CBSE Class 10 Sample Paper 2026 Solution for Maths Standard

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2026
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MATHEMATICS STANDARD – Code No.041
MARKING SCHEME
CLASS – X (2025-26)
Maximum Marks: 80 Time: 3 hours

Q.No. Section A Marks

1. (C) 3 1
LCM(𝑎, 𝑏, 𝑐) = 2 × 3 × 5 × 7 = 3780
140 × 3 = 3780
3 = 27 = 3
𝑥=3

2. (A) 2 1
As shortest distance from (2, 3) to y-axis is the 𝑥 coordinate, i.e., 2.

3. (B) k≠ 1

≠ , hence
k≠

4. (C) 6cm 1
AB+CD=AD+BC
AB+4=3+7
AB=6cm

5. (D) 1
( ) ( )
= = = secϴ-tanϴ
( )( )

6. (D) (𝑥 +2) ( 𝑥 +1) = 𝑥 2 +2 𝑥 +3, so, 𝑥 2 +3 𝑥 +2= 𝑥 2 +2 𝑥 +3 gives 𝑥 -1=0 1

It’s not a quadratic equation.

7. D) 8[ -
√
] cm2 1

Required Area=8 × area of one segment (with r = 1cm and 𝜃 = 60˚)
° √
=8x ( x 𝜋 x 12 − x 12)
°
√
= 8[ − ] cm2

Page 1 of 12

For Visually Impaired candidates:
(D) 9𝜋cm2
area of circle=𝜋(3 )
=9 𝜋 cm2

8. (B) 1

Probability of getting sum 8 is
Probability of not getting sum 8 is

9. (B) 12° 1
√
sin 5𝑥 =
So, 5𝑥 = 60°
And hence 𝑥 = 12°

10. (C) 4 1

Since HCF=81, the numbers can be 81𝑥 and 81𝑦
81𝑥 +81𝑦 =1215
𝑥 + 𝑦 =15
which gives four pairs as
(1,14), (2,13), (4,11), (7,8)

11. (D) 5cm 1

𝜋r2= 51
V= × 𝜋r2 × h
85 = × 51 × h
h= = 5𝑐𝑚

12. (D) 1

As for equal roots to the corresponding equation,
b2 =4ac
Hence ac =
And hence ac > 0⇒ c and a must have same signs

13. (C) 231 1

Area of sector
= ×𝑙×r
= × 22 × 21 = 231cm2

Page 2 of 12

14. (C) 18cm 1

∆𝐴𝐵𝐶 ~ ∆𝐷𝐸𝐹
∆
= = =
∆

∆
=

Perimeter of ∆ ABC= 18cm

15. (B) 1

Probability of getting vowels in the word Mathematics is ,
So, =
⇒𝑥=

16. (C) Parallelogram 1
By visualising the figure by plotting points in co-ordinate plane it can be concluded it
is a Parallelogram.

17. (A) median is increased by 2 1

18. (A) 40cm 1

Since, tangent is perpendicular to the radius at the point of contact
In ∆OPT, right angled at T
OP2=OT2+TP2
412=92+TP2
TP2= 1681-81=1600
TP=40cm

19. (A) Both assertion (A) and reason (R) are true and reason (R) is the correct 1
explanation of assertion (A)

20. (A) 1
cosA+cos2A=1 ---------(i)
gives cos A= sin2A------(ii) (using sin2A+ cos2A=1)
Substituting value of cos A from (ii) in (i)
sin2A +sin4A=1
⸫ Both assertion (A) and reason (R) are true and reason (R) is the correct explanation
of assertion (A)
Page 3 of 12

(Section – B)

21. n =60, a =8 and d=2 ½
(A) 𝑡 = 8 + 59(2) =126 ½
𝑡 = 108 1
Hence 𝑡 + 𝑡 +……..+ 𝑡 = (108 +126) =1170
OR
(B) 230 = 6 + (n -1)7 gives n=33 1
⸫ Middle Term = 𝑡 = 6 + (16)(7) = 118 1

22. A+B = 90o and A – B= 30o 1
A=60o and B =30o 1

23.

△ABC∼△DEF

⇒ = ½

= (AP and DQ are the medians) ½

=

In △ABP and △DEQ
=

∠B=∠E (△ABC∼△DEF)

⇒△ABP ~△DEQ ½

Hence, = ½

24.(A) area of grass field that can be grazed by them
= × 𝜋 𝑟2 + × 𝜋 𝑟2 + 𝑥 𝜋 𝑟2
° ° °

= (𝜃 + 𝜃 + 𝜃 )
°
= × 180˚ 1
°
= ×
=308 m2 1

Page 4 of 12

OR

(B) Area of minor segment= Area of sector − area of triangle
°
= 𝜋 r2 − × r2
°
=( 𝜋 − ) cm2
1
Area of major segment = Area of circle – Area of minor segment
= 𝜋 52 – ( 𝜋 − )
= 25𝜋 – 𝜋+
=( 𝜋+ ) cm2 1

25.

