Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 3 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Basic Set 3 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

Section A consists of 20 questions of 1 mark each

Question 1.

The median of the data 12, 15, 17, 19, 22, 27 is

(a) 17

(b) 18

(c) \(\frac{41}{2}\)

(d) 19

Answer:

(b) 18

For the given data, these are two middle terms, 17 and 19.

∴ Median = \(\frac{17+19}{2}\)

= \(\frac{36}{2}\) = 18

Question 2.

Two identical solid cubes of side 3a are joined end to end. Then, total surface area of the resulting cuboid is

(a) 8a^{2}

(b) 6a^{2}

(c) 90a^{2}

(d) 12a^{2}

Answer:

(c) 90a^{2}

Length of the resulting cuboid = 6a

breadth of the resulting cuboid = 3a

and height of the resulting cuboid = 3a

Total surface area = 2 (lb + bh + hl)

= 2 [6a(3a) + 3 a(3a) + 3a (6a)]

= 2 (18a^{2} + 9a^{2} + 18a^{2})

= 2 × 45a^{2}

= 90a^{2}

Question 3.

The HCF of two numbers is 116 and their LCM is 1740. If one number is 580, then the other number is

(a) 580

(b) 348

(c) 680

(d) 448

Answer:

(b) 348

Let the other number be y, then.

(1 st number × the other number) = HCF × LCM

580 × y = 116 × 1740

y = \(\frac{116 \times 1740}{580}\) = 348

Question 4.

If one zero of the quadratic polynomial kx^{2} + 5x + k is 3, then the value of k is 15

(a) – \(\frac{3}{2}\)

(b) \(\frac{}{}\)

(c) \(\frac{}{}\)

(d) \(\frac{}{}\)

Answer:

(a) – \(\frac{3}{2}\)

Since, 3 is a zero of kx^{2} + 5x + k.

∴ k (3)^{2} + 5(3) + k = 0

⇒ 9k + 15 + k = 0

⇒ 10k + 15 = 0

k = \(-\frac{15}{10}=-\frac{3}{2}\)

Question 5.

If we join two hemispheres of same radius along their bases, then we get a

(a) cone

(b) cylinder

(c) cuboid

(d) sphere

Answer:

(d) Sphere

Question 6.

If the pair of lines are coincident, then we say that pair of lines is consistent and it has a

(a) unique solution

(b) no solution

(c) infinite solutions

(d) None of the above

Answer:

(c) infinite solutions

If the pair of lines are coincident, then it has infinite number of solutions and hence, consistent.

Question 7.

If x cot 45° cos 60° = sin 30° sin 90°, then the value of x is

(a) \(\frac{1}{2}\)

(b) – 1

(c) 1

(d) 2

Answer:

(c) 1

We have, x cot45° cos 60° = sin30° sin90°

∴ x × 1 × \(\frac{1}{2}\) = \(\frac{1}{2}\) × 1

⇒ x = 1

Question 8.

The discriminant of the quadratic equation, x^{2} – 4x + 1= 0 is

(a) 2√3

(b) 12

(c) 15

(d) – 8

Answer:

(b) 12

Given, quadratic equation is x^{2} – 4x + 1 = 0.

On comparing with ax^{2} + bx + c = 0, we get

a = 1, b = – 4 and c = 1

∴ D = b^{2} – 4ac

= (- 4)^{2} – 4 (1) (1)

= 16 – 4

= 12

Question 9.

In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of ∠A, then BD : DC is

(a) 3 : 4

(b) 4 : 3

(c) 9 : 16

(d) √3 : 2

Answer:

(a) 3 : 4

We know that \(\frac{B D}{D C}=\frac{A B}{A C}\)

= \(\frac{6}{8}=\frac{3}{4}\)

Question 10.

The prime factor of 1584 is

(a) 2^{5} × 3 × 11

(b) 2^{3} × 3^{3} × 11

(c) 2^{3} × 3 × 11^{2}

(d) 2^{4} × 3^{2} × 11

Answer:

(d) 2^{4} × 3^{2} × 11

The prime factor of 1584 is,

∴ 1584 = 2^{4} × 3^{2} × 11

Question 11.

