6 + 2  3 + 4 5 + y
 ,  ,
2 2  =  2 2 


Comparing the co-ordinates, we get,

x +1 3+ 4
= So, x = 6
2 2
6+2 5+ y
Similarly, = 2 So, y = 3
2
∴ (x, y) = (6, 3)
30. (c) ΔACD ~ Δ ABC (AA)
AC AD

∴ AB = (CPST)
AC
8 3
=
AB 8
64
This gives AB = cm
3
64 55
So, −3 =
BD = AB – AD = cm
3 3
31. (d) Any point (x, y) of perpendicular bisector will be equidistant from A and B.

∴ ( x − 4 )2 + ( y − 5 )2 = ( x + 2)2 + ( y − 3)2
Solving we get– 12x – 4y + 28 = 0 or 3x + y – 7 = 0
AC
cot y° BC CD CD 1
32. (b) = AC = = =
cot x° BC 2CD 2
CD
1
33. (a) The smallest number by which should be multiplied so that its decimal expansion
13
13 1 13 1
terminates after two decimal points is as × = = 0.01
100 13 100 100
34. (b) ΔABE is a right triangle & FDGB is a square of side x cm

ΔAFD ~ Δ DGE (AA-similarity)
AF FD

∴ DG = (CPST)
GE
16 − x x
=
x 8− x
16
128 = 24x or x= cm
3

35. (a) Since P divides the line segment joining R (– 1, 3) and S (9, 8) in ratio k : 1
 9 k − 1 8k + 3 

∴ coordinates of P are  , .
 k + 1 k + 1 
9 k − 1 8k + 3
Since P lies on the line x – y + 2 = 0, then k + 1 − k + 1 + 2 = 0
9k – 1 – 8k – 3 + 2k + 2 = 0
2
which gives k =
.
3
36. (c)
Shaded area = A rea of semicircle + (Area of half square – Area of two

quadrants)
= Area of semicircle + (Area of half square – Area of semicircle)
= Area of half square
1
= × 14 × 14 = 98 cm²
2
37. (d) Let O be the center of the circle. OA = OB = AB = 1 cm
So, ΔOAB is an equilateral triangle and ∴ ∠AOB = 60°

Required Area = 8 × Area of one segment with r = 1 cm, q = 60°
 60° 3 2
= 8×  × π × 12 − ×1 
 360° 4 
π 3
= 8 − cm²
6 4 

1 5
38. (b) Sum of zeroes = 2 + =−
2 p
5 5
i.e.,
=− . So, p = – 2
2 p
1 r
Product of zeroes = 2 × 2 = p
r
i.e.
= 1 or r = p = – 2
p
100
39. (c) 2πr = 100. So, Diameter = 2r = = diagonal of the square.
π
100
side 2 = diagonal of square =
π
100 50 2

∴ side = =
2π π
40. (b) 3x + y = 243 = 35
So, x+y= 5 ...(1)
243x – y = 3
(35)x – y = 31
So, 5x – 5y = 1 ...(2)
a1 b1
Since, ≠ , so unique solution.
a2 b2
41. (c) Initially, at t = 0, Annie’s height is 48ft
So, at t = 0, h should be equal to 48
h(0) = – 16(0)² + 8(0) + k = 48
So, k = 48
42. (b) When Annie touches the pool, her height = 0 feet
i.e., – 16t² + 8t + 48 = 0 above water level

2t² – t – 6 = 0
2t² – 4t + 3t – 6 = 0
2t (t – 2) +3 (t – 2) = 0
(2t + 3) (t – 2) = 0
3
i.e.
t= 2 or t = –
2
Since time cannot be negative, so t = 2 seconds

43. (d) t = – 1 and t = 2 are the two zeroes of the polynomial p(t)
Then, p(t) = k {t – (–1)} (t – 2)
= k(t + 1) (t – 2) = k (t2 – t – 2)
When t = 0 (initially) h1 = 48 ft
p(0) = k (0² – 0 – 2) = 48
i.e. – 2k = 48
So the polynomial is – 24 (t² – t – 2) = – 24t² + 24t + 48
44. (c) A polynomial q(t) with sum of zeroes as 1 and the product as – 6 is given by
q(t) = k [t² – (sum of zeroes)t + product of zeroes]
= k [t² – 1t + (–6)] ...(1)
When t = 0 (initially) q(0) = 48 ft
q(0) = k (0² – 1(0) – 6) = 48
i.e. – 6k = 48 or k = – 8
Putting k = – 8 in equation (1), reqd. polynomial is – 8 [t² – 1t + (–6)]
= – 8t² + 8t + 48
45. (a) When the zeroes are negative of each other,
sum of the zeroes = 0
−b
So, = 0
a
− (k − 3)
= 0
−12
k −3
= 0
12
k – 3 = 0,
i.e. k= 3
46. (a) Centroid of ΔEHJ with E (2, 1), H (– 2, 4) and J (– 2, – 2) is
 2 + ( − 2) + ( − 2) 1 + 4 + ( − 2 )   2 

,  =  − , 1
3 3 3 
47. (c) If P needs to be at equal distance from A (3, 6) and G (1, – 3), such that A, P and G are
collinear, then P will be the mid-point of AG.
 3 + 1 6 + ( − 3)   2, 3 
So coordinates of P will be  ,  =  2 
2 2
48. (a) Let the point on x-axis equidistant from I (– 1, 1) and E (2, 1) be (x, 0)
Then ( x + 1)2 + (0 − 1)2 = ( x − 2)2 + (0 − 1)2
x2 + 1 + 2x +1 = x2 + 4 – 4x + 1
6x = 3

1
So, x= .
2
1 
∴ the required point is  ,
0
2 
49. (b) Let the coordinates of the position of a player Q such that his distance from K(– 4, 1) is
twice his distance from E(2, 1) be Q(x, y)
Then KQ : QE = 2 : 1
 2 × 2 + 1 × − 4 2 × 1 + 1 × 1
Q(x, y) =  ,  = (0, 1)
3 3
50. (d) Let the point on y-axis equidistant from B(4, 3) and C(4, – 1) be (0, y)
Then ( 4 − 0 )2 + ( 3 − y )2 = (0 − 4)2 + ( y + 1)2

16 + y2 + 9 – 6y = 16 + y2 + 1 + 2y
– 8y = – 8
So, y = 1.
∴ the required point is (0, 1)
❑❑❑

E:\Super_20_All-In-One-10_(27-09-2021)\Open_Files\4_Super_20_Maths_Standard\Unsolved Sample Paper-2\Unsolved Sample Paper-2
Sample Paper– 2
General Instructions: Same as Sample Paper-1
SECTION-A
1. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45
cm, respectively. Minimum distance each should walk so that each can cover the same distance
in complete steps is 1
(a) 2220 (b) 2520 cm (c) 2250 cm (d) 2260 cm
2. The value of k so that the following system of equations has no solution is given by
3x – y – 5 = 0, 6x – 2y + k = 0 1
(a) k ≠ –7 (b) k ≠ 5 (c) k ≠ 3 (d) k ≠ 10
3. For what value of k, (– 4) is a zero of p(x) = x2 – x – (2k – 2)? 1
(a) 4 (b) 3 (c) 1 (d) 11
4. If the zero of polynomials 3x2 – px + 2 and 4x2 – qx – 10 is 2, then value of 2p –3q is given
by 1
(a) 8 (b) 7 (c) 5 (d) 6
5. A card is selected from a deck of 52 cards. The probability of it being a red face card will be 1
11 5 3 1
(a) (b) (c) (d)
26 26 26 26
1
6. In an equilateral triangle ABC, D is a point on the side BC such that BD = BC. Value of
3
9AD2 equals 1
(a) 6AB 2 (b) 7AB 2 (c) 4AB 2 (d) 5AB 2
7. In the given figure, DABC is right angled at B, BC = 7 cm and AC – AB = 1 cm. The value
of cos A + sin A is given by 1
A
31 31
(a) (b)
26 25
31 31
(c) (d)
29 32
B C
8. 429 as a product of its prime factors can be express as 1

