Solved 2023 Question Paper ICSE Class 10 Mathematics

Solving,

$\Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2.x + 0.y \\ 0.x + 4.y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 0 \\ 0 + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2x = 2 \text{ and } 4y = -8 \\[1em] \Rightarrow x = \dfrac{2}{2} \text{ and } y = -\dfrac{8}{4} \\[1em] \Rightarrow x = 1 \text{ and } y = -2.$

Hence, Option 1 is the correct option.

By factor theorem,

If x - a is a factor of f(x), then f(a) = 0.

Given,

x - 2 is a factor of x³ - kx - 12

⇒ x - 2 = 0

⇒ x = 2.

∴ 2³ - 2k - 12 = 0

⇒ 8 - 2k - 12 = 0

⇒ -2k - 4 = 0

⇒ 2k = -4

⇒ k = $-\dfrac{4}{2}$ = -2.

Hence, Option 3 is the correct option.

We know that,

Angle between tangent and the chord at the point of contact is equal to angle of the alternate segment.

∴ ∠PQS = ∠PST = 30°

In △ PQS,

By angle sum property of triangle,

⇒ ∠PQS + ∠QPS + ∠PSQ = 180°

⇒ 30° + 60° + ∠PSQ = 180°

⇒ ∠PSQ + 90° = 180°

⇒ ∠PSQ = 180° - 90° = 90°.

Hence, Option 4 is the correct option.

All the letters of English alphabets have 5 vowels (a, e, i, o and u) and total number of English alphabets = 26.

Probability that letter chosen is a vowel = $\dfrac{\text{No. of vowels}}{\text{No. of alphabets}} = \dfrac{5}{26}$.

Hence, Option 2 is the correct option.

Given,

3 is a root of the quadratic equation x² - px + 3 = 0

∴ 3² - 3p + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 3p = 12

⇒ p = $\dfrac{12}{3}$ = 4.

Hence, Option 1 is the correct option.

Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB // CD.

From figure,

PA = PC + AC = 6 + 4 = 10 cm.

In △ PAB and △ PCD,

⇒ ∠PAB = ∠PCD (Both equal to 70°)

⇒ ∠BPA = ∠DPC (Both are equal)

∴ △ PAB ~ △ PCD (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{PB}{PD} = \dfrac{PA}{PC} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{10}{6} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{5}{3}.$

Let PB = 5x and PD = 3x.

From figure,

DB = PB - PD = 5x - 3x = 2x.

$\Rightarrow \dfrac{PD}{DB} = \dfrac{3x}{2x} = \dfrac{3}{2}$

⇒ PD : DB = 3 : 2.

Hence, Option 3 is the correct option.

GST charged = GST (%) × M.P.

= $\dfrac{10}{100} \times 3080$

= ₹ 308.

Hence, Option 2 is the correct option.

Solving,

⇒ (1 + sin A)(1 - sin A)

⇒ 1 - sin A + sin A - sin² A

⇒ 1 - sin² A

⇒ cos² A.

Hence, Option 4 is the correct option.

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Centroid of triangle ABC (G)

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

Hence, Option 1 is the correct option.

Given,

⇒ a_n = 2n + 5

⇒ a₁₀ = 2(10) + 5 = 20 + 5 = 25.

Hence, Option 3 is the correct option.

Let x be the mean proportional between 4 and 9 then,

$\Rightarrow \dfrac{4}{x} = \dfrac{x}{9} \\[1em] \Rightarrow x^2 = 4 \times 9 \\[1em] \Rightarrow x^2 = 36 \\[1em] \Rightarrow x = \sqrt{36} = 6.$

Hence, Option 2 is the correct option.

We know that,

To determine mean we do not need any graphical method.

Hence, Option 4 is the correct option.

Let radius of cylinder be r cm.

Given,

Height of cylinder (h) = 3 cm

By formula,

⇒ Volume of cylinder = πr²h

⇒ 3πr² = 48π

⇒ 3r² = 48

⇒ r² = $\dfrac{48}{3}$

⇒ r² = $\dfrac{48}{3}$

⇒ r² = 16

⇒ r = $\sqrt{16}$ = 4 cm.

Hence, Option 3 is the correct option.

By formula,

Interest earned = Maturity value - Sum deposited

= ₹ 4884 - ₹ (800 × 6)

= ₹ 4884 - ₹ 4800

= ₹ 84.

Hence, Option 1 is the correct option.

