ICSE Class 10 Mathematics — 2025 Paper

Comparing the given quadratic equation with ax² + bx + c = 0, we get :

a = 3, b = $\sqrt{7}$ and c = 2

Discriminant = b² - 4ac

= $(\sqrt{7})^2$ - 4 x 3 x 2

= 7 - 24

= -17.

Since, D < 0, there are no real roots.

Hence, option 4 is the correct option.

Given,

Monthly deposit = ₹500

Number of months = 18

Interest earned = ₹570

By formula,

Matured amount = Total deposit + Interest

= Monthly deposit x number of months + Interest

= ₹(500 x 18 + 570)

Hence, option 1 is the correct option.

Probability of any event is always between 0 and 1.

$\frac{5}{4}$ = 1.25, which is greater than 1, which is not possible.

Hence, option 1 is the correct option.

Given,

⇒ 3x + 4y + 7 = 0

⇒ 4y = -3x - 7

⇒ y = $-\frac{3}{4}x - \frac{7}{4}$

Comparing above equation with y = mx + c, we get :

m = $-\frac{3}{4}$.

By point-slope formula,

Equation of line : y - y₁ = m(x - x₁)

Thus, equation of line parallel to 3x + 4y + 7 = 0 and passing through origin (0, 0) is :

⇒ y - 0 = $-\frac{3}{4}$ (x - 0)

⇒ y = $-\frac{3x}{4}$

⇒ 4y = -3x

⇒ 4y + 3x = 0.

Hence, option 4 is the correct option.

Given,

$$A = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}$$

Substituting values we get :

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

Hence, option 3 is the correct option.

From figure,

∠ACD and ∠ACB forms linear pairs [BD is a straight line].

⇒ ∠ACD + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180° - 120°

⇒ ∠ACB = 60°.

In ΔABC,

⇒ BC = AC [Given]

⇒ ∠ABC = ∠BAC = x° (let) [Angles opposite to equal sides in triangle are equal]

According to angle sum property in ΔABC,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 60° + x° + x° = 180°

⇒ 2x° = 180° - 60°

⇒ 2x° = 120°

⇒ x° = $\frac{120°}{2}$ = 60°.

From figure,

ABCE is cyclic quadrilateral.

We know that,

Opposite angles of cyclic quadrilateral are supplementary.

⇒ ∠ABC + ∠AEC = 180°

⇒ 60° + ∠AEC = 180°

⇒ ∠AEC = 180° - 60°

⇒ ∠AEC = 120°.

Hence, option 4 is the correct option.

Factorize :

⇒ x² - 4

⇒ (x - 2)(x + 2)

Factorize :

⇒ x³ - x² - 4x + 4

⇒ (x³ - x²) - (4x - 4)

⇒ x²(x - 1) - 4(x - 1)

⇒ (x - 1)(x² - 4)

⇒ (x - 1)(x² - 2²)

⇒ (x - 1)(x - 2)(x + 2).

Common factors between x² - 4 and x³ - x² - 4x + 4 = (x - 2) and (x + 2).

Hence, option 3 is the correct option.

Given,

Dividend rate = 12% of face value.

So, if face value = ₹100, dividend = ₹12.

Return = 10%

By formula,

Return % = $\frac{\text{Dividend on one share}}{\text{Investment on one share}} \times 100%$

Investment on one share equals to the market value of the share.

Substituting values we get :

$$\begin{aligned}\Rightarrow 10 = \frac{\text{12}}{\text{Market value}} \times 100 \\[1em] \Rightarrow \text{Market value} = \frac{12}{10} \times 100 \\[1em] \Rightarrow \text{Market value} = ₹120.\end{aligned}$$

Since market value > face value, the shares are said to be above par.

Hence, option 3 is the correct option.

The circumcenter of a triangle is defined as the point where the perpendicular bisectors of the sides intersect. It is equidistant from the three vertices (D, E, and F) of the triangle.

∴ Statement 1 is true.

The incenter of a triangle is defined as the point of concurrency of the three angle bisectors of the triangle. This point is also the center of the inscribed circle and is equidistant from the three sides of the triangle.

∴ Statement 2 is true.

∴ Both statements are true.

Hence, option 1 is the correct option.

Given,

⇒ sin² A + sin A = 1

⇒ sin A = 1 - sin² A

⇒ sin A = cos² A

Squaring both the sides, we get :

⇒ (sin A)² = (cos² A)²

⇒ sin² A = cos⁴ A ......................(1)

Solving L.H.S. of cos⁴ A + cos² A = 1, we get :

⇒ cos⁴ A + cos² A

⇒ sin² A + cos² A [From equation (1)]

⇒ 1.

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

∴ Assertion (A) is true.

As we know that, sin² A + cos² A = 1

⇒ 1 - sin² A = cos² A

∴ Reason (R) is true.

∴ Both (A) and (R) are true, and (R) is the correct reason for (A).

Hence, option 3 is the correct option.

Given, ΔABC ∼ ΔEFG

So, the corresponding sides are in proportion.

$$\begin{aligned}\Rightarrow \frac{AB}{EF} = \frac{BC}{FG}\\[1em] \Rightarrow \frac{15}{75} = \frac{3}{FG}\\[1em] \Rightarrow \frac{1}{5} = \frac{3}{FG}\\[1em] \Rightarrow FG = 3 \times 5\\[1em] \Rightarrow FG = 15\text{ cm}\end{aligned}$$

Hence, option 1 is the correct option.

