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,

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

Geometric Progression

Find the geometric progression (GP) for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.

Let the first term of GP $= t_1 = a$; $\;$ common ratio $= r$

$n^{th}$ term of GP $= t_n = ar^{n-1}$

$2^{nd}$ term of GP $= t_2 = ar^{2 - 1} = ar$

$3^{rd}$ term of GP $= t_3 = ar^{3 - 1} = ar^2$

$5^{th}$ term of GP $= t_5 = ar^{5 - 1} = ar^4$

Now, $\;$ $t_1 + t_2 = -4$ $\;\;\;$ [given]

i.e. $\;$ $a + ar = -4$ $\implies$ $a = \dfrac{-4}{1 + r}$ $\;\;\; \cdots \; (1)$

$t_5 = 4 \times t_3$ $\;\;\;$ [given]

i.e. $\;$ $a r^4 = 4 \times ar^2$

i.e. $\;$ $r^2 = 4$ $\implies$ $r = \pm 2$

Substituting $r = +2$ in equation $(1)$ gives

$a = \dfrac{-4}{1 + 2} = \dfrac{-4}{3}$

Then the GP is $\;$ $\dfrac{-4}{3}, \; \dfrac{-8}{3}, \; \dfrac{-16}{3}, \; \dfrac{-32}{3}, \; \dfrac{-64}{3}, \cdots$

Substituting $r = -2$ in equation $(1)$ gives

$a = \dfrac{-4}{1 - 2} = 4$

Then the GP is $\;$ $4, \; -8, \; 16, \; -32, \; 64, \cdots$

Mensuration

A solid metallic sphere of radius $12 \; cm$ is melted and recast into a solid cylinder of height $36 \; cm$. Find the radius and the curved surface area of the cylinder. Take $\pi = 3.14$

Radius of sphere $= R = 12 \; cm$

Height of cylinder $= h = 36 \; cm$

Let radius of cylinder $= r \; cm$

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

Volume of cylinder $= V_C = \pi r^2 h$

Since the sphere is melted and recast into a solid cylinder,

Volume of sphere $= $ Volume of cylinder

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

i.e. $\;$ $r^2 = \dfrac{4 R^3}{3 h} = \dfrac{4 \times 12^3}{3 \times 36} = 64$

$\implies$ $r = \pm 8$

$\therefore \;$ Radius of cylinder $= r = 8 \; cm$ $\;$ [negative value is discarded as radius cannot be negative]

Curved surface area of the cylinder $= CSA = 2 \pi r h$

i.e. $\;$ $CSA = 2 \times 3.14 \times 8 \times 36 = 1808.64 \; cm^2$

Trigonometry

Show that $\;$ $\sqrt{\dfrac{1 - \cos A}{1 + \cos A}} = \dfrac{1 - \cos A}{\sin A}$

$\begin{aligned} LHS & = \sqrt{\dfrac{1 - \cos A}{1 + \cos A}} \\\\ & = \sqrt{\dfrac{\left(1 - \cos A\right)^2}{\left(1 + \cos A\right) \left(1 - \cos A\right)}} \\\\ & = \dfrac{1 - \cos A}{\sqrt{1 - \cos^2 A}} \\\\ & = \dfrac{1 - \cos A}{\sqrt{\sin^2 A}} \\\\ & = \dfrac{1 - \cos A}{\sin A} = RHS \end{aligned}$

Hence proved.

Statistics - Mean, Median

The median of the following observations $\;$ $9$, $\;$ $10$, $\;$ $12$, $\;$ $\left(x - 4\right)$, $\;$ $\left(x + 2\right)$, $\;$ $\left(x + 7\right)$, $\;$ $30$, $\;$ $36$, $\;$ $45$ $\;$ arranged in ascending order is $22$.

Find the value of $x$ and hence find the mean of the observations.

