Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 1 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions:

1. This question paper has 5 Sections A-E.

2. Section A has 20 MCQs carrying 1 mark each.

3. Section B has 5 questions carrying 2 marks each.

4. Section C has 6 questions carrying 3 marks each.

5. Section D has 4 questions carrying 5 marks each.

6. Section E has 3 Case Based integrated units of assessment (4 marks each).

7. All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks and 2 questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.

8. 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 two positive integers a and b are written as a = x³y² and b = xy³, where .r, y are prime numbers, then the result obtained by dividing the product of the positive integers by the LCM (a, b) is [1]

(a) xy

(b) xy²

(c) x³y³

(d) x²y²

Answer:

(b) xy²

Given, a = x³y²

ana b = xy³

⇒ ab = (x³y²)(xy³)

= x^{4}y^{5}

and LCM (a, b) = x³y³

∴ \(\frac{a b}{{LCM}(a, b)}=\frac{x^4 y^5}{x^3 y^3}=x y^2\)

Question 2.

The given linear polynomial y = f(x) has [1]

(a) 2 zeroes

(b) 1 zero and the zero is ‘3 ’

(c) 1 zero and the zero is ‘4’

(d) No zero

Answer:

(b) 1 zero and the zero is ‘3’

Since, the given polynomial y = f(x) intersect the X-axis at point (3, 0) only. So, the polynomial has 1 zero and the zero is 3.

Question 3.

The given pair of linear equations is non-intersecting. Which of the following statements is true? [1]

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

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

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

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

Answer:

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

Since, the given pair of linear equation is non-intersecting i.e. parallel.

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

Question 4.

Write the nature of roots of the quadratic equation 9x² – 6x – 2 = 0 [1]

(a) no real roots

(b) 2 equal real roots

(c) 2 distinct real roots

(d) more than 2 real roots

Answer:

(c) 2 distinct real roots

Given, 9x² – 6x – 2 = 0

Discriminant, D = (-6)² – 4(9) (- 2)

= 36 + 72 = 108 > 0

∴ The given equation has 2 distinct real roots.

Question 5.

Two APs have the same common differ-ence. The first term of one of these is -1 and that of the other is – 8. Then, the difference between their 4th terms is [1]

(a) 1

(b) 8

(c) 7

(d) 9

Answer:

(c) 7

Let the common difference of two APs be d.

Given, first term of one AP(a_{1}) = -1

and first term of second AP(b_{1}) = – 8

Fourth term of 1 st AP(a_{4}) = a_{1} + (4 – 1)d

= -1 + 3d

and fourth term of 2nd AP(b_{4}) = b_{1} + (4 – 1)d

= – 8 + 3d

∴ a_{4} – b_{4} = (-1 + 3d) – (- 8 + 3d)

= -1 + 3d + 8 – 3d = 7

Question 6.

The ratio in which the line segment joining (2, -3) and (5, 6) is divided by X-axis [1]

(a) 1 : 2

(b) 2 : 1

(c) 2 : 5

(d) 5 : 2

Answer:

(a) 1 : 2

Let the point on X-axis which divides the join of (2, – 3) and (5, 6) be (x, 0).

Let the line segment be divided by (x, 0) in the ratio k : 1.

Using section formula,

0 = \(\frac{k(6)+1(-3)}{k+1}\)

⇒ 6k = 3

⇒ k = \(\frac{1}{2}\)

∴ The ratio is k : 1 = \(\frac{1}{2}\) : 1 = 1 : 2

Question 7.

(x, y) is 5 unit away from the origin. How many such points lie in the third quadrant? [1]

(a) 0

(b) 1

(c) 2

(d) infinitely many

Answer:

(d) infinitely many

Let the point (- x, -y) lie in the third quadrant and it is at a distance of 5 units from the origin.

By distance formula, distance of (- x, -y)from origin is

\(\sqrt{(-x-0)^2+(-y-0)^2}=\sqrt{x^2+y^2}\)

⇒ \(\sqrt{x^2+y^2}\) = 5

On squaring both the sides, we get x² + y² = 25

Since, infinitely many values of (x, y) satisfy this equation, therefore infinitely many points lie in the third quadrant.

Question 8.

In ∆ABC, DE || AB. If AB = a, DE = x, BE = b and EC = c. Express x in terms of a, band c [1]

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

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

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

(d) \(\frac{a b}{b+c}\)

Answer:

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

In ∆CED and ∆CBA,

∠CED = ∠CBA [corresponding angles, ∵ DE || AB and CB is the transversal]

∠C is common angle.

