SAMPLE PAPER TEST 01 FOR BOARD EXAM 2025

(ANSWERS)
SUBJECT: MATHEMATICS MAX. MARKS : 80
CLASS : X DURATION : 3 HRS
General Instruction:
1. This Question Paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each.
3. Section B has 5 questions carrying 02 marks each.
4. Section C has 6 questions carrying 03 marks each.
5. Section D has 4 questions carrying 05 marks each.
6. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the
values of 1, 1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and
2 Questions of 2 marks has been provided. An internal choice has been provided in the 2marks
questions of Section E
8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated.
SECTION – A
Questions 1 to 20 carry 1 mark each.

1. Two circles touch each other externally at C and AB is common tangent of circles, then ∠ACB is
(a) 70° (b) 60° (c) 100° (d) 90°
Ans: (d) 90°
Draw CM perpendicular to AB.

Now, AM = MC and MB = MC (tangents drawn from external point are equal).

⇒ AM = MC
⇒ ∠MAC = ∠MCA = 45°
(Since Δ AMC is right triangle)
∴ Also, MB = MC ⇒ ∠MBC = ∠MCB = 45° (Since Δ MBC is right angle triangle)
∴ ∠ACB = ∠MCA + ∠MCB = 45° + 45° = 90° ⇒ ∠ACB = 90°
2. The pair of linear equations 2x + 3y = 5 and 4x + 6y = 10 is
(a) inconsistent (b) consistent (c) dependent consistent (d) none of these
Ans: (c) dependent consistent
3. Nature of roots of quadratic equation 2x2 – 4x + 3 = 0 is
(a) real (b) equal (c) not real (d) none of them
Ans: (c) not real
D = b2 – 4ac = 42 – 4 × 2 × 3 = 16 – 24 = –8 < 0
Since D < 0
Hence, roots are not real.
4. If ∆ABC ~ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true?
(a) BC. EF = AC. FD (b) AB. EF = AC. DE
(c) BC. DE = AB. EF (d) BC. DE = AB. FD
Ans: (c) BC. DE = AB. EF
Since, ∆ABC ~ ∆EDF
Therefore, the ratio of their corresponding sides is proportional.
BC AB
   BC.DE  AB.EF
EF DE

5. If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle > Area of the square
(c) Area of the circle < Area of the square
(d) Nothing definite can be said about the relation between the areas of the circle and square.
Ans: (b) Area of the circle > Area of the square

6. The sum of the lower limit of median class and the upper limit of the modal class of the following
data is:
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students 8 10 12 22 30 18
(a) 70 (b) 80 (c) 90 (d) 100
Ans: (b) 80
7. The radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is
(a) 2.1 cm (b) 4.2 cm (c) 3.1 cm (d) 2.2 cm
Ans: (a) 2.1 cm
The diameter of the largest right circular cone that can be cut out from a cube of edge 4.2 cm.
∴ 2r = 4.2 cm ⇒ r = 2.1 cm
⇒ Radius of the largest right circular cone = 2.1 cm.

8. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of
hemisphere?
(a) 9 units (b) 6 units (c) 4.5 units (d) 18 units
Ans: (a) 9 units
Volume of hemisphere = Surface area of hemisphere
2 9
  r 3  3 r 2  r  units
3 2
∴ d = 9 units

9. In the ∆ABC, DE ∥ BC and AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3, then find the

value of x
(a) 1 (b) 7/10 (c) both (a) & (b) (d) none of these
Ans: (c) both (a) & (b)
 Given that, AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3
AD AE
By Basic Proportionality theorem, we have 
BD EC
3x  2 5 x  4
   (3 x  2)(5 x  3)  (5 x  4)(7 x  5)
7 x  5 5x  3
 15 x 2  19 x  6  35 x 2  53 x  20
 20 x 2  34 x  14  0  10 x 2  17 x  7  0
7
 ( x  1)(10 x  7)  0  x  1, x 
10

