CBSE Class 10 Mathematics Practice Paper 1 (2026-27)

A full-length Class 10 Mathematics practice paper in the CBSE 2026-27 pattern, with its answer key. Free to read and to save as a PDF.

Time: 3 hours | Maximum Marks: 80

General Instructions

1. This question paper consists of 5 sections—A, B, C, D and E.

2. Section A has 18 questions of 1 mark each.

3. Section B has 5 questions of 2 marks each.

4. Section C has 6 questions of 3 marks each.

5. Section D has 4 questions of 5 marks each.

6. Section E has 3 case-based questions of 4 marks each.

7. There is no overall choice in any section.

8. However, internal choices have been provided in 2 questions of 2 marks each, 2 questions of 3 marks each, 2 questions of 5 marks each and in the 2-mark sub-part of each case-based question.

AMultiple Choice Questions

The decimal expansion of the rational number

\frac{147}{120}

will terminate after how many places of decimals?(a) 1(b) 2(c) 3(d) 4

α

and

β

are the zeroes of the quadratic polynomial

p (x) = x^{2} - 5 x + 4

, then the value of

α + β

is:(a) 4(b) 5(c) -5(d) -4

The pair of linear equations

x + 2 y + 5 = 0

and

- 3 x - 6 y - 15 = 0

represents:(a) Intersecting lines(b) Parallel lines(c) Coincident lines(d) Lines intersecting only at one point

If the roots of the quadratic equation

2 x^{2} + k x + 2 = 0

are equal, then the value of k is:(a) 4(b) -4(c) ±4(d) ±2

The nth term of an AP is given by

a_{n} = 3 n - 2

. The common difference is:(a) 1(b) 2(c) 3(d) 5

△ A B C

, if

∠ A = 90^{\circ}

, then

B C^{2}

is equal to:(a)

A B^{2} + A C^{2}

(b)

A B^{2} - A C^{2}

(c)

A C^{2} - A B^{2}

(d)

A B^{2} + A C^{2} + 2 A B \cdot A C

The distance between the points

(a, b)

and

(- a, - b)

is:(a)

\sqrt{a^{2} + b^{2}}

(b)

2 \sqrt{a^{2} + b^{2}}

(c)

\sqrt{a^{2} - b^{2}}

(d)

2 \sqrt{a^{2} - b^{2}}

\sin A = \frac{3}{4}

, then

\cos A

is:(a)

\frac{4}{3}

(b)

\frac{\sqrt{7}}{4}

(c)

\frac{4}{\sqrt{7}}

(d)

\frac{3}{4}

The angle of elevation of the top of a tower from a point on the ground, 30m away from the foot of the tower is 30°. The height of the tower is:(a)

10 \sqrt{3}

m(b)

30 \sqrt{3}

m(c)

20 \sqrt{3}

m(d)

10 / \sqrt{3}

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. The length of PQ is:(a)

\sqrt{119}

cm(b)

\sqrt{179}

cm(c)

\sqrt{199}

cm(d)

\sqrt{299}

The area of a sector of a circle with radius 6 cm and angle 60° is:(a)

6 π

cm²(b)

3 π

cm²(c)

12 π

cm²(d)

36 π

cm²

A solid is in the shape of a cone mounted on a hemisphere of the same radius. If the radius of the cone and hemisphere is 3 cm and the height of the cone is 4 cm, then the volume of the solid is:(a)

100 π

cm³(b)

36 π

cm³(c)

48 π

cm³(d)

108 π

cm³

The mean of the first n natural numbers is:(a) n/2(b) (n+1)/2(c) n(n+1)/2(d) (n+1)/4

If the mean of 5 observations

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

\overset{ˉ}{x}

, then the mean of

x_{1} + a, x_{2} + a, x_{3} + a, x_{4} + a, x_{5} + a

is:(a)

\overset{ˉ}{x}

(b)

\overset{ˉ}{x} + a

(c)