Solving,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\dfrac{10}{2}$

⇒ x ≤ 5.

Since, x ∈ W,

∴ x = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.

By factor theorem,

If x - a is a factor of f(x), then f(a) = 0.

Given,

x - a is a factor of the polynomial 3x³ + x² - ax - 81.

⇒ x - a = 0

⇒ x = a.

∴ 3a³ + a² - a.a - 81 = 0

⇒ 3a³ + a² - a² - 81 = 0

⇒ 3a³ - 81 = 0

⇒ 3a³ = 81

⇒ a³ = $\dfrac{81}{3}$

⇒ a³ = 27

⇒ a³ = 3³

⇒ a = 3.

Hence, value of a = 3.

(a) Given,

Salman deposits ₹ 1000 every month in a recurring deposit account for 2 years.

Total deposit = ₹ 1,000 × 2 × 12 = ₹ 24,000.

By formula,

Total interest earned = Maturity value - Total deposit

= ₹ 26,000 - ₹ 24,000

= ₹ 2,000.

Hence, interest earned = ₹ 2,000.

(b) Let rate of interest be r%. Time (n) = 24 months

By formula,

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

Substituting values we get :

$\Rightarrow 2000 = 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \dfrac{2000}{250} \\[1em] \Rightarrow r = 8$

Hence, rate of interest earned = 8%.

(a) We know that,

Angle in a semi-circle is a right angle.

∴ ∠BDA = 90°.

Hence, ∠BDA = 90°.

(b) In △ BAD,

By angle sum property of triangle,

⇒ ∠BDA + ∠BAD + ∠ABD = 180°

⇒ 90° + ∠BAD + 26° = 180°

⇒ ∠BAD + 116° = 180°

⇒ ∠BAD = 180° - 116° = 64°.

Hence, ∠BAD = 64°.

(c) From figure,

CE is tangent to the circle.

Angle between tangent and radius of the circle is 90°.

From figure,

⇒ ∠CAD + ∠BAD = 90°

⇒ ∠CAD + 64° = 90°

⇒ ∠CAD = 90° - 64° = 26°.

Hence, ∠CAD = 26°.

(d) Join OD.

In △ OBD,

⇒ OD = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

⇒ ∠ODB = ∠OBD = 26°.

Hence, ∠ODB = 26°.

Comparing x² + 4x - 8 = 0 with ax² + bx + c, we get :

a = 1, b = 4 and c = -8.

By formula,

$\Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-8)}}{2 \times 1} \\[1em] = \dfrac{-4 \pm \sqrt{16 + 32}}{2} \\[1em] = \dfrac{-4 \pm \sqrt{48}}{2} \\[1em] = \dfrac{-4 \pm 4\sqrt{3}}{2} \\[1em] = -2 \pm 2\sqrt{3} \\[1em] = -2 \pm 2 \times 1.732 \\[1em] = -2 \pm 3.464 \\[1em] = -2 + 3.464 \text{ or } -2 - 3.464 \\[1em] = 1.464 \text{ or } -5.464 \\[1em] = 1.5 \text{ or } -5.5$

Hence, x = 1.5 or 5.5

Solving L.H.S. of the given equation :

⇒ (sin² θ - 1)(tan² θ + 1) + 1

⇒ (1 - cos² θ - 1).sec² θ + 1

⇒ -cos² θ. sec² θ + 1

⇒ -cos² θ $\times \dfrac{1}{\text{cos}^2 \text{θ}}$ + 1

⇒ -1 + 1

⇒ 0.

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

Hence, proved that (sin² θ - 1)(tan² θ + 1) + 1 = 0.

The points are shown in the graph below:

(c) From graph,

F = (-1, -1)

Hence, co-ordinates of F = (-1, -1).

(e) From graph,

The closed figure formed is a star.

B + C = $\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}$

Substituting values we get :

$A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$

Hence, A(B + C) - 14I = $\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$

(a) By formula,

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

Given,

D is the mid-point of BC.

$\therefore \text{Co-ordinates of D} = \Big(\dfrac{0 + (-6)}{2}, \dfrac{4 + 4}{2}\Big) \\[1em] = \Big(\dfrac{-6}{2}, \dfrac{8}{2}\Big) \\[1em] = (-3, 4).$

Hence, co-ordinates of D = (-3, 4).

(b) By two-point form :

Equation of a line :

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

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $\dfrac{4 - (-1)}{(-3) - 1}(x - 1)$

⇒ y + 1 = $\dfrac{5}{-4}(x - 1)$

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y = 1.