Number of observations $= n = 9 $ $\;$ [odd number of observations]

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

Given: $\;$ Median $= 22$

i.e. $\;$ $x + 2 = 22$ $\implies$ $x = 20$

$\therefore \;$ The given observations are:

$9, \; 10, \; 12, \; 16, \; 22, \; 27, \; 30, \; 36, \; 45$

Mean of observations $= \dfrac{\Sigma x}{n}$

i.e. $\;$ Mean $= \dfrac{9 + 10 + 12 + 16 + 22 + 27 + 30 + 36 + 45}{9} = \dfrac{207}{9} = 23$

Arithmetic Progression

In an arithmetic progression, the third term is $8$ and the seventh term is $12$. Find the

1. first term;
2. common difference;
3. sum of first $20$ terms.

Let

first term of the A.P be $= t_1 = a$

common difference $= d$

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

$\therefore \;$ third term of AP $= t_3 = a + 2d = 8$ $\;\;\;$ [given] $\;\;\; \cdots \; (1)$

seventh term of AP $= t_7 = a + 6d = 12$ $\;\;\;$ [given] $\;\;\; \cdots \; (2)$

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

$4d = 4$ $\implies$ $d = 1$

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

$a = 8 - 2d = 8 - 2 = 6$

$\therefore \;$ First term $= a = 6$; $\;$ common difference $= d = 1$

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

$\therefore \;$ Sum of first $20$ terms $= S_{20} = \dfrac{20}{2} \left[2 \times 6 + \left(20 - 1\right) \times 1\right]$

i.e. $S_{20} = 10 \left[12 + 19\right] = 310$

Factor and Remainder Theorems

$\left(x - 1\right)$ is a factor of the expression $2x^3 + ax^2 + bx - 14$.

When the expression is divided by $\left(x - 3\right)$, it leaves a remainder $52$.

Find the values of $a$ and $b$.

Let $\;$ $f \left(x\right) = 2x^3 + ax^2 + bx -14$

Given: $\;$ $\left(x - 1\right)$ is a factor of $f \left(x\right)$

Then, by factor theorem, $\;$ $f \left(1\right) = 0$

i.e. $\;$ $f \left(1\right) = 2 \times 1^3 + a \times 1^2 + b \times 1 - 14 = 0$

i.e. $\;$ $a + b = 12$ $\;\;\; \cdots \; (1)$

Given: $\;$ When $f \left(x\right)$ is divided by $\left(x - 3\right)$, it leaves a remainder $52$

Then, by remainder theorem, $\;$ $f \left(3\right) = 52$

i.e. $\;$ $f \left(3\right) = 2 \times 3^3 + a \times 3^2 + b \times 3 - 14 = 52$

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

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

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

$2a = -8$ $\implies$ $a = -4$

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

$b = 12 - a = 12 - \left(-4\right) = 16$

Trigonometry

Prove that $\;$ $\left(\text{cosec} \theta - \sin \theta\right) \left(\sec \theta - \cos \theta\right) \left(\tan \theta + \cot \theta\right) = 1$

$\begin{aligned} LHS & = \left(\text{cosec} \theta - \sin \theta\right) \left(\sec \theta - \cos \theta\right) \left(\tan \theta + \cot \theta\right) \\\\ & = \left(\dfrac{1}{\sin \theta} - \sin \theta\right) \left(\dfrac{1}{\cos \theta} - \cos \theta\right) \left(\tan \theta + \dfrac{1}{\tan \theta}\right) \\\\ & = \left(\dfrac{1 - \sin^2 \theta}{\sin \theta}\right) \left(\dfrac{1 - \cos^2 \theta}{\cos \theta}\right) \left(\dfrac{1 + \tan^2 \theta}{\tan \theta}\right) \\\\ & = \dfrac{\cos^2 \theta}{\sin \theta} \times \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sec^2 \theta}{\tan \theta} \\\\ & = \dfrac{\cos \theta \times \sin \theta}{\cos^2 \theta \times \tan \theta} \\\\ & = \dfrac{\sin \theta}{\cos \theta} \times \dfrac{1}{\tan \theta} \\\\ & = \dfrac{\tan \theta}{\tan \theta} = 1 = RHS \end{aligned}$

Hence proved.

