Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 4 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Basic Set 4 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.

In the adjoining figure ∆APQ is similar to ∆ABC, AQ = 10 cm, BC = 3.5 cm and PQ = 7 cm, then AC is equals to

(a) 5 cm

(b) 3 cm

(c) 4 cm

(d) 2 cm

Answer:

(a) 5 cm

∆APQ ~ ∆ABC

∴ \(\frac{A Q}{A C}=\frac{P Q}{B C}\)

⇒ \(\frac{10}{A C}=\frac{7}{3.5}\)

AC = \(\frac{10 \times 3.5}{7}\)

⇒ AC = 5

Question 2.

A point P is 13 cm from the centre of the circle. The length of the tangent drawn from P to the circle is 12 cm. Then, the radius of the circle is

(a) 3 cm

(b) 5 cm

(c) 9 cm

(d) 7 cm

Answer:

(b) 5 cm

Since, tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠OTP = 90°

In right angled ∆OTP, we have

OP^{2} = OT^{2} + PT^{2}

113^{2} = OT^{2} + 12^{2}

⇒ OT^{2} = 13^{2} – 12^{2}

= (13 – 12) (13 + 12)

= 25

⇒ OT = 5 cm

Question 3.

The mean of the following distribution is

(a) 19.5

(b) 18

(c) 16.96

(d) 15.24

Answer:

(c) 16.96

Σ f_{i}x_{i} = 3 × 12 + 5 × 14 + 8 × 18 + 7 × 20

= 36 + 70 + 144 + 140

= 390

∴ Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)

= \(\frac{390}{3+5+8+7}\)

= \(\frac{390}{23}\)

= 1696

Question 4.

A fair dice is rolled. Probability of getting a number greater than 3 is

(a) 0

(b) \(\frac{1}{3}\)

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

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

Answer:

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

Favourable outcomes = 4, 5, 6

Total number of outcomes = 6

∴ P = Number of favourable outcomes / Total number of outcomes

= \(\frac{3}{6}=\frac{1}{2}\)

Question 5.

If d_{i} = x_{i} – 20, Σf_{i}d_{i} = 300 and Σf_{i} = 40, then the value of \(\bar{x}\) is

(a) 20.3

(b) 32

(c) 27.5

(d) 22.9

Answer:

(c) 27.5

Given, Σf_{i}d_{i} = 300,

Σf_{i} = 40

and d_{i} = x_{i} – 20

We know that d_{i} = x_{i} – a

On comparing, we get a = 20

∴ \(\bar{x}\) = a + \(\left\{\frac{\Sigma f_i d_i}{\Sigma f_i}\right\}\)

= 20 + \(\frac{300}{40}\)

= 20 + \(\frac{30}{4}\)

= 20 + 7.5 = 27.5

Question 6.

If 2 sin 3θ = √3, then the value of θ is

(a) 60°

(b) 30°

(c) 20°

(d) 15°

Answer:

(c) 20°

we have, 2 sin 3θ = √3

⇒ sin 3θ = \(\frac{\sqrt{3}}{2}\)

⇒ sin 3θ = sin 60°

⇒ θ = \(\frac{60^{\circ}}{3}\)

∴ θ = 20°

Question 7.

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

(a) 6

(b) 2

(c) 4

(d) 8

Answer:

(a) 6

Given, polynomial is f(x) = 4x^{2} + 4x – 4a.

Since, 2 is a zero of the given polynomial.

∴ f(2) = 0

⇒ 4 (2)^{2} + 4(2) – 4a = 0

⇒ 16 + 8 – 4a = 0

⇒ 4a = 24

⇒ a = \(\frac{24}{4}\) = 6

Question 8.

The roots of given equation (x + 4) (3x – 5)= 0 are

(a) (\(\frac{5}{2}\), 1)

(b) (- 4, \(\frac{5}{3}\))

(c) (4, \(\frac{5}{3}\))

(d) (- 4, – \(\frac{5}{3}\))

Answer:

(b) (- 4, \(\frac{5}{3}\))

(x + 4) (3x – 5) = 0

⇒ x + 4 = 0 or 3x – 5 = 0

⇒ x = – 4 0r x = \(\frac{5}{3}\)

Hence, the roots of the given equation are – 4 and \(\frac{5}{3}\).

Question 9.

The radius of a wheel is 0.25 m. The number of approximate revolutions it will make to travel a distance of 11 km, is

(a) 5000

(b) 7000

(c) 6000

(d) 1000

Answer:

(b) 7000

The distance covered in one revolution is equal to the circumference of the wheel.

