Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 6 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Basic Set 6 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.

If the product of the zeroes of the polynomial mx^{2} – 6x – 6 is – 3, then the value of m is

(a) 6

(b) 2

(c) – 2

(d) – 3

Answer:

(b) 2

Let α and β be the zeroes of mx^{2} – 6x – 6.

Here, a = m,

b = – 6

and c = – 6

Given, αβ = – 3

∴ \(\frac{c}{a}\) = – 3

⇒ \(\frac{-6}{m}\) = – 3

⇒ m = 2.

Question 2.

Which term of the AP 3, 15, 27, 39, ……………… will be 120 more than 21st term?

(a) 20

(b) 37

(c) 31

(d) 32 3

Answer:

(c) 31

Here, a = 3

and d = (15 – 3) = 12.

∴ 21st term is given by

T_{21} = a + (21 – 1) d

= a + 20d

= (3 + 20 × 12) = 243

∴ Required term = (243 + 120) = 363

Let it be nth term.

Then,

T_{n} = 363

⇒ a + (n – 1) d = 363

3 + (n – 1) × 12 = 363

12n = 372

⇒ n = 31

Hence, 31st term is the required term.

Question 3.

The value of cos^{2} θ + \(\frac{1}{1+\cot ^2 \theta}\)

(a) 2 sin^{2} θ

(b) 1

(c) 0

(d) cos^{2} θ

Answer:

(b) 1

We have,

cos^{2} θ +\(\frac{1}{1+\cot ^2 \theta}\)

[∵ 1 + cot^{2} A = cosec^{2} A]

= cos^{2} θ + sin^{2} θ

[∵ sin^{2} A + cos^{2} A = 1]

= 1

Question 4.

Which term of the AP 5, 9, 13, 17 ……………,… is 81?

(a) 80

(b) 10

(c) 40

(d) 20

Answer:

(d) 20

In the given AP, we have

First term (a) = 5

and common difference (d) = 9 – 5 = 4

Let its nth term be 81.

Then,

T_{n} = 81

⇒ a + (n – 1) d = 81

⇒ 5 + (n – 1) 4 = 81

⇒ 4n = 80

⇒ n = 20 [∵ a = 5]

Hence, the 20th term of the given AP is 81.

Question 5.

If a cylinder is covered by two hemispherical lid of equal shape, then the total curved surface area of the new object will be (where r is the radius and h is the height of cylinder)

(a) 4πrh + 2πr^{2}

(b) 4πrh – 2πr^{2}

(c) 2πrh + 4πr^{2}

(d) 2πrh + 4πr

Answer:

(c) 2πrh + 4πr^{2}

Total curved surface area = Curved surface area of cylinder + 2 × Curved surface area of hemispheres

= 2πrh + 2 × (2πr^{2})

= 2πrh + 4πr^{2}

Question 6.

If in ∆ABC and ∆DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\), then they will be similar, when

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

Answer:

(c) ∠B = ∠D

Given, in ∆ABC and ∆DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\)

∆ABC and ∆EDF will be similar, if

∠B = ∠D [by SAS similarity criterion]

Question 7.

Suppose mean of 10 observations is 12.5 and each observation is multiplied by 5, then what is the new mean?

(a) 50

(b) 62.5

(c) 60

(d) 48.2

Answer:

(b) 62.5

Let 10 observations be x_{1}, x_{2}, x_{3}, …………, x_{10}.

Given, \(\frac{x_1+x_2+x_3+\ldots+x_{10}}{10}\) = 12.5 ………(i)

Now, if each observation multiphed by 5, then new

mean = \(\frac{5 x_1+5 x_2+\ldots+5 x_{10}}{10}\)

= \(\frac{5\left(x_1+x_2+\ldots+x_{10}\right)}{10}\) [using Eq. (1)]

= 5 × 12.5

= 62.5

Question 8.