10. A card is selected at random from a well shuffled deck of 52 cards. The probability of its being a
face card is
(a) 3/26 (b) 3/13 (c) 2/13 (d) 1/2
Ans: (b) 3/13
7
11. In ABC right angled at B, sin A = , then the value of cos C is ………….
25
7 24 7 24
(a) (b) (c) (d)
25 25 24 7
7
Ans: (a)
25
5sin   3cos 
12. If 5 tan θ = 4, then the value of is
5sin   2 cos 
(a) 1/6 (b) 1/7 (c) 1/4 (d) 1/5
Ans: (a) 1/6

13. Given that sin α = 1/2 and cos β = 1/2, then the value of (β – α) is
(a) 0° (b) 30° (c) 60° (d) 90°
Ans: (b) 30°

14. Two identical solid hemispheres of equal base radius are stuck along their bases. The total surface
area of the combination is
(a) πr2 (b) 2πr2 (c) 3πr2 (d) 4πr2
2
Ans: (d) 4πr
The resultant solid will be a sphere of radius r whose total surface area is 4πr2.

15. If two positive integers p and q can be expressed as p = ab2 and q = a3b; a, b being prime numbers,
then LCM (p, q) is
(a) ab (b) a2b2 (c) a3b2 (d) a3b3
3 2
Ans: (c) a b

16. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5 units (b) 12 units (c) 11 units (d) (7 + √5) units
Ans: (b) 12 units

17. The zeroes of the polynomial x2 – 3x – m(m + 3) are

(a) m, m + 3 (b) –m, m + 3 (c) m, – (m + 3) (d) –m, – (m + 3)
Ans: (b) –m, m + 3
Let p(x) = x2 – 3x – m (m + 3)
⇒ p(x) = x2 – (m + 3) x +mx – m (m + 3)
= x{x – (m + 3)} + m {x – (m + 3)}
For zeros of p(x)
 ⇒ p(x) = (x + m) {(x – (m + 3)} = 0 ⇒ x = – m, m + 3
∴ Its zeros are – m, m + 3.

18. The area of a quadrant of a circle, whose circumference is 22 cm, is

11 77 77 77
(a) cm2 (b) cm2 (c) cm2 (d) cm2
8 8 2 4
77
Ans: (b) cm2
8
Direction : In the question number 19 & 20 , A statement of Assertion (A) is followed by a
statement of Reason(R) . Choose the correct option

19. Assertion (A): 6n never ends with the digit zero, where n is natural number.
Reason (R): Any number ends with digit zero, if its prime factor is of the form 2m × 5n, where m, n
are natural numbers.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of
Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of
Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Ans: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of
Assertion (A).
6n = (2 × 3)n = 2n × 3n, Its prime factors do not contain 5 i.e., of the form 2m × 5n,
where m, n are natural numbers. Hence, 6n never ends with the digit zero.

20. Assertion (A): The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.
Reason (R): Distance between two points is given by ( x2  x1 )2  ( y2  y1 )2
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of
Assertion (A)
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of
Assertion (A)
(c) Assertion (A) is true but reason(R) is false.
(d) Assertion (A) is false but reason(R) is true.
Ans: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of
Assertion (A)

SECTION-B
Questions 21 to 25 carry 2M each
21. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle
which touches the smaller circle.
Ans: Let O be the centre of the concentric circle of radii 5 cm and 3 cm respectively. Let AB be a
chord of the larger circle touching the smaller circle at P
 Then AP = PB and OP⊥AB
Applying Pythagoras theorem in △OPA, we have
OA2 = OP2 + AP2 ⇒ 25 = 9 + AP2
⇒ AP2 = 16 ⇒ AP = 4 cm
∴ AB = 2AP = 8 cm

1
22. If sin (A + B) = √3/2 and sin (A – B) = , 0 ≤ A + B ≤ 90° and A > B, then find A and B.
2
Ans: sin(A + B) = √3/2 = sin 60⁰
⇒ A + B = 60⁰ ........(i)
sin (A - B) = 1/2 = sin 30⁰
⇒ A - B = 30⁰ ........(ii)
Solving eq. (i) and (ii), A = 45⁰ and B = 15⁰
OR
(1  sin  )(1  sin  )
If tan θ =3/4, evaluate
(1  cos  )(1  cos  )
3 4
Ans: tan    cot  
4 3
2
(1  sin  )(1  sin  ) 1  sin 2  cos 2  2  4  16
   cot     
(1  cos  )(1  cos  ) 1  cos2  sin 2  3 9

23. Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the
corresponding major sector. (Use π = 3.14)
 r 2 3.14  42  300
Ans: Area of sec tor AOB  0
 0
 4.19cm 2
360 360
Area of major sector = Area of circle – Area of sector AOB
= r2 – 4.19 = 3.14 × 16 – 4.19 = 46.1 cm2
OR
What is the angle subtended at the centre of a circle of radius 10 cm by an arc of length 5π cm?

