Solved 2024 Question Paper ICSE Class 10 Mathematics

CGST% = $\dfrac{CGST}{M.P.} \times 100$

Substituting values we get :

CGST% = $\dfrac{120}{2000} \times 100$ = 6%

GST% = 2 × CGST% = 2 × 6% = 12%.

Hence, Option 3 is the correct option.

Polynomial : x³ + x² - 2x + 1

Division by x - 3

⇒ x - 3 = 0

⇒ x = 3.

Let k be subtracted from the polynomial, so resulting polynomial is x³ + x² - 2x + 1 - k.

Resulting polynomial should be exactly divisible by x - 3,

Thus, substituting x = 3, in polynomial x³ + x² - 2x + 1 - k, remainder = 0.

⇒ 3³ + 3² - 2(3) + 1 - k = 0

⇒ 27 + 9 - 6 + 1 - k = 0

⇒ 31 - k = 0

⇒ k = 31.

Hence, Option 4 is the correct option.

By formula,

D = b² - 4ac

For equation, px² - qx + r = 0

D = (-q)² - 4 × p × r

We know that,

Roots of a quadratic equation are real and equal if discriminant = 0.

⇒ q² - 4pr = 0

⇒ q² = 4pr.

Hence, Option 2 is the correct option.

$\phantom{\Rightarrow} A^2 = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow A^2 = \begin{bmatrix*}[r] 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \\ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 4 + 4 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 8 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow x = 8.$

Hence, Option 3 is the correct option.

No. of observations = 10, which is even.

Median = $\dfrac{n}{2}\text{th term} = \dfrac{10}{2}$ = 5th term.

Given,

Median = 64

∴ x + 4 = 64

⇒ x = 64 - 4 = 60.

Hence, Option 1 is the correct option.

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC.

$\Rightarrow \dfrac{-2 - y}{3 - x} = \dfrac{-5 - (-2)}{4 - 3} \\[1em] \Rightarrow \dfrac{-2 - y}{3 - x} = \dfrac{-5 + 2}{1} \\[1em] \Rightarrow \dfrac{-2 - y}{3 - x} = -3 \\[1em] \Rightarrow -2 - y = -3(3 - x) \\[1em] \Rightarrow -2 - y = -9 + 3x \\[1em] \Rightarrow y = -2 + 9 - 3x \\[1em] \Rightarrow y = 7 - 3x.$

Hence, Option 4 is the correct option.

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

Average speed = $\dfrac{60}{2}$ = 30 km/hr

$\Rightarrow \dfrac{90}{x} = 30 \\[1em] \Rightarrow x = \dfrac{90}{30} = 3. \\[1em] \phantom{\Rightarrow} \dfrac{y}{5} = 30 \\[1em] \Rightarrow y = 30 \times 5 = 150.$

Hence, Option 4 is the correct option.

In the A.P. :

$\dfrac{1}{a}, \Big(\dfrac{1}{a} + 1\Big), \Big(\dfrac{1}{a} + 2\Big), ......$

First term (x) = $\dfrac{1}{a}$

Common difference (d) = $\dfrac{1}{a} + 1 - \dfrac{1}{a}$ = 1.

x₇ = x + (7 - 1)d

= $\dfrac{1}{a} + 6 \times 1$

= $\Big(\dfrac{1}{a} + 6\Big)$.

Hence, Option 1 is the correct option.

N.V. = ₹ 75

Discount = 20%

M.V. = ₹ 75 - $\dfrac{20}{100} \times 75$

= ₹ 75 - ₹ 15

= ₹ 60.

Cost of 15 shares = 15 × ₹ 60 = ₹ 900.

Hence, Option 4 is the correct option.

The circumcenter of a triangle is defined as the point equidistant from the three vertices of the triangle.

Hence, Option 2 is the correct option.

Trigonometry identity :

sin² θ + cos² θ = 1

cosec² θ - cot² θ = 1

∴ Only statement (i) is correct.

Hence, Option 1 is the correct option.

In △ PST,

⇒ PS = PT (Tangents from an external point to a circle are equal in length)

⇒ ∠PST = ∠PTS = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠PST + ∠PTS + ∠SPT = 180°

⇒ a + a + 80° = 180°

⇒ 2a = 180° - 80°

⇒ 2a = 100°

⇒ a = $\dfrac{100°}{2}$ = 50°.