In ∆ABC, right angled at A and AB = 5, AC = 12 and BC = 13, then the value of sin B and tan B is

(a) \(\frac{12}{5} \text { and } \frac{5}{13}\)

(b) \(\frac{5}{13} \text { and } \frac{5}{12}\)

(c) \(\frac{12}{13} \text { and } \frac{12}{5}\)

(d) \(\frac{13}{5} \text { and } \frac{12}{13}\)

Answer:

(c) \(\frac{12}{13} \text { and } \frac{12}{5}\)

With reference to ∠B, we have

base = AB = 5,

perpendicular = AC = 12

and hypotenuse = BC = 13

∴ sin B = \(\frac{A C}{B C}=\frac{12}{13}\)

and tan B = \(\frac{A C}{A B}=\frac{12}{5}\)

Question 9.

The value of 3 sin 30° – 4 sin^{3} 60° is

(a) 3 – √3

(b) \(\frac{3(1-\sqrt{3})}{2}\)

(c) \(\frac{3 \sqrt{3}-1}{2}\)

(d) – \(\frac{3}{2}\)

Answer:

(b) \(\frac{3(1-\sqrt{3})}{2}\)

3 sin 30° – 4 sin^{3} 60°

= 3 × \(\frac{1}{2}\) – 4 \(\left(\frac{\sqrt{3}}{2}\right)^3\)

= \(\frac{3}{2}-4 \times \frac{3 \sqrt{3}}{8}\)

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

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

= \(\frac{3(1-\sqrt{3})}{2}\)

Question 10.

The solution of 2x^{2} – 5x – 3 = 0 is

(a) – \(\frac{1}{2}\), 3

(b) 1, – 3

(c) – 1, 3

(d) – 1, \(\frac{1}{2}\)

Answer:

(a) – \(\frac{1}{2}\), 3

Given, 2x^{2} – 5x – 3 = 0

By splitting the middle tem, we get

2x^{2} – 6x + x – 3 = 0

⇒ 2x (x – 3) + 1 (x – 3) = 0

⇒ (x – 3) (2x + 1) = 0

⇒ x = \(\frac{1}{2}\), 3

Question 11.

If the lines given by 4x + ky = 1 and 6x – 10y = 14 has unique solution, then the value of k is

(a) \(\frac{20}{3}\)

(b) – \(\frac{5}{7}\)

(c) – 15

(d) all real values except – \(\frac{20}{3}\)

Answer:

(d) all real values except – \(\frac{20}{3}\)

The given equations can be re-written as

4x + ky – 1= 0

and 6x – 10y – 14 = 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} = 4,

b_{1} = k,

c_{1} = – 1

and a_{2} = 6,

b_{2} = – 10,

c_{2} = – 14

For unique solution,

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

⇒ \(\frac{4}{6} \neq \frac{k}{-10}\)

⇒ k ≠ \(-\frac{20}{3}\)

Thus, given lines have a unique solution for all real values of k, except \(-\frac{20}{3}\).

Question 12.

The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm, is

(a) 16.5 cm^{2}

(b) 17.5 cm^{2}

(c) 19.62 cm^{2}

(d) 8.75 cm^{2}

Answer:

(d) 8.75 cm^{2}

We know that area of sector A of radius r and length of arc l is given by

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

A = \(\frac{1}{2}\) × 3.5 × 5

= 8.75 cm^{2}

Question 13.

The distance between the points A (7, 13) and B (10, 9) is

(a) 5 units

(b) 6 units

(c) 9 units

(d) 10 units

Answer:

(a) 5 units

The given points are A (7, 13) and B (10, 9)

Then, x_{1}= 7,

y_{1} = 13

and x_{2} = 10,

y_{2} = 9

∴ AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

= \(\sqrt{(10-7)^2+(9-13)^2}\)

= \(\sqrt{3^2+(-4)^2}\)

= \(\sqrt{9+16}\)

= \( \sqrt{25}\)

= 5 units

Question 14.

A die is thrown once. The probability of getting a number greater than 4 is

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

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

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

(d) 1

Answer:

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

On a die, there are six numbers 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Number on dice which are greater than 4 = 5, 6

Favourable number of elementary events = 2

Required probability = \(\frac{2}{6}=\frac{1}{3}\).

Question 15.

