Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 2 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Basic Set 2 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.

which of the following equation has root as 3?

(a) x^{2} – 5x + 6 = 0

(b) – x^{2} + 3x – 3 = 0

(c) √2 x^{2} – \(\frac{3}{\sqrt{2}}\) x + 1 = 0

(d) 3x^{2} – 3x + 3 = 0

Answer:

(a) x^{2} – 5x + 6 = 0

Given that x^{2} – 5x + 6 = 0

On putting x = 3, we get

(3)^{2} – 5 (3) + 6 = 9 – 15 + 6 = 0

Hence, x = 3 is a root of the equation.

From option (b),

– x2 + 3x – 3 = 0

On putting x = 3, we get

– (3)^{2} + 3 (3) – 3 = – 9 + 9 – 3

= – 3 ≠ 0

Hence, x = 3 is not a root of the equation.

From option (c),

√2 x^{2} – \(\frac{3}{\sqrt{2}}\) x + 1 = 0

On putting x = 3, we get

√2 (3)^{2} – \(\frac{3(3)}{\sqrt{2}}\) + 1

9√2 – \(\frac{9}{\sqrt{2}}\) + 1 ≠ 0

Hence, x = 3 is not a root of the equation.

From option (d),

3x^{2} – 3x + 3 = 0

On putting x = 3, we get

3 (3)^{2} – 3 (3) + 3 = 37 – 9 + 3

= 21 ≠ 0

Hence, x = 3 is not a root of the equation.

Question 2.

The prime factorisation of 1250 is

(a) 2 × 5^{4}

(b) 2 × 3 × 5^{4}

(c) 2 × 5^{6}

(d) 5^{4} × 3 × 5

Answer:

(a) 2 × 5^{4}

1250 = 2 × 5 × 5 × 5 × 5 = 2 × 5^{4}

Question 3.

If a number x is added to twice its square, then the resultant is 21. Then, the quadratic representation of this statement is

(a) 2x^{2} – 2 + 21 = 0

(b) 2x^{2} + x – 21 = 0

(c) 2x^{2} – x – 20 = 0

(d) None of these

Answer:

(b) 2x^{2} + x – 21 = 0

Let the number be x.

Then, according to the given condition,

2x^{2} + x = 21

⇒ 2x^{2} + x – 21 = 0

Question 4.

If the product of the zeroes of the polynomial ax^{2} – 6x – 6 is 4, then the value of a is

(a) – \(\frac{2}{3}\)

(b) – \(\frac{3}{2}\)

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

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

Answer:

(b) – \(\frac{3}{2}\)

Let α and β be the zeroes of the polynomial ax^{2} – 6x – 6

Then, αβ = \(\frac{c}{a}=\frac{-6}{a}\)

⇒ αβ = 4

= \(\frac{-6}{a}\)

a = \(\frac{-6}{4}=\frac{-3}{2}\)

Question 5.

If the lines given by 2x + ky = l and 3x – 5y = 7 has unique solution, then the value of k is

(a) – \(\frac{10}{3}\)

(b) All real values

(c) \(\frac{10}{3}\)

(d) All real values, except

Answer:

(d) All real values, except

(d) The given equations can be rewritten as,

2x + ky – 1 = 0 and 3x – 5y – 7 = 0

On comparing with a_{1}x + b_{1}y + c_{1} = 0

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

we get a_{1} = 2, b_{1} =/r, c_{1} = – 1

and a_{2} = 3, b_{2} = – 5, c_{2} = – 7

For unique solution,

\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{2}{3} \neq \frac{k}{-5}\)

⇒ k ≠ – \(\frac{10}{3}\)

Thus, the given lines have a unique solution for all real values of k, except – \(\frac{10}{3}\).

Question 6.

If a = 10 and d = 7, then first four terms will be

(a) 10, 3, – 4, – 11

(b) 3, 10, 17, 24

(c) 10, 17, 24, 31

(d) 7, 17, 27, 37

Answer:

(c) 10, 17, 24, 31

(c) Given, a = 10 and d = 7

We know that,

a_{n} = a + (n – 1) d

∴ a_{1} = 10 + (1 – 1) (7) = 10,

a_{2} = 10 + (1) (7) = 17,

a_{3} = 10 + (2) (7) = 24

and a_{4} = 10 + (3) (7) = 31

Question 7.

The distance of the point (12, 5) from the origin is

(a) 25 units

(b) 13 units

(c) \(\sqrt{13}\) units

(d) \(\sqrt{35}\) units

Answer:

(b) 13 units

Let the point be P(12, 5) and origin (0, 0).

