# CBSE Class 10 Maths Question Paper 2023 (Series: WX1YZ/6) with Solutions

Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2023 (Series: WX1YZ/6) to familiarize themselves with the exam format and marking scheme.

## CBSE Class 10 Maths Question Paper 2023 (Series: WX1YZ/6) with Solutions

Time allowed: 3 hours

Maximum marks: 80

General Instructions:

Read the following instructions carefully and follow them:

- This question paper contains 38 questions. All questions are compulsory.
- Question paper is divided into FIVE Sections – Section A, B, C, D and E.
- In Section A – Question no. 1 to 18 are Multiple Choice Questions (MCQ) and Question number 19 and 20 are Assertion-Reason based questions of 1 mark each.
- In Section B – Question number 21 to 25 are Very Short Answer (VSA) type questions of 2 marks each.
- In Section C – Question number 26 to 31 are Short Answer (SA) type questions, carrying 3 marks each.
- In Section D – Question number 32 to 35 are Long Answer (LA) type questions carrying 5 marks each.
- In Section E – Question number 36 to 38 are case study based integrated units of assessment questions

carrying 4 marks each. Internal choice is provided in 2 mark questions in each Case-study. - There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
- Use of calculators is NOT allowed.

SET I Code No. 30/6/1

Section – A (Multiple choice questions)

Section A consists of 20 questions of 1 mark each.

Question 1.

If P^{2} = \(\frac{32}{50}\), then P is a/an

(a) whole number

(b) integer

(c) rational number

(d) irrational number

Answer:

(c) rational number

We have, p^{2} = \(\frac{32}{50}\) = \(\frac{16}{25}\)

∴ p = ± \(\frac{4}{5}\) (Rational Number)

Question 2.

The distance of the point (-6, 8) from x-axis is

(a) 6 units

(b) -6 units

(c) 8 units

(d) 10 units

Answer:

(c) 8 units

8 units (ordinate)

Question 3.

The number of quadratic polynomials having zeroes -5 and -3 is 1

(a) 1

(b) 2

(c) 3

(d) more than 3

Answer:

(d) more than 3

Sum of zeroes, S = -5 + (-3) = -8

Product of zeroes, P = (-5) x (-3) = 15

∴ Quadratic Polynomial = k (x^{2} – Sx + P)

= k(x^{2} + 8x +15) …[Where k is non-zero real number

∴ Infinitely many Quadratic Polynomials are possible.

Question 4.

The point of intersection of the line represented by 3x – y = 3 and y-axis is given by 1

(a) (0,-3)

(b) (0, 3)

(c) (2, 0)

(d) (-2, 0)

Answer:

(a) (0,-3)

(a) Given. 3x – y = 3

⇒ 0 – y = 3 (At Y-axis Put x = 0)

∴ y = -3 ∴ Point = (x, y) = (0, -3)

Question 5.

The circumferences of two circles are in the ratio 4 : 5. What is the ratio of their radii? 1

(a) 16 : 25

(b) 25 : 16

(c) 2 : √5

(d) 4 : 5

Answer:

(d) 4 : 5

Let c_{1} and c_{2} be the circumferences of two circles respectively.

Given. \(\frac{c_1}{c_2}\) = \(\frac{1}{2}\) ⇒ \(\frac{2 \pi r_1}{2 \pi r_2}\) = \(\frac{1}{2}\)

∴ r_{1} : r_{2} = 4 : 5

Question 6.

If α and β are the zeroes of the polynomial x^{2} – 1, then the value of (α + β) is 1

(a) 2

(b) 1

(c) -1

(d) 0

Answer:

(d) 0

Given, x^{2} – 1 = 0

Here, a = 1, b = 0, c = -1

∴ α + β = \(\frac{-b}{a}=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}\)

= \(\frac{-0}{1}\) = 0

Question 7.

\(\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}\), in simplified form, is: 1

(a) tan^{2} θ

(b) sec^{2} θ

(c) 1

(d) -1

Answer:

(d) -1

1^{st} Method: \(\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}\)

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

2^{nd} Method: \(\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}\)

= cot^{2} θ – cosec^{2} θ

= -(cosec^{2} θ – cot^{2} θ) = -1

Question 8.

If ΔPQR ~ ΔABC; PQ = 6 cm , AB = 8 cm and the perimeter of ΔABC is 36 cm, then the perimeter of ΔPQR is 1

(a) 20.25 cm

(b) 27 cm

(c) 48 cm

(d) 64 cm

Answer:

(b) 27 cm

ΔPQR ~ ΔABC

∴ Peri. (ΔPQR) = \(\frac{1}{2}\) × 36 = 27 cm

Question 9.

If the quadratic equation ax^{2} + bx + c = 0 has two real and equal roots, then ‘c’ is equal to’

(a) \(\frac{-b}{2 a}\)

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

(c) \(\frac{-b^2}{4 a}\)

(d) \(\frac{b^2}{4 a}\)

Answer:

(d) \(\frac{b^2}{4 a}\)

D = 0 …[Since Roots are real and equal

⇒ b^{2} – 4ac = 0 ⇒ b^{2} = 4ac ∴ c = \(\frac{b^2}{4 a}\)

Question 10.