Given, the volume of two spheres = 27: 64

By formula, volume of a sphere = $\frac{4πr^3}{3}$, where r is the radius of the sphere.

Let r and R be the radii of the two spheres.

$$\begin{aligned}\Rightarrow \frac{\frac{4πr^3}{3}}{\frac{4πR^3}{3}} = \frac{27}{64} \\[1em] \Rightarrow \frac{r^3}{R^3} = \frac{27}{64} \\[1em] \Rightarrow \frac{r}{R} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}} \\[1em] \Rightarrow \frac{r}{R} = \frac{3}{4}.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

Marked price of the article = ₹1,375.

The CGST (Central Goods and Services Tax) = 4%.

As we know that the total GST is composed of CGST and SGST (State Goods and Services Tax) at equal rates, the total GST rate is 8% (4% CGST + 4% SGST).

GST = 8% of marked price of the article

= $\frac{8}{100} \times 1375 = \frac{8 \times 1375}{100} = \frac{11000}{100}$ = ₹110

Final Price = Marked Price + GST Amount

= ₹1,375 + ₹110 = ₹1,485

Hence, option 4 is the correct option.

Given, 0 < $-\frac{x}{3}$ < 2

Solving L.H.S. of equation :

⇒ 0 < $-\frac{x}{3}$

⇒ $\frac{x}{3}$ < 0

⇒ x < 0 × 3

⇒ x < 0................... (1)

Solving R.H.S. of equation :

⇒ $-\frac{x}{3}$ < 2

⇒ -x < 2 × 3

⇒ -x < 6

⇒ x > -6 .................... (2)

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

-6 < x < 0 and x ∈ Z.

Solution set for x = {-5, -4, -3, -2, -1}.

Hence, option 1 is the correct option.

First 9 natural numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9

By formula,

$$\begin{aligned}\Rightarrow \text{Mean } = \frac{\text{Sum of all observations}}{\text{Total number of observations}}\\[1em] = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{9}\\[1em] = \frac{45}{9}\\[1em] = 5.\end{aligned}$$

∴ (A) is false, (R) is true.

Hence, option 2 is the correct option.

Given,

⇒ 2x² − 5x − 4 = 0

Comparing equation 2x² − 5x − 4 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -5 and c = -4.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times (2) \times (-4)}}{2 \times (2)} \\[1em] = \frac{5 \pm \sqrt{25 - (-32)}}{4} \\[1em] = \frac{5 \pm \sqrt{25 + 32}}{4} \\[1em] = \frac{5 \pm \sqrt{57}}{4} \\[1em] = \frac{5 + \sqrt{57}}{4} \text{ or } \frac{5 - \sqrt{57}}{4} \\[1em] = \frac{5 + 7.55}{4} \text{ or } \frac{5 - 7.55}{4} \\[1em] = \frac{12.55}{4} \text{ or } \frac{-2.55}{4} \\[1em] = 3.14 \text{ or } -0.638\end{aligned}$$

Hence, x = 3.14, -0.638.

(a) Given,

Mrs. Rao deposited ₹ 250 per month in a recurring deposit account for a period of 3 years.

Total deposit = ₹ 250 × 3 × 12 = ₹ 9,000.

By formula,

Interest = Maturity value - Total deposit = ₹ 10,110 - ₹ 9,000 = ₹ 1,110.

Let rate of interest be r%.

Time (n) = 36 months

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1110 = \frac{250 \times 36 \times (36 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 1110 = \frac{9000 \times 37}{24} \times \frac{r}{100} \\[1em] \Rightarrow 1110 = \frac{333000}{24} \times \frac{r}{100} \\[1em] \Rightarrow r = \frac{24 \times 1110 \times 100}{333000} \\[1em] \Rightarrow r = \frac{2664000}{333000} \\[1em] \Rightarrow r = 8%.\end{aligned}$$

Hence, rate of interest = 8%.

(b) If per month ₹ 50 more is deposited, then :

P = ₹ 250 + ₹ 50 = ₹ 300.

P = ₹ 300, r = 8%, n = 36 months

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Interest} = \frac{300 \times 36 \times (36 + 1)}{2 \times 12} \times \frac{8}{100} \\[1em] \Rightarrow \text{Interest} = \frac{10800 \times 37}{24} \times \frac{8}{100} \\[1em] \Rightarrow \text{Interest} = \frac{399600}{24} \times \frac{8}{100} \\[1em] \Rightarrow \text{Interest} = \frac{399600 \times 8}{24 \times 100} \\[1em] \Rightarrow \text{Interest} = \frac{3996}{3} \\[1em] \Rightarrow \text{Interest} = ₹1,332\end{aligned}$$

Additional Interest = New Interest - Old Interest

= ₹1,332 - ₹1,110

= ₹222.

Hence, Mrs. Rao would receive ₹222 more as interest if she had deposited ₹50 more per month at the same rate of interest and for the same time.

(a) In ΔABC and ΔAED,

⇒ ∠ABC = ∠AED [Both = 90°]

⇒ ∠BAC = ∠DAE [Common angles]

∴ ΔABC ~ ΔAED (By AA similarity postulate)

Hence, proved that ΔABC ~ ΔAED.

(b) Given,

AB = 20 cm, AC = 25 cm, DE = 3 cm

In ΔABC,

By pythagoras theorem,

⇒ AB² + BC² = AC²

⇒ (20)² + BC² = (25)²

⇒ 400 + BC² = 625

⇒ BC² = 625 - 400

⇒ BC² = 225

⇒ BC = $\sqrt{225}$

⇒ BC = 15 cm.