Distance between two points = \(\sqrt{(0-12)^2+(0-5)^2}\)

= \(\sqrt{144+25}\)

= \(\sqrt{169}\)

= 13 units

Question 8.

Name the criteria of similarity by which following triangles are similar.

(a) SSS

(b) SAS

(c) AAA

(d) ASA

Answer:

(a) SSS

In ∆ABC and ∆PQR,

∵ \(\frac{3}{4.5}=\frac{5}{7.5}=\frac{4}{6}=\frac{2}{3}\)

∴ \(\frac{A C}{P R}=\frac{A B}{P Q}=\frac{B C}{R Q}\)

So, by SSS similarity criteria, ∆ ABC ~ ∆ PQR.

Question 9.

If 2 is a zero of polynomial p(x) = 4x^{2} + 2x – 5a, then the value of a is

(a) – \(\frac{5}{2}\)

(b) 4

(c) 2

(d) 6

Answer:

(b) 4

Given polynomial is p(x) = 4x^{2} + 2x – 5a

Since, 2 is a zero of polynomial.

∴ P(2) = 0

⇒ 4(2)^{2} + 2(2) – 5a = 0

⇒ 16 + 4 – 5a = 0

⇒ 5a = 20

⇒ a = 4

Question 10.

The value of 2 sin^{2} 30° – 4 cos^{2} 45° + tan^{2} 45° + cot^{2} 45° is

(a) 2

(b) \(\frac{3}{4}\)

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

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

Answer:

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

2 sin^{2} 30° – 4 cos^{2} 45° + tan^{2} 45° + cot^{2} 45°

= \(2\left(\frac{1}{2}\right)^2-4\left(\frac{1}{\sqrt{2}}\right)^2\) + (1)^{2} + (1)^{2}

= 2 × \(\frac{1}{4}\) – 4 × \(\frac{1}{2}\) + 1 + 1

= \(\frac{1}{2}\) – 2 + 2

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

Question 11.

The radius of the wheel of a bus is 20cm. If the speed of the bus is 30 km/h, then how many revolutions will the wheel make in 1 mm?

(a) 390

(b) 398

(c) 326

(d) 450

Answer:

(b) 398

In 1 h, distance covered by wheel = 30 km

In 1 mm, distance covered by wheel = \(\frac{30 \times 1000}{60}\) = 500 m

Number of revolutions made in 1 min = Distance covered by wheel / Circumference of the wheel

= \(\frac{500}{2 \times \frac{22}{7} \times \frac{20}{100}}\)

= \(\frac{500 \times 7 \times 100}{2 \times 22 \times 20}\)

= \(\frac{50 \times 25 \times 7}{22}\)

= \(\frac{625 \times 7}{11}\)

= 398 (approx)

Question 12.

In ∆ ABC, it is given that \(\frac{A B}{A C}=\frac{B D}{D C}\)

If ∠B = 70° and ∠C = 50°, then ∠BAD is

(a) 30°

(b) 40°

(c) 45°

(d) 50°

Answer:

(a) 30°

∠A = 180° – (70° + 50°) = 60°

Since, \(\frac{B D}{D C}=\frac{A B}{A C}\)

[it means AD is the bisector of ∠A]

⇒ ∠BAD = \(\frac{1}{2}\) × 60° = 30°

Question 13.

The mid-point of the line segment joining the points (- 7, 9) and (- 3, 5) is

(a) \(\left(\frac{5}{2}, \frac{7}{2}\right)\)

(b) (2, 7)

(c) (- 5, 7)

(d) (5, 2)

Answer:

(c) (- 5, 7)

Given, (x_{1}, y_{1}) = (- 7, 9) and (x_{2}, y_{2}) = (- 3, 5)

Let the mid point be (x, y).

∴ x = \(\frac{x_1+x_2}{2}\)

= \(\frac{-7-3}{2}=\frac{-10}{2}\)

= – 5

and y = \(\frac{y_1+y_2}{2}\)

= \(\frac{9+5}{2}=\frac{14}{2}\) = 7

∴ Mid-point is (- 5, 7).

Question 14.

The ratio of the length of a rod and its shadow is √3 : 1. Then, the angle of elevation of the Sun, is

(a) 45°

(b) 40°

(e) 30°

(d) 60°

Answer:

(d) 60°

Let the length of rod be BC and length of shadow be AB.

Then, \(\frac{B C}{A B}=\frac{\sqrt{3}}{1}\) [given]

Let θ be the angle of elevation of Sun.

Then, tan θ = \(\frac{B C}{A B}=\frac{\sqrt{3}}{1}\)

θ = 60°

Question 15.

