CBSE Class 10 Maths Basic Question Paper 2025 with Solutions

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CBSE Class 10 Maths Basic Board Question Paper 2025 with Solutions

Time : 3 Hrs.
Max. Marks : 80

General Instructions :

• This question paper has 5 Section 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 ( 4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
• All questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Qs 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 whenever required if not stated.

SECTION-A
(Attempt all questions from this section)

Question 1.
In the given figure, graph of polynomial p(x) is shown. Number of zeroes of p(x) is……….

(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(a) ∵ The graph touches the X-axis at 3 points.
∵ Number of zeroes of p(x) is 3

Question 2.
22nd term of the AP \(\frac{3}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}, \ldots\) is……………
(a) \(\frac{45}{2}\)
(b) – 9
(c) \(\frac{-39}{2}\)
(d) -21
Answer:

Question 3.
The line 2 x-3 y=6 intersects X-axis at
(a) (0,-2)
(b) (0,3)
(c) (-2,0)
(d) (3,0)
Answer:
(d) ∵ On X-axis, y-coordinate is zero.
Then, options (c) and (d) satisfy this rule.
Now, check points in the given equation.
For (-2, 0),
2 ×-2-3× 0=6
-4-0=6
-4=6 [not satisfy]
For (3, 0),
2 × 3-3 ×0=6
6-0=6
6=6 [satisfy]

Question 4.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm then height of the solid is …………..

(a) 8 cm
(b) 4 √5cm
(c) 2 √5cm
(d) 12 cm
Answer:
(b) Given, radius of the base of cone (r)=4 cm
and slant height of cone (l)=6 cm
Now, height of one cone,

Question 5.
The value of k for which the system of equations 3x-7y = 1 and kx + 14y = 6 is inconsistent, is
(a) -6
(b) \(\frac{2}{3}\)
(c) 6
(d) \(\frac{-3}{2}\)
Answer:

Question 6.
Two dice are rolled together. The probability of getting a sum more than 9 is
(a) \(\frac{5}{6}\)
(b) \(\frac{5}{18}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{2}\)
Answer:
(c) Two dice are rolled together. So, the sample space of two dice,
S= [(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1),  (2, 2), (2, 3), (2, 4), (2,  5), (2, 6),
(3, 1),  (3, 2), (3, 3), (3, 4), (3,  5), (3, 6),
(4, 1),  (4, 2), (4,  3), (4, 4), (4,  5), (4, 6),
(5, 1),  (5, 2), (5,  3), (5, 4), (5,  5), (5, 6),
(6, 1),  (6, 2), (6, 3), (6, 4), (6,  5), (6, 6)}
Total sample space n(S) = 36 Now, sample space of getting a sum more than 9 = ((4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6))
∴ Probability of getting a sum more than = \(\frac{6}{36}=\frac{1}{6}\)

Question 7.
ABCD is a rectangle with its vertices at (2, -2), (8, 4), (4, 8) and (-2, 2) taken in order. Length of its diagonal is
(a) \(\frac{5}{96}\)
(b) \(\frac{5}{18}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{2}\)
Answer:
(d) ABCD is rectangle with its vertices at (2,-2), (8,4),(4,8) and (-2,2).

Question 8.
In the given figure, PA is tangent to a circle with centre O. If ∠APO = 30° and OA = 5 cm then OP is equal to

(a) 2.5 cm
(b) 5 cm
(c) \(\frac{5}{\sqrt{3}}\) cm
(d) 2 cm
Answer:
(b) In the given figure, PA is tangent to a circle with centre O.

Question 9.
If probability of happening of an event is 57% then probability of non-happening of the event is
(a) 0.43
(b) 0.57
(c) 53
(d) \(\frac{1}{57}\)
Answer:
(a) The probability of happening of an event is 57 %. Convert the percentage to a decimal
\(=\frac{57}{100}\)=0.57
Now, the probability of non-happening
=1-0.57=0.43

Question 10.
O A B is a sector of a circle with centre O and radius 7cm. If length of \(\overparen{A B}=\frac{22}{3}\)cm then ∠AOB is equal to
(a) \(\left(\frac{120}{7}\right)\)°
(b) 45°
(c) 60°
(d) 30°
Answer:
(c) Given, radius =7 cm and length of arc= \(\frac{22}{3}\) cm

Question 11.
In Δ ABC, DE || BC. If A E=(2 x+1)cm, E C=4cm, A D=(x+1)cm and D B=3cm then the value of x is …………….

