CBSE Sample Papers for Class 10 Maths Basic Set 1 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Basic Set 1 with Solutions

Time : 3 hrs
Max. Marks : 80

General Instructions:

• This Question Paper has 5 Sections A, B, C, D and E.
• Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E has 3 Case Based integrated units of assessment questions (4 marks each).
• All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks question of Section E.
• Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION-A
(This section comprises of multiple choice questions (MCQs) ofl mark each)

Question 1.
The exponent of 3 in the prime factorisation of 2025 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) We have, the number 2025.
By prime factorisation, 2025 = 3⁴ x 5²

So, the exponent of 3 in the prime factorisation of 2025 is 4.

Question 2.
If 2024x + 2025y = 1; 12025x + 2024y = -1 then x – y is equal to
(a) 0
(b) -2
(c) 2
(d) -1
Answer:
(b) Given, equation are
2024x + 2025y = 1  ……………….(i)
and 2025x + 2024y = – 1   ……………. (ii)
On subtracting Eq. (i) from Eq. (ii), we get
x-y = -1-1 ⇒ x-y = -2

Question 3.
The number of polynomials having -2 and 5 as its zeroes is
(a) one
(b) two
(c) three
(d) Infinitely many
Answer:
(d) We have, the zeroes of two polynomial are -2 and 5.
Therefore, the polynomial is given by
P(x) = K(x + 2)(x -5)
⇒ P(x) = k(x² -3x- 10), k≠ 0
Since, k can be any real number.
So, there are infinitely many such polynomial.

Question 4.
Which of the following is not a quadratic equation?
(a) (x + 2)² = 2 (x + 3)
(b) x² + 3x= (-1)(1 – 3x²)
(c) (x + 2)(x – 1) = x² – 2x – 3
(d) x³ – x² + 2x + 1 = (x + 1)³
Answer:
(c) By option (a), we have
(x + 2)² = 2{x + 3)
⇒ x² + 4+4x = 2x + 6 ⇒ x² + 2x-2 = 0,
which is quadratic equation.
By option (b), we have,
x² + 3x =(-1)(1-3x²)
⇒ x² + 3x = -1 + 3x²
⇒ 2x² – 3x – 1 = 0,
which is a quadratic equation.
By option (c), we have
(x + 2)(x – 1) = x² – 2x – 3
⇒ x²-x + 2x-2 = x²-2x-3
⇒ 3x + 1 = 0, which is not a quadratic equation.
By option (d), we have,
x³ -x² + 2x + 1 = (x + 1)³
⇒  x³-x² + 2x + 1 = x³ + 1+ 3x(x + 1)
⇒ x³-x² + 2x+1=x³ + 1+ 3x² + 3x
⇒ 4x² + x = 0,
which is also a quadratic equation,
Hence, option (c) is correct.

Question 5.
The value of x for which 2x, (x +10) and (3x + 2)are the three consecutive terms of an AP is
(a) 6
(b) -6
(c) -2
(d) 2
Answer:
(a) Given that, 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP.
∴ 2(x + 10) = 2x + (3x + 2)
[∵ If a, b and c are in AP then, 2b = a + c]
⇒ 2x + 20 = 2x + 3x + 2
⇒ 3x = 18
⇒ x = 6

Question 6.
If 1+ 2+ 3 + 4 + … + 50 = 25A: then k is equal to
(a) 50
(b) 51
(c) 49
(d) 26
Answer:
(b) Given, 1+2+3+4+. + 50 = 25k …(i)
We know that
Sum of n natural numbers = \(\frac{n(n+1)}{2}\)
Therefore, form Eq. (i), we get
\(\frac{50 \times(50+1)}{2}=25 k\)
⇒ 25 x 51 = 25k ⇒ k = 51

Question 7.
The distance between the points (cos30°, sin30°) and (cos60°, – sin60°)is
(a) 0 unit
(b) √3 units
(c) 1 unit
(d) -√2 units
Answer:

Question 8.
The coordinates of the point, which is the mirror image of the point (-3, 5) about X-axis are
(a) (3,5)
(b) (3, -5)
(c) (-3,-5)
(d) (-3,5)
Answer:
(c) Given, point is (-3, 5).
We know that for the coordinates of a mirror image of a point in X-axis, abscissa remains the same and ordinate will be of opposite sign of the ordinate of the given point.
So, the mirror image of the point (-3, 5) about X-axis is (-3,-5).