Hence, equation of median AD is 5x + 4y = 1.

(a) We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

Since, PT is tangent from point P and PAB is secant.

∴ PT² = PA × PB

⇒ PT² = (9 + 16) × 16

⇒ PT² = 25 × 16

⇒ PT² = 400

⇒ PT = $\sqrt{400}$ = 20 cm.

Hence, PT = 20 cm.

(b) Since, angle between tangent and chord at point of contact is equal to angle of alternative segment.

∴ ∠BAT = ∠PTB = 50°.

Hence, ∠BAT = 50°.

(c) We know that,

Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at some point on the remaining circumference.

∴ ∠BOT = 2 × ∠BAT = 2 × 50° = 100°.

Hence, ∠BOT = 100°.

(d) From figure,

⇒ OB = OT (Radii of same circle)

⇒ ∠OBT = ∠OTB = x (let) (Angles opposite to equal sides are equal)

In △ BOT,

By angle sum property of triangle,

⇒ ∠OBT + ∠OTB + ∠BOT = 180°

⇒ x + x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

∴ ∠OBT = 40°

From figure,

⇒ ∠ABT = ∠ABO + ∠OBT = 45° + 40° = 85°.

Hence, ∠ABT = 85°.

(a) By formula,

GST on an article = (Price of item - Discount) × Rate of GST

GST on hair oil = (1200 - 100) × 18%

= 1100 × $\dfrac{18}{100}$

= 11 × 18

= ₹ 198.

GST on cashew nuts = (600 - 0) × 12%

= 600 × $\dfrac{12}{100}$

= 6 × 12

= ₹ 72.

Total GST paid = ₹ 198 + ₹ 72 = ₹ 270.

Hence, total GST paid = ₹ 270.

(b) Total bill amount including GST = Total cost price + Total GST paid

= [(₹ 1200 - ₹ 100) + ₹ 600] + ₹ 270

= ₹ 1100 + ₹ 600 + ₹ 270

= ₹ 1970.

Hence, total bill amount = ₹ 1970.

Given, equation :

-5(x - 9) ≥ 17 - 9x > x + 2

Solving L.H.S. of the given equation :

⇒ -5(x - 9) ≥ 17 - 9x

⇒ -5x + 45 ≥ 17 - 9x

⇒ -5x + 9x ≥ 17 - 45

⇒ 4x ≥ -28

⇒ x ≥ $-\dfrac{28}{4}$

⇒ x ≥ -7 ............(1)

Solving L.H.S. of the given equation :

⇒ 17 - 9x > x + 2

⇒ x + 9x < 17 - 2

⇒ 10x < 15

⇒ x < $\dfrac{15}{10}$

⇒ x < $\dfrac{3}{2}$

⇒ x < 1.5 .............(2)

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

Solution set : {x : -7 ≤ x < 1.5}

Solution set on the number line :

(a) Given,

DE // BF

∴ ∠DEG = ∠GBF (Alternate angles are equal)

∠EDG = ∠GFB (Alternate angles are equal)

⇒ △GED ~ △ GBF (By A.A. axiom)

Hence, proved that △GED ~ △ GBF.

(b) We know that,

Corresponding sides of similar triangle are proportional.

Since, △ GED ~ △ GBF

$\Rightarrow \dfrac{DE}{BF} = \dfrac{EG}{BG} \\[1em] \Rightarrow \dfrac{DE}{30} = \dfrac{8}{16} \\[1em] \Rightarrow DE = 30 \times \dfrac{8}{16} \\[1em] \Rightarrow DE = 15 \text{ cm}.$

Hence, DE = 15 cm.

(c) In △ BAC and △ BDE,

⇒ ∠BAC = ∠BDE (Corresponding angles are equal)

⇒ ∠ABC = ∠DBE (Common angle)

⇒ △ BAC ~ △ BDE (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional.

Since, △ BAC ~ △ BDE

$\Rightarrow \dfrac{AB}{BD} = \dfrac{AC}{DE} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{DE}{AC} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{15}{24} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{5}{8}$

⇒ BD : AB = 5 : 8.

Hence, BD : AB = 5 : 8.

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 = ₹ 530

Hence, required mode = ₹ 530.

If first term is a and common difference is d of the A.P.

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th term = -12

⇒ a₉ = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\dfrac{16}{4}$

⇒ d = -4.

