Matrices

Solve for matrices $A$ and $B$:
$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ and $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$

Given matrix equations:

$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ $\;\;\; \cdots \; (1)$; $\;\;$ $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$

$A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + 2 B$ $\;\;\; \cdots \; (3)$

$\implies$ $2A = \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes

$\begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$

i.e. $\;$ $5 B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 3 - 8 & -4 - 6 \\ 2 - 2 & 7 - 2 \end{bmatrix} = \begin{bmatrix} -5 & -10 \\ 0 & 5 \end{bmatrix}$

$\implies$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (5)$

Substituting the value of $B$ from equation $(5)$ in equation $(3)$ gives

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

i.e. $\;$ $A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} -2 & -4 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} 4 - 2 & 3 - 4 \\ 1 + 0 & 1 + 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

$\therefore \;$ $A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$; $\;$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$

Algebra - Airthmetic Progressions

The eighth term of an A.P is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{th}$ term of the A.P.

Let the first term of the A.P be $\;$ $t_1 = a$ $\;$ and the common difference $= d$

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

$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$;

$4^{th}$ term $= t_4 = a + 3d$;

$8^{th}$ term $= t_8 = a + 7d$; $\;\;$ and

$11^{th}$ term $= t_{11} = a + 10d$

As per the question,

$t_8 = \dfrac{1}{2} \times t_2$

i.e. $\;$ $a + 7d = \dfrac{1}{2} \times \left(a + d\right)$

i.e. $\;$ $2a + 14 d = a + d$ $\;\;\;$

i.e. $\;$ $a + 13d = 0$ $\implies$ $a = -13d$ $\;\;\; \cdots \;\;\; (1)$

Also, as per the problem

$t_{11} = \left(\dfrac{1}{3} \times t_4\right) + 1$

i.e. $\;$ $a + 10 d = \dfrac{1}{3} \times \left(a + 3d\right) + 1$

i.e. $\;$ $3a + 30d = a + 3d + 3$

i.e. $\;$ $2a + 27d = 3$ $\;\;\; \cdots \;\;\; (2)$

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

$2 \times \left(-13 d\right) + 27 d = 3$ $\;\;\;$ $\implies$ $d = 3$

Substituting the value of $d$ in equation $(1)$ gives

$a = -13 \times 3 = -39$

Now, $15^{th}$ term of A.P $= t_{15} = a + 14d$

i.e. $\;$ $t_{15} = -39 + \left(14 \times 3\right) = 3$

Statistics - Ogive

Draw an ogive for the given frequency distribution:

Class Interval	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$3$	$4$	$6$	$9$	$7$	$1$

Hence find the lower quartile, upper quartile and the inter-quartile range.

Class Interval	Frequency	Cumulative Frequency
$5 - 10$	$3$	$3$
$10 - 15$	$4$	$7$
$15 - 20$	$6$	$13$
$20 - 25$	$9$	$22$
$25 - 30$	$7$	$29$
$30 - 35$	$1$	$30$

Total number of observations $= N = 30$

Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th}$ term $= \left(\dfrac{30}{4}\right)^{th}$ term $= 7.5^{th}$ term $= 15.5$

Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th}$ term $= \left(\dfrac{3 \times 30}{4}\right)^{th}$ term $= 22.5^{th}$ term $= 25.5$

Inter-quartile range $= Q_3 - Q_1 = 25.5 - 15.5 = 10$

Commercial Mathematics - Shares and Dividends

A person invests ₹ $9000$ in shares of a company which is paying $8 \%$ dividend. If ₹ $100$ shares are available at a discount of $10 \%$, find the number of shares purchased and the annual income.

