ICSE Class 10 Mathematics — 2026 Specimen

Given,

(x - 2) and (x + 2) are the factors of x³ + x² - 4x - 4.

⇒ (x - 2)(x + 2) = x² - 4

Thus, x² - 4 is also the factor of x³ + x² - 4x - 4.

$$\begin{array}{l} \phantom{x - 3x)}{\quad x + 1} \\ x^2 - 4\overline{\smash{\big)}\quad x^3 + x^2 - 4x - 4 } \\ \phantom{x^2 + n}\phantom(\underline{\underset{-}{+}x^3 \underset{+}{-}4x} \\ \phantom{x^2 + 3x -])7 + o)()} x^2 - 4 \\ \phantom{x^2 + 3x - ((5) + )}\underline{\underset{-}{+}x^2 \underset{+}{-}4} \\ \phantom{x^2 + 3x - 5) + 24x )}\times \end{array}$$

Thus, x³ + x² - 4x - 4 = (x² - 4)(x + 1)

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

Hence, option 3 is the correct option.

Given,

P = ₹ 400

n = 18 months

By formula,

Qualifying Sum = $P \times \frac{n(n + 1)}{2}$

= $400 \times \frac{18(18 + 1)}{2}$

= 400 × 9 × 19

= 400 × 171

= ₹ 68,400

Hence, option 3 is the correct option.

Given,

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In ∆APQ and ∆ABC,

⇒ ∠APQ = ∠ABC [Both equal to 90°]

⇒ ∠BAC = ∠PAQ [Common angles]

∴ ∆APQ ∼ ∆ABC (By A.A. axiom)

We know that,

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

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆APQ}}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{18}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC}= \frac{(4 + 3)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{49}{9} \times 18 = 98 \text{ cm}^2.\end{aligned}$$

From figure,

Area QPBC = Area of ∆ABC − Area of ∆APQ

= 98 - 18 = 80 cm².

Hence, option 3 is the correct option.

Join OA and OP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

Given,

⇒ sec θ + tan θ = a

⇒ sec θ − tan θ = b

⇒ ab = (sec θ + tan θ)(sec θ - tan θ)

⇒ ab = sec² θ - tan² θ

⇒ ab = 1

So assertion (A) is true.

We know that,

⇒ sec² θ - tan² θ = 1

This is a fundamental trigonometric identity.

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

Hence, option 3 is the correct option.

By formula,

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

Given,

A solid sphere is cut into two identical hemispheres.

Volume of hemisphere = $\frac{2}{3} \pi r^3$

Volume of two identical hemispheres = $2 \times \frac{2}{3} \pi r^3$

= $\frac{4}{3} \pi r^3$

Thus, volume of a sphere = volume of two identical hemispheres.

So assertion (A) is true.

We know that,

Surface area of sphere = 4πr²

When a sphere is cut into two hemispheres, two new flat circular surfaces are created,

Surface area of a single hemisphere = Curved surface area + Area of its flat circular face

= 2πr² + πr²

= 3πr²

Total surface area of two hemispheres = 2 × 3πr² = 6πr².

Thus, surface area of a single hemisphere ≠ surface area of two hemispheres.

So reason (R) is false.

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

Hence, option 1 is the correct option.

Given,

N = 7

Sum of the squares of the first seven natural numbers = 140.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

= $\frac{140}{7}$

= 20.

Hence, option 1 is the correct option.

Given,

Number of red marbles = 3

Number of blue marbles = 2

The total number of marbles in the bag is 3 + 2 = 5.

Number of black marbles in the bag = 0

Probability = $\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

= $\frac{0}{5}$

= 0.

Hence, option 1 is the correct option.

Calculating,

$$\Rightarrow AB = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 3 + 2 \times 4 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 + 8 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \end{bmatrix*}.$$

Hence, option 3 is the correct option.

Given,

2 parts of red pigments is mixed with 5 parts of the base.

$$\begin{aligned}\therefore \frac{4}{a} = \frac{2}{5} \\[1em] \Rightarrow a = \frac{4 \times 5}{2} = \frac{20}{2} = 10. \\[1em] \therefore \frac{b}{12.5} = \frac{2}{5} \\[1em] \Rightarrow b = \frac{2}{5} \times 12.5 = 5. \\[1em] \therefore \frac{6}{c} = \frac{2}{5} \\[1em] \Rightarrow c = \frac{6 \times 5}{2} = 3 \times 5 = 15.\end{aligned}$$

Hence, option 4 is the correct option.

