aekaakee: November 2020

Trigonometry

Prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

To prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

i.e. $\;$ To prove that $\;$ $\dfrac{1}{\text{cosec }A - \cot A} + \dfrac{1}{\text{cosec }A + \cot A} = \dfrac{1}{\sin A} + \dfrac{1}{\sin A}$

i.e. $\;$ To prove that $\;$ $\dfrac{\text{cosec }A + \cot A + \text{cosec }A - \cot A}{\left(\text{cosec }A + \cot A\right) \left(\text{cosec }A - \cot A\right)} = \dfrac{2}{\sin A}$

i.e. $\;$ To prove that $\;$ $\dfrac{2 \; \text{cosec }A}{\text{cosec}^2 A - \cot^2 A} = 2 \; \text{cosec }A$

Now, $\;$ $1 + \cot^2 A = \text{cosec}^2 A$

$\implies$ $\text{cosec}^2 A - \cot^2 A = 1$

$\therefore \;$ To prove that $\;$ $2 \; \text{cosec }A = 2 \; \text{cosec A}$ $\;\;$ which is true.

$\therefore \;$ $\dfrac{1}{\text{cosec }A - \cot A} - \dfrac{1}{\sin A} = \dfrac{1}{\sin A} - \dfrac{1}{\text{cosec }A + \cot A}$

Hence proved.

Constructions

Construct a regular hexagon of side $5.5 \; cm$, with an inscribed circle. Measure the radius of the incircle.

Construct a regular hexagon $ABCDEF$ of side $5.5 \; cm$.

Construct the angle bisectors of the interior angles $A$ and $B$. The angle bisectors meet at point $I$.

From point $I$, draw $IP$ perpendicular to $AP$.

With $I$ as center and $IP$ as radius, draw a circle which touches all the sides of the regular hexagon.

Radius of the incircle $= 4.7 \; cm$

Similarity

In the figure, $\;$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\triangle TSR\right)} = \dfrac{3}{1}$, $\;$ $TS \parallel QR$.

Find

$PS : SR$
$TS : QR$
$\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle PQR\right)$
$\text{area } \left(\triangle PTS\right) : \text{area} \left(\text{quadrilateral } TQRS\right)$

Let height of $\triangle PTS = $ height of $\triangle TSR = h$

From the figure,

$\text{area} \left(\triangle PTS\right) = \dfrac{1}{2} \times PS \times h$

$\text{area} \left(\triangle TSR\right) = \dfrac{1}{2} \times SR \times h$

Given: $\;$ $\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle TSR\right) = 3 : 1$

i.e. $\;$ $\dfrac{1}{2} \times PS \times h: \dfrac{1}{2} \times SR \times h = 3 : 1$

i.e. $\;$ $PS : SR = 3 : 1$

Given: $\;$ $TS \parallel QR$

$PR$ is a transversal

$\therefore \;$ $\angle PST = \angle PRQ$ $\;\;\;$ [corresponding angles] $\;\;\; \cdots \; (1)$

In $\triangle PTS$ and $\triangle PQR$,

$\angle P = \angle P$ $\;\;\;$ [common angle]

$\angle PST = \angle PRQ$ $\;\;\;$ [by (1)]

$\therefore \;$ $\triangle PTS \sim \triangle PQR$ $\;\;\;$ [by AA axiom of similarity]

$\implies$ $\dfrac{TS}{QR}= \dfrac{PS}{PR} = \dfrac{PT}{PQ}$ $\;\;\; \cdots (2)$

$\;\;\;$ [corresponding sides of similar triangles are in proportion]

$\because \;$ $\dfrac{PS}{SR} = \dfrac{3}{1}$ $\;\;\;$ [part (a)]

$\therefore \;$ $\dfrac{PS}{PR} = \dfrac{PS}{PS + SR} = \dfrac{3}{3 + 1} = \dfrac{3}{4}$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(2)$ becomes

$TS : QR = 3 : 4$

$\because \;$ $\triangle PTS \sim \triangle PQR$

$\implies$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \triangle PQR} = \dfrac{TS^2}{QR^2} = \dfrac{PS^2}{PR^2} = \dfrac{PT^2}{PQ^2}$

[areas of two similar triangles are proportional to the squares of their corresponding sides]

$\therefore \;$ $\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle PQR\right) = 3^2 : 4^2 = 9 : 16$

$\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\text{quadrilateral } TQRS\right)} = \dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\triangle PQR\right) - \text{area } \left(\triangle PTS\right)}$

i.e. $\;$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\text{quadrilateral } TQRS\right)} = \dfrac{9}{16 - 9}$

$\therefore \;$ $\text{area } \left(\triangle PTS\right) : \text{area} \left(\text{quadrilateral } TQRS\right) = 9 : 7$

Locus

Construct $\triangle PQR$ where $QR = 5 \; cm$, $\angle Q =45^\circ$ and $PQ = 6.5 \; cm$.
Find by construction points $X$ and $Y$ which are equidistant from $Q$ and $R$, and are $4 \; cm$ from $Q$.

$\triangle PQR$ is constructed with $QR = 5 \; cm$, $\angle Q =45^\circ$ and $PQ = 6.5 \; cm$.

Locus $\left(L_1\right)$ of points which are equidistant from $Q$ and $R$ lie on the perpendicular bisector $AB$ of $QR$.

Locus $\left(L_2\right)$ of points which are at a distance of $4 \; cm$ from $Q$ lie on the circle $C$ with center at $Q$ and radius $4 \; cm$.

Points $X$ and $Y$ lie at the intersection of $L_1$ and $L_2$.

Circle

$AB$ is the diameter of a circle. $\angle PAB = 35^\circ$, $\angle PQB = 25^\circ$.

Find $\angle RAB$, $\angle PRA$, $\angle ARB$, $\angle PRB$

Given: $AB$ is the diameter of the circle.

$\therefore \;$ $\angle APB = 90^\circ$ $\;\;\;$ [angle in a semi-circle]

$\angle APB + \angle BPQ = 180^\circ$ $\;\;\;$ [linear pair]

$\therefore \;$ $\angle BPQ = 180^\circ - \angle APB$

i.e. $\;$ $\angle BPQ = 180^\circ - 90^\circ = 90^\circ$

In $\triangle BPQ$,

$\angle BPQ + \angle PQB + \angle QBP = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $\angle QBP = 180^\circ - \left(\angle BPQ + \angle PQB\right)$

Given: $\angle PQB = 25^\circ$

$\therefore \;$ $\angle QBP = 180^\circ - \left(90^\circ + 25^\circ\right) = 65^\circ$

$ARBP$ is a cyclic quadrilateral.

$\therefore \;$ $\angle PAR = \angle QBP$

[external angle of a cyclic quadrilateral is equal to internal opposite angle]

$\therefore \;$ $\angle PAR = 65^\circ$

But $\;$ $\angle PAR = \angle PAB + \angle RAB$

$\therefore \;$ $\angle RAB = \angle PAR - \angle PAB$

Given: $\;$ $\angle PAB = 35^\circ$

i.e. $\;$ $\angle RAB = 65^\circ - 35^\circ = 30^\circ$

$\angle PRB = \angle PAB$ $\;\;\;$ [angles in the same segment of a circle are equal]

$\therefore \;$ $\angle PRB = 35^\circ$

$\because \;$ $AB$ is the diameter of the circle.

$\therefore \;$ $\angle ARB = 90^\circ$ $\;\;\;$ [angle in a semi-circle]

$\angle ARB = \angle PRA + \angle PRB$

$\therefore \;$ $\angle PRA = \angle ARB - \angle PRB$

i.e. $\;$ $\angle PRA = 90^\circ - 35^\circ = 55^\circ$

Statistics

Plot a cumulative distribution curve for the following data:

Weight in kg	$40 - 45$	$45 - 50$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$
Pupils	$7$	$15$	$28$	$20$	$12$	$10$	$6$	$2$

Estimate the median and the quartiles.

If $50 \; kg$ is standard weight, find the number of pupils that are overweight.

Weight in kg (Class Interval)	Pupils (Frequency)	Cumulative Frequency
$40 - 45$	$7$	$7$
$45 - 50$	$15$	$22$
$50 - 55$	$28$	$50$
$55 - 60$	$20$	$70$
$60 - 65$	$12$	$82$
$65 - 70$	$10$	$92$
$70 - 75$	$6$	$98$
$75 - 80$	$2$	$100$

Number of students $= N = \Sigma f_i = 100$

Taking weight (in kg) (class intervals) along X-axis and cumulative frequency along Y-axis, draw an ogive.