Money invested $= $ ₹ $9000$

Face value (FV) of each share $= $ ₹ $100$

Market value (MV) of each share $= $ ₹ $100 - 10 \%$ of ₹ $100$

$ = $ ₹ $\left(100 - \dfrac{10}{100} \times 100\right) = $ ₹ $90$

$\therefore \;$ Number of shares bought $= \dfrac{\text{money invested}}{\text{MV of each share}} = \dfrac{9000}{90} = 100$

Dividend (income) on one share $= 8 \%$ of FV $= \dfrac{8}{100} \times $ ₹ $100 = $ ₹ $8$

$\therefore \;$ Total income from the shares $= 100 \times $ ₹ $8 = $ ₹ $800$

Probability

In a single throw of two dice, what is the probability of getting

1. a total of $9$;
2. a doublet;
3. $5$ on one die and $6$ on the other.

Let $S =$ event of throwing two dice.

Number of elements in sample space $= n \left(S\right) = 36$

1. Let $A =$ event of getting a total of $9$ when two dice are thrown
   
   i.e. $\;$ $A = \left\{\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(6,3\right) \right\}$
   
   $\therefore \;$ Number of elements in $A = n \left(A\right) = 4$
   
   $\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{4}{36} = \dfrac{1}{9}$



2. Let $B =$ event of getting a doublet when two dice are thrown
   
   i.e. $\;$ $B = \left\{\left(1,1\right), \left(2,2\right), \left(3,3\right), \left(4,4\right), \left(5,5\right), \left(6,6\right) \right\}$
   
   $\therefore \;$ Number of elements in $B = n \left(B\right) = 6$
   
   $\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$



3. Let $C =$ event of getting a $5$ on one die and $6$ on the other
   
   i.e. $\;$ $C = \left\{\left(5,6\right), \left(6,5\right) \right\}$
   
   $\therefore \;$ Number of elements in $C = n \left(C\right) = 2$
   
   $\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{2}{36} = \dfrac{1}{18}$

Circle

In the figure, $\angle BAD = 60^\circ$, $\;$ $\angle ABD = 70^\circ$, $\;$ $\angle BDC = 40^\circ$

1. Prove that $AC$ is the diameter of the circle.
2. Find $\angle ACB$.
3. Find $\angle BCD$.

$ABCD$ is a cyclic quadrilateral.

$\therefore \;$ $\angle BAD + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \;$ $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 60^\circ = 120^\circ$

In $\triangle BAD$, $\;$ $\angle BAD + \angle ABD + \angle ADB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

$\therefore \;$ $\angle ADB = 180^\circ - \left(\angle BAD + \angle ABD\right) = 180^\circ - \left(60^\circ + 70^\circ\right) = 50^\circ$

Now, $\angle ADC = \angle ADB + \angle BDC = 50^\circ + 40^\circ = 90^\circ$

But angle in a semicircle is a right angle.

$\therefore \;$ $\angle ADC$ is an angle in a semicircle.

$\implies$ $AC$ is the diameter of the circle.

Now, $\;$ $\angle ADB = \angle ACB$ $\;\;\;$ [angles of same segment are equal]

$\therefore \;$ $\angle ACB = 50^\circ$

Similarity

In the figure, $AB$ and $DE$ are perpendicular to $BC$.

1. Prove that $\triangle ABC \sim \triangle DEC$.
2. If $AB = 6 \; cm$, $DE = 4 \; cm$, $AC = 15 \; cm$, calculate $CD$.
3. Find the ratio of $\;\;$ $\dfrac{\text{Area}\left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)}$

1. Given: $\;\;$ $AB \perp BC$, $\;$ $DE \perp BC$
   
   $\therefore \;$ $\angle ABC = \angle DEC = 90^\circ$ $\;\;\; \cdots \; (1)$
   
   In triangles $ABC$ and $DEC$,
   
   $\angle ACB = \angle DCE$ $\;$ [common angle] $\;\;\; \cdots \; (2)$
   
   $\therefore \;$ From equations $(1)$ and $(2)$, $\;$ $\triangle ABC \sim \triangle DEC$ $\;$ [by Angle-Angle postulate]



2. $\because \;$ $\triangle ABC \sim \triangle DEC$
   
   $\therefore \;$ $\dfrac{AB}{DE} = \dfrac{AC}{DC}$ $\;\;\;$ [corresponding sides of similar triangles are in proportion]
   
   $\therefore \;$ $DC = \dfrac{AC \times DE}{AB} = \dfrac{15 \times 4}{6} = 10 \; cm$



3. $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{AB^2}{DE^2}$
   
   [areas of two similar triangles are proportional to the squares on their corresponding sides]
   
   i.e. $\;$ $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{6^2}{4^2} = \dfrac{9}{4}$

Circle

In the figure, $O$ is the center of a circle.

