Solved 2024 Specimen Paper ICSE Class 10 Mathematics

Order of A : 1 × 2 and order of B : 2 × 2

Since, no. of columns in A is same as the no. of rows in B.

∴ AB is possible.

Hence, Option 3 is the correct option.

Given,

⇒ x² + kx + 6 = (x - 2)(x - 3)

⇒ x² + kx + 6 = x² - 3x - 2x + 6

⇒ x² + kx + 6 = x² - 5x + 6

⇒ x² - x² + kx + 6 - 6 = -5x

⇒ kx = -5x

⇒ k = -5.

Hence, Option 1 is the correct option.

C.P. for retailer = ₹ 1500

S.P. for retailer = ₹ 1500 + 10% of ₹ 1500

= ₹ 1500 + $\dfrac{10}{100} \times 1500$

= ₹ 1500 + ₹ 150

= ₹ 1650.

GST paid = 10%

= $\dfrac{10}{100} \times$ ₹ 1650

= ₹ 165.

Hence, Option 4 is the correct option.

Given,

Roots of equation x² - 6x + k = 0 are real and distinct.

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-6)² - 4 × 1 × k > 0

⇒ 36 - 4k > 0

⇒ 4k < 36

⇒ k < $\dfrac{36}{4}$

⇒ k < 9.

Hence, Option 4 is the correct option.

In first series :

1, 4, 9, 16, ...........

16 - 9 = 7, 9 - 4 = 5.

Since, difference between consecutive terms are not equal.

∴ It is not an A.P.

In second series :

$\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3},$ ............

$4\sqrt{3} - 3\sqrt{3} = 3\sqrt{3} - 2\sqrt{3} = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$.

Since, difference between consecutive terms are equal.

∴ It is an A.P.

In third series :

8, 6, 4, 2, ..........

2 - 4 = 4 - 6 = 6 - 8 = -2.

Since, difference between consecutive terms are equal.

∴ It is an A.P.

Hence, Option 3 is the correct option.

Given, x is proportional to y.

$\Rightarrow \dfrac{6}{M} = \dfrac{12}{18} \\[1em] \Rightarrow M = \dfrac{6 \times 18}{12} \\[1em] \Rightarrow M = \dfrac{108}{12} \\[1em] \Rightarrow M = 9. \\[1.5em] \dfrac{12}{18} = \dfrac{N}{6} \\[1em] \Rightarrow N = \dfrac{12 \times 6}{18} \\[1em] \Rightarrow N = \dfrac{72}{18} \\[1em] \Rightarrow N = 4.$

Hence, Option 3 is the correct option.

Given,

△ ABC ~ △ PQR

⇒ ∠B = ∠Q (Corresponding angles of similar triangle are equal)

In △ ABD and △ PQS,

⇒ ∠B = ∠Q (Proved above)

⇒ ∠D = ∠S (Both equal to 90°)

∴ △ ABD ~ △ PQS (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AB}{PQ} = \dfrac{AD}{PS} = \dfrac{3}{8}$.

∴ AB : PQ = 3 : 8.

Hence, Option 3 is the correct option.

On rotating the right angle triangle figure formed is :

Hence, Option 2 is the correct option.

The sun always rises from east and also in a leap year February has 29 days.

∴ Event A and Event C has probability equal to 1.

Hence, Option 4 is the correct option.

Let ABC be the triangle and O is the circumcenter of triangle.

Circumcenter is the center of circle which passes through all vertices of triangle.

∴ OA = OB = OC (Radius of circle)

Hence, Option 3 is the correct option.

Let l and m be tangents to circle with center O, touching the circle at point A and B.

From figure,

Shortest distance between tangents = OA + OB = R + R = 2R.

Hence, Option 2 is the correct option.

From figure,

⇒ tan α = $\dfrac{AB}{BC}$

⇒ tan β = $\dfrac{AB}{BD}$

⇒ tan γ = $\dfrac{AB}{BE}$

Since, BC < BD < BE.

∴ tan α > tan β > tan γ.

∴ tan α > tan β.

Hence, Option 1 is the correct option.

Companies in which Market value is greater than nominal value, there shares are at premium.