The product of the HCF and LCM of two prime numbers a and b is

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

(b) a – b

(c) a + b

(d) a × b

Answer:

(d) a × b

HCF (a, b) = 1

LCM (a, b) = ab

∴ HCF (a,b) × LCM(a, b) = 1 × ab = ab

Question 16.

The probability of an impossible event is

(a) 0

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

(c) 1

(d) None of these

Answer:

(a) 0

The probability of an impossible event is 0.

Question 17.

If median = 143 and mean = 143.06, then the mode is

(a) 143.18

(b) 142.94

(c) 142.88

(d) 143

Answer:

(c) 142.88

We know that

Mode = 3 Median – 2 Mean

= 3 (143) – 2(143.06)

= 429 – 286.12

= 142.88.

Question 18.

If radius of circle is 8 cm and tangent is drawn from an external point to the circle is 15 cm, then the distance from centre of circle to the external point is

(a) \(\sqrt{241}\) cm

(b) 17 cm

(c) 10 cm

(d) None of these

Answer:

(b) 17 cm

In right angled ∆OPQ,

using Pythagoras theorem

OP = \(\sqrt{O Q^2+Q P^2}\)

= \(\sqrt{8^2+15^2}\)

= \(\sqrt{64+225}\)

= \(\sqrt{289}\)

= 17 cm

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

Question 19.

Assertion (A) : The value of sin 60° cos 30° + sin30° cos60° is 1.

Reason (R) : sin 90° = 1 and cos 90° = 0

(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).

Assertion

sin 60° cos 30° + sin 30° cos 60°

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

= \(\frac{3}{4}+\frac{1}{4}\)

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

So, Assertion (A) is true.

Reason

We know, sin 90° = 1

and cos 90° = 0

So, Reason (R) is true.

But Reason (R) is not the correct explanation of Assertion (A).

Question 20.

Assertion (A) : The system of equations 2x + 3y + 5 = 0 and 4x + ky + 7 = 0 is inconsistent when k = 6.

Reason (R) : The system of equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 is inconsistent when \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

(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:

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

Assertion

Here, a_{1} = 2,

b_{1} = 3,

c_{1} = 5

and a_{2} = 4,

b_{2} = 6,

c_{2} = 7

[∵ k = 6]

So, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

[∵ \(\frac{2}{4}=\frac{3}{6} \neq \frac{5}{7}\)]

So, the given system of equations has no solution (i.e. inconsistent).

So, the Assertion (A) is true.

Reason:

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

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

We know, for the system of equations to be inconsistent,

\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

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

Section – B

Section B consists of 5 questions of 2 marks each

Question 21.

In the given figure, a circle is inscribed in a ∆PQR. If PQ = 10 cm, QR = 8 cm and PR = 12 cm, then find the lengths of QM,RN and PL.

Answer:

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

Let PL = PN = x;

QL = QM = y

and RM = RN = z

Now, PL + QL = PQ

⇒ x + y = 10 ………….. (i)

QM + RM = QR

⇒ y + z = 8 …………….(ii)

RN + PN = PR

⇒ 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

On substituting y = 3m Eq. (ii), we get

z = 5

∴ QM = y = 3 cm

RN = z = 5 cm

and PL = x = 7 cm.

Question 22.

What is the probability that a number selected from the numbers 1,2, 3, …………., 25 is a prime number, when each of the given numbers is equally likely to be selected?

Answer:

Out of 25 numbers, 1, 2,3, ………….., 25 one number can be chosen in 25 ways.

∴ Total number of elementary events = 25

The number selected will be a prime number, if it is chosen from the numbers

2, 3, 5, 7, 11, 13, 17, 19, 23.

Favourable number of elementary events = 9

Hence, required probability = \(\frac{9}{25}\)

Or

A letter is chosen at random from the letters of word ‘ASSASSINATION’. Find the probability that the letter chosen is a

(i) vowel

(ii) consonant.

Answer:

There are 13 letters in the word ‘ASSASSINATION’ out of which one letter can be chosen in 13 ways.

Total number of elementary events = 13

(i) There are 6 vowels in the word ‘ASSASSINATION’.