The product of the zeroes of the polynomial, 6x^{2} – 4x + 9 is

(a) – \(\frac{1}{2}\)

(b) – \(\frac{3}{2}\)

(c) \(\frac{3}{2}\)

(d) \(\frac{1}{2}\)

Answer:

(c) \(\frac{3}{2}\)

Let P(x) = 6x^{2} – 4x + 9

Now product of zeroes = \(\frac{\text { Constant term }}{\text { Coetficient of } x^2}\)

= \(\frac{9}{6}=\frac{3}{2}\)

Question 12.

The centroid of a ∆ABC, whose vertices are A (1, 2), B (- 1, 3) and C (3, – 1) are

(a) (\(\frac{3}{2}\), 2)

(b) (3, 4)

(c) (1, \(\frac{4}{3}\))

(d) None of these

Answer:

(c) (1, \(\frac{4}{3}\))

The centroid of a ∆ABC having vertices A (1, 2), B (- 1, 3) and C (3, – 1) is

G\(\left(\frac{1-1+3}{3}, \frac{2+3-1}{3}\right)\) or G(1, \(\frac{4}{3}\))

Question 13.

Coordinates of A and B are (- 3, a) and (1, a + 4). If mid-point of AB is (- 1, 1), then the value of a is

(a) 0

(b) – 1

(c) – \(\frac{1}{2}\)

(d) 1

Answer:

(b) – 1

Mid-point of AB = \(\left(\frac{-3+1}{2}, \frac{a+a+4}{2}\right)\)

= (- 1, 1)

⇒ (- 1, a + 2) = (- 1, 1)

⇒ a + 2 = 1

⇒ a = – 1

Question 14.

Which of the following pairs of lines in a circle cannot be parallel?

(a) 2 chord

(b) a chord and a tangent

(c) 2 tangent

(d) 2 diameter

Answer:

(d) 2 diameter

Every diameter passes through the centre and so no two diameter of a circle can be parallel.

Question 15.

The distance between the points A (9, 3) and B (15, 15) is

(a) 6√5 units

(b) 2√5 units

(c) 3√5 units

(d) 3√3 units

Answer:

(a) 6√5 units

We know that d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

= \(\sqrt{(15-9)^2+(15-3)^2}\)

= \(\sqrt{(6)^2+(12)^2}\)

= \(\sqrt{36+144}\)

= \(\sqrt{180}\)

= 6√5 units

Question 16.

A card is drawn from a well-shuffled deck of 52 cards. Then, the probability that the card drawn is a black ace is

(a) \(\frac{1}{52}\)

(b) \(\frac{1}{26}\)

(c) \(\frac{1}{13}\)

(d) \(\frac{3}{52}\)

Answer:

(b) \(\frac{1}{26}\)

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52

There are two black aces one of each suit.

∴ The number of favourable outcomes to the event drawing an ace = 2

∴ P(an ace) = \(\frac{2}{52}=\frac{1}{26}\)

Question 17.

Two concentric circles are of radii 17 cm and 8 cm. Then, the length of the chord of the larger circle which touches the smaller circle is

(a) 16 cm

(b) 36 cm

(c) 15 cm

(d) None of these

Answer:

(d) None of these

Let O be the centre of concentric circles of radii 17 cm and 8 cm, respectively.

Let AB be a chord of the larger circle touching the smaller circle at P.

Then, AP = PS and OP ⊥ AB.

Now, in ∆OPA, we have

OA^{2} = OP^{2} + AP^{2} [by Pythagoras theorem]

⇒ (17)^{2} = (8)^{2} + AP^{2}

⇒ 289 = 64 + AP^{2}

⇒ AP^{2} = 225

⇒ P = 15 cm

∴ AB = 2AP

= 2 × 15 = 30 cm

Question 18.

The pair of equations 3x + 2y + 7 = 0 and 6x + 4y + 14 = 0 has

(a) unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

Answer:

(c) infinitely many solutions

Given, 3x + 2y + 7 = 0

and 6x + 4y +14 = 0

Here, \(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\)

\(\frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}\)

Hence, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}\)

So, the lines representing the given pair of linear equation will coincide, therefore there are infinitely many solutions.

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

Question 19.