∵ Circumference of wheel = 2πr

= 2 × \(\frac{22}{7}\) × 0.25

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

∴ Number of revolutions = Distance covered by wheel / Circumference of wheel

= \(\frac{11 \times 1000}{11 / 7}\)

= 7000

Question 10.

The first term of an AP is – 7 and the common difference is 5. Its 18th term is

(a) 97

(b) 92

(c) 83

(d) 78

Answer:

(d) 78

We have, a = first term = – 7

and d = 5

∴ a_{18} = a + (18 – 1) d

[∵ a_{n} = a + (n – 1) d]

⇒ a_{18} = a + 17d

= – 7 + 17 × 5

= – 7 + 85 = 78

Question 11.

If HCF of 306 and 657 is 9, then the LCM is

(a) 19428

(b) 27352

(c) 22338

(d) None of these

Answer:

(c) 22338

Given, HCF of 306 and 657 is 9.

We know that

⇒ LCM × HCF = Product of two numbers

⇒ LCM × 9 = 306 × 657

⇒ LCM = \(\frac{306 \times 657}{9}\)

= 34 × 657 = 22338

∴ LCM of 306 and 657 = 22338.

Question 12.

The coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 1 : 3 internally are

(a) (4, 3)

(b) (7, 3)

(c) (3, 5)

(d) (5, – 1)

Answer:

(d) (5, – 1)

Let (x_{1}, y_{1}) = (4, – 3)

and (x_{2}, y_{2}) = (8, 5)

Let (x, y) be the coordinates of the point which divides the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) in ratio m : n = 1 : 3 internally.

So, (x, y) = \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)

= \(\left(\frac{1(8)+3(4)}{1+3}, \frac{1(5)+3(-3)}{1+3}\right)\)

= \(\left(\frac{20}{4}, \frac{-4}{4}\right)\)

= (5, – 1)

Question 13.

If tan (3x + 30°) = 1, then the value of x is

(a) 3°

(b) 5°

(c) 10°

(d) 0°

Answer:

(b) 5°

We have, tan(3x + 30°) = 1

⇒ tan(3x + 30°) = tan 45°

[∵ tan 45° = 1]

⇒ 3x + 30° = 45°

⇒ 3x = 45° – 30° = 15°

⇒ x = 5°

Question 14.

C is the mid-point of PQ, if P is (4, x), C is (y, – 1) and Q is (- 2, 4), then x and y respectively, are

(a) – 6 and 1

(b) – 6 and 2

(c) 6 and – 1

(d) 6 and – 2

Answer:

(a) – 6 and 1

Given, C is the mid-point of PQ

i.e. P (4, x) and Q (- 2, 4).

Therefore, (y, – 1) = \(\left(\frac{4-2}{2}, \frac{x+4}{2}\right)\)

⇒ (y, – 1) = \(\left(1, \frac{x+4}{2}\right)\)

On equating the coordinates, we get

y = 1

and – 1 = \(\frac{x+4}{2}\)

∴ x = – 6 and y = 1

Question 15.

The LCM and HCF of two non-zero positive numbers are equal, then the numbers must be

(a) prime

(b) coprime

(c) composite

(d) equal

Answer:

(d) equal

Given, LCM and HCF are equal.

Let two non-zero positive numbers are p and q.

Then, HCF(p, q) = LCM (p, q) [given]

Let HCF (p, q) = k

⇒ p = ka

and q = kb

where, a and b are natural numbers,

∵ HCF × LCM = Product of numbers

k × k = ka × kb

⇒ a × b = 1

∴ a = b = 1, as they are natural numbers.

Hence, p = q or the number must be equal.

Question 16.

The volume of a hemisphere is 2425\(\frac{1}{2}\) cm^{3}, then its curved surface area is

(a) 696 cm^{2}

(b) 936 cm^{2}

(c) 684 cm^{2}

(d) 693 cm^{2}

Answer:

(d) 693 cm^{2}

Given, volume of a hemisphere = 2425 \(\frac{1}{2}\) cm^{3}

\(\frac{2}{3}\) πr^{3} = 2425.5

⇒ r^{3} = \(\frac{50935.5}{44}\)

⇒ r^{3} = 1157.625

⇒ r = 10.5 cm

∴ Curved surface area = 2πr²

= 2 × \(\frac{22}{7}\) × 10.5 × 10.5

= 693 cm^{2}

Question 17.