Substituting value of d in equation (1), we get :

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\dfrac{n}{2}(a + l)$

Sum of 16 terms of A.P. = $\dfrac{n}{2}(a + a_{16})$

$= \dfrac{16}{2}[a + a + (16 - 1)d] \\[1em] = 8(2a + 15d) \\[1em] = 8 \times (2 \times 20 + 15 \times -4) \\[1em] = 8 \times (40 - 60) \\[1em] = 8 \times -20 \\[1em] = -160.$

Hence, sum of 16 terms of A.P. = -160.

(a) From figure,

A = (4, 0) and B = (0, 4).

(b) Let coordinates of P 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 :

$\Rightarrow (x, y) = \Big(\dfrac{3 \times 0 + 1 \times 4}{3 + 1}, \dfrac{3 \times 4 + 1 \times 0}{3 + 1}\Big) \\[1em] = \Big(\dfrac{0 + 4}{4}, \dfrac{12 + 0}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{12}{4}\Big) \\[1em] = (1, 3).$

Hence, coordinates of P = (1, 3).

(c) By formula,

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

Substituting values we get :

Slope of AB = $\dfrac{4 - 0}{0 - 4} = \dfrac{4}{-4}$ = -1.

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of line perpendicular to AB = -1

⇒ -1 × Slope of line perpendicular to AB = -1

⇒ Slope of line perpendicular to AB = $\dfrac{-1}{-1}$ = 1.

Line passing through P and perpendicular to AB :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 1)

⇒ y - 3 = x - 1

⇒ y = x - 1 + 3

⇒ y = x + 2.

Hence, required equation is y = x + 2.

Total number of outcomes = 25

(a) Number of multiples of 5 between 1 to 25 = 5 (5, 10, 15, 20, 25).

P(that number on card is a multiples of 5)

= $\dfrac{\text{No. of multiples of 5}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the required probability is $\dfrac{1}{5}$.

(b) Number of perfect squares between 1 to 25 = 5 (1, 4, 9, 16, 25).

P(that number on card is a perfect square)

= $\dfrac{\text{No. of perfect squares}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the required probability is $\dfrac{1}{5}$.

(c) Number of prime numbers between 1 to 25 = 9 (2, 3, 5, 7, 11, 13, 17, 19, 23).

P(that number is a prime number)

= $\dfrac{\text{No. of prime numbers}}{\text{Total number of outcomes}} = \dfrac{9}{25}$.

Hence, the required probability is $\dfrac{9}{25}$.

In first case :

Distance = 100 km

Speed = x km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x}$

In second case :

Distance = 100 km

Speed = (x + 5) km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x + 5}$

According to question,

The difference between the time taken between first and second case is 1 hour.

$\therefore \dfrac{100}{x} - \dfrac{100}{x + 5} = 1 \\[1em] \Rightarrow \dfrac{100(x + 5) - 100x}{x(x + 5)} = 1 \\[1em] \Rightarrow \dfrac{100x + 500 - 100x}{x^2 + 5x} = 1 \\[1em] \Rightarrow \dfrac{500}{x^2 + 5x} = 1 \\[1em] \Rightarrow 500 = x^2 + 5x \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x - 20)(x + 25) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 25 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -25.$

Since, speed cannot be negative.

∴ x = 20 km/hr.

Hence, original speed = 20 km/hr.

Let the rise in water level be x cm.

∴ Volume of water that rises by x cm in the cylindrical container = Volume of hemisphere submerged + Volume of cone submerged

⇒ π × (14)² × x = $\dfrac{2}{3} \times π \times 7^3 + \dfrac{1}{3} \times π \times 7^2 \times 4$

⇒ 196πx = $\dfrac{2}{3} \times π \times 343 + \dfrac{1}{3} \times π \times 196$

⇒ 196x = $\dfrac{2}{3} \times 343 + \dfrac{1}{3} \times 196$

⇒ 196x = $\dfrac{686 + 196}{3}$

⇒ 196x = $\dfrac{882}{3}$

⇒ x = $\dfrac{882}{3 \times 196} = \dfrac{882}{588}$ = 1.5 cm.

Hence, rise in water level = 1.5 cm

By formula,

Mean = A + $\dfrac{Σfd}{Σf} = 25 + \dfrac{90}{40}$

= 25 + 2.25

= 27.25

Hence, required mean = 27.25

Let x be added to each number.

So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.