$\angle ABC = 100^\circ$, $\angle ACD = 40^\circ$

$CT$ is a tangent to the circle at $C$.

Find $\angle ADC$ and $\angle DCT$.

$O$ is the center of the circle; $\;$ $\angle ABC = 100^\circ$; $\;$ $\angle ACD = 40^\circ$;

$CT$ is a tangent to the circle at a point $C$

In the figure, $\;$ $ABCD$ is a cyclic quadrilateral.

$\therefore \;$ $\angle ABC + \angle ADC = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\implies$ $\angle ADC = 180^\circ - \angle ABC = 180^\circ - 100^\circ = 80^\circ$

Now, $\;$ $\angle AOC = 2 \angle ADC$

[angle at the center of a circle is twice the angle at the remaining circumference]

$\therefore \;$ $\angle AOC = 2 \times 80^\circ = 160^\circ$

Join $OA$ and $OC$.

Then, $\;$ $OA = OC$ $\;\;\;$ [radii of the same circle]

$\implies$ $\angle ACO = \angle CAO$ $\;\;\;$ [angles opposite equal sides are equal]

In $\triangle AOC$, $\;$ $\angle AOC + \angle ACO + \angle CAO = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

i.e. $\;$ $\angle AOC + 2 \angle ACO = 180^\circ$

$\implies$ $\angle ACO = \dfrac{180^\circ - \angle AOC}{2} = \dfrac{180^\circ - 160^\circ}{2} = 10^\circ$

From the figure, $\;$ $\angle ACD = \angle ACO + \angle OCD$

$\therefore \;$ $\angle OCD = \angle ACD - \angle ACO = 40^\circ - 10^\circ = 30^\circ$

Also, $\;$ $\angle OCT = 90^\circ$

[tangent at any point of a circle and the radius through this point are perpendicular to each other]

But, $\;$ $\angle OCT = \angle OCD + \angle DCT$

$\therefore \;$ $\angle DCT = \angle OCT - \angle OCD = 90^\circ - 30^\circ = 60^\circ$

Mensuration

A hemispherical and a conical hole are scooped out of a solid wooden cylinder. The height of the solid cylinder is $7 \; cm$, radius of each of hemisphere, cone and cylinder is $3 \; cm$. The height of the cone is $3 \; cm$. Find the volume of the remaining solid. Give your answer correct to the nearest whole number. Take $\pi = 3.142$

Height of cylinder $= H = 7 \; cm$

Radius of cylinder $= R = 3 \; cm$

Radius of hemisphere $= R_S = 3 \; cm$

Radius of cone $= r = 3 \; cm$

Height of cone $= h = 3 \; cm$

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

Volume of hemisphere $= V_S = \dfrac{2}{3} \pi R_S^3$

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

Volume of remaining solid $= V - \left(V_S + V_C\right)$

$= \pi R^2 H - \left(\dfrac{2}{3} \pi R_S^3 + \dfrac{1}{3} \pi r^2 h\right)$

$= \pi \left[3^2 \times 7 - \left(\dfrac{2}{3} \times 3^3 + \dfrac{1}{3} \times 3^2 \times 3\right)\right]$

$= \pi \left[63 - \left(18 + 9\right)\right] = 3.142 \times 36 = 113.112$

$\therefore \;$ Volume of the remaining solid $= 113 \; cm^3$ $\;\;\;$ [correct to the nearest whole number]

Constructions

Use a ruler and compass only for this question.

Draw a circle of radius $4 \; cm$.

Mark the center as $O$. Mark a point $P$ outside the circle at a distance of $7 \; cm$ from the center.

Construct two tangents to the circle from the external point $P$.

