Solved 2026 Improvement Paper ICSE Class 10 Mathematics

The marked price of a smart TV is ₹ 18,000. The rate of GST is 18%. The price paid by the customer for the smart TV is:

1. ₹12,960

2. ₹19,620

3. ₹18,000

4. ₹21,240

Answer

Given,

Marked price = ₹ 18,000

Rate of GST = 18%

GST = $\dfrac{18}{100} \times 18000$ = ₹ 3,240

Price paid by customer = Marked price + GST

= ₹ 18,000 + ₹ 3,240

= ₹ 21,240.

Hence, option 4 is the correct option.

Two right circular cones, C₁ and C₂, have their heights in the ratio 2 : 1 and radii in the ratio 1 : 2. The ratio of their volumes is:

1. 1 : 4

2. 1 : 2

3. 2 : 1

4. 4 : 1

Answer

Given,

Ratio of heights ($h_1 : h_2$) = 2 : 1

Ratio of radii ($r_1 : r_2$) = 1 : 2

By formula,

Volume of cone = $\dfrac{1}{3} \pi r^2 h$

$\therefore \dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{3} \pi r_1^2 h_1}{\dfrac{1}{3} \pi r_2^2 h_2} \\[1em] = \dfrac{r_1^2 h_1}{r_2^2 h_2} \\[1em] = \dfrac{1^2 \times 2}{2^2 \times 1} \\[1em] = \dfrac{2}{4} \\[1em] = \dfrac{1}{2}.$

Ratio of volumes = 1 : 2.

Hence, option 2 is the correct option.

A person deposited ₹500 per month in a recurring deposit account for one year at $r$% per annum. The rate of interest is revised to $2r$%. The interest earned by the person compared to the original amount of interest will be:

1. one-fourth

2. unchanged

3. half

4. double

Answer

By formula,

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

In this formula, the interest is directly proportional to the rate of interest, as the principal (P) and the number of months ($n$) remain unchanged.

When the rate is revised from $r$% to $2r$%, the rate becomes double.

∴ The interest earned also becomes double.

Hence, option 4 is the correct option.

A company declares 9% dividend on its shares of ₹100 available at ₹120. It means a person will receive annual dividend which is 9% of:

1. market value

2. face value

3. the difference of market value and face value

4. company's profit

Answer

Dividend is always calculated on the face value (nominal value) of the shares.

Here, the face value of the share is ₹ 100.

∴ The person will receive an annual dividend which is 9% of the face value.

Hence, option 2 is the correct option.

The discriminant of the quadratic equation $2x^2 + 5x + c = 0$ is 1. The value of $c$ is:

1. -3

2. 0

3. 2

4. 3

Answer

Given,

Quadratic equation : $2x^2 + 5x + c = 0$

Comparing with $ax^2 + bx + c = 0$,

$a = 2, b = 5, c = c$

By formula,

Discriminant (D) = $b^2 - 4ac$

$\Rightarrow 1 = 5^2 - 4 \times 2 \times c \\[1em] \Rightarrow 1 = 25 - 8c \\[1em] \Rightarrow 8c = 25 - 1 \\[1em] \Rightarrow 8c = 24 \\[1em] \Rightarrow c = 3.$

Hence, option 4 is the correct option.

$(x + 3), (x + 6)$ and $(x + 10)$ are in continued proportion. The value of $x$ is:

1. 3

2. 6

3. 9

4. 12

Answer

Given,

$(x + 3), (x + 6)$ and $(x + 10)$ are in continued proportion.

We know that,

Three quantities $a, b$ and $c$ are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third.

i.e. $a : b = b : c$

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

So, the square of the middle term is equal to the product of the first and the last terms.

Here,

$a = (x + 3), b = (x + 6)$ and $c = (x + 10)$

∴ $(x + 6)^2 = (x + 3)(x + 10)$

Expanding both sides,

$\Rightarrow x^2 + 12x + 36 = x^2 + 13x + 30$

Cancelling $x^2$ from both sides,

$\Rightarrow 12x + 36 = 13x + 30$

Transposing the like terms,

$\Rightarrow 36 - 30 = 13x - 12x \\[1em] \Rightarrow 6 = x \\[1em] \Rightarrow x = 6.$

Hence, option 2 is the correct option.

$(x – 1)$ is a factor of the polynomial $x^3 + 2x^2 – x – k$. The value of $k$ is:

1. – 2

2. – 1

3. 1

4. 2

Answer

Given,

$(x – 1)$ is a factor of $x^3 + 2x^2 – x – k$.

∴ On putting $x = 1$, the value of the polynomial is zero.

$\Rightarrow (1)^3 + 2(1)^2 - (1) - k = 0 \\[1em] \Rightarrow 1 + 2 - 1 - k = 0 \\[1em] \Rightarrow 2 - k = 0 \\[1em] \Rightarrow k = 2.$

Hence, option 4 is the correct option.

Matrix $A = \begin{bmatrix} -1 & 2 \\ 3 & 0 \end{bmatrix}$ and the Matrix $mA = \begin{bmatrix} 3 & -6 \\ -9 & 0 \end{bmatrix}$. The value of $m$ is:

1. $-\dfrac{1}{3}$

2. – 3

3. $\dfrac{1}{3}$

4. 3

Answer

Given,

$mA = m\begin{bmatrix} -1 & 2 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} -m & 2m \\ 3m & 0 \end{bmatrix}$

Comparing this with the given matrix $\begin{bmatrix} 3 & -6 \\ -9 & 0 \end{bmatrix}$,

$\Rightarrow -m = 3\\[1em] \Rightarrow m = -3.$

Hence, option 2 is the correct option.