\overset{ˉ}{x} - a

(d)

a \overset{ˉ}{x}

The mode of the data set 2, 3, 5, 2, 3, 5, 2, 3, 5, 2, 3, 5 is:(a) 2(b) 3(c) 5(d) 2, 3, 5

If P(E) = 0.05, then the probability of ‘not E’ is:(a) 0.05(b) 0.95(c) 0.00(d) 1.00

A die is thrown once. The probability of getting an odd number is:(a) 1/2(b) 1/3(c) 1/4(d) 1/6

A card is drawn from a well-shuffled deck of 52 cards. The probability of drawing a red face card is:(a) 1/26(b) 3/26(c) 5/26(d) 1/13

Assertion-Reason Questions:

Assertion (A): The number

2 \times 3 \times 5 \times 7 + 5

is a composite number.Reason (R): If a number can be expressed as a product of two or more primes, it is composite.

Assertion (A): The pair of linear equations

x + y = 5

and

2 x + 2 y = 10

has a unique solution.Reason (R): For a pair of linear equations

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

, if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

, then the pair has infinitely many solutions.

BShort Answer Questions

A boat in a river moves away from the bank such that the distance covered is proportional to the square of the time elapsed. If the boat covers 15 km in 2 hours, find the distance covered by the boat in 6 hours.

(A) Find the sum of the first 15 terms of the AP: 8, 5, 2, ...OR(B) Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

In the given figure, O is the centre of the circle. PQ is a chord and PT is the tangent at P such that ∠TPQ = 50°. Find ∠POQ.

(A) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use

π = 3.14

)OR(B) The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

From a point P, two tangents PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP.

CShort Answer Questions

(A) Prove that

\sqrt{7}

is an irrational number.OR(B) The sum of two numbers is 27 and their product is 182. Find the numbers.

Solve the following pair of linear equations:

2 x + 3 y = 8

4 x + 6 y = 7

(A) Find the roots of the quadratic equation

x^{2} - 2 a x + (a^{2} - b^{2}) = 0

for

x

.OR(B) If one root of the quadratic equation

x^{2} + λ x + 3 = 0

is 1, find the value of

λ

and then find the other root.

△ A B C

, D and E are points on sides AB and AC respectively such that AD = 4 cm, BD = 5 cm, AE = 6 cm and EC = 7.5 cm. Prove that

△ A B C

is similar to

△ A D E

A hemispherical bowl is made of stainless steel, which is

\frac{1}{2}

cm thick. If the inner radius of the bowl is 4 cm, find the volume of steel used to make the bowl.

A die is thrown twice. Find the probability of getting:(i) a number greater than 4 on the first die, or(ii) a total of 8 on the two throws.

DLong Answer Questions

(A) The bottom of a rectangular parallelepiped is a square. If the total surface area is 150 sq. cm and the area of the base is 25 sq. cm, find the volume of the parallelepiped.OR(B) A solid metallic sphere of radius 10.5 cm is melted and recast into a right circular cone of height 21 cm. Find the radius of the cone. (Use

π = \frac{22}{7}

)

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

(A) 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.OR(B) Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.

(A) If the sum of the first 'm' terms of an AP is equal to the sum of the first 'n' terms, show that the sum of the first (m + n) terms is zero.OR(B) Find the sum of all 3-digit numbers which are divisible by 7.