Assertion (A) : In a circle of radius 6 cm, the angle of a sector is 60°, then the area of the sector is \(\frac{132}{7}\) cm^{2}.

Reason (R) : Area of the circle with radius r is πr^{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 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.

Answer:

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

Area of a sector = \(\frac{\theta}{360^{\circ}}\) × πr²

Here, θ = 60° and r = 6 cm

∴ Area of given sector = \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 6 × 6

= \(\frac{22 \times 6}{7}\)

= \(\frac{132}{7}\) cm^{2}

So, the Assertion (A) is true.

Area of a circle with radius, r = πr²

Clearly, Reason (R) is also true but it is not the correct explanation of Assertion (A).

Question 20.

Assertion (A) : x^{2} + 5x + 7 has no real zeroes.

Reason (R) : A quadratic polynomial can have at the most two zeroes.

(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.

Answer:

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

Let p(x) = x^{2} + 5x +1

On putting, p(x) = 0

⇒ x^{2} + 5x + 7 = 0

On comparing ax^{2} + bx + c = 0, we get

a = 1, b = 5 and c = 7

∴ Discriminant, D = b^{2} – 4ac

= (5)^{2} – 4 × 1 × 7

= 25 – 28

= – 3 < 0

So, the given polynomial has no real roots.

Hence, the given Assertion (A) is true.

Clearly, Reason (R) is also true but it is not the correct explanation of given Assertion (A).

Section – B

Section B consists of 5 questions of 2 marks each

Question 21.

The length of minute hand of a clock is a 28 cm. Find the area swept by the minute hand in 1 min.

Answer:

The minute hand of a clock describes a circle of radius equal to its length

i.e. 28 cm in 1 h.

So, the angle described by minute hand in 60 min = 360°

∴ Angle described by minute hand in 1 min = \(\frac{360}{60}\) = 6°

So, the area swept by the minute hand in 1 min is the area of a sector of angle 6° in a circle of radius 28 cm

Required area = \(\frac{\theta}{360^{\circ}}\) πr²

= \(\frac{6^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × (28)^{2}

= \(\frac{6^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 28 × 28

= 41.06 cm

Question 22.

Use elimination method to find all possible solutions of the following pair of equation ax + by – a + b = 0 and bx – ay – a – b = 0

Answer:

Given, pair of linear equations is

ax + by = a – b …………….(i)

and bx – ay = a + b ……………(ii)

On multiplying Eq. (i) by a, Eq. (ii) by b and then adding them, we get

a^{2}x + aby + b^{2}x – aby = a^{2} – ab + ba + b^{2}

⇒ (a^{2} + b^{2}) x = (a^{2} + b^{2})

⇒ x = \(\frac{a^2+b^2}{a^2+b^2}\) = 1

On putting x = 1 in Eq. (i), we get

a + by = a – b

⇒ y = – \(\frac{b}{b}\) = – 1

Hence, x = 1 and y = – 1, which is the required unique solution.

Or

Find the solution of the following system of equations by substitution method. 1.1 x + 1.5 y + 2.3 = 0 and 0.7 x – 0.2y = 2

Answer

Given, pair of linear equations are

1.1 x + 1.5 y + 2.3 = 0

and 0.7 x – 0.2y = 2

On multiplying these equations by 10, we get

11x + 15y + 23 = 0 ………….(i)

and 7x – 2y = 20 ……………(ii)

From Eq. (i),

15y = – 23 – 11x

⇒ y = \(\frac{-23-11 x}{15}\)

On putting the value of y in Eq. (ii), we get

7x – 2 \(\left(\frac{-23-11 x}{15}\right)\) = 20

⇒ 105 x + (46 + 22 x) = 15 × 20

⇒ 127x = 300 – 46

⇒ 127x = 254

⇒ x = 2

On putting x = 2 in Eq. (ii), we get

7 (2) – 2y = 20

⇒ 2y = 14 – 20

⇒ y = 7 – 10

⇒ y = – 3

Question 23.

Without solving the following quadratic equation, find the value of p for which the roots are equal px^{2} – 9x – 6 = 0

Answer:

Given, quadratic equation is px^{2} – 9x – 6 = 0

On comparing it with ax^{2} + bx + c = 0, we get

a = p,

b = – 9

and c = – 6

Since, the given equation has equal roots.