The points A ($x$, 0), B (−1, −1) and C (0, $y$) are collinear. The value of $\dfrac{1}{x} + \dfrac{1}{y}$ is:

1. – 1

2. 0

3. 1

4. $xy$

Answer

Given,

The points A ($x$, 0), B (−1, −1) and C (0, $y$) are collinear.

We know that,

Collinear points lie on the same straight line, so they cannot enclose a triangle.

∴ Area of the triangle formed by them = 0.

By formula,

Area of triangle = $\dfrac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

Here,

$\Rightarrow x_1 = x, y_1 = 0 \\[1em] \Rightarrow x_2 = −1, y_2 = −1 \\[1em] \Rightarrow x_3 = 0, y_3 = y$

Substituting values we get :

$\Rightarrow \dfrac{1}{2}[x(-1 - y) + (-1)(y - 0) + 0(0 - (-1))] = 0$

Multiplying both sides by 2,

$\Rightarrow x(-1 - y) + (-1)(y - 0) + 0(0 - (-1)) = 0 \\[1em] \Rightarrow -x - xy - y + 0 = 0 \\[1em] \Rightarrow -(x + y + xy) = 0 \\[1em] \Rightarrow x + y + xy = 0 \\[1em] \Rightarrow x + y = -xy.$

Since $\dfrac{1}{x} + \dfrac{1}{y}$ is required, we divide both sides by $xy$ (where $x ≠ 0$ and $y ≠ 0$),

$\Rightarrow \dfrac{x + y}{xy} = \dfrac{-xy}{xy} \\[1em] \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = -1 \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = -1 \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = -1.$

Hence, option 1 is the correct option.

In the given diagram, PQ // BC and ar∆APQ = ar quad BPQC. The value of AP : PB is:

1. 1 : 2

2. 2 : 1

3. 1 : $\sqrt{2}$

4. $(\sqrt{2} + 1)$ : 1

Answer

Given,

PQ // BC and ar ∆APQ = ar quad BPQC.

From figure,

ar ∆ABC = ar ∆APQ + ar quad BPQC = ar ∆APQ + ar ∆APQ = 2 ar ∆APQ.

$\Rightarrow \dfrac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \dfrac{1}{2}$

In ∆APQ and ∆ABC,

⇒ ∠PAQ = ∠BAC [Common]

⇒ ∠APQ = ∠ABC [Corresponding angles as PQ // BC]

∴ ∆APQ ~ ∆ABC (By A.A. axiom)

We know that,

The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

$\therefore \dfrac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{AP^2}{AB^2} \\[1em] \Rightarrow \dfrac{AP}{AB} = \dfrac{1}{\sqrt{2}}.$

Let AP = $k$, then AB = $\sqrt{2}k$.

$\Rightarrow PB = AB - AP = \sqrt{2}k - k = (\sqrt{2} - 1)k \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{k}{(\sqrt{2} - 1)k} = \dfrac{1}{\sqrt{2} - 1}.$

Rationalizing the denominator,

$\Rightarrow \dfrac{AP}{PB} = \dfrac{1}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} = \dfrac{\sqrt{2} + 1}{2 - 1} = \dfrac{\sqrt{2} + 1}{1}.$

AP : PB = $(\sqrt{2} + 1)$ : 1.

Hence, option 4 is the correct option.

In the given diagram, PA and PB are the tangents to the circle with centre O, such that PA = PB = AB. The value of ∠ACB is:

1. 30°

2. 45°

3. 60°

4. 90°

Answer

Given,

PA = PB = AB.

Since all three sides of ∆PAB are equal, ∆PAB is an equilateral triangle.

∴ ∠APB = 60°.

We know that,

Radius is perpendicular to the tangent at the point of contact.

∴ ∠OAP = ∠OBP = 90°.

In quadrilateral OAPB,

⇒ ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

⇒ ∠AOB + 90° + 60° + 90° = 360°

⇒ ∠AOB = 360° - 240° = 120°.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

$\Rightarrow \angle \text{ ACB }= \dfrac{1}{2} \angle \text{ AOB }= \dfrac{1}{2} \times 120° = 60°.$

Hence, option 3 is the correct option.

A semi-circular sheet of paper having radius 7 cm is folded to form a right circular cone, as shown in the diagram. The slant height of the cone so formed is:

1. 3.5 cm

2. 7 cm

3. $7\sqrt{2}$ cm

4. 14 cm

Answer

When a semi-circular sheet is folded to form a right circular cone, the radius of the semi-circle becomes the slant height of the cone.

Radius of semi-circular sheet = 7 cm.

∴ Slant height of the cone = 7 cm.

Hence, option 2 is the correct option.

Assertion (A): $\sec θ(1 + \sin θ)(\sec θ + \tan θ) = 1$

Reason (R): $\sec^2 θ - \tan^2 θ = 1$, for any value of $θ$.

1. (A) is true, (R) is false.

2. (A) is false, (R) is true.