So, there are 6 ways of selecting a vowel.

∴ Probability of selecting a vowel = \(\frac{6}{13}\)

(ii) We have, probability of selecting a consonant = 1 – Probabihty of selecting a vowel

=1 – \(\frac{6}{13}\) = \(\frac{7}{13}\)

Question 23.

Find the sum of first 15 even natural numbers.

Answer:

The sequence goes like this 2, 4, 6, 8, ……………..

Here, 4 – 2 = 6 – 4

= 8 – 6 = 2

So, it is an AP with first term, a = 2,

common difference, d = 4 – 2 = 2

and total number of terms, n = 15

∴ Sum of first 15 even natural numbers

S_{15} = \(\frac{n}{2}\) [2a + (n – 1) d]

= \(\frac{15}{2}\) [2 × 2 + (15 – 1) 2]

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

= \(\frac{15}{2}\) [4 + 28]

= \(\frac{15}{2}\) × 32 = 240

Question 24.

If one root of the quadratic equation 2x^{2} + kx – 6 = 0 is 2, then find the value of k. Also, find the other root.

Answer:

Since, x = 2 is a root of the equation of 2x^{2} + kx – 6 = 0

∴ 2×2^{2} + 2k – 6 = 0

⇒ 8 + 2k – 6 = 0

⇒ 2k + 2 = 0

⇒ k = – 1

On putting k = – 1 in the equation 2x^{2} + kx – 6 = 0, we get

2x^{2} – x – 6 = 0

⇒ 2x^{2} – 4x + 3x – 6 = 0

⇒ 2x (x – 2) + 3 (x – 2) = 0

⇒ (x – 2) (2x + 3) = 0

⇒ x – 2 = 0 or 2x + 3 = 0

⇒ x = 2 or – \(\frac{3}{2}\)

Hence, the other root is – \(\frac{3}{2}\).

Question 25.

Prove that cot A + tan A = sec A cosec A.

Answer:

LHS = cot A + tan A

Or

Prove that \(\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\ {cosec} A-1}{\ {cosec} A+1}\).

Answer:

LHS = \(\frac{\cot A-\cos A}{\cot A+\cos A}\)

= \(\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)

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

= \(\frac{\sin A(\ {cosec} A-1)}{\sin A(\ {cosec} A+1)}\)

= \(\frac{\ {cosec} A-1}{\ {cosec} A+1}\)

Hence proved.

Section – C

Section C consists of 6 questions of 3 marks each

Question 26.

The Resident Welfare Association of a colony decided to build two straight paths in their neighbourhood park such that they do not cross each other, to plant trees along the boundary lines of each path. One of the members of association, Sarika suggested that the paths should be constructed represented by the two linear equations x – 3 y = 2 and – 2 x + 6 y = 5. Check whether the two paths will cross each other or not.

Answer:

Given, linear equations are

x – 3y = 2

⇒ x – 3y – 2 = 0 …………..(i)

and – 2x + 6y = 5

⇒ – 2x + 6y – 5= 0 ……….(ii)

Here, a_{1} = 1,

b_{1} = – 3

c_{1} = – 2

and a_{2} = – 2,

b_{2} =6

c_{2} = – 5

Now \(\frac{a_1}{a_2}=\frac{1}{-2}=-\frac{1}{2}\)

and \(\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

So, the paths represented by the equations are parallel i.e. not cross (intersect) each other.

Question 27.

A coircle is inscribed in a ∆ABC having sides 8 cm, 10 cm and 12 cm as shown in figure. Find AD, BE and CF.

Answer:

Given, a circle is inscribed in the triangle, wtose sides are

BC = 8 cm,

AC = 10 cm

and AB = 12 cm.