(a) 7 × 11 × 13 (b) 2 × 7 × 13 (c) 11 × 13 × 5 (d) 3 × 11 × 13
9. If the pair of equations x sin q + y cos q = 1 and x + y = 2 has infinitely many solutions,
then what is the value of q? 1
(a) 75° (b) 30° (c) 45° (d) 60°
D–16
10. The point on the y-axis which is equidistant from (2, –5) and (–2, 9) is given by 1
(a) (0, 2) (b) (0, 3) (c) (0, 5) (d) (0, 4)
11. In a seminar the number of participants in Hindi, English and Mathematics are 60, 84 and
108 respectively. The minimum number of rooms required if in each room the same number
of participants are to be seated and all of them being in the same subject is given by 1
(a) 17 (b) 18 (c) 21 (d) 20
441
12. The rational number 2 7 2 has a 1
2 ⋅5 ⋅7
(a) terminating decimal (b) non-terminating decimal
(c) terminating with repeating decimal (d) none of these
13. The coordinates of the points P and Q are respectively (4, –3) and (–1, 7). The x-coordinate
PR 3
(abscissa) of a point R on the line segment PQ such that = is given by 1
PQ 5
(a) 2 (b) 1 (c) 3 (d) 4
14. Let P and Q be the points of trisection of the line segment joining the points A (2, – 2) and B (–7, 4)
such that P is nearer to A. The coordinates of P and Q are given by 1
(a) P(2, 3), Q(–4, 0) (b) P(–1, 0), Q(–4, 2) (c) P(5, 3), Q(–1, 2) (d) P(1, 0), Q(2, 4)
15. The circumference of a circle whose area is equal to the sum of areas of the circles with
diameters 10 cm and 24 cm is given by 1
1 1 5
(a) 82 cm (b) 80 cm (c) 83 cm (d) 81 cm
2 2 7
16. Two poles of height a and b (b > a) are c metres apart. The height h (in metres) of the point
of intersection of the lines joining the top of each pole to the foot of the opposite pole is 1
ab ab 2 a 2b ab
(a) (b) (c) (d)
a+b a−b a+b a−b
17. If the sum of the zeroes of the polynomial x2 – (k + 6) x + 2 (2k – 1) is half their product, then
value of k is 1
(a) 4 (b) 5 (c) 7 (d) 8
18. If sin q = x and sec q = y then value of cot q is given by 1
x 1 1 1
(a) (b) (c) 2 (d) xy
y xy 2
x y
19. The value of m and n so that the following pair of linear equations has infinite number of
solutions (2m – 1)x + 3y = 5; 3x + (n – 1)y = 2 is given by 1
17 11 7 1 1 3 1
(a) m = ,n= (b) m = , n = − (c) m = , n = (d) m = , n = 4
4 5 4 2 2 2 4
20. A jar contains 24 marbles, some are green and other are blue. If a marble is drawn random from
2
the jar, the probability that it is green is . The number of blue marbles in the jar is given by 1
3
(a) 8 (b) 9 (c) 10 (d) 11

SECTION-B
21. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is 1
(a) 4 (b) 2 (c) 1 (d) 3
22. For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident
lines? 1
1 1
(a) (b) − (c) 2 (d) –2
2 2
23. The distance between the points A(0, 6) and B(0, –2) is 1
(a) 6 (b) 8 (c) 4 (d) 2
24. The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely
many solutions is 1
(a) 3 (b) –3 (c) –12 (d) no value
25. If two positive integers p and q are written as p = ab2 and q = a3b; a, b are prime numbers,
then LCM (p, q) is 1
(a) ab 2
(b) a b 2 3
(c) a b 2 3
(d) a b 3
26.
A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from
house B, 5 from house C, 2 from house D and rest from house E. A single student is selected
at random to be the class monitor. The probability that the student is not from A, B and C is
1
4 6 8 17
(a) (b) (c) (d)
23 23 23 23
___ ___
27. If P (A) = 0.65, P(B) = 0.15, then P( A ) + P( B ) = 1
(a) 1.5 (b) 1.2 (c) 0.8 (d) None of these
28. The points (–4, 0), (4, 0), (0, 3) are the vertices of a 1
(a) right triangle (b) isosceles triangle (c) equilateral triangle (d) scalene triangle
a
29. Given that sin q = , then cos q is equal to 1
b
b b b2 − a 2 a
(a) (b) (c) (d)
b2 − a 2 a b b2 − a 2
30. If the polynomial x3 – 3x2 + kx + 42 is divisible by x + 3, then the value of k will be 1
(a) 4 (b) 14 (c) –4 (d) –14
31. The distance between the points (a cos q, a sin q) and (– a cos q, – a sin q) is 1
(a) 4a (b) 3a (c) 2a (d) a
32. The fourth vertex D of parallelogram ABCD whose three vertices are A(–2, 3), B(6, 7) and
C(8, 3) is 1
(a) (0, 1) (b) (0, –1) (c) (–1, 0) (d) (1, 0)

33. Four bells toll at an interval of 8, 12, 15 and 18 seconds, respectively. All the four begin to
toll together. How many times will they toll together in one hour excluding the one at the
start? 1
(a) 9 times (b) 10 times (c) 11 times (d) 12 times
34. If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is – 1, then the product of the
other two zeroes is 1
(a) b – a + 1 (b) b – a – 1 (c) a – b + 1 (d) a – b + 1
35. The coordinates of the point which is equidistant Y
from the three vertices of the ∆AOB as shown B(0, 2y)
in given figure is 1
(a) (x, y) M
(b) (y, x) X X
O A(2x, 0)
x y
(c)  , 
2 2
 y x
(d)  ,  Y
 2 2
36. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii
24 cm and 7 cm is 1
(a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm
37. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is 1
(a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14
38. The zeroes of the quadratic polynomial x2 + kx + k, k ≠ 0, 1
(a) cannot both be positive (b) cannot both be negative
(c) are always unequal (d) are always equal
39. The area of circle varies as the square of its radius. If the area of circle of radius 10 cm is 300
cm2, what is the area of circle with radius 12 cm? 1
(a) 360 cm2 (b) 423 cm2 (c) 432 cm2 (d) 452. 5 cm2
40. A pair of linear equations which has a unique solution x = 2, y = – 3 is 1
(a) x + y = –1 (b) 2x + 5y = –11
2x – 3y = – 5 4x + 10y = – 22
(c) 2x – y = –1 (d) x – 4y – 14 = 0
3x + 2y = 0 5x – y – 13 = 0

Case Study-1
A Frame House: A frame-house is a house constructed from a wooden skeleton, typically
covered with timber board. The concept of similar triangles is used to construct it. Look at the
following picture:

A
D
P
B C
Q R
House (i) House (ii)
41. The front view of house (i) is shown along side in which point
P on AB is joined with point Q on AC. 1
If PQ || BC, AP = x m, PB = 10 m, AQ = (x – 2) m, QC = 6 m,
then the value of x is
(a) 3 m
(b) 4 m
(c) 5 m
(d) 8 m
42. The side view of house (i) is shown below in which point F on
AC is joined with point G on DE. 1
If ACED is a trapezium with AD || CE, F and G are points on
non-parallel sides AC and DE respectively such that FG is
AF
parallel to AD, then =
FC
DG AD
(a) (b)
GE CE
AF DG
(c) (d)
GE FC
43. The front view of house (ii) is shown along side in which point
S on PQ is joined with point T on PR. 1
PS PT
If = and ∠PST = 70°, ∠QPR = 50°, then angle
QS TR
∠QRP =
(a) 70° (b) 50°
(c) 80° (d) 60°

44. Again consider the front view of house (ii). If S and T are
points on side PQ and PR respectively such that ST || QR and
PS : SQ = 3 : 1. Also TP = 6.6 m, then PR is 1
(a) 6.9 m
(b) 8.8 m
(c) 10.5 m
(d) 9.4 m
45. Sneha has also a frame house whose front view is shown below 1
If MN || AB, BC = 7.5 m, AM = 4 m and MC = 2 m, then length
of BN is
(a) 5 m
(b) 4 m
(c) 8 m
(d) 9 m