(a) 1
(b) \(\frac{1}{2}\)
(c) -1
(d) \(\frac{1}{3}\)
Answer:
(b) Given, in A ABC, DEW BC
If a line is parallel to one side of a triangle and intersects the other two sides. It divides the two sides in the same ratio.
Since, DE || BC, we have
\(\frac{A D}{D B}=\frac{AE}{EC}\)
On substituting the given values, we get
\(\frac{x+1}{3}=\frac{2 x+1}{4}\)
⇒ 4 (x + 1) = 3 (2x + 1) ⇒ 4a: + 4 = 6x + 3
⇒ 4x-6x = 3- 4 ⇒ -2x = -1
∴ x =  \(\frac{1}{2}\)
Hence, the required value of x is \(\frac{1}{2}\)

Question 12.
Three coins are tossed together. The probability that exactly one coin shows head, is
(a) \(\frac{1}{8}\)
(b) \(\frac{1}{4}\)
(c) 1
(d) \(\frac{3}{8}\)
Answer:
(d) Given, three coins are tossed together.
So, sample space,
S = {HHH, HHT, HTT, HTH, TTH, THT, THH, TTT}
Now, the space that exactly one coin shows head = {HTT; TTH, THT}
Total sample space = 8 and favourable space = 3
∴ Probability (getting exactly one head) = \(\frac{3}{8}\)

Question 13.
In two concentric circles centred at O, a chord A B of the larger circle touches the smaller circle at C. If  OA=3.5cm, OC=2.1cm then A B is equal to

Answer:
(a) We know that tangent at a point on a circle is perpendicular to the radius through the point of contact.
∴ OC is perpendicular to AB.
Δ OCA is a right angled triangle.
AC² + OC² = OA²
[by Pythagoras theorem]
⇒  AC² + (2.1)² = (3.5)²
⇒ AC² + 441=12.25
⇒ AC² = 12.25 – 4.41
⇒ AC² = 7.84
⇒ AC = √7.84 ⇒ AC = 2.8 cm
Since, OC is perpendicular to chord AB, C is the mid-point of AB.
∴ A B=2.A C
⇒  AB=2. 2.8
⇒  AB=5.6 cm
Hence, the length of AB is 5.6cm

Question 14.
If √3 sinθ =cos θ then the value of θ is
(a) √3
(b) 60°
(c) \(\frac{1}{\sqrt{3}}\)
(d) 30°
Answer:

Question 15.
To calculate mean of a grouped data, Rahul used assumed mean method. He used d=(x-A), where, A is assumed mean. Then, \(\bar{x}\) is equal to
(a) A+\(\bar{d}\)
(b) A+h \(\bar{d}\)
(c) h\((A+\bar{d})\)
(d) A-h \(\bar{d}\)
Answer:
(a) The formula for the mean using the assumed mean method is \(\bar{x}=A+\bar{d}\)
Substitute the given information into the formula for the mean using the assumed mean method.

Question 16.
If the sum of first n terms of an AP is given by S_n=\(\frac{n}{2}(3 n+1)\), then the first term of the AP is
(a) 2
(b) \(\frac{3}{2}\)
(c) 4
(d) \(\frac{5}{2}\)
Answer:

∵ S₁ represents the sum of the first term, which is the first term itself.
Therefore, the first term a₁ is 2.

Question 17.
In ΔABC, ∠B=90°. If \(\frac{A B}{AC}=\frac{1}{2}\) then cos C is equal to
(a) \(\frac{3}{2} \)
(b) \(\frac{1}{2} \)
(c) \(\frac{\sqrt{3}}{2} \)
(d) \(\frac{1}{\sqrt{3}} \)
Answer:

Question 18.
The volume of air in a hollow cylinder is 450cm ³. A cone of same height and radius as that of cylinder is kept inside it. The volume of empty space in the cylinder is

(a) 225 cm³
(b) 150 cm³
(c) 250 cm³
(d) 300 cm³
Answer:
(d) Given, volume of cylinder,

Directions (Q.Nos. 19 and 20) In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Question 19.
Assertion (A) (a+√b) .(a-√b) is a rational number, where a and b are positive integers.
Reason (R) Product of two irrationals is always rational.
(a) Both Assertion and Reason are true and Reason is correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Answer:
(c) Assertion We have, (a +√b) (a – √b) = (a)²-(S)²=a²-b
Since, a and b are integers. So, a² is also an integer.
The difference between two integers (a² and b) is also an integer. Therefore, (a² -b) is an integer and all integers are rational.
So, Assertion is true.
Reason The product of two irrationals is not always rational.
e.g. √2 . √3  = √6 , which is irrational.
However, √2 . √2 = 2, which is rational.
So, Reason is false.
Therefore, Assertion is true but Reason is false.