∴ Shares of company C are at premium.

Hence, Option 3 is the correct option.

Substituting x = 0 and y = 0 in L.H.S. of the equation 2x - 3y = 0, we get :

⇒ 2 × 0 - 3 × 0

⇒ 0 - 0

⇒ 0.

Since, L.H.S. = R.H.S.

∴ Line 2x - 3y = 0 represents a line passing through origin.

Hence, Option 2 is the correct option.

We know that,

To find the median of the given data, the variate needs to be arranged in ascending or descending order and also the median is the central most term of the arranged data.

∴ Both A and R are true.

Hence, Option 3 is the correct option.

From figure,

Radius of cylindrical part = Radius of hemispherical part = r = 7 m.

Length (Height h) of the cylindrical part = 34 - 7 - 7 = 20 m.

Surface of cylindrical part in contact of water

= $\dfrac{1}{2} \times 2πrh = \dfrac{1}{2} \times 2 \times \dfrac{22}{7} \times 7 \times 20$ = 440 m²

Surface of each hemisphere in contact of water

= $\dfrac{1}{2} \times 2πr^2$

= πr²

= $\dfrac{22}{7} \times 7^2$

= 22 × 7

= 154 m².

∴ Surface area of the tank in contact of water = 440 + 2 × 154

= 440 + 308

= 748 m².

Hence, surface area of tank in contact with water = 748 m².

(a) Given,

Interest earned by man = One-twelfth of the total deposit

= $\dfrac{1}{12} \times 9600$

= ₹ 800.

Hence, the interest earned = ₹ 800.

(b) Given,

In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is ₹ 9600.

Money deposited per month = $\dfrac{9600}{24}$ = ₹ 400.

Hence, monthly deposit = ₹ 400.

(c) By formula,

Rate of interest = $\dfrac{\text{Interest earned}}{\text{Amount invested}} \times 100$

$= \dfrac{800}{9600} \times 100$ %.

Hence, rate of interest = $8\dfrac{1}{3}$ %.

(a) Solving,

⇒ (sin θ + cosec θ)²

⇒ sin² θ + cosec² θ + 2 × sin θ × cosec θ

⇒ sin² θ + 1 + cot² θ + 2 × sin θ × $\dfrac{1}{\text{sin θ}}$

⇒ sin² θ + 1 + cot² θ + 2

⇒ sin² θ + cot² θ + 3.

(b) Solving,

⇒ (cos θ + sec θ)²

⇒ cos² θ + sec² θ + 2 × cos θ × sec θ

⇒ cos² θ + 1 + tan² θ + 2 × cos θ × $\dfrac{1}{\text{cos θ}}$

⇒ cos² θ + tan² θ + 1 + 2

⇒ cos² θ + tan² θ + 3.

Solving,

⇒ (sin θ + cosec θ)² + (cos θ + sec θ)²

⇒ sin² θ + cot² θ + 3 + cos² θ + tan² θ + 3.

⇒ sin² θ + cos² θ + cot² θ + tan² θ + 3 + 3

⇒ 1 + cot² θ + tan² θ + 3 + 3

⇒ 7 + tan² θ + cot² θ.

Hence, proved that (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan² θ + cot² θ.

Since, a, b and c are in continued proportion.

$\therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow b^2 = ac.$

Substituting, b² = ac in L.H.S. of equation $\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}$, we get :

$\Rightarrow \dfrac{3a^2 + 5ab + 7ac}{3ac + 5bc + 7c^2} \\[1em] \Rightarrow \dfrac{a(3a + 5b + 7c)}{c(3a + 5b + 7c)} \\[1em] \Rightarrow \dfrac{a}{c}.$

Hence, proved that $\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}$.

In △OAC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OCA = ∠OAC = 32° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OCA + ∠OAC + ∠AOC = 180°

⇒ 32° + 32° + ∠AOC = 180°

⇒ ∠AOC + 64° = 180°

⇒ ∠AOC = 180° - 64°

⇒ ∠AOC (x°) = 116°.

We know that,

Angle subtended by an arc at the center of the circle is twice the angle subtended at the remaining part of the circumference.