Let AD = AF = x,

BD = BE = y

and CE = CF = z

[∵ the length of two tangents drawn from an external point to a circle are equal)

We have, AB = 12

⇒ AD + DB = 12

⇒ x + y = 12 …………(i)

AC = 10

⇒ AF + FC = 10

⇒ x + z = 10

and BC = 8

⇒ CE + EB = 8

⇒ z + y = 8 ………….(iii)

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

2 (x + y + z) = 12 + 10 + 8

⇒ x + y + z = \(\frac{30}{2}\) = 15 ………….(iv)

On putting x + y = 12 from Eq. (i) in Eq. (iv), we get

12 + z = 15

⇒ z = 3

On putting z + y = 8 from Eq. (iii) in Eq. (iv), we get

x + 8 = 15

⇒ x = 7

On putting x + z =10 from Eq. (ii) in Eq. (iv). we get

10 + y = 15

⇒ y = 5

Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.

Question 28.

In a ∆ABC, let D be a point on BC such that \(\frac{B D}{D C}=\frac{A B}{A C}\). Prove that AD is the bisector of ∠A.

Answer:

Given : A ∆ABC in which D is mid-point on BC such that

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

Construction : Produce BA to E such that AE = AC.

Join EC.

Proof : \(\frac{B D}{D C}=\frac{A B}{A C}\) [givven]

⇒ \(\frac{B D}{D C}=\frac{A B}{A E}\) [∵ AC = CE]

⇒ DA || CE

[by the converse of Thales theorem]

∠2 = ∠3 …………..(i)

[alternate inteiior angles]

∠1 = ∠4 …………..(ii)

Also AE = AC

⇒ ∠3 = ∠4 ………….(iii)

∠1 = ∠2 [from Eqs. (i), (ii) and (iii)]

Hence, AD is the bisector of ∠A.

Hence proved.

Or

If the bisector of an angle of a triangle bisect the opposite side, then prove that the triangle is isosceles.

Answer:

Given : A ∆ABC in which D is mid-point on BC such that BD = DC

Proof :

Since, AD is the bisector of ∠A by the Angle Bisector theorem, we have

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

[∵ BD = DC]

⇒ \(\frac{A B}{A C}\) = 1

⇒ AB = AC

Hence, ∆ABC is an isosceles triangle.

Hence proved.

Question 29.

A tower is 50 m high. Its shadow is x m shorter when the Sun’s altitude is 45°, then when it is 30°. Find the value of x.

Answer:

Let AB = x m

From right angted ∆APQ, we have

\(\frac{A P}{P Q}\) = cot 45°

⇒ \(\frac{A P}{50}\) = 1

⇒ AP = 50

From right angled ∆BPQ, we have

\(\frac{b P}{P Q}\) = cot 30°

\(\frac{x+50}{50}\) = √3

[∵ BP = AP + AB]

⇒ x = 50 (√3 – 1)

⇒ x = 50 (1.732 – 1)

∴ x = 36.6

Question 30.

Find the roots of the following equation \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ – 4, 7.

Answer:

Given equation is \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ – 4, 7

⇒ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)

⇒ \(\frac{x-7-x-4}{x^2-7 x+4 x-28}=\frac{11}{30}\)

⇒ \(\frac{-11}{x^2-3 x-28}=\frac{11}{30}\)

⇒ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)

⇒ – 30 = x^{2} – 3x – 28

⇒ x^{2} – 3x + 2 = 0

On comparing with the standard quadratic equation

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

a = 1, b = – 3 and c = 2

By using quadratic formula, we get

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

= \(\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(2)}}{2 \times 1}\)

= \(\frac{3 \pm \sqrt{9-8}}{2}\)

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

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

⇒ x = \(\frac{4}{2}\) or x = \(\frac{2}{2}\)

∴ x = 2 or x = 1

Or

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

Answer:

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

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

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 [byfactorisation method]

⇒ 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. (j), we get

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

⇒ 5 x -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.

Question 31.

In the given figure, two chords AB and CD intersect each other at the point P. Prove that

(i) ∆APC ~ ∆DPB

(ii) AP . PB = CP . DP

Answer:

Given, In figure, two chords AB and CD intersect each other at point P.