∴ ∆CED ~ ∆CBA [by AA similarity criterion]

⇒ \(\frac{C E}{C B}=\frac{E D}{B A}\)

⇒ \(\frac{c}{c+b}=\frac{x}{a}\)

⇒ x = \(\frac{a c}{b+c}\)

Question 9.

If O is centre of a circle and chord PQ makes an angle 50° with the tangent PR at the point of contact P, then the angle made by the chord at the centre is [1]

(a) 130°

(b) 100°

(c) 50°

(d) 30°

Answer:

(b) 100°

In the given figure,

OP ⊥ PR [tangent to a circle is perpendicular to the radius through the point of contact]

⇒ ∠OPR = 90°

∠QPO = 90° – 50°= 40° ……(i)

Also, OQ = OP [radii of a circle]

⇒ ∠OQP = ∠OPQ [angles opposite to equal sides of a triangle]

⇒ ∠OQP = 40°

In ∆OPQ,

∠OQP + ∠OPQ + ∠POQ =180°

⇒ ∠POQ = 180° -2(40°)

= 100°

Question 10.

A quadrilateral PQRS is drawn to circumscribe a circle. If PQ =12 cm, QR = 15 cm and RS = 14 cm, then the length of SP is [1]

(a) 15 cm

(b) 14 cm

(c) 12 cm

(d) 11 cm

Answer:

(d) 11 cm

A quadrilateral PQRS is drawn to circumscribe a circle. As shown in figure.

Let T, U, V and W be the points of contact of the line segment SP, PQ, QP and RS, respectively.

Now, PT = PU ………..(i)

[tangents from an external point]

ST = WS ………….(ii)

RW = RV ⇒ PV = RW ………(iii)

QU = QV ⇒ QV = QU ………(iv)

Adding Eqs. (i), (ii), (iii) and (iv), we get

PT+ ST + RV+ OV = PU + WS + RW + QU

⇒ (PT + ST) + (RV + QV) = (PU + QU) + (WS + WR)

⇒ SP + RQ = PQ + RS

⇒ SP + 15 = 12 + 14

⇒ SP = 11 cm

Question 11.

Given that sin θ = \(\frac{a}{b}\), then cos θ is [1]

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

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

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

(d) \(\frac{a}{\sqrt{b^2-a^2}}\)

Answer:

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

Given, sin θ = \(\frac{a}{b}\)

We know that sin²θ + cos²θ = 1

⇒ cos²θ = 1 – \(\frac{a^2}{b^2}\)

⇒ cosθ = \(\frac{\sqrt{b^2-a^2}}{b}\)

Question 12.

(sec A + tan A) (1 – sin A) is equal to [1]

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Answer:

(d) cos A

We have, (sec A + tan A) (1 – sin A)

= sec A – sec A sin A + tan A – tan A sin A

= sec A – tan A + tan A – tan A sin A

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

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

Question 13.

A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is [1]

(a) 60°

(b) 45°

(c) 30°

(d) 90°

Answer:

(a) 60°

Let AS = 6 m be the height of the pole and length of the shadow SC = 2√3 m.

In right angled ∆ABC

tan C = \(\frac{A B}{B C}\) = \(\frac{6}{2√3}\) = √3

∠C = 60°

Question 14.

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is [1]

(a) 2 units

(b) n units

(c) 4 units

(d) 7 units

Answer:

(a) 2 units

Let the radius of a circle be r.

Given, 2πr = πr²

⇒ r² – 2r = 0

⇒ r(r – 2) = 0

r = 0 or r = 2

Since, radius can not be 0. Therefore, r = 2 units.

Question 15.

It is proposed to build a single circular park equal in area to the stun of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park is [1]

(a) 10 m

(b) 15 m

(c) 20 m

(d) 24 m

Answer:

(a) 10 m

Let the radius of new park be r m.

Let radius of two circular parks be r_{1} and r_{1}, respectively.

∴ πr_{1}² + πr_{2}² = πr²

⇒ r_{1}² + r_{2}² = r²

⇒ \(\left(\frac{16}{2}\right)^2+\left(\frac{12}{2}\right)^2=r^2\) = r²

⇒ 64+ 36 = r²

⇒ r² = 100 ⇒ r = 10 m

Question 16.

There is a green square board of side ‘2 a’ unit circumscribing a red circle. Jayadev is asked to keep a dot on the above said board. Then, the probability that he keeps the dot on the green region is [1]

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

(b) \(\frac{4-\pi}{4}\)

(c) \(\frac{\pi-4}{4}\)

(d) \(\frac{4}{\pi}\)

Answer:

(b) \(\frac{4-\pi}{4}\)

Given, the side of a green square board is 2a.