Matrices

Given $A = \begin{bmatrix} 4 & -12 \\ 4 & 0 \end{bmatrix}$, $B = \begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 4 & 0 \\ 0 & 0 \end{bmatrix}$, find matrix $X$ such that $A + 2X = 2B + C$

$A + 2X = 2B + C$

i.e. $\;$ $2X = 2B + C - A$

i.e. $\;$ $X = B + \dfrac{1}{2} \left(C - A\right)$

$\begin{aligned} i.e. \;\; X & = \begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix} + \dfrac{1}{2} \left\{\begin{bmatrix} 4 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 4 & -12 \\ 4 & 0 \end{bmatrix} \right\} \\\\ & = \begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix} + \dfrac{1}{2} \begin{bmatrix} 4 - 4 & 0 + 12 \\ 0 - 4 & 0 - 0 \end{bmatrix} \\\\ & = \begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix} + \dfrac{1}{2} \begin{bmatrix} 0 & 12 \\ -4 & 0 \end{bmatrix} \\\\ & = \begin{bmatrix} -3 & 2 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 6 \\ -2 & 0 \end{bmatrix} \\\\ & = \begin{bmatrix} -3 + 0 & 2 + 6 \\ 4 - 2 & 1 + 0 \end{bmatrix} \\\\ & = \begin{bmatrix} -3 & 8 \\ 2 & 1 \end{bmatrix} \end{aligned}$

Commercial Mathematics - Goods and Services Tax

A wholesaler buys a TV from a manufacturer for ₹ $25,000$. The wholesaler marks the price of the TV $20 \%$ above the cost price and sells it to a retailer at a discount of $10 \%$ on the marked price. If the rate of GST is $28 \%$, find

1. the marked price;
2. retailer's cost price inclusive of tax;
3. GST paid by the wholesaler.

1. Cost Price of TV for wholesaler $= $ ₹ $25,000$
   
   Marked price of TV $= $ ₹ $\left(25,000 + 20 \% \text{ of } 25,000\right)$
   
   i.e $\;$ Marked price $= $ ₹ $\left(25,000 + \dfrac{20}{100} \times 25,000\right) = $ ₹ $\left(25,000 + 5,000\right) = $ ₹ $30,000$



2. Discount given by wholesaler $= 10 \% \text{ of }$ ₹ $30,000$
   
   i.e. $\;$ Discount $= $ ₹ $\dfrac{10}{100} \times 30,000 = $ ₹ $3,000$
   
   $\therefore \;$ Amount paid by retailer without GST $= $ ₹ $\left(30,000 - 3,000\right) = $ ₹ $27,000$
   
   Rate of GST $= 28\%$
   
   $\therefore \;$ Amount of GST paid by retailer $= 28 \% \text{ of }$ ₹ $27,000$
   
   i.e. $\;$ Amount of GST paid by retailer $= $ ₹ $\left(\dfrac{28}{100} \times 27,000\right) =$ ₹ $7,560$
   
   $\therefore \;$ Retailer's cost price (inclusive of tax) $= $ ₹ $\left(27,000 + 7,560\right) = $ ₹ $34,560$



3. Cost price for wholesaler $= $ ₹ $25,000$
   
   $\therefore \;$ GST paid by wholesaler for purchase $= $ ₹ $28 \% \text{ of } 25,000$
   
   i.e. $\;$ GST paid by wholesaler $= $ ₹ $\left(\dfrac{28}{100} \times 25,000\right) = $ ₹ $7,000$
   
   Sale price for wholesaler $= $ ₹ $27,000$
   
   GST charged by wholesaler on selling of TV $= $ ₹ $28 \% \text{ of } 27,000 = $ ₹ $7,560$
   
   $\therefore \;$ GST paid by wholesaler $= $ GST charged on selling price $- $ GST paid against purchase price
   
   i.e. $\;$ GST paid by wholesaler $= $ ₹ $\left(7,560 - 7,000\right) = $ ₹ $560$

Constructions

Construct the circumcircle to triangle $ABC$ in which $AB = 5 \; cm$, $BC = 8 \; cm$ and $\angle ABC = 60^\circ$.

Construct cyclic quadrilateral $ABCD$ such that $D$ is equidistant from $B$ and $C$.