Question 20.
Assertion (A) ΔABC sim ΔPQR such that ∠A=65°, ∠C=60°. Hence, ∠Q=55°.
Reason (R) Sum of all angles of a triangle is 180°
(a) Both Assertion and Reason are true and Reason is correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Answer:
(b) Assertion Since, ΔABC − ΔPQR,
∠A = 65° and ∠C = 60°
We know that the sum of angles in ΔABC is 180°
∴ ∠A + ∠B + ∠C = 180°
⇒ 65° + ∠B + 60° = 180?
⇒ ∠B= 180° – 65° – 60°
⇒ ∠B = 55°
Since, ΔABC − ΔPQR, corresponding angles are equal.
∠Q=∠B
⇒ ∠Q = 55°
So, Assertion is true.
Reason We know that sum of all angles of triangle is 180°.
Reason is true but it does not correctly explain the Assertion.

SECTION-B
(Answer any four questions from this section)

Question 21.
(a) Solve the equation 4x²-9 x+3=0, using quadratic formula.
Or
(b) Find the nature of roots of the equation 3 x²-4 √3 x+4=0.
Answer:
(a) Given, the quadratic equation is 4x² – 9x + 3
Now, comparing with ax² + bx + c = 0, we get
a = 4, b = – 9 and c = 3

Or
(b) Given, the quadratic equation is 3x² – 4√3x + 4.
Now, comparing with ax² + bx + c = 0,
we get a = 3, b = – 4√3 and c = 4 Now, D = b² – 4ac
= (-4√3)² – 4 x 3 x 4 = 16 x 3 – 48 = 48 – 48 = 0
Since, D = 0 the equation has two equal real roots.

Question 22.
In a trapezium ABCD, AB || DC and its diagonals intersect at O. Prove that \(\frac{O A}{O C}=\frac{O B}{O D}\)
Answer:
Since, AB || DC, ∠OAB = ∠OCD because they are alternate interior angles.
Since, AB || DC, ∠OBA = ∠ODC, because they are alternate interior angles.

∠AOB = ∠COD [vertically opposite angles]
In ΔAOB and ΔCOD, we have
∠AOB = ∠COD [proved above]
and ∠OAB = ∠OCD [proved above]
∴ ΔAOB ~ Δ COD [by AA criterion]
Since, ΔAOB ~ ΔCOD, the corresponding sides are proportional.
Therefore \(\frac{OA}{OC}=\frac{OB}{OD}\)

Question 23.
A box contains 120 discs, which are numbered from 1 to 120. If one disc is drawn at random from the box, find the probability that
(i) it bears a 2 -digit number.
(ii) the number is a perfect square.
Answer:
(i) Given, total number of discs = 120
2-digit numbers range from 10 to 99.
The total number of 2-digit numbers = 99 – 10 + 1 = 90
Now, probability of getting 2-digit number,
P (2-digit number) = \(\frac{90}{120}=\frac{3}{4}\)

(ii) Perfect square between 1 to 120 are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
So, 10 perfect squares between 1 to 120.
∴ P (perfect square) =\(\frac{10}{120}=\frac{1}{12}\)

Question 24.
(a) Evaluate \(\frac{\cos 45^{\circ}}{\tan 30^{\circ}+\sin 60^{\circ}}\)………………….
Or
(b) Verify that, sin 2A= \(\frac{2 \tan A}{1+\tan ^2 A}\), for A=30°
Answer:

Question 25.
Using prime factorisation, find the HCF of 180,140 and 210.
Answer:
HCF of 180, 140 and 210.
By prime factorisation, 180 = 2 x 2 x 3 x 3 x 5
140 = 2 x 2 x 5 x 7 and 210 =2 x 3 x 5 x 7
∴ HCF of 180, 140 and 210 = 2 x 5 = 10

SECTION-C
Q. Nos. 26 to 31 are Short Answer Type Questions of 3 marks each

Question 26.
(a) If α and β are zeroes of the polynomial 8 x²-5 x-1 then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).
Or
(b) Find the zeroes of the polynomial p(x)=3 x²+x-10 and verify the relationship between zeroes and its coefficients.
Answer:
(a) Given, a quadratic polynomial is 8x² – 5x – 1
α and β are zeroes of the polynomial.