So, the discriminat D wiIP be zero.

∴ D = b^{2} – 4ac = 0

⇒ (- 9)^{2} – 4 × p × (- 6) = 0

⇒ 81 + 24p = 0

⇒ 24p = – 81

⇒ p = \(\frac{-81}{24}\)

= \(\frac{-27}{8}\)

Question 24.

If α and β are the zeroes of quadratic polynomial p(y) = 3y^{2} + 5y + 2, then find the value of α + αβ + β.

Answer:

Given, polynomial is p(y) = 3y^{2} + 5y + 2

Now, sum of zeroes, α + β = – \(\frac{\text { Coefficient of } y}{\text { Coefficient of } y^2}\)

= – \(\frac{5}{3}\)

and product of zeroes, = Constant term / Coefficient of $y^2$

= \(\frac{2}{3}\)

Now, α + αβ + β = (α + β) + αβ

= \(-\frac{5}{3}+\frac{2}{3}\)

= \(-\frac{3}{3}\) = – 1

Question 25.

Find the 25th term of the AP – 5, \(\frac{-5}{2}\), 0, \(\frac{5}{2}\), ……………

Answer:

Given, AP is – 5, \(\frac{-5}{2}\), 0, \(\frac{5}{2}\), ……………

Here, a = – 5

and common difference (d) = \(\frac{-5}{2}\) – (- 5)

= \(\frac{-5}{2}\) + 5 = \(\frac{5}{2}\)

25th term of the given AP is

a_{25} = a + (25 – 1) d

[∵ a_{n} = a + (n – 1) d]

⇒ a_{25} = – 5 + 24 (\(\frac{5}{2}\))

= – 5 + 60 = 55

Hence, 25th term of the given AP is 55.

Or

Determine the AP, whose 3rd term is 5 and the 7th term is 9.

Answer:

Let a and d be the first term and common difference of an AP

∴ T_{3} = 5

and T_{7} = 9

⇒ a + (3 – 1) d = 5

[∵ a_{n} = a + (n – 1) d]

and a + (7 – 1) d = 9

⇒ a + 2d = 5 …………..(i)

and a + 6d = 9 ………….(ii)

On subtracting Eq. (I) from Eq. (ii), we get

6d – 2d = 9 – 5

4d = 4

⇒ d = 1

On putting d = 1 in Eq. (I), we get

a + 2 × 1 = 5

⇒ a = 3

∴ The required AP is 3, 4, 5, 6, ………….

Section – C

Section C consists of 6 questions of 3 marks each

Question 26.

In the given figure, DE || BC and CD || EF, then prove that AD^{2} = AB × AF.

Answer:

In ∆ABC, we have

DE || BC

⇒ \(\frac{A B}{A D}=\frac{A C}{A E}\) …………(i)

In ∆ADC, we have

FE || DC

⇒ \(\frac{A D}{A F}=\frac{A C}{A E}\) …………..(ii)

From Eqs. (i) and (ii), we get

\(\frac{A B}{A D}=\frac{A D}{A F}\)

⇒ AD^{2} = AB × AF

Hence proved.

Question 27.

Solve (x – 3) (x – 4) = \(\frac{34}{(33)^2}\)

Answer:

(x – 3) (x – 4) = \(\frac{34}{(33)^2}\)

x^{2} – 7x + 12 – \(\frac{34}{(33)^2}\) = 0

x^{2} – 7x + \(\frac{13034}{(33)^2}\) = 0

Question 28.

A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, then he find the angle of elevation to be 30°. Calculate

(i) the width of the river.

(ii) the height of the tree.

Answer:

Let AB = h be the height of the tree

and BC = x be the width of the river.

(i) In ∆ABC, we have

tan 60° = \(\frac{AB}{BC}\)

⇒ √3 = \(\frac{h}{x}\)

[∵ tan 60° = √3]

⇒ h = √3x

Now, in ∆ABD, we have

tan 30° = \(\frac{AB}{BD}\)

⇒ tan 30° = \(\frac{AB}{BC+BD}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{x+50}\)

[∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]

On putting the value of h from Eq. (i) to Eq. (ii), we get

⇒ x + 50 = √3h …………..(ii)

On putting the value of h from Eq. (i) to Eq. (ii),we get

x + 50 = √3 × √3

⇒ x + 50 = 3x

⇒ 3x – x = 50

⇒ 2x = 50

⇒ x = \(\frac{50}{2}\) = 25

Hence, the width of the river is 25 m.