3. Both (A) and (R) are true and (R) is the correct explanation of (A).

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

Answer

Solving the L.H.S. of the Assertion,

$\Rightarrow \sec θ(1 + \sin θ)(\sec θ + \tan θ) \\[1em] \Rightarrow \sec θ(1 + \sin θ)\Big(\dfrac{1}{\cos θ} + \dfrac{\sin θ}{\cos θ}\Big) \\[1em] \Rightarrow \sec θ(1 + \sin θ) \times \dfrac{1 + \sin θ}{\cos θ} \\[1em] \Rightarrow \dfrac{1}{\cos θ} \times \dfrac{(1 + \sin θ)^2}{\cos θ} \\[1em] \Rightarrow \dfrac{(1 + \sin θ)^2}{\cos^2 θ} \\[1em] \Rightarrow \dfrac{(1 + \sin θ)^2}{1 - \sin^2 θ} \\[1em] \Rightarrow \dfrac{(1 + \sin θ)^2}{(1 - \sin θ)(1 + \sin θ)} \\[1em] \Rightarrow \dfrac{1 + \sin θ}{1 - \sin θ}.$

Since, $\dfrac{1 + \sin θ}{1 - \sin θ} \ne 1$, the Assertion (A) is false.

Now, $\sec^2 θ - \tan^2 θ = 1$ is a fundamental trigonometric identity which is true.

So, Reason (R) is true.

Thus, (A) is false, (R) is true.

Hence, option 2 is the correct option.

Assertion (A): The probability of Sun rising from the west is 0.

Reason (R): The probability of an impossible event is always 0.

1. (A) is true, (R) is false.

2. (A) is false, (R) is true.

3. Both (A) and (R) are true and (R) is the correct explanation of (A).

4. Both (A) and (R) are true but, (R) is not the correct explanation of (A).

Answer

The Sun rising from the west is an impossible event, so its probability is 0.

∴ Assertion (A) is true.

The probability of an impossible event is always 0.

∴ Reason (R) is true.

Also, Reason (R) correctly explains why the probability of the Sun rising from the west (an impossible event) is 0.

Thus, both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

The mean of $x_1, x_2, x_3, ……, x_n$ is $m$. The value of each variate is increased by 2. The mean of the new data is:

1. $m – 2$

2. $m$

3. $m + 2$

4. $2m$

Answer

Given,

Mean of $x_1, x_2, x_3, ……, x_n$ = $m$

$\Rightarrow \dfrac{x_1 + x_2 + \dots + x_n}{n} = m$

When each variate is increased by 2, the new mean is:

$\Rightarrow \dfrac{(x_1 + 2) + (x_2 + 2) + \dots + (x_n + 2)}{n} \\[1em] \Rightarrow \dfrac{(x_1 + x_2 + \dots + x_n) + 2n}{n} \\[1em] \Rightarrow \dfrac{x_1 + x_2 + \dots + x_n}{n} + \dfrac{2n}{n} \\[1em] \Rightarrow m + 2.$

Hence, option 3 is the correct option.

A solid sphere completely fits into a cylindrical container of radius 3.5 cm and height 7 cm as shown in the figure below. Find the volume of water required to fill the remaining (unshaded) space of the container to the nearest of a whole number. $\left(\pi = \dfrac{22}{7}\right)$

Answer

Given,

Radius of cylinder (R) = 3.5 cm

Height of cylinder (H) = 7 cm

Since the sphere completely fits into the cylinder,

Radius of sphere ($r$) = 3.5 cm

By formula,

Volume of cylinder = $\pi R^2 H$

$= \dfrac{22}{7} \times (3.5)^2 \times 7 \\[1em] = \dfrac{22}{7} \times 12.25 \times 7 \\[1em] = 22 \times 12.25 \\[1em] = 269.5 \text{ cm}^3.$

By formula,

Volume of sphere = $\dfrac{4}{3} \pi r^3$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times (3.5)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 42.875 \\[1em] = \dfrac{4312}{24} \\[1em] = 179.67 \text{ cm}^3.$

Volume of water required = Volume of cylinder − Volume of sphere

= 269.5 − 179.67

= 89.83

≈ 90 cm³.

Hence, the volume of water required = 90 cm³.

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the:

(a) original speed of the car.

(b) time taken with the increased speed.

Answer

(a) Given,

Distance covered = 400 km

Let the original speed of the car be $x$ km/h.

Since the speed is 12 km/h more in the second case,

Increased speed = $(x + 12)$ km/h.

By formula,

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

∴ Time taken with the original speed = $\dfrac{400}{x}$ hours

∴ Time taken with the increased speed = $\dfrac{400}{x + 12}$ hours

Converting the given difference of time into hours,

1 hour 40 minutes = $1\dfrac{40}{60}$ hours = $\dfrac{5}{3}$ hours.

The question says that with the increased speed the journey takes 1 hour 40 minutes less. So the time taken with the increased speed is smaller, and subtracting it from the time taken with the original speed must give $\dfrac{5}{3}$ hours.

∴ Time with original speed − Time with increased speed = $\dfrac{5}{3}$

According to the question,

$\Rightarrow \dfrac{400}{x} - \dfrac{400}{x + 12} = \dfrac{5}{3} \\[1em] \Rightarrow 400\Big[\dfrac{(x + 12) - x}{x(x + 12)}\Big] = \dfrac{5}{3} \\[1em] \Rightarrow 400 \times \dfrac{12}{x^2 + 12x} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{4800}{x^2 + 12x} = \dfrac{5}{3} \\[1em] \Rightarrow 5(x^2 + 12x) = 14400 \\[1em] \Rightarrow x^2 + 12x = 2880 \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x = 48 \text{ or } x = -60.$

Since speed cannot be negative, $x = 48$.

Hence, the original speed of the car = 48 km/h.

(b) Increased speed = $x + 12 = 48 + 12 = 60$ km/h.

By formula,

Time taken with increased speed = $\dfrac{\text{Distance}}{\text{Speed}}$

$= \dfrac{400}{60} \\[1em] = \dfrac{20}{3} \\[1em] = 6\dfrac{2}{3} \text{ hours}.$

= 6 hours 40 minutes.