Median $= \left(\dfrac{N}{2}\right)^{th} \text{value} = \left(\dfrac{100}{2}\right)^{th} \text{value} = 50^{th} \; \text{value} = 55$

$\therefore \;$ Median number of pupils $= 55$

Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th} \text{value} = \left(\dfrac{100}{4}\right)^{th} \text{value} = 25^{th} \; \text{value} = 51$

$\therefore \;$ Lower quartile $= 51$

Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th} \text{value} = \left(\dfrac{3 \times 100}{4}\right)^{th} \text{value} = 75^{th} \; \text{value} = 62$

$\therefore \;$ Upper quartile $= 62$

Given: Standard weight $= 50 \; kg$

From the ogive, number of pupils who weigh $50 \; kg = 22$

$\therefore \;$ Number of pupils who are overweight $= 100 - 22 = 78$

Similarity - Maps and Models

A map is drawn to a scale of $1:4000$.

Find the length on map (in cm) that represents $5 \; km$ of ground length.
Find the area on ground (in sq.m) that is represented by $4 \; cm^2$ on this map.

Scale of map $= 1 : 4000$

i.e. $\;$ $1 \; cm$ on map $\equiv$ $4000 \; cm$ on ground

i.e. $\;$ $1 \; cm$ on map $\equiv$ $4000 \times 10^{-5} \; km$ on ground $\;\;$ $\left[\because \; 1 \; cm = 10^{-5} \; km\right]$

i.e. $\;$ $1 \; cm$ on map $\equiv$ $4 \times 10^{-2} \; km$ on ground

$\therefore \;$ $x \; cm$ on map $\equiv$ $5 \; km$ on ground

$\implies$ $x = \dfrac{5}{4 \times 10^{-2}} \; cm = 125 \; cm$

$\therefore \;$ $125 \; cm$ on map $\equiv$ $5 \; km$ on ground

$1 \; cm$ on map $\equiv$ $4000 = 4 \times 10^3 \; cm$ on ground

$\therefore \;$ $1 \; cm^2$ on map $\equiv$ $16 \times 10^6 \; cm^2$ on ground

$\therefore \;$ $4 \; cm^2$ on map $\equiv$ $p \; cm^2$ on ground

i.e. $\;$ $p = 4 \times 16 \times 10^6 = 64 \times 10^6 \; cm^2$ on ground

Now, $100 \; cm = 1 \; m$

i.e. $\;$ $10^4 \; cm^2 = 1 \; m^2$

$\therefore \;$ $64 \times 10^6 \; cm^2 = \dfrac{64 \times 10^6}{10^4} = 6400 \; m^2$

$\therefore \;$ $4$ sq.cm on map $\equiv$ $6400$ sq.m on ground

Coordinate Geometry

Answer the entire question on a graph paper.

Take a scale of $1 \; cm = 1 $ unit on both the axes.

Plot point $A \left(6,6\right)$ and reflect it in the line $x = 0$ to obtain the point $A'$.

Plot point $B \left(-3, 3\right)$ and reflect it in the $Y$ axis to obtain the point $B'$.

Plot point $C \left(0,3\right)$ and reflect it in the line $y = -1$ to obtain the point $C'$.

Join $A, \; C, \; A', \; B, \; C', \; B', \; A$ to form a geometric figure. Assign a name to the figure.

Identify a point on the figure that is invariant on reflection in the line $x = 0$.

Point $A \left(6, 6\right)$ reflected in the line $x = 0$ i.e. $Y$ axis gives the point $A' \left(-6,6\right)$.

Point $B \left(-3, 3\right)$ reflected in the $Y$ axis gives the point $B' \left(3,3\right)$.

Point $C \left(0, 3\right)$ reflected in the line $y = -1$ gives the point $C' \left(0, 5\right)$.

The geometric figure $ACA'BC'B'A$ is an arrow.

The points $C \left(0, 3\right)$ and $C' \left(0, -5\right)$ on the figure are invariant on reflection in the line $x = 0$.

Circle

$BA$ and $BC$ are tangents to the given circle (with center $O$) from the point $B$.

$\angle AOB = 65^\circ$. Find $\angle AOC$, $\angle ADC$ and $\angle ABC$.

If two tangents $BA$ and $BC$ are drawn to a circle (with center $O$) from an exterior point ($B$), then

the two tangents are equal in length $\;$ i.e. $\;$ $BA = BC$

the tangents subtend equal angles at the center of the circle $\;$ i.e. $\;$ $\angle AOB = \angle COB$

the tangents are equally inclined to the line joining the point $\left(B\right)$ and the center of the circle $\left(O\right)$, $\;$ i.e. $\;$ $\angle ABO = \angle CBO$

$\angle COB = \angle AOB = 65^\circ$ $\;\;$ [by (2.)]

$\angle AOC = \angle AOB + \angle COB = 65^\circ + 65^\circ = 130^\circ$

Now, $\angle AOC = 2 \angle ADC$ $\;\;$ [angle at the center of a circle is twice the angle at remaining circumference]

$\therefore \;$ $\angle ADC = \dfrac{\angle AOC}{2} = \dfrac{130^\circ}{2} = 65^\circ$

$OA = OC = $ radii of the same circle

$OA \perp AB$; $\;$ $OC \perp CB$ $\;\;\;$ [the tangent at any point of a circle and the radius through this point are perpendicular to each other]

$\therefore \;$ $\angle OAB = \angle OCB = 90^\circ$

In right $\triangle OAB$, $\;$ $\angle OAB + \angle AOB + \angle ABO = 180^\circ$ $\;\;$ [sum of angles of a triangle]

i.e. $\;$ $90^\circ + 65^\circ + \angle ABO = 180^\circ$

$\implies$ $\angle ABO = 25^\circ$

Now, $\angle CBO = \angle ABO = 25^\circ$ $\;\;$ [by (3.)]

$\therefore \;$ $\angle ABC = \angle ABO + \angle CBO = 25^\circ + 25^\circ = 50^\circ$

$\therefore \;$ $\angle AOC = 130^\circ$, $\;$ $\angle ADC = 65^\circ$, $\;$ $\angle ABC = 50^\circ$

Section Formula

Find the ratio in which the line $x = 0$ divides line $AB$ where $A = \left(-4,4\right)$ and $B = \left(2, 8\right)$. Also find the coordinates of the point where $AB$ is divided by the Y axis.

Given: $\;$ $A \left(x_1, y_1\right) = \left(-4,4\right)$; $\;$ $B \left(x_2, y_2\right) = \left(2, 8\right)$

Any point on the line $x = 0$ (i.e. the Y axis) is $P \left(0,p\right)$

Let point $P$ divide line $AB$ in the ratio $k : 1$

Then by section formula, $\;$ $0 = \dfrac{2k - 4}{k + 1}$

i.e. $\;$ $2k - 4 = 0$

i.e. $\;$ $2k = 4$ $\implies$ $k = 2$

$\therefore \;$ Any point on the line $x = 0$ divides the line $AB$ in the ratio $2 : 1$

Now, by section formula, $\;$ $p = \dfrac{2 \times 8 + 1 \times 4}{2 + 1} = \dfrac{16 + 4}{3} = \dfrac{20}{3}$

$\therefore \;$ The coordinates of the point where $AB$ is divided by the Y axis is $\left(0, \dfrac{20}{3}\right)$

Statistics

Plot a histogram to represent the following data, using the class marks given in cm. Estimate the mode and list the modal class.

Height	$125$	$135$	$145$	$155$	$165$	$175$	$185$
Pupils	$8$	$14$	$24$	$30$	$20$	$10$	$6$

Height (Class Marks)	Class Interval	Pupils (Frequency)
$125$	$120 - 130$	$8$
$135$	$130 - 140$	$14$
$145$	$140 - 150$	$24$
$155$	$150 - 160$	$30$
$165$	$160 - 170$	$20$
$175$	$170 - 180$	$10$
$185$	$180 - 190$	$6$

In the histogram, the highest rectangle represents the maximum frequency (or modal class).

Inside this rectangle, draw lines $AC$ and $BD$ diagonally from the upper corners $A$ and $D$ of adjacent rectangles.

Let the point of intersection of $AC$ and $BD$ be $K$.

Draw $KL$ perpendicular to the horizontal axis.

The value of point $L$ on the horizontal axis represents the mode and the class interval in which point $L$ lies is the modal class.

From the figure, mode $= 154$; $\;$ modal class $= 150 - 160$

Trigonometry

From the top of a building $60 \; m$ high, angles of elevation and depression were observed to the top and base of a tower to be $30^\circ$ and $60^\circ$ respectively. Find the height of the tower, given both the structures are on the same level ground and opposite each other.