In an arithmetic progression, if a = 6, d = – 8 and n = 6 then a_{n} is

(a) – 17

(b) – 30

(c) – 34

(d) Cannot be determined

Answer:

(c) – 34

Given, a = 6,

d = – 8

and n = 6

a_{n} = a + (n – 1) d

= 6 + ( 6 – 1) × (- 8)

= 6 + 5 × (- 8)

= 6 – 40

= – 34

Question 16.

7 + 5√8 is

(a) rational

(b) irrational

(c) prime

(d) integer

Answer:

(b) irrational

We have, 7 + 5√8 = 7 + 10√2

Since, √2 is an irrational number.

Therefore, 7 + 10√2 is irrational number.

Question 17.

If the probability of Sania winning a tennis match is 0.63, then probability of her losing the match is

(a) 0.63

(b) 1

(c) 0

(d) 0.37

Answer:

(d) 0.37

Let E be the event Sania winning the match, then

P(E) = 0.63 [given]

∴ P (Sania losing the match) = P(not E)

= P(\((\bar{E}\))

= 1 – P (E)

= 1 – 0.63

= 0.37

Question 18.

If the HCF of 45 and 55 is 5, LCM of 45 and 55 is 99 x a, then the value of a is

(a) 25

(b) 49

(c) 5

(d) 7

Answer:

(c) 5

We know that,

HCF × LCM = Product of two numbers

⇒ 5 × 99 × a = 45 × 55

⇒ a = \(\frac{45 \times 55}{5 \times 99}\) = 5

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) : Three consecutive terms k, 2k + 1, 5k – 2 form an AP, then k is equal to 2.

Reason (R) : In an AP : a, a + d, a + 2d ……….., the sum of n terms of AP is

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

(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:

(c) Assertion (A) is true but Reason (R) is false.

Taking first two terms,

Common difference = 2k + 1 – k = k + 1

Taking second and third terms,

Common difference = 5k – 2 – 2k – 1 = 3k – 3

Since, the terms are in AP

∴ k + 1 = 3k – 3

⇒ – 2k = – 4

⇒ k = 2

So, the given Assertion (A) is true.

We know that sum of n terms of an AP

= \(\frac{n}{2}\) [2a + (n – 1) d] ≠ \(\frac{n}{2}\) [a + (n – 1)d]

So, the given Reason (R) is false.

Question 20.

Assertion (A) : x^{3} + x has only one real zero.

Reason (R) : A polynomial of nth degree must have n real 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:

(c) Assertion (A) is true but Reason (R) is false.

x^{3} + x = 0

⇒ x (x^{2} + 1) = 0

⇒ x = 0 or x = – \(\sqrt{- 1}\)

So, the polynomial has only one real zero i.e. 0.

∵ x^{3} + x is a cubic polynomial with degree 3 and has only one real zero.

So, the given Reason (R) is false.

Section B

Section B consists of 5 questions of 2 marks each

Question 21.

Find the radius of a circle, if the length of tangent from a point at distance of 13 cm from the centre of the circle is 12 cm.

Answer:

Let AB be a tangent drawn from point B to a circle with centre 0 such that AB = 12 cm and OB = 13 cm.

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

∴ In ∆AOB, OA ⊥ AB

Now, in right angled ∆ OAB,

OB^{2} = OA^{2} + AB^{2}

⇒ (13)^{2} = OA^{2} + (12)^{2}

⇒ 169 = OA^{2} + 144

⇒ OA^{2} = 169 – 144 = 25

⇒ OA = 5 cm

Hence, the radius of the required circle is 5 cm.

Question 22.

Solve the quadratic equation \(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\) by factorisation method.

Answer:

Given, equation is

\(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\)

⇒ \(\frac{x(x+3)-(1-x)(x-2)}{x(x-2)}=\frac{17}{4}\)

⇒ 4 [x^{2} + 3x – (x – 2 – x^{2} + 2x)] = 17 (x^{2} – 2x)

⇒ 4x^{2} + 12x – 12x + 8 + 4x^{2} = 17x^{2} – 34x

⇒ 8x^{2} – 17x^{2} + 34x + 8 = 0

⇒ 9x^{2} – 34x – 8 = 0

[dividing both sides by – 1]

⇒ 9x^{2} – 36x + 2x -8=0 [by factorisation]

⇒ 9x (x – 4) + 2 (x – 4) = 0

⇒ (x – 4) (9x + 2) = 0

⇒ x – 4 = 0 or 9x + 2 = 0

∴ x = 4 or x = – \(\frac{2}{9}\)

Hence, the required roots of the given equation are 4 and \(\frac{- 2}{9}\).