From figure,

⇒ ∠QST = ∠STP = 50° (Alternate angles are equal)

Hence, Option 4 is the correct option.

Sample space when a die is thrown = {1, 2, 3, 4, 5, 6}.

Sample space for even numbers = {2, 4, 6}

Probability of getting an even number = $\dfrac{3}{6} = \dfrac{1}{2}$.

∴ A is false and R is true.

Hence, Option 2 is the correct option.

In first case :

Height of cylinder (h) = 7 cm

Let radius be r cm

⇒ 2πr = 11

⇒ r = $\dfrac{11}{2π}$

In second case :

Height of cylinder (H) = 11 cm

Let radius be R cm

⇒ 2πR = 7

⇒ R = $\dfrac{7}{2π}$

$\therefore \dfrac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \dfrac{2πrh}{2πRH} \\[1em] = \dfrac{rh}{RH} \\[1em] = \dfrac{\dfrac{11}{2π} \times 7}{\dfrac{7}{2π} \times 11} \\[1em] = \dfrac{\dfrac{77}{2π}}{\dfrac{77}{2π}} \\[1em] = \dfrac{1}{1} \\[1em] = 1 : 1.$

Hence, Option 1 is the correct option.

Given,

∆ABC ∼ ∆PQR

∴ ∠A = ∠P

⇒ $\dfrac{∠A}{2} = \dfrac{∠P}{2}$

⇒ ∠BAD = ∠QPS

∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]

In ∆ABD ∼ ∆PQS,

⇒ ∠BAD = ∠QPS

⇒ ∠B = ∠Q

∴ ∆ABD ∼ ∆PQS (By A.A. axiom)

Hence, Option 2 is the correct option.

$\phantom{\Rightarrow} AB = C \\[1em] \Rightarrow \begin{bmatrix*}[r] x & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 0 \\ y & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ x & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 4 + 0 \times y & x \times 0 + 0 \times 1 \\ 1 \times 4 + 1 \times y & 1 \times 0 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ x & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4x + 0 & 0 + 0 \\ 4 + y & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ x & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4x & 0 \\ 4 + y & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ x & 1 \end{bmatrix*} \\[1em] \Rightarrow 4x = 4 \text{ and } 4 + y = x \\[1em] \Rightarrow x = \dfrac{4}{4} \text{ and } 4 + y = x \\[1em] \Rightarrow x = 1 \text{ and } 4 + y = 1 \\[1em] \Rightarrow x = 1 \text{ and } y = 1 - 4 = -3.$

Hence, x = 1 and y = -3.

(a) Given,

Diameter of cylinder (d) = 7 cm

Radius of cylinder (r) = $\dfrac{d}{2} = \dfrac{7}{2}$ = 3.5 cm

Height of cylinder (h) = 10 cm

Total surface area (both the halves) = Total surface area of cylinder + Area of two rectangles

= [2πr(h + r)] + [2 × (l × b)]

= [2πr(h + r)] + [2 × (h × d)]

= $\Big[2 \times \dfrac{22}{7} \times 3.5 \times (3.5 + 10) \Big] + [2 \times 10 \times 7]$

= (2 × 22 × 0.5 × 13.5) + 140

= 297 + 140

= 437 cm².

Hence, total surface area of both the halves = 437 cm².

(b) Total cost of painting the two halves = Total surface area × Rate

= 437 × 30

= ₹ 13,110.

Hence, total cost of painting the two halves = ₹ 13,110.

Given,

G.P. : 15, 30, 60, 120 ......

First term (a) = 15

Common ratio (r) = $\dfrac{30}{15}$ = 2

(a) nth term of G.P. = arⁿ = 15 x 2ⁿ

Hence, nth term of the given G.P. is 15 x 2ⁿ

(b) Let sum of n terms of G.P. is 945.

By formula,

Sum of n terms of G.P. = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow 945 = \dfrac{15.(2^n - 1)}{2 - 1} \\[1em] \Rightarrow 945 = \dfrac{15.(2^n - 1)}{1} \\[1em] \Rightarrow 2^n - 1 = \dfrac{945}{15} \\[1em] \Rightarrow 2^n - 1 = 63 \\[1em] \Rightarrow 2^n = 63 + 1 \\[1em] \Rightarrow 2^n = 64 \\[1em] \Rightarrow 2^n = 2^6 \\[1em] \Rightarrow n = 6.$

Hence, sum of 6 terms of G.P. = 945.