In the given figure, DE || BC. If AD = 3 cm, AB = 7 cm and EC = 3 cm, then the length of AE is 1

(a) 2 cm

(b) 2.25 cm

(c) 3.5 cm

(d) 4 cm

Answer:

(b) 2.25 cm

In ΔABC, DE || BC …[Given

\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) …[Thales theorem

⇒ \(\frac{3}{7-3}=\frac{\mathrm{AE}}{3}\) ⇒ 4 AE = 9

∴ AE = \(\frac{9}{4}\) = 2.25 cm

Question 11.

A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball? 1

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

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

(c) \(\frac{7}{20}\)

(d) \(\frac{13}{20}\)

Answer:

(c) \(\frac{7}{20}\)

Total Balls = 5 + 8 + 7 = 20

∴ P (neither a blue nor a pink ball)

= 1 – (\(\))

= 1 – (\(\frac{8+5}{20}\)) =1 – \(\frac{13}{20}\) = \(\frac{7}{20}\)

Question 12.

The volume of a right circular cone whose area of the base is 156 cm2 and the vertical height is 8 cm, is 1

(a) 2496 cm^{3}

(b) 1248 cm^{3}

(c) 1664 cm^{3}

(d) 416 cm^{3}

Answer:

(d) 416 cm^{3}

Volume of cone

= \(\frac{1}{3}\) × (Area of Base) × height

= \(\frac{1}{3}\) × 156 × 8 = 416 cm^{3}

Question 13.

3 chairs and 1 table cost ₹900, whereas 5 chairs and 3 tables cost ₹2,100. If the cost of 1 chair is ₹ x and the cost of 1 table is ₹ y, then the situation can be represented algebraically as 1

(a) 3x + y = 900, 3x + 5y = 2100

(b) x + 3y = 900, 3x + 5y = 2100

(c) 3x + y = 900,5x + 3y = 2100

(d) x + 3y = 900, 5x + 3y = 2100

Answer:

(c) 3x + y = 900,5x + 3y = 2100

3x + 1y = ₹900

5x + 3y = ₹2100

Question 14.

In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of ∠AQB is 1

(a) 621/2°

(b) 125°

(c) 55°

(d) 90°

Answer:

(a) 621/2°

AC & BC are radii and PA and PB are tangents.

∠PAC = ∠PBC = 90° …[Tangent is ⊥ to the radius through the point of contact

In quad. PACB,

55° + 90° + ∠ACB + 90° = 360° …[Angle-sum-property of a quadrilateral

⇒ ∠ACB = 360° – 235° = 125°

Now, ∠AQB = ∠ACB

= \(\frac{125}{2}\) = 62\(\frac{1}{2}\)° ….[The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of circle.

Question 15.

A card is drawn at random from a well shuffled deck of 52 playing cards. 1

The probability of getting a face card is

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

(b) \(\frac{3}{13}\)

(c) \(\frac{4}{13}\)

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

Answer:

(b) \(\frac{3}{13}\)

∴ P (Face Card) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

Question 16.

If θ is an acute angle of a right angled triangle, then which of the following equations is not true? 1

(a) sin θ cot θ = cos θ

(b) cos θ tan θ = sin θ

(c) cosec^{2} θ – cot^{2} θ = 1

(d) tan^{2} θ – sec^{2} θ = 1

Answer:

(d) tan^{2} θ – sec^{2} θ = 1

Identity: sec^{2} θ – tan^{2} θ = 1

so, tan^{2} θ – sec^{2} θ ≠ 1

Question 17.

If the zeroes of the quadratic polynomial x^{2} + (a + 1) x + b are 2 and -3, then 1

(a) a = -7, b = -1

(b) a = 5, b = -1

(c) a = 2, b = -6

(d) a = 0, b = -6

Answer:

(d) a = 0, b = -6

Given. Zeroes of quadratic polynomial,

x^{2} + (a + 1) x + b

α = 2 and β = -3

Here, A = 1, B = a + 1, C = b

∴ Sum of 0 = \(\frac{-B}{A}\)

⇒ 2 + (-3) = \(\frac{-(a+1)}{1}\)

⇒ -1 = -(a + 1)

∴ a = 1 – 1 = 0

∴ Product of 0 = \(\frac{C}{A}\)

⇒ 2(-3) = \(\frac{b}{1}\)

∴ b = -6

Question 18.

If the sum of the first n terms of an A.P be 3n^{2} + n and its common difference is 6, then its first term is 1

(a) 2

(b) 3

(c) 1

(d) 4

Answer:

(d) 4

S_{n} = 3n^{2} + n …[Given

Put n = 1, S_{1} = 3(1)^{2} + 1 = 3 + 1= 4

∴ First term = a = S_{1} = 4

Assertion – Reason Based Questions: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).

Choose the correct option out of the following:

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

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

(c) A is true but R is false.

(d) A is False but R is true.

Question 19.

Assertion (A): If 5 + √7 is a root of a quadratic equation with rational co-efficients, then its other root is 5 – √7

Reason (R): Square roots of a quadratic equation with rational co-efficients occur in conjugate pairs.

Answer:

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

Question 20.

Assertion (A): For 0 < θ ≤ 90°, cosec θ – cot θ and cosec θ + cot θ are reciprocal of each other. 1

Reason (R): cosec^{2} θ – cot^{2} θ = 1.

Answer:

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

(cosec θ – cot θ) × (cosec θ + cot θ)

= cosec^{2} θ – cot^{2} θ

= 1 R is True.

Section – B

Section B consists of Very Short Answer (VSA) type of questions of 2 marks each.

Question 21.