Or

If – 4 is a root of the quadratic equation x^{2} + px – 4 = 0 and the quadratic equation x^{2} + px + k = 0 has equal roots, then find the value of k.

Solution:

Since, – 4 is a root of the equation.

x^{2} + px – 4 = 0

∴ (- 4)^{2} + p × (- 4) – 4 = 0

[∵ a root always satisfies the equation]

⇒ 16 – 4p – 4 = 0

⇒ 4p = 12

⇒ p = 3 (1)

The equation x^{2} + px + k = 0 has equal roots.

Here, a = 1, b = p and c = k

∴ Discriminant = 0

⇒ b^{2} – 4ac = 0

⇒ p^{2} – 4k = 0

⇒ 9 – 4k = 0

∴ k = \(\frac{9}{4}\)

Question 23.

A teacher told his student to write a quadratic polynomial whose sum and product of the zeroes are – 6 and 8, then the students write the polynomial x^{2} + 6x + 8. Is it right? Justify your answer.

Answer:

Yes, we know that, if α and β are the zeroes of quadratic polynomial, then the quadratic polynomial

= x^{2} – (Sum of zeroes) x + Product of zeroes

∵ Required polynomial = x^{2} – (- 6) x + 8

= x^{2} + 6x + 8.

Question 24.

If the mean of the following distribution is 6, then find the value of p.

Solution:

Table for product of observation and frequency

X_{i} |
f_{j} |
f_{i}x_{i} |

2 | 3 | 6 |

4 | 2 | 8 |

6 | 3 | 18 |

9 | 1 | 10 |

p+5 | 2 | 2p+1O |

Total | N = Σf_{i} = 11 |
f_{i}x_{i} = 2p + 52 |

We have, N = Σf_{i} = 11

and Σf_{i}x_{i} = 2p + 52

∴ Mean = \(\frac{\Sigma f_i x_i}{N}\)

⇒ 6 = \(\frac{2 p+52}{11}\)

⇒ 66 = 2p + 52

⇒ 14 = 2p

∴ P = 7

Question 25.

Prove that tan^{2} θ + cot^{2} θ + 2 = sec^{2} θ cosec^{2} θ

Answer:

LHS = tan^{2} θ + cot^{2} θ + 2

= (1 + tan^{2} θ) + (1 + cot^{2} θ)

= sec^{2} θ + cosec^{2} θ (1)

= \(\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}\)

= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}\)

= \(\frac{1}{\cos ^2 \theta \sin ^2 \theta}\)

= \(\frac{1}{\cos ^2 \theta} \cdot \frac{1}{\sin ^2 \theta}\)

= sec^{2} θ cosec^{2} θ

= RHS

Hence proved.

Or

Prove that (cosec A – sin A) (sec A – cos A) = sin A cos A

Solution:

LHS = (cosec A – sin A) (sec A – cos A)

= \(\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\)

= \(\frac{1-\sin ^2 A}{\sin A} \times \frac{1-\cos ^2 A}{\cos A}\)

= \(\frac{\cos ^2 A}{\sin A} \times \frac{\sin ^2 A}{\cos A}\)

= cos A sin A

= RHS

Hence proved.

Section C

Section C consists of 6 questions of 3 marks each

Question 26.

Find the sum of series 7 + 10 \(\frac{1}{2}\) + 14 + …………… + 84.

Answer:

The given numbers are 7, 10 \(\frac{1}{2}\), 14, ………….., 84.

∵ 10 \(\frac{1}{2}\) – 7 = 14 – 10 \(\frac{1}{2}\)

= 7 \(\frac{1}{2}\)

∴ The given numbers forms an AP (1/2)

Here, first term, a = 7,

common difference, d = \(\frac{7}{2}\)

and last term, l = a_{n} = 84

∵ a_{n} = a + (n – 1) d

∵ 84 = 7 + (n – 1) \(\frac{7}{2}\) {∵ a_{7} = and d = \(\frac{7}{2}\)]

⇒ \(\frac{7}{2}\) (n – 1) = 84 – 7 (1)

⇒ \(\frac{7}{2}\) (n – 1) = 77

⇒ n – 1 = 77

⇒ n – 1 = 22

⇒ n = 23 (1/2)

∵ Sum of n terms of an Ap S_{n} = \(\frac{n}{2}\) (a + l)

∴ Sum of 23 terms S_{23} = \(\frac{23}{2}\) (7 + 84)

= \(\frac{23}{2}\) × 91

= \(\frac{2093}{2}\)

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

Question 27.