$\therefore \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \\[1em] \Rightarrow 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4.$

Hence, required number is 4.

Steps of construction :

1. Draw AB = 6 cm.

2. Draw AP so that angle BAP = 120°.

3. From AP cut AC = 5 cm ⇒ ∠BAC = 120°.

4. Join B and C to get triangle ABC.

5. Draw perpendicular bisectors of AB and AC which cut each other at point O.

6. With O as center and OA (or OB or OC) as radius draw circle which passes through points A, B and C.

Hence, radius of circle obtained = 5 cm.

Given : $\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2 \\[1em] \Rightarrow \sqrt{2x + 2} = 2\sqrt{2x - 1}$

Squaring both sides we get :

⇒ 2x + 2 = 4(2x - 1)

⇒ 2x + 2 = 8x - 4

⇒ 8x - 2x = 2 + 4

⇒ 6x = 6

⇒ x = $\dfrac{6}{6}$ = 1.

Hence, x = 1.

Given,

A.P. : 15, 30, 45, 60 .....

First term (a) = 15

Common difference (d) = 30 - 15 = 15

Let n^th term be 300.

⇒ a_n = 300

⇒ a + (n - 1)d = 300

⇒ 15 + 15(n - 1) = 300

⇒ 15 + 15n - 15 = 300

⇒ 15n = 300

⇒ n = $\dfrac{300}{15}$ = 20.

Sum of the terms of the given A.P. :

⇒ S_n = $\dfrac{n}{2}(a + a_n)$

⇒ S₂₀ = $\dfrac{20}{2}(15 + 300)$

= 10 × 315

= 3150.

Hence, sum of all terms of the given A.P. = 3150.

Let CD be the tower.

From figure,

⇒ ∠A = ∠EDA = 45° (Alternate angles are equal)

⇒ ∠B = ∠FDB = 38° (Alternate angles are equal)

In △ ACD,

⇒ tan A = $\dfrac{CD}{AC}$

⇒ tan 45° = $\dfrac{100}{AC}$

⇒ AC = $\dfrac{100}{\text{tan 45°}} = \dfrac{100}{1}$ = 100 m.

In △ BCD,

⇒ tan B = $\dfrac{CD}{BC}$

⇒ tan 38° = $\dfrac{100}{BC}$

⇒ BC = $\dfrac{100}{\text{tan 38°}} = \dfrac{100}{0.7813}$ = 127.99 m.

From figure,

The distance between ships A and B = AC + BC

= 100 + 127.99

= 227.99 m ≈ 228 m.

Hence, the distance between the two ships A and B = 228 m.

Substituting x = -2 in 2x³ - x² - 13x - 6, we get :

⇒ 2(-2)³ - (-2)² - 13(-2) - 6

⇒ 2(-8) - 4 + 26 - 6

⇒ -16 - 4 + 20

⇒ -20 + 20

⇒ 0.

∴ x + 2 is a factor of the polynomial 2x³ - x² - 13x - 6.

Dividing, 2x³ - x² - 13x - 6 by x + 2, we get :

$\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \\ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \\ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x + 2}x^3-2}-5x^2 - 13x \\ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \\ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 2x³ - x² - 13x - 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 - x² - 13x - 6 = (x + 2)(2x + 1)(x - 3).

(a)

Here, n = 60, which is even.

Median = $\dfrac{n}{2}$ th term = $\dfrac{60}{2}$ = 30th term.

Steps of construction :

1. Take 1 cm = 2 kg on x-axis.

2. Take 1 cm = 10 students on y-axis.

3. Since, x axis starts at 28 hence, a kink is drawn at the starting of x-axis. Plot the point (28, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60).

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

6. Draw a line parallel to x-axis from point A (no. of students) = 30, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 36.2

Hence, median = 36.2 kg

(ii) Here, n = 60, which is even.

By formula,

Upper quartile = $\dfrac{3n}{4} = \dfrac{3 \times 60}{4}$ = 45th term.

Draw a line parallel to x-axis from point D (no. of students) = 45, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 38.2 kg

Hence, upper quartile = 38.2 kg.

(iii) Draw a line parallel to y-axis from point G (weight) = 37 kg, touching the graph at point H. From point H draw a line parallel to x-axis touching y-axis at point I.

From graph, I = 38.

∴ 38 students have weight less than or equal to 37 kg.

No. of students whose weight is more than 37 kg = 60 - 38 = 22.

Hence, no. of students whose weight is more than 37 kg = 22.