$AB = $ Building of height $60 \; m$

$OT = $ Tower

$AO = $ Distance between the building and the tower

From $B$ draw $BP \perp OT$

Then, $\;$ $AB = OP = 60 \; m$ $\;$ and $\;$ $AO= BP$

In $\triangle ABO$, $\;$ $\dfrac{AB}{AO} = \tan 60^\circ$

$\implies$ $AO = \dfrac{AB}{\tan 60^\circ} = \dfrac{60}{\sqrt{3}} = 20 \sqrt{3} \; m$

$\therefore \;$ $BP = AO = 20 \sqrt{3} \; m$

In $\triangle PTB$, $\;$ $\dfrac{PT}{BP} = \tan 30^\circ$

$\implies$ $PT = BP \times \tan 30^\circ = 20 \sqrt{3} \times \dfrac{1}{\sqrt{3}} = 20 \; m$

Height of tower $= OP + PT = 60 + 20 = 80 \; m$

Arithmetic Progression

The fourth term of an A.P is $11$ and the $8^{th}$ term exceeds twice the $4^{th}$ term by $5$. Find the sum of the first fifty terms of the A.P.

Let the first term of A.P $= a$

Let the common difference of A.P $= d$

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$

Fourth term of A.P $= t_4 = a + 3d = 11$ (Given) $\;\;\; \cdots \; (1)$

Eighth term of A.P $= t_8 = a + 7d$

Given: $\;$ $t_8 = 2 \times t_4 + 5$

i.e. $\;$ $a + 7d = 2 \left(a + 3d\right) + 5$

i.e. $\;$ $a + 7d = 2a + 6d + 5$

i.e. $\;$ $a - d = -5$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously gives,

$4d = 16$ $\implies$ $d = 4$

Substituting the value of $d$ in equation $(2)$ gives $\;$ $a = -1$

Sum to $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

$\therefore \;$ Sum to $50$ terms of A.P $= S_{50} = \dfrac{50}{2} \left[2 \times \left(-1\right) + \left(50 - 1\right) \times 4\right]$

i.e. $\;$ $S_{50} = 25 \left[-2 + 196\right] = 4850$

Linear Inequations

Solve the linear inequation and represent the solution set on a number line:
$-3 \left(x - 7\right) \geq 15 - 7x > \left(\dfrac{x + 1}{3}\right), \;\; x \in R$

Consider $\;$ $-3 \left(x - 7\right) \geq 15 - 7x$

i.e. $\;$ $-3x + 21 \geq 15 - 7x$

i.e. $\;$ $4x \geq -6$

i.e. $\;$ $x \geq -\dfrac{3}{2}$ $\implies$ $- \dfrac{3}{2} \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $15 - 7x > \dfrac{x + 1}{3}$

i.e. $\;$ $45 - 21x > x + 1$

i.e. $\;$ $44 > 22 x$

i.e. $\;$ $2 > x$ $\implies$ $x < 2$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;\;$ $- \dfrac{3}{2} \leq x < 2$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -\dfrac{3}{2} \leq x < 2, \; x \in R \right\}$

Arithmetic Progression

Find the sum of the first $51$ terms of an arithmetic progression, whose $2^{nd}$ and $3^{rd}$ terms are $14$ and $18$ respectively.

Let $a = $ first term of A.P; $\;$ $d = $ common ratio

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$

$2^{nd}$ term of A.P $= t_2 = a + d = 14$ $\;\;\; \cdots \; (1)$

$3^{rd}$ term of A.P $= t_3 = a + 2d = 18$ $\;\;\; \cdots \; (2)$

Subtracting equations $(1)$ and $(2)$ we get, $\;\;\;$ $d = 4$

Substituting the value of $d$ in equation $(1)$ gives $\;\;\;$ $a = 14 - 4 = 10$

Sum of $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

$\therefore \;$ Sum of first $51$ terms $= S_{51} = \dfrac{51}{2} \left[2 \times 10 + \left(51 - 1\right) \times 4\right]$

i.e. $\;$ $S_{51} = \dfrac{51}{2} \left[20 + 200\right]$

i.e. $\;$ $S_{51} = \dfrac{51}{2} \times 220 = 5610$

Commercial Mathematics - Shares and Dividends

An investment of ₹ $8800$ is made on ₹ $100$ shares at $10\%$ premium paying $22\%$ dividend. Find the return percent on the investment and the dividend earned in one year.

Nominal value of $1$ share $= N.V = $ ₹ $100$

Market value of $1$ share $= M.V = $ ₹ $100 + $ ₹ $10 = $ ₹ $110$

Investment $=$ ₹ $8800$

$\therefore \;$ Number of shares bought $= \dfrac{\text{Investment}}{\text{M.V of 1 share}} = \dfrac{8800}{110}= 80$

Rate of dividend $= 22 \%$ (Given)

$\therefore \;$ Income (i.e. dividend) earned $=$ Number of shares $\times$ Rate of dividend $\times$ N.V

i.e. $\;$ Dividend earned $= 80 \times \dfrac{22}{100} \times 100 = $ ₹ $1760$

i.e. $\;$ ₹ $1760$ is the income obtained on investing ₹ $8800$

$\therefore \;$ Return $\%$ on investment $= \dfrac{1760}{8800} \times 100 = 20 \%$

Trigonometry

The angle of elevation, observed from ground to the top of a building $60 \; m$ high, is $30^\circ$. On walking '$x$' meters towards the building, the angle of elevation to the top changes to $60^\circ$. Find '$x$' to the nearest meter.

$AB =$ Building of height $60 \; m$

$D = $ Initial position of observation

$C = $ Final position of observation

$CD = x \;$ meter

In $\triangle ABC$,

$\tan 60^\circ = \dfrac{AB}{BC}$

i.e. $\;$ $BC = \dfrac{60}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$

In $\triangle ABD$,

$\tan30^\circ = \dfrac{AB}{BD} = \dfrac{AB}{BC + CD} = \dfrac{AB}{BC + x}$

i.e. $\;$ $BC + x = \dfrac{60}{1 / \sqrt{3}} = 60 \sqrt{3}$

i.e. $\; $ $x = 60 \sqrt{3} - BC$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$ equation $(2)$ becomes,

$x = 60 \sqrt{3} - \dfrac{60}{\sqrt{3}} = \dfrac{180 - 60}{\sqrt{3}} = \dfrac{120}{\sqrt{3}}= 40 \sqrt{3} = 69.28 \;$ meter

i.e. $\;$ $x = 69 \; m$ (to the nearest meter)

Ratio and Proportion

If $\;$ $\dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4}$, $\;$ find

$m : n$
$\dfrac{2m^2 + 11n^2}{2m^2 - 11n^2}$

Given: $\;$ $\dfrac{4m + 3n}{4m - 3n} = \dfrac{7}{4}$

By componendo and dividendo we have,

$\dfrac{4m + 3n + 4m - 3n}{4m + 3n - 4m + 3n} = \dfrac{7 + 4}{7 - 4}$

i.e. $\;$ $\dfrac{8m}{6n} = \dfrac{11}{3}$

i.e. $\;$ $\dfrac{m}{n} = \dfrac{11}{3} \times \dfrac{6}{8}$

i.e. $\;$ $m : n = 11 : 4$
By componendo and dividendo we have,

$\begin{aligned} \dfrac{2m^2 + 11n^2}{2m^2 - 11n^2} & = \dfrac{2m^2 + 11n^2 + 2m^2 - 11n^2}{2m^2 + 11n^2 - 2m^2 + 11n^2} \\\\ & = \dfrac{4m^2}{22 n^2} \\\\ & = \dfrac{2}{11} \times \left(\dfrac{m}{n}\right)^2 \\\\ & = \dfrac{2}{11} \times \left(\dfrac{11}{4}\right)^2 \;\;\; [\text{from part (1)}] \\\\ & = \dfrac{11}{8} \end{aligned}$

Geometric Progression

Find the sum of $\;$ $2 + 6 + 18 + 54 + \cdots + 4374$

$2, \; 6, \; 18, \; 54, \; \cdots, \; 4374$ is a geometric progression (G.P) with

first term $= a = 2$, $\;$ common ratio $= r = 3$, $\;$ $n^{th}$ term $= t_n = 4374$

$n^{th}$ term of a G.P $= t_n = ar^{n-1}$

$\therefore \;$ We have, $\;$ $4374 = 2 \times 3^{n-1}$

i.e. $\;$ $3^{n - 1} = \dfrac{4374}{2} = 2187 = 3^7$

$\implies$ $n - 1 = 7$ $\implies$ $n = 8$

i.e. $\;$ Number of terms in the given G.P $= n = 8$

Now, sum to n terms of a G.P $= S_n = \dfrac{a \left(r^n - 1\right)}{r - 1}$, $\;$ $r > 1$

$\therefore \;$ Sum of terms of given G.P is

$S_8 = \dfrac{2 \left(3^{8}-1\right)}{3-1} = \dfrac{2 \times \left(3^8 - 1\right)}{2} = 6560$

$\therefore \;$ $2 + 6 + 18 + 54 + \cdots + 4374 = 6560$

Ratio and Proportion

Find two numbers such that the mean proportion between them is $14$ and the third proportion to them is $112$.