⇒ ∠AOC = 2∠ABC

⇒ x° = 2y°

⇒ y° = $\dfrac{x°}{2} = \dfrac{116°}{2}$ = 58°.

In △ BDC,

⇒ ∠BDC + ∠BCD + ∠CBD = 180°

⇒ 90° + z° + y° = 180°

⇒ z° = 180° - 90° - y° = 90° - 58° = 32°.

Hence, x° = 116°, y° = 58° and z° = 32°.

(a) From graph,

The curve plotted is a cumulative frequency curve (ogive).

(b) From graph,

The total number of students = 40.

(c) From graph,

The median marks = 56.

(d) From graph,

No of students scoring below 80 = 37

No of students scoring below 50 = 12

∴ No. of students scoring between 50 and 80 = 37 - 12 = 25.

Hence, the no. of students scoring between 50 and 80 = 25.

Given,

⇒ A² = pA

$\Rightarrow \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \\ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \\ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 32 = 4p \\[1em] \Rightarrow p = \dfrac{32}{4} = 8.$

Hence, p = 8.

Given, equation : x² - 4x - 2 = 0.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1\times -2}}{2 \times 1} \\[1em] = \dfrac{4 \pm \sqrt{16 + 8}}{2} \\[1em] = \dfrac{4 \pm \sqrt{24}}{2} \\[1em] = \dfrac{4 \pm 2\sqrt{6}}{2} \\[1em] = 2 \pm \sqrt{6} \\[1em] = 2 + \sqrt{6}, 2 - \sqrt{6} \\[1em] = 2 + 2.449, 2 - 2.449 \\[1em] = 4.449, -0.449$

Hence, x = 4.449 or -0.449

(a) In △ADF,

⇒ ∠ADF = 90°

By angle sum property of triangle,

⇒ ∠ADF + ∠AFD + ∠A = 180°

⇒ 90° + ∠AFD + ∠A = 180°

⇒ ∠AFD + ∠A = 180° - 90°

⇒ ∠AFD + ∠A = 90° .........(1)

In △ ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 180° - 90°

⇒ ∠A + ∠C = 90° .........(2)

From equation (1) and (2), we get :

⇒ ∠AFD + ∠A = ∠A + ∠C

⇒ ∠AFD = ∠C

In △ ADF and △ FEC,

⇒ ∠ADF = ∠FEC (Both equal to 90°)

⇒ ∠AFD = ∠C (Proved above)

∴ △ ADF ~ △ FEC [By A.A. axiom]

Hence, proved that △ ADF ~ △ FEC.

(b) In △ ADF and △ ABC,

⇒ ∠DAF = ∠BAC (Common angle)

⇒ ∠ADF = ∠ABC (Both equal to 90°)

∴ △ ADF ~ △ ABC [By A.A. axiom]

Hence, proved that △ ADF ~ △ ABC.

(c) From figure,

⇒ DB = FE = x (let)

⇒ AB = AD + DB = (6 + x) cm

⇒ DF = BE = BC - CE = 12 - 4 = 8 cm.

△ ADF ~ △ ABC [proved above]

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{AD}{AB} = \dfrac{DF}{BC} \\[1em] \Rightarrow \dfrac{6}{6 + x} = \dfrac{8}{12} \\[1em] \Rightarrow 8(6 + x) = 6 \times 12 \\[1em] \Rightarrow 48 + 8x = 72 \\[1em] \Rightarrow 8x = 72 - 48 \\[1em] \Rightarrow 8x = 24 \\[1em] \Rightarrow x = \dfrac{24}{8} = 3 \text{ cm}.$

Hence, FE = 3 cm.

(d) We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{(6 + 3)^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{6^2}{9^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADF}}{\text{Area of △ ABC}} = \dfrac{36}{81} = \dfrac{4}{9}.$

Hence, area △ ADF : area △ ABC = 4 : 9.

In the given table,

Class size (i) = 4.