Radius of the circle = \(\frac{2 a}{2}\) = a units

∴ Area of the square board = (2a)² = 4a² sq units

∴ Area of red region = Area of circle = ita2 sq units

∴ Area of green region = Area of the square board – Area of red region

= (4a² – πa²)sq units

∴ Probability (keeping the dot on the green region)

= \(\frac{\text { Area of the green region }}{\text { Area of the square board }}\)

= \(\frac{4 a^2-\pi a^2}{4 a^2}\) = \(\frac{4-\pi}{4}\)

Question 17.

2 cards of heart and 4 cards of spade are missing from a pack of 52 cards. What is the probability of getting a black card from the remaining pack? [1]

(a) \(\frac{22}{52}\)

(b) \(\frac{22}{46}\)

(c) \(\frac{24}{52}\)

(d) \(\frac{24}{46}\)

Answer:

(a) \(\frac{22}{52}\)

Total cards left in the pack = 52 – 6 = 46

Black cards left in the pack = 26 – 4 = 22

[∵ total black cards in a complete pack = 26]

∴ P(getting a black card from the remaining pack) = \(\frac{22}{46}\)

Question 18.

The upper limit of the modal class from the given distribution. [1]

Height [in cm] | Below 140 | Below 145 | Below 150 | Below 155 | Below 160 | Below 165 |

Number of girls | 4 | 11 | 29 | 40 | 46 | 51 |

(a) 165

(b) 160

(c) 155

(d) 150

Answer:

(d) 150

Let us make the table for the given distribution

Height (in cm) | Class-interval (in cm) | Cumulative frequency | Frequency |

Below 140 | 0-140 | 4 | 4 |

Below 145 | 140-145 | 11 | 7 |

Below 150 | 145-150 | 29 | 18 |

Below 155 | 150-155 | 40 | 11 |

Below 160 | 155-160 | 46 | 6 |

Below 165 | 160-165 | 51 | 5 |

∴ Modal class is 145-150 and upper limit of model class is 150.

Question 19.

Direction In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. [1]

Assertion (A) : Total surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.

Reason (R) : Top is obtained by fixing the plane surfaces of the hemisphere and cone together.

(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 and 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).

Given,

Total surface area (TSA) of top

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

This is because the top is obtained by joining the plane surfaces of hemisphere and cone.

Question 20.

Assertion (A) : -5, \(\frac{-5}{2}\), 0, \(\frac{5}{2}\), …….. is in Arithmetic Progression.

Reason (R) : An Arithmetic Progression cannot have both positive and negative rational numbers. [1]

(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 and 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.

Given progression is -5, \(\frac{-5}{2}\), 0, \(\frac{5}{2}\), ……..

Here, first term a_{1} = -5,

second term a_{2} = \(\frac{-5}{2}\),

third term a_{3} = 0,

and fourth term a_{4} = \(\frac{5}{2}\)

Now,

a_{2} – a_{1} = \(\frac{-5}{2}\) – (-5)

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

a_{3} – a_{2} = 0 – \(\frac{-5}{2}\) = \(\frac{5}{2}\)

a_{4} – a_{3} = \(\frac{-5}{2}\) – 0 = \(\frac{5}{2}\)

∵ Common difference between each term is \(\frac{5}{2}\)

∴ Assertion (A) is true.

Section B

(Section-B consists of 5 Questions of 2 marks each)

Question 21.

Prove that √2 is an irrational number. [2]

Solution:

Let us assume that √2 is rational.

The √2 can be expressed as \(\frac{p}{q}\), where p and q are

co-primes and q 0.

∴ √2 = \(\frac{p}{q}\)

On squaring both sides, we get

2 = \(\frac{p^2}{q^2}\) ⇒ \(p^2=2 q^2\) ………(i) (1/2)

⇒ p^{2} is a multiple of 2

=> 2 divides p^{2}

∴ 2 divides p.

∴ p = 2a for some integer a.

Substituting p = 2a in Eq. (i), we get

(2a)^{2} = 2q^{2}

⇒ 4a^{2} = 2q^{2}

⇒ q^{2} = 2a^{2}

⇒ q^{2} is a multiple of 2

⇒ 2 divides q^{2}

∴ 2 divides q (1/2)

⇒ p and q have atleast 2 as a common factor.

But this contradicts the statement that p and q are co-primes.

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

∴ √2 is irrational. (1)

Question 22.

ABCD is a parallelogram. Point P divides AB in the ratio 2 : 3 and point Q divides DC in the ratio 4:1. [2]

Prove that OC is half of OA.

Solution:

Given,

In ∆OAP and ∆OCQ,

∴ ∠AOP = ∠COQ [vertically opposite angles]

∴ ∠OAP = ∠OCQ [alternate angles]

⇒ ∆OAP ~ ∆OCQ [by AA similarity criterion]

∴ \(\frac{O A}{O C}=\frac{A P}{C Q}\) ………….(i) (1)

Given, \(\frac{A P}{P B}=\frac{2}{3}\) and \(\frac{D Q}{Q C}=\frac{4}{1}\)

\(\frac{A P}{A B}=\frac{2}{5}\) and \(\frac{Q C}{D C}=\frac{1}{5}\)

AP = \(\frac{2}{5}\) AB and QC = \(\frac{1}{5}\) DC (1/2)

From Eq. (i),

\(\frac{O A}{O C}=\frac{\frac{2}{5} A B}{\frac{1}{5} D C}\) = 2

[∵ ABCD is a parallelogram ∴ AB = CD]

⇒ OC = \(\frac{1}{2}\) OA (1/2)

Question 23.

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. [2]

At a point E on the circle, a tangent is drawn to intersect PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of APCD.

Solution:

Given, PA = 10 cm

PB = PA = 10 cm

[∵ tangents from an external point to a circle are equal]

Similarly, DE = BD

and CE = AC

⇒ DE + CE = BD + AC

⇒ CD=BD + AC ..(i) (1)

∴ Perimeter of APCD = PC + PD + CD

= (PA – AC) + (PB – BD) + CD

= PA – AC + PB – BD + BD + AC [Using Eq. (i)]

= PA + PB = 10+ 10 =20 cm (1)

Question 24.

If tan(A + B) = √3 and tan(A – B) = \(\frac{1}{\sqrt{3}}\) ; 0°< A + B < 90°; A > B, find A and B. [2]

Or

Find the value of x

2 cosec²30° + x sin²60° – tan²30° = 10

Solution:

Given, tan(A + B) = √3 = tan 60°

and tan(A – B) = –\(\frac{1}{\sqrt{3}}\) = tan 30°

⇒ A + B = 60° and A – B = 30° (1)

⇒ 2A = 90°

⇒ A = 45°

and B = 15° (1)

Or

We have,

2 cosec²30° + x sin²60° – \(\frac{3}{4}\)tan²30° = 10

⇒ \(\frac{3}{4}\) =

⇒ 8 + \(\frac{3x}{4}\) – \(\frac{1}{4}\) = 10

⇒ \(\frac{3x}{4}\) = 2 + \(\frac{1}{4}\) = \(\frac{9}{4}\)

⇒ x = 3 (2)

Question 25.

With vertices A, B and C of ∆ABC as centres, arcs are drawn with radii 14 cm and the three portions of the triangle, so obtained are removed. Find the total area removed from the triangle. [2]

Or

Find the area of the unshaded region shown in the given figure.

Solution:

According to the given information, the figure is given below.

Or

The given figure is

Let the radius of each semi-circle = r

∴ From figure,

3 + r + 2r + r + 3 = 14

4r + 6 = 14

r = \(\frac{8}{4}\) = 2 cm (1)

Area of unshaded region

= 4 × Area of each semi-circle + Area of square with length of each side, 2r

= 4 × \(\frac{\pi r^2}{2}\) + (2r)^{2}

= 2πr^{2} + 4r^{2} = 2r^{2}(π + 2) cm^{2} (1)

Section C

(Section-C consists of 6 Questions of 3 marks each)

Question 26.

National Art convention got registrations from students from all parts of the country, of which 60 are interested in music, 84 are interested in dance and 108 students are interested in handcrafts. For optimum cultural exchange, organisers wish to keep them in minimum number of groups such that each group consists of students interested in the same artform and the number of students in each group is the same. Find the number of students in each group. Find the number of groups in each art form. How many rooms are required if each group will be allotted a room? [3]

Solution:

Students interested in music = 60

Students interested in dance = 84

Students interested in handcrafts = 108

Prime factorisation of 60, 84 and 108 are

60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

84 = 2 × 2 × 3 × 7 = 2² × 3 × 7 (1)

and 108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³ respectively.

HCF (60, 84, 108) = 2² × 3 = 12

Number of students in each group = HCF (60, 84, 108) =12

Number of groups in music = \(\frac{60}{12}\) = 5

Number of groups in dance = \(\frac{84}{12}\) = 7

Number of groups in handcrafts = \(\frac{108}{12}\) = 9

Rooms required if each group will be allotted a room

= Total number of groups = 5+7 + 9 = 21 (1)

Question 27.

If α, ß are zeroes of quadratic polynomial x² +5x + 1, find the value of [3]

(i) α² + ß²

(ii) α^{-1} + ß^{-1}

Solution:

Given, a and p are the roots of the quadratic polynomial x² + 5x + 1.

Sum of roots = α + ß = \(\frac{-5}{1}\) = -5 (1)

Product of roots = αß = 1

(i) α^{2} + ß^{2} = (α + ß)^{2} – 2αß = 25 – 2 = 23 (1)

(ii) α^{-1} + ß^{-1} = \(\frac{1}{α}\) + \(\frac{1}{ß}\)

= \(\frac{α + ß}{αß}\) = \(\frac{-5}{1}\) = -5 (1)

Question 28.

The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? [3]

Or

Solve the following pair of linear equation by using substitution method

2x + 3y = 11

2x-4y = -24

Solution:

Let the digit at unit place be x

and the digit at tens place be y.

∴ The two-digit number is 10y + x.

Now, the number obtained by reversing the digits = 10x + y (1)

According to the question,

(10y + x) + (10x + y) = 66

⇒ 11(x + y) = 66

⇒ x + y = 6 ……. (i)

Also, the digits of the number differ by 2

⇒ x – y = 2 ……….. (ii)

or y – x = 2 ……… (iii) (1)

On solving Eqs. (i) and (ii),

we get x = 4, y = 2

On solving Eqs. (i) and (iii),

we get y = 4, x = 2

∴ Two such numbers are there

They are 10(4) + 2 = 42

and 10(2) + 4 = 24 (1)

Or

Given, 2x + 3y = 11 ………..(i)

2x – 4y = – 24 ………..(ii)

From Eq. (i),

2x = 11 – 3y

Substituting the value of 2x in Eq. (ii), we get

11 – 3y – 4y = – 24

⇒ 11 – 7y = -24

⇒ 7y = 35

⇒ y = 5

Now, from 2x = 11 – 3y

Substituting y = 5 in above relation, we get

2x = 11 – 3 × 5

= 11 – 15

= – 4

⇒ x = -2

Hence, x = – 2 and y = 5

Question 29.

PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes an angle of 30° with the radius at the point of contact.

If length of the chord is 6 cm, find the length of the tangent PA and the length of the radius OA.

Or

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. [3]

Solution:

The given figure is

In ∆OAB

∠OAB + ∠OBA + ∠AOB = 180°

⇒ ∠AOB + 2 × 30° = 180°

∠AOB = 120°

Draw ON ⊥ AB.

⇒ AN = \(\frac{6}{2}\) = 3cm

[∵ perpendicular from centre to a chord bisects the chord]

In right angled ∆ONA,

cos 30° = \(\frac{A N}{A O}\)

⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{3}{A O}\)

⇒ AO = 2√3 cm

Now, OB ⊥ BP [tangent to a circle is perpendicular to the radius through the point of contact]

⇒ ∠OBP = 90°

⇒ ∠ABP = 90° – ∠OBA

= 90° – 30°

= 60°

Similarly,

∠BAP = 60°

∴ In ∆ABP

∠APB = 60°

∆ABP is an equilateral triangle

∴ AB = BP = PA = 6cm

Or

Given, two tangents TP and TQ drawn to a circle, as shown in the figure.

OP ⊥ PT

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

⇒ ∠OPT = 90°

⇒ ∠TPQ =90° -∠OPQ …(i)

In ATPQ,

TP = TQ

[tangents to a circle from an external point]

⇒ ∠TPQ = ∠TQP

[angles opposite to equal sides are equal]

and ∠TPQ + ∠TQP + ∠PTQ =180°

⇒ ∠PTQ = 180° – 2 ∠TPQ

⇒ ∠PTQ = 180° – 2(90° – ∠OPQ) [using Eq. (i)]

⇒ ∠PTQ = 2∠OPQ

Question 30.

If 1 + sin²θ = 3 sin θ cos θ, then prove that tan θ = 1 or \(\frac{1}{2}\). [3]

Solution:

Given, 1 + sin²θ = 3 sin θ cos θ

⇒ sin²θ + cos²θ + sin²θ = 3 sin θ cos θ

[∵ sin² A + cos² A = 1]

⇒ 2 sin²θ + cos²θ – 3 sin θ cos θ = 0

⇒ 2 sin²θ – 3 sin θ cos θ + cos²θ = 0

⇒ 2 sin²θ – 2 sin θ cos θ – sin θ cos θ + cos²θ = 0

⇒ 2 sin θ(sin θ – cos θ) – cos θ(sin θ – cos θ) = 0

⇒ (sin θ – cos θ) (2 sin θ – cos θ) = 0

⇒ sin θ – cos θ = 0 or 2 sin θ – cos θ = 0

⇒ sin θ = cos θ or 2 sin θ = cos θ

⇒ tan θ = 1 or tan θ = – 2

Question 31.

The length of 40 leaves of a plant are measured correct to nearest millimetre and the data obtained is represented in the following table. [3]

Length [in mm] | Number of leaves |

118-126 | 3 |

127-135 | 5 |

136-144 | 9 |

145-153 | 12 |

154-162 | 5 |

163-171 | 4 |

172-180 | 2 |

Find the average length of the leaves.

Solution:

Let us make the table for given data.

Section D

(Section-D consists of 4 Questions of 5 marks each)

Question 32.

A motor boat whose speed is 18 km/h in still water takes 1 h more to go 24 km upstream than to return downstream to the same spot. Find the speed of stream. [5]

Or

Two water taps together can fill a tank in \(9 \frac{3}{8}\) h. The tap of larger diameter takes 10 h less than the smaller one to fill the tank separately.

Find the time in which each tap can separately fill the tank.

Solution:

Let the speed of the stream be x km/h.

Then, speed of the boat downstream = (18 + x) km/h

and speed of the boat upstream = (18 – x) km/h

Then, according to the question,

\(\frac{24}{18-x}\) – \(\frac{24}{18+x}\) = 1

⇒ 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)

⇒ 24x + 24x = 324 – x²

⇒ x² + 48x – 324 = 0

⇒ x² + 54x – 6x – 324 = 0 (1)

⇒ x(x + 54) – 6(x + 54) = 0

⇒ (x + 54) (x – 6) = 0

∴ x = 6

[∵ x cannot be negative]

∴ Speed of stream = x = 6 km/h

Or

Let time taken by tap of smaller diameter be t h.

∴ Time taken by tap of larger diameter = (t -10) h

Given, time taken by both the taps to fill the tank

= \(9 \frac{3}{8}\) = \(\frac{75}{8}\)h

Portion of the tank filled by tap of smaller diameter in

1 h = \(\frac{1}{t}\)

Portion of the tank filled by tap of larger diameter in

1 h = \(\frac{1}{t-10}\)

Portion of the tank filled by both the taps in 1 h

= \(\frac{1}{\frac{75}{8}}\) = \(\frac{8}{75}\)

∴ \(\frac{1}{t}\) – \(\frac{1}{t-10}\) = \(\frac{8}{75}\)

⇒ \(\frac{(t-10)+t}{t(t-10)}\) = \(\frac{8}{75}\)

⇒ \(\frac{2t-10}{t^2-10t}\) = \(\frac{8}{75}\)

⇒ 150t – 750 = 8t² – 80t

⇒ 8t² – 230t + 750 = 0

⇒ 4t² – 115t + 375 = 0

⇒ 4t² – 100t – 15f + 375 = 0

⇒ 4t(t – 25) – 15(t – 25) = 0

⇒ (t – 25)(4t – 15) = 0

t = 25 or t = \(\frac{15}{4}\)

If t = 25, then time taken by tap of larger diameter

= 25 – 10 = 15 h

If t = \(\frac{15}{4}\), then time taken by tap of larger diameter

= \(\frac{15}{4}\) – 10 = \(– \frac{25}{4}\) which is not possible.

∴ Time taken by taps of smaller and larger diameter are 25 h and 15 h, respectively.

Question 33.

(a) State and prove Basic Proportionality theorem.

(b) In the given figure ∠CEF = ∠CFE. F is the mid-point of DC.

Prove that \(\frac{A B}{B D}=\frac{A E}{F D}\). [5]

Solution:

(a) Statement If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, the other two sides are divided in the same ratio.

Proof

Let ABC be a triangle in which a line parallel to side SC intersects the other two sides AB and AC at D and E, respectively.

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

Construction: Join CD, BE and draw DM ⊥ AC and EN ⊥ AB.

Since, ∆BDE and ∆CED are on the same base DE and between the same parallel lines DE and SC.

∴ ar(∆BDE) = ar(∆CED) ………(iii)

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

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

(b) The given figure is

Question 34.

Water’ is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?

What should be the speed of water, if the rise in water level is to be attained in 1 h?

Or

A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m, respectively and the total height of the tent is 13.5 m, find the area of the canvas required for making the tent, keeping a provision of 26 m² of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of ₹ 500 per m². [5]

Solution:

Given, diameter of cylindrical pipe = 14 cm

⇒ Radius of cylindrical pipe (r) = 7 cm

Also, length of the pond (l) = 50 m,

Breadth of the pond (b) = 44 m,

Height of the water level in the pond (h) = 21 cm

Lett h be the time taken to raise the level of water by 21 cm.

Length of the water column in f h = (15 × t) km

= 15000t m³

Volume of water column that come out from cylindrical pipe in t h

= π × \(\left(\frac{7}{100}\right)^2\) × 15000t

∴ Volume of water column =Volume of cuboid with height h

⇒ π × \(\left(\frac{7}{100}\right)^2\) × 15000t = 50 × 44 × \(\frac{21}{100}\)

⇒ \(\frac{22}{7}\) × \(\frac{7^2}{10}\) × 15t = 22 × 21

⇒ 7 × 15t = 210 ⇒ t = 2 h

Hence, in 2 h, level of water in pond rise by 21 cm.

Let v km/h be the speed of water.

Length of the water column in 1 h = v km = (v × 1000)m

∴ Volume of water column

= π × \(\left(\frac{7}{100}\right)^2\) × v × 1000

= \(\frac{22 \times 7 \times v}{10}\) m³

∴ \(\frac{22 \times 7 \times v}{10}\) = 50 × 44 × \(\frac{21}{100}\)

⇒ v = 30 km/h

Hence, speed of water is 30 km/h.

Or

Given, total height of the tent = 13.5 m

and radius of the base of the tent=14 m

As shown in figure,

Height of the cone, = Total height – Height of cylinder.

= 13.5 – 3 = 10.5m

Radius of cone and radius of cylinder r = 14 m

Slant height / of cone = \(\sqrt{h_1^2 + r^2}\)

=\(\sqrt{(105)^2 + (14)^2}\) = 17.5 m

Curved surface area of cone = πrl

= π × 14 × 17.5 = 245 π m²

Curved surface area of cylinder = 2πrh²,

where h² is height of cylinder

= 2π × 14 × 3 = 847t m²

∴ Area of canvas required = 245π + 84π + 26

= 245 × \(\frac{22}{7}\) + 84 × \(\frac{22}{7}\) + 26

= 770 + 264 + 26= 1060 m²

Cost of canvas = ₹ (500 x 1060) = ₹ 530000

Question 35.

The median of the following data is 50. Find the values of ‘p’ and ‘q’ , if the sum of all frequencies is 90. Also, find the mode. [5]

Marks obtained | Number of students |

20-30 | p |

30-40 | 15 |

40-50 | 25 |

50-60 | 20 |

60-70 | q |

70-80 | 8 |

80-90 | 10 |

Solution:

Table for cumulative frequency is given below

Marks obtained | Number of students (f) | Cumulative frequency |

20-30 | p | P |

30-40 | 15 | p+ 15 |

40-50 | 25 | p+ 40 |

50-60 | 20 | p+ 60 |

60-70 | q | p+ q + 60 |

70-80 | 8 | p+ q + 68 |

80-90 | 10 | p + q + 78 |

M = Σf = p + q + 78 = 90 | Cumulative frequency |

⇒ p + q = 12 …(i)

Since, median = 50, therefore median class is 50-60.

∴ Median = \(I+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)

Where, l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class

h = class size

Here, \(\frac{n}{2}\) = \(\frac{90}{2}\) = 45

l = 50, cf = p + 40, f = 20 and h = 10

∴ 50 = 50 + \(\left(\frac{45-(p+40)}{20}\right)\) × 10

⇒ 45 – (p + 40) = 0

⇒ p + 40 = 45

⇒ p = 5

From Eq. (i),

q = 12 – p = 12 – 5 = 7

Now, modal class = 40-50

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

Where, l = lower limit of modal class,

f_{1} = frequency of modal class,

f_{0} = frequency of class preceeding the modal class,

f_{2} = frequency of class succeeding the modal class,

h = size of the class interval.

Here, l = 40, f_{1} = 25, f_{0} = 15, f_{2} = 20 and h = 10

∴ Mode = 40 + \(\left(\frac{25-15}{2(25)-15-20}\right)\) × 10

= 40 + \(\frac{10}{15}\) × 10

= 40 + \(\frac{20}{3}\)

= 46.66

Section E

(Case study based questions are compulsory)

Question 36.