(a) Show that 6^{n} cannot end with digit 0 for any natural number ‘n’. 2

Or

(b) Find the HCF and LCM of 72 and 120.

Answer:

(a) If the number 6^{n}, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 6^{n} will contain the prime number 5. This is not possible because Prime factorisation of 6^{n} does not contain 5 as a factor.

6^{n} = (2 × 3)^{n} = 2^{n} × 3^{n}

So, there is no natural number (n) for which 6″ ends with the digit zero.

Or

(b) 72 = 2^{3} × 3^{2}

120 = 2^{3} × 3 × 5

∴ HCF = 2^{3} × 3^{1}

= 24

∴ LCM = 2^{3} × 3^{2} × 5 = 360

Question 22.

A line intersects y-axis and x-axis at point P and Q respectively. If R(2, 5) is the mid-point of line segment PQ, then find the coordinates of P and Q. 2

Answer:

Let P (0, y) and Q (x, 0).

Mid point of PQ = R

(\(\frac{0+x}{2}, \frac{y+0}{2}\)) (2, 5) …[mid-point theorem

⇒ \(\frac{x}{2}\) = 2

⇒ x = 4

⇒ \(\frac{y}{2}\) = 5

⇒ y = 10

∴ Coordinates of P (0, y) = (0, 10)

Coordinates of Q (x, 0) = (4, 0)

Question 23.

Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the sun is such that tan θ = \(\frac{6}{7}\).

Answer:

In a ΔABC, tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇒ \(\frac{6}{7}\) = \(\frac{18}{B C}\) …[ tan θ = \(\frac{6}{7}\) …[Given

⇒ 6 BC = 18 × 7

∴ Length of shadow, BC = \(\frac{18 \times 7}{6}\) = 21 m

Question 24.

In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter.

If ∠AOC = 130°, then find the measure of ∠APB, where O is the centre of the circle. 2

Answer:

Given. OA is the radius and PA is the tangent.

⇒∠PAO = 90° …[Tangent is ⊥ to the radius through the point of contact

∴ ∠APO + ∠PAO = ∠AOC …[Exterior angle theorem

⇒ ∠APB + 90° = 130°

∴ ∠APB = 130° – 90° = 40°

Question 25.

(a) In the given figure, ABC is a triangle in which DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then find the value of x. 2

Or

(b) Diagonals AC and BD of trapezium ABCD with AB || DC intersect each other at point O. Show that

\(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\)

Answer:

(a) In ΔABC, DE || BC

∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) …[Thales’ theorem

⇒ \(\frac{x}{x-2}=\frac{x+2}{x-1}\)

⇒ x (x – 1) = (x – 2) (x + 2)

⇒ x^{2} – x = x^{2} – 4

∴ x = 4 units

(b) In ΔAOB and ΔCOD,

AB || CD …[Given

∠1 = ∠3 …[Alternate interior angles

∠2 = ∠4

∴ ΔAOB ~ ΔCOD …[By AA corollary

\(\frac{\mathrm{BO}}{\mathrm{DO}}\) = \(\frac{\mathrm{AO}}{\mathrm{CO}}\) ……[In similar Δs corresponding sides are proportional.

∴ \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\) (Hence proved)

Section – C

Section C consists of Short Answer (SA) type of questions of 3 marks each.

Question 26.

Find the ratio in which the line segment joining the points A(6, 3) and B(-2, -5) is divided by x-axis. 3

Answer:

Let C (x, 0) be any point on x-axis.

Let AC : CB = m_{2} : m_{1} = k : 1

Using section formula, \(\frac{-5 k+3}{k+1}=\frac{0}{1}\)

⇒ -5k + 3 = 0 ⇒ -5k = -3

∴ k = \(\frac{3}{5}\)

∴ Required Ratio = k : 1 = \(\frac{3}{5}\) : 1 or = 3 : 5

Question 27.

(a) Find the HCF and LCM of 26, 65 and 117, using prime factorisation. 3

Or

(b) Prove that √2 is an irrational number.

Answer:

(a) Prime factorisation of 26 = 2 × 13

65 = 5 × 13

117 = 3^{2} × 13

∴ HCF = 13

LCM = 2 × 5 × 3^{2} × 13 = 1170

Or

(b) Let us assume, to the contrary, that √2 is rational. That is, we can find integers a

and b (≠ 0) that √2 = \(\frac{a}{b}\).

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are coprime.

So, b√2 = a

Squaring on both sides, & rearranging, we get 2b^{2} = a^{2} Therefore, a^{2} is divisible by 2 and a is also divisible by 2. So, we can write a = 2c for some integer c. Substituting for a, we get 2b^{2} = 4c^{2}, that is b^{2} = 2c^{2}.

This means that b2 is divisible by 2 and so b is also divisible by 2. Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √2 is rational.

So we conclude that √2 is irrational.

Question 28.

In the given figure, Eisapoint on the side CB produced of anisosceles triangle ABC with AB = AC. If AD⊥BC and EF⊥ AC, then prove that triangle ΔABD ~ ΔECF. 3

Answer:

In ΔABC, AB = AC …[Given

∴ ∠4 = ∠2 …(i) [Angles opposite equal sides

In ΔABD and ΔECF,

∠3 = ∠1 …[Each 90°

∠4 = ∠2 …[From (i)

∴ ΔABD ~ ΔECF …[AA similarity criterion

Question 29.

(a) The sum of two numbers is 15. If the sum of their reciprocals is \(\frac{3}{10}\), find the two numbers. 3

Or

(b) If α and β are roots of the quadratic equation x^{2} – 7x + 10 = 0, find the quadratic equation , whose roots are α^{2} and β^{2}. 3

Answer:

(a) Let two numbers be x and 15 – x. According to question,

⇒ \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}[latex] ⇒ [latex]\frac{15-x+x}{x(15-x)}=\frac{3}{10}[latex]

⇒ 150 = 3x(15 – x) ⇒ 50 = 15x – x^{2}

⇒ x^{2} – 15x + 50 = 0

⇒ x^{2} – 10x – 5x + 50 = 0

⇒ x(x – 10) -5 (x – 10) = 0

⇒ (x – 5) (x – 10) = 0

⇒ x – 5 = 0 or x – 10 = 0

⇒ x = 5 or x = 10

When x = 5, numbers are 5, 10

When x = 10, numbers are 10, 5

Or

(b) Given, x^{2} – 7x + 10 = 0

a = 1, b = -7, c = 10

Sum of roots, α + β = [latex]\frac{-b}{a}=\frac{-(-7)}{1}\) = 7 …(i)

Product of roots, αβ = \(\frac{c}{a}\) = 10 …(ii)

For required Polynomial,

Sum of roots, S = α^{2} + β^{2} = (α + β)^{2} – 2αβ

= (7)^{2} – 2(10) …[From (i) and (ii)

= 49 – 20 = 29

Product of roots, P = α^{2}β^{2}

= (αβ)^{2} …[From (ii)

= (10)^{2} = 100

∴ Required Quadratic Polynomial,

x^{2} – Sx + P = 0 ∴ x^{2} – 29x + 100 = 0

Question 30.

Prove that: \(\frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}\) 3

Answer:

Question 31.

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc. 3

Answer:

Here r = 21 cm, θ = 60°

Area of sector = \(\frac{\theta}{360^{\circ}}\)πr^{2}

= \(\frac{60^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 21 × 21

= 231 cm^{2}

Length of an arc = \(\frac{\theta}{360^{\circ}}\) (2πr)

= \(\frac{60^{\circ}}{360^{\circ}}\) × 2 × \(\frac{22}{7}\) × 21 = 22 cm

Section – D

Section D Consists of Long Answer (LA) type questions of 5 marks each.

Question 32.

(a) Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ. 5

Or

(b) A circle touches the side BC of a ΔABC at a point P and touches AB and AC when produced at Q and R respectively.

Show that AQ = \(\frac{1}{2}\) (Perimeter of ΔABC).

Answer:

(a) Given. A circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.

To Prove. ∠PTQ = 2∠OPQ

Proof: Let ∠PTQ = θ

Now, TP = TQ ……[∵ Lengths of tangents drawn from an external Point to a circle are equal

So, TPQ is an isosceles triangle

∴ ∠TPQ = ∠TQP = \(\frac{1}{2}\) (180° – θ)

= 90° – \(\frac{1}{2}\)θ …….. (i)

Also, ∠OPT = 90° ……[∵Tangent at any point of a circle is⊥to radius through the pt. of contact

So, ∠OPQ = ∠OPT – ∠TPQ = 90° – (90° – \(\frac{1}{2}\) θ) …….[From (i)

∠OPQ = \(\frac{1}{2}\) θ ⇒ ∠OPQ = \(\frac{1}{2}\) ∠PTQ

∴ ∠PTQ = 2∠OPQ (Hence Proved)

Or

(b) Since the tangents from an external point to a circle are equal in length.

Now, Perimeter of ΔABC = AB + BC + AC

= AQ – BQ + BP + PC + AR – CR

= AQ – BP + BP + CR + AQ – CR …[Using (i), (ii) and (iii)

∴ Perimeter of ΔABC = 2AQ

AQ = \(\frac{1}{2}\) (Perimeter of ΔABC)

(Hence Proved)

Question 33.

A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid. 5

Answer:

Given, r = 7 cm, h = 2r = 14 cm

Vol. of solid = Vol. of cone + Vol. of hemisphere

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

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 (14 + 14)

= \(\frac{22 \times 7 \times 28}{3}\) \(\frac{4312}{3}\)

= 1437.3 cm^{3}

Question 34.

(a) The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms. 5

Or

(b) If the sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Mass (in grams) 80-100 100-120 120-140 140-160 160-180

Number of apples 20 60 70 X 60

Answer:

(a) \(\frac{a_{11}}{a_{18}}\) = \(\frac{2}{3}\) …[Given

\(\frac{a+10 d}{a+17 d}\) = \(\frac{2}{3}\) …[∵ a_{n} = a + (n – 1)d

⇒ 3a + 30d = 2a + 34 d

⇒ 3a – 2a = 34d + -30d

a = 4d …… (i)

Or

(b) Given. S_{6} = 36

⇒ \(\frac{6}{2}\)(2a + 5d) = 36 … [∵ S_{n} = \(\frac{n}{2}\)(2a + (n – 1 )d

⇒ 2a + 5d = 12 ⇒ 2a = 12 – 5d ……. (i)

and S_{16} 256 …..[Given

⇒ \(\frac{16}{2}\) (2a + 15d) = 256

⇒ 2a + 15d = \(\frac{256}{2}\)

⇒ 12 – 5d + 15d = 32 …..[From (i)

10d = 32 – 12

∴ d \(\frac{20}{10}\) = 2

From (i), 2a = 12 – 5(2)

⇒ 2a = 2 :. a = 1

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

∴ S_{10} = \(\frac{10}{2}\)(2a + 9d)

= 5 (2(1) + 9(2))

= 5 (2 + 18) = 5 (20) = 100

Question 35.