Case Study-2
Skysails’ is that genre of engineering science
that uses extensive utilization of wind
energy to move a vessel in the sea water.
The ‘Skysails’ technology allows the towing
kite to gain a height of anything between
100 metres to 300 metres. The sailing kite is
made in such a way that it can be raised to its
proper elevation and then brought back with
the help of a ‘telescopic mast’ that enables
the kite to be raised properly and effectively.
46. In the given figures, if sin q = cos (3q – 30°), where q and 3q – 30° are acute angles, then the
value of q is 1
(a) 30° (b) 60° (c) 45° (d) None of these.
47. What should be the length of the rope of the kite sail in order to pull the ship at the angle q
(calculated in part (a) and be at a vertical height of 200 m? 1
(a) 300 m (b) 400 m (c) 500 m (d) 600 m
48. If BC = 15 m, q = 30°, then AB is 1
(a) 2 3 m (b) 15 m (c) 24 m (d) 5 3 m
49. Suppose AB = BC = 12 m, then q = 1
(a) 0° (b) 30° (c) 45° (d) 60°
50. Given that BC = 6 m and q = 45°. The values of AB and AC are respectively 1
(a) AB = 4 m, AC = 4 2 m (b) AB = 7 m, AC = 7 5 m
(c) AB = 9 m, AC = 9 3 m (d) AB = 6 m, AC = 6 2 m
❑❑❑

Sample Paper– 3
SECTION-A
1. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one
number is 280, then the other number will be 1
(a) 70 (b) 75 (c) 78 (d) 80
2. The value of k for which the following pair of linear equations have infinitely many solutions
2x + 3y = 7, (k + 1)x + (2k –1)y = 4k + 1 is given by 1
(a) 3 (b) 4 (c) 2 (d) 5
3. The zeroes of the polynomial p(y) = 5 5 y 2 + 30 y + 8 5 are 1
2 4 3 2 3 5 4 2
(a) − ,− (b) − ,− (c) ,− (d) ,
5 5 5 5 5 3 5 5
4. If 1 is a zero of polynomial p(x) = ax2 – 3(a – 1) – 1, then the value of a is 1
(a) 4 (b) 3 (c) 1 (d) 2
5. Two dice are thrown at the same time and the product of numbers appearing on them is noted.
The probability that the product is a prime number is given by 1
5 1 1 2
(a) (b) (c) (d)
6 6 3 3
6. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s.
If the lamp is 3.6 m above the ground, the length of her shadow after 4 seconds is 1
(a) 1.8 m (b) 1.4 m (c) 1.6 m (d) 2.6 m
7. The values of x, for which the distance between the points P (2, –3) and Q (x, 5) is 10 is given
by 1
(a) x = 2, 4 (b) x = –4, 8 (c) x = 4, –6 (d) x = 2, –3
8. If HCF of 45 and 105 is 15 then their LCM will be given by 1
(a) 315 (b) 305 (c) 302 (d) 310
9. If tan a = 5 , the value of sec a will be 1
12
11 13 10 13
(a) (b) (c) (d)
2 12 3 2
10. The value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7) is given
by 1
(a) 6 (b) 5 (c) 3 (d) 4
D–22
11. The largest number which on dividing 1251, 9377 and 15628 leaves ramainders 1, 2 and 3
respectively is given by 1
(a) 625 (b) 600 (c) 620 (d) 575
27
12. After how many places of decimal the decimal form of the number 3 4 2 will terminate?
2 .5 .3
1
(a) 5 places (b) 2 places (c) 4 places (d) 3 places
13. The ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, –2)
and B (3, 7) is given by 1
(a) 5 : 3 (b) 1 : 4 (c) 3 : 5 (d) 2 : 9
14. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, value of x
and y are given by 1
(a) x = 2, y = 3 (b) x = 6, y = 3 (c) x = 2, y = 4 (d) x = –1, y = 4
15. A square ABCD is inscribed in a circle of radius 10 units. The area of the circle, not included
in the square is given by 1
(a) 105 cm 2 (b) 102 cm 2 (c) 114 cm 2 (d) 110 cm 2
16. P is the mid-point of side BC of DABC, Q is the mid-point of AP, BQ when produced meets
AC at L. Then length of AL is equal to 1
1 2 1
(a) AC (b) 1 AC (c) AC (d) AC
4 5 3 3
17. If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3,
then values of p and q will be 1
(a) p = 4, q = –4 (b) p = 2, q = –3 (c) p = 4, q = –5 (d) p = –5, q = –6
2 2
18. Value of 3 cot 60° + sec 45° is equal to 1
(a) 3 (b) 4 (c) 5 (d) 6
19. If sin q + cos q = 3 , then tan q + cot q is equal to 1
(a) 3 (b) 2 (c) 1 (d) 4
20. A game consists of tossing a 10 rupee coin 3 times and noting its outcome each time. Sudhir
wins if all the tosses give the same result, i.e., three heads or three tails and loses otherwise.
The probability that Sudhir will not win the game will be 1
1 1 1 3
(a) (b) (c) (d)
4 3 4 4
SECTION-B
21. The (HCF × LCM) for the numbers 189 and 297 is 1
(a) 57633 (b) 56337 (c) 56733 (d) 65337
22. If cos (a + b) = 0, then sin (a – b) can be reduced to 1
(a) cos b (b) cos 2b (c) sin a (d) cos 2a

a 
23. If P  , 4 is the mid-point of the line segment joining the points Q(– 6, 5) and R(–2, 3),
3 
then the value of a is 1
(a) –4 (b) –12 (c) 12 (d) –6
p
24. If sin q = , then the value of tan q + sec q is 1
q
q− p q+ p q2 + p2 q2 − p2
(a) (b) (c) (d)
q+ p q− p q2 − p2 q2 + p2
25. The areas of three fields are 165 m2, 195 m2 and 285 m2 respectively. From these, flowers
beds of equal size are to be made. If the breadth of each bed be 3 metres, what will be the
maximum length of each bed? 1
(a) 4 m (b) 5 m (c) 6 m (d) 7 m
26. A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card
is not a red king, is 1
1 12 1 25
(a) (b) (c) (d)
13 13 26 26
27. Out of 600 bolts, 20% are too large and 10% are too small. The remaining are considered to
be suitable, if a bolt is selected at random, the probability that it will be suitable is 1
1 7 1 3
(a) (b) (c) (d)
10 10 5 10
28. The distance of the point (1, 2) from the mid-point of line segment joining the points (6, 8)
and (2, 4) is 1
(a) 10 (b) 6 (c) 7 (d) 5
29. If cos 9a = sin a, then the values of tan 5a is 1
1
(a) (b) 3 (c) 1 (d) 0
3
30. A quadratic polynomial whose zeroes are – 3 and 4, is 1
x2 x
(a) x2 – x + 12 (b) x2 + x + 12 (c) − −6 (d) 2x2 + 2x – 24
2 2
31. The point which divides the line joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally
lies in the 1
(a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant
32. If the perpendicular bisector of the line segment joining the points A(1, 5) and B(4, 6) cuts
the y-axis at 1
(a) (0, 13) (b) (0, –13) (c) (0, 12) (d) (13, 0)
33. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one
number is 280, then the other number is given by 1
(a) 40 (b) 60 (c) 80 (d) 100

34. The zeroes of the quadratic polynomial x2 + 99x + 127 are 1

(a) both positive (b) both negative
(c) one positive and one negative (d) both equal
35. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid-
point of PQ, then the coordinates of P and Q are, respectively 1
(a) (0, –5) and (2,0) (b) (0, 10) and (–4, 0)
(c) (0, 4) and (–10, 0) (d) (0, –10) and (4, 0)
36. The radii of two circles are 8 cm and 6 cm respectively. The diameter of the circle having
area equal to the sum of the areas of the two circles is : 1
(a) 10 cm (b) 14 cm (c) 20 cm (d) 28 cm
37. The circumference of a circle is 22 cm. The area of its quadrant (in cm2) is 1
77 77 77 77
(a) (b) (c) (d)
2 4 8 16
38. If a, b, g are zeroes of the polynomial 6x + 3x – 5x + 1, then the value of a–1 + b–1 + g–1 is
3 2
1
(a) 4 (b) 7 (c) 5 (d) 9
39. If the difference between the circumference and the radius of a circle is 37 cm, then using
22
p= the circumference (in cm) of the circle is 1
7
(a) 154 (b) 44 (c) 14 (d) 7
 4 sin θ − cos θ 
40. If 4 tan q = 3, then  is equal to 1
 4 sin θ + cos θ 
2 1 1 3
(a) 3 (b) 3 (c) 2 (d) 4