Hence, the time taken with the increased speed = 6 hours 40 minutes.

The coordinates of the vertex of ∆ABC are A ($a, b$), B (5, −3) and C (−5, 3). The coordinates of the centroid G of ∆ABC is $G\left(\dfrac{2}{3}, \dfrac{1}{3}\right)$. Find the:

(a) coordinates of the vertex A.

(b) equation of median through the vertex C.

Answer

(a) Given,

Vertices of ∆ABC are A ($a, b$), B (5, −3) and C (−5, 3).

Centroid G = $\Big(\dfrac{2}{3}, \dfrac{1}{3}\Big)$

By formula,

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

Here,

$x_1 = a, y_1 = b$

$x_2 = 5, y_2 = -3$

$x_3 = -5, y_3 = 3$

Substituting values we get :

$\Rightarrow \Big(\dfrac{2}{3}, \dfrac{1}{3}\Big) = \Big(\dfrac{a + 5 + (-5)}{3}, \dfrac{b + (-3) + 3}{3}\Big) \\[1em] \Rightarrow \Big(\dfrac{2}{3}, \dfrac{1}{3}\Big) = \Big(\dfrac{a}{3}, \dfrac{b}{3}\Big).$

Comparing the coordinates,

$\Rightarrow \dfrac{a}{3} = \dfrac{2}{3} \\[1em] \Rightarrow a = 2 \\[1em] \Rightarrow \dfrac{b}{3} = \dfrac{1}{3} \\[1em] \Rightarrow b = 1.$

Hence, the coordinates of vertex A are (2, 1).

(b) Let CM be the median, passing through C, here M is the mid-point of AB.

A = (2, 1) and B = (5, −3)

By formula,

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

Here,

$x_1 = 2, y_1 = 1$

$x_2 = 5, y_2 = -3$

Substituting values we get :

$\text{ M }= \Big(\dfrac{2 + 5}{2}, \dfrac{1 + (-3)}{2}\Big) \\[1em] = \Big(\dfrac{7}{2}, \dfrac{-2}{2}\Big) \\[1em] = \Big(\dfrac{7}{2}, -1\Big).$

Hence, M = $\Big(\dfrac{7}{2}, -1\Big)$.

The median through C is the line joining C (−5, 3) and M $\Big(\dfrac{7}{2}, -1\Big)$.

By formula,

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

Here,

$x_1 = -5, y_1 = 3$

$x_2 = \dfrac{7}{2}, y_2 = -1$

Substituting values we get :

$\text{Slope (m)} = \dfrac{-1 - 3}{\dfrac{7}{2} - (-5)} \\[1em] = \dfrac{-4}{\dfrac{7 + 10}{2}} \\[1em] = \dfrac{-4}{\dfrac{17}{2}} \\[1em] = \dfrac{-8}{17}.$

By point-slope form,

Equation of a line : $y - y_1 = m(x - x_1)$

Substituting values we get :

$\Rightarrow y - 3 = \dfrac{-8}{17}(x - (-5)) \\[1em] \Rightarrow 17(y - 3) = -8(x + 5) \\[1em] \Rightarrow 17y - 51 = -8x - 40 \\[1em] \Rightarrow 8x + 17y - 51 + 40 = 0 \\[1em] \Rightarrow 8x + 17y - 11 = 0.$

Hence, the equation of the median through vertex C is $\bm{8x + 17y - 11 = 0}$.

Anamika deposited ₹300 per month in a cumulative deposit account with a bank for 3 years. Her brother Sanjeev started depositing ₹500 per month for 2 years in the same scheme. The banks paid 10% simple interest per annum to both. At the time of maturity, find:

(a) the interest earned by Anamika.

(b) who got more money as interest and how much?

Answer

(a) For Anamika,

P = ₹ 300, $n$ = 3 years = 36 months, $r$ = 10%

By formula,

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

Substituting values we get :

$= 300 \times \dfrac{36(36 + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 300 \times \dfrac{36 \times 37}{24} \times \dfrac{10}{100} \\[1em] = 300 \times 55.5 \times \dfrac{10}{100} \\[1em] = 1665.$

Hence, the interest earned by Anamika = ₹ 1,665.

(b) For Sanjeev,

P = ₹ 500, $n$ = 2 years = 24 months, $r$ = 10%

By formula,

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

Substituting values we get :

$= 500 \times \dfrac{24(24 + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 500 \times \dfrac{24 \times 25}{24} \times \dfrac{10}{100} \\[1em] = 500 \times 25 \times \dfrac{10}{100} \\[1em] = 1250.$

Interest earned by Anamika = ₹ 1,665

Interest earned by Sanjeev = ₹ 1,250

Difference = ₹ 1,665 − ₹ 1,250 = ₹ 415.

Hence, Anamika got more interest, by an amount of ₹ 415.

$\sqrt{2}, \sqrt{8}, \sqrt{18}$ …… forms a progression.

(a) Identify the type of progression.

(b) The sum of its first 10 terms is $p\sqrt{2}$. Find the value of $p$.

Answer

(a) Given,

$\sqrt{2}, \sqrt{8}, \sqrt{18}$ ......

Simplifying the terms,

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

Second term − First term = $2\sqrt{2} - \sqrt{2} = \sqrt{2}$

Third term − Second term = $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (A.P.) with first term $a = \sqrt{2}$ and common difference $d = \sqrt{2}$.

Hence, the progression is an Arithmetic Progression.