250 apples of a box were weighed and the distribution of masses of the apples is given in Find the value of x and the mean mass of the apples.

(ii) Find the modal mass of the apples.

Answer:

Total Apples = 250 …..[Given

⇒ 20 + 60 + 70 + x + 60 = 250

⇒ 210 + x = 250

∴ x = 250 – 210 = 40

(i) Mean Mass = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

= \(\frac{33700}{250}\) = 134.8 gms

(ii) Since Maximum Frequency is 70

So Modal class in 120 – 140

∴ Modal mass = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h ..[Here, l = 120, f_{0} = 60, f_{1} = 70, f_{2} = 40, h = 20

= 120 + \(\frac{70-60}{140-60-40}\) × 20

= 120 + \(\frac{10 \times 20}{40}\)

h = 20

= 120 + 5 = 125 gms

Section – E

3 Case Study Based Questions. Each question is of 4 marks.

Question 36.

A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹9000 and ‘from batch II is ₹26,000. Assume that each” poor child

pays ₹ x per month and each rich child pays ₹ y per month.

Based on the above information, answer the following questions :

(i) Represent the information given above in terms of x and y. 1

(ii) Find the monthly fee paid by a poor child. 2

Or

Find the difference in the monthly fee paid by a poor child and a rich child. 2

(iii) If there are 10 poor and 20 rich children in batch III, what is the total monthly collection of fees from batch III? 1

Answer:

(i) A.T.Q. 20x + 5y = 9,000 ……(A)

5x + 25 y = 26,000 …….(B)

(ii) Multiplying (A) by 1 and (B) by \(\frac{1}{5}\) we have

∴ x = \(\frac{3800}{19}\) = ₹200

From (C), 200 + 5y = 5200

⇒ 5y = 5200 – 200

∴ y = \(\frac{5000}{5}\) = ₹1,000

∴ Monthly fee paid by a poor child x = ₹200

∴ Required difference = y – x

= ₹(1,000 – 200) = ₹800

(iii) Total collection of Fees from Batch III

= 10x + 20y

= 10(200) + 20(1,000) …[From Point (ii)

= 2,000 + 20,000 = ₹22,000

Question 37.

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O.

Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is 30° and the angle of elevation of the top of Section A is 45°.

Based on the above information, answer the following questions :

(i) Find the length of the wire from the point O to the top of Section B. 1

(ii) Find the distance AB. 2

Find the area of ΔOPB,

(iii) Find the height of the Section A from the base of the tower. 1

Answer:

∴ AB = AP – BP = 36 – 12√3

= 12(3 – √3)cm or 12√3(√3 – 1) cm

Or

ar. (ΔOPB) = \(\frac{1}{2}\) × Base × height

= \(\frac{1}{2}\) × PO × PB

= \(\frac{1}{2}\) × 36 × 12√3

= 216√3 cm^{3}

(iii) From (ii) above AP = 36 cm

Alternatively, AP = AB + BP

AP = AB + BP

= 12√3(√3 – 1) + 12√3

= 36 – 12√3 + 12√3 = 36 cm

Question 38.

“Eight Ball” is a game played on a pool table with 15 balls numbered 1 to 15 and a “cue ball” that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15.

The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random.

Based on the above information, answer the following questions :

(i) What is the probability that the drawn ball bears number 8? 1

(ii) What is the probability that the drawn ball bears an even number? 2

What is the probability that the drawn ball bears a number, which is a multiple of 3?

(in) What is the probability that the drawn ball is a solid coloured and bears an even number? 1

Answer:

Non-White balls = 8 (nos. 1 to 8)

Striped balls = 7 (nos. 9 to 15)

White solid cue ball = 1

Total number of balls = 8 + 7 = 15

(i) P(number 8 ball) = \(\frac{1}{15}\) …[There is one eight number ball

(ii) Even balls 2,4, 6,8,10,12,14 i.e 7 balls 7

∴ P(even number) = \(\frac{7}{15}\)

Or

“Multiples of 3” are 3, 6, 9, 12, 15 i.e 5 balls

∴ P(multiple of 3) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

(iii) Solid coloured and bear even number balls are 2, 4, 6, 8 i.e 4 balls

∴ Required probability = \(\frac{4}{15}\)

SET Code No. 30/6/2

Note: Except for the following questions, all the remaining questions have been asked in Set – I.

Question 6.

The LCM of smallest 2-digit number and smallest composite number is 1

(a) 12

(b) 4

(c) 20

(d) 40

Answer:

(c) 20

Smallest 2 digit number = 10 = 2 × 5

Smallest composite number = 4 = 2^{2}

∴ LCM (10,4) = 2^{2} × 5 = 20

Question 7.

The distance of the point (-4, 3) from y-axis is 1

(a) -4

(b) 4

(c) 3

(d) 5

Answer:

(b) 4

Distance of (-4, 3) from y-axis is 4.

Note : Distance is not negative.

Question 8.

If one zero of the polynomial x^{2} + 3x + k is 2, then the value of k. 1

(a) -10

(b) 10

(c) 5

(d) -5

Answer:

(a) -10

Let P(x) = x^{2} + 3x + k

∴ x = 2 …[ ∵ 2 is a zero of P(x)

⇒ (2)^{2} + 3(2) + k = 0

⇒ 4 + 6 + k = 0 ∴ k = -10

Question 16.