The prime factorisation of 352 is

(a) 2^{8}

(b) 2^{6} × 11

(c) 2^{5} × 11

(d) 2^{6} × 7

Answer:

(c) 2^{5} × 11

We have, 352 = 2 × 2 × 2 × 2 × 2 × 11

∴ 352 = 25 × 11

Question 18.

The discriminant of the quadratic equation 6x^{2} – 7x + 2 = 0 is

(a) \(\sqrt{41}\)

(b) \(\frac{1}{12}\)

(c) 5

(d) None of these

Answer:

(d) None of these

Given, 6x^{2} – 7x + 2 = 0 …………..(i)

On comparing Eq. (i) with ax^{2} + bx + c = 0, we get

a = 6, b = -7 and c = 2

We know that D = b^{2} – 4ac = (- 7)^{2} – 4 (6) (2)

= 49 – 48 = 1

Hence, the value of discriminant is 1.

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) : A coin is tossed 30 times and head appears 20 times. Then, the probability of getting a tail is 1 / 3.

Reason (R) : Probability of happening of an event

Number of trials in which = the event happened / Total number of trials

(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:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation , of Assertion (A).

Total number of times in which tail appeared = 30 – 20 = 10

∴ Probability of getting a tail = \(\frac{10}{30}=\frac{1}{3}\)

So, the given Assertion (A) is true.

Also, the given Reason (R) is also true and it is the correct explanation of Assertion (A).

Question 20.

Assertion (A) If Sn is the sum of the first n terms of an AP, then its nth term is given by a_{n} = S_{n} – S_{n-1}

Reason (R) The 8th term of the AP : 3, 8, 13, ……………, is 43.

(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.

nth term of an AP is given by a_{n} = a + (n – 1) d

Sum of n terms of an AP S_{n} = \(\frac{n}{2}\) [2a + (n – 1) d] ………….(i)

Sum of n – 1 terms of an AP

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

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

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

= na + \(\frac{n}{2}\) (n – 1) d – (n – 1) a – \(\frac{(n-1)(n-2) d}{2}\)

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

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

= a + (n – 1 )d

= a_{n}

∴ The given Assertion (A) is true.

Now, for AP 3, 8, 13, ……………. a = 3 and d = 5

a_{6} = 3 + (8 – 1) 5

= 3 + 7 × 5

= 38 ≠ 43

So, the given Reason (R) is false.

Section – B

Section B consists of 5 questions of 2 marks each.

Question 21.

In the given figure, a circle is inscribed in a ∆ABC. If AB = 10 cm BC = 8 cm, and AC = 12 cm, then find the lengths of BM, CN and AL.

Answer:

We know that the lengths of the tangents drawn from an external point to a circle are equal.

Let AL = AN = x ;

BL = BM = y ;

CM =CN = z

Now, AL+BL = AB, x + y = 10 ……….(i)

BM + CM = BC, y + z = 8 ………….. (ii)

CN+AN = AC,z + x = 12 ……………(iii)

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

x – y = 4 …………. (iv)

On solving Eqs. (i) and (iv), we get

x = 7, y = 3

Now, substituting y = 3 in Eq. (ii), we get

z = 5

BM = y = 3 cm,

CN = z = 5 cm

and AL = x = 7 cm

Question 22.

Prove that (1 + tan^{2} θ) (1 + sin θ) (1 – sin θ) = 1.

Answer:

LHS = (1 + tan^{2} θ) (1 + sin θ) (1 – sin θ)

= (1 + tan^{2} θ) (1 – sin^{2} θ)

= sec^{2} θ × cos^{2} θ

[∵ sec^{2} θ – tan^{2} θ = 1

and sin^{2} θ + cos^{2} θ = 1 ]

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

[∵ sec A = \(\frac{1}{\cos A}\)]

Hence proved.

Or

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

Answer:

We have,

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

= (\(\frac{1}{\sin \theta}\) – sin θ) (\(\frac{1}{\cos \theta}\) – cos θ)

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

= sin θ . cos θ

[∵ 1 = sin^{2} A + cos^{2} A]

RHS = \(\frac{1}{\tan \theta+\cot \theta}\)

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

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

= sin θ – cos θ

[∵ sin^{2} A + cos^{2} A = 1]

Question 23.

If the 3rd and 9th terms of an AP are 4 and – 8 respectively, then which term of this AP • is zero?