Using the given data, construct $\triangle ABC$ with $AB = 5 \; cm$, $BC = 8 \; cm$ and $\angle ABC = 60^\circ$.

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

With $O$ as center and $OA$ as radius draw a circle which passes through the vertices of $\triangle ABC$.

This circle is the required circumcircle to $\triangle ABC$.

Since point $D$ is equidistant from the points $B$ and $C$, $D$ lies on the perpendicular bisector of $BC$.

The point where the perpendicular bisector of $BC$ intersects the circumcircle is the point $D$.

$ABCD$ is the required cyclic quadrilateral.

Similarity

In the figure, $\triangle ABC$ and $\triangle CEF$ are two triangles where $BA$ is parallel to $CE$ and $\dfrac{AF}{AC} = \dfrac{5}{8}$.

1. Prove that $\;$ $\triangle ADF \sim \triangle CEF$.
2. Find $AD$ if $CE = 6 \; cm$.
3. If $DF \parallel BC$, find $\;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)}$.

1. Given: $\;$ $BA \parallel CE$
   
   $\implies$ $AD \parallel CE$ $\;\;\;$ [$B - D - A$ are collinear points]
   
   $AC$ is a transversal.
   
   $\therefore \;$ $\angle DAF = \angle FCE$ $\;\;\;$ [alternate angles]
   
   $\angle AFD = \angle CFE$ $\;\;\;$ [vertically opposite angles]
   
   $\therefore \;$ by AA (angle - angle) axiom, $\;$ $\triangle ADF \sim \triangle CEF$



2. $\because \;$ $\triangle ADF \sim \triangle CEF$,
   
   $\dfrac{AD}{CE} = \dfrac{AF}{CF}$ $\;\;\; \cdots \; (1)$ [corresponding sides of similar triangles are in proportion]
   
   Given: $\;$ $\dfrac{AF}{AC} = \dfrac{5}{8}$
   
   i.e. $\;$ $\dfrac{AF}{AF + CF} = \dfrac{5}{8}$
   
   i.e. $\;$ $\dfrac{AF + CF}{AF} = \dfrac{8}{5}$
   
   i.e. $\;$ $1 + \dfrac{CF}{AF} = \dfrac{8}{5}$
   
   i.e. $\;$ $\dfrac{CF}{AF} = \dfrac{8}{5} - 1 = \dfrac{3}{5}$
   
   $\therefore \;$ $\dfrac{AF}{CF} = \dfrac{5}{3}$ $\;\;\; \cdots \; (2)$
   
   In view of equation $(2)$, equation $(1)$ becomes
   
   $\dfrac{AD}{CE} = \dfrac{5}{3}$
   
   i.e. $\;$ $AD = CE \times \dfrac{5}{3} = 6 \times \dfrac{5}{3} = 10 \; cm$



3. $\because \;$ $DF \parallel BC$ and $AB$ is the transversal,
   
   in triangles $ADF$ and $ABC$,
   
   $\angle ADF = \angle ABC$ $\;\;\;$ [corresponding angles]
   
   $\angle DAF = \angle BAC$ $\;\;\;$ [common angle]
   
   $\therefore \;$ $\triangle ADF \sim \triangle ABC$ $\;\;\;$ [by A.A axiom]
   
   Now, ratio between the area of two similar triangles is equal to the ratio between the squares of its corresponding sides.
   
   $\therefore \;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ABC\right)} = \dfrac{AF^2}{AC^2} = \dfrac{5^2}{8^2} = \dfrac{25}{64}$
   
   i.e. $\;$ $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area} \left(\triangle ADF\right)} = \dfrac{64}{25}$ $\;\;\; \cdots \; (3)$
   
   Now, $\;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)} = \dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ABC\right) - \text{Area} \left(\triangle ADF\right)}$
   
   $\begin{aligned} i.e. \; \dfrac{\text{Area} \left(\square DFCB\right)}{\text{Area} \left(\triangle ADF\right)} & = \dfrac{\text{Area} \left(\triangle ABC\right) - \text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\triangle ADF\right)} \\\\ & = \dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area} \left(\triangle ADF\right)} - 1 \\\\ & = \dfrac{64}{25} - 1 \;\;\; [\text{in view of equation (3)}] \\\\ & = \dfrac{39}{25} \end{aligned}$
   
   $\therefore \;$ $\dfrac{\text{Area} \left(\triangle ADF\right)}{\text{Area} \left(\square DFCB\right)} = \dfrac{25}{39}$

Circle

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

$CD$ and $BE$ are parallel.