Let r be the radius of the inscribed circle

BD=BE=10cm
CD=CF=8cm
Let AF=AE= 𝑥 ½

ar(△ABC) =ar(△AOC) + ar(△BOC) + ar(△AOB)
½
= × r x AC + × r × BC + × r × AB
90 = × 4 (𝑥 +8+18+ 𝑥 +10)
𝑥 = 4.5cm
½
∴ AB=4.5+10=14.5cm
AC=4.5+8=12.5cm
½
For Visually Impaired candidates:

AC2=AB2+BC2= 242+72=625
AC=25cm ½
Area of ∆ABC= × 7× 24=84cm2 -----(i) ½
Let r=radius of circle
Also, Area of ∆ABC= (24r+25r+7r)
= × 56 r -------(ii) ½
From (i) and (ii), we get
r=3cm ½

Page 5 of 12

(Section – C)

26. In ∆APO and ∆ACO
AP=AC (Tangents from External Point)
AO=AO (common)
OP=OC (radii)
∆APO ≅ ∆ACO 1
∠POQ=180˚ (PQ is the diameter)
∠POA+∠COA+∠QOB+∠COB=180˚ 1
2∠COA+2∠COB=180˚
∠ AOB = 90° 1

For Visually Impaired candidates:

PA=PB (Tangents from external point to a circle)
∠PAB=∠PBA= 𝑥 (angles opposite to equal sides) ½
In ∆PAB, ∠PAB+∠PBA+∠APB=180°
𝑥 + 𝑥 +∠APB=180°
∠APB=180°−2 𝑥 ------(i)
1
Also,
∠PAB+∠OAB=90° (radius is perpendicular to the tangent at the point of contact)
𝑥 +∠OAB=90°
𝑥 =90°−∠OAB --------(ii)
Substituting (ii) in (i), we get 1
∠APB=180°-2(90°− ∠OAB)
∠APB=2∠OAB ½

27. HCF (36,60,84) =12 1½
Required number of rooms= + + 1
=3+5+7
=15 ½

28. 2 𝑥 2 − (1+2√2) 𝑥 + √2

= 2 𝑥 2 − 𝑥 − 2√2 𝑥 + √2 1

= (2 𝑥 − 1) ( 𝑥 − √2 ) Hence the zeroes are and √2 . 1

√ √
Now = = √2 + and = = × √2
1
Page 6 of 12

29. 𝑠𝑖𝑛𝛳 + 𝑐𝑜𝑠𝛳 = √3 gives (𝑠𝑖𝑛𝛳 + 𝑐𝑜𝑠𝛳) = 3. 1
Hence 1 + 2𝑠𝑖𝑛𝛳𝑐𝑜𝑠𝛳 = 3
So 2𝑠𝑖𝑛𝛳𝑐𝑜𝑠𝛳 = 2
⇒ 𝑠𝑖𝑛𝛳 𝑐𝑜𝑠𝛳 = 1 1

⸫ t𝑎𝑛 𝛳 + 𝑐𝑜𝑡𝛳 = =1 1

OR

( )( )
=
( )( ) 1

= 1

( )
= = = 𝑐𝑜𝑠𝑒𝑐𝐴 + 𝑐𝑜𝑡𝐴 1

30. P(Vidhi drives the car) = as favourable outcomes are HHT,THH,HHH 1

P(Unnati drives the car) = as favourable outcomes are THT,THH,HTH,TTH 1
As >
1
Unnati has greater probability to drive the car

31. Let the income of Aryan and Babban be 3𝑥 and 4𝑥 respectively 1
And let their expenditure be 5𝑦 and 7𝑦 respectively.
Since each saves ₹ 15,000, we get
3𝑥 – 5𝑦 = 15000
4𝑥 – 7𝑦 = 15000 1
Hence 𝑥 = 30000

Their income thus become ₹90,000 and ₹1,20,000 respectively. 1
OR

2 for
correct
Graph

Page 7 of 12

Hence, the solution is 𝑥 = 2, 𝑦 = 2 ½

Area= 2 sq. units ½

For Visually Impaired candidates

Let the present age of father be 𝑥 and son be 𝑦
So, (𝑥 + 5) = 3(𝑦 + 5) ⇒𝑥 − 3𝑦 = 10 1
𝑥 − 5 = 7(𝑦 − 5) ⇒ 𝑥 − 7𝑦 = − 30 1
1
So, 𝑥 = 40, 𝑦 = 10.
Hence the present ages of father and son are 40 years and 10 years
Respectively

Section D

32. Let the original speed of train be 𝑥 km/hr
Distance =63km, time(t1) = hrs 1
Faster speed = (𝑥 +6) km/hr
time (t2)= hrs
1
Now t1 + t2 = 3 hrs