Let the two numbers be $p$ and $q$.

Given: Mean proportion between $p$ and $q$ is $14$.

$\implies$ $p$, $14$ and $q$ are in continued proportion.

i.e. $\;$ $p : 14 = 14 : q$

i.e. $\;$ $14^2 = 196 = p \times q$ $\;\;\; \cdots \; (1)$

Given: Third proportion to $p$ and $q$ is $112$.

$\implies$ $p$, $q$ and $112$ are in continued proportion.

i.e. $\;$ $p : q = q : 112$

i.e. $\;$ $q^2 = 112 \times p$

$\implies$ $p = \dfrac{q^2}{112}$ $\;\;\; \cdots \; (2)$

Substituting the value of $p$ from equation $(2)$ in equation $(1)$ we have,

$196 = \dfrac{q^3}{112}$

i.e. $\;$ $q^3 = 196 \times 112 = 21952$ $\implies$ $q = \sqrt[3]{21952} = 28$

Substituting the value of $q$ in equation $(2)$ we get,

$p = \dfrac{28^2}{112} = \dfrac{784}{112} = 7$

$\therefore \;$ $p = 7$, $\;$ $q = 28$

Probability

In a pack of playing cards, all black kings and black queens are removed. If a card is drawn at random, find the probability of getting

a face card;

a black card;

a red card or a face card;

a black card or a face card

There are $2$ black kings and $2$ black queens which are removed from a pack of $52$ cards.

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 52 - 4 = 48$

Let $A =$ event of selecting a face card

There are a total of $8$ face cards from which $4$ face cards are removed.

$\therefore \;$ Number of elements in $A = n \left(A\right) = 4$

$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{4}{48} = \dfrac{1}{12}$

Let $B =$ event of selecting a black card

There are $26$ black cards from which $4$ black cards (face cards) are removed.

$\therefore \;$ Number of elements in $B = n \left(B\right) = 22$

$\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{22}{48} = \dfrac{11}{24}$

Let $C = $ event of selecting a red card

There are $26$ red cards.

$\therefore \;$ Probability of drawing a red card $= P \left(\text{red card}\right) = \dfrac{26}{48}$

$A =$ event of selecting a face card

$P \left(A\right) = \dfrac{4}{48}$

Let $X = $ event of selecting a red card OR a face card

Probability of event $X = P \left(\text{C}\right) + P \left(\text{A}\right) = \dfrac{26}{48} + \dfrac{4}{48} = \dfrac{30}{48} = \dfrac{5}{8}$

Let $D = $ event of selecting a black card OR a face card

Probability of event $D = P \left(\text{black card}\right) + P \left(\text{face card}\right) = \dfrac{22}{48} + \dfrac{4}{48} = \dfrac{26}{48} = \dfrac{13}{24}$

Statistics

Find the mean of the following distribution of marks using step deviation method.

Marks	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$	$80 - 90$	$90 - 100$
Number of students	$8$	$12$	$24$	$16$	$9$	$7$	$4$

Class size $= h = 10$

Let assumed mean $= A = 65$

Marks	Frequency $\left(f_i\right)$	Mid-value $\left(x_i\right)$	deviation $= x_i - A$	$ t_i = \dfrac{x_i - A}{h} $	$ f_i \times t_i $
$30 - 40$	$8$	$35$	$-30$	$-3$	$-24$
$40 - 50$	$12$	$45$	$-20$	$-2$	$-24$
$50 - 60$	$24$	$55$	$-10$	$-1$	$-24$
$60 - 70$	$16$	$65$	$0$	$0$	$0$
$70 - 80$	$9$	$75$	$10$	$1$	$9$
$80 - 90$	$7$	$85$	$20$	$2$	$14$
$90 - 100$	$4$	$95$	$30$	$3$	$12$

$\Sigma f_i = 80$, $\;$ $\Sigma f_i \times t_i = -37$

Mean $= A + \dfrac{\Sigma f_i \times t_i}{\Sigma f_i} \times h$

i.e. $\;$ Mean $= 65 + \dfrac{\left(-37\right)}{80} \times 10 = 65 - 4.625 = 60.375$

Quadratic Equations

Solve the quadratic equation $\;$ $3x^2 - 12 x - 1 = 0$, $\;$ giving the answer to three significant figures.

Given quadratic equation: $\;$ $3x^2 - 12 x - 1 = 0$

Comparing with the standard equation $\;$ $ax^2 + bx + c = 0$ gives

$a = 3$, $\;$ $b = -12$, $\;$ $c = -1$

By quadratic formula, $\;$ $x = \dfrac{- b \pm \sqrt{b^2 - 4a c}}{2 a}$

$\begin{aligned} i.e. \; \; x & = \dfrac{12 \pm \sqrt{\left(-12\right)^2 - 4 \times 3 \times \left(-1\right)}}{2 \times 3} \\\\ & = \dfrac{12 \pm \sqrt{156}}{6} \\\\ & = \dfrac{12 \pm 12.4899}{6} \\\\ & = \dfrac{24.4899}{6} \; \; or \; \; -\dfrac{0.4899}{6} \\\\ & = 4.0817 \;\; or \; \; -0.08165 \end{aligned}$

i.e. $\;$ $x = 4.08$ $\;$ or $\;$ $x = -0.0817$ $\;$ [correct to 3 significant figures]

Statistics

Find '$x$' from the following marks obtained by $9$ students if the mean mark is equal to the median:

$5, \; 7, \; 9, \; 10, \; x, \; 15, \; 21, \; 26,\; 27$.

The marks are arranged in ascending order.

Number of students $= N = 9$ (Odd)

The marks obtained by the $9$ students are (in ascending order): $\;$ $5, \; 7, \; 9, \; 10, \; x, \; 15, \; 21, \; 26,\; 27$

Median mark $= \left(\dfrac{N + 1}{2}\right)^{th}$ term $= \left(\dfrac{9 + 1}{2}\right)^{th}$ term $= 5^{th}$ term

i.e. $\;$ Median mark $= x$ $\;\;\; \cdots \; (1)$

Mean mark $= \dfrac{\Sigma x_i}{N} = \dfrac{5 + 7 + 9 + 10 + x + 15 + 21 + 26 + 27}{9} = \dfrac{120 + x}{9}$ $\;\;\; \cdots \; (2)$

Given: $\;$ Mean mark $=$ Median mark

$\therefore \;$ We have from equations $(1)$ and $(2)$,

$x = \dfrac{120 + x}{9}$

i.e. $\;$ $9x = 120 + x$

i.e. $\;$ $8x = 120$ $\implies$ $x = 15$

Equation of a Line

Find the equation of the line $PQ$ passing through the points $\left(5, 0\right)$ and $\left(0, -6\right)$. Also find the equation of another line $AB$ intersecting $PQ$ at right angles and passing through the point $\left(6,1\right)$.

Let $P \left(x_1, y_1\right) = \left(5,0\right)$; $\;$ $Q \left(x_2, y_2\right) = \left(0, -6\right)$

Slope of $PQ = m_1 = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-6 - 0}{0 - 5} = \dfrac{6}{5}$

Equation of $PQ$ is: $\;$ $y - y_1 = m_1 \left(x - x_1\right)$

i.e. $\;$ $y - 0 = \dfrac{6}{5} \left(x - 5\right)$

i.e. $5y = 6x - 30$

$\therefore \;$ Equation of $PQ$ is: $\;$ $6x - 5y = 30$

$\because \;$ $AB \perp PQ$ $\implies$ Slope of $AB = m_2 = - \dfrac{1}{m_1} = -\dfrac{5}{6}$

$AB$ passes through the point $\left(6, 1\right)$

$\therefore \;$ Equation of $AB$ is: $\;$ $y - 1 = - \dfrac{5}{6} \left(x - 6\right)$

i.e. $\;$ $6 y - 6 = - 5 x + 30$

$\therefore \;$ Equation of $AB$ is: $\;$ $5x + 6y = 36$

Geometric Progression

The first term of a G.P is $27$ and the eight term is $\dfrac{1}{81}$. Find the sum of the first five terms.