Case Study-1
Amit is planning to buy a house and the layout is given below. The design and the measurement
has been made such that areas of two bedrooms and kitchen together is 95 sq.m.
x 2 y
5m Bedroom 1 Bath
room Kitchen
2m
5m Bedroom 2 Living Room
15 m

Based on the above information, answer the following questions:

41. The pair of linear equations in two variables from above situation is 1
(a) x + y = 13, 2x + y = 19 (b) 2x + y = 13, x + y = 19
(c) x + 2y = 13, x – y = 19 (d) None of these
42. The length of the outer boundary of the layout is 1
(a) 35 m (b) 54 m (c) 42 m (d) 60 m
43. The area of each bedroom and kitchen in the layout respectively is 1
(a) 25 m, 35 m (b) 15 m, 25 m (c) 30 m, 35 m (d) 25 m, 30 m
44. The area of living room in the layout is 1
(a) 55 sq. m (b) 65 sq. m (c) 75 sq. m (d) 85 sq. m
45. The cost of laying tiles in kitchen at the rate of ` 50 per sq m is 1
(a) ` 1260 (b) ` 1750 (c) ` 1590 (d) ` 1810
Case Study-2
Rohan wants to measure the distance of a pond during the visit to his native. He marks points A
and B on the opposite edges of a pond as shown in the figure below. To find the distance between
the points, he makes a right-angled triangle using rope connecting B with another point C at a
distance of 12 m, connecting C to point D at a distance of 40 m from point C and connecting D
to the point at A which is at a distance of 30 m from D such that ADC = 90º
B 12 m
A C
30
m
40
m
D
46. Which property of geometry will be used to find the distance AC? 1
(a) Similarity of triangles (b) Thales Theorem
(c) Pythagoras Theorem (d) Congruency of triangles
47. What is the distance AC? 1
(a) 50 m (b) 12 m (c) 100 m (d) 70 m
48. Which is the following does not form a Pythagoras triplet? 1
(a) (7, 24, 25) (b) (15, 8, 17) (c) (5, 12, 13) (d) (21, 20, 28)
49. The length of AB is 1
(a) 12 m (b) 38 m (c) 50 m (d) 100 m
50. The length of the rope used the 1
(a) 120 m (b) 70 m (c) 82 m (d) 22 m
❑❑❑

Sample Paper– 4
SECTION-A
1. The HCF and the LCM of 12, 21, 15 respectively are: 1
(a) 3, 140 (b) 12, 420 (c) 3, 420 (d) 420, 3
2. The pair of linear equations 1
3x 5 y
+ = 7 and 9x + 10y = 14 is
2 3
(a) consistent (b) inconsistent
(c) consistent with one solution (d) consistent with many solutions
3. In DABC and DDEF, –B =–E, –F = –C and AB = 3DE. Then, the two triangles are 1
(a) congruent but not similar (b) similar but not congruent
(c) neither congruent nor similar (d) congruent as well as similar
AB BC CA
4. If in two triangles ABC and PQR, = = , then 1
QR PR PQ
(a) DPQR ~ DCAB (b) DPQR ~ DABC (c) DCBA ~ DPQR (d) DBCA ~ DPQR
5. Some one is asked to take a number from 1 to 100. The probability that it is a prime is 1
1 6 1 13
(a) (b) (c) (d)
5 25 4 15
6. If in ΔABC, AB = 9 cm, BC = 40 cm and AC = 41 cm, then the ΔABC is a/an 1
(a) Acute angled triangle (b) Right triangle
(c) Obtuse angled triangle (d) Isosceles triangle
1
7. If sin A = , then the value of cot A is 1
2
1 3
(a) 3 (b) (c) (d) 1
3 2
8. 180 can be expressed as a product of its prime factors as: 1
(a) 10 × 2 × 3 2 (b) 25 × 4 × 3 2
(c) 2 × 3 × 52 (d) 4 × 9 × 5
9. Which of the following is not a solution of the pair of equations 3x – 2y = 4 and 6x – 4y = 8?
1
(a) x = 2, y = 1 (b) x = 4, y = 4 (c) x = 6, y = 7 (d) x = 5, y = 3
10. If the point P(k, 0) divides the line segment joining the points A(2, –2) and B(–7, 4) in the ratio 1 : 2,
then the value of k is 1
(a) 1 (b) 2 (c) –2 (d) –1
D–27
11. A rational number can be expressed as a terminating decimal if the denominator has factors 1
(a) 2, 3 or 5 (b) 2 or 3 (c) 3 or 5 (d) 2 or 5
11
12. The decimal representation of 3 will
1 2 ×5
(a) terminate after 1 decimal place (b) terminate after 2 decimal places
(c) terminate after 3 decimal places (d) not terminate
a
13. Given that sin q = , then cos q is equal to 1
b
b b b2 − a 2 a
(a) (b) (c) (d)
b2 − a 2 a b b2 − a 2
3
14. Given that sin a = and cos b = 0, then the value of b – a is 1
2
(a) 0° (b) 90° (c) 60° (d) 30°
15. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii
24 cm and 7 cm is 1
(a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm
AB BC
16. If in DABC and DDEF, = , then they will be similar, when 1
DE FD
(a) ∠A = ∠F (b) ∠A = ∠D (c) ∠B = ∠D (d) ∠B = ∠E
17. A vertical stick 30 m long casts a shadow 15 m long on the ground. If the same time, a tower
costs a shadow 75 m long on the ground. The height of the tower is 1
(a) 150 m (b) 100 m (c) 25 m (d) 200 m
18. If sin A + sin2A = 1, then the value of the expression (cos2A + cos4 A) is 1
(a) 1 (b) 1 (c) 2 (d) 3
2
9
19. If the pair of equations 2x + 3y = 7 and kx + y = 12 have no solution, then the value of
k is: 2 1
2 3
(a) (b) (c) 3 (d) –3
3 2
20. Which of the following cannot be the probability of an event? 1
1 17
(a) (b) 0.1 (c) 3% (d)
3 16
SECTION-B
21. Given that LCM (91, 26) = 182, then HCF (91, 26) is: 1
(a) 13 (b) 26 (c) 17 (d) 9

22. The value of k for which the system of equations kx + 2y = 5, 3x + 4y = 1, has no solution, is
1
(a) 5 (b) 2/3 (c) 6 (d) 3/2
7
23. If cos A = , then cot2 A = 1
9
49 49 49 49
(a) (b) (c) (d)
72 52 32 62
24. Sunita has 10 paise and 50 paise coins in her purse. If the total number of coins is 17 and
their total values is ` 4.50, then the number of 10 paise and 50 paise coins in her purse is: 1
(a) 7, 10 (b) 10, 7 (c) 12, 5 (d) None of these
p
25. Rational number , q ≠ 0, will be terminating decimal if the prime factorisation of q is of
q
the form (m and n are non-negative integers) 1
m
(a) 2 × 3 n m
(b) 2 × 5 n m
(c) 3 × 5 n m
(d) 3 × 7 n
26. The probability of getting a king in a draw from a pack of 52 cards is 1
1 1 1 2
(a) (b) (c) (d) 13
7 11 13
27. A die is thrown once. The probability of getting a prime number is 1
2 1 1 1
(a) (b) (c) (d)
3 3 2 6
1 + cos 60o
28. = 1
2
3 1 1
(a) (b) 1 (c) (d)
2 2 4
29. The ratio in which the line segment joining (3, 4) and (–2, 1) divided by the y-axis is 1
(a) 2 : 5 (b) 2 : 3 (c) 3 : 2 (d) 1 : 3
30. ∆ABC ~ ∆DEF and their areas are respectively 625 cm2 and 64 cm2. If the altitude of ∆ABC
is 5 cm, then the corresponding altitude of ∆DEF is 1
(a) 5.3 cm (b) 1.6 cm (c) 3.8 cm (d) 5.8 cm
31. The point on the x-axis which is equidistant from the points (5, 4) and (–2, 3) is 1
(a) (3, 0) (b) (2, 0) (c) (4, 0) (d) (–1, 0)
32. 3 tan2 30° + sec4 45° – tan2 60° is equal to 1
(a) 0 (b) 1 (c) 2 (d) 3
33. If p, q are two consecutive natural numbers, then HCF (p, q) is 1
(a) q (b) p (c) 1 (d) pq