Measure and write down the length of the tangents.

Draw a circle with center $O$ and radius $4 \; cm$

Mark point $P$ at a distance of $7 \; cm$ from the center $O$.

Join $PO$.

Draw the perpendicular bisector of $OP$.

Draw a circle with $OP$ as diameter which cuts the given circle at points $A$ and $B$.

Join $PA$ and $PB$.

$PA$ and $PB$ are the required tangents.

Length of tangent $PA = PB = 5.7 \; cm$

Quadratic Equations

Solve the quadratic equation $\;$ $2x^2 - 5x - 10 = 0$ $\;$ giving your answer correct to two decimal places.

Given quadratic equation is $\;$ $2x^2 - 5x - 10 = 0$

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

$a = 2$, $\;$ $b = -5$, $\;$ $c = -10$

Solution of the standard quadratic equation is $\;$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\therefore \;$ Solution of the given quadratic equation is

$\begin{aligned} x & = \dfrac{5 \pm \sqrt{5^2 - 4 \times 2 \times \left(-10\right)}}{2 \times 2} \\\\ & = \dfrac{5 \pm \sqrt{25 + 80}}{4} \\\\ & = \dfrac{5 \pm \sqrt{105}}{4} \\\\ & = \dfrac{5 \pm 10.247}{4} \\\\ & = \dfrac{15.247}{4} \;\;\; or \;\;\; \dfrac{-5.247}{4} \\\\ & = 3.812 \;\;\; or \;\;\; -1.312 \end{aligned}$

i.e. $\;$ $x = 3.81$ $\;$ or $\;$ $x = -1.31$ $\;\;\;$ [correct to two decimal places]

Statistics - Histogram

Draw a histogram for the following frequency distribution and find the mode from the graph.

Class Interval	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$
Frequency	$2$	$5$	$10$	$14$	$8$	$5$

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 $= 17$

Trigonometry - Heights and Distances

A straight road leads to the foot of a tower $200 \; m$ high. From the top of the tower, the angles of depression of two cars standing on the road are observed to be $45^\circ$ and $60^\circ$ respectively. Find the distance between the two cars. Give your answer correct to two decimal places. Take $\sqrt{3} = 1.732$.

$OT$: Tower of height $200 \; m$

$C_1, \; C_2$: Positions of the two cars

$C_1C_2$: Distance between the two cars

In $\triangle OTC_1$,

$\dfrac{OT}{OC_1} = \tan 60^\circ = \sqrt{3}$

i.e. $\;$ $OC_1 = \dfrac{OT}{\sqrt{3}} = \dfrac{200}{\sqrt{3}} \; m$ $\;\;\; \cdots \; (1)$

In $\triangle OTC_2$,

$\dfrac{OT}{OC_2} = \tan 45^\circ = 1$

i.e. $\;$ $OC_2 = OT = 200 \; m$ $\;\;\; \cdots \; (2)$

Now, $\;$ $C_1 C_2 = OC_2 - OC1$

i.e. $\;$ $C_1 C_2 = 200 - \dfrac{200}{\sqrt{3}}$ $\;\;\;$ [in view of equations $(1)$ and $(2)$]

i.e. $\;$ $C_1 C_2 = 200 \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = 84.53 \; m$ $\;\;\;$ [correct to two decimal places]

Ratio and Proportion

If $x$, $y$ and $z$ are in continued proportion, prove that $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$

$x, \; y, \; z$ are in continued proportion.

$\implies$ $\dfrac{x}{y} = \dfrac{y}{z} = k \;$ (say)

$\implies$ $y = z k$ $\;\;\; \cdots \; (1)$

and $\;$ $x = yk = zk^2$ $\;\;\; \cdots \; (2)$

$\begin{aligned} LHS & = \dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} \\\\ & = \dfrac{\left(zk^2 + zk\right)^2}{\left(zk + z\right)^2} \;\;\; \left[\text{in view of equations (1) and (2)}\right] \\\\ & = \dfrac{\left[zk \left(k + 1\right)\right]^2}{\left[z \left(k + 1\right)\right]^2} \\\\ & = k^2 \;\;\; \cdots \; (3) \\\\ RHS & = \dfrac{x}{z} \\\\ & = \dfrac{zk^2}{z} = k^2 \;\;\; \cdots \; (4) \end{aligned}$

$\therefore \;$ From equations $(3)$ and $(4)$, $\;\;$ $LHS = RHS$

i.e. $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$

Hence proved.