By formula,

Mean = A + $\dfrac{Σfu}{Σf} \times i$

= 6 + $\dfrac{19}{100} \times 4$

= 6 + $\dfrac{76}{100}$

= 6 + 0.76

= 6.76

Hence, mean = 6.76

For LED TV set,

M.P. = ₹ 12,000

G.S.T. = 28%

G.S.T. amount = 28% of ₹ 12,000

= $\dfrac{28}{100} \times 12,000$

= 28 × 120

= ₹ 3360.

Total amount = ₹ 12,000 + ₹ 3360 = ₹ 15,360.

For MP4 player,

M.P. = ₹ 5,000

G.S.T. = 18%

G.S.T. amount = 18% of ₹ 5,000

= $\dfrac{18}{100} \times 5,000$

= 18 × 50

= ₹ 900.

Total amount = ₹ 5,000 + ₹ 900 = ₹ 5900.

Total bill = ₹ 15,360 + ₹ 5,900 = ₹ 21,260.

Hence, total bill = ₹ 21,260.

(a) We know that,

Angle between the radius and tangent at the point of contact is 90°.

∴ ∠PBA = ∠PBQ = 90°

In △ PBQ,

By angle sum property of triangle,

⇒ ∠PBQ + ∠BQP + ∠QPB = 180°

⇒ 90° + 55° + ∠QPB = 180°

⇒ ∠QPB = 180° - 90° - 55° = 35°.

From figure,

⇒ ∠SPB = ∠QPB = 35°

We know that,

Angles in same segment are equal.

⇒ ∠SRB (x°) = ∠SPB = 35°

⇒ x° = 35°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

∴ ∠SOB = 2∠SRB

⇒ y° = 2x° = 2 × 35° = 70°.

From figure,

⇒ z° = x° = 35° (Angles in alternate segment are equal)

Hence, x° = 35°, y° = 70° and z° = 35°.

(b) From figure,

⇒ OB = OR (Radius of the same circle)

⇒ ∠OBR = ∠ORB (Angles opposite to equal sides are equal)

⇒ ∠OBR = x° = 35°

∴ ∠OBR = ∠OPS

The above angles are alternate angles.

∴ RB // PS.

Hence, proved that RB // PS.

(a) Let the three positive numbers be $\dfrac{a}{r}, a, ar$.

Given,

Product of the numbers is 3375.

$\Rightarrow \dfrac{a}{r} \times a \times ar = 3375 \\[1em] \Rightarrow a^3 = 3375 \\[1em] \Rightarrow a^3 = (15)^3 \\[1em] \Rightarrow a = 15.$

(b) Given,

Result of the product of first and second number added to the product of second and third number is 750.

$\therefore \dfrac{a}{r} \times a + a \times ar = 750 \\[1em] \Rightarrow \dfrac{a^2}{r} + a^2r = 750 \\[1em] \Rightarrow a^2\Big(\dfrac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow 15^2\Big(\dfrac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{750}{15^2} \\[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{10}{3} \\[1em] \Rightarrow 3(1 + r^2) = 10r \\[1em] \Rightarrow 3 + 3r^2 = 10r \\[1em] \Rightarrow 3r^2 - 10r + 3 = 0 \\[1em] \Rightarrow 3r^2 - 9r - r + 3 = 0 \\[1em] \Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\[1em] \Rightarrow (3r - 1)(r - 3) = 0 \\[1em] \Rightarrow 3r - 1 = 0 \text{ or } r - 3 = 0 \\[1em] \Rightarrow 3r = 1 \text{ or } r = 3 \\[1em] \Rightarrow r = \dfrac{1}{3} \text{ or } r = 3.$

Let r = $\dfrac{1}{3}$

Numbers : $\dfrac{a}{r}, a, ar$

= $\dfrac{15}{\dfrac{1}{3}}, 15, 15 \times \dfrac{1}{3}$

= 45, 15, 5.

Let r = 3

Numbers : $\dfrac{a}{r}, a, ar$

= $\dfrac{15}{3}, 15, 15 \times 3$

= 5, 15, 45.

Hence, numbers are 5, 15 and 45 or 45, 15 and 5.

Steps of construction :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.

3. Through the point K (the point of intersection of diagonals AC and BD), draw KL perpendicular to the horizontal axis.

4. The value of point L on the horizontal axis represents the value of mode.

From graph,

L = 59 years

Hence, required mode = 59 years.