(b) Given,

$a = \sqrt{2}, d = \sqrt{2}, n = 10$

By formula,

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

Substituting values we get :

$= \dfrac{10}{2}[2\sqrt{2} + (10 - 1)\sqrt{2}] \\[1em] = 5[2\sqrt{2} + 9\sqrt{2}] \\[1em] = 5 \times 11\sqrt{2} \\[1em] = 55\sqrt{2}.$

Given, the sum of first 10 terms = $p\sqrt{2}$.

$\Rightarrow p\sqrt{2} = 55\sqrt{2} \\[1em] \Rightarrow p = 55.$

Hence, the value of $\bm{p}$ = 55.

Use graph sheet for this question. Take 2 cm = 1 unit along the axes. Plot and write coordinates of:

(a) A (2, 3) and B (4, 5).

(b) image of A and B in the x-axis as A′ and B′.

(c) image of A′ in the y-axis as A″.

(d) name the single transformation that maps A to A″.

Answer

(a) The points A (2, 3) and B (4, 5) are plotted on the graph :

(b) From figure,

On reflecting points A and B in the x-axis, we get :

A′ = (2, −3) and B′ = (4, −5).

(c) From figure,

On reflecting point A' in y-axis, we get :

A″ = (−2, −3).

(d) Point A (2, 3) is mapped to A″ (−2, −3).

Since both the coordinates change their signs, the single transformation that maps A to A″ is a reflection in the origin.

Hence, the single transformation that maps A to A″ is reflection in the origin.

A computer mechanic charges repairing cost of different components as tabulated below:

The rate of GST is 18%. Find the total:

(a) discount

(b) amount of GST

Answer

(a) For Component A,

Repairing cost = ₹ 5,000, Discount = 20%

Discount = $\dfrac{20}{100} \times 5000$ = ₹ 1,000

For Component B,

Repairing cost = ₹ 6,500, Discount = 30%

Discount = $\dfrac{30}{100} \times 6500$ = ₹ 1,950

Total discount = ₹ 1,000 + ₹ 1,950 = ₹ 2,950.

Hence, the total discount = ₹ 2,950.

(b) For Component A,

Cost after discount = ₹ 5,000 − ₹ 1,000 = ₹ 4,000

For Component B,

Cost after discount = ₹ 6,500 − ₹ 1,950 = ₹ 4,550

Total taxable amount = ₹ 4,000 + ₹ 4,550 = ₹ 8,550

GST = $\dfrac{18}{100} \times 8550$ = ₹ 1,539.

Hence, the total amount of GST = ₹ 1,539.

Given, matrix $A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$ and I is a unit matrix of order 2 × 2.

(a) Find the matrix, A² and 5A

(b) value of $k$, such that $A^2 - 5A = k.I$

Answer

(a) Calculating A²,

$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 3 + (-5)(-4) & 3 \times (-5) + (-5)(2) \\ (-4)(3) + 2(-4) & (-4)(-5) + 2 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}.$

Calculating 5A,

$5A = 5\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}.$

Hence, $A^2 = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$ and $5A = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}$.

(b) Calculating A² − 5A,

$A^2 - 5A = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \\[1em] = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = 14 \cdot I.$

Comparing with $A^2 - 5A = k.I$,

$\Rightarrow k = 14.$

Hence, the value of $\bm{k}$ = 14.

Prove that:

$\sqrt{\dfrac{\sec\theta + 1}{\sec\theta - 1}} + \sqrt{\dfrac{\sec\theta - 1}{\sec\theta + 1}} = 2,\text{cosec},\theta$

Answer

Solving the L.H.S.,

$\Rightarrow \sqrt{\dfrac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\dfrac{\sec θ - 1}{\sec θ + 1}} \\[1em] \Rightarrow \dfrac{\sqrt{\sec θ + 1}}{\sqrt{\sec θ - 1}} + \dfrac{\sqrt{\sec θ - 1}}{\sqrt{\sec θ + 1}} \\[1em] \Rightarrow \dfrac{\sqrt{\sec θ + 1} \times \sqrt{\sec θ + 1} + \sqrt{\sec θ - 1} \times \sqrt{\sec θ - 1}}{\sqrt{\sec θ - 1} \times \sqrt{\sec θ + 1}} \\[1em] \Rightarrow \dfrac{(\sec θ + 1) + (\sec θ - 1)}{\sqrt{(\sec θ + 1)(\sec θ - 1)}} \\[1em] \Rightarrow \dfrac{\sec θ + 1 + \sec θ - 1}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \dfrac{2\sec θ}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \dfrac{2\sec θ}{\sqrt{\tan^2 θ}} \\[1em] \Rightarrow \dfrac{2\sec θ}{\tan θ} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\cos θ}}{\dfrac{\sin θ}{\cos θ}} \\[1em] \Rightarrow 2 \times \dfrac{1}{\cos θ} \times \dfrac{\cos θ}{\sin θ} \\[1em] \Rightarrow \dfrac{2}{\sin θ} \\[1em] \Rightarrow \text{2 cosec θ}.$

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

Hence, proved that $\bm{\sqrt{\dfrac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\dfrac{\sec θ - 1}{\sec θ + 1}} = 2,\text{cosec},θ}$.

Prasoon invests ₹7500 in a company paying 6% dividend annually. ₹100 shares of this company are available at ₹150.

(a) Find the number of shares he purchased.

(b) Find his annual income.

(c) After one year, he sold 50% of his original shares at ₹175 each. Find his gain on the shares he sold.