Manpreet Kaur is the national record holder for women in the shot-put discipline. Her throw of 18.86 m at the Asian Grand Prix in 2017 is the biggest distance for an Indian female athlete.

Keeping her as a role model, Sanjitha is determined to earn gold in Olympics one day.

Initially her throw reached 7.56 m. only. Being an athlete in school, she regularly practiced both in the mornings and in the evenings and was able to improve the distance by 9 cm every week.

During the special camp for 15 days, she started with 40 throws and every day kept increasing the number of throws by 12 to achieve this remarkable progress

(i) Flow many throws Sanjitha practiced on 11th day of the camp? [1]

(ii) What would be Sanjitha’s throw distance at the end of 6 months? [2]

Or

When will she be able to achieve a throw of 11.16m? [2]

(iii) How many throws did she do during the entire camp of 15 days ? [1]

Solution:

(i) Number of throws on first day a = 40

Number of throws on second day = 40 + 12 = 52

AP formed by number of throws is 40, 52, (52 + 12)…

(I day) (II day) (III day)

Here, common difference,

d = 12

Number of throws she practiced on 11th day

a_{11}= a + (11 – 1)d

= 40 + 10 × 12 = 160

Distance of Sanjitha’s throw on first day,

a = 7.56 m

Distance of Sanjitha throw after a week

= (7.56 + 0.09) m = 7.65 m

AP formed by the distances is

7.56, 7.65, 7.65 + (0.09) ………

(I week) (II week) (III week)

Here, common difference, d = 0.09 m

Sanjitha’s throw distance at the end of 6 months (i.e. 26 weeks)

a_{26} = a + (26 – 1)d

= 7.56 + 25 × 0.09

= 7.56 + 2.25

= 9.81 m

Or

Here, a_{n} = 11.16 = a + (n – 1) d

⇒ 7.56 + (n – 1)0.09 = 11.16

⇒ n – 1 = \(\frac{3.6}{0.09}\) = n = 41

∴ On 41st week, she will be able to achieve a throw of 11.16 m.

(iii) Total throws done by her during the entire camp of

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

where n is total number of terms

⇒ S_{15} = \(\frac{15}{2}\) [2 × 40 + 14 × 12]

= 15[40 + 84] = 15 × 124 = 1860

Question 37.

Tharunya was thrilled to know that the football tournament is fixed with a monthly time frame from 20th July to 20th August 2023 and for the first time in the FIFA Women’s World Cup’s history, two nations host in 10 venues. Her father felt that the game can be better understood, if the position of players is represented as points on a coordinate plane.

(i) At an instance, the midfielders and

forward formed a parallelogram. Find the position of the central midfielder (D), if the position of other players who formed the parallelogram are A(1, 2), 5(4, 3) and C (6, 6). [1]

(ii) Check if the Goal keeper G(-3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line. [2]

Or

Check if the full-back J(5, – 3) and centre-back I(- 4, 6) are equidistant from forward C(0, 1) and if C is the mid-point of IJ. [2]

(iii) If Defensive midfielder A(1, 4), Attacking midfielder B(2, -3) and Striker E(a, b) lie on the same straight line and B is equidistant from A and E, find the position of E. [1]

Solution:

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

(ii) Distance between G and H,

Or

(iii) ∵ 6(2, – 3) is equidistant from A(1,4) and E(a,b)

∴ B is the mid-point of AE.

⇒ a + 1 = 4 and b + 4 = – 6

⇒ a = 3 and b = – 10

Question 38.

One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80m. He observed a bird on the tree at an angle of elevation of 45°.

When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 s, he observed the bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60°.

(i) At what distance from the foot of the tree was he observing the bird sitting on the tree? [1]

(ii) How far did the bird fly in the mentioned time? [2]

Or

After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? [2]

(iii) What is the speed of the bird in m/min, if it had flown 20(√3 + 1) m? [1]

Solution:

(i) In right angled ∆ABC,

tan 45° = \(\frac{A B}{B C}\)

⇒ BC = AB = 80m …(i)

[∵ AB = 80 m and tan 45° = 1]

∴ Required distance = 80 m

(ii) In right angled ∆DEC,

(iii) Speed of the bird = \(\frac{\text { Distance }}{\text { Time }}\)

\(\frac{20(\sqrt{3}+1)}{2}\) m/s

= 10(√3 + 1) m/s

= 600(√3 + 1) m/min

[∵1s = \(\frac{1}{60}\) min 60]