Ans: Arc length of a circle of radius r   2 r
3600
  5 1
 5  0
 2  10  0
 
360 360 20 4
0
   90

24. For what values of k will the following pair of linear equations have infinitely many solutions? kx
+ 3y – (k – 3) = 0 and 12x + ky – k = 0
Ans: Comparing with a1 x  b1 y  c1 and a2 x  b2 y  c2
a1  k , a2  12, b1  3, b2  k , c1  k  3, c2  k
a b c
For infinite solutions, 1  1  1
a2 b2 c2
k 3 k 3 k 3
      k 2  36  k  6
12 k k 12 k

25. In the given figure, AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm, AC = 10 cm. Find the length
of AD
 AP 3
Ans: In ∆ABC,  ……. (i)
AB 5
AQ 6 3
  ……. (ii)
AC 10 5
AP AQ

From (i) and (ii), we get ⇒ PQ || BC
AB AC
AP AR 3 4.5
In ∆ABD, PR || BD ⇒     AD  7.5cm
AB AD 5 AD

SECTION-C
Questions 26 to 31 carry 3 marks each

26. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth
is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area
increases by 67 square units. Find the dimensions of the rectangle.
Ans: Let length and breadth be x and y , Area = xy
1st condition: ( x – 5 ) ( y + 3 ) = xy – 9
⇒ 3x – 5y = 6
2nd condition: ( x + 3 ) ( y + 2 ) = xy + 67
⇒ 2x + 3y = 61
Solve 1st and 2nd equations, we get x = 17 and y = 9
Hence, Length of rectangle = 17 units and breadth of rectangle = 9 units

sin   cos   1
27. Prove that  sec   tan 
sin   cos   1
tan   1  sec 
Ans: LHS = tan   1  sec  (Dividing numerator and denominator by cos )

tan   sec   1

tan   1  sec 
tan   sec   (sec 2   tan 2  )

tan   1  sec 
(sec   tan  )(1  sec   tan  )

tan   1  sec 
 sec   tan  = RHS

28. Two dice are thrown at the same time. What is the probability that the sum of the two numbers
appearing on the top of the dice is (i) 5? (ii) 10? (iii) at least 9?
Ans: Total number of outcomes = 36
(i) Number of outcomes in which the sum of the two numbers is 5 = 4
∴ Required Probability = 4/36 = 1/9
(ii) Number of outcomes in which the sum of the two numbers is 10 = 3
 ∴ Required Probability = 3/36 = 1/12
(i) Number of outcomes in which the sum of the two numbers is at least 9 = 10
∴ Required Probability = 10/36 = 5/18

29. In the below figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another
tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

Ans: Join OC. Since, the tangents drawn to a circle from an external point are equal.
∴ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC
⇒ Δ PAO ≅ Δ AOC [SSS Congruency]
∴ ∠PAO = ∠CAO = ∠1
∠PAC = 2 ∠1 ...(1)
Similarly ∠CBQ = 2 ∠2 ...(2)
Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
⇒ 2 ∠1 + 2 ∠2 = 180° [From (1) and (2)]
⇒ ∠1 + ∠2 = 180°/2 = 90° ...(3)
Also ∠1 + ∠2 + ∠AOB = 180° [Sum of angles of a triangle]
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 180° − 90° ⇒ ∠AOB = 90°.
OR
In the below figure, two equal circles, with centres O and O', touch each other at X. OO' produced
meets the circle with centre O' at A. AC is tangent to the circle with centre O, at the point C. O'D is
DO '
perpendicular to AC. Find the value of .
CO
 Ans: AC is tangent to circle with centre O.
Thus ∠ACO = 90°
In ∆AO'D and ∆AOC, ∠ADO' = ∠ACO = 90°
∠A = ∠A (Common)
∴ ∆AO'D ~ ∆AOC (By AA similarity)
AO ' DO '
⇒ 
AO CO
Now, AO = AO' + O' X + XO = 3r
DO ' r 1
  