First term of G.P $= a = 27$ $\;\;\; \cdots \; (1)$

Let common ratio of G.P $= r$

$n^{th}$ term of G.P $= t_n = ar^{n-1}$

Eight term of G.P $= t_8 = ar^7 = \dfrac{1}{81}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ we get,

$\dfrac{ar^7}{a} = \dfrac{1}{81 \times 27}$

i.e. $\;$ $r^7 = \left(\dfrac{1}{3}\right)^7$ $\implies$ $r = \dfrac{1}{3}$

Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(1 - r^n\right)}{1 - r}$, $\;$ when $\;$ $r <0$

$\therefore \;$ Sum of first five terms $= S_5 = \dfrac{27 \left[1 - \left(\dfrac{1}{3}\right)^5\right]}{1 - \dfrac{1}{3}}$

i.e. $\;$ $S_5 = \dfrac{27 \times 242 \times 3}{243 \times 2} = \dfrac{121}{3}$

Arithmetic Progression

Divide $207$ into three parts such that these parts are in A.P and the product of the two smaller parts is $4623$.

$\because \;$ The three parts are in A.P, let the parts be $\;$ $a - d, \; a, \; a+d$

Product of the smaller parts $= 4623$

i.e. $\;$ $a \left(a - d\right) = 4623$ $\;\;\; \cdots \; (1)$

Sum of the parts $= 207$

i.e. $\;$ $a - d + a + a + d = 207$

i.e. $\;$ $3a = 207$

$\implies$ $a = 69$ $\;\;\; \cdots \; (2)$

Substituting the value of '$a$' from equation $(2)$ in equation $(1)$,

$69 \left(69 - d\right) = 4623$

i.e. $\;$ $69 - d = 67$ $\implies$ $d = 2$

$\therefore \;$ The three parts are $\;$ $67, \; 69, \; 71$

Quadratic Equations

A journey of $300 \; km$ would take $2$ hours less if the speed was increased by $5 \; kmph$. Find the original speed.

Distance covered $= d = 300 \; km$

Let original speed = $s \; kmph$

Then, original time $= t_1 = \dfrac{d}{s} = \dfrac{300}{s} \;$ hours

New speed $= s + 5 \; kmph$

$\therefore \;$ New time $= t_2 = \dfrac{300}{s + 5} \;$ hours

As per sum, $\;$ $t_2 = t_1 - 2$

i.e. $\;$ $\dfrac{300}{s + 5} = \dfrac{300}{s} - 2$

i.e. $\;$ $\dfrac{150}{s + 5} = \dfrac{150}{s} - 1$

i.e. $\;$ $150 s = 150 s + 750 - s \left(s + 5\right)$

i.e. $\;$ $s^2 + 5 s - 750 = 0$

i.e. $\;$ $s^2 + 30 s - 25 s - 750 = 0$

i.e. $\;$ $s \left(s + 30\right) - 25 \left(s + 30\right) = 0$

i.e. $\;$ $\left(s + 30\right) \left(s - 25\right)= 0$

i.e. $\;$ $s = -30$ $\;$ or $\;$ $s = 25$

$\because \;$ Original speed cannot be negative, original speed $= 25 \; kmph$

Mensuration

A solid metallic cone of slant height $13 \; cm$ and radius $5 \; cm$ is melted and recast into solid spheres each of radius $1 \; cm$. Find the number of spheres recast.

Slant height of cone $= \ell = 13 \; cm$

Radius of cone $= r = 5 \; cm$

Let height of cone $= h \; cm$

$h = \sqrt{\ell^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \; cm$

Volume of cone $= V_1 = \dfrac{1}{3}\pi r^2 h$

Let number of spheres made $= n$

Radius of each sphere $= R = 1 \; cm$

Volume of each sphere $= V_2 = \dfrac{4}{3} \pi R^3$

$\because \;$ The cone is melted and recast into $n$ spheres, we have,

$V_1 = n \times V_2$

i.e. $\;$ $\dfrac{1}{3} \pi r^2 h = n \times \dfrac{4}{3} \pi R^3$

i.e. $\;$ $n = \dfrac{r^2 h}{4 R^3}$

i.e. $\;$ $n = \dfrac{5^2 \times 12}{4 \times 1^3} = 75$

$\therefore \;$ Number of spheres made $= 75$

Factor and Remainder Theorems

The expression $x^3 + ax^2 + bx + 6$ has $\left(x - 2\right)$ as a factor and leaves a remainder $3$ when divided by $\left(x - 3\right)$. Find '$a$' and '$b$'.

Let $\;$ $f \left(x\right) = x^3 + ax^2 + bx + 6$

Given: $\;$ $\left(x - 2\right)$ is a factor of $f\left(x\right)$

Then, by factor theorem, $\;$ $f \left(2\right) = 0$

i.e. $\;$ $2^3 + a \times 2^2 + b \times 2 + 6 = 0$

i.e. $\;$ $4a + 2b + 14 = 0$

i.e. $\;$ $2a + b = - 7$ $\;\;\; \cdots \; (1)$

Given: $\;$ $f \left(x\right)$ when divided by $\left(x - 3\right)$ leaves a remainder $3$.

Then, by remainder theorem, $\;$ $f \left(3\right) = 3$

i.e. $\;$ $3^3 + a \times 3^2 + b \times 3 + 6 = 3$

i.e. $\;$ $9a + 3b + 12 = 0$

i.e. $\;$ $3a + b = - 4$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously we have,

$a = 3$ $\;$ and $\;$ $b = - 13$

Commercial Mathematics - Banking

If ₹ $\; 15,000$ is earned as interest on a monthly deposit of ₹ $\; 5,000$ for $2$ years. Find the rate of interest on the recurring deposit.

Money deposited each month $= P =$ ₹ $5,000$

Time for which money deposited $= n = 2 $ years $= 24 $ months

Let, rate of interest $= r \%$

Interest received $= $ ₹ $15,000$

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

i.e. $\;$ $15000 = 5000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100}$

i.e. $r = \dfrac{15000 \times 2 \times 12 \times 100}{5000 \times 24 \times 25} = 12 \%$

$\therefore \;$ Rate of interest on the recurring deposit $= 12 \%$

Matrices

Given $\;$ $\begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 1 & -4 \end{bmatrix} = A + 7 I$

Find matrix $A$ if $I$ is an unit matrix of order $2 \times 2$.

$I$ is an unit matrix of order $2 \times 2$

$\therefore \;$ $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Now, $\;$ $\begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 1 & -4 \end{bmatrix} = A + 7 I$

$\implies$ $\begin{bmatrix} 2 \times 0 + \left(-1\right) \times 1 & 2 \times 2 + \left(-1\right) \times \left(-4\right) \\ 0 \times 0 + 3 \times 1 & 0 \times 2 + 3 \times \left(-4\right) \end{bmatrix} = A + 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} -1 & 8 \\ 3 & -12 \end{bmatrix} = A + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} -1 & 8 \\ 3 & -12 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = A$

i.e. $\;$ $A = \begin{bmatrix} -1 - 7 & 8 - 0 \\ 3 - 0 & -12 - 7 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} -8 & 8 \\ 3 & -19 \end{bmatrix}$

Similarity

In the figure, $ABC$ and $CEF$ are two triangles where $BA$ is parallel to $CE$ and $AF : AC = 5 : 8$.

Prove that $\triangle ADF \sim \triangle CEF$

Find $AD$ if $CE = 6 \; cm$.

If $DE \parallel BC$, find $\;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)}$.

$BA \parallel CE$ $\;\;$ [Given]

$AC$ is a transversal.