Factorizing,

⇒ sin³ θ + cos³ θ = (sin θ + cos θ)(sin² θ + cos² θ - sin θ cos θ)

⇒ sin³ θ + cos³ θ = (sin θ + cos θ)(1 - sin θ cos θ) ...........(1)

To prove,

$\dfrac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}}$ + sin θ cos θ = 1

Substituting value of sin³ θ + cos³ θ from equation (1) in L.H.S. of above equation :

⇒ $\dfrac{\text{(sin θ + cos θ)(1 - sin θ cos θ)}}{\text{sin θ + cos θ}} + \text{sin θ cos θ}$

⇒ 1 - sin θ cos θ + sin θ cos θ

⇒ 1.

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

Hence, proved that $\dfrac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1$.

(a) From figure,

⇒ ∠OQP = 90° (Tangent is perpendicular to radius at the point of contact)

⇒ ∠OQM = ∠OQP - ∠PQM

⇒ ∠OQM = 90° - 42° = 48°.

Hence, ∠OQM = 48°.

(b) From figure,

⇒ ∠QNM = ∠PQM = 42° (By alternate segment theorem)

⇒ ∠SNM = ∠PSM = 25° (By alternate segment theorem)

⇒ ∠QNS = ∠QNM + ∠SNM

⇒ ∠QNS = 42° + 25° = 67°.

Hence, ∠QNS = 67°.

(c) We know that,

Angle subtended by an arc at the center is twice the angle subtended by the arc at any other point of the circle.

⇒ ∠QOS = 2∠QNS

⇒ ∠QOS = 2 × 67° = 134°.

Hence, ∠QOS = 134°.

(d) From figure,

QMSN is a cyclic quadrilateral.

We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠QMS + ∠QNS = 180°

⇒ ∠QMS + 67° = 180°

⇒ ∠QMS = 180° - 67° = 113°.

Hence, ∠QMS = 113°.

From figure,

Coordinates of point A' = (0, -1), B' = (-2, 1) and C' = (-4, -1).

The closed figure ABA'B' is a square.

Let time be n months.

By formula,

I = $P \times \dfrac{n(n + 1)}{2\times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \dfrac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.$

Since, no. of months cannot be negative.

∴ n = 12.

Hence, total time for which the account was held = 12 months.

In the given table i is the class interval which is equal to 10.

Mean = A + $\dfrac{Σfu}{Σf} \times i = 135 + \dfrac{-38}{50} \times 10$

= $135 + \dfrac{-380}{50}$

= 135 - 7.6

= 127.4

Hence, mean = 127.4

(a) Solving,

$\Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27} \\[1em] \Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{4^3}{3^3} \\[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = \Big(\dfrac{4}{3}\Big)^3 \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{4}{3} \\[1em]$

Hence, $\dfrac{a + b}{a - b} = \dfrac{4}{3}.$

(b) Solving further,

$\Rightarrow 3(a + b) = 4(a - b) \\[1em] \Rightarrow 3a + 3b = 4a - 4b \\[1em] \Rightarrow 4a - 3a = 3b + 4b \\[1em] \Rightarrow a = 7b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{7}{1} \\[1em] \Rightarrow a : b = 7 : 1.$

Hence, a : b = 7 : 1.

(a) Frequency table :

(b) From graph,

The modal class is 60-70.

(c) From graph,

The mode = 64.

(d) From graph,

The number of plants whose height range is between 80 cm to 90 cm are 3.

From figure,

⇒ tan 52° = $\dfrac{CD}{AC}$

⇒ 1.28 = $\dfrac{100}{AC}$

⇒ AC = $\dfrac{100}{1.28}$ = 78.125 m

⇒ tan 45° = $\dfrac{CD}{BC}$

⇒ 1 = $\dfrac{100}{BC}$

⇒ BC = 100 m

⇒ AB = AC + BC = 78.125 + 100 = 178.125 m

Hence, AB = 178.125 m

Comparing equation 2x² - 10x + 5 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -10 and c = 5

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{2(5 \pm \sqrt{15})}{4} \\[1em] = \dfrac{5 \pm \sqrt{15}}{2} \\[1em] = \dfrac{5 \pm 3.87}{2} \\[1em] = \dfrac{5 + 3.87}{2}, \dfrac{5 - 3.87}{2} \\[1em] = \dfrac{8.87}{2}, \dfrac{1.13}{2} \\[1em] = 4.44, 0.565$

Hence, x = 4.44, 0.565

(a) T_n = 6(7 - n)

Substituting n = 1, we get :

T₁ = 6(7 - 1) = 6 × 6 = 36.