CO 3r 3

30. Four bells toll at an interval of 8, 12, 15 and 18 seconds respectively. All the four begin to toll
together. Find the number of times they toll together in one hour excluding the one at the start.
Ans: Prime factorisation of the given numbers are:
8 = 2 × 2 × 2 = 23
12 = 2 × 2 × 3 = 22 × 31
15 = 3 × 5 = 31 × 51
18 = 2 × 3 × 3 = 21 × 32
LCM (8, 12, 15 and 18) = 23 × 32 × 51 = 8 × 9 × 5 = 360 sec = 6 min
∴ Four bells toll together in one hour = 60 ÷ 6 = 10 times.

31. Find the zeroes of the quadratic polynomial 6x2– 3 – 7x and verify the relationship between the
zeroes and the coefficients of the polynomial.
Ans: 6x2 – 7x – 3 = 0
6x2 – 9x + 2x – 3 = 0
3x(2x – 3) + 1(2x – 3) = 0
 (3x + 1) (2x – 3) = 0
1 3
x = ,
3 2
1 3 2  9 7 b 7 b
Now,        and     
3 2 6 6 a 6 a
1 3 1 c 1 c
    and    
3 2 2 a 2 a

OR
Find the quadratic polynomial sum and product of whose zeros are –1 and –20 respectively. Also
find the zeroes of the polynomial so obtained.
Ans: Let α and β be the zeros of the quadratic polynomial.
∴ Sum of zeros, α + β = – 1
and product of zeros, α. β = – 20
Now, quadratic polynomial be
x2 – (α + β) . x + α β = x2 –(–1) x – 20 = x2 + x – 20
Now, for zeroes of this polynomial
 x2 + x – 20 = 0 ⇒ x2 + 5x – 4x – 20 = 0
⇒ x(x + 5) – 4 (x + 5) = 0 ⇒ (x + 5) (x – 4) = 0
⇒ x = – 5, 4
∴ zeroes are – 5 and 4

SECTION-D
Questions 32 to 35 carry 5M each

32. If the median of the following distribution is 58 and sum of all the frequencies is 140. What is the
value of x and y?
Class 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65 65 – 75 75 – 85 85 – 95
Frequency 8 10 x 25 40 y 15 7
Ans:

33. A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as
that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the
volume of the hemisphere, calculate the height of the cone and the surface area of the toy.
Ans: We have, Radius of cone = Radius of hemisphere = 21 cm ⇒ r = 21 cm

2
According to question, Volume of cone = × volume of hemisphere
3
1 2 2 4 4
  r 2 h    r 3  h  r   21  28 cm
3 3 3 3 3
Slant height, l  r 2  h 2  212  282  35 cm
Total surface area = CSAcone  CSAhemisphere   rl  2 r 2   r (l  2r)
22
  21 (42  35)  22  3  77  5082 cm 2
7
 OR
A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top,
which is open, is 5 cm. 100 spherical lead balls are dropped into the vessel. One fourth of the water
flows out of the vessel. Find the radius of a spherical ball.
Ans: Height (h) of the cone = 8 cm and radius (r) of the cone = 5 cm
1
∴ Volume of water flows out = × volume of cone
4
1 1 2 1
   r h     25  8
4 3 12
∴ Volume of water flows out =100× volume of spherical ball
1 4
    25  8  100   R3
12 3
1 1
 R3   R  cm  0.5cm
8 2

34. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to
return downstream to the same spot. Find the speed of the stream.
Ans: Let the speed of the stream be x km/h.
Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat downstream =
(18 + x) km/h.
24
The time taken to go upstream = distance/speed =
18  x
24
Similarly, the time taken to go downstream =
18  x
24 24
According to the question,  1
18  x 18  x
24(18 + x) – 24(18 – x) = (18 – x) (18 + x) x2 + 48x – 324 = 0 x = 6 or – 54
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = – 54. Therefore,
x = 6 gives the speed of the stream as 6 km/h.