We know that,

Since, corresponding sides of similar triangles are proportional we have :

$\frac{AB}{AE} = \frac{BC}{DE} = \frac{AC}{AD}$

Solving,

$$\begin{aligned}\Rightarrow \frac{AB}{AE} = \frac{BC}{DE} \\[1em] \Rightarrow \frac{20}{AE} = \frac{15}{3} \\[1em] \Rightarrow \frac{20 \times 3}{15} = AE \\[1em] \Rightarrow AE = \frac{20}{5} \\[1em] \Rightarrow AE = 4\text{ cm}.\end{aligned}$$

Substituting values in $\frac{BC}{DE} = \frac{AC}{AD}$ we get :

$$\begin{aligned}\Rightarrow \frac{BC}{DE} = \frac{AC}{AD} \\[1em] \Rightarrow \frac{15}{3} = \frac{25}{AD} \\[1em] \Rightarrow 5 = \frac{25}{AD} \\[1em] \Rightarrow AD = \frac{25}{5} \\[1em] \Rightarrow AD = 5\text{ cm}.\end{aligned}$$

Hence, BC = 15 cm, AE = 4 cm, AD = 5 cm.

(c) Given,

Area on ground = 576 m²

By formula,

Area of triangle = $\frac{1}{2}$ × base × height

Area of ΔABC = $\frac{1}{2}$ × AB × BC

= $\frac{1}{2} \times 20 \times 15$

= 150 cm².

Area of ΔAED = $\frac{1}{2}$ × AE × DE

$= \frac{1}{2} \times 4 \times 3$

$= \frac{1}{2} \times 12$

= 6 cm².

From figure,

⇒ Area of Quadrilateral (BCED) = Area of ΔABC - Area of ΔAED

= 150 - 6 = 144 cm².

⇒ Actual ground area = 576 m²

= 576 × 10000 cm² = 5760000 cm²

Let scale factor be k.

By formula,

k² = $\frac{\text{Area of BCED}}{\text{Actual area of BCED on ground}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow k^2 = \frac{144}{5760000} \\[1em] \Rightarrow k^2 = \frac{1}{40000} \\[1em] \Rightarrow k = \sqrt{\frac{1}{40000}}\\[1em] \Rightarrow k = \frac{1}{200}\\[1em]\end{aligned}$$

Hence, scale factor equals 1 : 200.

Given,

AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°

Steps of construction,

1. Draw a line segment AB = 6 cm.

2. Construct ∠BAY = 120°.

3. With A as center and radius equal to 4.5 cm draw an arc intersecting AY at C.

4. Join points B and C.

5. Draw the perpendicular bisectors of AB and AC. The point where the two perpendicular bisectors intersect is the center of the circumscribing circle. Mark this point as O.

6. Measure OC, with O as center and OC as radius, construct a circle passing through points A, B and C.

On measuring, we get :

OC = 5.2 cm

Hence, radius of circle = 5.2 cm.

(a) Given,

$$\Rightarrow A + C = \begin{bmatrix*}[r] 1 & 2 \\ 3 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] -5 & 1 \\ 7 & -4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + (-5) & 2 + 1 \\ 3 + 7 & 4 + (-4) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -4 & 3 \\ 10 & 0 \end{bmatrix*}.$$

Hence, A + C = $$\begin{bmatrix*}[r] -4 & 3 \\ 10 & 0 \end{bmatrix*}$$.

(b) From part (a), we get :

A + C = $$\begin{bmatrix*}[r] -4 & 3 \\ 10 & 0 \end{bmatrix*}$$.

$$\Rightarrow B(A + C) = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*} . \begin{bmatrix*}[r] -4 & 3 \\ 10 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] (2 \times -4 + 1 \times 10) & (2 \times 3 + 1 \times 0) \\ (4 \times -4 + 2 \times 10) & (4 \times 3 + 2 \times 0) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -8 + 10 & 6 + 0 \\ -16 + 20 & 12 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 6 \\ 4 & 12 \end{bmatrix*}.$$

Hence, B(A + C) = $$\begin{bmatrix*}[r] 2 & 6 \\ 4 & 12 \end{bmatrix*}$$.

(c) Solving,

$$5B = 5 . \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 5 \\ 20 & 10 \end{bmatrix*}.$$

Hence, 5B = $$\begin{bmatrix*}[r] 10 & 5 \\ 20 & 10 \end{bmatrix*}$$.

(d) From part b,

B(A + C) = $$\begin{bmatrix*}[r] 2 & 6 \\ 4 & 12 \end{bmatrix*}$$.

From part c,

5B = $$\begin{bmatrix*}[r] 10 & 5 \\ 20 & 10 \end{bmatrix*}$$.

Solving,

$$\Rightarrow B(A + C) - 5B = \begin{bmatrix*}[r] 2 & 6 \\ 4 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 10 & 5 \\ 20 & 10 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 2 - 10 & 6 - 5 \\ 4 - 20 & 12 - 10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -8 & 1 \\ -16 & 2 \end{bmatrix*}.$$

Hence, B(A + C) − 5B = $$\begin{bmatrix*}[r] -8 & 1 \\ -16 & 2 \end{bmatrix*}$$.

(a) From graph,

A(3, 3), B(0, −2), C(−4, −2), D(−1, 3).