ECase Study Questions

Case Study 1: Navigation and TrigonometryA boat is moving away from a lighthouse. The angle of depression of the boat from the top of the lighthouse is observed to be 30°. After sailing 100 meters closer to the lighthouse, the angle of depression changes to 60°.(i) Draw a diagram to represent the situation. (1 mark)(ii) Find the initial distance of the boat from the lighthouse. (1.5 marks)(A) Find the height of the lighthouse. (1.5 marks)OR(B) Find the distance of the boat from the lighthouse after moving 100 meters closer. (1.5 marks)

Case Study 2: Coordinate Geometry - CollinearityThree friends Alex, Ben, and Chris are sitting at points A, B, and C respectively in a park. Their positions are represented by the coordinates given below: A(2, 3), B(4, y), C(6, -3). They are sitting in a straight line.(i) What condition must be satisfied for the points to be collinear? (1 mark)(ii) Find the value of 'y'. (1.5 marks)(A) Find the relation between x and y if a point P(x,y) is equidistant from A and C. (1.5 marks)OR(B) If A(2, 3), B(4, y) and C(6, -3) are collinear, find the ratio in which B divides AC. (1.5 marks)

Case Study 3: Statistics - A Survey on School CommuteA survey was conducted in a locality to find the mode of transport used by students to commute to school. The data collected is given in the table below:Mode of Transport Frequency (Number of Students) Bus 20 Cycle 45 Car 10 Walking 25(i) What is the most preferred mode of transport among the students? (1 mark)(ii) Find the number of students who use a cycle or walk to school. (1.5 marks)(A) Find the total number of students surveyed.OR(B) Calculate the probability that a randomly selected student uses a bus to commute to school. (1.5 marks)

Answer Key

(b) 5

(a)

A B^{2} + A C^{2}

(b)

2 \sqrt{a^{2} + b^{2}}

(b)

\frac{\sqrt{7}}{4}

(a)

10 \sqrt{3}

(a)

\sqrt{119}

(b)

3 π

cm²

(a)

100 π

cm³

(b) (n+1)/2

(b)

\overset{ˉ}{x} + a

(b) 0.95

(a) 1/2

(b) 3/26

(A) True (R) True. Reason (R) is a correct explanation for Assertion (A).

(A) False (R) True. Reason (R) is correct, but Assertion (A) is false because the lines are coincident, meaning infinitely many solutions.

Distance =

k t^{2}

15 = k (2^{2}) \Rightarrow k = 15 / 4

. Distance in 6 hours =

(15 / 4) (6^{2}) = (15 / 4) (36) = 15 \times 9 = 135

km.

(A)

a = 8, d = - 3, n = 15

S_{15} = \frac{15}{2} [2 (8) + (15 - 1) (- 3)] = \frac{15}{2} [16 - 42] = \frac{15}{2} [- 26] = 15 \times (- 13) = - 195

.OR (B)

a_{11} = a + 10 d = 38

a_{16} = a + 15 d = 73

. Subtracting gives

5 d = 35 \Rightarrow d = 7

. Then

a + 10 (7) = 38 \Rightarrow a = - 32

a_{31} = a + 30 d = - 32 + 30 (7) = - 32 + 210 = 178

△ O P Q

O P = O Q

(radii).

∠ O P Q = ∠ O Q P = 50^{\circ}

∠ P O Q = 180^{\circ} - (50^{\circ} + 50^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}

. (Note:

O P ⊥ P Q

is not given, this is a mistake in the question setup). Assuming the diagram implies

△ O P Q

is isosceles with base PQ. Correct interpretation:

O P = O Q

(radii). If

∠ O P Q = 50^{\circ}

, then

∠ O Q P = 50^{\circ}

(angles opposite equal sides).

∠ P O Q = 180^{\circ} - (50^{\circ} + 50^{\circ}) = 80^{\circ}

. However, if PT is tangent at P, then

∠ O P T = 90^{\circ}

∠ O P Q = ∠ O P T - ∠ T P Q = 90^{\circ} - 50^{\circ} = 40^{\circ}

. Since

△ O P Q

is isosceles,

∠ O Q P = ∠ O P Q = 40^{\circ}

∠ P O Q = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 100^{\circ}

(A) Minor segment area = Area of sector - Area of

△

. Area of sector =

\frac{90}{360} π (10^{2}) = \frac{1}{4} \times 3.14 \times 100 = 78.5

cm². Area of

△ = \frac{1}{2} \times 10 \times 10 = 50

cm². Area of segment =

78.5 - 50 = 28.5

cm².OR (B) Angle swept in 5 min =

(360 / 60) \times 5 = 30^{\circ}

. Area =

\frac{30}{360} π (14^{2}) = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{1}{12} \times 22 \times 28 = \frac{22 \times 7}{3} = \frac{154}{3} \approx 51.33

cm².