$\angle AOB = 80^\circ$ and $\angle ACE = 10^\circ$. Find:

1. $\angle BEC$;
2. $\angle BCD$;
3. $\angle CED$.

Given:

$AC =$ diameter of circle

$O =$ center of circle

$CD \parallel BE$

$\angle AOB = 80^\circ$; $\;$ $\angle ACE = 10^\circ$

1. $\angle AOB + \angle BOC = 180^\circ$ $\;\;\;$ [linear pair]
   
   $\therefore \;$ $\angle BOC = 180^\circ - \angle AOB = 180^\circ - 80^\circ = 100^\circ$
   
   Now, $\;$ $\angle BEC = \dfrac{1}{2} \angle BOC$ $\;\;\;$ [angle at center is twice the angle at the remaining circumference]
   
   i.e. $\;$ $\angle BEC = \dfrac{1}{2} \times 100^\circ = 50^\circ$



2. $CD \parallel BE$ and $EC$ is a transversal.
   
   $\therefore \;$ $\angle ECD = 50^\circ$
   
   Now, $\;$ $OB = OC$ $\;\;\;$ [radii of the circle]
   
   $\therefore \;$ $\angle OBC = \angle OCB$ $\;\;\;$ [angles opposite equal sides are equal]
   
   Now, $\;$ in $\triangle BOC$,
   
   $\angle OBC + \angle OCB + \angle BOC = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
   
   i.e. $\;$ $2 \angle OCB + \angle BOC = 180^\circ$
   
   i.e. $\;$ $2 \angle OCB = 180^\circ - \angle BOC = 180^\circ - 100^\circ = 80^\circ$
   
   $\implies$ $\angle OCB = 40^\circ$
   
   Now, $\;$ $\angle BCD = \angle ECD + \angle ACE + \angle OCB$
   
   i.e. $\angle BCD = 50^\circ + 10^\circ + 40^\circ = 100^\circ$



3. $BCDE$ is a cyclic quadrilateral.
   
   Then, $\;$ $\angle BED + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
   
   $\therefore \;$ $\angle BED = 180^\circ - \angle BCD = 180^\circ - 100^\circ = 80^\circ$
   
   Now, $\;$ $\angle BED = \angle BEC + \angle CED$
   
   $\therefore \;$ $\angle CED = \angle BED - \angle BEC = 80^\circ - 50^\circ = 30^\circ$

Statistics - Ogive

The following table gives the daily wages of workers in a factory:

Wages (₹)	Number of workers
$400 - 450$	$2$
$450 - 500$	$6$
$500 - 550$	$12$
$550 - 600$	$18$
$600 - 650$	$24$
$650 - 700$	$13$
$700 - 750$	$5$

Use a graph paper to draw an ogive for the given data. Use a scale of $2 \; cm = $ ₹ $100$ on the X-axis and $2 \; cm = 20$ workers on the Y-axis.

From the ogive estimate

1. the median wage of the workers;
2. the inter-quartile range;
3. the number of workers who earn more than ₹ $625$ daily.

Wages (₹) (Class Interval)	Number of workers (frequency)	Cumulative Frequency
$400 - 450$	$2$	$2$
$450 - 500$	$6$	$8$
$500 - 550$	$12$	$20$
$550 - 600$	$18$	$38$
$600 - 650$	$24$	$62$
$650 - 700$	$13$	$75$
$700 - 750$	$5$	$80$