So + =3
1

63(𝑥 + 6) + 72𝑥 = 3(𝑥 + 6)𝑥
135𝑥 +378=3𝑥 2 +18𝑥
3𝑥2 −117𝑥 − 378 = 0
𝑥 2 −39𝑥 − 126 = 0 1
𝑥 2 −42𝑥 +3𝑥 − 126 = 0 gives (𝑥 + 3)(𝑥 − 42) = 0
As 𝑥 can’t be negative, so 𝑥 = 42 km/hr 1

The original speed of train=42 km/hr

33. Correct given, figure and construction 2
Correct Proof 2
since LM is parallel to QR
Let PM= 𝑥
=
. ½
=
. .
𝑥 =PM=3.3cm ½

Page 8 of 12

34. (A)
Slant height of the cone 𝐿 = √𝑅 + 𝐻 = √12 + 6
= 3√20 𝑐𝑚 ½
Curved Surface area of cone= 𝜋𝑅𝐿 = 𝜋 × 12 × 3√20
= (36√20) 𝜋 𝑐𝑚 1
Area of base circle of cone (= area of outer circle –
area of inner circle + top circular area of cylinder)
= 𝜋𝑅 = 𝜋 × (12)
= 144𝜋 𝑐𝑚 1
Curved Surface area of cylinder= 2𝜋𝑟ℎ = 2𝜋 × 4 × 3
= 24 𝜋 𝑐𝑚 1

Surface area of the remaining solid= Curved surface of cone
+ area of base circle of cone 1
+ curved surface area of cylinder
= 36√20 𝜋 + 144𝜋 + 24𝜋
= 168 + 36√20 𝜋 𝑐𝑚 ½

OR

(B) Volume of cone= 𝜋r2h= 𝜋 × 3×3×12= 36𝜋cm3
Volume of ice-cream in the cone= × 36𝜋cm3 = 30𝜋 cm3 2

Volume of ice-cream in the hemispherical part= 𝜋r3= 𝜋 ×3×3×3=18𝜋 cm3 1+½
Total volume of the ice-cream = (30𝜋+18𝜋 ) =48 𝜋=150.86cm3 (approx.) 1+½

35. (A) Mode of the frequency distribution = 55
Modal class is 45-60. Lower limit is 45 Class Interval (h) =15 ½
Now, Mode = 𝑙 + ( ) ×h
55 = 45 + ×5 1
So, 𝑥 = 5 1

CI 𝒇𝒊 𝒙𝒊 𝒇𝒊 𝒙𝒊
0-15 10 7.5 75
15-30 7 22.5 157.5
30-45 5 37.5 187.5 1½
45-60 15 52.5 787.5
60-75 10 67.5 675
75-90 12 82.5 990
59 2872.5

.
Mean= 𝑥̅ = = 48.68 1

Page 9 of 12

OR
(B)
Height (in cm) Number Class Interval frequency
of girls
less than 140 04 135-140 4

less than 145 11 140-145 7

less than 150 29 145-150 18
1
less than 155 40 150-155 11

less than 160 46 155-160 6

less than 165 51 160-165 5

Median = l +( )𝑥h 1

=145 + ( )×5 1
=149.03
Median height = 149.03cm
1
3×Median= Mode +2× Mean 1
3×149.03=148.05+2× Mean
Mean=149.52

Section E

36.
(i) Common difference of first progression= 3

Common difference of first progression= −3
Sum of common difference=0. 1

(ii) t34 = 187 +(34-1) (−3)
So, t34 =88 1

(iii) (A) Sum = [2(−5) + (10 − 1)(3)] 1
= 85 1
OR

(B) −5 +(n−1)3 = 187 +(n−1) (−3)
n = 33 1
1

Page 10 of 12

37. (i)
PR= (8 − 2) + (3 − 5) = 2√10 1

(ii) Co-ordinates of Q (4,4). ½
The mid-point of PR is (5,4) ½
⸫Q is not the mid-point of PR

(iii) (A) Let the point be (𝑥,0)

So, (2 − 𝑥) + 25 = (4 − 𝑥) + 16
1
Hence 𝑥 = .Therefore the point is ( ,0). 1
OR
(B) The coordinates of S will be

× ×
,
× × 1

= , 1

38. (i) Distance from India gate = 41m,
Height of monument = 42m,
Shreya’s height =1m
So, tan 𝜃= =1 ½
½
Angle of elevation = 𝜃 = 45o.

(ii) Angle of elevation =60o
Perpendicular = 41m
Let the distance from the India Gate be 𝑥 m
Hence tan 60o= ½
⟹𝑥= ½
√
√
∴ Shreya is standing at a distance of m

Page 11 of 12

(iii) (A)
Distance from the India Gate = 41 m
Let the distance moved back be 𝑥 m
Then, tan30o = 1
𝑥 = (41√3 −41) m = 41(√3-1) m 1
∴ The distance moved back = 41(√3-1) m

OR

(B) Let the angle of elevation of be 𝜃
Now, tan 𝜃 = = √3 1
√
This gives 𝛳 =60o 1

Page 12 of 12

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