34. A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If
the lamp is 3.6 m above the ground, then the length of her shadow after 4 seconds is 1
(a) 1. 9 m (b) 1. 7 m (c) 1. 5 m (d) 1. 6 m
35. If (1, 1) is the mid-point of AB and the point B is (4, 3), the coordinates of the point A are 1
(a) (–1, –2) (b) (–2, –1) (c) (1, –3) (d) (–3, 1)
36. The diameter of a cycle wheel is 28 cm. The number of revolution will it make in moving 11
km is 1
(a) 12000 (b) 12200 (c) 12500 (d) 12400
37. The area of a sector is one-twelfth that of a complete circle. Then the angle of the sector is
1
(a) 30° (b) 60° (c) 36° (d) 45°
3 2
38. If the sum of the zeroes of the cubic polynomial kx – 5x – 11x – 3 is , then k = 1
(a) 1 (b) 2 (c) 3 (d) –3
39. A horse, left for grazing inside a rectangular enclosure 40 m × 36 m, is tethered to one corner
of the field by a rope 14 m long. The area of the quadrant over which the horse can graze 1
(a) 144 sq. m (b) 154 sq.m (c) 164 sq. m (d) 135 sq. m
40. The value of k for which the system of equations x + 2y + 7 = 0 and 2x + ky + 14 = 0 will
have infinitely many solutions is: 1
(a) 2 (b) 4 (c) 6 (d) 8

Case Study-1
Due to heavy storm an electric wire got
Y
bent as shown in the figure. It followed
a mathematical shape. Answer the 6
following questions below: 5
41. Name the shape in which the wire is 4
bent. 1 3
(a) spiral (b) ellipse 2
(c) linear (d) parabola 1
42. How many zeroes are there for the X'–6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7

X
polynomial (shape of the wire)? 1 –1
(a) 2 (b) 3 –2
(c) 1 (d) 0 –3
43. The zeroes of the polynomial are 1 –4
(a) –1, 5 (b) –1, 3 –5
(c) 3, 5 (d) –4, 2 Y'

44. What will be the expression of the polynomial? 1

(a) x2 + 2x – 3 (b) x2 – 2x + 3 (c) x2 – 2x – 3 (d) x2 + 2x + 3
45. What is the value of the polynomial if x = –1? 1
(a) 6 (b) –18 (c) 18 (d) 0
Case Study-2
In the sport of cricket the Captain sets the field according to a plan. He instructs the players to
take a position at a particular place. There are two reasons to set a cricket field—to take wickets
and to stop runs being scored.
The following graph shows the position of players during a cricket match.
Slip Wicketkeeper
Third Man
11
Gully Square
2 1 Leg
3
Point 4 10
Cover 5
9 Mid
6 Wicket
Extra
7 8
Cover
Mid Off Mid On

46. If the distance between the points showing the players at Gully A(1, 0) and wicketkeeper at
B(4, p) is 5 m, then p = 1
(a) 4 m (b) 8 m (c) 6 m (d) 9 m
47. Suppose the length of a line segment joining the players of Mid-off and Mid-on be 10 units.
If the coordinates of its one end are (2, –3) and the abscissa of the other end is 10 units, then
its ordinate is 1
(a) 9, 6 (b) 3, –9 (c) –3, 9 (d) 9, –6
48. The coordinate of the point on x-axis which are equidistant from the points representing the
players at Cover P(–3, 4) and Mid-wicket Q(2, 5) are 1
4
(a) (20, 0) (b) (–23, 0) (c)  , 0 (d) None of these.
5 
49. The ratio in which (4, 5) divides the line segment joining the points Extra Cover S(2, 3) and
Fine Leg (7, 8) is 1
(a) 4 : 3 (b) 5 : 2 (c) 3 : 2 (d) 2 : 3
50. If the points (4, 3) and (x, 5) are on the circular field with centre (2, 4), then the value of x is
1
(a) 0 (b) 1 (c) 2 (d) 3
❑❑❑

Time Allowed: 90 Minutes
Sample Paper– 5 Maximum Marks: 40
SECTION-A
1. The total number of factors of a prime number is 1
(a) 1 (b) 0 (c) 2 (d) 3
2. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent
is 1
14 2
(a) − (b) (c) 5 (d) 10
3 5
AD 2
3. In given figure, DE || BC. If = and AE = 2.7 cm, then EC is equal to 1
DB 3
(a) 2.0 cm (b) 1.8 cm (c) 4.0 cm (d) 2.7 cm

4. If DPQR ~ DXYZ, ∠Q = 50° and ∠R = 70°, then ∠X + ∠Y is equal to 1
(a) 70° (b) 110° (c) 120° (d) 50°
5. One card is drawn from a well shuffled deck of 52 cards. The probability that it is black queen
is 1
1 1 1 2
(a) (b) (c) (d)
26 13 52 13
6. In ΔABC, ∠B = 90° and BD ⊥ AC. If AC = 9 cm and AD = 3 cm, then BD is equal to 1
(a) 2 2 cm (b) 3 2 cm (c) 2 3 cm (d) 3 3 cm

7. The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1 : 2
internally lies in the 1
(a) Ist quadrant (b) IInd quadrant
(c) IIIrd quadrant (d) IVth quadrant
D–32
8. 225 can be expressed as: 1

(a) 5 × 32 (b) 52 ×3 (c) 52 × 32 (d) 53 ×3
9. The area of the square that can be inscribed in a circle of radius 8 cm is 1
(a) 256 cm2 (b) 128 cm2 (c) 64 2 cm2 (d) 64 cm2
10. The length of an altitude of an equilateral triangle of side a is 1
2 2 2 3 a
(a) a (b) a (c) a (d)
5 3 2 2
14570
11. The decimal expansion of the rational number will terminate after: 1
1250
(a) one decimal place (b) two decimal places
(c) three decimal places (d) four decimal places
p
12. Rational number , q ≠ 0, will be terminating decimal if the prime factorisation of q is of
q
the form (m and n are non-negative integers) 1
(a) 2m × 3n (b) 2m × 5n (c) 3m × 5n (d) 3m × 7n
13. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4 ), then
1 1 1
(a) AP = AB (b) AP = PB (c) PB = AB (d) AP = AB
3 3 2
m 
14. If A  , 5 is the mid-point of the line segment joining the points Q (–6, 7) and R (–2, 3),
3 
then the value of m is 1
(a) –12 (b) –4 (c) 12 (d) –6
15. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the
arc is 1
(a) 11 cm (b) 22 cm (c) 27 cm (d) 44 cm
16. In DABC if AB = 4 cm, BC = 8 cm and AC = 4 3 cm, then the measure of ∠A is 1
(a) 60° (b) 90° (c) 45° (d) 30°

17. The centre of a circle whose end points of a diameter are (–6, 3) and (6, 4) is 1
 7  7
(a) (8, –1) (b) (4, 7) (c)  0,  (d)  4, 
 2  2
18. The quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is 1
(a) x2 + 5x + 6 (b) x2 – 5x + 6
(c) x2 – 5x – 6 (d) –x2 + 5x + 6