Total number of workers $= N = 80$

1. Median wage of workers $= \left(\dfrac{N}{2}\right)^{th}$ term
   
   $\begin{aligned} \text{i.e.} \; \text{Median wage} & = \left(\dfrac{80}{2}\right)^{th} \text{ term} \\\\ & = 40^{th} \text{ term} = 605 \end{aligned}$
   
   i.e. $\;$ Median wage of workers $= $ ₹ $605$



2. Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th}$ term
   
   i.e. $\;$ $Q_1 = \left(\dfrac{80}{4}\right)^{th}$ term $= 20^{th}$ term $= 550$
   
   Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th}$ term
   
   i.e. $\;$ $Q_3 = \left(\dfrac{3 \times 80}{4}\right)^{th}$ term $= 60^{th}$ term $= 640$
   
   $\therefore \;$ Inter-quartile range $= Q_3 - Q_1 = 640 - 550 = 90$



3. Number of workers who earn ₹ $625$ daily $= 50$
   
   $\therefore \;$ Number of workers who earn more than ₹ $625$ daily $= 80 - 50 = 30$

Factor and Remainder Theorems

Use the remainder theorem to factorize the expression $\;$ $2x^3 + x^2 - 13x + 6$

Let $\;$ $f \left(x\right) = 2x^3 + x^2 - 13x + 6$

For $x = 2$, the value of the given expression is:

$f \left(2\right) = 2 \times 2^3 + 2^2 - 13 \times 2 + 6 = 16 + 4 - 26 + 6 = 0$

$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$.

$\begin{array}{lllll} x - 2 & ) & 2x^3 + x^2 - 13x + 6 & ( & 2x^2 + 5x - 3 \\ & & 2x^3 - 4x^2 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ 5x^2 - 13x & & \\ & & \ \ \ \ \ \ \ \ 5x^2 - 10x & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{array}$

$\begin{aligned} \therefore \; 2x^3 + x^2 - 13x + 6 & = \left(x - 2\right) \left[2x^2 + 5x - 3\right] \\\\ & = \left(x - 2\right) \left[2x^2 + 6x - x - 3\right] \\\\ & = \left(x - 2\right) \left[2x \left(x + 3\right) - 1 \left(x + 3\right)\right] \\\\ & = \left(x - 2\right) \left(2x - 1\right) \left(x + 3\right) \end{aligned}$

Circle

In the figure, $\;$ $PQ = QR$, $\;$ $\angle RQP = 68^\circ$, $\;$ $PC$ and $CQ$ are tangents to the circle with center $O$.

Calculate the values of

1. $\angle QOP$
2. $\angle QCP$

Given:

$PQ = QR$, $\;$ $\angle RQP = 68^\circ$, $\;$ $PC, \; CQ$ are tangents, $\;$ $O $ is the center of the circle

1. In $\triangle PQR$, $\;$ $\angle PRQ = \angle RQP = 68^\circ$ $\;\;\;$ [$\because \;$ $PQ = QR$, angles opposite equal sides are equal]
   
   $\angle QOP = 2 \angle PRQ$ $\;\;\;$ [angle at the center is twice the angle at remaining circumference]
   
   $\therefore \;$ $\angle QOP = 2 \times 68^\circ = 136^\circ$



2. $OP = OQ$ $\;\;\;$ [radii of the same circle]
   
   $\angle OPC = \angle OQC = 90^\circ$ $\;\;\;$ [angle between radius and tangent is $90^\circ$]
   
   $POQC$ is a quadrilateral.
   
   $\therefore \;$ $\angle QOP + \angle OPC + \angle OQC + \angle QCP = 360^\circ$ $\;\;\;$ [sum of angles of a quadrilateral]
   
   $\begin{aligned} \therefore \; \angle QCP & = 360^\circ - \left(\angle QOP + \angle OPC + \angle OQC\right) \\\\ & = 360^\circ - \left(136^\circ + 90^\circ + 90^\circ\right) \\\\ & = 44^\circ \end{aligned}$

Circle

In the figure, $PT$ is a tangent to the circle at point $T$. $CT$ is a diameter.

$\angle ABC = 70^\circ$, $\;$ $\angle ACB = 50^\circ$.