$\therefore \;$ $\angle DAF = \angle ECF$ $\;\;$ [Alternate angles] $\;\;\; \cdots \; (1)$

Also, $\;$ $\angle AFD = \angle EFC$ $\;\;$ [Vertically opposite angles] $\;\;\; \cdots \; (2)$

$\therefore \;$ From $(1)$ and $(2)$, $\;$ $\triangle ADF \sim \triangle CEF$ $\;\;$ [By Angle-Angle postulate]

$\because \;$ $\triangle ADF \sim \triangle CEF$,

$\dfrac{AD}{CE} = \dfrac{AF}{CF}$ $\;\;$ [corresponding sides of similar triangles are proportional]

$\implies$ $AD = CE \times \dfrac{AF}{CF}$ $\;\;\; \cdots \; (3)$

Given: $\;$ $\dfrac{AF}{AC} = \dfrac{5}{8}$

$\implies$ $1 - \dfrac{AF}{AC} = 1 - \dfrac{5}{8}$

i.e. $\;$ $\dfrac{AC - AF}{AC} = \dfrac{CF}{AC} = \dfrac{3}{8}$

i.e. $\;$ $\dfrac{AC}{CF} = \dfrac{8}{3}$

$\therefore \;$ $\dfrac{AF}{AC} \times \dfrac{AC}{CF} = \dfrac{5}{8} \times \dfrac{8}{3}$

i.e. $\;$ $\dfrac{AF}{CF} = \dfrac{5}{3}$ $\;\;\; \cdots \; (4)$

Given: $\;$ $CE = 6 \; cm$ $\;\;\; \cdots \; (5)$

In view of equations $(4)$ and $(5)$, equation $(3)$ becomes

$AD = 6 \times \dfrac{5}{3} = 10 \; cm$

Given: $\;$ $DE \parallel BC$ $\implies$ $DF \parallel BC$

$AB$ is a transversal.

$\therefore \;$ $\angle ADF = \angle ABC$ $\;\;$ [Corresponding angles] $\;\;\; \cdots \; (6)$

$\angle BAC = \angle DAF$ $\;\;$ [Common angle] $\;\;\; \cdots \; (7)$

$\therefore \;$ From $(6)$ and $(7)$, $\;$ $\triangle ADF \sim \triangle ABC$ $\;\;$ [By Angle-Angle postulate]

$\therefore \;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)} = \dfrac{AF^2}{AC^2}$

[Areas of two similar triangles are proportional to the squares of their corresponding sides]

$\therefore \;$ $\dfrac{\text{Area } \left(\triangle ADF\right)}{\text{Area } \left(\triangle ABC\right)} = \left(\dfrac{5}{8}\right)^2 = \dfrac{25}{64}$

Locus

Taking a scale of $2 \; cm$ as $2$ units on both the axes, on a graph paper plot the points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$.

Complete the quadrilateral $ABCD$.

Find the point $P$ inside the quadrilateral such that $AP = BP$ and $P$ is also equidistant from $AB$ and $AD$.

Record the length $PB$.

The points $A \left(1, 2\right)$, $B \left(8, 3\right)$, $C \left(7, 8\right)$ and $D \left(3, 7\right)$ are plotted on a graph paper and the quadrilateral $ABCD$ is drawn.

Given: $\;$ $AP = BP$

$\implies$ $P$ lies on the perpendicular bisector of the line $AB$.

[The locus of a point, which is equidistant from two fixed points, is the perpendicular bisector of the line segment joining the two fixed points.]

$EF$ is the perpendicular bisector of the line segment $AB$.

Given: $\;$ $P$ is equidistant from $AB$ and $AD$.

$\implies$ $P$ lies on the angle bisector of lines segments $AB$ and $AD$.

[The locus of a point equidistant from two intersecting lines is the bisector of the angle between the lines.]

$AG$ is the angle bisector of $\angle DAB$.

$EF$ and $AG$ intersect at point $P$, which is the the point inside the quadrilateral such that $AP = BP$ and is equidistant from $AB$ and $AD$.

Length of $PB = 6.5 \; cm$

Circle

In the figure, tangent $CT$ and chord $BA$ extended meet at $T$. If $\angle CTA = 36^\circ$ and $AT = AC$, find $\angle ABC$.
If length of tangent $CT = 12 \; cm$ and $AT = 8 \; cm$, find the length of chord $AB$.

Given: $\;$ $\angle CTA = 36^\circ$; $\;$ $AT = AC$

$\because \;$ $AT = AC$ $\implies$ $\angle TCA = \angle TAC$ $\;\;$ [angles opposite equal sides are equal]

Now, in $\triangle ATC$,

$\angle TCA + \angle TAC + \angle CTA = 180^\circ$ $\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $2 \angle TCA + 36^\circ = 180^\circ$

i.e. $\;$ $2 \angle TCA = 144^\circ$ $\implies$ $\angle TCA = 72^\circ$

Now, $\;$ $\angle TBC = \angle TCA = 72^\circ$ $\;\;$ [angles in alternate segments of the circle]

But points $B$, $A$ and $T$ are collinear.

$\therefore \;$ $\angle ABC = 72^\circ$

From the figure, $\;$ $BT \times AT = CT^2$

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

i.e. $\;$ $BT = \dfrac{CT^2}{AT} = \dfrac{12^2}{8} = 18 \; cm$

But $\;$ $BT = AB + AT$

$\therefore \;$ $AB = BT - AT = 18 - 8 = 10 \; cm$

Constructions

Construct a regular hexagon of side $4.3$ cm and construct a circle circumscribing the hexagon.

Construct a regular hexagon $ABCDEF$ with each side equal to $4.3$ cm.

Draw the perpendicular bisectors of sides $AB$ and $BC$ which intersect each other at point $O$.

With $O$ as center and $OA$ as radius, draw a circle passing through all the vertices of the regular hexagon $ABCDEF$.
The circle so obtained is the required circle circumscribing the given regular hexagon.

Probability

Some cards are numbered from $10$ to $40$. They are well shuffled and one card is drawn at random. What is the probability that the card drawn is:

a prime number;

divisible by $2$ and $5$;

a perfect square.

Sample space $= S = \left\{10, 11, 12, \cdots, 40 \right\}$

$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 31$

Let $A =$ event that the number on the card is a prime number

Then $A = \left\{11, 13, 17, 19, 23, 29, 31, 37 \right\}$

$\therefore \;$ Number of elements in $A = n \left(A\right) = 8$

$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{8}{31}$

Let $B =$ event that the number on the card drawn is divisible by $2$ and $5$

Then $B = \left\{10, 20, 30, 40 \right\}$

$\therefore \;$ Number of elements in $B = n \left(B\right) = 4$

$\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{4}{31}$

Let $C =$ event that the number on the card is a perfect square

Then $C = \left\{16, 25, 36 \right\}$

$\therefore \;$ Number of elements in $C = n \left(C\right) = 3$

$\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{3}{31}$

Statistics

The following table shows the marks of $120$ students obtained in an examination. Draw an ogive for the table.

Marks	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$	$80 - 90$	$90 - 100$
Number of students	$1$	$3$	$11$	$21$	$43$	$32$	$9$

Use the ogive to estimate:

the median;

the upper quartile;

the number of students who got more than $95 \%$ marks;

the marks obtained by top $20 \%$ students in the examination.

Marks (Class Interval)	Number of students (frequency $f_i$)	Cumulative Frequency
$30 - 40$	$1$	$1$
$40 - 50$	$3$	$4$
$50 - 60$	$11$	$15$
$60 - 70$	$21$	$36$
$70 - 80$	$43$	$79$
$80 - 90$	$32$	$111$
$90 - 100$	$9$	$120$

Number of students $= N = \Sigma f_i = 120$

Taking marks (class intervals) along X-axis and cumulative frequency along Y-axis, draw an ogive.

Median $= \left(\dfrac{N}{2}\right)^{th} \text{term} = \left(\dfrac{120}{2}\right)^{th} \text{term} = 60^{th} \text{term} = 75.5$

$\therefore \;$ Median value $= 75.5$

Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th} \text{term} = \left(\dfrac{3 \times 120}{4}\right)^{th} \text{term} = 90^{th} \text{term} = 83$

$\therefore \;$ Upper quartile $= 83$

Number of students who got $95 \%$ marks $= 117$

$\therefore \;$ Number of students who got more than $95 \%$ marks $= 120 - 117 = 3$
Number of top $20 \%$ students $= \dfrac{20}{100} \text{ of } 120 = 24$ students

$\therefore \;$ Marks obtained by top $20 \%$ students $= 85 \text{ to } 100$

Commercial Mathematics - Banking

A person deposited ₹ $ 400$ at the beginning of every month in a recurring deposit account and received ₹ $ 16,398$ at the end of $3$ years. Find the rate of interest given by the bank.