19. The graph of a polynomial is shown in figure, then the number of its zeroes is 1
(a) 3 (b) 1 (c) 2 (d) 4

2
20. The sum of the zeroes of the polynomial 2x – 8x + 6 is 1
(a) –3 (b) 3 (c) –4 (d) 4
SECTION-B
21. If d is HCF of two positive integers a and b, then there exist two integers k and l such that 1
(a) a = kd + lb (b) b = ka + ld
(c) d = ka + lb (d) None of these
22. A card is drawn at random from a well shuffled pack of 52 playing cards. The probability of
getting a red face card is 1
(a) 3 (b) 3 (c) 3 (d) None of these
25 26 28
23. If (2, 4) is the mid-point of the join of (6, a) and (b, 5) then a and b are 1
(a) a = 2, b = – 2 (b) a = 3, b = – 2
(c) a = – 2, b = – 2 (d) a = –2, b = 3
3 sin θ tan θ + 1
24. If cos q = , then the value of is 1
5 2 tan 2 θ
88 91 92 93
(a) (b) (c) (d)
160 160 160 160
25. If d = HCF (48, 72), the value of d is 1
(a) 24 (b) 48 (c) 12 (d) 72
26. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. The probability
of getting a black ball is 1
3 2 4 1
(a) (b) (c) (d)
5 5 5 5
27. A card is drawn at random from a well-shuffled deck of playing cards. The probability that
card drawn is neither a king nor a queen 1
11 11 11
(a) (b) (c) (d) None of these
12 13 15
28 If A (– 4, 2), B (2, 0), C (8, 6) and D (a, b) are the vertices of a parallelogram ABCD then a
and b are 1
(a) (2, 5) (b) (2, 8) (c) (3, 8) (d) (2, 3)

29. The three vertices of a parallelogram are (1, 1), (4, 4), and (4, 8). The fourth vertex is 1
(a) (2, 4) (b) (3, 5) (c) (1, 5) (d) (5, 1)
30. In an isosceles triangle ABC if AC = BC and AB2 = 2 AC2, then ∠C, 1
(a) 60° (ii) 90° (c) 60° (d) 30°
31. The coordinates of a point equidistant from three given points A(5, 1), B(–3, –7) and
C(7, –1) are 1
(a) (1, –4) (b) (3, –4) (c) (2, –4) (d) (4, –2)
1
32. If sin A = , then the value of (9 cot2 A + 9) is 1
3
1
(a) 1 (b) 81 (c) 9 (d)
81
33. If two positive integers a and b are written as a = x2y2 and b = xy2, where x, y are prime
numbers, then HCF (a, b) is 1
(a) xy (b) xy 2 2
(c) x y 2 2
(d) x y 3
34. A and B are the mid-points on the sides RP and RQ respectively of ΔPQR right angled at R.
Then 4(PB2 + QA2) = 1
(a) 3 PQ2 (b) 5PQ2 (c) 6PQ2 (d) None of these
35. The points (2a, 4a), (2a, 6a) and {(2 + 3 ) a, 5a} are the vertices of an 1
(a) isosceles triangle (b) scalene triangle
(c) equilateral triangle (d) None of these
36. A rectangular park is 70m by 56m. It is surrounded by semicircular flower beds all round.
The cost of semicircular flower beds at 60 paise per m2 is 1
(a) ` 3788.40 (b) ` 2587.50
(c) ` 3500 (d) None of these
37. The area of the largest circle that can be drawn in a square of area 196 cm2 is 1
(a) 160 cm 2 (b) 154 cm 2 (c) 165 cm 2 (d) 162 cm 2
38. Sum of the zeroes of the polynomial x2 + 7x + 10 are 1

(a) 7 (b) – 7 (c) 10 (d) – 10
39. A quadratic polynomial whose zeroes are 7 and 5 is 1
2
(a) x + 2x – 35 2
(b) x – 12x + 35 2
(c) x + 12x + 53 (d) None of these
40. If α, β are the zeroes of the polynomial x2 – 6x + 6, then the value of α2 + β2 is 1
(a) 36 (b) 24 (c) 12 (d) 6

Case Study-1
The state governments revise fares from time to time based on various factors such as inflation, fuel
price, demand from various quarters, etc. The government notifies different fares for different types
of vehicles like Auto Rickshaws, Taxis, Radio Cab, etc.

The auto charges in a city comprise of a fixed charge together with the charge for the distance
covered. Study the following situations:
Situation-I: In city A, for a journey of 10 km, the charge paid is ` 75 and for a journey of 15 km,
the charge paid is ` 110.
Situation-II: In city B, for a journey of 8 km, the charge paid is ` 91 and for a journey of 14 km,
the charge paid is ` 145.
Refer Situation I
41. If the fixed charges of auto rickshaw be ` x and the running charges be ` y km/hr, the pair of
linear equations representing the situation is 1
(a) x + 10y = 110, x + 15y = 75 (b) x + 10y = 75, x + 15y = 110
(c) 10x + y = 110, 15x + y = 75 (d) 10x + y = 75, 15x + y = 110
42. What will a person have to pay for travelling a distance of 25 km? 1
(a) ` 160 (b) ` 280 (c) ` 180 (d) ` 260
43. A person travels a distance of 50 km. The amount he has to pay is 1
(a) ` 155 (b) ` 255 (c) ` 355 (d) ` 455
Refer Situation II
44. What will a person have to pay for travelling a distance of 30 km? 1
(a) ` 185 (b) ` 289 (c) ` 275 (d) ` 305
45. The graphs of lines representing the conditions are 1
(a) (b)

(c) (d)

Case Study-2
Three children were playing with sticks. As they had one stick each A
of them, they put all the three sticks together. Finding all the three
sticks equal, they pick up the sticks and put them in a triangular form
in such a way that the ends of each stick touch the other. They were 30°30°
surprised. Now they thought of a plane. They took another stick and 2a 2a
put it as in the adjacent figure. The stick AD is just touching the stick
BC. Somehow, they measured each angle. Finding that each angle.
∠A = ∠B = ∠C = 60° (equal) and ∠BAD = ∠CAD = 30°. Likewise, 60°
they measured BD = CD, and ∠ADB = ∠ADC = 90°. Taking AB a a
B D C
= BC = CA = 2a, you are required to answer the following questions:
46. The length of AD is 1
(a) a (b) 2a (c) 2a (d) 3a
47. Using the above figure, the value of sin 30° is 1
1 1 3
(a) (b) (c) (d) 1
2 2 2
48. Using the above figure, the value of cos 60° is 1
1 3
(a) 0 (b) 1 (c) (d)
2 2
49. Using the above figure, the value of tan 30° is 1
1
(a) 3 (b) 1 (c) 0 (d)
3
50. Using the above figure, the value of cosec 60° is 1
2 3 1 1
(a) (b) (c) (d)
3 2 2 2
❑❑❑

Sample Paper– 6
SECTION-A
1. The LCM of smallest two digit composite number and smallest composite number is: 1
(a) 12 (b) 4 (c) 20 (d) 44
2. If a pair of linear equations is consistent, then the lines will be: 1
(a) parallel (b) always coincident
(c) intersecting or coincident (d) always intersecting
3. It is given that DABC ~ DDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm.
Then, the following is true: 1
(a) DE = 12 cm, ∠F = 50° (b) DE = 12 cm, ∠F = 100°
(c) EF = 12 cm, ∠D = 100° (d) EF = 12 cm, ∠D = 30°
4. In DABC, DE || BC, the value of x will be 1
A
x x+3
D E
x+1 x+5
B C
(a) 2 (b) 4 (c) 1 (d) 3
5. Someone is asked to take a number from 1 to 100. The probability that it is a prime is: 1
1 6 1 13
(a) (b) (c) (d)
5 25 4 50
AB BC CA
6. If in two triangles ABC and PQR, = = , then: 1
QR PR PQ
(a) ∆PQR ~ ∆CAB (b) ∆PQR ~ ∆ABC
(c) ∆CBA ~ ∆PQR (d) ∆BCA ~ ∆PQR
7. The value of the expression [cosec (75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q)]
is: 1
3
(a) –1 (b) 0 (c) 1 (d)
2
D–38
8. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is:
1
(a) 13 (b) 65 (c) 875 (d) 1,750
9. The pair of equations y = 0 and y = –7 has: 1
(a) one solution (b) two solutions
(c) infinitely many solutions (d) no solution
10. The points (–4, 0), (4, 0) and (0, 3) are the vertices of a/an: 1
(a) right triangle (b) isosceles triangle (c) equilateral triangle (d) scalene triangle
11. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then
HCF (a, b) is: 1
(a) xy (b) xy2 3
(c) x y 3 2
(d) x y 2
14587
12. The decimal expansion of the rational number will terminate after: 1
1250
(a) one decimal place (b) two decimal places
(c) three decimal places (d) four decimal places
13. If cos (a + b) = 0, then sin (a – b) can be reduced to 1
(a) cos b (b) cos 2b (c) sin a (d) sin 2a
14. sin (45° + q) – cos (45° – q) is equal to ______ . 1
(a) 3 (b) 2 (c) 0 (d) 5
15. It is proposed to build a single circular park equal in area to the sum of areas of two circular
parks of diameters 16 m and 12 m in a locality. The radius of the new park would be: 1
(a) 10 m (b) 15 m (c) 20 m (d) 24 m
16. In the given Figure, DE || BC, AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm, the length of side
AD is 1
A
1.8 cm
D E
7.2 cm
5.4 cm
B
C
(a) 1.8 cm (b) 2.2 cm (c) 2.4 cm (d) 3.4 cm
17. In the given figure, if AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm and ∠ADE = 48°,
then ∠ABC is given by 1
A
D
E
C
(a) 52° (b) 48° (c) 60° (d) 70°