(a) From figure,

Radius of cylindrical part = Radius of conical part = r = 7 m

Height of cylindrical part (H) = 7 m

Height of conical part (h) = 14 - 7 = 7 m.

Quantity of air inside the tent = Volume of cylindrical part + Volume of conical part

$= πr^2H + \dfrac{1}{3}πr^2h \\[1em] = \dfrac{22}{7} \times 7^2 \times 7 + \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 7 \\[1em] = 22 \times 49 + \dfrac{1}{3} \times 22 \times 49 \\[1em] = 1078 + 359.33 \\[1em] = 1437.33 \text{ m}^3.$

Hence, the quantity of air inside the tent = 1437.33 m³.

(b) Let radius of required sphere be R m.

According to question,

Volume of sphere = Quantity of air inside the tent

$\therefore \dfrac{4}{3}πR^3 = πr^2H + \dfrac{1}{3}πr^2h \\[1em] \Rightarrow \dfrac{4}{3}R^3 = r^2H + \dfrac{1}{3}r^2h \\[1em] \Rightarrow \dfrac{4}{3}R^3 = \dfrac{3r^2H + r^2h}{3} \\[1em] \Rightarrow 4R^3 = 3r^2H + r^2h \\[1em] \Rightarrow R^3 = \dfrac{3 \times 7^2 \times 7 + 7^2 \times 7}{4} \\[1em] \Rightarrow R^3 = \dfrac{3 \times 49 \times 7 + 49 \times 7}{4} \\[1em] \Rightarrow R^3 = \dfrac{1029 + 343}{4} \\[1em] \Rightarrow R^3 = \dfrac{1372}{4} \\[1em] \Rightarrow R^3 = 343 \\[1em] \Rightarrow R^3 = 7^3 \\[1em] \Rightarrow R = 7 \text{ m}.$

Hence, radius of sphere = 7 m.

(a) From figure,

Coordinates of M = (-1, 0) and coordinates of N = (0, -1).

(b) From figure,

Coordinates of S = (0, 6).

(c) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AB = $\dfrac{2 - (-3)}{-3 - 2} = \dfrac{2 + 3}{-5} = \dfrac{5}{-5}$ = -1.

Hence, slope of AB = -1.

(d) We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of PQ = -1

⇒ -1 × Slope of PQ = -1

⇒ Slope of PQ = $\dfrac{-1}{-1}$ = 1.

Since, PQ passes through point S.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

⇒ y - 6 = 1(x - 0)

⇒ y - 6 = x

⇒ y = x + 6.

Hence, equation of line PQ is y = x + 6.

Let ∠BCP = α and ∠ACP = β

From figure,

⇒ α + 54° = 90°

⇒ α = 90° - 54° = 36°.

⇒ β + 42° = 90°

⇒ β = 90° - 42° = 48°.

⇒ tan α = $\dfrac{BP}{CP}$

⇒ tan 36° = $\dfrac{BP}{100}$

⇒ 0.7265 = $\dfrac{BP}{100}$

⇒ BP = 0.7265 × 100 = 72.65 m

⇒ tan β = $\dfrac{AP}{CP}$

⇒ tan 48° = $\dfrac{AP}{100}$

⇒ 1.1106 = $\dfrac{AP}{100}$

⇒ AP = 1.1106 × 100 = 111.06 m

(a) From figure,

AB = AP + BP = 72.65 + 111.06 = 183.71 m

Hence, the distance between two ships = 183.71 m.

(b) On rounding off,

AB = 184 m.

Hence, the distance between two ships = 184 m.