Money deposited each month $= P =$ ₹ $ 400$

Time for which money deposited $= n = 3 $ years $= 36 $ months

Let, rate of interest $= r \%$

Amount received on maturity $= $ ₹ $ 16,398$

$\begin{aligned} \text{Interest} & = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100} \\\\ & = 400 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{r}{100} \\\\ & = 222 \; r \end{aligned}$

i.e. $\;$ Interest $= $ ₹ $ 222 \; r$

Total money deposited $= 400 \times 36 = $ ₹ $ 14,400$

$\text{Amount received on maturity} = \text{Money deposited} + \text{Interest}$

i.e. $\;$ $16398 = 14400 + 222 \; r$

i.e. $\;$ $222 \; r = 1998$

i.e. $\;$ $r = 9$

$\therefore \;$ Rate of interest given by the bank $= 9 \%$

Commercial Mathematics - Shares and Dividends

A person bought ₹ $\; 100$ shares of dividend $9 \%$ selling at a certain price. If the rate of return is $7.5 \%$, calculate:

the market value of each share;
the amount to be invested to obtain an annual income of ₹ $\; 1260$;
how many more shares should be bought to increase the income to ₹ $\; 1890$.

Nominal value (N.V) of each share $= $ ₹ $ 100$

Rate of dividend $= 9 \%$

Rate of return $= 7.5 \%$

Let market value (M.V) of each share $= $ ₹ $ x$

Now, $\;$ $\text{Rate of return} \times M.V = \text{Rate of dividend} \times N.V$

i.e. $\;$ $\dfrac{7.5}{100} \times x = \dfrac{9}{100} \times 100$

$\implies$ $x = \dfrac{900}{7.5} = 120$

i.e. $\;$ Market value of each share $= $ ₹ $ 120$
Total annual income $= $ ₹ $ 1260$

Annual income on $1$ share $= 9 \%$ of ₹ $ 100 = $ ₹ $ 9$

$\begin{aligned} \text{Number of shares bought} & = \dfrac{\text{Total annual income}}{\text{Annual income on 1 share}} \\\\ & = \dfrac{1260}{9} \\\\ & = 140 \end{aligned}$

$\therefore \;$ Amount to be invested $=$ Number of shares bought $\times$ M.V of 1 share

i.e. $\;$ Amount to be invested $= 140 \times $ ₹ $ 120 = $ ₹ $ 16,800$
New annual income $= $ ₹ $ 1890$

$\begin{aligned} \therefore \; \text{Number of shares} & = \dfrac{\text{Total annual income}}{\text{Annual income on 1 share}} \\\\ & = \dfrac{1890}{9} \\\\ & = 210 \end{aligned}$

$\therefore \;$ Number of extra shares $= 210 - 140 = 70$

Coordinate Geometry

Taking a scale of $1 \; cm = 1$ unit on both the axes on a graph paper,

plot the points $A \left(-3, 3\right)$ and $B \left(2,2\right)$;

write the coordinates of $A'$ and $B'$, the images of $A$ and $B$ respectively on reflection in origin;

write the geometrical name of the figure $ABA'B'$;

write the coordinates of two invariant points in Y axis in the figure.

Points $A \left(-3,3\right)$ and $B \left(2,2\right)$ are plotted.

$A' \left(3,-3\right)$ and $B' \left(-2, -2\right)$ are the images of $A$ and $B$ on reflection in origin.

$ABA'B'$ is a parallelogram.

$P \left(0, 2.4\right)$ and $Q \left(0, -2.4\right)$ are two invariant points in Y axis in the figure.

Statistics

Draw a histogram to determine the mode of the following frequency distribution. Also state the modal class.

Class Interval	$80 - 90$	$90 - 100$	$100 - 110$	$110 - 120$	$120 - 130$	$130 - 140$
Frequency	$8$	$10$	$12$	$14$	$7$	$4$

Circle

In the given figure, O is the center of the circle. If $\angle POR = 150^\circ$ and $\angle ORQ = 60^\circ$, find the measure of:

$\angle PQR$
$\angle OPR$
$\angle QRP$
$\angle RPQ$

Given: $\angle POR = 150^\circ$

$\therefore \;$ Reflex $\angle POR = 360^\circ - 150^\circ = 210^\circ$

Now, $\;$ $\angle PQR = \dfrac{1}{2} \times \text{ Reflex } \angle POR$

[$\because \;$ angle at the center is twice the angle at the remaining circumference]

$\therefore \;$ $\angle PQR = \dfrac{210^\circ}{2} = 105^\circ$
Join $PR$.

In $\triangle POR$, $\;$ $OP = OR = $ radii of the same circle

$\therefore \;$ $\angle OPR = \angle ORP$ $\;\;\;$ [$\because \;$ angles opposite to equal sides are equal]

Now, $\;$ $\angle POR + \angle OPR + \angle ORP = 180^\circ$ $\;\;\;$ [$\because \;$ sum of angles of a $\triangle$ equal $180^\circ$]

i.e. $\;$ $150^\circ + 2 \angle OPR = 180^\circ$

i.e. $\;$ $2 \angle OPR = 30^\circ$ $\implies$ $\angle OPR = 15^\circ$
$\because \;$ $\angle OPR = \angle ORP$ $\implies$ $\angle ORP = 15^\circ$

From the figure, $\;$ $\angle ORQ = \angle ORP + \angle QRP$

Given: $\;$ $\angle ORQ = 60^\circ$

$\therefore \;$ We have, $\;$ $60^\circ = 15^\circ + \angle QRP$

$\implies$ $\angle QRP = 60^\circ - 15^\circ = 45^\circ$
In $\triangle PQR$, $\;$ $\angle RPQ + \angle PQR + \angle QRP = 180^\circ$ $\;\;$ [$\because \;$ sum of angles of a $\triangle$ equal $180^\circ$]

i.e. $\;$ $\angle RPQ + 105^\circ + 45^\circ = 180^\circ$

$\implies$ $\angle RPQ = 30^\circ$

Linear Inequations

Solve the following inequation: $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}, \; x \in R$
Write the solution set and represent the solution set on a number line.

Consider $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1$

i.e. $\;$ $\dfrac{1}{5} \leq \dfrac{3x + 10}{10}$

i.e. $\;$ $1 \leq \dfrac{3x + 10}{2}$

i.e. $\;$ $2 \leq 3x + 10$

i.e. $\;$ $- 8 \leq 3x$

i.e. $- \dfrac{8}{3} \leq x$ $\;\;\; \cdots \; (1)$

$\implies$ $x \geq - \dfrac{8}{3}$

Consider $\;$ $\dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}$

i.e. $\;$ $\dfrac{3x + 10}{10} < \dfrac{8}{5}$

i.e. $\;$ $\dfrac{3x + 10}{2} < 8$

i.e. $\;$ $3x + 10 < 16$

i.e. $\;$ $3x < 6$ $\implies$ $x < 2$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- \dfrac{8}{3} \leq x < 2$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -\dfrac{8}{3} \leq x < 2, \; x \in R \right\}$

Mensuration

The total surface area of a hollow metal cylinder, open at both ends, of external radius $8 \; cm$ and height $10 \; cm$ is $338 \pi \; cm^2$. Taking $r$ to be the internal radius, write down an equation in $r$ and use it to find the thickness of the metal in the cylinder.

External radius of the hollow cylinder $= R = 8 \; cm$

Internal radius of the hollow cylinder $= r \; cm$

Height of the hollow cylinder $= h = 10 \; cm$

Total surface area of the hollow cylinder $= A = 2 \pi R h + 2 \pi r h + 2 \pi \left(R^2 - r^2\right)$

i.e. $\;$ $2 \pi \left(Rh + rh + R^2 - r^2\right) = 338 \pi$

i.e. $\;$ $Rh + rh + R^2 - r^2 = 169$

i.e. $\;$ $8 \times 10 + 10 r + 8^2 - r^2 = 169$

i.e. $\;$ $10 r - r^2 = 25$

i.e. $\;$ $r^2 - 10r + 25 = 0$

i.e. $\;$ $\left(r - 5\right)^2 = 0$ $\implies$ $r = 5 \; cm$

$\therefore \;$ Thickness of the metal in the cylinder $= R - r = 8 - 5 = 3 \; cm$

Similarity - Maps and Models

A map of a square plot of land is drawn to a scale of $1 : 2500$. If the area of the plot on the map is $72 \; cm^2$, find

the actual area of the land in square kilometer;
the length of the diagonal in the actual plot of land in meters.