OR
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and
Bangalore (without taking into consideration the time they stop at intermediate stations). If the
average speed of the express train is 11km/h more than that of the passenger train, find the average
speed of the two trains.
Ans: Let the average speed of the passenger train = x km/hour
And the average speed of the express train = (x + 11) km/hour.
132
The time taken by the passenger train = hour
x
132
and the time taken by the express train = hour
x  11
132 132
According to the question,  1
x  11 x
⇒ x2 + 11x −1452 = 0
⇒ (x − 33)(x + 44) = 0 ⇒ x = 33, x = − 44
The speed cannot be negative, so the speed of the passenger train is 33 km/hour and the speed of the express
train is 33 + 11 = 44 km/ hour.
35. State and prove Basic Proportional Theorem.
Ans: Statement – 1 mark
Given, To Prove, Construction and Figure – 2 marks
Correct Proof – 2 marks
 SECTION-E (Case Study Based Questions)
Questions 36 to 38 carry 4M each

36. Ananya saves Rs. 24 during the first month Rs. 30 in the second month and Rs. 36 in the third
month. She continues to save in this manner.

On the basis of above information answer the following questions.

(i) Whether the monthly savings of Ananya form an AP or not? If yes then write the first term and
common difference.
(ii) What is the amount that she will save in 15th month?
(iii) In which month, will she save Rs. 66?
OR
What is the common difference of an AP whose nth term is 8 – 5n?
Ans: (i) Savings of Ananya are Rs. 24, Rs. 30, Rs. 36, ...
Since it is uniformly increasing by Rs. 6, therefore it forms an AP.
Here, a = 24, d = 30 – 24 = 6
(ii) a15 = a + 14d = 24 +14 × 6 = 24 + 84 = Rs. 108
(iii) an = 66 ⇒ a + (n – 1)d = 66
⇒ 24 + (n – 1)6 = 66 ⇒ n – 1 = 42/6 = 7 ⇒ n = 8
OR
an = 8 – 5n
a1 = 8 – 5 = 3
a2 = 8 – 10 = –2 ⇒ d = a2 – a1 = –2 – 3 = – 5

37. A person/observer on the sea coast observes two ships in the sea, both the ships are in same straight
path one behind the other.
If the observer is on his building of height 20 meters (including observer) and he observes the angle
of depression of two ships as 45° and 60° respectively.

On the basis of above information answer the following questions.

 (i) If a person observes a ship whose angle of depression is 60° then how much distance is the ship
away from the building?
(ii) If a person observes another ship whose angle of depression is 45° then how much distance that
ship is away from the building?
(iii) If a person observes the ship whose angle of depression changes from 60° to 30° then how far
be ship from the building if the observer is at 20 m of height (including him)?
OR
At a time when a person observes two ships whose angle of depressions are 60° and 45° the
distance between the ships is (in meter).
OC 20 20 20 3
Ans: (i) tan 600   3  AC   m  11.55m
AC AC 3 3

OC 20
(ii) tan 450  1  BC  20m
BC BC

OB 1 20
(iii) tan 300     OA  20 3m
OA 3 OA

OR
Distance between two ships 20 m = BC – AC = 20 – 11.55 = 8.45 m

38. The top of a table is shown in the figure given below:

On the basis of above information answer the following questions.
(i) Find the distance between points A and B.
(ii) Write the co-ordinates of the mid point of line segment joining points M and Q.
(iii) If G is taken as the origin, and x, y axis put along GF and GB, then find the point denoted by
coordinates (4, 2) and (8, 4).
OR
Find the coordinates of H, G and also find the distance between them.
Ans: (i) Distance between A(1, 9) and B(5, 13) is
 (5  1)2  (13  9)2  16  16  32  4 2units
(ii) Midpoint of the line segment joining M(5, 11) and Q(9, 3) is given by
 5  9 11  3   14 14 
 ,    ,   (7, 7)
 2 2  2 2
(iii) If G is (0, 0) then Q is (4, 2) and E is (8, 4).
OR
As per graph the coordinate of H is (1, 5) and of G is (5, 1).
Distance HG  (5  1)2  (1  5)2  16  16  32  4 2units