Now, a new quadratic polynomial,
\(x^2-\left(\frac{2}{\alpha}+\frac{2}{\beta}\right) x+\left(\frac{2}{\alpha} \cdot \frac{2}{\beta}\right)\)
= x² – (-10) x + (-32) = x² + 10x – 32
Hence, the required quadratic polynomial is x² +10x – 32
Or
(b) Given polynomial is p (x) = 3x² + x – 10
= 3x² + x – 10 = 3x² + (6 – 5) x – 10
= 3x² + 6x – 5x – 10 = 3x (x + 2) – 5 (x + 2)
= (x + 2) (3x – 5)
∴   x=-2, \(\frac{5}{3}\)
The zeroes of the polynomial are \( \frac{5}{3}\) and -2

Verification
Now, comparing with ax²+b x+c=0, we get
a=3, b=1, c=-10

Hence, the relationship between the zeroes and coefficients is verified.

Question 27.
Find length and breadth of a rectangular park, whose perimeter is 100 m and area is 600m².
Answer:
Given, perimeter of rectangle, P = 100 m
and area of rectangle, A = 600 m²
∴ Perimeter = 2(l + b)
⇒ 100 = 2(l + b) ⇒ 50 = l+ b
⇒ l = 50 – b …………..(i)
Now, Area = l x b
⇒ 600 = (50 – b) b  [from Eq. (i)]
⇒ 600 = 50b – b²
⇒ b² – 50b + 600 = 0
⇒ b² – (20 + 30) b + 600 = 0
⇒ b² – 20b – 30b + 600 = 0
⇒ b (b – 20) – 30(b – 20) = 0
⇒ (b – 20) (b – 30) = 0
∴  b – 20 = 0 or b – 30 = 0
So, b = 20, 30
If b = 20, then l = 50 – 20 = 30
If b = 30, then l = 50 – 30 = 20
Hence, the length and breadth of the rectangular park are 30 m and 20 m, respectively.

Question 28.
Three measuring rods are of lengths 120cm, 100cm and 150cm. Find the least length of a fence that can be measured an exact number of times, using any of the rods. How many times each rod will be used to measure the length of the fence?
Answer:
The prime factorisation of each number
120 = 2³ x 3 x 5, 100 = 2² x 5 ²and 150 = 2 x 3 x 5²
LCM is the product of highest powers of all prime factors that appear in the factorisation.
LCM (120, 100, 150) = 2³ x 3 x 5² = 600
The least length of the fence is equal to the LCM that is 600 cm.
Now, how many times each rod measures the fence.
Rod 1 (120 cm) : = \(\frac{600}{120}\) 5 times 120
Rod 2 (100 cm) : = \(\frac{600}{100}\) 6 times
Rod 3 (150 cm) : = \(\frac{600}{150}\)4 times
The least length of fence is 600 cm and the 120 cm rod can measure it 5 times, the 100 cm rod 6 times and the 150 cm rod 4 times.

Question 29.
AB and CD are diameter of a circle with centre O and radius 7 cm. If ∠BOD=30° then find the area and perimeter of the shaded region.

Answer:
Given, AB and CD are diameters of a circle with centre O
and radius, r = 7 cm and ∠BOD = 30°,
∴ ∠AOC = ∠BOD=30° [vertically opposite angle]

Question 30.
Prove that \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) =secθ cosecθ+1.
Answer:

Question 31.
(a) Find the AP whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the AP.
Or
(b) Find the sum of first 20 terms of an AP whose nth term is given by a_n=5+ n. Can 52 be a term of this AP ?
Answer:
(a) Let a be the first term and d be the common difference of given AP.
Now, according to given condition,

SECTION-D
Q. Nos. 32 to 35 are Long Answer Type Questions of 5 marks each

Question 32.
(a) Solve the following pair of linear equations by graphical method 2 x+y=9 and x-2 y=2.
Or
(b) Nidhi received simple interest of ₹ 1200 when invested ₹ x at 6% p.a. and ₹y at 5%p.a. for 1 year. Had she invested ₹ x at 3% p.a. and ₹ y at  8% p.a. for that year, she would have received simple interest of ₹ 1260. Find the values of x and y.
Answer:


Clearly, lines (i) and (ii) intersect each other at point A(4,1).
Hence, x=4 and y=1 is the solution of the given system of equations.
Or
(b) Nidhi received ₹ 1200 when invested ₹ x at 6% p.a. and ₹ y at 5 % p.a. for 1 year.
We know that

On multiplying Eq. (ii) by 2, we get
6 x+16 y=252000
On subtracting Eq. (i) from Eq. (iii), we get
(6 x+16 y)-(6 x+5 y) =252000-120000
11 y =132000
y =12000
On substituting y=12000 in Eq. (i), we get
6x+5 × 12000 =120000
6x =120000-60000
6x =60000
x =10000
Hence, the values of x and y are 10000 and 12000, respectively.