Calculate the measure of

1. $\angle CBT$
2. $\angle BAT$
3. $\angle APT$

$\angle ABC = 70^\circ$, $\;$ $\angle ACB = 50^\circ$

$PT =$ tangent to the circle at point $T$

$CT =$ diameter of the circle

1. $\angle CBT = 90^\circ$ $\;\;\;$ [angle in a semicircle]



2. In $\;$ $\triangle ABC$, $\;$ $\angle CAB + \angle ABC + \angle ACB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
   
   $\therefore \;$ $\angle CAB = 180^\circ - \left(\angle ABC + \angle ACB\right) = 180^\circ - \left(70^\circ + 50^\circ\right) = 60^\circ$
   
   $\angle CAT = 90^\circ$ $\;\;\;$ [angle in a semicircle]
   
   Now, $\;$ $\angle CAT = \angle CAB + \angle BAT$
   
   $\therefore \;$ $\angle BAT = \angle CAT - \angle CAB = 90^\circ - 60^\circ = 30^\circ$



3. $\angle CTB = \angle CAB = 60^\circ$ $\;\;\;$ [angles of the same segment]
   
   In the figure, $\;$ $\angle CBT = \angle ABC + \angle PBT$
   
   $\therefore \;$ $\angle PBT = \angle CBT - \angle ABC = 90^\circ - 70^\circ = 20^\circ$
   
   Let $PB$ and $CT$ intersect at point $X$.
   
   $\because \;$ $P - X - B$ are collinear points, $\;$ $\angle XBT = \angle PBT = 20^\circ$
   
   $\because \;$ $C - X - T$ are collinear points, $\;$ $\angle XTB = \angle CTB = 60^\circ$
   
   In $\;$ $\triangle XTB$, $\;$ $\angle XTB + \angle XBT + \angle TXB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
   
   $\therefore \;$ $\angle TXB = 180^\circ - \left(\angle XTB + \angle XBT\right) = 180^\circ - \left(60^\circ + 20^\circ\right) = 100^\circ$
   
   Now, $\;$ $\angle TXB + \angle CXB = 180^\circ$ $\;\;\;$ [linear pair]
   
   $\therefore \;$ $\angle CXB = 180^\circ - \angle TXB = 180^\circ - 100^\circ = 80^\circ$
   
   $\angle PXT = \angle CXB = 80^\circ$ $\;\;\;$ [vertically opposite angles]
   
   $\angle PTX = 90^\circ$ $\;\;\;$ [angle between diameter and tangent is $90^\circ$]
   
   In $\triangle PXT$, $\;$ $\angle PXT + \angle PTX + \angle XPT = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
   
   $\therefore \;$ $\angle XPT = 180^\circ - \left(\angle PXT + \angle PTX\right) = 180^\circ - \left(80^\circ + 90^\circ\right) = 10^\circ$
   
   $\because \;$ $P - A - X$ are collinear points, $\;$ $\angle APT = \angle XPT = 10^\circ$

Commercial Mathematics - Shares and Dividends

A person invested ₹ $10,000$ in $8 \%$, ₹ $25$ shares at ₹ $40$. After a year, the shares were sold at ₹ $42$ each and the proceeds (including the dividend) were invested in $9 \%$, $\;$ ₹ $10$ shares at ₹ $11$. Find:

1. the dividend for the first year;
2. the new number of shares bought;
3. the percentage increase in the return on the original investment.

Investment $= $ ₹ $10,000$

Dividend $\% = 8 \%$

Nominal Value (N.V) of each share $= $ ₹ $25$

Market Value (M.V) of each share $= $ ₹ $40$

1. Number of shares bought $= \dfrac{\text{Investment}}{\text{M.V of each share}} = \dfrac{10,000}{40} = 250$
   
   Dividend on $1$ share $= 8 \% $ of ₹ $25 = \dfrac{8}{100} \times 25 = $ ₹ $2$
   
   $\therefore \;$ Dividend for first year $= $ ₹ $\left(2 \times 250\right) = $ ₹ $500$



2. Since each share is sold for ₹ $42$,
   
   $\therefore \;$ Proceeds (including dividend) $= 250 \times 42 + 500 = $ ₹ $11,000$
   