Find the largest positive integer that will divide 444, 486 and 604 leaving remainders 7, 11 and 15, respectively.

Solution:

It is given that on dividing 444 by the required number, there is a remainder of 7.

This means that 444 – 7 = 437 is exactly divisible by the required number.

In other words, required number is a factor of 437.

Similarly, required positive integer is a factor of

486 – 11 = 475

and 604 – 15 = 589.

Clearly, required number is the HCF of 437, 475 and 589.

Using the factor tree method, the prime factorisations of 437, 475 and 589 are as follows

437 = 19 × 23

475 = 5^{2} × 19

and 589 = 19 × 31

Hence, required number = 19.

Question 28.

Prove that sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ = 1

Answer:

L.H.S. = sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ

= (sin^{2} θ)^{3} θ + (cos^{2} θ)^{3} θ + 3 sin^{2} θ cos^{2} θ

= (sin^{2} θ + cos^{2} θ)^{3} – 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ) + 3 sin^{2} θ cos^{2} θ

[∵ a^{3} + b^{3} = (a + b)^{3} – 3 ab (a + b)]

= (1)^{3} – 3 sin^{2} θ cos^{2} θ (1) + 3 sin^{2} θ cos^{2} θ

= 1 – 3 sin^{2} θ cos^{2} θ + 3 sin^{2} θ cos^{2} θ = 1

= RHS

Hence proved.

Or

Prove that \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ

Answer:

L.H.S. = \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\)

= sec θ cosec θ + 1

= RHS

Hence proved.

Question 29.

A Rafree is standing on the side of the athletic track and kept a stopwatch, which is used to find the time that it took a group of students to run 200 m.

In different time intervals, different number of students completed the race are given in the table as shown below

Find mode of the given data.

Answer:

The highest frequency in the given data is 15 whose modal class is 50-100.

Here, l = 50,

f = 15,

f_{1} = 10,

f_{2} = 7

and h = 50

∵ Mode = l + \(\frac{f-f_1}{2 f-f_1-f_2}\) × h

= 50 + \(\frac{15-10}{2 \times 15-10-7}\) × 50

= 50 + \(\frac{5 \times 50}{30-17}\)

= 50 + \(\frac{250}{13}\)

= 50 + 19.23

= 69.23

Question 30.

In the given figure, PT and PS are tangents to a circle from a point P such that PT =4 cm and ∠TPS = 60°. Find the length of chord TS. How many line segment of same length TS can be drawn in the circle?

Answer:

We know that tangents drawn from external point to the circle are equal in length.

Here, P is an external point.

∴ PS = PT = 4 cm

So, ∠PTS = ∠PST

[∵ angles opposite to equal sides are equal]

In ∆PTS, we have

⇒ ∠PTS + ∠PST + ∠TPS = 180°

[by ange sum property of triangle]

⇒ ∠PTS + ∠PTS + 60° = 180°

[∵ ∠PST = ∠PTS and ∠TPS = 60°]

⇒ 2 ∠PTS = 180° – 60°

⇒ 2 ∠PTS = 120°

⇒ ∠PTS = \(\frac{120^{\circ}}{2}\) = 60°

∴ ∆PTS is an equilateral triangle.

Hence, TS = 4rm

Here, infinite lines of same length TS can be drawn in a circle.

Or

AB is a diameter and AC is a chord of a circle such that ∠BAC = 30°. If the tangent at C intersects AB produced in D, then prove that BC = BD.

Answer:

Given :AB is a diameter of the circle with centre 0

and DC is the tangent of circle and ∠BAC = 30°.

To prove : BC = BD

Construction : Join 0 to C.

Proof : Since, OC ⊥ CD

[∵ the tangent at any point of a circle is perpendicular to the radius through the point of contact]

∴ ∠OCB + ∠BCD = 90°

Now, OC = QA [radii]

⇒ ∠OCA = ∠OAC

[angles opposite to equal sides are equal]

∴ ∠OCA = 30°

Now, ∠ACB = 90° [anglIe in a semi-circle]

∴ ∠OCA + ∠OCB = 90°

⇒ ∠OCB = 60°

and ∠BCD = 30°

In ∆ACD,

∠ACD + ∠CAD + ∠AOC = 180°

⇒ 120° + 30° + ∠ADC =180°

⇒ ∠ADC = 30°

∴ In ∆ACD,

∠BCD = ∠BDC = 30°

∴ BC = BD

Hence proved.

Question 31.