△ O A P

∠ O P A = 90^{\circ}

(radius

⊥

tangent). In

△ O A P

and

△ O B P

O P = O P

(common),

O A = O B

(radii),

∠ O A P = ∠ O B P = 90^{\circ}

△ O A P ≅ △ O B P

(RHS). So

A P = B P

. Also

∠ A P O = ∠ B P O = 120^{\circ} / 2 = 60^{\circ}

. In right

△ O A P

\sin (∠ A P O) = O A / O P \Rightarrow \sin 60^{\circ} = O A / O P \Rightarrow \sqrt{3} / 2 = O A / O P

. Also

\tan (∠ A P O) = O A / A P \Rightarrow \tan 60^{\circ} = O A / A P \Rightarrow \sqrt{3} = O A / A P

O P = \sqrt{O A^{2} + A P^{2}}

. Let

A P = x

. Then

O A = x \sqrt{3}

O P = \sqrt{(x \sqrt{3})^{2} + x^{2}} = \sqrt{3 x^{2} + x^{2}} = \sqrt{4 x^{2}} = 2 x

. So

O P = 2 A P

(A) Assume

\sqrt{7}

is rational,

\sqrt{7} = p / q

p, q

coprime.

7 = p^{2} / q^{2} \Rightarrow 7 q^{2} = p^{2}

. So

7 ∣ p^{2} \Rightarrow 7 ∣ p

. Let

p = 7 k

7 q^{2} = (7 k)^{2} = 49 k^{2} \Rightarrow q^{2} = 7 k^{2}

. So

7 ∣ q^{2} \Rightarrow 7 ∣ q

. Thus

p, q

have common factor 7, contradiction. Hence

\sqrt{7}

is irrational.OR (B) Let the numbers be

x

and

y

x + y = 27

x y = 182

. Consider

t^{2} - (x + y) t + x y = 0 \Rightarrow t^{2} - 27 t + 182 = 0

(t - 13) (t - 14) = 0

. Numbers are 13 and 14.

Multiply first eqn by 2:

4 x + 6 y = 16

. Subtracting this from the second eqn

4 x + 6 y = 7

gives

0 = 9

, which is false. Hence, the pair of equations has no solution (parallel lines).

(A) Roots are

x = \frac{- (- 2 a) \pm \sqrt{(- 2 a)^{2} - 4 (1) (a^{2} - b^{2})}}{2 (1)} = \frac{2 a \pm \sqrt{4 a^{2} - 4 a^{2} + 4 b^{2}}}{2} = \frac{2 a \pm \sqrt{4 b^{2}}}{2} = \frac{2 a \pm 2 b}{2} = a \pm b

.OR (B) If 1 is a root,

1^{2} + λ (1) + 3 = 0 \Rightarrow 1 + λ + 3 = 0 \Rightarrow λ = - 4

. The equation is

x^{2} - 4 x + 3 = 0

(x - 1) (x - 3) = 0

. The other root is 3.

\frac{A D}{D B} = \frac{4}{5}

and

\frac{A E}{E C} = \frac{6}{7.5} = \frac{60}{75} = \frac{12}{15} = \frac{4}{5}

. Since

\frac{A D}{D B} = \frac{A E}{E C}

and

∠ A

is common, by SAS similarity criterion (if angles are equal and sides proportional), or more precisely if

\frac{A D}{A B} = \frac{A E}{A C}

(which is

\frac{4}{9} = \frac{6}{13.5}

- False). If

\frac{A D}{A B} = \frac{A E}{A C}

we need to check proportionality of sides.