Substituting n = 2, we get :

T₂ = 6(7 - 2) = 6 × 5 = 30.

Common difference (d) = T₂ - T₁

= 30 - 36 = -6.

Hence, common difference (d) = -6 and first term = 36.

(b) By formula,

Sum upto n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

$\text{Sum upto 25 terms} = \dfrac{25}{2}[2 \times 36 + (25 - 1) \times -6] \\[1em] = 12.5 \times [72 + 24 \times -6] \\[1em] = 12.5 \times [72 - 144] \\[1em] = 12.5 \times -72 \\[1em] = -900.$

Hence, sum of its first 25 terms = -900.

(a) In ∆APD and ∆BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Both equal to 90°)

Hence, proved that ∆APD ∼ ∆BPC.

(b) In ∆ADB,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = $\sqrt{64}$ = 8 cm.

⇒ PB = BD - PD = 8 - 4.5 = 3.5 cm

Hence, BD = 8 cm and PB = 3.5 cm.

(c) In ∆APD,

By pythagoras theorem,

⇒ AP² = AD² + DP²

⇒ AP² = 6² + (4.5)²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = $\sqrt{56.25}$ = 7.5 cm

Hence, length of AP = 7.5 cm.

(d) We know that,

Ratio of area of similar triangles is equal to the square of the corresponding sides.

$\therefore \dfrac{\text{Area of △APD}}{\text{Area of △BPC}} = \dfrac{AD^2}{BC^2} \\[1em] = \dfrac{6^2}{(2.4)^2} \\[1em] = \dfrac{6 \times 6}{2.4 \times 2.4} \\[1em] = \dfrac{1 \times 1}{0.4 \times 0.4} \\[1em] = \dfrac{10 \times 10}{4 \times 4} \\[1em] = \dfrac{100}{16} \\[1em] = \dfrac{25}{4} \\[1em] = 25 : 4.$

Hence, area ∆APD : area ∆BPC = 25 : 4.

(a) From figure,

∠AOC = ∠AOM + ∠COM = 65° + 65° = 130°.

We know that,

Angle at the center is twice the angle formed by the same arc at any other point of the circle.

⇒ ∠AOC = 2∠ABC

⇒ ∠ABC = $\dfrac{1}{2} \times ∠AOC = \dfrac{1}{2} \times 130°$ = 65°.

Hence, ∠ABC = 65°.

(b) In △ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC = 65° (Opposite angles of equal sides are equal)

By angle sum property of triangle,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 65° + 65° + ∠BAC = 180°

⇒ ∠BAC = 180° - 65° - 65° = 50°.

Hence, ∠BAC = 50°.

(c) We know that,

The angle formed between the tangent and the chord through the point of contact of the tangent is equal to the angle formed by the chord in the alternate segment.

∴ ∠BCQ = ∠BAC = 50°.

Hence, ∠BCQ = 50°.

Given,

-3 + x ≤ $\dfrac{7x}{2} + 2$ < 8 + 2x

Solving L.H.S. of the above inequation :

$\Rightarrow -3 + x \le \dfrac{7x}{2} + 2 \\[1em] \Rightarrow \dfrac{7x}{2} -x \ge -3 - 2 \\[1em] \Rightarrow \dfrac{7x - 2x}{2} \ge -5 \\[1em] \Rightarrow \dfrac{5x}{2} \ge -5 \\[1em] \Rightarrow x \ge \dfrac{-5 \times 2}{5} \\[1em] \Rightarrow x \ge -2 \text{ ..........(1)}$

Solving R.H.S. of the above inequation :

$\Rightarrow \dfrac{7x}{2} + 2 \lt 8 + 2x \\[1em] \Rightarrow \dfrac{7x}{2} - 2x \lt 8 - 2 \\[1em] \Rightarrow \dfrac{7x - 4x}{2} \lt 6 \\[1em] \Rightarrow \dfrac{3x}{2} \lt 6 \\[1em] \Rightarrow x \lt \dfrac{6 \times 2}{3} \\[1em] \Rightarrow x \lt 4 \text{ ..........(2)}$

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

-2 ≤ x < 4.

Since, x ∈ I.

x = {-2, -1, 0, 1, 2, 3}.