18. If 3 cos q = 1, then cosec q is equal to 1

5 3 3 3
(a) (b) (c) (d)
2 2 2 2 2 2
19. For which value (s) of p, will the lines represented by the following pair of linear equations
be parallel. 1
3x – y – 5 = 0
6x – 2y – p = 0
(a) all real values except 10 (b) 10
5 1
(c) (d)
2 2
20. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the
box, the probability that it bears a prime-number less than 23, is: 1
7 10 4 9
(a) (b) (c) (d)
90 90 45 89
SECTION-B
21. The product of the HCF and the LCM of the smallest prime number and the smallest composite
number is 1
(a) 2 (b) 4 (c) 6 (d) 8
22. The value of k for which the system of equations x + 2y + 7 = 0 and 2x + ky + 14 = 0 will
have infinitely many solutions is: 1
(a) 2 (b) 4 (c) 6 (d) 8
1 + cos 60°
23. = 1
2
3 1
(a) (b) 1 (c) 1 (d)
2 2 4
24. In two numbers, if 3 is added to each number, then their ratio is 1 : 2. However, if 3 is
subtracted from each number, then their ratio is 2 : 5. One of the numbers is: 1
(a) 21 (b) 18 (c) 15 (d) 12
25. If HCF (a, 8) = 4, LCM (a, 8) = 24, then a is 1
(a) 8 (b) 10 (c) 12 (d) 14
26. A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from
the box. The probability that the ball drawn is neither red nor white is 1
1 1 1 1
(a) (b) (c) (d)
3 2 4 6

27. A box contains 19 balls bearing numbers 1, 2, 3, ...... 19. A ball is drawn at random from the
box. The probability that the number of the ball is divisible by 3 or 5 is 1
5 7 8 11
(a) (b) (c) (d)
19 19 19 19
3 sin θ tan θ + 1
28. If cos θ = , then the value of 1
5 2 tan 2 θ
88 91 92 93
(a) (b) (c) (d)
160 160 160 160
29. If the distance between the points (4, p) and (1, 0) is 5, then the value of p is 1
(a) 4 only (b) ±4 (c) –4 only (d) 0
30. ∆ABC ~ ∆DEF. If BC = 4 cm, EF = 5 cm and area (∆ABC) = 32 cm2, then the area of ∆DEF
is 1
(a) 60 cm 2 (b) 50 cm 2 (c) 40 cm 2 (d) 80 cm 2
31. If C(1, 1) divides the line segment joining A(–2, 7) and B in the ratio 3 : 2 internally, then
coordinates of B are 1
(a) (2, –3) (b) (–2, 3) (c) (3, –3) (d) (–3, –3)
7
32. If cos A = , then cot2 A = 1
9
49 49 49 49
(a) (b) (c) (d)
72 52 32 62
33. A rational number can be expressed as a terminating decimal if the denominator has factors
1
(a) 2, 3 or 5 (b) 2 or 3 (c) 3 or 5 (d) 2 or 5
34. In an equilateral triangle ABC, AD is altitude drawn from A on the side BC. Then 3 AB2 =
1
(a) 2 AD2 (b) 5 AD2 (c) 4 AD2 (d) 6 AD2
35. Point A is on the y-axis at a distance of 4 units from origin. If coordinates of point B are
(–3, 0), the length of AB is 1
(a) 7 units (b) 5 units (c) 49 units (d) 25 units
36. A rectangular park is 70 m by 56 m. It is surrounded by semicircular flower beds all around.
The cost of semicircular flower beds at 60 paise per m2 is 1
(a) ` 3788.40 (b) ` 2587.50 (c) ` 3500 (d) None of these
37. The area of the largest circle that can be drawn in a square of area 196 cm2 is 1
(a) 160 cm2 (b) 154 cm2 (c) 165 cm2 (d) 162 cm2
38. If 2 of the zeroes of the polynomial x3 + 2x2 – x – 2 are 1 and – 1, then all the zeroes of the
polynomial are 1
(a) 1, –1, 3 (b) 1, –1, 2 (c) 1, –1, –2 (d) None of these
39. The area of the sector of a circle when the angle of the sector is 63° and the diameter of the
circle is 18 cm is 1
(a) 44 cm2 (b) 44.55 cm2 (c) 66 cm2 (d) 88 cm2
40. A number consists of two digits whose sum is 15. If 9 is added to the number, then the digits
change their places. The number is: 1
(a) 69 (b) 78 (c) 87 (d) 96

Case Study-1
Rainbow is an arch of colours that is visible in the sky, caused by the refraction and dispersion
of the sun’s light after rain or other water droplets in the atmosphere. The colours of the rainbow
are generally said to be red, orange, yellow, green, blue, indigo and violet.
Each colour of rainbow makes a parabola. We know that for any quadratic polynomial
ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two
shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or
a < 0. These curves are called parabolas.
41. A rainbow is represented by the quadratic polynomial x2 + (a + 1)x + b whose zeroes are 2
and –3. Then 1
(a) a = –7, b = –1 (b) a = 5, b = –1 (c) a = 2, b = –6 (d) a = 0, b = –6
42. The polynomial x2 – 2x – (7p + 3) represents a rainbow. If –4 is zero of it, then the value
of p is 1
(a) 1 (b) 2 (c) 3 (d) 4
43. The graph of a rainbow y = f (x) is shown below. 1
The number of zeroes of f (x) is

(a) 0 (b) 1 (c) 2 (d) 3
44. If graph of a rainbow does not intersect the x-axis but intersects y-axis in one point, then
number of zeroes of the polynomial is equal to 1
(a) 0 (b) 1 (c) 0 or 1 (d) none of these
45. The representation of a rainbow is a quadratic polynomial. The sum and the product of its
zeroes are 3 and –2 respectively. The polynomial is 1
2
(a) k(x – 2x – 3), for some real k. 2
(b) k(x – 5x – 9), for some real k.
(c) k(x2 – 3x – 2), for some real k. (d) k(x2 – 8x + 2), for some real k.