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23 is 1

(a) \(\frac{7}{90}\)

(b) \(\frac{1}{9}\)

(c) \(\frac{4}{45}\)

(d) \(\frac{9}{89}\)

Answer:

(c) \(\frac{4}{45}\)

Let x be the Prime number less than 23 Such numbers are 2, 3, 5, 7, 11, 13, 17, 19 i.e 8

∴ P(x) = \(\frac{8}{90}\) = \(\frac{4}{45}\)

Question 17.

The coordinates of the point where the line 2y = 4x + 5 crosses ar-axis is 1

(a) (o, \(\frac{-5}{4}\))

(b) (o, \(\frac{5}{2}\))

(c) (\(\frac{-5}{4}\), 0)

(d) (\(\frac{-5}{2}\), o)

Answer:

(c) (\(\frac{-5}{4}\), 0)

Given. 2y = 4x + 5

∴ 2(0) – 5 = 4x …[To cut x-axis, y = 0

∴ x = \(\frac{-5}{4}\)

∴ Point (x, y) = (\(\frac{-5}{4}\), 0)

Question 18.

(cos^{4} A – sin^{4} A) on simplification, gives 1

(a) 2 sin^{2} A – 1

(b) 2 sin^{2} A + 1

(c) 2 cos^{2} A + 1

(d) 2 cos^{2} A – 1

Answer:

(d) 2 cos^{2} A – 1

Given, cos^{4} A – sin^{4} A

= (cos^{2} A)^{2} – (sin^{2} A)^{2}

= (cos^{2} A – sin^{2} A) (cos^{2} A + sin^{2} A)

= [(cos^{2} A – (1 – cos^{2} A)]. (1)

= [(cos^{2} A – 1 + cos^{2} A]. 1 = 2 cos^{2} A – 1

Question 24.

Find the points on the x-axis, each of which is at a distance of 10 units from the point A(11, -8). 2

Answer:

Let B(x, 0) be any point on x-axis.

AB = 10 units …[Given

(AB)^{2} = 100 …[Squaring both sides

⇒ (x – 11)^{2} + (0 + 8)^{2} = 100

⇒ (x – 11)^{2} + 64 = 100

⇒ (x – 11)^{2} = 100 – 64

⇒ (x – 11)^{2} = 36

⇒ x – 11 = ±6 …[Taking Square-root on both sides

⇒ x = 11 ±6

⇒ x = 11 + 6 or x = 11 – 6

∴ x = 17 or x = 5

∴ B = (17, 0) or (5,0)

Question 26.

In the given figure, AB and CD are diameters of a circle with centre O perpendicular to each other. If OA = 7 cm, find the area of shaded region. 3

Answer:

Area of shaded region

= Area of Semi-circle – Ar. (ΔABC)

= \(\frac{1}{2}\)πr^{2} – \(\frac{1}{2}\) × AB × OC

= \(\frac{1}{2}\) × \(\frac{22}{7}\) × 7 × 7 – \(\frac{1}{2}\) × 14 × 7 [∵ AB = 7 + 7 = 14cm OC = r = 7 cm ]

= 77 – 49 = 28 cm^{2}

Question 27.

If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p^{2} – 1) = 2p. 3

Answer:

We have, p = sin θ + cos θ

q = sec θ + cosec θ

L.H.S. q (p^{2} – 1)

= (sec θ + cosec θ) [(sin θ + cos θ)^{2} – 1]

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

= (\(\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}\)) (2 sin θ cos θ)…[∵ sin^{2} θ + cos^{2} θ = 1

= 2 (sin θ + cos θ)

= 2p = R.H.S (Hence Proved)

Question 34.

A solid is in the shape of a right-cirular cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the soild. 5

Answer:

r = 3.5 cm or \(\frac{7}{2}\) cm

Height of cone, h =9.5 – 3.5

= 6 cm

∴ Vol. of solid = Vol. of cone + Vol. of hemisphere

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

= \(\frac{1}{2}\) πr^{2} (h + 2r)

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × (6 + 2 × \(\frac{7}{2}\))

= \(\frac{77}{3}\) × 13 = \(\frac{1001}{6}\) = \(166.8 \overline{3}\) cm^{3}

Question 35.

(a) Find the sum of integers between 100 and 200 which are 5

(i) divisible by 9

(ii) not divisible by 9.

Or

(b) Solve the equation : -4 + (-1) + 2 + 5 + ………. + x = 437.

Answer:

(a) (i) Nos. divisible by 9 between 100 & 200 are

108, 117, 126, ………… a_{n}

Here a = 108, d = 117 – 108 = 9, a_{n} – 198

∴ a + (n – 1) d = a_{n}

⇒ 108 + (n – 1) 9 = 198

⇒ (n – 1)9 = 198 – 108

⇒ (n – 1) = \(\frac{90}{9}\) ∴ n = 11

Now, S_{n} = \(\frac{n}{2}\) (a_{1} + a_{n})

S_{11} = \(\frac{11}{2}\) (108 + 198)

= \(\frac{11}{2}\) × 306 = 11 × 153 = 1683

(ii) Let S be the sum of integers between 100 and 200.