\frac{A D}{A B} = \frac{4}{4 + 5} = \frac{4}{9}

\frac{A E}{A C} = \frac{6}{6 + 7.5} = \frac{6}{13.5} = \frac{60}{135} = \frac{12}{27} = \frac{4}{9}

. Since

\frac{A D}{A B} = \frac{A E}{A C}

and

∠ A

is common, by SAS similarity,

△ A D E \sim △ A B C

Volume of steel = Volume of outer hemisphere - Volume of inner hemisphere. Outer radius

r_{1} = 4 + 1 / 2 = 4.5

cm. Inner radius

r_{2} = 4

cm. Volume =

\frac{2}{3} π (r_{1}^{3} - r_{2}^{3}) = \frac{2}{3} π ({4.5}^{3} - 4^{3}) = \frac{2}{3} π (91.125 - 64) = \frac{2}{3} π (27.125) \approx 56.77

cm³.

Total outcomes = 36.(i) Favourable outcomes for first die > 4 are 5, 6 (2 outcomes). Favourable outcomes for second die can be any (6 outcomes). Total favourable outcomes =

2 \times 6 = 12

. P(first die > 4) =

12 / 36 = 1 / 3

.(ii) Pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes. P(sum=8) = 5/36.(iii) The question asks for P(event 1 OR event 2). These are not mutually exclusive (e.g., (5,5) is not >4 on first but sum is 10). Let E1 be event (i), E2 be event (ii).

P (E 1 \cup E 2) = P (E 1) + P (E 2) - P (E 1 \cap E 2)

E 1 \cap E 2

means first die > 4 AND sum = 8. Pairs are (5,3), (6,2). So 2 outcomes.

P (E 1 \cap E 2) = 2 / 36 = 1 / 18

P (E 1 \cup E 2) = 1 / 3 + 5 / 36 - 1 / 18 = 12 / 36 + 5 / 36 - 2 / 36 = 15 / 36 = 5 / 12

(A) Let side of square base be 's' and height be 'h'.

s^{2} = 25 \Rightarrow s = 5

cm. TSA = 2(base area) + 4(area of a side face) =

2 s^{2} + 4 (s \times h) = 2 (25) + 4 (5 h) = 50 + 20 h

. Given TSA = 150.

50 + 20 h = 150 \Rightarrow 20 h = 100 \Rightarrow h = 5

cm. Volume = base area

\times

height =

s^{2} \times h = 25 \times 5 = 125

cm³.OR (B) Volume of sphere =

\frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times (10.5)^{3}

. Volume of cone =

\frac{1}{3} π R^{2} h = \frac{1}{3} π R^{2} (21)

. Equating volumes:

\frac{4}{3} \times \frac{22}{7} \times (10.5)^{3} = \frac{1}{3} π R^{2} (21)

4 \times \frac{22}{7} \times (10.5)^{3} = π R^{2} (21)

4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} = π R^{2} (21)

4 \times \frac{22}{7} \times \frac{9261}{8} = π R^{2} (21)

\frac{22}{7} \times \frac{9261}{2} = π R^{2} (21)

\frac{22 \times 1323}{2} = π R^{2} (21)

11 \times 1323 = π R^{2} (21)

14553 = π R^{2} (21)

R^{2} = \frac{14553}{21 π} = \frac{14553}{21 \times 22 / 7} = \frac{14553}{66} = 220.5

R = \sqrt{220.5} \approx 14.85

cm.