(ii) On putting x = 25 in Eq. (i), we get

h = √3 × 25

= 1.732 × 25

= 43.3 m

Hence, the height of the tree is 43.3 m Or

Or

A vertical tower is 20 m hight. A man standing at some distance from the tower knows that the cosine of the angle of elevation 0 of the top of the tower is 0.53 i.e. cos θ = 0.53 How far is he standing from the foot of the tower?

Answer:

Given, height of tower, AB = 20 m

Let a man standing at the point C.

The angle of elevation of the top of tower is 6 and we have cos θ = 0.53

Let the distance between foot of the tower and man be BC = x m.

Hence the distance between man and foot of tower is 12.47 m.

Question 29.

Ram started work in year 1995 at an annual salary of ₹ 5000 and received a ₹ 200 raise each year. In what year did his annual salary will reach ₹ 7000?

Answer:

Annual salary received by Ram in 1995, 1996, 1997 ……….. is ₹ 5000, ₹ 5200, ₹ 5400.

Clearly, it is AP with first term, a = 5000

and common difference, d = 5200 – 5000 = 200

Suppose Ram’s annual salary reaches to ₹ 7000 in nth yr.

Then, nth term of the above AP

T_{n} = ₹ 7000

∴ a + (n – 1) d = T_{n}

⇒ 5000 + (n – 1) × 200 = 7000

⇒ (n – 1) × 200 = 2000

⇒ n – 1 = \(\frac{2000}{200}\) = 10

⇒ n = 10 + 1 = 11

Thus, 11th annual salary received by Ram will be ₹ 7000.

This means that after 10yr i.e. in the year 2005 his annual salary will reach to ₹ 7000.

Question 30.

If tan θ = \(\frac{12}{5}\), then find the value of \(\frac{1+\sin \theta}{1-\sin \theta}\).

Answer:

We have tan θ = \(\frac{12}{5}\)

∴ sec θ = \(\)

[∵ sec^{2} A = 1 + tan^{2} A]

= \(\sqrt{1+\tan ^2 \theta}\)

= \(\sqrt{1+\left(\frac{12}{5}\right)^2}\)

= \(\sqrt{\frac{169}{25}}=\frac{13}{5}\)

∴ \(\frac{1+\sin \theta}{1-\sin \theta}=\frac{\sec \theta+\tan \theta}{\sec \theta-\tan \theta}\)

[divide numerator and denominator by cos θ]

= \(\frac{\frac{13}{5}+\frac{12}{5}}{\frac{13}{5}-\frac{12}{5}}\)

= 25

Or

If sin (A + B) = 1 and cos (A – B) = \(\frac{\sqrt{3}}{2}\), 0° < A + B ≤ 90°, A > B, then find the values of A and B.

Answer:

We have, sin (A + B) = 1

sin (A + B) = sin 90°

⇒ A + B = 90° …………(i)

and cos (A – B) = \(\frac{\sqrt{3}}{2}\)

cos (A – B) = cos 30°

⇒ A – B = 30° …………(ii)

On adding Eqs. (i) and (ii), we get

(A + B) + (A – B) = 90° + 30°

⇒ 2A = 120°

⇒ A = 60°

On putting A = 60° in Eq. (i), we get

60° + B = 90°

⇒ B = 30°

Hence. A = 60° and B = 30°.

Question 31.

Find the ratio in which the point P (x, 2) divides the line segment joining the points A (12, 5) and B(4, – 3). Also, find the value of x.

Answer:

Let the required ratFo be k : 1.

Then, by section formula,

P = \(\left(\frac{4 k+12}{k+1}, \frac{-3 k+5}{k+1}\right)\)

But this point is given as P(x, 2).