To prove :

(i) ∆APC ~ ∆DPB

(ii) AP . PB = CP DP

Proof:

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB [vertically opposite angles]

and ∠CAP = ∠BDP [angles in the same segment]

∴ ∆APC ~ ∆DPB [by AA similarity criterion]

(ii) We have,

∆APC ~ ∆DPB [proved in part (i)]

∴ \(\frac{A P}{D P}=\frac{C P}{B P}\)

[∵ if two triangles are similar, then the ratio of their corresponding sides are equal]

∴ AP . BP = CP . DP

or AP . PB = CP . OP

Hence proved.

Section – D

Section D consists of 4 questions of 5 marks each

Question 32.

Prove that, if a, b, c and d are positive rationals such that, a + √b = c + √d, then either a = c and b = d or b and d are squares of rationals.

Answer:

Given, a + √b = c + √d

If a = c, then a + √b = a + √d

⇒ √b = √d

⇒ b = d [squaring on both sides]

If a ≠ C, then there exists a positive rational number x such that,

a = c + x

Now, a + √b = c + √d

⇒ c + x + √d = c + √d [∵ a = c + x]

⇒ x + √b = √d ……………(i)

⇒ (x + √b)^{2} = (√d)^{2} [squaring on both sides]

= x^{2} + b + 2x√b = d

[∵ (A + B)^{2} = A^{2} + B^{2} + 2AB]

= x^{2} + 2x√b + b – d = 0

2x√b = d – b – x^{2}

∴ √b = \(\frac{d-b-x^2}{2 x}\)

Since, d, x and b are rational numbers and x > 0.

So, \(\frac{d-b-x^2}{2 x}\) is a rational.

Then, √b is a rational number.

Hence, b is the square of a rational number.

From Eq. (i), we get

√d = x + √b

Also, √d is a rational.

So, d is the square of a rational number.

Hence, either a = c and b = d or b and d are the squares of rationals.

Or

In a seminar, the number of participants, in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the number of rooms required, if in each ‘room the same number of participants are to be seated and all of them bring the same subject.

Answer:

The number of participants ¡n each room must be the HCF of 60, 84 and 108.

Now, prîme factors of numbers 60, 84 and 108 are

60 = 2^{2} × 3 × 5,

84 = 2^{2} × 3 × 7

and 108 = 2^{2} × 3^{3}

HCF of (60, 84, 108) = 2^{2} × 3 = 12

Therefore, in each room maximum 12 participants can be seated.

∴ Total number of participants = 60 + 84 + 108 = 252

∴ Numberof rooms required = \(\frac{252}{12}\) = 21

Question 33.

The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Answer:

Table for cumulative frequency is given below

Class interval | Frequency | Cumulative frequency |

0 – 6 | 4 | 4 + 0 = 4 |

6 – 12 | x | 4 + x = (4 + x) (cf) |

12 – 18 | 5(f) | 5 +(4 + x) = 9 + x |

18 – 24 | y | y + (9 + x) = 9 + x + y |

24 – 30 | 1 | 1 + (9 + x + y) = 10 + x + y |

Since, N = 20

∴ 10 + x + y = 20

x + y = 20 – 10

x + y = 10

Also, we have, median = 14.4 …………..(i)

which lies in the class interval 12 – 18.

∴ The median class is 12 – 18, such that

l = 12,

f = cf = 4 + x

and h = 6

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

⇒ 14.4 = 12 + \(\left[\frac{10-(4+x)}{5}\right]\) × 6

⇒ 144 – 12 = \(\frac{6-x}{5}\) × 6

⇒ 2.4 = \(\frac{36-6 x}{5}\)

⇒ 12 = 36 – 6x

⇒ 6x = 24

⇒ x = 4

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

4 + y = 10

⇒ y = 10 – 4 = 6

Thus, x = 4 and y = 6.

Question 34.

Find the length of the median drawn through A on BC of a ∆ABC, whose vertices are A (7, – 3), B (5, 3) and C (3, – 1) and also find the distance of the point A (7, – 3) from the origin.

Answer:

The median from a vertex of a triangle bisects the opposite side, to that vertex.

So, let AD be the median through A then D be the mid-point of the side BC.