Question 33.
(a) The given figure shows a circle with centre O and radius 4 cm circumscribed by ΔABC. BC touches the circle at D such that BD=6 cm, DC=10 cm. Find the length of AE.

Or
(b) PA and PB are tangents drawn to a circle with centre O.

If ∠AOB=120° and OA=10 cm then
(i) Find ∠OPA
(ii) Find the perimeter of ΔOAP
(iii) Find the length of chord AB
Answer:
(a) Given, BD=6cm and DC=10 cm
We know that tangents drawn from an external point to the circle are equal in length.
∴ BD =B E=6cm
CD =CF=10cm
and AF =A E=x cm (say)
Now, area of ΔOBC

Question 34.
The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are 30° and 60°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings. [use √3=1.73 ]
Answer:
Let the multi-storied building AB be h m and distance between the two buildings be d m and CD be the tall building of 9 m.
According to the question,

So, the height of the multi-storied building is 13.5 m and the distance between the two buildings is 7.785 m.

Question 35.
Find mean and mode of the following data

Class	15-20	20-25	25-30	30-35	35-40	40-45
Frequency	12	10	15	11	7	5

Answer:
Table for given data is

Here, the maximum frequency is 15 and the corresponding class is 25-30. So, this is the modal class.

Hence, mean and mode of the given data is 28 and 27.78 , respectively.

SECTION-E
Q. Nos. 36 to 38 are Case Study Based Questions of 4 marks each

Question 36.
A triangular window of a building is shown above. Its diagram represents a ΔABC with ∠A=90° and AB=AC. Points P and R trisect AB and PQ||RS|| AC.

Based on the above information, answer the following questions.
(i) Show that ΔBPQ sim ΔBAC.
(ii) Prove that PQ= \(\frac{1}{3}\) AC.
(iii) (a) If A B=3m, find the length BQ and BS. Verify that B Q= \(\frac{1}{2}\) BS.
Or
(b) Prove that BR²+RS²=\(\frac{4}{9}\) BC².
Answer:

Question 37.
A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm.Outer radius of the bowl is 10.5 cm.

Based on the above information, answer the following questions.
(i) Find the dimensions of the cuboidal box.
(ii) Find the total outer surface area of the box.
(iii) (a) Find the difference between the capacity of the bowl and the volume of the box. [use π=3.14 ]
Or
(b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.
Answer:
Let r₁ be the inner radius and r₂ be the outer radius of hemispherical bowl.
r₁ = 10 cm and r₂ = 10.5 cm
(i) ∵ The hemispherical bowl fits perfectly inside the cuboidal box.
∴ The length and breadth of the box will be equal to the outer diameter of the sphere and the height of the box will be equal to the outer radius of the hemisphere.
∴ Length (l) = 2x 10.5 = 21cm Breadth (b) = 2x 10.5 = 21cm Height (h) = 10.5 cm
∴ The dimensions of the cuboidal box are 21 cm x 21 cm x 10.5 cm.

(ii) Total surface area of the cuboidal box
= 2(lb+bh+hl)
= 2(21×21+21×10.5+ 10.5×21)
= 2(441+ 220.5+ 220.5)=2x 882 = 1764 cm² 

(iii) (a) Capacity of the bowl = \(\frac{2}{3} \pi r_1^3\)
= \(\frac{2}{3}\) × 3.14 x 10³ = 2093.33 cm³
Volume of the box = l × b × h
= 21 × 21 × 10.5 = 4630.50 cm³
∴ Required difference = 4630.50 – 2093.33
= 2537.17cm³
Or
(b) Inner surface area of the bowl
= 2πr₁² = 2 × 3.14 ×(10)²=628 cm²
Area of the circular ring formed by thickness of the bowl = πr₁² – πr₂²
= π[(10.5)²-(10)²] = 3.14(110.25-100)
=3.14 × 10.25=32.185 cm²
∴ Total area to be painted =628 + 32185
=660.165 cm²

Question 38.
Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some coordinate of points are marked on it. It starts from (0,0) moves to the points listed in order (in straight lines) and ends at (0,0).

Arushi entered the points P(8,6), Q(12,2) and S(-6,6) in order. The path drawn by robot is shown in the figure.
Based on the above information, answer the following questions.
(i) Determine the distance O P.
(ii) QS is represented by equation 2 x+9 y=42. Find the coordinates of the point where it intersects Y-axis.
(iii) (a) Point R(4.8, y) divides the line segment OP in a certain ratio, find the ratio. Hence, find the value of y.
Or
(b) Using distance formula, show that \(\frac{P Q}{O S}=\frac{2}{3}\).
Answer: