Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 12 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 12 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions

- This question paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment (4 marks each).
- 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.
- 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 \(\sqrt{3}\)tanθ = 3sinθ. Then, the value of sin^{2}θ – cos^{2}θ is (1)

(a) 3

(b) 1

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

(d) \(\sqrt{3}\)

Answer:

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

Question 2.

If tan^{2} 45° – cos^{2} 30° = x sin 30°cos 60°, then the value of x is (1)

(a) 0

(b) 2

(c) 1

(d) -1

Answer:

(c) 1

We have, tan^{2} 45° – cos^{2} 30° = xsin 30°cos 60°

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

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

⇒ x = 1

Question 3.

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (1)

(a) perpendicular

(b) coincident

(c) neither coincident nor perpendicular

(d) None of the above

Answer:

(b) coincident

If a pair of linear equations in two variable is consistent, then the lines represented by two equations are coincident.

Question 4.

If x – 2 and x = 0 are the zeroes of the polynomial f(x) = 5x^{2} + ax + b. Then, the values of a and b are (1)

(a) a = 2, b = 1

(b) a = 10, b = 1

(c) a = 1, b = 2

(d) a = -10, b = 0

Answer:

(d) a = -10, b = 0

Given, f(x) = 5x^{2} + ax + b

Since, x = 2 and x = 0 are the zeroes of f(x).

f(2) = 0 and f(0) = 0 ⇒ 5(2)^{2} + 2a + b = 0

and 5(0) + a(0) + b = 0

⇒ 20 + 2a + b = 0 and b = 0

⇒ 20 + 2a + 0 = 0 and b = 0

⇒ 2a = -20 and b = 0

a = -10 and b = 0

Question 5.

The value of k, for which one root of the quadratic equation kx^{2} – 14x + 8 = 0 is six times the other, is (1)

(a) 0

(b) 1

(c) 2

(d) 3

Answer:

(d) 3

Let a and 6a be the roots of the quadratic equation kx^{2} -14x + 8 = 0, then Sum of roots,

α + 6α = –\(\left(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\right)\) = \(\frac{-(-14)}{k}\)

⇒ 7α = \(\frac{14}{k}\) ⇒ α = \(\frac{2}{k}\) ……….(i)

and product of roots

α × 6α = –\(\left(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\right)\) = \(\frac{8}{k}\)

⇒ 6α^{2} = \(\frac{8}{k}\) ….. (ii)

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

6\(\left(\frac{2}{k}\right)^3\) = \(\frac{8}{k}\) ⇒ \(\frac{6 \times 4}{k^2}\) = \(\frac{8}{k}\)

k = \(\frac{24}{8}\) = 3

Question 6.

The ratio in which the X-axis divides the line segment joining A(3, 6) and B(12, -3) is (1)

(a) 1 : 2

(b) 2 : 1

(c) -2 : 1

(d) -1 : 2

Answer:

(b) 2 : 1

Let P(x, 0) be the point of intersection of X-axis with the line segment joining A(3, 6) and B(12, -3) which divides the line segment AB in the ratio λ : 1.

Then, by using distance formula,

(x, 0) = \(\left(\frac{12 \lambda+3}{\lambda+1}, \frac{-3 \lambda+6}{\lambda+1}\right)\)

Now, equating the y component of both sides, we get

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

⇒ -3λ + 6 = 0 ⇒ λ = \(\frac{2}{1}\)

So, X-axis divides AB in the ratio 2: 1.

Question 7.

The area of a sector of circle of radius 21 cm and central angle 90° is (1)

[take π = \(\frac{22}{7}\)]

(a) 346.5 cm^{2}

(b) 340 cm^{2}

(c) 341.5 cm^{2}

(d) None of these

Answer:

(a) 346.5 cm^{2}

Let O be the centre of the circe of radius 21 cm such that \(\widehat{A P B}\) subtends 90° at the centre.

Question 8.

The smallest number by which \(\sqrt{27}\) should be multiplied, so as to get a rational number is (1)

(a) \(\sqrt{27}\)

(b) 3\(\sqrt{3}\)

(c) \(\sqrt{3}\)

(d) 3

Answer:

(c) \(\sqrt{3}\)

As \(\sqrt{27}\) = \(\sqrt{3 \times 3 \times 3}\) = 3\(\sqrt{3}\)

If we multiply it by \(\sqrt{3}\), then it will become 3\(\sqrt{3}\) × \(\sqrt{3}\) = 3 × 3 = 9

i.e. a rational number.

Question 9.

It is given that ΔABC ~ ΔDFE, ∠A = 50°, ∠C = 30°, AB = 10 cm, AC = 15 cm and DF = 8 cm. Then, which of the following is true? (1)

(a) DE = 12 cm and ∠F = 50°

(b) DE = 12 cm and ∠F = 100°

(c) EF = 12 cm and ∠D = 100°

(d) EF = 12 cm and ∠D = 30°

Answer:

(b) DE = 12 cm and ∠F = 100°

Given, ΔABC ~ ΔDFE

Then, ∠A = ∠D = 50°

∠C = ∠E = 30°

∴ ∠B = ∠F = 180° – (30° + 50°) = 100°

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

⇒ \(\frac{10}{8}\) = \(\frac{15}{D E}\)

⇒ DE = \(\frac{15 \times 8}{10}\) = 12 cm

Hence, DE = 12 cm and ∠F = 100°

Question 10.

The tangents drawn at the extremities of the diameter of a circle are (1)

(a) parallel

(b) perpendicular

(c) neither parallel nor perpendicular

(d) None of the above

Answer:

(a) parallel

Since, OP ⊥ AB and OQ ⊥ CD

∴ ∠1 = 90° and ∠2 = 90°

⇒ ∠1 = ∠2, which are alternate angles.

∴ AB || CD

Question 11.

Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, then the distance between their tops is (1)

(a) 14 m

(b) 16 m

(c) 13 m

(d) 15 m

Answer:

(c) 13 m

Let AB and CD be the vertical poles.

AB = 6 m, CD = 11 m and AC = 12 m

Draw BE || AC,

DE = (CD – CE) = (11 – 6)m = 5 m

In rightangled ΔBED, using Pythagoras theorem,

(BD)^{2} = (BE)^{2} + (DE)^{2}

= 12^{2} + 5^{2} = 169

∴ BD = \(\sqrt{169}\)m = 13 m

Hence, distance between their tops = 13 m

Question 12.

A toy is the shape of cone over a hemisphere and the radius of the hemisphere is 3.5 cm. If the height of the toy is 15.5 cm, then the total area of the toy is (1)

(a) 214.5 cm^{2}

(b) 215.4 cm^{2}

(c) 216.5 cm^{2}

(d) 210 cm^{2}

Answer:

(a) 214.5 cm^{2}

Height of the cone = (15.5— 3.5)cm = 12 cm

Slant height, l = \(\sqrt{h^2+r^2}\) = \(\sqrt{12^2+(3.5)^2}\)

= \(\sqrt{144+12.25}\) = 12.5 cm

Curved surface area of the cone

= πrl = \(\frac{22}{7}\) × 3.5 × 12.5 = 137.5 cm^{2}

Curved surface area of the hemisphere = 2πr^{2}

= \(\frac{2 \times 22}{7}\) × 3.5 × 3.5 cm^{2} = 77 cm^{2}

∴ Total area of the toy = (137.5 + 77) cm^{2}

= 214.5 cm^{2}

Question 13.

If the nth term of an AP is 3n – 8, then its 16th term is (1)

(a) 30

(b) 20

(c) 10

(d) 40

Answer:

(d) 40

Given, a_{n} = 3n – 8

∴ a_{16} = 3 × 16 – 8 = 48 – 8 = 40

Question 14.

In an isosceles right angled triangle, if the hypotenuse is 5\(\sqrt{2}\) cm, then the length of the sides of triangle is (1)

(a) 4 cm

(b) 6 cm

(c) 5 cm

(d) None of these

Answer:

(c) 5 cm

Let ABC is an isosceles right angled triangle, right angled at B with AB = BC and AC = 5\(\sqrt{2}\)cm.

By pythagoras theorem,

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

⇒ (AC)^{2} = 2(AB)^{2}

⇒ (5\(\sqrt{2}\))^{2} = 2(AB)^{2}

⇒ \(\frac{50}{2}\) = (AB)^{2}

⇒ (AB)^{2} = 25

⇒ AB = 5 cm

Hence, the length of the equal sides of a triangle is 5 cm.

Question 15.

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

(a) – 6 and 1

(b) – 6 and 2

(c) 6 and – 1

(d) 6 and – 2

Answer:

(a) – 6 and 1

Given, C is the mid-point of PO i.e. P(4, x) and Q(-2, 4)

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

⇒ (y, -1) = (1, \(\frac{x+4}{2}\))

On equating the coordinates, we get

y = 1 and -1 = \(\frac{x+4}{2}\)

∴ x = -6 and y = 1

Question 16.

The mean and median of a distribution are 16 and 17, respectively. Then, the value of mode is (1)

(a) 17

(b) 19

(c) 16

(d) 15

Answer:

(b) 19

Given, mean =16 and median = 17

By empirical relationship, we have

Mode = 3 median – 2 mean

= 3 × 17 – 2 × 16

= 51 – 32 = 19

Question 17.

For grouped data, if Σf_{i} =20, Σf_{i}x_{i} = 2p + 20 and mean of distribution is 12, then the value of p is (1)

(a) 110

(b) 100

(c) 90

(d) 120

Answer:

(a) 110

Given, Σf_{i} = 20, Σf_{i}x_{i} = 2p + 20 and mean = 12

We know that

mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

∴ 12 = \(\frac{2 p+20}{20}\)

⇒ 12 × 20 = 2p + 20

⇒ 240 = 2p + 20

⇒ 2p = 240 – 20

⇒ 2p = 220

⇒ p = \(\frac{220}{2}\) = 110

Question 18.

The difference between the circumference and the radius of a circle is 37 cm. The area of the circle is (1)

(a) 149 cm^{2}

(b) 154 cm^{2}

(c) 121 cm^{2}

(d) 169 cm^{2}

Answer:

(b) 154 cm^{2}

Let r be the radius of the circle.

Given, circumference of circle – radius of the circle = 37

⇒ 2πr – r = 37

⇒ r(2π -1) = 37

⇒ r = \(\frac{37}{2 \pi-1}\)

= \(\frac{37}{2\left(\frac{22}{7}\right)-1}\) = \(\frac{37 \times 7}{37}\) = 7 cm

∴ Area of circle = πr^{2} = \(\frac{22}{7}\) × (7)^{2} = 154 cm^{2}

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

Question 19.

Assertion (A) Points (3, 2), (-2, -3) and (2, 3) form a right triangle.

Reason (R) If (x, y) is equidistant from (3, 6) and (-3, 4), then 3x + y = 5. (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 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).

Let A(3, 2), B(-2, -3) and C(2, 3).

∴ AB = \(\sqrt{(-2-3)^2+(-3-2)^2}\) = \(\sqrt{50}\) units

BC = \(\sqrt{(-2-2)^2+(-3-3)^2}\) = \(\sqrt{52}\) units and

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

∴ BC^{2} = AB^{2} + CA^{2}

⇒ ΔABC is a right triangle.

Let A'(3, 6), B'(-3, 4) and P (x, y)

Since, P is equidistant from A’ and B’.

∴ PA’ = PB’

⇒ PA’^{2} = PB’^{2}

⇒ (x – 3)^{2} + (y – 6)^{2} = (x + 3)^{2} + (y – 4)^{2}

⇒ x^{2} – 6x + 9 + y^{2} – 12y + 36 = x^{2} + 6x + 9 + y^{2} – 8y + 16

⇒ 12x + 4y = 20 ⇒ 3x+y = 5

Both Assertion (A) and Reason (R) are true but the Reason (R) is not correct explanation of the Assertion – (A).

Question 20.

Assertion (A) 2 is a rational number.

Reason (R) The square roots of all positive integers are irrational. (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 but Reason (R) is not the correct explanation of Assertion (A).

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

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

Answer:

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

∵ \(\sqrt{4}\) = ±2, which is not an irrational number.

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

Section B

(Section B consists of 5 questions of 2 marks each.)

Question 21.

If two tangents inclined at an angle of 60° are drawn to a circle of radius 5 cm, then find the length of each tangent. (2)

Answer:

In the given figure, OA = 5 cm and ∠APB = 60°

In ΔAOP, ∠PAO = 90°

[∵ the tangent at any point of a circle is perpendicular to the radius through the point of contact] (1)

Since, ∠APB = 60°

⇒ ∠APO = \(\frac{60^{\circ}}{2}\) = 30°

∴ tan 30° = \(\frac{A O}{A P}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{5}{A P}\)

⇒ AP = 5\(\sqrt{3}\) cm

Or

In the given figure, if the angle between two radii of a circle is 130°, then find the angle between the tangent at the ends of the radii.

Answer:

From figure, ∠AOB = 130° ………. (i)

Clearly, ∠PAO = ∠PBO = 90° …….. (ii)

[∵ radius is perpendicular to the tangent at the point of contact]

Now, applying angle sum properly of quadrilateral in OAPB, we have

∠PAQ + ∠AOB + ∠OBP + ∠APB = 360°

90° + 130° + 90° + ∠APB = 360°

[from Eqs. (i) and (ii)]

∴ ∠APB = 360° – 310° = 50°

Hence, the angle between the tangent at the ends of the radii is 50°.

Question 22.

In a right-angled ΔABC, right-angled at B, if the ratio of AB to AC is 1 : \(\sqrt{2}\), then find the values of

(i) \(\frac{2 \tan A}{1-\tan ^2 A}\)

(ii) \(\frac{2 \tan A}{1+\tan ^2 A}\) (2)

Answer:

We have, AB : AC = 1 : \(\sqrt{2}\) ⇒ \(\frac{A B}{A C}\) = \(\frac{1}{\sqrt{2}}\)

Let AB = x and AC = \(\sqrt{2}\)x, for some x

By Pythagoras theorem, we have

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

⇒ \((\sqrt{2} x)^2\) = x^{2} + (BC)^{2}

⇒ (BC)^{2} = 2x^{2} – x^{2} = x^{2}

⇒ BC = x [taking positive square roots]

∴ tan A = \(\frac{B C}{A B}\) = \(\frac{x}{x}\) = 1 [∵ tanθ = \(\frac{\text { perpendicular }}{\text { base }}\)]

(i) We have,

\(\frac{2 \tan A}{1-\tan ^2 A}\) = \(\frac{2 \times 1}{1-1}\) = \(\frac{2}{0}\), which is not defined

(ii) We have,

\(\frac{2 \tan A}{1+\tan ^2 A}\) = \(\frac{2 \times 1}{1+1^2}\) = \(\frac{2}{2}\) = 1

Question 23.

If ΔABC ~ ΔPQR, AB = 6.5 cm. PQ =10.4 cm and perimeter of ΔABC = 60 cm, then find the perimeter of ΔPQR. (2)

Answer:

Given, ΔABC ~ ΔPQR

We know that, the ratio of perimeter of two similar triangles is equal to the ratio of their corresponding sides.

Question 24.

Find the value of K for given equation has real and equal roots.

Kx^{2} – 5x + K = 0 (2)

Answer:

Given, quadratic equation is

kx^{2} – 5x + k = 0

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

Here, a = K . b = – 5 and C = K

∴ D = b^{2} – 4ac = (-5)^{2} – 4 × K × (K)

= 25 – 4K^{2}

∴ The given equation have real and equal roots,

So, D = 0

⇒ 25 – 4K^{2} = 0 ⇒ 4K^{2} = 25

K = ±\(\frac{5}{2}\)

Question 25.

Find the sum of the series

7 + 10\(\frac{1}{2}\) + 14 + … + 84. (2)

Answer:

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

∴ 10\(\frac{1}{2}\) – 7 = 14 – 10\(\frac{1}{2}\) = ……. = \(\frac{7}{2}\)

∴ The given numbers form an AP

Here, first term, a = 7,

common difference, d = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\) = \(\frac{7}{2}\)

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

∵ a_{n} = a + (n – 1)d

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

⇒ \(\frac{7}{2}\)(n – 1) = 84 – 7

⇒ \(\frac{7}{2}\)(n – 1) = 77

⇒ n – 1 = 77 × \(\frac{2}{7}\)

⇒ n – 1 = 22

⇒ n = 23

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

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

= \(\frac{23}{2}\) × 91

= \(\frac{2093}{2}\) = 1046\(\frac{1}{2}\)

Or

Sarita saved ₹ 5 in the first week of the year and then increased her weekly savings by ₹ 1.75 each week. In which week will her weekly savings be ₹ 20.75? (2)

Answer:

Suppose, amount of weekly savings will be ₹ 20.75 in the nth week.

Clearly, amount of weekly savings form an AP with first term (a) = 5

and common difference (d) = 1.75

∵ nth term = 20.75

∴ a + (n – 1)d = 20.75 [∵ a_{n} = a + (n – 1)d]

⇒ 5 + (n – 1) × 1.75 = 20.75

⇒ (n – 1) × 1.75 = 20.75 – 5

⇒ (n – 1) × (1.75) = 15.75

⇒ n – 1 = \(\frac{15.75}{1.75}\)

⇒ n – 1 = 9

⇒ n = 10

Hence, Sarita’s weekly savings will be ₹ 20.75 in 10th week.

Section C

(Section C consists of 6 questions of 3 marks each.)

Question 26.

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel. (3)

Answer:

Let the radius of the wheel be r cm.

Distance covered by the wheel in one revolution

\(=\frac{\text { Distance moved }}{\text { Number of revolutions }}\) = \(\frac{11}{5000}\)km

= \(\frac{11}{5000}\) × 1000 × 100 cm = 220 cm

∴ Circumference of the wheel = 220 cm

⇒ 2πr = 220 cm

⇒ 2 × \(\frac{22}{7}\) × r = 220 ⇒ r = 35 cm

∴ Distance = 2r = (2 × 35) = 70 cm

Hence, the diameter of the wheel is 70 cm.

Or

If the perimeter of a semi-circular protractor is 108 cm. Find the diameter of the protractor. [take π = \(\frac{22}{7}\)] (3)

Answer:

Let the radius of the protractor be r cm.

Then, perimeter = 108 cm

⇒ \(\frac{1}{2}\)(2πr) + 2r = 108

[∴ perimeter of a semi-circle = \(\frac{1}{2}\)(2πr)]

⇒ πr + 2r = 108

⇒ \(\frac{22}{7}\) × r + 2r = 108

⇒ 36r = 108 × 7

⇒ r = 3 × 7 = 21 cm

∴ Distance of the protractor = 2r = (2 × 21) cm

= 42 cm

Question 27.

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO i.e. BO = 2 BC. (3)

Answer:

Two tangents BD and BC are drawn from an external point B.

To prove BO = 2BC

Given, ∠DBC = 120°

Join OC, OD and BO.

Since, BC and BD are tangents.

∴ OC ⊥ BC and OD ⊥ BD

We know, OB is an angle bisector of ∠DBC.

∴ ∠OBC = ∠DBO = 60°

In right angled ΔOBC,

cos 60° = \(\frac{B C}{O B}\)

⇒ \(\frac{1}{2}\) = \(\frac{B C}{O B}\)

⇒ OB = 2BC

Also, BC = BD

[tangent drawn from external point to circle are equal]

OB = BC + BC

⇒ OB = BC + BD

Question 28.

The polynomial x^{2} – (k + 6)x + 2(2k – 1) has sum of its zeroes equal to half of their product. Find the value of k. (3)

Answer:

Let α and β are the roots of given quadratic equation

x^{2} – (k + 6)x + 2(2k – 1) = 0

Now, sum of roots = α + β

= –\(\left\{\frac{-(k+6)}{1}\right\}\)

= k + 6

Product of roots = αβ = \(\frac{2(2 k-1)}{1}\)

= 2(2k – 1)

According to the question,

Sum of roots (zeroes)

= \(\frac{1}{2}\) × products of roots (zeroes)

⇒ k + 6 = \(\frac{1}{2}\) × 2(2k – 1)

⇒ k + 6 = 2k – 1

⇒ 6 + 1 = 2k – k

⇒ k = 7

Question 29.

If we add 1 to the numerator and subtract 1 from denominator, a function reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction? (3)

Answer:

Let x and y be the numerator and denominator of a fraction, respectively.

Then, the fraction = \(\frac{x}{y}\)

According to the question,

\(\frac{x+1}{y-1}\) = 1

⇒ x + 1 = y – 1

⇒ x – y = -2

⇒ -x + y = 2 …… (i)

and \(\frac{x}{y+1}\) = \(\frac{1}{2}\)

⇒ 2x = y + 1

⇒ 2x – y = 1 …… (ii)

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

x = 3

From Eqs. (i), putting x = 3, we get

-3 + y = 2 ⇒ y = 5

Hence, the function is \(\frac{3}{5}\).

Question 30.

Find the value of

\(\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}}\) (3)

Answer:

Or

Given, cos θ = \(\frac{21}{29}\), determine the value of \(\frac{\sec \theta}{\tan \theta-\sin \theta}\). (3)

Answer:

Given, cos θ = \(\frac{21}{29}\)

Thus, base and hypotenuse are in the ratio 21:29.

Let us take a right ΔABC in which ∠ACB = θ.

Let BC = 21 k and AC = 29k, where k is any positive integers. Using Pythagoras theorem, we have

Question 31.

Find the points on the X-axis, which are at a distance of 2\(\sqrt{5}\) from the point (7, -4). How many such points are there? (3)

Answer:

At X-axis, y-coordinate is zero. So, let A (x, 0) be the point on X-axis, which is at a distance of 2\(\sqrt{5}\) from the point S (7, – 4).

AB = 2\(\sqrt{5}\)

⇒ \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\) = 2\(\sqrt{5}\)

[by distance formula]

⇒ \(\sqrt{(x-7)^2+(0+4)^2}\) = 2\(\sqrt{5}\)

On squaring both sides, we get

(x – 7)^{2} + (0 + 4)^{2} = (2\(\sqrt{5}\))^{2}

⇒ x^{2} + 49 – 14x + 16 = 20

[∵ (a – b)^{2} = a^{2} + b^{2} – 2ab]

⇒ x^{2} -14x + 65 – 20 = 0

⇒ x^{2} -14x + 45 = 0 (1)

⇒ x^{2} – 9x – 5x + 45 = 0 [by factorisation]

⇒ x(x – 9) – 5 (x – 9) = 0

⇒ (x – 9) (x – 5) = 0

x = 5 or 9

Hence, the required points are P_{1} (5, 0) and P_{2} (9, 0).

Also, note that we have only two such points.

Section D

(Section D consists of 4 questions of 5 marks each.)

Question 32.

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, then find the value of length of her shadow after 4s. (5)

Answer:

Let AB be the lamp-post, CD be the girl and D be the position of girl after 4s.

Again, let DE = x m be the length of shadow of the girl.

Given, CD = 90 cm = 0.9 m, AB = 3.6 m

and speed of the girl = 1.2 m/s

∴ Distance of the girl from lamp-post after 4s,

BD = 1.2 × 4 = 4.8 m

[∵ distance = speed × time]

In ΔABE and ΔCDE,

∠B = ∠D [Each 90°]

∠E = ∠E [common angle]

∴ ΔABE ~ ΔCDE [by AA similarity criterion]

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

[since, corresponding sides of similar triangles are pcoportional]

On substituting all the values in Eq. (i), we get

\(\frac{4.8+x}{x}\) = \(\frac{3.6}{0.9}\) [∵ BE = BD + DE = 4.8 + x]

⇒ \(\frac{4.8}{3}\) = 1.6 m

⇒ 4.8 + x = 4x

⇒ 3x = 4.8

⇒ x = \(\frac{4.8}{3}\)

Hence, the length of her shadow after 4s is 1.6 m.

Question 33.

The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70. (5)

Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |

Frequency | 12 | a | 12 | 15 | b | 6 | 6 | 4 |

Answer:

The cumulative frequency table is given below

Here, 55 + a + b = 70

⇒ a + b = 70 – 55

⇒ a + b = 15 ……… (i)

Given, median of the data is 16, which lies in the interval 15-20.

Here, l = 15, cf = 24 + a, f = 15, h = 5

∴ Median = l + \(\frac{\frac{70}{2}-24-a}{15}\) × h

⇒ 16 = 15 + \(\frac{\frac{70}{2}-24-a}{15}\) × 5

⇒ 1 = (11 – a)/ 3

⇒ 11 – a = 3

⇒ a = 8 (1)

From Eq. (i), we get

8 + b = 15

⇒ b = 15 – 8 = 7

Question 34.

Find the sum of first 51 terms of an AP -Those second and third terms are 14 and 18, respectively. (5)

Answer:

Here, n = 51, T_{2} = 14 and T_{3} = 18

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

We have, T_{2} ⇒ a + d = a + d = 14 ……(i)

T_{3} = a + 2d ⇒ a + 2d = 18 ……..(ii)

Subtracting Eq. (i) from Eq. (ii), we get

a + 2d – a – d = 18 – 14 ⇒ d = 4 (2)

From Eq. (i), we get a + d = 14

⇒ a + 4 = 14 ⇒ a = 14 – 4 = 10

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

S_{51} = \(\frac{51}{2}\)[(2 × 10) + (51 – 1) × 4]

= \(\frac{51}{2}\)[20 + 200] = \(\frac{51}{2}\) [220]

= 51 × 110 = 5610

Thus, the sum of 51 terms is 5610.

Or

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of first n terms. (5)

Answer:

Here, we have S_{7} = 49 and S_{17} = 289

Let the first term of the AP be ‘a’ and ‘d’ be the common difference, then

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

⇒ S_{7} = \(\frac{7}{2}\)[2a + (7 – 1)d] = 49

⇒ 7(2a + 6d) = 2 × 49 = 98

⇒ 2a + 6d = \(\frac{98}{7}\) = 14 ⇒ 2[a + 3d] = 14

⇒ a + 3d = \(\frac{14}{2}\) = 7 ⇒ a + 3d = 7 …… (i)

Also, S_{17} = \(\frac{17}{2}\)[2a + (17 – 1)d] = 289

⇒ \(\frac{17}{2}\)(2a + 16d) = 289

⇒ a + 8d = \(\frac{298}{17}\) = 17 ⇒ a + 8d = 17 …….. (ii)

Subtracting Eq. (i) a from Eq. (ii), we have

a + 8d – a – 3d = 17 – 7

⇒ 5d = 10 ⇒ d = 2

Now, from Eq. (i), we have

a + 3(2) = 7

⇒ a = 7 – 6 = 1

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

= \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\)[2n] = n × n = n^{2}

Thus, the required sum of n terms = n^{2}

Question 35.

A wooden article was made by scooping out a hemisphere from one end of a cylinder and cone from the other end as shown in the figure. If the height of the cylinder is 40 cm, radius of the cylinder is 7 cm and height of the cone is 24 cm. (5)

(i) Find the slant height of the cone and the volume of hemisphere.

(ii) Find the total volume of the article.

Answer:

Given, height of the cylinder (H) = 40 cm

Radius of the cylinder (r) = 7 cm

Radius of the hemisphere (r) = 7 cm

Radius of the cone (r) = 7 cm

Height of the cone (h) = 24 cm

(ii) Volume of the article = Volume of the cylinder – Volume of the cone – Volume of the hemisphere

Or

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 4 cm and the diameter of the base is 8 cm. If a right circular cylinder circumscribes the solid. Find, how much more space it will cover? (5)

Answer:

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

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

We have, height of cone OA = 4 cm

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

∴ Radius of the base of cone,

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

Here, AP = AO + OP

= 4 + 4 = 8 cm

Now, volume of the right circular cylinder = πr^{2}h

= π(4)^{2} × 8

= 128π cm^{3}

Volume of the solid toy = Volume of cone + Volume of hemisphere

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

= \(\frac{2}{3}\)π × (4)^{3} + \(\frac{1}{3}\)π(4)^{2} × 4

= \(\frac{128}{3}\)π + = \(\frac{64}{3}\)π

= \(\frac{192 \pi}{3}\)π

∴ Required space = Volume of the right circular cylinder – Volume of the toy

= 128π – \(\frac{192 \pi}{3}\)

= \(\frac{192 \pi}{3}\) cm^{3}

Hence, the right circular cylinder covers \(\frac{192 \pi}{3}\) cm^{3} more space than the solid toy.

Section E

(Case Study Based Questions)

(Section E consists of 3 questions. All are compulsory.)

Question 36.

Morning Walk

In a morning walk, Pankaj, Saksham and Mayank step off together, their steps measuring 240 cm, 90 cm and 120 cm

respectively. They want to go for a juice shop for a health issue, which is situated near by them.

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

(i) If a and b are two number, then find the relation between LCM and HCF. (1)

(ii) Find the minimum distance of shop from where they start to walk together, so that one can cover the distance in complete steps? (2)

Answer:

(i) The relation between LCM and HCF is LCM(a, b) × HCF (a, d) = a × b

(ii) Minimum required distance to reach the juice shop = LCM (240, 90, 120)

∴ 240 = 2 × 2 × 2 × 2 × 3 × 5

90 = 2 × 3 × 3 × 5

and 120 = 2 × 2 × 2 × 3 × 5

Now, LCM = 2^{4} × 3^{2} × 5

= 16 × 9 × 5 = 720

Hence, required distance is 720 cm.

Or

Find the number of common steps cover by all of them to reach the juice shop. (2)

(iii) A largest positive integer that divides given two positive integers is called (1)

Answer:

The number of common steps cover by all of them = HCF (240, 90, 120)

Now, 240 = 2 × 2 × 2 × 2 × 3 × 5

90 = 2 × 3 × 3 × 5 and 120 = 2 × 2 × 2 × 3 × 5

So, HCF = 2 × 3 × 5 = 30

(iii) A largest positive integer that divides given two positive integers is called HCF.

Question 37.

Play Cards

There are three friends and they want to play some interesting game. Firstly, they consider some cards and marked with the numbers 2 to 101 are placed in a box and mixed throughly. One card is drawn from this box.

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

(i) Find the probability that the drawn card is an even number. (1)

(ii) Find the probability that the drawn card is a number less than 14. (1)

(iii) Find the probability that the drawn card is a number which is a perfect square. (2)

Answer:

(i) There are 100 cards in the box out of which one card can be drawn in 100 ways.

∴ Total number of elementary events = 100

From numbers 2 to 101, there are 50 even numbers, namely, 2, 4, 6, 8,…… 100. Out of these 50 even numbered cards, one card can be chosen in 50 ways.

∴ Favourable number of elementary events = 50 Hence, P (getting an even number card)

= \(\frac{50}{100}\) = \(\frac{1}{2}\)

(ii) There are 12 cards bearing number less than 14, i.e. numbers 2, 3, 4, 5, …. 13.

∴ Favourable number of elementary events = 12

∴ P (getting a number less than 14)

= \(\frac{12}{100}\) = \(\frac{3}{25}\)

(iii) Those numbers from 2 to 101 which are perfect squares are 4, 9, 16, 25, 36, 49, 64, 81, 100, i.e. squares of 2, 3, 4, 5,…….. and 10 respectively.

Therefore, there are 9 cards marked with the numbers which are perfect squares.

∴ Favourable number of elementary events = 9

P(getting a number which is a perfect square) = \(\frac{9}{100}\)

Or

Find the probability that the drawn card is a prime number and it is less than 20. (2)

Answer:

Prime numbers less than 20 in the numbers from 2 to 101 are 2, 3, 5, 7, 11, 13, 17 and 19.

Thus, there are 8 cards marked with prime numbers less than 20. Out of these 8 cards one card can be choose in 8 ways.

∴ Favourable number of elementary events = 8

Hence, P (getting a prime number less than 20)

= \(\frac{8}{100}\) = \(\frac{2}{25}\)

Question 38.

Telecasting Tower

A straight highway leads to the foot of a national communication and telecasting tower. A watchman standing at the top of the tower observes a car at an angle of depression of 300 which is approaching the foot of the tower with a uniform speed.

Two minutes later, the angle of depression was found to be 60°. The watchman suspects that some terrorist are approaching the tower. It needs half a minute for the watchman to inform the security staff so that it may alert.

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

(i) Find the angle of depression from object to the point on the ground and the angle of elevation of the same point on the ground to the same object. (1)

(ii) The angle of an object viewed, is the angle formed by the line of sight with the horizontal. (1)

(iii) How much time the car will take to reach the foot of the tower? (2)

Answer:

(i) The angle of depression from object to the point ‘ on the ground and the angle of elevation of the same point on the ground to the same object is equal.

(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal.

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

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

Then, CD = 2x m

[∵ time taken to cover distance CD is 2 min and distance = speed × time]

and BD = xy m

⇒ 2 + y = 3y ⇒ 2 = 2y

∴ y = 1 min

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

Or

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of the tower. (2)

Answer:

Let h be the height of transmission tower and x be the distance between the point and foot of the building.