Market price = ₹ 1,200

Discount = 20% = $\frac{20}{100} \times 1200$ = ₹ 240

Price after discount = ₹ 1200 - ₹ 240 = ₹ 960.

GST = 18%

SGST = $\frac{GST}{2} = \frac{18}{2} = 9%$

= $\frac{9}{100} \times 960$

= 9 × 9.6

= ₹ 86.40

Hence, option 4 is the correct option.

M.V of share = ₹ 38.50

No of shares bought = 50

Money required = ₹ 38.50 × 50 = ₹ 1925.

Hence, option 3 is the correct option.

⇒ x² – 1 = 0

⇒ x² – 1² = 0

⇒ (x + 1)(x - 1) = 0

⇒ (x + 1) = 0 or (x - 1) = 0 [Using zero product rule]

⇒ x = -1 or x = 1.

Hence, option 4 is the correct option.

Equation:

⇒ x - y = 7

⇒ y = x - 7

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

m = 1.

A line is equally inclined to the axes if slope = 1.

Hence, option 2 is the correct option.

Solving the given equation :

$$\begin{aligned}\Rightarrow x + 2 \le \frac{x}{3} + 3 \\[1em] \Rightarrow x - \frac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \frac{3x - x}{3} \le 1 \\[1em] \Rightarrow \frac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \frac{3}{2} \\[1em] \Rightarrow x \le 1.5\end{aligned}$$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, option 1 is the correct option.

Given,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\frac{1}{2}$

Substituting x = $-\frac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$$\begin{aligned}\Rightarrow 2 \times \Big(-\frac{1}{2}\Big)^3 + 7 \times \Big(-\frac{1}{2}\Big)^2 + 2 \times \Big(-\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\frac{1}{8} + 7 \times \frac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\frac{1}{4} + \frac{7}{4} - 4 \\[1em] \Rightarrow \frac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \frac{6}{4} - 4 \\[1em] \Rightarrow \frac{6 - 16}{4} \\[1em] \Rightarrow \frac{-10}{4} \\[1em] \Rightarrow \frac{-5}{2}.\end{aligned}$$

Since, remainder is not equal to zero.

Hence, (2x + 1) is not a factor of the given polynomial.

Substituting x = $\frac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{1}{2}\Big)^3 + 7 \times \Big(\frac{1}{2}\Big)^2 + 2 \times \Big(\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times \frac{1}{8} + 7 \times \frac{1}{4} + 1 - 3 \\[1em] \Rightarrow \frac{1}{4} + \frac{7}{4} - 2 \\[1em] \Rightarrow \frac{1 + 7}{4} - 2 \\[1em] \Rightarrow \frac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.\end{aligned}$$

Since, remainder is equal to zero. Thus, x - $\frac{1}{2}$ is the factor of the polynomial.

$$\begin{aligned}\Rightarrow x - \frac{1}{2} = 0 \\[1em] \Rightarrow x = \frac{1}{2} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow 2x - 1.\end{aligned}$$

2x - 1 is factor of polynomial.

Dividing 2x³ + 7x² + 2x – 3 by 2x - 1, we get:

$$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x – 3} \\ \phantom{x^2 +o 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)[][]} 8x^2 + 2x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{-}{+}8x^2 \underset{+}{-}4x} \\ \phantom{x^2 + 3x - 5) + (24)}6x - 3 \\ \phantom{{x^2 + 3x - 5) + 24}}\underline{\underset{-}{+}6x \underset{+}{-}3} \\ \phantom{{x^2 + 3x - 54)} + 2x ,k}\times\ \end{array}$$

2x³ + 7x² + 2x – 3 by 2x - 1 = (2x - 1)(x² + 4x + 3)

= (2x - 1)[x² + 3x + x + 3]

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

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

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

(a) Let the point P(x, 0) divide the line segment joining A(-6, 2) and B(9, -4) in the ratio m : n.

By section formula,

$y = \frac{my_2 + ny_1}{m + n}$

Substituting the values we get :

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

Hence, the ratio in which P divides the line segment AB is 1 : 2.

(b) From part (a),

A (-6, 2) and B (9, - 4)

m : n = 1 : 2

By section-formula,

$x = \frac{mx_2 + nx_1}{m + n}$

$$\begin{aligned}\Rightarrow x = \frac{1(9) + 2(-6)}{1+2} \\[1em] \Rightarrow x = \frac{9 - 12}{3} \\[1em] \Rightarrow x = \frac{-3}{3} \\[1em] \Rightarrow x = -1.\end{aligned}$$

P = (x, 0) = (-1, 0).

Hence, the coordinates of the point P are (-1, 0).

(c) By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow m_{AB} = \frac{-4 - 2}{9 - (-6)} \\[1em] \Rightarrow m_{AB} = \frac{-6}{15} \\[1em] \Rightarrow m_{AB} = \frac{-2}{5}\end{aligned}$$

Line parallel to AB will have the same slope, so the slope of the new line is also m = $\frac{-2}{5}$.

By point-slope formula,

y - y₁ = m(x - x₁)

Line passing through (-3, -2) and parallel to AB is :

$$\begin{aligned}\Rightarrow y - (-2) = \frac{-2}{5}(x - (-3)) \\[1em] \Rightarrow y + 2 = \frac{-2}{5}(x + 3) \\[1em] \Rightarrow 5(y + 2) = -2(x + 3) \\[1em] \Rightarrow 5y + 10 = -2x - 6 \\[1em] \Rightarrow 5y + 10 + 2x + 6 = 0 \\[1em] \Rightarrow 2x + 5y + 16 = 0 \\[1em]\end{aligned}$$

Hence, the equation of the line is 2x + 5y + 16 = 0.

(a) Join AE.

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

∴ ∠AOB = 2∠AEB

⇒ 80° = 2∠AEB

⇒ ∠AEB = $\frac{80^\circ}{2} = 40°$

We know that,

Angle in a semi-circle is a right angle.

⇒ ∠AEC = 90°

From figure,

⇒ ∠BEC = ∠AEC - ∠AEB = 90° - 40° = 50°.

Hence, ∠BEC = 50°.

(b) From figure,

⇒ ∠ECD = ∠CEB = 50° (Alternate angles are equal)

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

⇒ ∠AOB = 2∠BCA

⇒ 80° = 2∠BCA

⇒ ∠BCA = $\frac{80^\circ}{2}$ = 40°.

From figure,

⇒ ∠BCD = ∠BCA + ∠ACE + ∠ECD = 40° + 20° + 50° = 110°.

Hence, ∠BCD = 110°.

(c) As sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° - ∠BCD = 180° - 110° = 70°.

From figure,

⇒ ∠BED = ∠BEC + ∠CED

⇒ 70° = 50° + ∠CED

⇒ ∠CED = 70° - 50° = 20°.

Hence, ∠CED = 20°.

(a) Given,

-11, -7, -3, ......,49, 53

Second term - First term : −7 − (−11) = 4

Third term - Second term : −3 − (−7) = 4

Last term - Second last term : 53 − 49 = 4

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (AP) with first term (a) = -11 and common difference (d) = 4.

Hence, the progression is an Arithmetic Progression.

(b) Let 53 be the nth term.

By formula,

⇒ t_n = a + (n - 1)d

⇒ 53 = -11 + (n - 1)4

⇒ 53 = -11 + 4n - 4

⇒ 53 = -15 + 4n

⇒ 4n = 53 + 15

⇒ 4n = 68

⇒ n = $\frac{68}{4}$

⇒ n = 17.

Hence, there are a total of 17 terms in the progression.

(c) Since, n = 17, is odd.

By formula,

Middle term = $\frac{n + 1}{2}$

= $\frac{17 + 1}{2}$

= $\frac{18}{2}$

= 9.

Thus, 9th term is middle term.

By formula,

⇒ t₉ = a + (n - 1)d

= -11 + (9 - 1) × 4

= -11 + 8 × 4

= -11 + 32

= 21.

Hence, the value of the middle most term is 21.

When vertically placed,

Height of liquid (h) = $\frac{1 + 6}{2} = \frac{7}{2}$ cm

Diameter of base = 7 cm

Radius (r) = $\frac{7}{2}$ cm

Area of wet surface = πr² + 2πrh

= πr(r + 2h)

= $\frac{22}{7} \times \frac{7}{2} \Big(\frac{7}{2} + 2 \times \frac{7}{2}\Big)$

= 11 × (3.5 + 7)

= 11 × 10.5

= 115.5 cm².

Hence, area of wet surface = 115.5 cm².

Steps of construction :

1. With O as center draw a circle of radius 4.5 cm.

2. Take a point A on the circumference with A as center cut an arc of radius 6.5 cm, intersecting circumference at point B.

3. Construct ∠CAB = 75°, where C lies on the circumference of the circle.

4. Draw XY, the perpendicular bisector of AB.

5. Draw AZ, the angular bisector of angle A.

6. Mark point P as the intersection of AZ and XY.

7. Measure CP.

Hence, the length of CP = 5.2 cm.

Given,

For initial investment,

Investment = ₹ 8,000

Face Value = ₹ 100

Market Value = ₹ 80

Dividend Rate = 6%

By formula,

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

= $\frac{8000}{80}$

= 100

By formula,

Dividend for the first year = No. of shares × Rate of div. × N.V. of 1 share

= 100 × $\frac{6}{100}$ × 100

= ₹ 600

Hence, dividend for first year = ₹ 600.

(b) Given,

Number of shares sold = 100

Selling price per share = ₹ 75

Proceeds from sale = Number of shares × selling price

= 100 × 75

= ₹ 7,500

Total proceeds = Proceeds from sale + Dividend received = 7500 + 600 = ₹ 8,100

Total investment = ₹ 8,100

Market value per share = ₹ 27

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

= $\frac{8100}{27}$

= 300.

Hence, number of shares purchased by Ms. Kaur = 300.

Solving L.H.S of inequation : 5x - 21 < $\frac{5x}{7}$ - 6 ≤ $-3\frac{3}{7}$ + x, we get:

$$\begin{aligned}\Rightarrow 5x - 21 \lt \frac{5x}{7} - 6 \\[1em] \Rightarrow 5x - \frac{5x}{7} \lt 21 - 6 \\[1em] \Rightarrow \frac{35x - 5x}{7} \lt 15 \\[1em] \Rightarrow \frac{30x}{7} \lt 15 \\[1em] \Rightarrow x \lt \frac{7 \times 15}{30} \\[1em] \Rightarrow x \lt \frac{7}{2}\text{ ......(1)}\end{aligned}$$

Solving R.H.S of inequation 5x - 21 < $\frac{5x}{7}$ - 6 ≤ $-3\frac{3}{7}$ + x, we get:

$$\begin{aligned}\Rightarrow \frac{5x}{7} - 6 \le -3\frac{3}{7} + x \\[1em] \Rightarrow \frac{5x}{7} - 6 \le -\frac{24}{7} + x \\[1em] \Rightarrow x - \frac{5x}{7} \ge \frac{24}{7} - 6 \\[1em] \Rightarrow \frac{7x - 5x}{7} \ge \frac{24 - 42}{7} \\[1em] \Rightarrow \frac{2x}{7} \ge \frac{-18}{7} \\[1em] \Rightarrow 2x \ge -18 \\[1em] \Rightarrow x \ge \frac{-18}{2} \\[1em] \Rightarrow x \ge -9 \text{ .....(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = {x : -9 ≤ x < $\frac{7}{2}$ , x ∈ R}

Hence, solution set = {x : -9 ≤ x < $\frac{7}{2}$ , x ∈ R}.

Solving,

(sin θ + cos θ)(cosec θ - sec θ)

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

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.

(a) In ∆APB and ∆CPD,

⇒ ∠APB = ∠CPD (Vertically opposite angles are equal)

⇒ ∠BAP = ∠DCP (Angles in same segment are equal)

∴ ∆APB ~ ∆CPD (By A.A. axiom)

Hence, proved that ∆APB ~ ∆CPD.

(b) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{CD}{AB} = \frac{PD}{PB} \\[1em] \Rightarrow \frac{CD}{9} = \frac{2}{3} \\[1em] \Rightarrow CD = \frac{2}{3} \times 9 \\[1em] \Rightarrow CD = 6 \text{ cm}.\end{aligned}$$

Hence, CD = 6 cm.

(c) We know that,

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

$$\begin{aligned}\therefore \frac{\text{Area of ∆APB}}{\text{Area of ∆CDP}} = \frac{PB^2}{PD^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4} \\[1em] = 9 : 4.\end{aligned}$$

Hence, area ∆APB : area ∆CPD = 9 : 4.

Let rate of interest be r%.

Given,

P = ₹ 600/month

n = 2 years or 24 months

M.V. = ₹ 15,600

By formula,

M.V = $P \times n + P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$\Rightarrow 15600 = 600 \times 24 + 600 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100}$

⇒ 15600 = 14400 + 6 × 25 × r

⇒ 15600 − 14400 = 6 × 25 × r

⇒ 150r = 1200

⇒ r = $\frac{1200}{150}$ = 8%.

Hence, rate of interest = 8%.

In the given table i is the class interval which is equal to 15.

Class	Class Mark (x)	d = (x - A)	u = d/i	Frequency (f)	fu
0 - 15	7.5	-45	-3	3	-9
15 - 30	22.5	-30	-2	4	-8
30 - 45	37.5	-15	-1	7	-7
45 - 60	A = 52.5	0	0	6	0
60 - 75	67.5	15	1	8	8
75 - 90	82.5	30	2	2	4
Total				Σf = 30	Σfu = -12

By formula,

Mean = A + $\frac{Σfu}{Σf} \times i$

= 52.5 + $\frac{-12}{30}$ × 15

= = 52.5 - $\frac{180}{30}$

= 52.5 - 6

= 46.50

Hence, mean = 46.50.

By formula,

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

Substituting values we get:

⇒ Centroid of △ ABC = $\Big(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\Big)$

⇒ P = $\Big(\frac{2}{3}, \frac{2}{3}\Big)$

By formula,

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

$$\begin{aligned}\text{Slope of AB} = \frac{-1 - 3}{3 - (-1)} \\[1em] = \frac{-4}{4} \\[1em] = -1.\end{aligned}$$

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁ )

Substituting values we get :

Equation of line passing through P and parallel to AB :

$$\begin{aligned}\Rightarrow y - \frac{2}{3} = -1 \Big(x - \frac{2}{3}\Big) \\[1em] \Rightarrow \frac{3y - 2}{3} = -1 \times \Big(\frac{3x - 2}{3}\Big) \\[1em] \Rightarrow 3y - 2 = -1 \times (3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3x + 3y = 2 + 2 \\[1em] \Rightarrow 3x + 3y = 4.\end{aligned}$$

Hence, equation of a line passing through P and parallel to AB is 3x + 3y = 4.

(a) We know that,

If two tangents are drawn to a circle from an exterior point the tangents are equal length.

AP = AS, BP = BQ, CQ = CR, DR = DS

From figure,

⇒ AB + CD = (AP + BP) + (DR + CR)

⇒ AB + CD = (AS + BQ) + (DS + CQ)

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC .......(1)

We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD and AD = BC

Substituting above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) Since, AB = BC, AB = CD and AD = BC.

∴ AB = BC = CD = AD.

A parallelogram with all four sides equal is a rhombus.

Hence, ABCD is a rhombus.

For Dry fruits :

M.P. = ₹ 1,200

Discount = ₹ 100

Discounted price = ₹ 1200 - ₹ 100 = ₹ 1,100

GST = 12%

Tax = $\frac{12}{100} \times 1100$ = ₹ 132

Price including tax = ₹ 1,100 + ₹ 132 = ₹ 1,232.

For Wheat flour :

M.P. = ₹ 286

Discount = Nil

GST = 5%

Tax = $\frac{5}{100} \times 286$ = ₹ 14.30

Price including tax = ₹ 286 + ₹ 14.30 = 300.30

For bakery products :

M.P. = ₹ 500

Discount = 10% = $\frac{10}{100} \times 50$ = ₹ 50.

Discounted price = ₹ 500 - ₹ 50 = ₹ 450

GST = 12%

Tax = $\frac{12}{100} \times 450$ = ₹ 54

Price including tax = ₹ 450 + ₹ 54 = ₹ 504.

Total price = ₹ 1,232 + ₹ 300.30 + ₹ 504 = ₹ 2,036.30

Hence, total amount to be paid = ₹ 2,036.30.

From figure,

Let the top of the lighthouse be T, the base of the lighthouse be L, and the two ships be A and B.

Given,

The height of the lighthouse, TL, is 80 m.

The angles of depression are 30° and 50°.

These are equal to the alternate interior angles of elevation at the ships.

∠TBL = 50° and ∠TAL = 30°

In triangle TLA,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 30° = \frac{TL}{LA} \\[1em] \Rightarrow LA = \frac{TL}{\tan 30°} \\[1em] \Rightarrow LA = \frac{80}{0.577} \\[1em] \Rightarrow LA = 138.65 \text{ m}.\end{aligned}$$

In triangle TLB,

$$\begin{aligned}\Rightarrow \tan 50° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 50° = \frac{TL}{LB} \\[1em] \Rightarrow LB = \frac{TL}{\tan 50°} \\[1em] \Rightarrow LB = \frac{80}{1.192} \\[1em] \Rightarrow LB = 67.11 \text{ m}.\end{aligned}$$

Distance between ships = LA + LB

= 138.65 m + 67.11 m = 205.76 m.

Hence, the distance between the two ships is 206 m.

Marks %	No.of students	cf
0 - 10	5	5
10 - 20	7	12
20 - 30	11	23
30 - 40	20	43
40 - 50	40	83
50 - 60	52	135
60 - 70	36	171
70 - 80	15	186
80 - 90	9	195
90 - 100	5	200

Steps :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Plot the points (10, 5), (20, 12), (30, 23), (40, 43), (50, 83), (60, 135), (70, 171), (80, 186), (90, 195) and (100, 200).

4. Join the points by free hand curve.

(a) Here, n (no. of students) = 200, which is even.

Median = $\frac{n}{2} = \frac{200}{2}$ = 100^th term.

Through point L = 100 draw a horizontal line parallel to x-axis touching the graph at point M, through M draw a vertical line parallel to y-axis touching x-axis at point N = 53.

Hence, median = 53.

(b) Total marks = 100

65% of 100 = 65

Through point O = 65 draw a vertical line parallel to y-axis touching the graph at point P, through P draw a horizontal line parallel to x-axis touching y-axis at point Q = 154.

∴ 154 students score less than or equal to 65%.

∴ 46 (200 - 154) students score more than 65%.

Hence, 46 students score more than 65%.

(c) Total marks = 100

35% of 100 = 35

Through point R = 35 draw a vertical line parallel to y-axis touching the graph at point S, through S draw a horizontal line parallel to x-axis touching y-axis at point T = 33.

Hence, 33 students did not pass the exam.

The cards are numbered from 0 to 100.

Total number of cards = 101 (as there is no. zero card also)

(a) Favorable outcome = Getting a card having a number which is divisible by 6.

{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

Probability $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{101}$.

Hence, probability of getting a card which is divisible by 6 = $\frac{17}{101}$.

(b) Favorable outcome = Getting a card having a number which is not divisible by 6.

P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)

= $1 - \frac{17}{101}$

= $\frac{101 - 17}{101}$

= $\frac{84}{101}$.

Hence, probability of getting a card which is not divisible by 6 = $\frac{84}{101}$.

Given,

x, y and z are in continued proportion.

$\therefore \frac{x}{y} = \frac{y}{z}$

⇒ y² = xz

To prove :

$\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$

Solving L.H.S,

$$\begin{aligned}\Rightarrow \frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{x^3}{x^3.z^3} + \frac{y^3}{x^3.z^3} + \frac{z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(xz)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(y^2)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{y^6} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3}.\end{aligned}$$

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

Hence, proved that $\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$.

(a) Given,

Radius of ball bearings = 7 mm

Volume of box = 2156 cm³ = 2156 × 10³ mm³

Number of ball bearings that each box can have (N)

= $\frac{\text{Volume of box}}{\text{Volume of each ball}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow N = \frac{2156 \times 10^3}{\frac{4}{3} \times \frac{22}{7} \times 7^3} \\[1em] = \frac{2156 \times 10^3}{\frac{88}{3} \times 7^2} \\[1em] = \frac{2156 \times 3 \times 1000}{88 \times 49} \\[1em] = \frac{6468000}{4312} \\[1em] = 1500.\end{aligned}$$

Hence, maximum no. of ball bearings in a box = 1500.

(b) Mass of each box = No. of balls × Mass of each ball

= 1500 × 4 gm

= 6000 gm

= $\frac{6000}{1000}$

= 6 kg.

Hence, mass of each box = 6 kg.

(a) From histogram,

The number of batsmen who scored between 500 and 600 runs is 3.

The number of batsmen who scored between 600 and 700 runs is 2.

Thus, the total number of batsmen who scored 500 to 700 runs is 3 + 2 = 5.

Hence, number of batsmen who scored 500 to 700 runs = 5.

(b) The modal class is the class with the highest frequency.

Hence, the modal class interval is 400 - 500.

(c) From graph,

Mode = 430.

Hence, mode = 430.

Let common difference be d.

First term (a) = 3

By formula,

Sum of first n terms of an A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

⇒ S₈ = 2 × S₅

$\therefore \frac{8}{2}[2 \times a + (8 - 1) \times d] = \frac{5}{2}[2 \times a + (5 - 1) \times d]$

⇒ 4[2a + 7d] = 5[2a + 4d]

⇒ 4[2 × 3 + 7d] = 5[2 × 3 + 4d]

⇒ 4[6 + 7d] = 5[6 + 4d]

⇒ 24 + 28d = 30 + 20d

⇒ 28d - 20d = 30 - 24

⇒ 8d = 6

⇒ d = $\frac{6}{8} = \frac{3}{4}$.

Hence, common difference = $\frac{3}{4}$.

Given,

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (r - p)² - 4 × (q - r) × (p - q) = 0

⇒ r² + p² - 2pr - 4(qp - q² - rp + qr) = 0

⇒ r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0

⇒ r² + p² + 2pr - 4qp - 4qr + 4q² = 0

⇒ (r + p)² - 4q(p + r) + 4q² = 0

Let p + r = y,

⇒ y² - 4qy + 4q² = 0

⇒ (y - 2q)² = 0

⇒ y - 2q = 0

⇒ y = 2q

⇒ p + r = 2q.

Hence, proved that p + r = 2q.

Let the three consecutive even numbers be represented as x, x + 2, and x + 4.

Given,

The sum of their squares is 596.

⇒ x² + (x + 2)² + (x + 4)² = 596

⇒ x² + x² + 4x + 4 + x² + 8x + 16 = 596

⇒ 3x² + 12x + 20 = 596

⇒ 3x² + 12x + 20 - 596 = 0

⇒ 3x² + 12x - 576 = 0

⇒ 3(x² + 4x − 192) = 0

⇒ x² + 4x − 192 = 0

⇒ x² + 16x - 12x − 192 = 0

⇒ x(x + 16) - 12(x + 16) = 0

⇒ (x - 12)(x + 16) = 0

⇒ (x - 12) = 0 or (x + 16) = 0 [Using zero product rule]

⇒ x = 12 or x = -16.

Case 1 : x = 12

The three consecutive even numbers are:

⇒ x = 12

⇒ x + 2 = 12 + 2 = 14

⇒ x + 4 = 12 + 4 = 16

Case 2: x = −16

The three consecutive even numbers are:

⇒ x = −16

⇒ x + 2 = −16 + 2 = −14

⇒ x + 4 = −16 + 4 = −12

Hence, the two possible sets of three consecutive even numbers are 12, 14, 16 and -16, -14, -12.

Given,

X² = 4X + 5I

Solving for L.H.S.,

$$X^2 = \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1 \times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

Solving for R.H.S.,

$$4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 4 \\ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

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

Hence, proved that X² = 4X + 5I.

Steps of Construction:

1. Plot the points : A(4, 0), B(4, 2), C(2, 2), D(2,4) and E(0, 4).

2. On reflecting the points A, B, C and D on the y-axis, they become F, G, H and I respectively.

3. Join the points ABCDEIHGF.

4. Reflect the figure ABCDEIHGF in x-axis.

5. On reflecting the points A and F in x-axis, they remain same as they lie on x-axis.

6. On reflecting points B, C, D, E, I, H and G on the x-axis, they become M, N, P, Q, R, S and T respectively.

7. Join the points A, E, F and Q.

In figure AEFQ,

There are four congruent isosceles right angle triangles.

Thus, all the sides AE, EF, FQ and QA are equal.

Also each interior angle is equal to 90°.

Hence, the figure AEFQ is square.