To prove:

3 - 2x ≥ x + $\dfrac{1 - x}{3}$ > $\dfrac{2x}{5}$

Solving L.H.S. of the above inequation, we get :

⇒ 3 - 2x ≥ x + $\dfrac{1 - x}{3}$

⇒ 3 - 2x ≥ $\dfrac{3x + 1 - x}{3}$

⇒ 3(3 - 2x) ≥ 2x + 1

⇒ 9 - 6x ≥ 2x + 1

⇒ 2x + 6x ≤ 9 - 1

⇒ 8x ≤ 8

⇒ x ≤ $\dfrac{8}{8}$

⇒ x ≤ 1 ............(1)

Solving R.H.S. of the above equation, we get :

⇒ x + $\dfrac{1 - x}{3} \gt \dfrac{2x}{5}$

⇒ $\dfrac{3x + 1 - x}{3} \gt \dfrac{2x}{5}$

⇒ 5(2x + 1) > 3 × 2x

⇒ 10x + 5 > 6x

⇒ 10x - 6x > -5

⇒ 4x > -5

⇒ x > $-\dfrac{5}{4}$ ...........(2)

From equation (1) and (2), we get :

Solution set = {x : $-\dfrac{5}{4}$ < x ≤ 1, x ∈ R}

Representation of solution set on real number line is :

(a) In △ BDC,

⇒ BC = CD (Equal sides)

⇒ ∠BDC = ∠DBC = 43° (Angles opposite to equal sides are equal)

From figure,

⇒ ∠ADC = ∠ADB + ∠BDC = 62° + 43° = 105°.

Hence, ∠ADC = 105°.

(b) We know that,

Opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 105° = 180°

⇒ ∠ABC = 180° - 105° = 75°.

From figure,

⇒ ∠ABD = ∠ABC - ∠DBC = 75° - 43° = 32°.

Hence, ∠ABD = 32°.

(c) From figure,

⇒ ∠FAD = ∠ABD = 32°. (Angles in alternate segment are equal)

Hence, ∠FAD = 32°.

(a) Given,

BD : DC = 2 : 1.

Let coordinates of D be (x, y)

By section-formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values, we get :

$\Rightarrow (x, y) = \Big(\dfrac{2 \times 8 + 1 \times (-4)}{2 + 1}, \dfrac{2 \times -6 + 1 \times 3}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{16 - 4}{3}, \dfrac{-12 + 3}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{12}{3}, \dfrac{-9}{3}\Big) \\[1em] \Rightarrow (x, y) = (4, -3).$

Hence, coordinates of D = (4, -3).

(b) By mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Given,

M(6, 0) is the mid-point of AD.

$\therefore (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b + (-3)}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b - 3}{2}\Big) \\[1em] \Rightarrow \dfrac{a + 4}{2} = 6 \text{ and } \dfrac{b - 3}{2} = 0 \\[1em] \Rightarrow a + 4 = 12 \text{ and } b - 3 = 0 \\[1em] \Rightarrow a = 12 - 4 = 8 \text{ and } b = 3.$

A = (a, b) = (8, 3).

Hence, coordinates of A = (8, 3).

(c) By formula,

Slope of line = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of line BC = $\dfrac{-6 - 3}{8 - (-4)} = \dfrac{-9}{12} = -\dfrac{3}{4}$.

We know that,

Slope of parallel lines are equal.

Slope of line parallel to BC = $-\dfrac{3}{4}$.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Equation of line parallel to BC and passing through M is :

⇒ y - 0 = $-\dfrac{3}{4}(x - 6)$

⇒ y = $-\dfrac{3}{4}(x - 6)$

⇒ 4y = -3(x - 6)

⇒ 4y = -3x + 18

⇒ 3x + 4y = 18.

Hence, equation of the required line is 3x + 4y = 18.

Given,

$\dfrac{x^2 + 3x}{3x^2 + 1} = \dfrac{14}{13}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - (3x^2 + 1)} = \dfrac{14 + 13}{14 - 13} \\[1em] \Rightarrow \dfrac{x^2 + 3x + 3x^2 + 1}{x^2 + 3x - 3x^2 - 1}= \dfrac{27}{1} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{3^3}{1^3} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{3}{1}\Big)^3 \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = 3 \\[1em] \Rightarrow x + 1 = 3(x - 1) \\[1em] \Rightarrow x + 1 = 3x - 3 \\[1em] \Rightarrow 3x - x = 1 + 3 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} = 2.$

Hence, x = 2.

Let no. of people be x.