Now, coordinates of D = \(\left(\frac{5+3}{2}, \frac{3-1}{2}\right)\)

= (4, 1)

[∵ coordinates of mid – point of line segment joining (x1, Yi) and (x2. Y2) = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)]

and length of median AD is given by

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

= \(\sqrt{(4-7)^2+(1+3)^2}\)

= \(\sqrt{(-3)^2+(4)^2}\)

= \(\sqrt{9+16}\)

= \(\sqrt{25}\)

= 5 units

Also. OA = \(\sqrt{(0-7)^2+(0+3)^2}\)

= \(\sqrt{49+9}\)

= \(\sqrt{58}\) units

Or

Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Answer:

Let C (x, y) be the centre of the circle passing through the points P (6, – 6),Q (3, – 7) and R(3, 3)

Then, PC = QC = CR [radii of circle]

Now, PC = OC

⇒ PC^{2} = QC^{2} [squaring on both sides]

⇒ (x – 6)^{2} + (y + 6)^{2} = (x – 3)^{2} + (y + 7)^{2}

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

⇒ x^{2} – 12x + 36 + y^{2} + 12y + 36 = x^{2} – 6x + 9 + y^{2} + 14y + 49

[‘.. (a – b)^{2} = a^{2} + b^{2} – 2ab]

⇒ – 12x + 6x + 12y – 14y + 72 – 58 = 0

⇒ – 6x – 2y + 14 = 0

⇒ 3x + y – 7 = 0 [dividing by – 2] …………. (i)

and QC = CR

QC^{2} = CR^{2} [squaring on both sides]

⇒ (x – 3)^{2} + (y + 7)^{2} = (x – 3)^{2} + (y – 3)^{2}

⇒ (y + 7)^{2} = (y – 3)^{2}

⇒ y^{2} + 14y + 49 = y^{2} – 6y + 9

⇒ 20y + 40 = 0

⇒ y = – \(\frac{40}{20}\) = – 2

On putting y = – 2 in Eq. (i), we get

⇒ 3x – 2 – 7 = 0

⇒ 3x = 9

⇒ x = 3

Hence, the centre of circle is (3 – 2).

Question 35.

A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

Answer:

Given, diameter of hemispherical bowl = 36 cm

Radius (r) = 18 cm

Volume of liquid in the bowl = \(\frac{2}{3}\) πr^{3}

= \(\frac{2}{3}\) × π × 18 × 18 × 18

= 3888π cm^{3}

Amount of the liquid wasted = 3888π × \(\frac{10}{100}\)

= \(\frac{3888 \pi}{10}\) cm^{3}

Liquid transferred into the bottles = 3888π – \(\frac{3888 \pi}{10}\) cm^{3}

= \(\frac{34992 \pi}{10}\) cm^{3} ………………(i)

Also, given diameter of bottle = 6 cm

⇒ Radius(r) = 3 cm

Let h be the height of the bottle,

Volume of bottle = πr²h

= π × 3 × 3 × h

= 9πh cm^{3}

Volume of 72 such bottles = 72 × 9πh cm^{3}

= 648πh cm^{3}

Now, volume of 72 bottles = Volume of liquid transferred into the bottles

⇒ 648πh = \(\frac{34992 \pi}{10}\) [using Eq. (i)]

⇒ h = 5.4 cm

∴ Height of each bottle is 5.4 cm.

Section – E

Case study based questions are compulsory

Question 36.

Roller Coaster Polynomials

Polynomials are everywhere. They play a key role in the study of algebra in analysis and on the whole many mathematical problems involving them.

Since, polynomial are used to describe curves of various types engineers use polynomials to graph the curves on roller coasters.

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

(i) If the roller coaster is represented by the following graph y = p (x), then name the type of the polynomial it traces.

Answer:

Since, graph of given polynomial intersect X-axis at 3 points.

So, it has three zeroes.

Hence, it is a cubic polynomial.

(ii) If p(x) = 2x^{3} – 5x^{2} – 14x + 8, then find the sum and product of zeroes.