Answer:

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

∵ The n th term of an AP is a_{n} = a + (n – 1) d

∴ a_{3} = a + 2d

⇒ a + 2d = 4

[∵ a_{3} = 4 (given)] …………(i)

and a_{9} = a + 8d

⇒ a + 8d = – 8

[∵ ag = – 8 (given)]… (ii)

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

6d = – 12

⇒ d = \(\frac{- 12}{6}\) = – 2

∴ From Eq. (i),

a + 2 × (- 2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let the nth term of an AP is zero.

Then, a_{n} = 0

⇒ a + (n – 1) d = 0

⇒ 8 + (n – 1) (- 2) = 0

⇒ (n – 1) (- 2) = – 8

⇒ n – 1 = \(\frac{- 8}{- 2}\) = 4

⇒ n = 4 + 1 = 5

Hence, 5th term of an AP is zero.

Question 24.

The quadratic equation x^{2} – 3x + p = 0 has distinct real roots, then find the value of p.

Answer:

Given quadratic equation is x^{2} – 3x + p = 0

On comparing with ax2 + bx + c = 0, we get

a = 1,

b = – 3

and c = p

The condition for distinct real roots is b^{2} – 4ac > 0

⇒ (- 3)^{2} – 4 (1) (p) > 0

⇒ 9 – 4p > 0

⇒ 9 > 4p

⇒ 4p < 9

⇒ p < \(\frac{9}{4}\)

Question 25.

Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die.

Answer:

Here, two dice are thrown, so possible outcomes are

(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) and (6, 6)

∴ Total number of outcomes = 36

Let E = Event of getting a multiple of 2 on one die and a multiple of 3 on the other die.

Here, multiples of 2 are 2, 4 and 6 and multiples of 3 are 3 and 6.

So, favourable outcomes for event E are (2, 3), (4, 3), (6,3), (2,6), (4, 6), (6, 6), (3,2), (3,4), (3,6), (6,2) and (6, 4).

∴ Number of outcomes favourable to E = 11

∴ Required probability = P(E)

= \(\frac{\text { Number of favourable outcomes to } E}{\text { Total number of outcomes }}\)

= \(\frac{11}{36}\)

Or

Suppose you drop a die 5 m at random on the rectangular region shown in figure. What is the probability that it will land inside the circle of diameter 2 m?

Answer:

Area of rectangle = 5 × 3 = 15 m^{2}

and area of circle of radius 1 m = π (1)^{2}

= π m^{2}

[∵ diameter = 2m

⇒ radius = \(\frac{2}{2}\) = 1 m]

Now, probability that the die land inside the circle = \(\frac{\pi}{15}\)

Section – C

Section C consists of 6 questions of 3 marks each

Question 26.

In the given figure, AOB is a flower bed in the shape of a sector of a circle of radius 80 m and ∠AOB = 60°. Also, a 30 m wide concrete track is made as shown in the figure.

Two friends Vicky and Love went to the concrete track. Love said that area of the track is 10205 m^{2}. Is he right? Explain.

Solution:

Given, radius of circle = 80 m

and angle of sector = 60°

∴ Angles for the major sectors of both the circles at θ is same i.e. 300°.

Radius of inner circle = 80 – 30 = 50 m

∴ Area of concrete track = \(\frac{300^{\circ}}{360^{\circ}}\) × π × (80)^{2} – \(\frac{300^{\circ}}{360^{\circ}}\) × π × (50)^{2}

[∵ Area of track Area of outer sector – Area of inner sector]

= \(\frac{5}{6}\) × π × (6400 – 2500)

= \(\frac{5}{6}\) × 3.14 × 3900

= 10205 m^{2}

Yes, Love is right.

Question 27.

In the given figure, PP’ and QQ’ are the two common tangents of the two circles. Show that PP’ = QQ’.

Answer:

Produce PP’ and QQ’ to meet at point R (say)

[∵ R is an external point and lengths of tangents drawn from an external point to a circle are equal]

⇒ RP’ + P’P = RQ’ + Q’Q

⇒ P’P = Q’Q

[∵ RP’ = RQ’]

Question 28.

Prove that √3 is an irrational.

Answer:

Let √3 be rational in the simplest form of \(\frac{p}{q}\),

where p and q are integers and having no common factor otherthan 1 and q ≠ 0

i.e. √3 = \(\frac{p}{q}\)

On squaring both sides, we get

3 = \(\frac{p^2}{q^2}\)

⇒ 3q^{2} = p^{2} ……………..(i)

Since, 3q^{2} is divisible by 3.

∴ p^{2} is also divisible by 3.

⇒ p is divisible by 3 ……………..(ii)

Let p = 3c for some integer c.