∴ S = 101 + 102 + 103 + ………… + 199

Here a_{1} = 101, a_{n} = 199,

n = 199 – 101 + 1 = 99

∴ S = \(\frac{99}{2}\) (101 + 199)

= \(\frac{99}{2}\) (300) = 99 × 150 = 14,850

Now, Sum not divisible by 9 = 14850 – 1683 = 13167

Or

(b) Here, a = -4, d = -1 – (-4) = 3, a_{n} = x

Given. S_{n} = 437

∴ \(\frac{n}{2}\) [2a + (n – 1) (d)] = 437

⇒ \(\frac{n}{2}\) [2 (-4) + (n – 1)(3)] = 437

⇒ n (-8 + 3n – 3) = 874

⇒ n (3n – 11) = 874

⇒ 3n^{2} – 11n – 874 = 0

⇒ 3n^{2} – 57n + 46n – 874 = 0

⇒ 3n(n – 19) + 46 (n – 19) = 0

⇒ (n – 19) (3n + 46) = 0

⇒ n – 19 = 0 or 3n + 46 = 0

∴ n = 19 or n = \(\frac{-46}{3}\) (Not possible) …[n must be a rational number

Now, x = a_{n}

x = a + (n – 1)d

= -4 + (19 – 1)(3) = -4 + 54 = 50

*Note: 2^{nd} Method using x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

SET III Code. 30/6/3

Note: Except for the following questions, all tire remaining questions have been asked in Set – I and Set II.

Question 1.

The distance between the points (0, 5) and (-3, 1) is :

(a) 8 units

(b) 5 units

(c) 3 units

(d) 25 units

Answer:

(b) 5 units

Given. A (0, 5) and B (-3, 1)

∴ AB = \(\sqrt{(-3-0)^2+(1-5)^2}\) …[Using distance formula

= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5 units

Question 2.

If tan θ = \(\frac{x}{y}\), then cos θ is equal to

Answer:

(a) \(\frac{x}{\sqrt{x^2+y^2}}\)

(b) \(\frac{y}{\sqrt{x^2+y^2}}\)

(c) \(\frac{x}{\sqrt{x^2-y^2}}\)

(d) \(\frac{y}{\sqrt{x^2-y^2}}\)

Answer:

(b) \(\frac{y}{\sqrt{x^2+y^2}}\)

Given, tan θ = \(\frac{x}{y}\)

As we know, sec^{2} θ = 1 + tan^{2} θ

sec^{2} θ = 1 + \(\frac{x^2}{y^2}\)

⇒ sec^{2} θ = \(\frac{y^2+x^2}{y^2}\)

⇒ sec θ = \(\sqrt{\frac{x^2+y^2}{y^2}}=\frac{\sqrt{x^2+y^2}}{y}\)

∴ cos θ = \(\frac{1}{\sec \theta}=\frac{y}{\sqrt{x^2+y^2}}\)

Alternatively, tan θ = \(\frac{x}{y}=\frac{\mathrm{P}}{\mathrm{B}}\)

In rt. Δ, H^{2} = P^{2} + B^{2}

= k^{2}x^{2} + k^{2}y^{2} …[Pythagoras’ theorem

⇒ H^{2} = k^{2} (x^{2} + y^{2})

⇒ H = k\(\sqrt{x^2+y^2}\)

∴ cos θ \(\frac{\mathrm{B}}{\mathrm{H}}=\frac{k y}{k \sqrt{x^2+y^2}}=\frac{y}{\sqrt{x^2+y^2}}\)

Question 3.

The zeroes of the polynomial 3x^{2} + 11x -4 are:

(a) \(\frac{1}{3}\), -4

(b) \(\frac{-1}{3}\), 4

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

(d) \(\frac{-1}{3}\), 4

Answer:

(a) \(\frac{1}{3}\), -4

Given. 3x^{2} + 11x – 4

⇒ 3x^{2} + 12x – x – 4

⇒ 3x (x + 4) – 1(x + 4)

⇒ (x + 4) (3x – 1)

∴ Zeroes are: x + 4 = 0 or 3x – 1 = 0

x = -4 or x = 1/3

Question 16.

If ‘p’ is a root of the quadratic equation x^{2} – (p + q) x + k = 0 then the value of ‘k’ is

(a) p

(b) q

(c) p + q

(d) pq

Answer:

(d) pq

Given. x^{2} – (p + q)x + k = 0

Since p is one of the root, …[Given

∴ p^{2} – (p + q)p + k = 0

⇒ p^{2} – p^{2} – pq + k = 0 ∴ k = pq

Question 17.

Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out of the bag at random. What is the probability that the number on the card taken out is an even number? 1

(a) \(\frac{9}{17}\)

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

(c) \(\frac{5}{9}\)

(d) \(\frac{7}{18}\)

Answer:

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

We have, 3, 4, 5 ………….. 20

Total cards = 18

Even number cards are 4, 6, 8, 10, 12, 14, 16, 18, 20 i.e., 9

∴ P(even numbers) = \(\frac{9}{18}\) = \(\frac{1}{2}\)

Question 18.

The condition for the system of linear equations ax + by = c; lx + my = n have a unique solution is 1

(a) am ≠ bl

(b) al ≠ bm

(c) al = bm

(d) am = bl

Answer:

(a) am ≠ bl

Given, ax + by = c

lx + my = n

For unique solution: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

\(\frac{a}{l} \neq \frac{b}{m}\)

∴ am ≠ bl

Question 21.

Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). 2

Answer:

Let C(0,y) be any point on y-axis.