Hence, A(3, 3), B(0, −2), C(−4, −2), D(−1, 3).

(b) Given,

The diagonals of a parallelogram bisect each other.

Hence, the intersection point P is the midpoint of both AC and BD.

By mid-point formula,

$\Rightarrow M = \Big(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\Big)$

Mid-point of AC :

$$\begin{aligned}\Rightarrow P = \Big(\frac{3 + (-4)}{2}\ , \frac{3 + (-2)}{2}\Big) \\[1em] \Rightarrow P = \Big(\frac{-1}{2}\ , \frac{1}{2}\Big) \\[1em] \Rightarrow P = (-0.5, 0.5).\end{aligned}$$

Hence, P = (-0.5, 0.5).

(c) By formula,

$\text{Slope (m)} = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values for line CB,

$$\begin{aligned}\Rightarrow m_{CB} = \frac{-2-(-2)}{0-(-4)} \\[1em] = \frac{0}{4} \\[1em] = 0.\end{aligned}$$

Substituting values for line DA,

$$\begin{aligned}\Rightarrow m_{DA} = \frac{3-3}{3-(-1)} \\[1em] = \frac{0}{4} \\[1em] = 0 .\end{aligned}$$

Since, slope of parallel lines are equal, thus CB || DA.

Hence, proved that CB || DA.

(d) By formula,

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

Slope of AC = $\frac{-2 - 3}{-4 - 3} = \frac{-5}{-7} = \frac{5}{7}$.

Using point-slope formula,

y - y₁ = m(x - x₁)

Equation of AC :

$$\begin{aligned}\Rightarrow y - 3 = \frac{5}{7}(x - 3) \\[1em] \Rightarrow 7(y - 3) = 5(x - 3) \\[1em] \Rightarrow 7y - 21 = 5x - 15 \\[1em] \Rightarrow 5x - 7y - 15 + 21 = 0 \\[1em] \Rightarrow 5x - 7y + 6 = 0.\end{aligned}$$

Hence, equation of the diagonal AC is 5x - 7y + 6 = 0.

Solving L.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow 2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \\[1em] \Rightarrow \frac{6x - 5}{3} \lt \frac{3x + 50}{5} \\[1em] \text{Multiplying both side by 15, we get:} \\[1em] \Rightarrow 15\Big(\frac{6x - 5}{3}\Big) \lt 15\Big(\frac{3x + 50}{5}\Big) \\[1em] \Rightarrow 5(6x - 5) \lt 3(3x + 50) \\[1em] \Rightarrow 30x - 25 \lt 9x + 150 \\[1em] \Rightarrow 30x - 9x \lt 25 + 150 \\[1em] \Rightarrow 21x \lt 175 \\[1em] \Rightarrow x \lt \frac{175}{21} \\[1em] \Rightarrow x \lt \frac{25}{3} \text{ ........(1)}\end{aligned}$$

Solving R.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 \\[1em] \Rightarrow \frac{3x + 50}{5} \le \frac{4x + 55}{5} \\[1em] \text{Multiplying both side by 5, we get:} \\[1em] \Rightarrow 5\Big(\frac{3x + 50}{5}\Big) \le 5\Big(\frac{4x + 55}{5}\Big) \\[1em] \Rightarrow 3x + 50 \le 4x + 55 \\[1em] \Rightarrow 50 - 55 \le 4x - 3x \\[1em] \Rightarrow -5 \le x \\[1em] \Rightarrow x \ge -5 \text{ ........(2)}\end{aligned}$$

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

-5 ≤ x < $\frac{25}{3}$.

Hence, solution set equals to -5 ≤ x < $\frac{25}{3}$, x ∈ R.

Given,

First term a = 5, last term l = 50, and sum of terms S_n = 440

(a) Let the Number of terms be (n),

Formula for sum of an A.P.:

$$\begin{aligned}S_n = \frac{n}{2}(a + l) \\[1em] \text{Substituting values:} \\[1em] \Rightarrow 440 = \frac{n}{2}(5 + 50) \\[1em] \Rightarrow 440 = \frac{n}{2}(55) \\[1em] \Rightarrow 440 = \frac{55n}{2} \\[1em] \Rightarrow 440 \times 2 = {55n} \\[1em] \Rightarrow n = \frac{880}{55} \\[1em] \Rightarrow n = 16.\end{aligned}$$

Hence, the number of terms is 16.

(b) Let the Common difference be (d),

Formula for nth term:

l = a + (n-1)d

Substitute values:

50 = 5 + (16-1)d

50 = 5 + 15d

50 - 5 = 15d

15d = 45

d = $\frac{45}{15}$

d = 3

Hence, common difference = 3.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{(\cot A+\tan A-1)(\sin A+\cos A)}{\sin^{3}A+\cos^{3}A} \\[1em] \Rightarrow\frac{\Big(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}-1\Big)(\sin A+\cos A)}{(\sin A+\cos A)(\sin^{2}A-\sin A\cos A+\cos^{2}A)} \\[1em] \Rightarrow \frac{\Big(\frac{\cos^{2}A+\sin^{2}A}{\sin A\cos A}-1\Big)(\sin A+\cos A)}{(\sin A+\cos A)(1-\sin A\cos A)} \\[1em] \Rightarrow \frac{\Big(\frac{1}{\sin A\cos A}-1\Big)(\sin A+\cos A)}{(\sin A+\cos A)(1-\sin A\cos A)} \\[1em] \Rightarrow \frac{\frac{1-\sin A\cos A}{\sin A\cos A}(\sin A+\cos A)}{(\sin A+\cos A)(1-\sin A\cos A)} \\[1em] \Rightarrow \frac{\text{(1 - sin A cos A)}}{\text{sin A cos A(1 - sin A cos A)}} \\[1em] \Rightarrow \frac{1}{\sin A\cos A} \\[1em] \Rightarrow \sec A \times \cosec A\end{aligned}$$

Hence, $\frac{(\cot A + \tan A - 1)(\sin A + \cos A)}{\sin^{3}A + \cos^{3}A} = \sec A \times \cosec A$.