Let height of tower be H and its shadow length be

L_{T}

. Let height of pole be

h = 6

m and its shadow length be

l_{p} = 4

m. Angle of elevation of sun,

θ

\tan θ = h / l_{p} = 6 / 4 = 3 / 2

\tan θ = H / L_{T} \Rightarrow H = L_{T} \tan θ

. We are given

L_{T} = l_{p} + 28 = 4 + 28 = 32

(This interpretation is wrong). Re-read: "shadow of a tower is 40 m longer when the sun's altitude is 30° than when it is 60°". Let height of tower be h. Let shadow at 60° be

x

\tan 60^{\circ} = h / x \Rightarrow \sqrt{3} = h / x \Rightarrow x = h / \sqrt{3}

. Shadow at 30° is

x + 40

\tan 30^{\circ} = h / (x + 40) \Rightarrow 1 / \sqrt{3} = h / (x + 40) \Rightarrow x + 40 = h \sqrt{3}

. Substitute

x = h / \sqrt{3}

h / \sqrt{3} + 40 = h \sqrt{3} \Rightarrow 40 = h \sqrt{3} - h / \sqrt{3} = h (\sqrt{3} - 1 / \sqrt{3}) = h (\frac{3 - 1}{\sqrt{3}}) = h (2 / \sqrt{3})

h = 40 \times \frac{\sqrt{3}}{2} = 20 \sqrt{3}

(A) Let the chord be AB. Let O be the centre, P the point of contact, and T the point where tangent at P meets the chord of the larger circle. AB is tangent to smaller circle at P. OP

⊥

AB. Since OP is perpendicular to the chord AB of the larger circle, OP bisects AB. So AP = PB. Let radius of smaller circle be

r = 3

cm and radius of larger circle be

R = 5

cm. In

△ O P A

O A^{2} = O P^{2} + A P^{2} \Rightarrow 5^{2} = 3^{2} + A P^{2} \Rightarrow 25 = 9 + A P^{2} \Rightarrow A P^{2} = 16 \Rightarrow A P = 4

cm. Length of chord AB = 2 * AP = 8 cm.OR (B) In

△ O A P

∠ O P A = 90^{\circ}

O P = 13

O A = 5

A P^{2} = O P^{2} - O A^{2} = 13^{2} - 5^{2} = 169 - 25 = 144

A P = 12

cm. So

T P = 12

cm.In

△ O Q C

∠ O Q C = 90^{\circ}

O Q

is a common tangent to the inner circle at Q. Let OQ = r (radius of inner circle). But OQ is not given. Let's assume the diagram implies O is center of the larger circle too and is common. Also, BC is a tangent to circle at B.

T P

and

T Q

are tangents from T to circle O.

P A

is tangent at A.

B C

is tangent at B.

O

is center.

T P = T Q

∠ T P O = 90^{\circ}

O P = r a d i u s

T P^{2} = T O^{2} - O P^{2}

.Let's re-interpret the question for (B): Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that

∠ P T Q = 2 ∠ O P Q

. (This is a standard proof).Assume the question meant: TP and TQ are tangents from T. PA is tangent at A. BC is tangent at B. This is too complex.Let's assume the question implies: TP and TQ are tangents to circle with center O. OP=5cm. TO=13cm. Find TP. TP =

\sqrt{T O^{2} - O P^{2}} = \sqrt{13^{2} - 5^{2}} = \sqrt{169 - 25} = \sqrt{144} = 12

cm. The mention of BC tangent at B is likely a typo or from a different problem. If we assume BC is also a tangent from T, then BC = TP = 12 cm.

(A) Let

S_{m} = S_{n}

\frac{m}{2} [2 a + (m - 1) d] = \frac{n}{2} [2 a + (n - 1) d]

m [2 a + m d - d] = n [2 a + n d - d]

2 a m + m^{2} d - m d = 2 a n + n^{2} d - n d

2 a (m - n) + d (m^{2} - n^{2}) - d (m - n) = 0

2 a (m - n) + d (m - n) (m + n) - d (m - n) = 0

. Since

m \neq n

, divide by

(m - n)

2 a + d (m + n) - d = 0

2 a + d (m + n - 1) = 0

. Sum of first

(m + n)

terms:

S_{m + n} = \frac{m + n}{2} [2 a + (m + n - 1) d] = \frac{m + n}{2} [0] = 0

.OR (B) 3-digit numbers divisible by 7 are: 105, 112, ..., 994. This is an AP with

a = 105

d = 7

a_{n} = 994 \Rightarrow 105 + (n - 1) 7 = 994 \Rightarrow (n - 1) 7 = 889 \Rightarrow n - 1 = 127 \Rightarrow n = 128

. Sum

S_{128} = \frac{128}{2} [105 + 994] = 64 [1099] = 70336

Case Study 1: Navigation and Trigonometry(i) Diagram: Lighthouse AB, point P on ground, boat at C initially.