On putting p = 3c in Eq. (i), we get

3q^{2} = (3c)^{2}

⇒ 3q^{2} = 9c^{2}

⇒ q^{2} = 3c^{2}

Since, 3c^{2} is divisible by 3.

∴ q^{2} is divisible by 3 ……………(iii)

From Eqs. (ii) and (iii), we get 3 is a common factor of p and q.

But this contradict our assumption that p and q are having no common factor other than 1.

Therefore, our assumption that √3 is rational, is wrong.

Hence, √3 is an irrational.

Hence proved.

Or

There is a circular path around a sports field, Sania takes 18 min to drive one round of the field, while Ravi takes 12 min for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Answer:

Time taken by Sania to drive one round of the field = 18 min

lime taken by Ravi to drive one round of the field = 12 min

The LCM of 18 and 12 gives the exact number of minutes after which they will meet at the starting point again.

Now, 18 = 2 × 3 × 3

= 2 × 3^{2}

and 12 = 2 × 2 × 3

= 2^{2} × 3

∴ LCM of 18 and 12 = 2^{2} × 3^{2}

= 4 × 9 = 36

Hence, Sania and Ravi will meet again at the starting point after 36 min.

Question 29.

In ∆ABC, ∠B = 90° and BD ⊥ AC. If AC = 10 cm and AD = 4 cm, then find the value of BD.

Answer:

Given, AC = 10 cm,

AD = 4 cm

CD = AC – AD

= 10 – 4 = 6 cm

In ∆ABC and ∆ADB

∠BAC = ∠BAD

∠ABC = ∠ADB [each 90° angle]

∴ ∆ABC ~ ∆ADB

[by AA similarity criterion] …………….. (i)

In ∆ABC and ∆BDC,

∠ABC = ∠BDC [each 90° angle]

∠ACB = ∠BCD

∴ ∆ABC ~ ∆BDC

[by AA similarity criterion] ……………. (ii)

From Eqs. (i) and (ii),

∆ADB ~ ∆BDC

∴ \(\frac{B D}{C D}=\frac{A D}{B D}\)

⇒ BD^{2} = AD . CD

= 4 × 6

= 24 cm

⇒ BD = 2√6 cm

Question 30.

Prove that sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ

Answer:

L.H.S. = sin θ (1 + tan θ) + cos θ (1 + cot θ)

Or

If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Answer:

Given, sin θ + cos θ = √3

⇒ (sin θ + cos θ)^{2} = (√3)^{2}

[squaring both sides]

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

⇒ 1 + 2 sin θ cos θ = 3

[∵ sin^{2} A + cos^{2} A = 1]

⇒ 2 sin θ cos θ = 2

⇒ sin θ cos θ = 1

⇒ sin θ cos θ = sin^{2} θ + cos^{2} θ

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

[dividing both sides by sin θ cos θ]

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

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

⇒ 1 = tan θ + cot θ

[∵ \(\frac{\sin A}{\cos A}\) = tan A,

\(\frac{\cos A}{\sin A}\) = cot A]

Hence proved.

Question 31.

Solve for x, \(2\left(\frac{2 x+3}{x-3}\right)-25\left(\frac{x-3}{2 x+3}\right)\) = 5.

Answer:

Given \(2\left(\frac{2 x+3}{x-3}\right)-25\left(\frac{x-3}{2 x+3}\right)\) = 5

Let \(\frac{2 x+3}{x-3}\) = y ……………(i)

Then, \(\frac{x-3}{2 x+3}=\frac{1}{y}\) = \(\frac{1}{y}\)

Therefore, the given equation reduces to

2y – 25 \(\frac{1}{y}\) = 5

⇒ 2y^{2} – 25 = 5y

⇒ 2y^{2} – 5y – 25 = 0

⇒ 2y^{2} – 10y + 5y – 25 = 0

⇒ 2y (y – 5) + 5 (y – 5) = 0

⇒ (y – 5) (2y + 5) = 0

⇒ y = 5 or y = \(-\frac{5}{2}\)

Now, on putting y = 5 in Eq. (i), we get

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

⇒ 5x – 15 = 2x + 3

⇒ 3x = 18

⇒ x = 6

Again, on putting y = – \(\frac{5}{2}\) in Eq. (i), we get

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

⇒ – 5x + 15 = 4x + 6

∴ 9x = 9

⇒ x = 1

Hence, the values of x are 1 and 6.

Section – D

Section D consists of 4 questions of 5 marks each

Question 32.

A tent is in the form of a cylinder on which a cone is surmounted. If height and diameter of cylindrical portion are 2.1 m and 4 m and slant height of cone is 2.8 m, then find the area of used canvas in making this tent. Also, find the volume of air within tent.

Answer:

Given, a tent which is combination of a cylinder and a cone.

Also, we have slant height (l) of the cone = 2.8 m

Radius of the cone, r = Radius of cylinder

= \(\frac{\text { Diameter }}{2}\)

= \(\frac{4}{2}\)

= 2 m

and height of the cylinder, h_{1} = 2.1 m

∴ Required surface area of the tent = Surface area of cone + Surface area of cylinder

= πrl + 2πrh_{1}

= πr (l + 2h_{1})

= \(\frac{22}{7}\) × 2 × (2.8 + 2 × 21)

= \(\frac{44}{7}\) (2.8 + 4.2)

= \(\frac{44}{7}\) × 7 = 44 m

Now, height of cone,

h_{2} = \(\sqrt{P C^2-O^{\prime} C^2}\)

= \(\sqrt{(2.8)^2-(2)^2}\)

= \(\sqrt{7.84-4}\)

= \(\sqrt{3.84}\)

= 1.96 m

Now, the volume of air within tent = Volume of tent

= Volume of cone + Volume of cylinder

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

= π [\(\frac{1}{3}\) × (2)^{2} × 1.96 + (2)^{2} × 2.1]

= \(\frac{22}{7}\left[\frac{7.84}{3}+8.4\right]\)

= \(\frac{22}{7}\) [2.61 + 8.4]

= \(\frac{11.01 \times 22}{7}=\frac{242.22}{7}\)

= 34.60 cm^{3}

Or

From a solid cylinder whose height is 12 cm and diameter is 10 cm, a conical cavity of same height and same diameter is hollowed out. Find the volume and total surface area of the remaining solid.

Answer:

Given, diameter of the cylinder = 10 cm

Radius of the cylinder r = \(\frac{10}{2}\) = 5 cm

and height of the cylinder, h = 12 cm

∴ Volume of the cylinder = πr²h

= \(\frac{22}{7}\) × 5 × 5 × 12

= \(\frac{6600}{7}\) cm^{3}

and voiume of the cone = \(\frac{1}{3}\) πr²h

= \(\frac{1}{3} \times \frac{22}{7}\) × 5 × 5 × 12

= \(\frac{2200}{7}\) cm^{3}

Now, volume of remaining solid = Volume of the cylinder – Volume of the cone

= \(\frac{6600}{7}-\frac{2200}{7}=\frac{4400}{7}\)

= 628.57 cm^{3}

Since, slant height_of the cone,

l = \(\sqrt{r^2+h^2}\)

= \(\sqrt{(5)^2+(12)^2}\)

= \(\sqrt{169}=\)

= 13 cm

Curved surface area of the cone = πrl

= \(\frac{22}{7}\) × 5 × 13

= \(\frac{1430}{7}\) cm^{2}

Clearly, curved surface area of the cylinder = 2πrh

= 2 × \(\frac{22}{7}\) × 5 × 12

= \(\frac{2640}{7}\) cm^{2}

and area of upper base of the cylinder = πr²

= \(\frac{22}{7}\) × 5 × 5

= \(\frac{550}{7}\) cm^{2}

Now, total surface area of the remaining solid = Curved surface area of the cylinder + Curved surface area of the cone + Area of upper base of the cylinder

= \(\frac{2640}{7}+\frac{1430}{7}+\frac{550}{7}\)

= \(\frac{4620}{7}\)

= 660 cm^{2}

Question 33.

From the top of a 7 m high building the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45° . Determine thc height of the tower.

Answer:

Let AD = 7m be the height of the building

and BC = h be the height of the cable tower.

From the top of the building D, the angles of elevation and depression are ∠CDE = 60° and ∠EDB = 45°, respectively.

From the point D, draw a line DE || AB,

Then, ∠EDB = ∠48° = 45° (alternate angles)

Also, let AB = DE = x be the distance between building and tower.

In right angled ∆BAD,

tan 45° = \(\frac{P}{B}=\frac{A D}{A B}\)

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

[∵ tan 45° = 1]

⇒ x = 7 m

and in right angled ∆CED,

tan 60° = \(\frac{C E}{D E}\)

= \(\frac{C B-B E}{A B}\)

[∵ CE = CB – BE]

⇒ √3 = \(\frac{h-7}{x}\)

[∵ tan 60° = √3]

⇒ h – 7 = x√3

⇒ h = x√3 + 7

⇒ h = 7√3 + 7 [from Eq.(i)]

⇒ h = 7(√3 + 1) m

Hence, the height of the tower is 7 (√3 + 1) m.