Given,

$\Rightarrow \frac{6x^{2} + 3x - 5}{3x - 5} = \frac{9x^{2} + 2x + 5}{2x + 5}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{6x^{2} + 3x - 5 + (3x - 5)}{6x^{2} + 3x - 5 -(3x - 5)} = \frac{9x^{2} + 2x + 5 + (2x + 5)}{9x^{2} + 2x + 5 - (2x + 5)} \\[1em] \Rightarrow \frac{6x^{2} + 3x - 5 + 3x - 5}{6x^{2} + 3x - 5 - 3x + 5} = \frac{9x^{2} + 2x + 5 + 2x + 5}{9x^{2} + 2x + 5 - 2x - 5} \\[1em] \Rightarrow \frac{6x^{2} + 6x - 10}{6x^{2}} = \frac{9x^{2} + 4x + 10}{9x^{2}} \\[1em] \Rightarrow \frac{6x^{2} + 6x - 10}{6} = \frac{9x^{2} + 4x + 10}{9} \\[1em] \Rightarrow 9 \times (6x^{2} + 6x - 10) = 6 \times (9x^{2} + 4x + 10) \\[1em] \Rightarrow 54x^{2} + 54x - 90 = 54x^{2} + 24x + 60 \\[1em] \Rightarrow 54x^{2} + 54x - 90 - 54x^{2} - 24x - 60 = 0 \\[1em] \Rightarrow 54x^{2} - 54x^{2} + 54x - 24x - 90 - 60 = 0 \\[1em] \Rightarrow 30x - 150 = 0 \\[1em] \Rightarrow 30x = 150 \\[1em] \Rightarrow x = \frac{150}{30} \\[1em] \Rightarrow x = 5.\end{aligned}$$

Hence, x = 5.

(a) Since (x − 2) is a factor of 2x³ − 7x² + kx − 2.

Thus, on substituting x = 2, in 2x³ − 7x² + kx − 2, the remainder will be equal to zero.

⇒ 2(2)³ − 7(2)² + k(2) − 2 = 0

⇒ 16 − 28 + 2k − 2 = 0

⇒ −14 + 2k = 0

⇒ 2k = 14

⇒ k = $\frac{14}{2}$

⇒ k = 7.

Hence, k = 7.

(b) Substituting k = 7 in 2x³ − 7x² + kx − 2, we get :

Polynomial : 2x³ − 7x² + 7x − 2.

Dividing 2x³ − 7x² + 7x − 2 by x - 2, we get :

$$\begin{array}{l} \phantom{x - 31}{\quad2x^2 - 3x + 1} \\ x - 2\overline{\smash{\big)}\quad 2x^3 − 7x^2 + 7x − 2} \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \\ \phantom{x^2 + 3x - 7)} - 3x^2 + 7x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}3x^2 \underset{-}{+}6x} \\ \phantom{x^2 + 3x - 5) + 24x ())}x - 2 \\ \phantom{{x^2 + 3x - 5) + 24 +)}}\underline{\underset{-}{+}x \underset{+}{-}2} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times \end{array}$$

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)(2x² - 3x + 1)

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)[2x² - 2x - x + 1]

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)[2x(x - 1) - 1(x - 1)]

⇒ 2x³ − 7x² + 7x − 2 = (x - 2)(2x - 1)(x - 1).

Hence, 2x³ − 7x² + 7x − 2 = (x - 2)(2x - 1)(x - 1).

From figure,

Radius of cylindrical block = radius of hemispheres = r = 3.5 cm

Height of cylindrical block = h = 14 cm.

By formula,

Total surface area of cylinder = 2πr(h + r)

Total surface area of a hemisphere = 3πr²

Total surface area = Total surface area of cylinder + Total surface area of 2 hemispheres

= 2πr(h + r) + 2 × 3πr²

= 2πrh + 2πr² + 6πr²

= 2πrh + 8πr²

= 2πr(h + 4r).

Substituting values, we get :

$$\begin{aligned}\text{Total surface area} = 2 \times \frac{22}{7} \times 3.5 \times (14 + 4 \times 3.5) \\[1em] = 2 \times 22 \times 0.5 \times (14 + 14) \\[1em] = 22 \times 28 \\[1em] = 616 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area = 616 cm².

Steps of Construction:

1. Plot the given points: A(1, 3), B(1, 2) and C(3, 0).

2. Reflect points A and B in x-axis.The image of A under reflection in x-axis is E(1, -3) and image of B under reflection in x-axis is D(1, -2).

3. Reflect points A and B through the origin.The image of A under reflection through origin is F(-1, -3) and image of B under reflection through origin is G(-1, -2).

4. Reflect points A, B, and C in y-axis. The image of A under reflection in y-axis is is J(-1, 3),the image of B under reflection in y-axis is I(-1, 2) and the image of C under reflection in y-axis is H(-3, 0).