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:

We have, a ∆ABC in which AB = BC = CA and AD ⊥ BC

To prove 3AB^{2} = 4AD^{2}

Proof In ∆ADB and ∆ADC,

we have

AB = AC [given]

∠B = ∠C = 60°

and ∠ADB = ∠ADC = 90°

∆ADB = ∆ADC [MS congruence rule]

BD = DC = \(\frac{1}{2}\) BC

From right angled ∆ADB, we have

AB^{2} = AD^{2} + BD^{2} [by Pythagoras theorem]

= AD^{2} + (\(\frac{1}{2}\) BC)^{2}

= AD^{2} + \(\frac{1}{4}\) BC^{2}

⇒ 4AB^{2} = 4AD^{2} + BC^{2}

⇒ 3AB^{2} = 4AD^{2} [∵ BC = AB]

⇒ 3AB^{2} = 4AD^{2}

Section D

Section D consists of 4 questions of 5 marks each

Question 32.

In a single throw of two dice, find the probability of

(i) a doublet.

(ii) a number less than 3 on each die.

(iii) an odd number as a sum.

(iv) an odd number on a die and a number less than or equal to 4 on other die.

Answer:

Total number of outcomes in a single throw of two dice = 6 x 6 = 36

(i) Possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Number of favourable outcomes = 6

∴ P (getting a doublet) = \(\frac{6}{36}=\frac{1}{6}\)

(ii) Possible outcomes for a number less than 3 on each die, are (1, 1), (1, 2), (2, 1) and (2, 2)

Number of favourable outcomes = 4

∴ P (getting a number less than 3 on each die) = \(\frac{4}{36}=\frac{1}{9}\)

(iii) Possible outcomes for getting an odd number as a sum are (1, 2), (2, 1), (1, 4), (4, 1), (1, 6), (6, 1), (2, 3), (3, 2), (2, 5), (5, 2), (3, 4), (4, 3), (3, 6), (6, 3), (4, 5), (5, 4), (5, 6) and (6, 5)

Number of favourable outcomes = 18

∴ P getting (an odd number as a sum) = \(\frac{18}{36}=\frac{1}{2}\)

(iv) Possible outcomes for getting an odd number on a die and a number less than or equal to 4 on other die = (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 3), (4, 5), (5,1), (5, 2), (5, 3) and (5, 4)

Number of favourable outcomes = 20

∴ Required probability = \(\frac{20}{36}=\frac{5}{9}\)

Or

From a pack of 52 playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

(i) a black queen.

(ii) a red card.

(iii) a ten.

(iv) a picture card (jacks, queens and kings are picture cards).

Answer:

Total number of cards = 52

∴ Total number of possible outcomes = 52

Number of cards that removed = 4 × 2 = 8

∴ Total number of remaining cards = (52 – 8) = 44

(i) ∵ Number of black queens is 2.

∴ P (getting a black queen) = \(\frac{2}{44}=\frac{1}{22}\)

(ii) ∵ Number of red cards = (26 – 8) = 18

∴ P (getting a red card) = \(\frac{18}{44}=\frac{9}{22}\)

(iii) ∵ Number of tens = 4

∴ P (getting a ten) = \(\frac{4}{44}=\frac{1}{11}\)

(iv) We know that jacks, queens and kings are picture cards.

∵ Remaining number of picture cards = 3 × 2 = 6

∴ P (getting a picture card) = \(\frac{6}{44}=\frac{3}{22}\)

Question 33.

ABC is an isosceles triangle with AB = AC and D is a point on AC such that BC^{2} = AC × DC, then prove that BD = BC.

Answer:

We have, BC^{2} = AC × DC

⇒ BC × BC = AC × DC

⇒ \(\frac{B C}{D C}=\frac{A C}{B C}\)

In ∆ABC and ∆BDC,

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

and ∠C = ∠C [common]

∴ ∆ABC ~ ∆BDC

[by SAS similarity critetian] (1)

⇒ \(\frac{A C}{B C}=\frac{A B}{B D}\)

⇒ \(\frac{A C}{B C}=\frac{A C}{B D}\)

[∵ AB = AC]

⇒ BD = BC

Hence proved.

Question 34.

A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

[take, π = 3.14]

Answer:

Let BAC be a right-angled triang’e such that

AB = 15cm

and AC = 20 cm.