Let AC : CB = k : 1

Coordinates of C = Coordinates of C

\(\frac{-k+5}{k+1}, \frac{-4 k-6}{k+1}\) = (0, y) ..[Using formula

⇒ \(\frac{-k+5}{k+1}\) = 0 ⇒ -k + 5

∴ k = 5

∴ Required Ratio = k : 1 = 5: 1

Question 26.

Prove that (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ. 3

Answer:

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

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

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

= \(\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\) ….[∵ sin^{2} θ + cos^{2} θ = 1

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

= sec θ+ cosec θ = R.H.S (Hence Proved)

Question 27.

(a) A natural number, when increased by 12, equals 160 times its reciprocal. Find the number. 3

Or

(b) If one root of the quadratic equation x^{2} + 12x – k = 0 is thrice the other root, then find the value of k.

Answer:

(a) Let the number be x.

A.T.Q., (x + 12) = 16o(\(\frac{1}{x}\))

x (x + 12) = 160

⇒ x^{2} + 12x – 160 = 0

⇒ x^{2} + 20x – 8x – 160 = 0

⇒ (x – 8) (x + 20) = 0

⇒ x – 8 = 0 or x + 20 = 0

⇒ x = 8 or x = -20 (not possible)

Natural number must be positive

∴ Required number, x = 8.

Or

(b) Given, x^{2} + 12x + k = 0. Let roots be α and 3α

Here a = 1, b = 12, c = -k

Sum of roots = \(\frac{-b}{a}\)

α + 3α = \(\frac{-12}{1}\)

4α = -12

∴ α = -3 ….. (i)

Product of roots = \(\frac{c}{a}\)

α(3α) = \(\frac{-k}{1}\)

3α^{2} = -k

3(-3)^{2} = -k ….[From (i)

∴ k = -27

Question 32.

(a) The sum of first seven terms of an A.P. is 182. If its 4^{th} term and the 17^{th} term are in the ratio 1: 5, find the A.P. 5

Or

(b) The sum of first q terms of an A.P. is 63q – 3q^{2}. If its p^{th} term is – 60, find the value of p. Also, find the 11^{th} term of this A.P.

Answer:

(a) S_{7} = 182

⇒ \(\frac{7}{2}\) (2a + 6d) = 182 …[∵ S_{n} = \(\frac{n}{2}\) (2a + (n – 1)d

⇒ (a + 3d) = \(\frac{182}{7}\) ⇒ a + 3d = 26

⇒ a = 26 – 3d …… (i)

From (i), a = 26 – 3 (8) = 26 – 24 = 2

∴ AP is a, a + d, a + 2d, a + 3d, 2, 10, 18, 26..

Or

(b) Given. Sq = 63q – 3q^{2}

Put q = 1, S_{1} = 63(1) – 3(1)^{2} = 63 – 3 = 60

Put q = 2, S_{2} = 63(2) – 3(2)^{2} = 126 – 12 = 114

a_{1} = S_{1} = 60

a_{2} = S_{2} – S_{1} = 114 – 60 = 54

d = a_{2} – a_{1}

= 54 – 60 = -6

Now, a_{p} = -60

⇒ a + (p – 1 )d = -60 …[∵ a = a_{1} = 60

⇒ 60 + (p – 1)(-6) = -60

⇒ (p -1 )(-6) = -60 – 60

⇒ P – 1 = \(\frac{-120}{-6}\)

∴ p = 20 + 1 = 21

Now, a_{11} = a + 10d = 60 + 10(-6) = 60 – 60 = 0

Question 33.

(a) Prove that a parallelogram circumscribing a circle is a rhombus. 5

Or

(b) In the given figure, tangents PQ and PR are drawn to a circle such that ∠RPQ 30°. A chord RS is drawn parallel to the tangent PQ. Find the measure of ∠RQS.

Answer:

(a)

Given. ABCD is a ||^{gm}.

To prove. ABCD is a rhombus.

Proof. In || Sm, opposite sides are equal

∴ AB = CD

and AD = BC ……. (i)

By adding these tangents,

(AP + PB) + (CR + DR) = AS + BQ + CQ + DS

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC ……[From (i)

⇒ 2AB = 2BC

⇒ AB = BC …….(ii)

From (i) and (ii), AB = BC = CD = DA

∴ ||^{gm} ABCD is a rhombus.

Or

(b)

Given. ∠RPQ = 30°

To find the value of = ∠RQS

Proof: PQ = PR …[Tangents drawn from an external point are equal

∴ In ΔPQR, ∠3 = ∠4 …[Angles opposite to equal sides

∴ In ΔPQR, ∠3 + ∠4 + ∠30° = 180° …….(i)

…[Angles sum Property of a Δ

⇒ ∠3 + ∠3 + 30° = 180° …[From (i)

⇒ 2∠3 = 180° – 30° = 150°

⇒ ∠3 = \(\frac{150}{2}\) = 75°

∴ ∠3 = ∠4 = 75° ……(ii)

RS || PQ …[Given

∠5 = ∠3 = 75° …(iii) …[Alternate Interior angles

PQ is a tangent and QR is a chord.

∠1 = ∠3 = 75° …[Angles in the alternate segments

In ΔQRS, ∠1 + ∠2 + ∠5 = 180° …[Angles-sum-Property of a Δ

⇒ 75° + ∠2 + 75° = 180° …[From (ii) and (iii)

⇒ ∠2 = 180° – 75° – 75°

∴ ∠2 or ∠RQS = 30°