Total expense = ₹ 18,000

Expense per person = ₹ $\dfrac{18000}{x}$

Given,

If three more people join them, then the share of each reduces by ₹ 3,000.

No, of people now = x + 3

Expense per person = ₹ $\dfrac{18000}{x + 3}$

According to question,

$\Rightarrow \dfrac{18000}{x} - \dfrac{18000}{x + 3} = 3000 \\[1em] \Rightarrow \dfrac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \\[1em] \Rightarrow \dfrac{18000x + 54000 - 18000x}{x^2 + 3x} = 3000 \\[1em] \Rightarrow \dfrac{54000}{x^2 + 3x} = 3000 \\[1em] \Rightarrow x^2 + 3x = \dfrac{54000}{3000} \\[1em] \Rightarrow x^2 + 3x = 18 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -6.$

Since, no. of people cannot be negative.

∴ x = 3.

Hence, original number of people = 3.

Steps of construction :

1. Draw a line segment BC = 5 cm.

2. Draw ∠DBC = 120°.

3. From DB cut off AB = 6 cm.

4. Draw BE the angle bisector of ∠ABC.

5. Draw RS, the perpendicular bisector of AB.

6. Mark point P, the intersection point of RS and BE.

7. Taking P as center and PA or PB as radius draw a circle.

(a) We know that,

The locus of points which are equidistant from two sides is the angular bisector of the angle between two sides.

Hence, required locus is BE.

(b) We know that,

The locus of points which are equidistant from two points is the perpendicular bisector of the line joining the two points.

Hence, required locus is RS.

(c) From figure,

Point P satisfies both the conditions (a) and (b).

Substituting x = 2, in the given polynomial, we get :

⇒ 2(2)³ - 9(2)² + 7(2) + 6

⇒ 2 × 8 - 9 × 4 + 14 + 6

⇒ 16 - 36 + 20

⇒ -20 + 20

⇒ 0.

∴ x - 2 is the factor of the given polynomial.

On dividing 2x³ - 9x² + 7x + 6 by x - 2, we get :

$\begin{array}{l} \phantom{x - 2)}{\quad 2x^2 - 5x - 3} \\ x - 2\overline{\smash{\big)}\quad 2x^3 - 9x^2 + 7x + 6} \\ \phantom{x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}2x^3-4}-5x^2 + 7x \\ \phantom{{x - 2)}2x^3-4}\underline{\underset{+}{-}5x^2 \underset{-}{+} 10x} \\ \phantom{{x - 2)}31x^3-2+1}-3x + 6 \\ \phantom{{x - 2)}31x^3-2+11}\underline{\underset{+}{-}3x \underset{-}{+} 6} \\ \phantom{{x - 2)}31x^3-2+11+1}\times \end{array}$

∴ 2x³ - 9x² + 7x + 6 = (x - 2)(2x² - 5x - 3)

= (x - 2)[2x² - 6x + x - 3]

= (x - 2)[2x(x - 3) + 1(x - 3)]

= (x - 2)(2x + 1)(x - 3).

Hence, 2x³ - 9x² + 7x + 6 = (x - 2)(2x + 1)(x - 3).

Total no. of outcomes = No. of different letters in the word "HOUSEWARMING" = 12.

(a) No. of vowels in the given word = 5 (o, u, e, a, i)

∴ No. of favourable outcomes = 5

P(that the letter on card is a vowel) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{12}$.

Hence, the probability that the letter on the card is a vowel = $\dfrac{5}{12}$.

(b) No. of different letters in the word SEWING that are also present in the word HOUSEWARMING = 6

∴ No. of favourable outcomes = 6

P(that the letter on card is a letter of the word SEWING)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{12} = \dfrac{1}{2}$.

Hence, the probability that the letter on the card is a letter of the word SEWING = $\dfrac{1}{2}$.

(c) No. of different letters in the word WEAR that are also present in the word HOUSEWARMING = 4

No. of letters that are not in word WEAR = 12 - 4 = 8

P(that the letter on the card is not a letter from the word)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{12} = \dfrac{2}{3}$.

Hence, the probability that the letter on the card is not a letter of the word WEAR = $\dfrac{2}{3}$.

Below graph shows all the required points: