Model Test Paper 2

Let x = $0.2\overline{3}$ = 0.23333......... (1)

Multiplying equation (1) by 10, we get :

⇒ 10x = 2.3333......... (2)

Multiplying equation (1) by 100, we get :

⇒ 100x = 23.3333......... (3)

Subtracting equation (2) from equation (3), we get :

⇒ 100x - 10x = 23.3333... - 2.3333...

⇒ 90x = 21

⇒ x = $\frac{21}{90} = \frac{7}{30}$.

Hence, option 4 is the correct option.

Solving the given expression :

⇒ (2 - a)(6 - a)

⇒ 2(6 - a) - a(6 - a)

⇒ 12 - 2a - 6a + a²

⇒ a² - 8a + 12.

Hence, option 1 is the correct option.

Factorising the given expression :

⇒ 7 - 12x - 4x²

⇒ 7 - 14x + 2x - 4x²

⇒ 7(1 - 2x) + 2x(1 - 2x)

⇒ (1 - 2x)(7 + 2x).

Hence, option 3 is the correct option.

Solving the given expression :

$$\begin{aligned}\Rightarrow (0.01)^{-1/2} \\[1em] = \left(\frac{1}{100}\right)^{-1/2} \\[1em] = (100)^{1/2} \\[1em] = (10^2)^{1/2} \\[1em] = 10^{2 \times \frac{1}{2}} \\[1em] = 10.\end{aligned}$$

Hence, option 1 is the correct option.

We know that, the logarithm of a number to its own base is 1.

i.e. log_a a = 1.

∴ log_2x 2x = 1.

Hence, option 1 is the correct option.

Given,

In △ABC, BC = AC.

We know that, angles opposite to equal sides are equal.

Since BC = AC, the angles opposite to them are equal.

⇒ ∠A = ∠B = 64°

In △ABC, by angle sum property,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 64° + 64° + ∠C = 180°

⇒ ∠C = 180° - 128°

⇒ ∠C = 52°.

Hence, option 3 is the correct option.

Given,

Sides of the rectangular field = 30 m and 40 m.

By Pythagoras theorem,

Diagonal = $\sqrt{(\text{length})^2 + (\text{breadth})^2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Diagonal} = \sqrt{40^2 + 30^2} \\[1em] = \sqrt{1600 + 900} \\[1em] = \sqrt{2500} \\[1em] = 50 \text{ m}.\end{aligned}$$

Hence, option 1 is the correct option.

In a parallelogram, both pairs of opposite sides are equal and both pairs of opposite angles are equal.

∴ Both the statements (a) and (b) are true.

Hence, option 3 is the correct option.

Given,

Radius (r) = 2.5 cm and length of chord = 4.8 cm.

We know that, the perpendicular from the centre to a chord bisects the chord.

∴ Half the length of chord = $\frac{4.8}{2}$ = 2.4 cm.

By Pythagoras theorem,

Distance of chord from centre = $\sqrt{r^2 - \left(\frac{\text{chord}}{2}\right)^2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Distance} = \sqrt{(2.5)^2 - (2.4)^2} \\[1em] = \sqrt{6.25 - 5.76} \\[1em] = \sqrt{0.49} \\[1em] = 0.7 \text{ cm}.\end{aligned}$$

Hence, option 3 is the correct option.

The first five multiples of 7 are 7, 14, 21, 28 and 35.

By formula,

Mean = $\frac{\text{sum of observations}}{\text{number of observations}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{7 + 14 + 21 + 28 + 35}{5} \\[1em] = \frac{105}{5} \\[1em] = 21.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Base = 18 cm and height = 15 cm.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Area} = \frac{1}{2} \times 18 \times 15 \\[1em] = 9 \times 15 \\[1em] = 135 \text{ cm}^2.\end{aligned}$$

Hence, option 4 is the correct option.

Let the radius of the protractor be r cm.