Coordinate Geometry: Reflection

Use a graph sheet for this question. Take $1 \; cm = 1$ unit on both the axes.

1. Plot the points $A \left(0, 5\right)$, $B \left(3,0\right)$, $C \left(1, 0\right)$ and $D \left(1, -5\right)$.
2. Reflect the points $B$, $C$ and $D$ on the $Y$ axis and name them as $B'$, $C'$ and $D'$ respectively.
3. Write the coordinates of $B'$, $C'$ and $D'$.
4. Join the points $ABCDD'C'B'A$, in order, and give a name to the closed figure formed.

The points $A \left(0, 5\right)$, $B \left(3,0\right)$, $C \left(1, 0\right)$ and $D \left(1, -5\right)$ are plotted.

Points $B$, $C$ and $D$ are reflected in the $Y$ axis giving the points $B' \left(-3,0\right)$, $C' \left(-1, 0\right)$ and $D' \left(-1, -5\right)$ respectively.

Points $ABCDD'C'B'A$ joined in order give an arrow.

Coordinate Geometry: Equation of a Line

The coordinates of $P \left(2, 6\right)$ and $Q \left(-3, 5\right)$ are given. Find

1. the gradient of $PQ$;
2. the equation of $PQ$;
3. the coordinates of the point where $PQ$ intersects the $X$ axis.

Given: $\;\;$ $P \left(x_1, y_1\right) = \left(2, 6\right)$, $\;\;$ $Q \left(x_2, y_2\right) = \left(-3, 5\right)$

1. Gradient of $PQ$ $= m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{5 - 6}{-3 - 2} = \dfrac{-1}{-5} = \dfrac{1}{5}$



2. Equation of $PQ$ is
   
   $y - y_1 = m \left(x - x_1\right)$
   
   i.e. $\;$ $y - 6 = \dfrac{1}{5} \left(x - 2\right)$
   
   i.e. $\;$ $5y - 30 = x - 2$
   
   i.e. $\;$ $x - 5y + 28 = 0$



3. When $PQ$ intersects the $X$ axis, the $Y$ coordinate of the point of intersection is $0$.
   
   Let $PQ$ intersect the $X$ axis at the point $A\left(a, 0\right)$.
   
   Substituting the coordinates of point $A$ in the equation of $PQ$ gives,
   
   $a - 5\times 0 + 28 = 0$ $\implies$ $a = -28$
   
   $\therefore \;$ $PQ$ intersects the $X$ axis at the point $\left(-28, 0\right)$.

Linear Inequations

Solve the inequation $\;\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}, \; x \in R$
Write the solution set and represent it on a number line.

Consider $\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{x}{2} + \dfrac{x}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{5x}{6}$

i.e. $\;$ $8 \leq 5x$

i.e. $\;$ $\dfrac{8}{5} \leq x$

i.e. $\;$ $1.6 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{1}{6} + \dfrac{4}{3}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{9}{6}$

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

$\therefore \;$ From equations $(1)$ and $(2)$, the solution set of the given inequation is

$\left\{x \; \bigg | \; 1.6 \leq x < 3, \; \; x \in R \right\}$

Commercial Mathematics - Banking

A person deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is $8 \%$ per annum and the person gets ₹ $8088$ from the bank after $3$ years, find the value of the monthly installment.

Let money deposited per month $= P = $ ₹ $x$

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

Rate of interest $= r = 8\%$ per annum

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

i.e. $\;$ $I =$ ₹ $\left[\dfrac{x \times 36 \times 37}{2 \times 12} \times \dfrac{8}{100}\right]$

i.e. $\;$ $I =$ ₹ $\left[\dfrac{111 x}{25}\right] = $ ₹ $4.44 x$

Total money deposited in $36$ months $= $ ₹ $36 x$

Maturity value (MV) $=$ Total money deposited $+ $ Interest

i.e. $\;$ $MV = $ ₹ $\left(36 x + 4.44 x\right) = $ ₹ $40.44 x$

Given: $\;$ $MV =$ ₹ $8088$

$\implies$ $40.44x = 8088$ $\implies$ $x = 200$

i.e. $\;$ money deposited per month $= $ ₹ $200$

Quadratic Equations

A shopkeeper purchases certain number of books for ₹ $960$. If the cost per book was ₹ $8$ less, the number of books that could be purchased for ₹ $960$ would be $4$ more. Write an equation, taking the original cost of each book to be ₹ $x$. Solve the equation to find the original cost of the books.

Let the original cost of each book $= $ ₹ $x$

Amount for which books purchased $= $ ₹ $960$

$\therefore \;$ Original number of books purchased $= \dfrac{960}{x}$

New cost of each book $= $ ₹ $\left(x - 8\right)$

New number of books that can be bought for ₹ $960$ $= \dfrac{960}{x - 8}$

As per question,

new number of books bought $= $ original number of books bought $+ \; 4$

i.e. $\;$ $\dfrac{960}{x - 8} = \dfrac{960}{x} + 4$

i.e. $\;$ $240 \left[\dfrac{1}{x - 8} - \dfrac{1}{x}\right] = 1$

i.e. $\;$ $240 \left[\dfrac{x - x + 8}{x^2 - 8x}\right] = 1$

i.e. $\;$ $x^2 - 8x - 1920 = 0$

i.e. $\;$ $x^2 - 48x + 40x - 1920 = 0$

i.e. $\;$ $x \left(x - 48\right) + 40 \left(x - 48\right) = 0$

i.e. $\;$ $\left(x - 48\right) \left(x + 40\right) = 0$

i.e. $\;$ $x - 48 = 0$ $\;$ or $\;$ $x + 40 = 0$

i.e. $\;$ $x = 48$ $\;$ or $\;$ $x = -40$

$\because \;$ cost of books cannot be negative,

$\therefore \;$ original cost of each book $= $ ₹ $x =$ ₹ $48$

Coordinate Geometry: Section Formula

Calculate the ratio in which the line joining $A \left(-4, 2\right)$ and $B \left(3, 6\right)$ is divided by the point $P \left(x, 3\right)$.

Also find the coordinates of point $P$ and the length of $AP$.

Given: $\;$ $A \left(-4, 2\right)$, $\;$ $B \left(3, 6\right)$

Let the point $P \left(x,3\right)$ divide the line joining the points $A$ and $B$ in the ratio $k : 1$

Then, by section formula [for $Y$ coordinate of point $P$]

$\dfrac{6k + 2 \times 1}{k + 1} = 3$

i.e. $\;$ $6k + 2 = 3k +3$

i.e. $\;$ $3k = 1$ $\implies$ $k = \dfrac{1}{3}$

i.e. $\;$ the point $P$ divides $AB$ in the ratio $1 : 3$

Again, by section formula [for $X$ coordinate of point $P$]

$\dfrac{-4 \times 3 + 3 \times 1}{1 + 3} = x$

i.e. $\;$ $x = \dfrac{-12 + 3}{4} = \dfrac{-9}{4}$

$\therefore \;$ Point $\;$ $P = \left(x, 3\right) = \left(\dfrac{-9}{4}, 3\right)$

By distance formula,

$\begin{aligned} \text{length of AP} & = \sqrt{\left[-4 - \left(\dfrac{-9}{4}\right)\right]^2 + \left[2 - 3\right]^2} \\\\ & = \sqrt{\left(\dfrac{-7}{4}\right)^2 + \left(-1\right)^2} \\\\ & = \sqrt{\dfrac{49}{16} + 1} \\\\ & = \dfrac{\sqrt{65}}{4} \end{aligned}$