Question 34.

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital.

Find the modal age.

Answer:

As, the distribution is discontinuous, so firstly we will convert it into a continuous distribution by using adjustment factor of \(\frac{15-14}{2}\) = 0.5,

hence we get the following table

Age (in years) | Number of cases |

4.5 – 14.5 | 6 |

14.5 – 24.5 | 11 |

24.5- 34.5 | 21 |

34.5 – 44.5 | 23 |

44.5 – 54.5 | 14 |

54.5 – 64.5 | 5 |

Total | 80 |

As, the class 34.5 – 44.5 has maximum frequency, so it is the modal class.

Here, l = 34.5

f_{1} = 23

f_{2} = 14,

f_{0} = 2l

and h = 10

∴ Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h

= 34.5 + \(\left(\frac{23-21}{2 \times 23-21-14}\right)\) × 10

= 34.5 + \(\frac{2}{46-35}\) × 10

= 34.5 + \(\frac{20}{11}\)

= 34.5 + 1.82

= 36.32 (approx)

Hence, the modal age is 36.32 yr.

Or

If the median of the distribution given below is 30, then find the values of x and y.

Answer:

Cumulative frequency table

Class interval | Frequency (f) | Cumulative frequency (cf) |

0 – 10 | 5 | 5 |

10 – 20 | x | 5 + x (cf) |

20 – 30 | 20(f) | 25 + x |

30 – 40 | 15 | 40 + x |

40 – 50 | y | 40 + x + y |

50 – 60 | 5 | 45 + x + y |

Total | Σf_{i} = 60 |

∵ Median = 30

So, it lies in the interval 20 – 30.

Thus, 20 – 30 is the median class.

∴ l = 20,

h = 10,

f = 20,

cf = 5 + x

and N = 60

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

30 = 20 + \(\frac{\frac{60}{2}-(5+x)}{20}\) × 10

⇒ 10 = \(\frac{30-5-x}{2}\)

⇒ 20 = 25 – x

⇒ x = 5

Also, we have

60 = 45 + x + y

⇒ 15 = x + y

⇒ 15 = 5 + y

⇒ y = 10

Hence, the values of x and y are 5 and 10.

Question 35.

In the given figure, AD is the bisector of ∠BAC of AABC.

Prove that \(\frac{B D}{D C}=\frac{A B}{A C}\).

Answer:

Given : In the given figure, AD is the bisector of ∠BAC of AABC.

∴ ∠BAC = ∠ CAD

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

Construction : Now, draw CT || DA to meet BA produced at A.

Proof CT|| DA

and transversal AC intersects them.

∴ ∠1 = ∠2 [alternate interior angles] ………….. (i)

Also, CT || DA

and transversal BT intersects them.

∴ ∠3 = ∠4 [corresponding angles] ……………. (ii)

But ∠1 = ∠3 [given]

⇒ ∠1 = ∠4 [from Eq. (ii)] …………… (iii)

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

∠2 = ∠4

⇒ AT = AC

[since, sides opposite to equal angles of a triangle are also equal] ……………… (iv)

In ∆BCT,

AD || CT

∴ \(\frac{B D}{D C}=\frac{A B}{A T}\) [by BPT)

⇒ \(\frac{B D}{D C}=\frac{A B}{A C}\) [from Eq. (iv)]

DC Hence proved.

Section – E

Case study based questions are compulsory

Question 36.

Vikas is working with TCS and he is sincere and dedicated to his work. He pay all his taxes on time and invest the some amount of his salary in funds for his future.

He invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. He received yearly interest of ₹ 130. But, if he interchange the amounts invested, he would have received ₹ 4 more as interest.

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

(i) Consider the amount invested at 12% be p and at 10% be q. Then, formulate the required linear equation for first condition.

Answer:

Vikas received ₹ 130 as profit.

∴ According to the situation,

\(\frac{12}{100}\) p + \(\frac{10}{100}\) q = 130

⇒ 12p + 10q = 13000

(ii) Formulate the linear equation for the second condition?

Answer:

Vikas received ₹ 4 extra, if he interchange the investment amount,

∴ According to the situation,

\(\frac{10}{100}\) p + \(\frac{12}{100}\) q = 130 + 4

⇒ 10p + 12q = 13400

(iii) Find the value of p and q.