Case Study-2
The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four
walls of the sun room are square clear glass panels. The roof is made using
• four clear glass panels, trapezium in shape, all the same size
• one tinted glass panel, half a regular octagon in shape

Refer to Top View
46. The mid-point of the segment joining the points J (6, 17) and I (9, 16) is given by 1
 33 15   3 1  15 33   1 3
(a)  ,  (b)  ,  (c)  ,  (d)  , 
 2 2  2 2  2 2  2 2
Refer to Front View
47. The distance of the point P from the y-axis is 1
(a) 4 (b) 15 (c) 19 (d) 25
Refer to Front View
48. The distance between the points A and S is 1
(a) 4 (b) 8 (c) 16 (d) 20
Refer to Front View
49. The coordinates of the point which divides the line segment joining the points A and B in the
ratio 1 : 3 internally is given by 1
(a) (8.5, 2.0) (b) (2.0, 9.5) (c) (3.0, 7.5) (d) (2.0, 8.5)
Refer to Front View
50. If a point (x, y) is equidistant from the Q(9, 8) and S(17, 8), then 1
(a) x + y = 13 (b) x – 13 = 0 (c) y – 13 = 0 (d) x – y = 13
❑❑❑

ANSWERS
Sample Paper–2
1. (b) 2. (d) 3. (d) 4. (c) 5. (c) 6. (b) 7. (b) 8. (d)
9. (c) 10. (a) 11. (c) 12. (b) 13. (b) 14. (b) 15. (d) 16. (a)
17. (c) 18. (d) 19. (a) 20. (a) 21. (b) 22. (c) 23. (b) 24. (d)
25. (c) 26. (b) 27. (b) 28. (b) 29. (c) 30. (c) 31. (c) 32. (b)
33. (b) 34. (a) 35. (a) 36. (d) 37. (c) 38. (a) 39. (c) 40. (b)
41. (c) 42. (a) 43. (d) 44. (b) 45. (a) 46. (a) 47. (b) 48. (d)
49. (c) 50. (d)
Sample Paper–3
1. (d) 2. (d) 3. (a) 4. (c) 5. (b) 6. (c) 7. (b) 8. (a)
9. (b) 10. (c) 11. (a) 12. (c) 13. (d) 14. (b) 15. (c) 16. (d)
17. (d) 18. (a) 19. (c) 20. (d) 21. (c) 22. (b) 23. (b) 24. (b)
25. (b) 26. (d) 27. (b) 28. (d) 29. (c) 30. (c) 31. (d) 32. (a)
33. (c) 34. (b) 35. (d) 36. (c) 37. (c) 38. (c) 39. (b) 40. (c)
41. (a) 42. (b) 43. (c) 44. (c) 45. (b) 46. (c) 47. (a) 48. (d)
49. (b) 50. (c)
Sample Paper–4
1. (c) 2. (b) 3. (b) 4. (a) 5. (c) 6. (b) 7. (a) 8. (c)
9. (d) 10. (d) 11. (d) 12. (c) 13. (c) 14. (d) 15. (d) 16. (c)
17. (a) 18. (a) 19. (c) 20. (d) 21. (a) 22. (d) 23. (c) 24. (b)
25. (b) 26. (c) 27. (c) 28. (a) 29. (c) 30. (b) 31. (b) 32. (c)
33. (c) 34. (d) 35. (b) 36. (c) 37. (a) 38. (c) 39. (b) 40. (b)
41. (d) 42. (a) 43. (b) 44. (c) 45. (d) 46. (a) 47. (b) 48. (d)
49. (d) 50. (a)
Sample Paper–5
1. (c) 2. (d) 3. (b) 4. (b) 5. (a) 6. (b) 7. (d) 8. (c)
9. (b) 10. (c) 11. (d) 12. (b) 13. (d) 14. (a) 15. (b) 16. (b)
17. (c) 18. (a) 19. (a) 20. (d) 21. (c) 22. (b) 23. (b) 24. (d)
25. (a) 26. (a) 27. (b) 28. (b) 29. (c) 30. (b) 31. (c) 32. (b)
33. (b) 34. (b) 35. (c) 36. (a) 37. (b) 38. (c) 39. (b) 40. (b)
41. (a) 42. (c) 43. (c) 44. (b) 45. (c) 46. (d) 47. (a) 48. (c)
49. (d) 50. (a)
Sample Paper–6
1. (c) 2. (c) 3. (b) 4. (d) 5. (c) 6. (a) 7. (b) 8. (a)
9. (d) 10. (b) 11. (b) 12. (d) 13. (b) 14. (c) 15. (a) 16. (c)
17. (b) 18. (c) 19. (a) 20. (c) 21. (d) 22. (b) 23. (a) 24. (a)
25. (c) 26. (c) 27. (c) 28. (d) 29. (b) 30. (b) 31. (c) 32. (c)
33. (d) 34. (c) 35. (b) 36. (a) 37. (b) 38. (c) 39. (b) 40. (b)
41. (d) 42. (c) 43. (c) 44. (a) 45. (c) 46. (c) 47. (a) 48. (c)
49. (d) 50. (b)

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TERM - 1 10

Full Marks Pvt Ltd

An ISO : 9001-2015 Company

9, Daryaganj, New Delhi-110002

Note from the Publishers

Mathematics (Standard) Sample Papers-X (Term-1) is a combined book

Features of Sample Papers

2. Sample Paper 2.............................................................................................................D–16

3. Sample Paper 3.............................................................................................................D–22

4. Sample Paper 4.............................................................................................................D–27

5. Sample Paper 5.............................................................................................................D–32

6. Sample Paper 6.............................................................................................................D–38

l Answers to Sample Papers 2 to 6.................................................................................. D–44

10. The distance of point A (–5, 6) from the origin is 1

(a) 7 cm (b) 6 cm (c) 4 cm (d) 3 cm

D–2 n Mathematics (Standard) – X

Sample Paper–1 n D–3

30. In the given figure, ∠ACB = ∠CDA, AC = 8 cm, AD = 3 cm, then BD is 1

D–4 n Mathematics (Standard) – X

Sample Paper–1 n D–5

SECTION-C (Case Study Based Questions)

41. What is the value of k? 48 1

D–6 n Mathematics (Standard) – X

46. The coordinates of the centroid of ΔEHJ are 1

Sample Paper–1 n D–7

Then,2 sin2 β – (1 – sin2 β) = 2

D–8 n Mathematics (Standard) – X

Sample Paper–1 n D–9

D–10 n Mathematics (Standard) – X

Comparing the co-ordinates, we get,

34. (b) ΔABE is a right triangle & FDGB is a square of side x cm

Sample Paper–1 n D–11

Shaded area = A  rea of semicircle + (Area of half square – Area of two

So, ΔOAB is an equilateral triangle and ∴ ∠AOB = 60°

D–12 n Mathematics (Standard) – X

Sample Paper–1 n D–13

D–14 n Mathematics (Standard) – X

Then ( 4 − 0 )2 + ( 3 − y )2 = (0 − 4)2 + ( y + 1)2

Sample Paper–1 n D–15

General Instructions: Same as Sample Paper-1

8. 429 as a product of its prime factors can be express as 1

Sample Paper–2 n D–17

D–18 n Mathematics (Standard) – X

SECTION-C (Case Study Based Questions)

Sample Paper–2 n D–19

House (i) House (ii)

D–20 n Mathematics (Standard) – X

Q46 – Q50 are based on Case Study-2

Sample Paper–2 n D–21

General Instructions: Same as Sample Paper-1

Sample Paper–3 n D–23

D–24 n Mathematics (Standard) – X

34. The zeroes of the quadratic polynomial x2 + 99x + 127 are 1

SECTION-C (Case Study Based Questions)

5m Bedroom 2 Living Room

Sample Paper–3 n D–25

Based on the above information, answer the following questions:

D–26 n Mathematics (Standard) – X

General Instructions: Same as Sample Paper-1

D–28 n Mathematics (Standard) – X

2. Sample Paper 2.............................................................................................................D–16

3. Sample Paper 3.............................................................................................................D–22

4. Sample Paper 4.............................................................................................................D–27

5. Sample Paper 5.............................................................................................................D–32

6. Sample Paper 6.............................................................................................................D–38

10. The distance of point A (–5, 6) from the origin is 1

Shaded area = A rea of semicircle + (Area of half square – Area of two

8. 429 as a product of its prime factors can be express as 1

34. The zeroes of the quadratic polynomial x2 + 99x + 127 are 1

43. The zeroes of the polynomial are 1 –4

44. What will be the expression of the polynomial? 1

8. 225 can be expressed as: 1

38. Sum of the zeroes of the polynomial x2 + 7x + 10 are 1

18. If 3 cos q = 1, then cosec q is equal to 1

The number of zeroes of f (x) is