∠ A C B = 30^{\circ}

. Boat moves to D,

∠ A D B = 60^{\circ}

. CD = 100 m. AB = h, CB = x.(ii) In

△ A B C

\tan 30^{\circ} = A B / C B \Rightarrow 1 / \sqrt{3} = h / x \Rightarrow x = h \sqrt{3}

.In

△ A B D

\tan 60^{\circ} = A B / D B \Rightarrow \sqrt{3} = h / (x - 100) \Rightarrow x - 100 = h / \sqrt{3}

.Substitute

x = h \sqrt{3}

h \sqrt{3} - 100 = h / \sqrt{3} \Rightarrow 100 = h \sqrt{3} - h / \sqrt{3} = h (2 / \sqrt{3}) \Rightarrow h = 50 \sqrt{3}

x = h \sqrt{3} = (50 \sqrt{3}) \sqrt{3} = 50 \times 3 = 150

m.(iii) (A) Height of lighthouse

h = 50 \sqrt{3}

\approx 50 \times 1.732 = 86.6

m.OR (B) The distance after moving 100m closer is DB =

x - 100 = 150 - 100 = 50

Case Study 2: Coordinate Geometry - Collinearity(i) Points A, B, C are collinear if the slope of AB equals the slope of BC (or AC). Slope

m = (y_{2} - y_{1}) / (x_{2} - x_{1})

.(ii) Slope AB =

(y - 3) / (4 - 2) = (y - 3) / 2

. Slope BC =

(- 3 - y) / (6 - 4) = (- 3 - y) / 2

.Equating slopes:

(y - 3) / 2 = (- 3 - y) / 2 \Rightarrow y - 3 = - 3 - y \Rightarrow 2 y = 0 \Rightarrow y = 0

.(iii) (A) P(x, y) is equidistant from A(2, 3) and C(6, -3).

P A^{2} = P C^{2}

(x - 2)^{2} + (y - 3)^{2} = (x - 6)^{2} + (y + 3)^{2}

x^{2} - 4 x + 4 + y^{2} - 6 y + 9 = x^{2} - 12 x + 36 + y^{2} + 6 y + 9

- 4 x - 6 y + 13 = - 12 x + 6 y + 45

8 x - 12 y - 32 = 0

2 x - 3 y - 8 = 0

.OR (B) B divides AC in ratio

k : 1

B = (\frac{k x_{C} + 1 x_{A}}{k + 1}, \frac{k y_{C} + 1 y_{A}}{k + 1})

4 = \frac{k (6) + 1 (2)}{k + 1} \Rightarrow 4 (k + 1) = 6 k + 2 \Rightarrow 4 k + 4 = 6 k + 2 \Rightarrow 2 k = 2 \Rightarrow k = 1

.The ratio is

1 : 1

Case Study 3: Statistics - A Survey on School Commute(i) The most preferred mode of transport is the one with the highest frequency. This is Cycle with 45 students.(ii) Number of students who use a cycle or walk = Frequency of Cycle + Frequency of Walking =

45 + 25 = 70

.(iii) (A) Total number of students surveyed =

20 + 45 + 10 + 25 = 100

.OR (B) Probability (student uses bus) = (Number of students using bus) / (Total students) =

20 / 100 = 1 / 5 = 0.2

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Class 10 Mathematics — Practice Paper 1

Answer Key