Hence, solution set = {-2, -1, 0, 1, 2, 3}.

(a) From graph,

Co-ordinates of A = (0, 6) and D = (-3, 0)

(b) By formula,

Co-ordinates of centroid = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

$= \Big(\dfrac{0 + 4 + (-4)}{3}, \dfrac{6 + (-4) + (-2)}{3}\Big) \\[1em] = \Big(\dfrac{0}{3}, \dfrac{0}{3}\Big) \\[1em] = (0, 0).$

Hence, co-ordinates of centroid of ∆ABC = (0, 0).

(c) By section-formula,

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

Given,

D divides AC in the ratio k : 1.

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

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y₁ = $\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of BD :

⇒ y - (-4) = $\dfrac{0 - (-4)}{-3 - 4}(x - 4)$

⇒ y + 4 = $\dfrac{4}{-7}(x - 4)$

⇒ -7(y + 4) = 4(x - 4)

⇒ -7y - 28 = 4x - 16

⇒ 4x + 7y - 16 + 28 = 0

⇒ 4x + 7y + 12 = 0.

Hence, equation of BD is 4x + 7y + 12 = 0.

⇒ x - 1 = 0

⇒ x = 1.

Given, (x - 1) is a factor of 3x³ + 8x² - 15x + k.

Thus, on substituting x = 1 in 3x³ + 8x² - 15x + k, the remainder will be zero.

⇒ 3.(1)³ + 8.(1)² - 15(1) + k = 0

⇒ 3.1 + 8.1 - 15 + k = 0

⇒ 3 + 8 - 15 + k = 0

⇒ 11 - 15 + k = 0

⇒ k - 4 = 0

⇒ k = 4.

Polynomial = 3x³ + 8x² - 15x + 4

On dividing (3x³ + 8x² - 15x + 4) by (x - 1), we get :

$\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \\ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \\ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \\ \phantom{{x - 1}31x^3-2}11x^2 - 15x \\ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \\ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \\ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \\ \phantom{{x - 1)}31x^3-2+11+1}\times \end{array}$

⇒ 3x³ + 8x² - 15x + 4 = (x - 1)(3x² + 11x - 4)

= (x - 1)[3x² + 12x - x - 4]

= (x - 1)[3x(x + 4) - 1(x + 4)]

= (x - 1)(3x - 1)(x + 4).

Hence, 3x³ + 8x² - 15x + 4 = (x - 1)(3x - 1)(x + 4).

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}

No. of cards = 8

(a) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}

Probability that the card drawn is a letter of the word MONDAY

= $\dfrac{\text{Letters of MONDAY present}}{\text{No. of cards}} = \dfrac{6}{8} = \dfrac{3}{4}$.

Hence, required probability = $\dfrac{3}{4}$.

(b) Letters of the word not present in MONDAY = {'S', 'U'}

Probability that the card drawn is not a letter of the word MONDAY

= $\dfrac{\text{Letters not present in MONDAY}}{\text{No. of cards}} = \dfrac{2}{8} = \dfrac{1}{4}$.

Hence, required probability = $\dfrac{1}{4}$.

(c) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.

Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY

= $\dfrac{\text{Letters present in both MONDAY and SUNDAY}}{\text{No. of cards}} = \dfrac{4}{8} = \dfrac{1}{2}$.

Hence, required probability = $\dfrac{1}{2}$.

Given,

Radius of spherical vessel (r) = 28 cm

Volume of spherical vessel (v) = $\dfrac{4}{3}πr^3$

Volume of oil in vessel = $\dfrac{3}{4}v$

Substituting values we get :

$\Rightarrow v = \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \\[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{3}{4} \times \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \\[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{22}{7} \times 28^3.$

Radius of cylindrical vessel (R) = 21 cm

Let height of oil in cylindrical vessel be h cm.

Volume of oil = Volume of cylinder upto which oil is filled (πR²h)

$\Rightarrow \dfrac{22}{7} \times 28^3 = \dfrac{22}{7} \times 21^2 \times h \\[1em] \Rightarrow 28^3 = 21^2 \times h \\[1em] \Rightarrow h = \dfrac{28^3}{21^2}\\[1em] \Rightarrow h = \dfrac{21952}{441} \\[1em] \Rightarrow h = 49.77 ≈ 50 \text{ cm}.$

Hence, height of the oil in the cylindrical vessel = 50 cm.