The perimeter of a semi-circular protractor = πr + 2r = r(π + 2)

= $r\left(\frac{22}{7} + 2\right)$

Substituting values we get :

$$\begin{aligned}\Rightarrow 32.4 = r\left(\frac{22}{7} + 2\right) \\[1em] \Rightarrow 32.4 = r\left(\frac{22 + 14}{7}\right) \\[1em] \Rightarrow 32.4 = r \times \frac{36}{7} \\[1em] \Rightarrow r = \frac{32.4 \times 7}{36} \\[1em] \Rightarrow r = 6.3 \text{ cm}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Dimensions of the room = 12 m × 9 m × 8 m.

The longest rod that can be placed in a room is along the diagonal of the cuboid.

By formula,

Length of diagonal = $\sqrt{l^2 + b^2 + h^2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Diagonal} = \sqrt{12^2 + 9^2 + 8^2} \\[1em] = \sqrt{144 + 81 + 64} \\[1em] = \sqrt{289} \\[1em] = 17 \text{ m}.\end{aligned}$$

Hence, option 1 is the correct option.

We know that, for any angle θ,

⇒ sin²θ + cos²θ = 1.

∴ sin²30° + cos²30° = 1.

Hence, option 2 is the correct option.

The perpendicular distance of a point from the x-axis is given by the absolute value of its y-coordinate.

For point P(3, 5), the y-coordinate is 5.

∴ Perpendicular distance from x-axis = 5. So, Assertion (A) is true.

The x-coordinate of a point gives its perpendicular distance from the y-axis, not the x-axis. So, Reason (R) is false.

∴ A is true but R is false.

Hence, option 1 is the correct option.

Let ABCD be a cyclic quadrilateral in which the opposite sides AB and CD are equal, i.e. AB = CD.

To prove : AC = BD.

We know that, equal chords of a circle subtend equal arcs.

Since AB = CD,

⇒ arc AB = arc CD ......... (1)

Adding arc BC to both sides of equation (1), we get :

⇒ arc AB + arc BC = arc CD + arc BC

⇒ arc ABC = arc DCB

We know that, equal arcs of a circle have equal chords.

The chord of arc ABC is AC and the chord of arc DCB is DB.

⇒ AC = DB.

Hence, proved that the diagonals AC and BD are equal.

Given,

Principal (P) = ₹50000, Rate (r) = 12% p.a., Time = 2 years 5 months.

First we find the amount for 2 years (compounded annually).

By formula,

Amount = $P\left(1 + \frac{r}{100}\right)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 50000\left(1 + \frac{12}{100}\right)^2 \\[1em] = 50000\left(\frac{112}{100}\right)^2 \\[1em] = 50000 \times \frac{112}{100} \times \frac{112}{100} \\[1em] = ₹62720.\end{aligned}$$

Now, for the remaining 5 months, the principal becomes ₹62720.

By formula,

Simple Interest = $\frac{P \times r \times t}{100}$

For 5 months, time = $\frac{5}{12}$ year.

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{S.I.} = \frac{62720 \times 12 \times \frac{5}{12}}{100} \\[1em] = \frac{62720 \times 5}{100} \\[1em] = ₹3136.\end{aligned}$$

∴ Final Amount = 62720 + 3136 = ₹65856.

By formula,

Compound Interest = Amount - Principal

Substituting values we get :

⇒ C.I. = 65856 - 50000 = ₹15856.

Hence, the amount = ₹65,856 and the compound interest = ₹15,856.

Solving L.H.S.,

We know that,

⇒ sin(90° - A) = cos A

⇒ cos(90° - A) = sin A

Substituting these values in L.H.S., we get :

$$\begin{aligned}\Rightarrow \frac{\cos A}{\sin(90^\circ - A)} + \frac{\sin A}{\cos(90^\circ - A)} \\[1em] \Rightarrow \frac{\cos A}{\cos A} + \frac{\sin A}{\sin A} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

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

Hence, proved that $\frac{\cos A}{\sin(90^\circ - A)} + \frac{\sin A}{\cos(90^\circ - A)} = 2$.