Answer:

On comparing given polynomial 2x^{3} – 5x^{2} – 14x + B with ax^{3} + bx^{2} + cx + d, we get

a = 2, b = – 5, c = – 14 and d = 8

Sum of roots, α + β + γ = \(\frac{-b}{a}\)

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

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

Product of roots, αβγ = \(\frac{-d}{a}\)

= \( \frac{-8}{2}\)

= – 4

Or

Find a quadratic polynomial, the sum and product of whose roots are – 3 and 2, respectively.

Answer:

Sum of roots, α + β = – 3

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

and product of roots. αβ = 2

= \(\frac{c}{a}\)

∴ Quadratic polynomial is x^{2} + 3x +2.

(iii) In the graph shown above, find the zeroes of y = p (x).

Answer:

The graph of given polynomial intersect X-axis at points – 4, – 1 and 2.

So, zeroes of given polynomial are – 4, – 1 and 2.

Question 37.

Ritu is studying in X standard. She observes two poles AB and DC and the heights of these poles are a m and b m respectively, shown as figure below

These poles are p m apart and O is the point of intersection of the lines joining the top of each pole to the foot of opposite pole and the distance between point O and L is h m. Few questions came to his mind while observing the poles. Give answer to his questions by looking at figure.

(i) If CL = x, then find x in terms of a, b and h.

Answer:

In ∆ABC and ∆LOC, we have

∠CAB = ∠CLO = 90° [common]

∴ ∆CAB ~ ∆CLO

∴ \(\frac{C A}{C L}=\frac{A B}{L O}\) [by AA similarity criterion]

⇒ \(\frac{p}{x}=\frac{a}{h}\)

⇒ x = \(\frac{p h}{a}\) ………….(i)

(ii) If AL = y, then find y in terms of a, b and h.3

Answer:

In ∆ALO and ∆ACD, we have

∠ALO = ∠ACD = 90°

∴ ∠A = ∠A [common]

∠ ∆ALO ~ ∆ACD [by AA similarity criterion]

⇒ \(\frac{A L}{A C}=\frac{O L}{D C}\)

⇒ \(\frac{y}{p}=\frac{h}{b}\)

⇒ y = \(\frac{p h}{b}\) ………..(ii)

(iii) Find h in terms of a and b.

Answer:

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

x + y = \(\frac{p h}{a}\) + \(\frac{p h}{b}\)

= ph \(\left(\frac{1}{a}+\frac{1}{b}\right)\)

⇒ p = ph \(\left(\frac{a+b}{a b}\right)\)

[∵ x + y = CL + LA = p]

⇒ h = \(\frac{a b}{a+b}\) m

Or

If a = 5 m and b = 10 m, then find h.

Answer:

If a = 5 m

and b = 10 m then

h = \(\frac{a b}{a+b}\)

= \(\frac{5 \times 10}{5+10}\)

= \(\frac{50}{15}=\frac{10}{3}\) m.

Question 38.

Sara is studying in X standard. She is making a figure to understand trigonometric ratio as shown below

Given answers to her questions by looking at the figure.

(i) What is the length of AS?

Answer:

We have, AR = AB – RB

= 18 – 6

= 12 cm

In ∆ARS by using Pythagoras theorem

AS^{2} = AR^{2} + RS^{2}

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

= 144 + 25 = 169

⇒ AS = \(\sqrt{169}\) = 13 cm

(ii) Find tan x.

Answer:

In ∆BRS,

tan x = \(\frac{\text { Perpendicular }}{\text { Base }}\)

= \(\frac{B R}{R S}\)

= \(\frac{6}{5}\)

(iii) Find sec x.

Answer:

We know that

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

= \(\sqrt{1+\left(\frac{6}{5}\right)^2}\)

= \(\sqrt{1+\frac{36}{25}}\)

= \(\sqrt{\frac{61}{25}}=\frac{\sqrt{61}}{5}\)

Or

Find cosec x.

Answer:

We know that

cosec x = \(\sqrt{1+\cot ^2 x}\)

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

= \(\sqrt{1+\left(\frac{5}{6}\right)^2}\)

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

= \(\sqrt{\frac{61}{36}}=\frac{\sqrt{61}}{6}\)