Given,

Radius of circle (r) = OA = OB = 9 cm

From figure,

AB = OA + OB = 9 + 9 = 18 cm

PB = PA - AB = 24 - 18 = 6 cm

We know that,

If a secant segment and tangent segment are drawn to a circle from the same external point, the length of the tangent segment is the geometric mean between the length of the secant segment and the length of the external part of the secant segment.

⇒ PQ² = PB × PA

⇒ PQ² = 6 × 24

⇒ PQ² = 144

⇒ PQ = $\sqrt{144}$ = 12 cm.

Hence, length of tangent PQ = 12 cm.

Money invested = ₹33000

N.V. of share = ₹100

M.V. = N.V + Premium

= ₹100 + $\dfrac{10}{100} \times 100$

= ₹100 + ₹10

= ₹110.

(a) Number of shares = $\dfrac{\text{Money invested}}{\text{M.V.}} = \dfrac{33000}{110}$ = 300.

Hence, no. of shares purchased = 300.

(b) By formula,

Annual dividend = Number of shares × Rate of dividend × N.V.

= 300 × $\dfrac{12}{100} \times 100$

= ₹3600.

Hence, annual dividend = ₹3600.

Steps of construction :

1. Take 2 cm = 5 years on x-axis.

2. Take 1 cm = 10 policy holders on y-axis.

3. Plot the points (20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96) and (60, 100).

4. Join the points by a free-hand curve.

Here, n = 100

Median = $\dfrac{n}{2} = \dfrac{100}{2}$ = 50th term

(a) Through J = 50 draw a horizontal line to meet the ogive at K. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 42.

Hence, the median age = 42 years.

(b) Through M = 52 draw a vertical line to meet the ogive at N. Through N, draw a horizontal line to meet the y-axis at O. The ordinate of the point O represents 91.

Hence, 91 people have their age less than or equal to 52.

∴ No. of people whose age is greater than 52 = 100 - 91 = 9.

Hence, number of policy holders whose age is above 52 years equal to 9.

For Laddu :

Total cost = ₹ 500 × 2 = ₹ 1000

GST = $\dfrac{5}{100} \times 1000$ = ₹50

Total amount (including GST) = 1000 + 50 = ₹ 1050.

For pastries :

Total cost = ₹ 100 × 12 = ₹ 1200

GST = $\dfrac{18}{100} \times 1200$ = ₹ 216

Total amount (including GST) = 1200 + 216 = ₹ 1416.

(a) Total GST paid = ₹50 + ₹216 = ₹266.

Hence, total GST paid = ₹266.

(b) Total bill (including GST) = ₹ 1050 + ₹ 1416 = ₹ 2466.

Hence, total bill = ₹ 2466.

(a) 1st equation :

⇒ kx - y + 4 = 0

⇒ y = kx + 4

Slope (s₁) : k

2nd equation :

⇒ 2y = 6x + 7

⇒ y = $\dfrac{6}{2}x + \dfrac{7}{2}$

⇒ y = 3x + $\dfrac{7}{2}$

Slope (s₂) : 3

We know that,

Product of slope of perpendicular lines = -1

⇒ k × 3 = -1

⇒ k = $-\dfrac{1}{3}$

Hence, k = $-\dfrac{1}{3}$.

(b) We know that,

Slope of parallel lines are equal.

Slope of line parallel to line 2y = 6x + 7 is 3.

By point-slope form :

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = 3[x - (-1)]

⇒ y - 1 = 3[x + 1]

⇒ y - 1 = 3x + 3

⇒ y = 3x + 3 + 1

⇒ y = 3x + 4.

Hence, equation of line parallel to 2y = 6x + 7 and passing through (–1, 1) is y = 3x + 4.

Steps of construction :

1. Draw a line segment BC = 8 cm

2. Construct ∠ABC = 90°, such that AB = 6 cm.

3. Draw XY, the perpendicular bisector of BC.

4. Draw PQ, the perpendicular bisector of AB.

5. Mark point O, the intersection of segment XY and PQ.

6. Draw BZ, the angle bisector of AB and BC.

We know that,

Locus of points equidistant from two points is the perpendicular bisector of the line joining the two points segment.

(a) Locus of points equidistant from B and C is XY.

(b) Locus of points equidistant from A and B is PQ.

We know that,

Locus of points equidistant from two sides is the angular bisector of angle between them.

(d) Locus of points which are equidistant from BA and BC is BZ.