Answer:

From above situations, we have

12p + 10q = 13000 ………….(i)

and 10p + 12p = 13400 ……………..(ii)

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

22p + 22q = 26400

⇒ p + q = 1200 ………….(iii)

Now, subtracting Eq. (ii) from Eq. (i), we get

2p – 2q = – 400

⇒ p – q = – 200 …………..(iv)

Now, on adding Eqs. (iii) and (iv), we get

2p = 1000

⇒ p = ₹ 500

On putting the value of p in Eq. (iii)

q = 1200 – 500 = ₹700

Or

Solve the following pair of equations

2x + y = 8

3x + y = 20

Answer:

We have, 2x + y = 8 …………(i)

3x + y = 20 ………………..(ii)

⇒ y = 20 – 3x

On substituting the value of y in Eq. (i), we get

2x + 20 – 3x = 8

⇒ – x = 8 – 20 = – 12

⇒ x = 12

On putting the value of x in Eq. (i), we get

2 × 12 + y = 8

⇒ y = 8 – 24

⇒ y = – 16

Question 37.

A teacher of class X draw the three graphs of a quadratic polynomials on the board as shown below.

Answer the following questions which are based on the above graphs.

(i) Write the zeroes of polynomial P(x) in graph I.

Answer:

From the graph I, we have

Zeroes of P(x) = – 3, – 1

(ii) Write the zeroes of polynomial P(x) in graph II.

Answer:

From the graph II, we have

Zeroes of P(x) = – 2, 2

(iii) Write the polynomial P(x) for graph (I).

Answer:

In graph I, sum of zeroes = – 3 -1 = – 4

and product of zeroes = (- 3) (- 1) = 3

∴ Polynomial P(x) will be x^{2} – (Sum of zeroes) x + Product of zeroes = x^{2} + 4x + 3

Or

Find the value of (a + b) for which p(x) = x2 + ax – b in graph III. (2)

Answer:

From the graph III, we have

Zeroes of p(x) = – 1, 3

Sum of zeroes = 2 = \(\frac{-a}{1}\)

⇒ a = – 2

and product of zeroes = – 3 = – b

⇒ b = 3

∴ a + b = – 2 + 3 = 1

Question 38.

Walking is a good habit for human beings to improve health and stamina. In this order Ayush starts walking from his house to office. Instead of going to the office directly he goes to bank first, from there he leaves his daughter to school and then reaches the office, [assume that all distances covered are in straight lines] if the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and the coordinates are in kilometre.

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

(i) Find the distance between bank and office.

Answer:

Required distance = \(\sqrt{(13-5)^2+(26-8)^2}\)

= \(\sqrt{8^2+18^2}\)

= \(\sqrt{64+324}\)

= \(\sqrt{388}\)

= 2\(\sqrt{97}\) km

(ii) Find the shortest distance between house and office.

Answer:

Required distance = \(\sqrt{(13-2)^2+(26-4)^2}\)

= \(\sqrt{11^2+22^2}\)

= \(\sqrt{121+484}\)

= \(\sqrt{605}\)

= 24.6 km

(iii) Find the extra distance travelled by Ayush in reaching his office.

Answer:

Extra distance = Distance from house to bank + Distance from bank to school + Distance from school to office – Distance from house to office

= \(\sqrt{(5-2)^2+(8-4)^2}\) + \(\sqrt{(13-5)^2+(14-8)^2}\) + \(\sqrt{(13-13)^2+(26-14)^2}\) – \(\sqrt{(13-2)^2+(26-4)^2}\)

= \(\sqrt{9+16}+\sqrt{64+36}\) + \(\sqrt{0+144}-\sqrt{121+484}\)

= \(\sqrt{25}+\sqrt{100}+\sqrt{144}-\sqrt{605}\)

= 5 + 10 + 12 – 24.6

= 27-24.6

= 2.4 km

Or

If on the shortest path from house to office, a point divides the path in the ratio 4 : 3. Find the coordinates of that point. (2)

Answer:

Let the coordinates of the required point be (x, y).

Then, by section formula,

(x, y) = \(\left(\frac{4 \times 13+3 \times 2}{4+3}, \frac{4 \times 26+3 \times 4}{4+3}\right)\)

= \(\left(\frac{52+6}{7}, \frac{104+12}{7}\right)\)

⇒ (x, y) = \(\left(\frac{58}{7}, \frac{116}{7}\right)\)

The coordinates of the point are \(\left(\frac{58}{7}, \frac{116}{7}\right)\).