Using Pythagoras theorem, we have

BC^{2} = AB^{2} + AC^{2}

BC^{2} = 15^{2} – 20^{2}

BC^{2} = 225 + 400 = 625

∴ BC = 25cm

[taking positive square root]

Let OB = x

and OA = y,

Again, using Pythagoras theorem in ∆OAB and ∆OAC, we have

AB^{2} = OB^{2} + OA^{2}

⇒ 15^{2} = x^{2} + y^{2}

⇒ x^{2} + y^{2} = 225 …………..(i)

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

⇒ 20^{2} = y^{2} + (25 – x)2^{2}

⇒ (25 – x)^{2} + y^{2} = 400 ……………..(ii)

From Eq. (ii) subtracting by Eq. (i), we get

⇒ {(25 – x)^{2} + y^{2}} – {x^{2} + y^{2}} = 400 – 225

⇒ (25 – x)^{2} – x^{2} = 175

⇒ (25 – x – x) (25 – x + x) = 175

[∵ a^{2} – b^{2} = (a – b) (a + b)]

⇒ (25 – 2x) × 25 = 175

⇒ 25 – 2x = 7

⇒ 2x = 18

⇒ x = 9

On putting x = 9 in Eq. (i), we get

81 + y^{2} = 225

y^{2} = 144

y = 12 [taking positive square root]

Thus,we have OA = 12 cm and OB = 9 cm

∴ Volume of the double cone = Volume of cone CAA’ + Volume of cone BAA’

= \(\frac{1}{3}\) π (OA)^{2} × OC + \(\frac{1}{3}\) π (OA)^{2} × OB

= \(\frac{1}{3}\) π × 12^{2} × 16 + \(\frac{1}{3}\) π × 12^{2} × 9

= \(\frac{1}{3}\) π × 144 (16 + 9)

= \(\frac{1}{3}\) × 3.14 × 144 × 25 cm^{3}

= 3768 cm^{3}

∴ Total surface area of the double cone = Curved surface area of cone CAA’ + Curved surface area of cone BAA’

= π × QA × AC + π × OA × AB

= (π × 12 × 20 + π × 12 × 15)

= 420 π

=420 × 3.14

= 1318.8 cm^{2}

Or

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 4 cm and the diameter of the base is 8 cm. If a right circular cylinder circumscribes the solid, then find how much more space it will cover?

Answer:

Let BPC be the hemisphere and ABC be the cone mounted on the base of the hemisphere.

Let EFGH be the right circular cylinder circumscribing the given toy.

We have, height of cone, OA = 4 cm

Diameter of the base of the cone, d = 8 cm

∴ Radius of the base of cone, r = \(\frac{d}{2}=\frac{8}{2}\) = 4 cm

Here, AP = AO + OP

= 4 + 4 = 8 cm

Required space = Volume of cylinder – (Volume of cone + Volume of hemisphere)

= πr²H – [\(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr²]

= πr² [H – \(\frac{1}{3}\) h – \(\frac{2}{3}\) r]

= π (4)^{2} [8 – \(\frac{1}{3}\) × 4 – \(\frac{2}{3}\) × 4]

[Here, H = AP = 8 cm

and h = AO = 4 cm]

= 16π \(\left[8-\frac{4}{3}-\frac{8}{3}\right]\)

= 16π \(\left[\frac{24-4-8}{3}\right]\)

= 16π × \(\frac{12}{3}\)

= 64π cm^{3}

Hence, the right circular cylinder covecs 64π cm^{3} more space than the solid toy.

Question 35.

A straight highway leads to the foot of a national communication and telecasting tower. A watchman standing at the top of the tower observes a car at an angle of

depression of 30° which is approaching the foot of the tower with a uniform speed. Two minutes later, the angle of depression was found to be 60°.

Find the time taken by the car to reach the foot of the tower from this point. The watchman suspects that some terrorist are approaching the tower. It needs half a minute for the watchman to inform the security staff to be on the alert.

Answer:

Let AB be the tower of height h m, C and D be the position of car at an angle of depression of 30° and 60°, respectively.

Also, let the speed of car be x km/min and y min will be taken by the car to cover distance BD.

Then, CD = 2x km

[∵ time taken to cover distance CD is 2 min and

Distance = Speed × Time]

and BD = xy km

Now, in ∆ABC, we have

tan 30° = \(\frac{A B}{B C}\)

[∵ tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\)]

\(\frac{1}{\sqrt{3}}=\frac{A B}{B D+C D}\)

[∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]

⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{2 x+x y}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{x(2+y)}\) …………..(i)

and in ∆ABD, we have

tan 60° = \(\frac{A B}{B D}\)

⇒ √3 = \(\frac{h}{x y}\)

⇒ \(\frac{h}{x}\) = √3 y

[∵ tan 60° = √3]

On substituting the value of \(\frac{h}{x}\) in Eq. (i), we get

\(\frac{1}{\sqrt{3}}=\frac{\sqrt{3} y}{2+y}\)

⇒ 2 + y = 3y

⇒ 2 = 2y

⇒ y = 1

Hence, my car will take 1 min to reach the toot of the tower.