Question 9.
If in ΔABC and ΔDEF, \(\frac{AB}{EF}=\frac{AC}{DE}\) then they will be similar when
(a) ∠A = ∠D
(b) ∠A = ∠E
(c) ∠C = ∠F
(d) ∠B = ∠E
Answer:

Question 10.
If ΔABC ~ ΔPQR then perimeter of the triangle PQR (in cm) is

(a) 12
(b) 24
(c) 18
(d) 20
Answer:

Question 11.
In the figure given below, radius r of the circle which touches the sides of the triangle is

Answer:
(a) From the given figure, we have,
AB = 25 cm, AC = 24 cm and BC = 7 cm
Now, AE = 24 – r = AF
[tangents from an external point to the circle are equal in length]
⇒ AF = 24 – r …(i)
and BF = AB – AF
= 25 – (24 – r) [using Eq. (i)]
⇒ BF = 1+r     …(ii)
But from figure, BF= BD
[tangents from an external point to the circle are equal in length]
⇒  BF=BC-DC
BF = 7 – r  …………. (iii) [∵ BC = 7 cm and DC = r]
From Eqs. (ii) and (iii), we get
1+ r = 7-r
⇒  2r = 6
⇒ r = 3 cm

Question 12.
Which one of the following is not equal to Unity?
(a) sin² x + cos² x
(b) cot² x – cosec² x
(c) sec² x – tan² x
(d) tan x – cot x
Answer:

Question 13.
Consider the following frequency distribution.

Class	0-5	5-10	10-15	15-20	20-25
Frequency	11	12	13	9	11

The upper limit of median class is
(a) 10
(b) 13
(c) 15
(d) 20
Answer:

Since, the cumulative frequency just greater than 28 is 36 and the corresponding class interval is 10-15. So, its upper limit is 15.

Question 14.
Let empirical relationship between the three measures of central tendency be a (Median) = Mode + b (Mean), then (2b + 3a) is equal to
(a) 11
(b) 12
(c) 13
(d) 14
Answer:
(c) We have, a (median) =mode + b (mean)
Since, 3 median = mode + 2 mean
So, a = 3 and b = 2
Therefore, 2b+3a = 2×2+3×3=4+9=13

Question 15.
From an external point Q the length of tangent to a circle is 12 cm and the distance of Q from the centre of circle is 13 cm. The radius of circle (in cm) is
(a) 10
(b) 5
(c) 12
(d) 7
Answer:
(b) We have,

Question 16.
In the given figure, P A is a tangent from an external point P to a circle with centre O and diameter AB. If ∠POB=115° then measure of  ∠APO is

Answer:
(a) In the given figure, we have ∠POB = 115°
Also, ∠PAO = 90°
[angle between tangent and radius at the point of contact]
and ∠POB + ∠POA = 180°        [linear pair]
⇒ ∠POA = 180° – 115° [∵ ∠POB = 115°]
⇒ ∠POA = 65°  …(ii)
Now, in APAO,
∠APO+ ∠POA + ∠PAO = lSCP
[by angle sum property of a triangle]
∠APO + 65° + 90° = 180° [using Eq. (i) and (ii)]
⇒∠APO = 180° – 90° – 65° = 25°

Question 17.
The circumferences of two circles are in the ratio 3: 4. The ratio of their areas is
(a) 3 : 4
(b) 4 : 3
(c) 9 :16
(d) 16 : 9
Answer:
(c) Let the radius of two circle be r₁ and r₂, respectively.
∵ Circumference of circle with radius r is given by 27t r.
∴ According to given condition,

[∵ area of circle = πr²] = 9:16

Question 18.
An event is most unlikely to happen. Its probability is
(a) 0001
(b) 0.001
(c) 0.01
(d) 0.1
Answer:
(a) Since, the event is most unlikely to happen. Therefore, its probability is 0001.