On comparing, we get

\(\frac{4 k+12}{k+1}\) = x ………….(i)

and \(\frac{-3 k+5}{k+1}\) = 2 …………(ii)

From Eq( ii),

\(\frac{-3 k+5}{k+1}\) = 2

⇒ – 3k + 5 = 2k + 2

⇒ 5k = 3

⇒ k = \(\frac{3}{5}\)

So, the rabo is 3 : 5,

On putting, k = \(\frac{3}{5}\) in Eq. (i), we get

x = \(\frac{\left(4 \times \frac{3}{5}+12\right)}{\left(\frac{3}{5}+1\right)}\)

= \(\frac{72}{8}\) = 9

Hence, x = 9.

Section – D

Section D consists of 4 questions of 5 marks each

Question 32.

ABCD is a trapezium, in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).

Answer:

Given ABCD is a trapezium in which AB || CD.

Diagonals AC and BD intersect each other at O.

To prove : \(\frac{A O}{B O}=\frac{C O}{D O}\)

Construction : Draw EOF || AB, also parallel to CD.

Proof : In ∆ACD, OE || CD [by construction]

⇒ \(\frac{A E}{E D}=\frac{A O}{O C}\) ………(i)

[by Basic Proportionality theorem]

ln ∆ABD, OE || BA [by construction]

⇒ \(\frac{E D}{A E}=\frac{O D}{O B}\)

⇒ [by Basic Proportionality theorem]

\(\frac{A E}{E D}=\frac{O B}{O D}\)

[taking reciprocal of the terms] ……………….(ii)

From Eqs (i) and (ii), we get

\(\frac{A O}{O C}=\frac{O B}{O D}\)

∴ \(\frac{A O}{B O}=\frac{C O}{D O}\)

Hence proved.

Or

In the given figure, altitudes AD and CE of ∆ ABC intersect each other at the point P.

Show that

(i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

Answer:

Given, AD and CE are altitudes which intersect each other at the point P.

(i) In ∆AEP and ∆CDP,

⇒ ∠AEP = ∠CDP [each 90°]

and ∠APE = ∠CPD [vertically opposite angles]

∴ ∆AEP ~ ∆CDP [by AA similarity criterion]

(ii) In ∆ABD and ∆CBE.

∠ADB = ∠CEB [each 90°]

and ∠ABD = ∠CBE [common angle]

∴ ∆ABO ~ ∆CBE [by AA similarity criterion]

(iii) In ∆AEP and ∆ADB,

∠AEP = ∠ADB [each 90°]

and ∠PAE = ∠BAD [common angle]

∴ ∆AEP ~ ∆ADB [by AA similarity criterion]

(iv) In ∆PDC and ABEC,

∠PDC = ∠BEC [each 90°]

and ∠PCD = ∠BCE [common angle]

∴ ∆PDC ~ ∆BEC [by AA similarity criterion]

Hence proved.

Question 33.

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is

(i) 5 of heart or diamond

(ii) jack or queen

(iii) ace and king

(iv) a red or a king

Answer:

A card is drawn from a well-shuffled pack of 52 cards.

∴ Number of possible outcomes = 52

(i) There are 2 such cards (5 of heart and 5 of diamond).

∴ P (drawn card is 5 of heart or diamond) = \(\frac{2}{52}=\frac{1}{26}\)

(ii) There are total 8 such cards (4 jack and 4 queen).

∴ P (drawn card is a jack or queen) = \(\frac{8}{52}=\frac{2}{13}\)

(iii) There is no card, which is an ace as well as king.

∴ P (drawn card is ace and king) = \(\frac{0}{52}\) = 0

(iv) There are total 28 such cards

(26 red cards and 2 black kings).

∴ P (drawn card is a red or a king) = \(\frac{28}{52}=\frac{7}{13}\)

Question 34.

An aeroplane flies horizontally in a fixed direction at a height of 1500V3m from ground. At any time the angle of elevation from a point on a ground is 60° and after 15 s, the angle of measurement of an aeroplane becomes 30°. Find the speed of the aeroplane.

Answer:

Let A be the position of an aeroplane which is flying at a height 1500√3 m from the ground

i.e. AB = 1500√3 m and angle of elevation from point P be 60°

i.e. ∠APB = 60° ………….(1)

After 15 s, aeroplane reaches at point A’, then angle of elevation from point P be ∠A’PB’ = 30°

and AB = A’B’ = 1500√3 m.