Section – E

Case study based questions are compulsory

Question 36.

To conduct Sports Day activities in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each, 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure.

Niharika runs \(\frac{1}{4}\)th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac{1}{5}\)th the distance AD on the 8th line and posts a red flag.

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

(i) Write the coordinates of green flag.

Answer:

From the given figure, the position of green flag posted by Niharika is M (2 × 1, \(\frac{1}{4}\) × 100 i.e. M(2, 25),

(ii) Find the distance between both the flags.

Answer:

∵ M = (2, 25)

and N = (8, 20)

Now, MN = \(\sqrt{(8-2)^2+(20-25)^2}\)

[∵ distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)]

= \(\sqrt{(6)^2+(-5)^2}\)

= \(\sqrt{36+25}=\sqrt{61}\)

Hence, the distance between flags is \(\sqrt{61}\) m.

(iii) Write the coordinates of red flag.

Answer:

From the given figure, the position of red flag posted by Preet is N (8 × 1, \(\frac{1}{5}\) × 100) i.e, N (8, 20),

Or

If Rashmi has to post a blue flag exactly half way between the line segment joining the flags, then find the coordinates of that point. (2)

Answer:

Let P be the position of the blue flag posted by Rashmi in the half way of line segment MN.

Then coordinates of P = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)

= \(\left(\frac{10}{2}, \frac{45}{2}\right)\)

= (5, 22.5)

Hence, the blue flag is on the 5th line at a distance of 22.5 m above it.

Question 37.

A regular morning walk, schedule can help you to reduce weight. You can take care of many health problems concerning your heart, joints, mental health, lung capacity, immune system and body strength just by walking. To improve the health two friends Vicky and Love went to the walk on concrete track. (Shown in figure). In the figure, AOB is a flower bed in the shape of a sector of a circle of radius 40 m and ∠AOB = 60°. Also, a 15 m wide concrete track is made as shown in the figure.

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

(i) Find the radius of inner circle. (1)

Answer:

Let radius of inner circle be r_{2} and outer circle be r_{1}

and given concrete track width = 15 m

∵ r_{1} = 40 [given]

∵ r_{2} = r_{1} – 15

= 40 – 15 = 25 m

(ii) Write the formula to find the area of sector. (1)

Answer:

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

(iii) Find the area of the sector AOB. (2)

Answer:

Area of sector AOB = \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 40 × 40

= \(\frac{1}{6} \times \frac{22}{7}\) × 40 × 40

= 838.09 m^{2}

Or

If instead of flower bed there had been a complete circular concrete track, then find the area of the concrete region. (2)

Answer:

∴ Required area = Area of complete circle -Area of inner circle = π × (40)^{2} – π × (25)^{2}

= π (40^{2} – 25^{2})

= π (40 + 25) (40 – 25)

= 65 × 15 × π

= 975 π m^{2}

Question 38.

Anil, a student of class X goes to Ganga rivers with his father. When he saw a boat in the river, he wanted to sit in the boat. So, he made his father Ramesh is ready to sit with him on the boat. In this order, Anil is sitting on a boat which is upstream at a speed of 6 km/h and downstream at a speed of 14 km/h. When Anil is on the boat, some questions arise in his mind.

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

(i) What is the speed (in km/h) of the boat in still water? (1)

Answer:

Let the speed of the boat in still water be x km/h

and speed of the stream be y km/h.

Then, x – y = 6 …………….(i)

x + y = 14 …………….(ii)

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

2x = 20

⇒ x = 10

∴ Speed of the boat in still water is 10 km/h.

(ii) What is the speed (in km/h) of stream? (1)

Answer:

∵ x = 10

On putting the value of x in equation x – y = 6, we get

10 – y = 6

⇒ – y = 6 – 10

⇒ y = 4

∴ Speed of the stream is 4 km/h.

(iii) Find the average speed of Anil, if he travels 5 km upstream and 10 km downstream. (2)

Answer:

Time taken by Anil to travel 5 km upstream = \(\frac{5}{6}\) h

Time taken by Anil to travel 10 km downstream = \(\frac{10}{14}\) h

∴ Average speed = \(\frac{\text { Total distance }}{\text { Total time }}\)

= \(\frac{5+10}{\frac{5}{6}+\frac{5}{7}}\)

= 9.69 km/h

Or

What is the ratio between the speed of the boat and speed of the water current respectively? (2)

Answer:

Speed of the boat = 10 km/h

Speed of the stream = 4 km/h

∴ Required ratio = \(\frac{10}{4}=\frac{5}{2}\)