Let us assume, to the contrary, that $\sqrt{3}$ is rational.

Then, $\sqrt{3}$ can be written in the form $\frac{a}{b}$, where a and b are co-prime integers and b ≠ 0.

⇒ $\sqrt{3} = \frac{a}{b}$

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow 3 = \frac{a^2}{b^2} \\[1em] \Rightarrow a^2 = 3b^2 \text{ ......... (1)}\end{aligned}$$

So, 3 divides a², which means 3 divides a.

Let a = 3c for some integer c.

Substituting a = 3c in equation (1), we get :

⇒ (3c)² = 3b²

⇒ 9c² = 3b²

⇒ b² = 3c²

So, 3 divides b², which means 3 divides b.

Thus, 3 divides both a and b. But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.

Hence, $\sqrt{3}$ is irrational.

Given,

⇒ $\left(x + \frac{1}{x}\right)^2 = 3$

Taking square root on both sides, we get :

⇒ $x + \frac{1}{x} = \sqrt{3}$ ......... (1) or $x + \frac{1}{x} = -\sqrt{3}$ ......... (2)

We know that,

⇒ $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$

Substituting value from equation (1), we get :

$$\begin{aligned}\Rightarrow x^3 + \frac{1}{x^3} = (\sqrt{3})^3 - 3(\sqrt{3}) \\[1em] = 3\sqrt{3} - 3\sqrt{3} \\[1em] = 0.\end{aligned}$$

Substituting value from equation (2), we get :

$$\begin{aligned}\Rightarrow x^3 + \frac{1}{x^3} = (-\sqrt{3})^3 - 3(-\sqrt{3}) \\[1em] = -3\sqrt{3} + 3\sqrt{3} \\[1em] = 0.\end{aligned}$$

Hence, proved that $\left(x^3 + \frac{1}{x^3}\right) = 0$.

Class-interval	Class mark	Frequency
-9 – 0	-4.5	0
0 – 9	4.5	9
9 – 18	13.5	15
18 – 27	22.5	6
27 – 36	31.5	12
36 – 45	40.5	18
45 – 54	49.5	0

Steps:

1. Find the class mark (mid-value) of each class.

2. Class mark = $\frac{\text{lower limit} + \text{upper limit}}{2}$

3. On a graph paper, mark class-marks along x-axis and frequencies along y-axis.

4. Take 1 cm along the x-axis = 5 units.

5. Take 1 cm along the y-axis = 2 units.

6. On the graph, mark points taking values of class-marks along x-axis and the values of their corresponding frequencies along y-axis.

7. Plot the points (-4.5, 0), (4.5, 9), (13.5, 15), (22.5, 6), (31.5, 12), (40.5, 18) and (49.5, 0). The classes -9 - 0 and 45 - 54 with frequency 0 are taken at both the ends to complete the polygon.

8. Join the plotted points in order by straight line segments to obtain the required frequency polygon.

Given,

⇒ $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}$

Rationalising the denominator of L.H.S. by multiplying the numerator and denominator by (2 + √3), we get :

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

Comparing $7 + 4\sqrt{3}$ with $a + b\sqrt{3}$, we get :

⇒ a = 7 and b = 4.

Hence, a = 7 and b = 4.

Given,

ABCD is a rhombus whose diagonals intersect at O and ∠OAB : ∠OBA = 2 : 3.

We know that, the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°.

Let ∠OAB = 2x and ∠OBA = 3x.

In △OAB, by angle sum property,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 2x + 3x + 90° = 180°

⇒ 5x = 90°

⇒ x = 18°.

∴ ∠OAB = 2x = 2 × 18° = 36°.

∴ ∠OBA = 3x = 3 × 18° = 54°.

Hence, the angles of △OAB are ∠OAB = 36°, ∠OBA = 54° and ∠AOB = 90°.