5. Join the points in order A, B, C, D, E, F, G, H, I, J and finally back to A to obtain the required closed figure ABCDEFGHIJ. The figure formed is a decagon.

Hence, the figure formed is a decagon.

From figure,

Let height of the building CD be H meters and height of tower AB be h meters.

Given,

BD = 100 m

E is the midpoint of CD

∴ CE = $\frac{CD}{2} = \frac{H}{2}$

From figure,

⇒ ∠CAP = ∠XCA = 35° [Alternate angles are equal]

⇒ ∠EAP = ∠YEA = 14° [Alternate angles are equal]

From figure,

PD = AB = h

AP = BD = 100 m

In ΔACP,

$$\begin{aligned}\Rightarrow \text{tan 35°} = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \text{tan 35°} = \frac{CP}{AP} \\[1em] \Rightarrow \text{tan 35°} = \frac{CD - PD}{AP} \\[1em] \Rightarrow 0.70 = \frac{H - h}{100} \\[1em] \Rightarrow 0.70 × 100 = H - h \\[1em] \Rightarrow H - h = 70 .....(1)\end{aligned}$$

In ΔAEP,

$$\begin{aligned}\Rightarrow \text{tan 14°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 14°} = \frac{EP}{AP}\\[1em] \Rightarrow \text{tan 14°} = \frac{ED - PD}{AP} \\[1em] \Rightarrow 0.25 = \frac{\frac{H}{2} - h}{100} \\[1em] \Rightarrow \frac{H}{2} - h = 0.25 × 100 \\[1em] \Rightarrow \frac{H}{2} - h = 25 .....(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow H - h - \Big({\frac{H}{2} - h}\Big) = 70 - 25 \\[1em] \Rightarrow H - h - {\frac{H}{2} + h} = 45 \\[1em] \Rightarrow H - \frac{H}{2} = 45 \\[1em] \Rightarrow \frac{H}{2} = 45 \\[1em] \Rightarrow H = 45 \times 2 = 90 \text{ m}.\end{aligned}$$

⇒ H - h = 70

⇒ 90 - h = 70

⇒ h = 90 - 70

⇒ h = 20 m.

Hence, height of the tower AB = 20 m and height of the building CD = 90 m.

Steps of Construction:

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (the modal class 60–70), draw two lines AC and BD diagonally from the upper corners C and D of the adjacent rectangles to the top corners of the modal rectangle.

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

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

From graph,

The Mode = L = 64 marks (approx.)

Hence, mode = 64.

(a) Given,

PQ cuts off intercepts of 5 units and 3 units along the x-axis and y-axis respectively.

Thus,

Coordinates of P = (5, 0) and coordinates of Q = (0, -3).

Hence, coordinates of P = (5, 0) and coordinates of Q = (0, -3).

(b) By formula,

Slope (m) = $-\frac{y_2 - y_1}{x_2 - x_1}$

⇒ Slope of PQ = m_PQ = $\frac{-3 - 0}{0 - 5} = \frac{-3}{-5} = \frac{3}{5}$.

We know that,

Product of slopes of perpendicular lines = -1.

⇒ Slope of RS × Slope of PQ = -1

⇒ Slope of RS × $\frac{3}{5}$ = -1

⇒ Slope of RS = $-\frac{5}{3}$.

By point-slope formula,

Equation of line : $y - y_{1} = m(x - x_{1})$

Since, slope of RS = $-\frac{5}{3}$ and it passes through the origin.

Equation of RS :

⇒ $y - 0 = -\frac{5}{3}(x - 0)$

⇒ y = $-\frac{5}{3}x$

⇒ 3y = -5x

⇒ 5x + 3y = 0.

Hence, the equation of line RS is 5x + 3y = 0.

For Article A,

M.P. = ₹ 190, Quantity = 6

Total M.P. = ₹ 190 × 6 = ₹ 1,140

G.S.T. = 12%

G.S.T. amount = $12%$ of ₹ 1,140

$= \frac{12}{100} \times 1140$ = ₹ 136.80

Total amount = ₹ 1,140 + ₹ 136.80 = ₹ 1,276.80

For Article B,

M.P. = ₹ 50, Quantity = 12

(Discount 5% applicable as quantity > 10)

Total M.P. = ₹ 50 × 12 = ₹ 600

Discount = 5% of ₹ 600

$= \frac{5}{100} \times 600$ = ₹ 30

Taxable value = ₹ 600 - ₹ 30 = ₹ 570

G.S.T. = 18%

G.S.T. amount = 18% of ₹ 570

$= \frac{18}{100} \times 570$ = ₹ 102.60

Total amount = ₹ 570 + ₹ 102.60 = ₹ 672.60

Total bill = ₹ 1,276.80 + ₹ 672.60 = ₹ 1,949.40

Rounded off to the nearest ₹10 then bill= ₹ 1,950

The customer paid ₹ 2,000, which is ₹ 50 extra (₹ 2,000 - ₹ 1,950).

Hence, the customer paid ₹ 50 extra.

For one share:

Face value = ₹200

Premium = 25% of Face value

= $\frac{25}{100} \times 200$

= ₹50

(a) By formula,

M.V. = Face value + Premium

= ₹200 + ₹50

= ₹250.

Hence, market Value = ₹ 250.

(b) Given,

Return = 5%

Return on 1 share = $\frac{5}{100} \times 250$

= ₹ 12.50

By formula,

Dividend earned = No. of shares × rate of dividend × F.V. of 1 share

Let rate of dividend be r%.

Substituting values we get :

⇒ 12.50 = 1 × $\frac{r}{100}$ × 200

⇒ r = $\frac{12.50}{200} \times 100$

⇒ r = 6.25%

Hence, dividend percent = 6.25%.

(c) By formula,

Annual dividend = Number of shares × Dividend% × Face value of 1 share

⇒ 1000 = Number of shares × $\frac{6.25}{100}$ × 200

⇒ 1000 = Number of shares × 12.5

⇒ Number of shares = $\frac{1000}{12.5}$ = 80.

Hence, number of shares purchased = 80.

x	f	fx
10	3	30
11	2	22
12	2	24
13	6	78
14	3	42
15	5	75
16	3	48
Total	Σf = 24	Σfx = 319

(a) By formula,

$\text{Mean} = \frac{\sum fx}{\sum f} = \frac{319}{24} = 13.291...\approx 13.$

Hence, required mean = 13.

(b) Cumulative frequency table :

x	f	cf
10	3	3
11	2	5 (3 + 2)
12	2	7 (5 + 2)
13	6	13 (7 + 6)
14	3	16 (13 + 3)
15	5	21 (16 + 5)
16	3	24 (21 + 3)

Median

Here, total frequency (N) = 24 and $\frac{N}{2} = 12$.

From the cumulative column,

For 8th to 13th term, the value of x = 13.

Hence, median = 13.

(a) Let the original speed be x km/hr.

Distance = 350 km

In first case (original speed):

Time $=\frac{\text{Distance}}{\text{Speed}}=\frac{350}{x}$ hrs

In second case (reduced speed):

Speed = (x - 20) km/hr

Time $=\frac{350}{x-20}$ hrs

According to the question,

Time taken to reach destination increases by 2 hours in the second case.

$$\begin{aligned}\therefore \frac{350}{x-20}-\frac{350}{x}=2 \\[1em] \Rightarrow \frac{350x-350(x-20)}{x(x-20)}=2 \\[1em] \Rightarrow \frac{350x - 350x + 7000}{x(x - 20)} = 2 \\[1em] \Rightarrow \frac{7000}{x(x-20)}=2 \\[1em] \Rightarrow x(x-20) = \frac{7000}{2} \\[1em] \Rightarrow x^2 - 20x = 3500 \\[1em] \Rightarrow x^{2} - 20x - 3500 = 0 \\[1em] \Rightarrow x^{2} + 50x - 70x - 3500 = 0 \\[1em] \Rightarrow x(x + 50) - 70(x + 50) = 0 \\[1em] \Rightarrow (x + 50)(x - 70) = 0 \\[1em] \Rightarrow (x + 50) = 0 \text{ or } (x - 70) = 0 \\[1em] \Rightarrow x = -50 \text{ or } x = 70.\end{aligned}$$

Since speed cannot be negative, x = 70 km/hr.

Hence, original speed of the car = 70 km/hr.

(b) Reduced Speed = x - 20 = 70 - 20 = 50 km/hr.

Time taken with reduced speed $= \frac{350}{50}$ = 7 hours.

Hence, with reduced speed the time taken to reach destination = 7 hours.

Given,

External radius of hollow sphere (R) = 5 cm, internal radius of hollow sphere (r) = 3 cm, height of cone (h) = 8 cm.

By formula,

Volume of metal in the hollow sphere V = $\frac{4}{3} \pi (R^3 − r^3)$

Substituting values we get :

$$\begin{aligned}\Rightarrow V_{\text{sphere}}=\frac{4}{3} \times \pi \times \Big(5^{3}-3^{3}\Big) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times (125-27) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times 98 \\[1em] \Rightarrow V_{\text{sphere}} =\frac{392}{3} \pi\end{aligned}$$

By formula,

Volume of cone = $\frac{1}{3} \pi \times \text{radius}^2h$

Let radius of cone formed be m cm.

$V_{\text{cone}}=\frac{1}{3}\pi m^{2}\times 8=\frac{8}{3}\pi m^{2}$

Since, the sphere is melted to form, the cone the volume of both objects will be equal.

$$\begin{aligned}\Rightarrow \frac{392}{3}\pi=\frac{8}{3}\pi m^{2} \\[1em] \Rightarrow 392 = 8m^{2} \\[1em] \Rightarrow m^2 = \frac{392}{8} \\[1em] \Rightarrow m^{2} = 49 \\[1em] \Rightarrow m = 7 \text{ cm}.\end{aligned}$$

Hence, the radius of the cone formed = 7 cm.

Given,

Total number of outcomes = 30

(a) Given,

Numbers that are perfect squares (between 1 to 30) are 1, 4, 9, 16, 25 (5 numbers).

∴ No. of favourable outcomes = 5 perfect squares

P(perfect square) $=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{30} = \frac{1}{6}$.

Hence, probability of numbers that are perfect squares = $\frac{1}{6}$.

(b) Given,

Even numbers that are also multiples of 3 (between 1 to 30) are 6, 12, 18, 24, 30 (5 numbers)

∴ No. of favourable outcomes = 5 Even numbers that are also multiples of 3

P(even and multiple of 3) $=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{30} = \frac{1}{6}$

Hence, probability of numbers that are multiples of 3 and even number = $\frac{1}{6}$.

(c) Given,

Prime numbers (1 to 30) are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (10 numbers)

∴ No. of favourable outcomes = 10 Prime numbers

P(prime)

$=\frac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \frac{10}{30} = \frac{1}{3}$

Hence, probability of prime numbers = $\frac{1}{3}$.

(a) Given,

In ΔBOX, OX ⟂ BX (radius ⟂ tangent at its point of contact)

⇒ ∠OXB = 90°.

By angle‑sum property of triangle,

⇒ ∠BOX + ∠OBX + ∠OXB = 180°

⇒ ∠BOX + 32° + 90° = 180°

⇒ ∠BOX + 122° = 180°

⇒ ∠BOX = 180° - 122°

⇒ ∠BOX = 58°.

Hence, ∠BOX = 58°.

(b) From figure,

∠COX = ∠BOX = 58°

We know that,

The angle which, an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

$$\begin{aligned}\Rightarrow ∠CYX = \frac{1}{2}∠COX \\[1em] \Rightarrow ∠CYX = \frac{58°}{2} \\[1em] \Rightarrow ∠CYX = 29°\end{aligned}$$

Hence, ∠CYX = 29°.

(c) We know that,

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

∠ZYX = ∠AXZ = 66°

Hence, ∠ZYX = 66°.

(d) From figure,

In isosceles ΔZXY,

ZY = XY

We know that,

The angles opposite to equal side of a triangle are equal.

∠ZXY = ∠XZY

By angle sum property in ΔXYZ,

$$\begin{aligned}\Rightarrow ∠XZY + ∠ZXY + ∠ZYX = 180° \\[1em] \Rightarrow ∠ZYX + 2∠XZY = 180° \\[1em] \Rightarrow 2∠XZY = 180° - ∠ZYX \\[1em] \Rightarrow ∠XZY = \frac{180° - ∠ZYX}{2} \\[1em] = \frac{180° - 66°}{2} \\[1em] = \frac{114°}{2} \\[1em] = 57° .\end{aligned}$$

We know that,

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

∠YXB = ∠XZY = 57°.

Also,

∠ZXY = ∠XZY = 57°.

From figure,

⇒ ∠OXY = ∠OXB - ∠YXB

⇒ ∠OXY = 90° - 57° = 33°.

Hence, ∠OXY = 33°.

Given,

G.P. : 7, 21, 63,.......

First term (a) = 7

Common ratio (r) = $\frac{21}{7} = 3$

(a) By formula,

⇒ T_n = a × r^(n-1)

⇒ 1701 = 7 × 3^(n-1)

$\Rightarrow 3^{n-1} = \frac{1701}{7}$

⇒ 3^(n-1) = 243

⇒ 3^(n-1) = 3⁵

⇒ n - 1 = 5

⇒ n = 6.

Hence, n = 6.

(b) By formula,

$S_{n} = a \times \Big(\frac{r^{,n}-1}{r-1}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow S_{6} = 7 \times \Big(\frac{3^{6}-1}{3-1}\Big) \\[1em] = 7 \times \Big(\frac{729-1}{2}\Big) \\[1em] = 7 \times \frac{728}{2} \\[1em] = 7 \times 364 \\[1em] = 2548.\end{aligned}$$

Hence, the sum of the n terms (here n = 6) of the G.P. is 2548.

(a) We know that,

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

∴ ∠PTR = ∠PST

⇒ ∠RPT = ∠TPS [Common angles]

Therefore, by AA similarity, ΔPTR ~ ΔPST.

Hence, proved that ΔPTR ~ ΔPST.

(b) Since, corresponding sides of similar triangles are proportional we have :

⇒ $\frac{PT}{PS} = \frac{PR}{PT}$

⇒ PT² = PR × PS.

Hence, proved that PT² = PR × PS.

(c) Given,

PR = 4 cm and PS = 16 cm.

$$\begin{aligned}\Rightarrow PT^2 = PR \times PS \\[1em] \Rightarrow PT = \sqrt{PR \times PS} \\[1em] \Rightarrow PT = \sqrt{4 \times 16} \\[1em] \Rightarrow PT = \sqrt{64} \\[1em] \Rightarrow PT = 8 \text{ cm}.\end{aligned}$$

Hence, PT = 8 cm.

Monthly salary	cf
0-2000	10
2000 - 4000	20
4000 - 6000	35
6000 - 8000	55
8000 - 10000	70
10000 - 12000	75

(a) From the graph, the cumulative frequency at ₹12,000 is 75.

Hence, the total number of workers = 75.

(b) By formula,

Median = $\frac{N + 1}{2}$

Total workers = 75

$\frac{75 + 1}{2} = \frac{76}{2} = 38$

From table,

36 to 55 worker lies in the class 6000 - 8000.

Thus, 38th worker lies in the class 6000 - 8000.

Hence, the median class = 6000 – 8000.

(c) Lower quartile = $\frac{N + 1}{4} = \frac{76}{4} = 19$

From table,

11 to 20 worker lies in the class 2000 - 4000.

Thus, 19th worker lies in the class 2000 - 4000.

Hence, the lower-quartile class = 2000 – 4000.

(d) From table,

Number of workers having salary less than ₹ 10,000 = 70

Number of workers having salary less than ₹ 6,000 = 35

Number of workers in this range = 70 − 35 = 35 workers.

Hence, the number of workers having monthly salary more than or equal to ₹ 6,000 but less than ₹ 10,000 = 35.