Scale of square plot of land $= 1 : 2500$

On Map Actual

$1 \; cm$ $2500 \; cm$

$1 \; cm^2$ $625 \times 10^4 \; cm^2$

$72 \; cm^2$ $x \; cm^2$

$\therefore \;$ $x = 72 \times 625 \times 10^4 = 45000 \times 10^4 = 45 \times 10^7 \; cm^2$

i.e. $\;$ Actual area of the land $= 45 \times 10^7 \; cm^2$

Now, $\;$ $1000 \; m \equiv 1 \; km$

$\therefore \;$ $10^6 \; m^2 \equiv 1 \; km^2$

$\left[\text{Note: } 1 \; m \equiv 100 \; cm \implies 1 \; m^2 = 10^4 \; cm^2 \right]$

i.e. $\;$ $10^6 \times 10^4 \; cm^2 \equiv 1 \; km^2$

i.e. $\;$ $10^{10} \; cm^2 \equiv 1 \; km^2$

$\therefore \;$ $45 \times 10^7 \; cm^2 \equiv \dfrac{45 \times 10^7}{10^{10}} = 45 \times 10^{-3} = 0.045 \; km^2$

$\therefore \;$ Actual area of the land in square kilometer $= 0.045 \; km^2$

Let the length of the side of the plot on the map be $= s \; cm$

Then, area of the square plot on the map $= s^2 \; cm^2$

Given: Area of the plot on the map $= 72 \; cm^2$

i.e. $\;$ $s^2 = 72$ $\implies$ $s = \sqrt{72} \; cm$

$\therefore \;$ Length of the diagonal of the square plot on the map

$= \ell = \sqrt{\left(\sqrt{72}\right)^2 + \left(\sqrt{72}\right)^2} = \sqrt{72 + 72} = \sqrt{144} = 12 \; cm$

On Map Actual

$1 \; cm$ $2500 \; cm$

$12 \; cm$ $x \; cm$

$\therefore \;$ $x = 12 \times 2500 = 30000 = 3 \times 10^4 \; cm$

i.e. $\;$ length of the diagonal in the actual plot of land $= 3 \times 10^4 \; cm$

Now, $\;$ $100 \; cm \equiv 1 \; m$

$\therefore \;$ $3 \times 10^4 \; cm \equiv \dfrac{3 \times 10^4}{10^2} = 3 \times 10^2 = 300 \; m$

$\therefore \;$ length of the diagonal (in meter) in the actual plot of land $= 300 \; m$

On Map	Actual
$1 \; cm$	$2500 \; cm$
$1 \; cm^2$	$625 \times 10^4 \; cm^2$
$72 \; cm^2$	$x \; cm^2$

On Map	Actual
$1 \; cm$	$2500 \; cm$
$12 \; cm$	$x \; cm$

Trigonometry

A man in a boat rowing away from a lighthouse $150 \; m$ high takes $1$ minute to change the angle of elevation of the top of the lighthouse from $60^\circ$ to $45^\circ$. Find the speed of the boat in meter per minute. [Given: $\sqrt{3} = 1.732$]

$AB =$ Lighthouse of height $150 \; m$

$C =$ First point of observation

$D =$ Second point of observation

$CD =$ Distance traveled by the boat in $1$ minute

In $\triangle ABD$, $\;$ $\dfrac{AB}{BD} = \tan 45^\circ = 1$

$\implies$ $BD = AB = 150 \; m$

In $\triangle ABC$, $\;$ $\dfrac{AB}{BC} = \tan 60^\circ = \sqrt{3}$

$\implies$ $BC = \dfrac{AB}{\sqrt{3}} = \dfrac{150}{\sqrt{3}} = 50 \sqrt{3} = 86.6 \; m$

Now, $\;$ $CD = BD - BC = 150 - 86.6 = 63.4 \; m$

$\therefore \;$ Distance covered by the boat in $1$ minute $= 63.4 \; m$

$\therefore \;$ Speed of the boat $= 63.4 \; m/\text{min}$

Statistics

The mean of the following distribution is $24$. Find the missing frequency '$a$'.

Class Interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Frequency	$10$	$6$	$a$	$12$	$5$

Class Interval	Mid value $x_i$	Frequency $f_i$	$f_i x_i$
$0 - 10$	$5$	$10$	$50$
$10 - 20$	$15$	$6$	$90$
$20 - 30$	$25$	$a$	$25 a$
$30 - 40$	$35$	$12$	$420$
$40 - 50$	$45$	$5$	$225$
		$\Sigma f_i = 33 + a$	$\Sigma f_i x_i =785 + 25 a$

Mean $= \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{785 + 25 a}{33 + a}$

Given: $\;$ Mean $= 24$

i.e. $\;$ $\dfrac{785 + 25 a}{33 + a} = 24$

i.e. $\;$ $785 + 25 a = 792 + 24 a$

$\implies$ $a = 7$

Matrices

Given: $\;$ $A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}$, $\;$ $B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}$ $\;$ and $\;$ $C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$. Find matrix $X$ such that $A + 2 X = 2B + C$.

Let $X = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$

Given: $\;$ $A + 2 X = 2 B + C$

i.e. $\;$ $\begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} + 2 \begin{bmatrix} p & q \\ r & s \end{bmatrix} = 2 \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} 2 + 2p & -6 + 2q \\ 2 + 2r & 0 + 2s \end{bmatrix} = \begin{bmatrix} -6 + 4 & 4 + 0 \\ 8 + 0 & 0 + 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} 2 + 2p & -6 + 2q \\ 2 + 2r & 2s \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix}$

Two matrices are equal when all their corresponding elements are equal.

i.e. $\;$ $2 + 2p = -2$ $\implies$ $2 p = -4$ $\implies$ $p = -2$

$-6 + 2q = 4$ $\implies$ $2 q = 10$ $\implies$ $q = 5$

$2 + 2 r = 8$ $\implies$ $2 r = 6$ $\implies$ $r = 3$

$2 s = 2$ $\implies$ $s = 1$

$\therefore \;$ $X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix}$

Arithmetic Progression

Find the sum of the terms of the A.P $\;$ $4, \; 9, \; 14, \; \cdots \; 89$. Also find the $4^{th}$ term from the end.

Given A.P is: $\;$ $4, \; 9, \; 14, \; \cdots \; 89$

First term $= a = 4$

Last term $= n^{th}$ term $= t_n = \ell = 89$

Common difference $= d = 5$

Let the number of terms in the given A.P be $= n$

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$

i.e. $\;$ $89 = 4 + \left(n - 1\right) \times 5$

i.e. $\;$ $n - 1 = \dfrac{85}{5} = 17$ $\implies$ $n = 18$

$\therefore \;$ Number of terms in the given A.P is $= n = 18$

Sum of $n$ terms of an A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

Here, $\;$ $S_{18} = \dfrac{18}{2} \left[2 \times 4 + \left(18 - 1\right) \times 5\right]$

i.e. $\;$ $S_{18} = 9 \left[8 + 85\right] = 837$

$n^{th}$ term from the end of an A.P $= \ell - \left(n - 1\right)d$

Here, $4^{th}$ term from the end of the A.P $= 89 - \left(4 - 1\right) \times 5 = 89 - 15 = 74$

Trigonometry

Prove that $\;$ $\dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} = 2 \cot A$

$\begin{aligned} LHS & = \dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{\cot^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & \left[\text{Note: }1 + \cot^2 \theta = \text{cosec}^2 \theta\right] \\\\ & = \dfrac{\text{cosec}^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec }A \left(\text{cosec }A - \cot A\right)}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec }A}{\sec A} \\\\ & = \dfrac{2 / \sin A}{1 / \cos A} \\\\ & = \dfrac{2 \; \cos A}{\sin A} \\\\ & = 2 \cot A = RHS \end{aligned}$

Hence proved.

Quadratic Equations

For what value of '$p$' does the equation $x^2 - 2px + \left(7p - 12\right) = 0$ have equal roots?

Given quadratic equation: $\;$ $x^2 - 2px + \left(7p - 12\right) = 0$

Comparing with the standard equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 1$, $\;$ $b = -2 p$, $\;$ $c = 7p - 12$

A quadratic equation has equal roots when Discriminant $= \Delta = b^2 - 4ac = 0$

Here, discriminant $\;$ $\Delta = \left(-2p\right)^2 - 4 \times 1 \times \left(7p - 12\right)$

i.e. $\;$ $\Delta = 4 \left(p^2 - 7 p + 12\right)$

$\therefore \;$ For equal roots,

$4 \left(p^2 - 7p + 12\right) = 0$

i.e. $\;$ $p^2 - 7p + 12 = 0$

i.e. $\;$ $p^2 - 3p - 4p + 12 = 0$

i.e. $\;$ $p \left(p - 3\right) - 4 \left(p - 3\right) = 0$

i.e. $\;$ $\left(p - 3\right) \left(p - 4\right) = 0$

$\implies$ $p = 3$ $\;$ or $\;$ $p = 4$

$\therefore \;$ The given quadratic equation has equal roots when $p = 3$ or $p = 4$.