In right angled ∆A’B’P

tan 30° = latex]\frac{A^{\prime} B^{\prime}}{P B^{\prime}}=\frac{A B}{P B^{\prime}}[/latex]

[∵ A’B’ = AB]

⇒ \(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{P B^{\prime}}\)

[∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]

⇒ PB’ = 1500√3 × √3

[using cross-multiplication]

⇒ PB’ = 4500 m

In right angled ∆APB,

tan 60° = \(\frac{AB}{PB}\)

⇒ √3 = \(\frac{1500 \sqrt{3}}{P B}\)

[∵ tan 60° = √3]

PB = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\)

= 1500 m

Distance travelled by aeroplane in 15 s,

BB’ = PS’ – PS

= (4500 – 1500)m = 3000 m

Now, speed of aeroplane = Distance covered in 15s / Time taken to cover the distance

= \(\frac{3000}{15}\)

= 200 m/s

Hence, the speed of aeroplane is 200 m/s.

Or

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

Let AB = 30 m be the height of the building,

DC = 1.5 m be the length of the boy.

The point D be the boy eyes.

The angle of elevations are ∠BDF = 30°

and ∠BEF = 60°.

Let DE = x

and EF = y

Now, BF = AB – AF

= 30 -1.5 = 28.5

In right angled ∆BDF,

tan 30° = \(\frac{B F}{D F}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{28.5}{x+y}\)

⇒ x + y = 28.5 √3 ……………..(i)

Again, in right angled ∆BEF,

tan 60° = \(\frac{B F}{E F}\)

√3 = \(\frac{28.5}{y}\)

⇒ y = \(\frac{28.5}{\sqrt{3}}\) m …………(ii)

On putting y = \(\frac{28.5}{\sqrt{3}}\) in Eq. (i), we get

x + \(\frac{28.5}{\sqrt{3}}\) = 28.5 √3

⇒ x = 28.5 (√3 – \(\frac{1}{\sqrt{3}}\))

= 28.5 \(\left(\frac{3-1}{\sqrt{3}}\right)\)

= \(=\frac{28.5 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{57 \sqrt{3}}{3}\)

= 19√3 m

Hence the required distance which he walked towards the building is 19√3 m.

Question 35.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 2 cm and the height of the cone is equal to its radius. Find the volume and surface area of the solid.

Answer:

Given, radius of cone, r_{1} = 2 cm

Height of cone, h_{1} = 2 cm

Radius of hemisphere, r_{2} = 2 cm

Let V and S be the volume and surface area of the solid.

Then V = volume of cone + Volume of hemisphere

= \(\frac{1}{3}\) πr_{1}^{2} h_{1} + \(\frac{2}{3}\) πr_{2}^{3}

= \(\frac{1}{3}\) π [(2)^{2} × 2 + 2 × (2)^{3}]

= \(\frac{1}{3}\) π (8 + 16)

= \(\frac{24}{3}\) π

= 8π cm^{3}

and S = Curved surface area of cone + Curved surface area of hemisphere

= πr_{1}l + 2πr_{1}^{2}

= π (r_{1}l + 2r_{1}^{2})

= π (2 + 2 × (2)^{2}]

= π (2 × \(\sqrt{2^2+2^2}\) + 8)

= π (2 × 2√2 + 8)^{2}

= 4π (√2 + 2) cm^{2}

Section – E

Case study based questions are compulsory

Question 36.

People of village Road want to construct a road nearest to a circular village Rampur. The road cannot pass through the village. But the people want the road should be at the shortest distance from the centre of the village. Suppose, the road start from point O, which is outside the circular village and touch the boundary of the circular village at point A such that OA = 20 cm and also, the straight distance of the point O from the centre C of the village is 25 cm.

On the basis of above information, answer the following questions.

(i) If a point is inside the circle, how many tangents can be drawn from that point?

(ii) If two circles are externally and they do not touch, then find the number of common tangents.

(iii) Find the shortest distance of the road from the centre of the village.

Answer:

(i) If a point is inside the circle, then no tangent can be drawn.

(ii)

It is clear that four common tangents can be drawn, when the two circles does not touch externally.

(iii) Given, OA =20 cm and OC =25 cm

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ In ∆OAC,

AC ⊥ OA

In right angled ∆OAC,

OC^{2} = OA^{2} + AC^{2}

[by Pythagoras theorem]

⇒ (25)^{2} = (20)^{2} + AC^{2}

625 = 400 + AC^{2}

⇒ AC^{2} = 225

⇒ AC = 15 cm

Or

If DE and DF are tangents from the external point D to the circular village. If DE ⊥ DF, then find the length of DE.

Answer:

∵ DE ⊥ DF,

∠EDF = 90°

and we know CE ⊥ ED and CF ⊥ FD

[∵ tangent is perpendicular to the radius through the point of contact]

Also, CF = CE

∴ FCED is a square.

⇒ CE = CF = FD = DE

∴ DE = 15 cm

[∵ radius = 15 cm]

Question 37.

Air pollution refers to the release of pollutants into the air that are detrimental to human health and the plant as a whole. In this order Rajasthan government, conducted a awareness programme related to environment localities of a city, the concentration of sulphur dioxide (SO 2) in the air (in parts per million i.e. ppm) in 24 are as following

Concentration of SO_{2} (in ppm) |
Frequency |

0.00 – 0.02 | 2 |

0.02 – 0.04 | 5 |

0.04 – 0.06 | 4 |

0.06 – 0.08 | 3 |

0.08 – 0.10 | 4 |

0.10 – 0.12 | 6 |

On the basis of above information, answer the following questions.

(i) Write formula to find the mean.

Answer:

Mean \((\bar{x})\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

(ii) Write the formula to find the median.

Answer:

Median = l + \(\left\{\frac{\frac{N}{2}-c f}{f}\right\}\) × h

(iii) Find the mean concentration of SO_{2}.

Answer:

the table for given distribution is,

Here, Σf_{i}x_{i} = 1.60

and Σf_{i} = 24

∴ Mean of concentration of SO_{2} = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

= \(\frac{1.60}{24}\)

= 0.066 (approx)

Or

Find the median concentration of SO_{2}.

Answer:

Here, N = 24

⇒ \(\frac{N}{2}\) = 12

The cumulative frequency just greater than 12 is 14

and corresponding class interval is 0.06 – 0.08.

Here, l = 0.06,

cf = 11,

f = 3

and h = 0.02

∴ Median = l + \(\left\{\frac{\frac{N}{2}-c f}{f}\right\}\) × h

= 0.06 + \(\left\{\frac{\frac{24}{2}-11}{3}\right\}\) × 0.02

= 0.06 + \(\left\{\frac{12-11}{3}\right\}\) × 0.02

= 0.06 + \(\frac{0.02}{3}\)

= 0.06 + 0.006

= 0.066 (approx.)

Question 38.

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- Section A and Section B of grade X. There are 32 students in Section A and 36 students in Section B.

On the basis of above information answer the following questions.

(i) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Answer:

Given, number of students in Section A = 32

Number of students in Section B = 36

The minimum number of books acquire for the class library = LCM of (32, 36)

= 2 × 2 × 2 × 2 × 2 × 3 × 3

= 2^{5} × 3^{2}

= 32 × 9

= 288

(ii) If the product of two positive integers is equal to the product of their HCF and LCM is true, then find the HCF (32, 36).

Answer:

Given, product of the two numbers = LCM × HCF

∴ 32 × 36 = LCM (32, 36) × HCF (32, 36)

⇒ 32 × 36 = 288 × HCF (32, 36)

⇒ HCF (32, 36) = \(\frac{32 \times 36}{288}\) = 4

(iii) Express 5310 as product of primes.

Answer:

Factor tree of 5310 is

∴ 5310 = 2 × 3^{2} × 5 × 59

Or

If p and q are positive integers, such that p = ab^{2} and q = a^{2} b, where a and b are prime numbers, then find the LCM (p, q).

Answer:

Given, p = ab^{2}

and q = a^{2} b

LCM (p, q) = Product of the greatest power of each prime factor involved in the numbers, with highest power

= a^{2} × b^{2}