Given,

Length of the rectangular plot = 30 m and width = 18 m.

A margin of 2.5 m is left all around the plot.

∴ Length of the inner grassy region = 30 - 2(2.5) = 30 - 5 = 25 m.

∴ Width of the inner grassy region = 18 - 2(2.5) = 18 - 5 = 13 m.

By formula,

Area = length × width

Substituting values we get :

$\Rightarrow \text{Area} = 25 \times 13 = 325 \text{ m}^2.$

Hence, the area to be laid with grass = 325 m².

Given,

OAB is a quadrant of a circle with radius OA = 3.5 cm and OD = 2 cm.

The shaded portion is the area of the quadrant excluding the triangle OAD, where OA is the radius and OD is the base of the triangle.

By formula,

Area of quadrant = $\frac{1}{4}\pi r^2$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Area of quadrant} = \frac{1}{4} \times \frac{22}{7} \times (3.5)^2 \\[1em] = \frac{1}{4} \times \frac{22}{7} \times 12.25 \\[1em] = 9.625 \text{ cm}^2.\end{aligned}$$

Area of △OAD = $\frac{1}{2} \times \text{OA} \times \text{OD}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Area of △OAD} = \frac{1}{2} \times 3.5 \times 2 \\[1em] = 3.5 \text{ cm}^2.\end{aligned}$$

∴ Area of shaded portion = Area of quadrant - Area of △OAD

⇒ Area of shaded portion = 9.625 - 3.5 = 6.125 cm².

Hence, the area of the shaded portion = 6.125 cm².

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \sqrt{a^{-1}b} \cdot \sqrt{b^{-1}c} \cdot \sqrt{c^{-1}a} \\[1em] = \sqrt{\frac{b}{a}} \cdot \sqrt{\frac{c}{b}} \cdot \sqrt{\frac{a}{c}} \\[1em] = \sqrt{\frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}} \\[1em] = \sqrt{1} \\[1em] = 1.\end{aligned}$$

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

Hence, proved.

Given,

ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.

To prove : PQRS is a parallelogram.

Construction : Join the diagonal AC.

In △ABC, P and Q are the mid-points of AB and BC respectively.

By mid-point theorem,

⇒ PQ ∥ AC and PQ = $\frac{1}{2}$ AC ......... (1)

In △ADC, S and R are the mid-points of DA and CD respectively.

By mid-point theorem,

⇒ SR ∥ AC and SR = $\frac{1}{2}$ AC ......... (2)

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

⇒ PQ ∥ SR and PQ = SR.

Since one pair of opposite sides of quadrilateral PQRS is equal and parallel, PQRS is a parallelogram.

Hence, proved that the figure obtained by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram.

Let p(x) = x³ - 6x² + 12x - 7.

By trial, substituting x = 1, we get :

$\Rightarrow p(1) = (1)^3 - 6(1)^2 + 12(1) - 7$

= 1 - 6 + 12 - 7

= 0.

Since p(1) = 0, (x - 1) is a factor of p(x).

Dividing p(x) by (x - 1),

$$\begin{array}{l} x - 1\overline{\smash{\big)}\quad x^3- 6x^2 + 12x - 7\smash{\big(}} x^2 - 5x + 7 \\ \phantom{ac ++}\phantom{)}\underline{\underset{-}{}{x^3} \underset{+}{-}x^2} \\ \phantom{ac +)}\phantom{+++}{-5x^2 + 12x} \\ \phantom{ac +)}\phantom{+++)}\underline{\underset{+}{-}5x^2 \underset{-}{+} 5x} \\ \phantom{ac +)}\phantom{++++++)}{ 7x - 7} \\ \phantom{ac +}\phantom{++++++)}\underline{\underset{-}{}7x \underset{+}{-} 7} \\ \phantom{ac +)}\phantom{++++++++}{ 0 } \\ \end{array}$$

∴ x³ - 6x² + 12x - 7 = (x - 1)(x² - 5x + 7).

Hence, x³ - 6x² + 12x - 7 = (x - 1)(x² - 5x + 7).

Given,

⇒ log₁₀ y + 2 log₁₀ x = 2

Using the power rule of logarithms, 2 log₁₀ x = log₁₀ x².

⇒ log₁₀ y + log₁₀ x² = 2

Using the product rule of logarithms,

⇒ log₁₀ (yx²) = 2

Writing in exponential form,

⇒ yx² = 10²

⇒ yx² = 100

⇒ y = $\frac{100}{x^2}$.

Hence, y = $\frac{100}{x^2}$.

Given,

Edge of the large cube = 12 cm.

Edges of two smaller cubes = 6 cm and 8 cm.

Let the edge of the third smaller cube be a cm.

Since the large cube is melted and recast into three smaller cubes, the volume remains the same.

⇒ Volume of large cube = Sum of volumes of three smaller cubes

By formula,

Volume of cube = (edge)³

Substituting values we get :

$$\begin{aligned}\Rightarrow 12^3 = 6^3 + 8^3 + a^3 \\[1em] \Rightarrow 1728 = 216 + 512 + a^3 \\[1em] \Rightarrow 1728 = 728 + a^3 \\[1em] \Rightarrow a^3 = 1000 \\[1em] \Rightarrow a = \sqrt[3]{1000} \\[1em] \Rightarrow a = 10 \text{ cm}.\end{aligned}$$

Hence, the edge of the third smaller cube = 10 cm.

Given,

PQRS and PXYZ are two parallelograms of equal area.

To prove: YR ∥ QZ.

Since the two parallelograms have equal areas,

ar(PQRS) = ar(PXYZ).

Subtracting the area of the common quadrilateral PQOZ from both sides, we get:

⇒ ar(PQRS) − ar(PQOZ) = ar(PXYZ) − ar(PQOZ)

⇒ ar(ZORS) = ar(QXYO).

Let,

ar(ZORS) = ar(QXYO) = x.

Since QX lies on PX, OY lies on ZY and PX ∥ ZY,

⇒ QX ∥ OY.

Also, QO lies on QR, XY ∥ PZ and QR ∥ PS, where P, Z and S are collinear.

⇒ QO ∥ XY.

Therefore, QXYO is a parallelogram.

Similarly, ZO lies on ZY, RS ∥ PQ and PQ lies on PX.

Since PX ∥ ZY,

⇒ ZO ∥ RS.

Also, OR lies on QR, ZS lies on PS and QR ∥ PS.

⇒ OR ∥ ZS.

Therefore, ZORS is a parallelogram.

The diagonal QY divides parallelogram QXYO into two triangles of equal area.

⇒ ar(△QOY) = $\frac{1}{2}$ ar(QXYO)

⇒ ar(△QOY) = $\frac{x}{2}$.

Similarly, the diagonal RZ divides parallelogram ZORS into two triangles of equal area.

⇒ ar(△ROZ) = $\frac{1}{2}$ ar(ZORS)

⇒ ar(△ROZ) = $\frac{x}{2}$.

Therefore,

⇒ ar(△QOY) = ar(△ROZ).

Adding ar(△ORY) to both sides, we get:

⇒ ar(△QOY) + ar(△ORY) = ar(△ROZ) + ar(△ORY)

⇒ ar(△QYR) = ar(△ZYR).

The triangles QYR and ZYR are on the same base YR, lie on the same side of YR and have equal areas.

Therefore, they lie between the same parallels.

∴ YR ∥ QZ.

Hence, proved that YR ∥ QZ.

Let the length of the room be l m and the breadth be b m.

∴ Original area = lb m².

According to the first condition, when length and breadth are each increased by 1 m :

⇒ (l + 1)(b + 1) = lb + 21

⇒ lb + l + b + 1 = lb + 21

⇒ l + b = 20 ......... (1)

According to the second condition, when length is increased by 1 m and breadth decreased by 1 m :

⇒ (l + 1)(b - 1) = lb - 5

⇒ lb - l + b - 1 = lb - 5

⇒ -l + b = -4

⇒ l - b = 4 ......... (2)

Adding equations (1) and (2), we get :

⇒ 2l = 24

⇒ l = 12.

Substituting l = 12 in equation (1), we get :

⇒ 12 + b = 20

⇒ b = 8.

∴ Area of the room = lb = 12 × 8 = 96 m².

Hence, the area of the room = 96 m².

Given,

Population three years ago = 50000.

Annual increase during three successive years = 5%, 8% and 10%.

By formula,

Present population = $P\left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right)\left(1 + \frac{r_3}{100}\right)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Present population} = 50000\left(1 + \frac{5}{100}\right)\left(1 + \frac{8}{100}\right)\left(1 + \frac{10}{100}\right) \\[1em] = 50000 \times \frac{105}{100} \times \frac{108}{100} \times \frac{110}{100} \\[1em] = 50000 \times 1.05 \times 1.08 \times 1.10 \\[1em] = 62370.\end{aligned}$$

Hence, the present population of the city = 62,370.

Given,

AY ⊥ ZY, BY ⊥ XY, AY = ZY and BY = XY.

Since AY ⊥ ZY, ∠AYZ = 90°.

Since BY ⊥ XY, ∠BYX = 90°.

∴ ∠AYZ = ∠BYX ......... (1)

Adding ∠AYX to both sides of equation (1), we get :

⇒ ∠AYZ + ∠AYX = ∠BYX + ∠AYX

⇒ ∠ZYX = ∠AYB ......... (2)

Now, in △ZYX and △AYB :

⇒ ZY = AY [Given]

⇒ ∠ZYX = ∠AYB [From equation (2)]

⇒ XY = BY [Given]

∴ △ZYX ≅ △AYB [By SAS congruence criterion]

⇒ ZX = AB [By C.P.C.T.]

Hence, proved that AB = ZX.

Given,

⇒ 13 sin θ = 5

⇒ sin θ = $\frac{5}{13}$.

We know that, sin θ = $\frac{\text{perpendicular}}{\text{hypotenuse}}$.

∴ perpendicular = 5 and hypotenuse = 13.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow \text{base} = \sqrt{13^2 - 5^2} \\[1em] = \sqrt{169 - 25} \\[1em] = \sqrt{144} \\[1em] = 12.\end{aligned}$$

∴ cos θ = $\frac{12}{13}$ and tan θ = $\frac{5}{12}$.

Substituting these values in the given expression, we get :

$$\begin{aligned}\Rightarrow \frac{5\sin\theta - 2\cos\theta}{\tan\theta} = \frac{5 \times \frac{5}{13} - 2 \times \frac{12}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{25}{13} - \frac{24}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{1}{13}}{\frac{5}{12}} \\[1em] = \frac{1}{13} \times \frac{12}{5} \\[1em] = \frac{12}{65}.\end{aligned}$$

Hence, the value of $\frac{5\sin\theta - 2\cos\theta}{\tan\theta} = \frac{12}{65}$.

Given,

Radius of the circle (r) = 17 cm.

AB and CD are two chords with the lengths AB = 30 cm and CD = 16 cm, lying on opposite sides of the centre.

We know that, the perpendicular from the centre to a chord bisects the chord.

OE is perpendicular to AB,

AE = EB = $\frac{AB}{2} = \frac{30}{2}$ = 15 cm.

Let distance of this chord from centre be d₁.

In right triangle AEO,

By Pythagoras theorem,

⇒ AO² = AE² + EO²

⇒ EO² = AO² - AE²

⇒ (d₁)² = (17)² - (15)²

$$\begin{aligned}\Rightarrow d_1 = \sqrt{17^2 - 15^2} \\[1em] = \sqrt{289 - 225} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ cm}.\end{aligned}$$

OF is perpendicular to CD.

CF = DF = $\frac{CD}{2} = \frac{16}{2}$ = 8 cm.

Let distance of this chord from centre be d₂

In right triangle CFO,

By Pythagoras theorem,

⇒ OC² = CF² + OF²

⇒ OF² = OC² - CF²

⇒ (d₂)² = (17)² - (8)²

$$\begin{aligned}\Rightarrow d_2 = \sqrt{17^2 - 8^2} \\[1em] = \sqrt{289 - 64} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ cm}.\end{aligned}$$

Since the chords lie on opposite sides of the centre, the distance between them = d₁ + d₂.

⇒ Distance between the chords = 8 + 15 = 23 cm.

Hence, the distance between the chords = 23 cm.

Given,

In △ABC, AC > AB and AD is the bisector of ∠A.

To prove : ∠ADC > ∠ADB.

Since AC > AB, the angle opposite to the larger side is larger.

⇒ ∠B > ∠C ......... (1)

Since AD is the bisector of ∠A,

⇒ ∠BAD = ∠CAD ......... (2)

In △ABD, by angle sum property,

⇒ ∠ADB = 180° - ∠B - ∠BAD ......... (3)

In △ACD, by angle sum property,

⇒ ∠ADC = 180° - ∠C - ∠CAD ......... (4)

From equation (2), ∠BAD = ∠CAD.

From equation (1), ∠B > ∠C, so -∠B < -∠C.

∴ From equations (3) and (4), since ∠B > ∠C,

⇒ 180° - ∠C - ∠CAD > 180° - ∠B - ∠BAD

⇒ ∠ADC > ∠ADB.

Hence, proved that ∠ADC > ∠ADB.

Let the required point on the x-axis be P(x, 0), since the y-coordinate of any point on the x-axis is 0.

Given, P is equidistant from A(2, -5) and B(-2, 9).

⇒ PA = PB

⇒ PA² = PB²

By distance formula,

$$\begin{aligned}\Rightarrow (x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2 \\[1em] \Rightarrow (x - 2)^2 + 25 = (x + 2)^2 + 81\end{aligned}$$

Expanding both sides, we get :

$$\begin{aligned}\Rightarrow x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \\[1em] \Rightarrow -4x + 29 = 4x + 85 \\[1em] \Rightarrow -8x = 56 \\[1em] \Rightarrow x = -7.\end{aligned}$$

∴ The required point is (-7, 0).

Hence, the point on the x-axis equidistant from A and B is (-7, 0).

Given,

In △ABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC in F.

To prove : AF = $\frac{1}{3}$ AC.

Construction : Through D, draw DG ∥ BF, meeting AC at G.

In △ADG, E is the mid-point of AD and EF ∥ DG.

By converse of mid-point theorem, F is the mid-point of AG.

⇒ AF = FG ......... (1)

In △BCF, D is the mid-point of BC and DG ∥ BF.

By converse of mid-point theorem, G is the mid-point of CF.

⇒ FG = GC ......... (2)

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

⇒ AF = FG = GC ......... (3)

Now, AC = AF + FG + GC.

From equation (3), AC = AF + AF + AF = 3 AF.

⇒ AF = $\frac{1}{3}$ AC.

Hence, proved that AF = $\frac{1}{3}$ AC.

Given, one class-interval is 9 - 12, so the class size = 12 - 9 = 3.

Since 12 is not included, we use the exclusive method of classification.

The minimum value is 6 and the maximum value is 23, so the classes are 6 - 9, 9 - 12, 12 - 15, 15 - 18, 18 - 21 and 21 - 24.

Counting the number of observations falling in each class, we get the following frequency table :

Class-interval (Age in years)	Tally marks	Frequency
6 – 9	IIII	5
9 – 12	IIII	4
12 – 15	IIII	4
15 – 18	IIII II	7
18 – 21	III	3
21 – 24	IIII II	7
Total		30

Hence, the required frequency table is constructed as shown above.