Answer

(a) Given,

Investment = ₹ 7,500, Market value of each share = ₹ 150

By formula,

Number of shares = $\dfrac{\text{Investment}}{\text{Market value of each share}}$

$= \dfrac{7500}{150}$

= 50.

Hence, the number of shares purchased = 50.

(b) Given,

Face value = ₹ 100, Dividend rate = 6%, Number of shares = 50

By formula,

Annual income = Number of shares × Rate of dividend × Face value

$= 50 \times \dfrac{6}{100} \times 100$

= ₹ 300.

Hence, his annual income = ₹ 300.

(c) Number of shares sold = 50% of 50 = $\dfrac{50}{100} \times 50$ = 25 shares.

Selling price of each share = ₹ 175

Cost price of each share = ₹ 150

Gain on each share = ₹ 175 − ₹ 150 = ₹ 25

Total gain = 25 × ₹ 25 = ₹ 625.

Hence, his gain on the shares he sold = ₹ 625.

Solve the following inequation, write the solution set and represent it on the real number line.

$3x + \dfrac{14}{3} \gt \dfrac{4x}{3} - 2 \geq 3x - 4, ; x \in R$

Answer

Given,

$3x + \dfrac{14}{3} \gt \dfrac{4x}{3} - 2 \geq 3x - 4, ; x \in R$

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

$\Rightarrow 3x + \dfrac{14}{3} \gt \dfrac{4x}{3} - 2 \\[1em] \Rightarrow 3x - \dfrac{4x}{3} \gt -2 - \dfrac{14}{3} \\[1em] \Rightarrow \dfrac{9x - 4x}{3} \gt \dfrac{-6 - 14}{3} \\[1em] \Rightarrow \dfrac{5x}{3} \gt \dfrac{-20}{3} \\[1em] \Rightarrow 5x \gt \dfrac{-20}{3} \times 3 \\[1em] \Rightarrow 5x \gt -20 \\[1em] \Rightarrow x \gt \dfrac{-20}{5} \\[1em] \Rightarrow x \gt -4 \text{ ......(1)}$

Solving the R.H.S. of the inequation, we get:

$\Rightarrow \dfrac{4x}{3} - 2 \geq 3x - 4 \\[1em] \Rightarrow \dfrac{4x}{3} - 3x \ge -4 + 2 \\[1em] \Rightarrow \dfrac{4x - 9x}{3} \ge -2 \\[1em] \Rightarrow \dfrac{-5x}{3} \ge -2 \\[1em] \Rightarrow \dfrac{-5x}{3} \times 3 \ge -2 \times 3 \\[1em] \Rightarrow -5x \ge -6 \\[1em] \Rightarrow -5x \times -1 \le -6 \times (-1) \\[1em] \Rightarrow 5x \le 6 \\[1em] \Rightarrow x \le \dfrac{6}{5} \text{ ......(2)}$

From equations (1) and (2),

Solution set = $\Big\lbrace x : -4 \lt x \le \dfrac{6}{5}, x \in R\Big\rbrace$

Representing on the real number line :

Hence, solution set = $\bm{\Big\lbrace x : -4 \lt x \le \dfrac{6}{5}, x \in R\Big\rbrace}$.

In the given diagram, lines AE and BD intersect each other at C such that ∆ABC ~ ∆DEC.

(a) Complete the equation with corresponding sides of ∆ABC and ∆DEC.

$\dfrac{AB}{DE} = \dfrac{AC}{?} = \dfrac{?}{EC}$

(b) ∠ABC is equal to which angle of ∆DEC?

(c) If, AB : DE = 2 : 3, find the ratio of ar (∆ABC) : ar (∆DEC).

Answer

(a) Since ∆ABC ~ ∆DEC, their corresponding sides are proportional.

$\dfrac{AB}{DE} = \dfrac{AC}{DC} = \dfrac{BC}{EC}$

Hence, $\mathbf{\dfrac{AB}{DE} = \dfrac{AC}{DC} = \dfrac{BC}{EC}}$.

(b) Since ∆ABC ~ ∆DEC, their corresponding angles are equal.

∴ ∠ABC = ∠DEC.

Hence, ∠ABC is equal to ∠DEC.

(c) We know that,

The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

$\therefore \dfrac{\text{ar (∆ABC)}}{\text{ar (∆DEC)}} = \dfrac{AB^2}{DE^2} \\[1em] = \Big(\dfrac{2}{3}\Big)^2 \\[1em] = \dfrac{4}{9} \\[1em] = 4 : 9.$

Hence, ar (∆ABC) : ar (∆DEC) = 4 : 9.

In the given diagram, AB is a tangent to the circle at P. If ∠PQR = 112° and PS = SR, find:

(a) ∠PSR

(b) ∠BPR

(c) ∠PTS

Answer

(a) PQRS is a cyclic quadrilateral.

We know that,

The sum of opposite angles of a cyclic quadrilateral is 180°.

$\Rightarrow \angle PQR + \angle PSR = 180° \\[1em] \Rightarrow 112° + \angle PSR = 180° \\[1em] \Rightarrow \angle PSR = 180° - 112° = 68°.$

Hence, ∠PSR = 68°.

(b) We know that,

By the alternate segment theorem, the angle between the tangent and a chord equals the angle subtended by the chord in the alternate segment.

∠BPR is the angle between tangent PB and chord PR, and its alternate segment contains Q.

∴ ∠BPR = ∠PQR = 112°.

Hence, ∠BPR = 112°.

(c) Given, PS = SR.

Equal chords subtend equal arcs.

∴ arc PS = arc SR.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it on the circumference.

∠PQR is the inscribed angle standing on arc PSR.

$\Rightarrow \text{arc PSR} = 2 \times \angle PQR = 2 \times 112° = 224°.$

Since arc PS = arc SR,

$\Rightarrow \text{arc PS} = \dfrac{224°}{2} = 112°.$

∠PTS is the inscribed angle standing on arc PS.

$\Rightarrow \angle PTS = \dfrac{1}{2} \times \text{arc PS} = \dfrac{1}{2} \times 112° = 56°.$

Hence, ∠PTS = 56°.

Using step – deviation method, find mean for the following data. Give your answer to the nearest of a whole number.

Answer

Let assumed mean A = 32.5 and class size $i = 5$.

By formula,

Mean = A + $\dfrac{Σfu}{Σf} \times i$

$= 32.5 + \dfrac{14}{50} \times 5 \\[1em] = 32.5 + \dfrac{70}{50} \\[1em] = 32.5 + 1.4 \\[1em] = 33.9 \\[1em] ≈ 34.$

Hence, the mean = 34.

The remainder obtained by dividing the polynomial $mx^2 - 3x + 6$ by $(x - 2)$ is twice the remainder obtained by dividing the polynomial $3x^2 + 5x - m$ by $(x + 3)$.

(a) Find the value of '$m$'.

(b) Using the value of '$m$', find the remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$.

Answer

(a) By the Remainder Theorem, the remainder when a polynomial is divided by $(x − a)$ is the value of the polynomial at $x = a$.

Let R₁ be the remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$. Put $x = 2$ :

$\Rightarrow R_1 = m(2)^2 - 3(2) + 6 \\[1em] \Rightarrow R_1 = 4m - 6 + 6 \\[1em] \Rightarrow R_1 = 4m.$

Let R₂ be the remainder when $3x^2 + 5x − m$ is divided by $(x + 3)$. Put $x = −3$ :

$\Rightarrow R_2 = 3(-3)^2 + 5(-3) - m \\[1em] \Rightarrow R_2 = 27 - 15 - m \\[1em] \Rightarrow R_2 = 12 - m.$

According to the question, R₁ = 2R₂.

$\Rightarrow 4m = 2(12 - m) \\[1em] \Rightarrow 4m = 24 - 2m \\[1em] \Rightarrow 4m + 2m = 24 \\[1em] \Rightarrow 6m = 24 \\[1em] \Rightarrow m = 4.$

Hence, the value of $\bm{m}$ = 4.

(b) The remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$ is R₁ = $4m$.

$\Rightarrow R_1 = 4 \times 4 = 16.$

Hence, the remainder = 16.

Each of the letters of the word 'BINOCULARS' is written on identical cards and put in a bag. They are well shuffled. If a card is drawn at random from the bag, find the probability that the letter is:

(a) a vowel

(b) one of the first 8 letters of the English alphabet which appears in the given word.

(Support your working with at least one sample space)

Answer

The word 'BINOCULARS' has 10 letters, all distinct.

Sample space = {B, I, N, O, C, U, L, A, R, S}

Total number of outcomes = 10.

(a) The vowels in the word are I, O, U, A.

Number of vowels = 4.

By formula,

Probability = $\dfrac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

$\Rightarrow P(\text{vowel}) = \dfrac{4}{10} = \dfrac{2}{5}.$

Hence, the probability that the letter is a vowel = $\mathbf{\dfrac{2}{5}}$.

(b) The first 8 letters of the English alphabet are A, B, C, D, E, F, G, H.

The letters among these which appear in 'BINOCULARS' are B, C and A.

Number of favourable outcomes = 3.

$\Rightarrow P = \dfrac{3}{10}.$

Hence, the required probability = $\mathbf{\dfrac{3}{10}}$.

Using properties of proportion, solve for $x$:

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

Answer

Given,

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

Applying componendo and dividendo,

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

Squaring both sides,

$\Rightarrow \dfrac{2x + 3}{2x - 3} = 4 \\[1em] \Rightarrow 2x + 3 = 4(2x - 3) \\[1em] \Rightarrow 2x + 3 = 8x - 12 \\[1em] \Rightarrow 3 + 12 = 8x - 2x \\[1em] \Rightarrow 15 = 6x \\[1em] \Rightarrow x = \dfrac{15}{6} = \dfrac{5}{2}.$

Hence, $\bm{x = \dfrac{5}{2}}$.

In the given diagram, PT is a tangent to the circle and QT is a secant meeting the circle at R, such that QR = 9 cm and RT = 16 cm.

(a) Prove that: ∆PTR ~ ∆QTP.

(b) Find the length of PT.

Answer

(a) In ∆PTR and ∆QTP,

⇒ ∠PTR = ∠QTP [Common angle]

⇒ ∠TPR = ∠TQP [Angle between tangent and chord is equal to the angle in the alternate segment]

∴ ∆PTR ~ ∆QTP (By A.A. axiom)

Hence, proved that ∆PTR ~ ∆QTP.

(b) Since, ∆PTR ~ ∆QTP, their corresponding sides are proportional.

$\Rightarrow \dfrac{PT}{QT} = \dfrac{TR}{TP} \\[1em] \Rightarrow PT^2 = QT \times TR.$

From figure,

QT = QR + RT = 9 + 16 = 25 cm.

Substituting the values,

$\Rightarrow PT^2 = 25 \times 16 \\[1em] \Rightarrow PT^2 = 400 \\[1em] \Rightarrow PT = \sqrt{400} \\[1em] \Rightarrow PT = 20 \text{ cm}.$

Hence, the length of PT = 20 cm.

Use graph paper for this question; draw the histogram and estimate mode for the following data. Use 2 cm = 10 units along the x-axis, and 2 cm = 2 units along the y-axis.

Answer

Steps of construction :

1. Take 2 cm = 10 units along the x-axis and 2 cm = 2 units along the y-axis.

2. Draw the rectangles (bars) for each class with heights equal to their frequencies.

3. The modal class is the class with the highest frequency, i.e. 50 – 60 (frequency 18).

4. In the highest rectangle, draw two lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

5. From the point of intersection of AC and BD, draw a vertical line to meet the x-axis. The point where it meets the x-axis gives the mode.

From the graph, the mode = 54.

Hence, the mode = 54.

Solve the quadratic equation $x^2 - 6x - 8 = 0$ and give your answer correct to three significant figures.

Answer

Given,

$x^2 - 6x - 8 = 0$

Comparing with $ax^2 + bx + c = 0$,

$a = 1, b = -6, c = -8$

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-8)}}{2(1)} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{36 + 32}}{2} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{68}}{2} \\[1em] \Rightarrow x = \dfrac{6 \pm 8.2462}{2}.$

Case 1:

$\Rightarrow x = \dfrac{6 + 8.2462}{2} = \dfrac{14.2462}{2} = 7.1231 ≈ 7.12.$

Case 2:

$\Rightarrow x = \dfrac{6 - 8.2462}{2} = \dfrac{-2.2462}{2} = -1.1231 ≈ -1.12.$

Hence, $\bm{x}$ = 7.12 or $\bm{x}$ = −1.12.

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into solid right circular cones of diameter 3.5 cm and height 3.5 cm. Find the:

(a) radius of the sphere.

(b) number of cones formed.

$\left(\pi = \dfrac{22}{7}\right)$

Answer

(a) Given,

Surface area of the solid metallic sphere = 616 cm²

Let the radius of the sphere be $r$ cm.

By formula,

Surface area of sphere = $4\pi r^2$

Equating this to the given surface area and substituting $\pi = \dfrac{22}{7}$, we get :

$\Rightarrow 4 \times \dfrac{22}{7} \times r^2 = 616 \\[1em] \Rightarrow \dfrac{88}{7} \times r^2 = 616 \\[1em] \Rightarrow r^2 = \dfrac{616 \times 7}{88} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.$

Hence, the radius of the sphere = 7 cm.

(b) We know that,

When a solid is melted and recast into a number of smaller solids, the material only changes its shape and no material is lost.

∴ Volume of the sphere = Number of cones × Volume of one cone

$\Rightarrow \text{Number of cones} = \dfrac{\text{Volume of sphere}}{\text{Volume of one cone}}$

So, we first find the volume of the sphere and the volume of one cone.

Volume of the sphere :

From part (a), radius of the sphere ($r$) = 7 cm.

By formula,

Volume of sphere = $\dfrac{4}{3} \pi r^3$

Substituting values we get :

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 343 \\[1em] = \dfrac{4}{3} \times 22 \times 49 \\[1em] = \dfrac{4312}{3} \text{ cm}^3.$

Volume of one cone :

Given,

Diameter of cone = 3.5 cm

$\Rightarrow \text{Radius of cone (r)} = \dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2} = \dfrac{7}{4} \text{ cm}$

Height of cone ($h$) = 3.5 cm = $\dfrac{7}{2}$ cm

By formula,

Volume of cone = $\dfrac{1}{3} \pi r^2 h$

Substituting values we get :

$= \dfrac{1}{3} \times \dfrac{22}{7} \times \Big(\dfrac{7}{4}\Big)^2 \times \dfrac{7}{2} \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{49}{16} \times \dfrac{7}{2} \\[1em] = \dfrac{22 \times 49 \times 7}{3 \times 7 \times 16 \times 2} \\[1em] = \dfrac{539}{48} \text{ cm}^3.$

Number of cones :

$\text{Number of cones} = \dfrac{\text{Volume of sphere}}{\text{Volume of one cone}} \\[1em] = \dfrac{\dfrac{4312}{3}}{\dfrac{539}{48}} \\[1em] = \dfrac{4312}{3} \times \dfrac{48}{539} \\[1em] = \dfrac{4312}{539} \times \dfrac{48}{3} \\[1em] = 8 \times 16 \\[1em] = 128.$

Hence, the number of cones formed = 128.

As observed from the top of a light house, 80 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 35° to 55°.

(a) Draw a free hand sketch representing the given information.

(b) Find the distance travelled by the ship during the period of observation. Give your answer to the nearest meter.

Answer

(a) Let T be the top of the lighthouse and L be its foot. Let A and B be the two positions of the ship, where the angles of depression are 35° and 55° respectively.

(b) From figure,

∠TAL = ∠ATX = 35° (alternate angles are equal)

∠TBL = ∠BTX = 55° (alternate angles are equal)

Height of lighthouse, TL = 80 m.

In right-angled triangle TLA,

$\Rightarrow \tan 35° = \dfrac{TL}{LA} \\[1em] \Rightarrow LA = \dfrac{TL}{\tan 35°} \\[1em] \Rightarrow LA = \dfrac{80}{0.7002} \\[1em] \Rightarrow LA = 114.25 \text{ m}.$

In right-angled triangle TLB,

$\Rightarrow \tan 55° = \dfrac{TL}{LB} \\[1em] \Rightarrow LB = \dfrac{TL}{\tan 55°} \\[1em] \Rightarrow LB = \dfrac{80}{1.4281} \\[1em] \Rightarrow LB = 56.02 \text{ m}.$