ASSERTION-REASON BASED QUESTIONS
(For question number 13 to 16, two statements are given-one labelled Assertion (A) and other labelled Reason (R). Select the correct answer of these questions from the codes (a), (b), (c) and (d) as given below.)

• Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
• Both Assertion (A) and Reason(R) are true but Reason (R) is not the correct explanation of Assertion (A).
• Assertion (A) is true but Reason (R) is false.
• Assertion (A) is false but Reason (R) is true.

Question 19.
Assertion (A) Line joining the midpoints of two sides of triangle is parallel to the third side.
Reason (R) If a line divides two sides of a triangle in the same ratio then it is parallel to the third side.
Answer:
(a) Clearly, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 20.
Assertion (A) Two coins are tossed simultaneously. Possible outcomes are two heads, one head and one tail, two tails. Hence, the probability of getting two heads is
Reason (R) Probabilities of ‘equally likely’ outcomes of an experiment are always equal.
Answer:
(d) When two coins are tossed, the possible outcomes are HH, HT, TH, TT.
So, total outcomes = 4
Favourable outcomes for getting two head = HH = 1
∴ Required probability = \(\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\)
= \(\frac{1}{4}\)
∴ Assertion is false.
Clearly, Reason is true.

SECTION-B
(This section comprises of 5 very short answer (VSA) type questions of 2 marks each)

Question 21.
(a) Show that the number 2 x 5 x 7 x 11 +11 x 13 is a composite number.
Or
Find the smallest number, which is divisible by both 306 and 657.
Answer:
(a) We have, 2 x 5 x 7 x 11÷ 11 x 13
It can be observed that,
2 x 5 x 7 x 11+11 x 13 = 11x(70+13)
= 11 x 83
Which is the product of two factors other than 1.
Therefore, it is a composite number. Hence Proved.
Or
(b) Given two numbers are 306 and 657.
We know that the smallest number, which is divisible by any two numbers is their LCM.
So, number which is divisible by both 306 and 657 = LCM (306,657)
Since, 306 = 2¹ x 3² x 17¹ and 657 = 3² x 73
∴ LCM (306, 657) = 2¹ x 3² x 17¹ x 73 = 22338

Question 22.
Find the radius of the circle with centre at origin, if line  given by x + y = 5 is tangent to the circle at

Answer:
Given, the equation of line l is
x+ 3y = 5 ….. (i)

Since, point P(3, a) lies on the line l.
Therefore, 3 + a = 5 [putting x = 3 and y = a in Eq. (i)]
⇒ a = 5 – 3 = 2
Now, the radius of the circle,
CP = Distance between point C(0,0) and P( 3, 2).
∴ CP = \(\sqrt{(3-0)^2+(2-0)^2}\)
[using distance formula]
= \(\sqrt{9+4}=\sqrt{13}\) units

Question 23.
If the zeroes of the quadratic polynomial x² +(a + 1)x + b are 2 and -3 then find the values of a and b.
Answer:
Given, quadratic polynomial is
x² + (a + l)x + b and its zeroes are 2 and -3.
∵ Sum of the zeroes = 2 + (-3) = a + 1
⇒ a+1=1
⇒ a = 0 and product of zeroes = 2 x (-3) = b
∴ b = – 6
Hence, a = 0 and b = – 6

Question 24.
Find the nature of roots of the quadratic equation x² + 4x – 3√2 = 0.
Answer:
Given, quadratic equation is
x² + 4x – 3√2 = 0    …(i)
On comparing Eq. (i)
with standard quadratic equation ax² + bx + c = 0,
we get a = 1, b = 4 and c = – 3-√2
Discriminant D = b² – Aac
= (4)² – 4 x 1 x (-3√2)
= 16 + 12√2 > 0
As, discriminant is positive.
So, roots are real and distinct.

Question 25.
(a) Evaluate 2 sin30° tan60° – 3 cos² 60° sec² 30°.
Or
(b) If sin \(x=\frac{7}{25}\) where x is an acute angle then find the value of sin x. cos x (tan x + cot x).
Answer:
(a) We have, 2sin 30° tan 60° – 3cos² 60° sec² 30°

SECTION-C
This section comprises of SA type questions of 3 marks each.

Question 26.
Show that, √2 – √5 is an irrational number.
Answer:

Question 27.
(a) The frequency distribution table of agriculture holdings in a village is given below.

Area of land (in hectares)	1-3	3-5	5-7	7-9	9-11	11-13
Number of families	20	45	80	55	40	12

Find the modal agriculture holdings of the village.
Or
(b) If the mean of the following distribution is 54, find the value of p.

Class interval	0-20	20-40	40-60	60-80	80-100
Frequency	7	p	10	9

Answer:
(a) The frequency distribution table is given below


Hence, modal agriculture holding of the village is 6.17 hectare (approx).
Or
(b) Distribution table for the given data is

Question 28.
A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the given figure.

Show that \(\frac{A B+C D}{A D+B C}=1\)
Answer:
As quadrilateral ABCD is drawn to circumscribe a circle.

We know that tangents drawn to a circle from an external point are equal in length.
Therefore, AP = AS
PB= BQ and CR = CQ DR=DS
On adding the above equations, we get (AP+ PB) + (CR + RD)
= (AS + BQ) + (CQ+ DS)
⇒ AB+CD=AD+BC
[∵ AP+ PB= AB, CR+ RD= CD,
AS+ DS = AD and BQ+ CQ= BC]
⇒ \(\frac{A B+C D}{A D+B C}=1\)
Hence proved.

Question 29.
(a) On a particular day, 50000 people attended a Cricket Test Match between India and Australia in Sydney Cricket Ground. Let x be the number of adults attended the cricket match and y be the number of children attended the cricket match. Cost of an adult ticket was ₹1000 while cost of a child ticket was ₹200. On that day revenue earned by selling all 50000 tickets, was ₹ 42000000. Find how many adults and how many children attended the cricket match?
Or
(b) Solve for x and y, graphically 2x + y = 6, x + y = 5
Answer:
(a) Let the required number of adult and children be x and y, respectively.
Then, given that,
Total people, x + y = 50000 … (i)
Adult ticket =₹ 1000 ,
Child ticket =₹ 200
Total revenue =₹ 42000000
According to the question,
1000x + 200y = 42000000
⇒ 5x + y = 210000   …(ii)
On subtracting Eq. (i) from Eq. (ii), we get 4x = 160000
⇒ x = 40000 On substituting value of x in Eq. (i), we get y =10000
∴ Number of adults attended the match is 40000 and number of children attended is 10000.
Or
(b) Given, equations are 2x + y = 6 and x + y = 5
Table for 2x + y = 6

x	1	3	0
y	4	0	6

Hence, points are (1, 4), (3,0) and (0, 6)
Table for x + y = 5

X	1	5	0
y	4	0	5

So, points are (1, 4), (5, 0) and (0, 5).
Now, plot all the points on two graph paper and join them to get lines l₁ and l₂, respectively.

It is clear from the graph that two lines l₁ and l₂ intersect at point C(1, 4).
Hence, required solution is x = 1, y = 4

Question 30.
Prove that: (sinx – cosx + l)-(secx – tanx) = (sinx + cosx -1)
Answer:

Question 31.
The sum of first n terms of an AP is 5n² – n. Find the nth term of the AP.
Answer:
Given that,
Sum of first n times of an AP = 5n² – n
∴ S_n=5n²-n  …(i)
Therefore S_n-1 =5(n- 1)² -(n – 1) …(ii)
Now, nth term is given by

SECTION-D
This section comprises of LA type questions of 5 marks each.

Question 32.
Prove that a line drawn parallel to one side of a triangle intersecting other two sides in distinct points, divides the other two sides in the same ratio.
Answer:
Given In ΔABC, a line l drawn parallel to side BC intersects AB and AC at D and E, respectively.
To prove \(\frac{A D}{D B}=\frac{A E}{E C}\)
Construction Draw perpendicular from D and E to AC and AB
i.e., DM ⊥ AC and EN ⊥ AB. Join DC and BE.

Question 33.
(a) The numerator of a fraction is 3 less than its denominator. If 2 is added to both of its numerator and denominator then the sum of the new fraction and original fraction is \(\frac{29}{20}\). Find the original fraction.
Or
(b) A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hr, it takes 2 hours less in the journey. Find the original speed of the train.
Answer:
(a) Let the denominator of the required fraction be x.
Then, its numerator = x – 3
So, the original fraction is \(\frac{x-3}{x}\)
According to given condition,
\(\frac{(x-3)+2}{x+2}+\frac{(x-3)}{x}=\frac{29}{20}\)

Or
(b) Let the original speed of the train be x km/hr.
Given, distance travelled by train = 300 km.
∴ Original time (t₀) = \(\frac{300}{x}\) hr
New speed of the train = (x + 5) km/hr

[not possible as speed cannot be negative]
∴ x =25
Hence, the original speed of the train is 25 km/hr.

Question 34.
(a) The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 meters, find the height of the chimney. Also, find the length of the wire tied from the top of the chimney to the top of tower.
Or
(b) The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60°, respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use √3 = 1.73)
Answer:
(a) Let BA be the Chimney and CD be the tower.

Thus, length of wire fixed from the top of the Chimney to the top of the tower is 40√7 m.
Or
(b) Given, height of building = 50 m
Let EC be the tower and AB be the building.
Then, AB = 50 m, ∠PEA = 45° and ∠PEB = 60⁰
Since, Angle of elevation = Angle of depression
∴ ∠EAD = 45° and ∠EBC = 60°


[horizontal distance between the tower and building, using Eq. (i)]
Hence, height of the tower = 68.3 + 50 = 118.3 m

Question 35.
A solid toy is in the form of a hemisphere surmounted by a right circular cone of height 2 cm and diameter of base 4 cm. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. [use π = 3.14]
Answer:
Given, height of cone, h = 2 cm
Diameter of base of cone,
2r = 4cm ⇒ r = 2cm.

Now, height of cylinder, H = h+ r = 2+2 = 4 cm
Volume of circumscribing cylinder
= πr²H = 3.14 x 2 x 2 x 4 = 50.24cm³
Now, difference in the volumes of circumscribing cylinder and the toy
= Volume of cylinder – Volume of toy
= (50.24 – 25.12)cm³ = 25.12cm³
Hence, difference in the volumes of circumscribing cylinder and the toy is 25.12 cm³.

SECTION-E
This section comprises of 3 case-study based questions of 4 marks each with three sub-parts.

Question 36.
Carpooling is the sharing of car journeys so that more than one person travels in a car and prevents the need for others to have to drive to a location themselves. By having more people using one vehicle, carpooling reduces each person’s travel costs such as fuel costs, tolls and the stress of driving. Carpooling is also a more environmentally friendly and sustainable way to travel as sharing journeys reduces air pollution, carbon emissions, traffic congestion on the roads and the need for parking spaces.

Three friends Amar, Bhavin and Chetanya live in societies represented by the points A (4,5), B (6,2) and C (2,6), respectively. They all work in offices located in a same building represented by the point 0(0,0). Since they all go to same building every day, they decided to do carpooling to save money on petrol. Based on the above information, answer the following questions.
(i) What is the distance between B and C?
(ii) If Bhavin and Chetanya planned to meet at a club situated at the mid-point of the line joining the points B and C, find the coordinates of this point.
(iii) Which society is farthest from the office? Also find its distance from the office.
Or
(b) Out of B and C, which society is nearer to A? Also find their distances.
Answer:
Given, points are A(4, 5), B( 6, 2) and C(2, 6).
Office location, 0(0, 0)

Question 37.
A water sprinkler is a device used to irrigate agricultural crops, lawns, landscapes, golf courses and other areas. Water sprinklers can be used for residential, industrial and agricultural usage.

A water sprinkler is set to shoot a stream of water a distance of 21 m and rotate through an angle which is equal to complementary angle of 10°.

(i) What is the area of sector in terms of arc length?
(ii) What is the area of the watered region (in terms of n)?
(iii) (a) If the radius (r) changes to 28m, find the angle θ so that the area of the watered region remains the same.
Or
(b) If the radius (r) is increased from 21 m to 28 m and the angle remains the same, what is the increase in the area of the watered region?
Answer:
Given, Radius (r) = 21 m
Angle (θ) = complement of 10° = 90° – 10° = 80°
Let length of arc = l

Question 38.
One of four main blood types can be found in a human body. They are known as A, B, AB and O. Each blood type can be further classified as either a Rhesus positive (+) or Rhesus negative (-). For example, a possible combination is blood type O and Rhesus negative, which is written as O^–.

The data below shows the distribution of the blood types and Rhesus types of given blood type for a Blood Donation Center recorded (in percentages) for the year 2023.

(i) Find the value of
(ii) Find the probability that a randomly selected person has a Rhesus negative blood type.
(iii) (a) What is the probability that the person selected from the record is Rhesus positive but neither blood type A nor B?
Or
(b) People with blood type AB positive (AB⁺) are known as the universal recipient and with blood type O negative (0^–) are known as universal donor. Find the probability of a selected person to be neither universal recipient nor universal donor.
Answer:
Given, table is

All precentage should add up to 100.
∴ x + 30 + 8 + 24 + 6 + 18 + 1 + 3 = 100
⇒  x + 90 = 100
⇒ x = 10
(ii) Negative blood type 0^–(10), A^–(8), B(6), AB^–(1)
Total = 10 + 8 + 6 + 1 = 25%
So, the required probability = \(\frac{25}{100}=\frac{1}{4}\) = 0.25

(iii) (a) Positive groups
0⁺ =30
A⁺ =24
B⁺ =18
AB⁺ =3
Rhesus positive but neither A nor B are only 0⁺ and AB⁺.
∴ Total favourable outcomes = 30 + 3 = 33
∵ Total outcomes = 100
∴ Probability that person is Rhesus positive, but neither A nor B type blood =\(\frac{33}{100}\)
Or
(b) AB⁺ = 3%, O^– = 10%
Total = 3 + 10 = 13%
So, remaining = 100 – 13 = 87%
∴ P(Person is neither universal recipient nor Universal Donar) =\(\frac{87}{100}\)

The post CBSE Sample Papers for Class 10 Maths Basic Set 1 with Solutions appeared first on Learn CBSE.

📚 NCsolve - Your Global Education Partner 🌍

Empowering Students with AI-Driven Learning Solutions

Welcome to NCsolve — your trusted educational platform designed to support students worldwide. Whether you're preparing for Class 10, Class 11, or Class 12, NCsolve offers a wide range of learning resources powered by AI Education.

Our platform is committed to providing detailed solutions, effective study techniques, and reliable content to help you achieve academic success. With our AI-driven tools, you can now access personalized study guides, practice tests, and interactive learning experiences from anywhere in the world.

🔎 Why Choose NCsolve?

At NCsolve, we believe in smart learning. Our platform offers:

• ✅ AI-powered solutions for faster and accurate learning.
• ✅ Step-by-step NCERT Solutions for all subjects.
• ✅ Access to Sample Papers and Previous Year Questions.
• ✅ Detailed explanations to strengthen your concepts.
• ✅ Regular updates on exams, syllabus changes, and study tips.
• ✅ Support for students worldwide with multi-language content.

🌐 Explore Our Websites:

🔹 ncsolve.blogspot.com
🔹 ncsolve-global.blogspot.com
🔹 edu-ai.blogspot.com

📲 Connect With Us:

👍 Facebook: NCsolve
📧 Email: ncsolve@yopmail.com

#NCsolve #EducationForAll #AIeducation #WorldWideLearning #Class10 #Class11 #Class12 #BoardExams #StudySmart #CBSE #ICSE #SamplePapers #NCERTSolutions #ExamTips #SuccessWithNCsolve #GlobalEducation