   $\therefore \;$ Sum invested $= $ proceeds $= $ ₹ $11,000$
   
   Nominal value (N.V) of each new share $= $ ₹ $10$
   
   Market value (M.V) of each new share $= $ ₹ $11$
   
   $\therefore \;$ Number of new shares bought $= \dfrac{\text{Investment}}{\text{M.V of each share}} = \dfrac{11,000}{11} = 1000$



3. New dividend $= 9 \%$
   
   Dividend on $1$ share $= 9 \% $ of ₹ $10 = \dfrac{9}{100} \times 10 = $ ₹ $0.90$
   
   $\therefore \;$ Annual dividend (income) in the second year $= $ ₹ $\left(0.90 \times 1000\right) = $ ₹ $900$
   
   $\therefore \;$ Increase in return $= $ ₹ $\left(900 - 500\right) = $ ₹ $400$
   
   $\therefore \;$ $\%$ Increase in return (on the original investment)
   
   $= \dfrac{\text{Increase}}{\text{Original investment}} \times 100 \% = \dfrac{400}{10,000} \times 100 = 4 \%$

Trigonometry - Heights and Distances

An airplane at a height of $6 \; km$ passes vertically above another plane at an instant when their angles of elevation at the same observing point are $60^\circ$ and $45^\circ$ respectively. Find the difference between the heights of the planes. Take $\sqrt{3} = 1.732$

$P_1, \; P_2 =$ Airplanes

$O =$ Point of observation

$AP_1 =$ Height of airplane $P_1 = 6 \; km$

$P_1 P_2 =$ Difference between the heights of the planes

In $\triangle OP_1A$, $\;$ $\dfrac{AP_1}{OA} = \tan 60^\circ$

$\implies$ $OA = \dfrac{AP_1}{\tan 60^\circ} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3} \; km$

In $\triangle OP_2A$, $\;$ $\dfrac{AP_2}{OA} = \tan 45^\circ$

$\implies$ $AP_2 = OA \times \tan 45^\circ = 2 \sqrt{3} \times 1 = 2 \sqrt{3} \; km$

Difference between the heights of the planes $= P_1P_2$

$P_1 P_2 = AP_1 - AP_2 = 6 - 2 \sqrt{3} = 2.536 \; km$

Similarity - Maps and Models

The dimensions of a model of multi-storied building are $1 \; m$ by $60 \; cm$ by $1.20 \; m$. If the scale factor is $1 : 50$, find:

1. the floor area of a room of the building in $m^2$, if the floor area of the corresponding room in the model is $50 \; cm^2$;
2. the space inside a room of the model in $cm^3$, if the space inside the corresponding room of the building is $90 \; m^3$.

Scale factor $= k = 1 : 50 = \dfrac{1}{50}$

1. Floor area of room in the model $= 50 \; cm^2 = 50 \times 10^{-4} \; m^2$
   
   Floor area of room in model $= k^2 \times$ Floor area of room of building
   
   i.e. $\;$ $50 \times 10^{-4} = \left(\dfrac{1}{50}\right)^2 \times $ Floor area of room of building
   
   i.e. $\;$ Floor area of room of building $= 50 \times 10^{-4} \times 2500 = 7.5 \; m^2$



2. Volume of a room of the model $= k^3 \times $ Volume of a room of the building
   
   $\begin{aligned} i.e. \; \text{Volume of a room of the model} & = \left(\dfrac{1}{50}\right)^3 \times 90 \\\\ & = \dfrac{90}{125 \times 10^3} \\\\ & = 0.72 \times 10^{-3} \; m^3 \\\\ & = 0.72 \times 10^{-3} \times 10^6 \; cm^3 \\\\ & = 720 \; cm^3 \end{aligned}$

Coordinate Geometry

Answer the entire question on a graph paper. Use a scale of $1 \; cm = 1 $ unit for both the $X$ and $Y$ axes.

Plot the points $A \left(-2, 2\right)$ and $B \left(4, 4\right)$.

Reflect $A$ and $B$ in the origin to get the points $A'$ and $B'$.

Write down the coordinates of points $A'$ and $B'$.

Give the geometrical name for the figure $ABA'B'$.

Points $A \left(-2, 2\right)$ and $B \left(4, 4\right)$ are reflected in the origin to get the points $A' \left(2, -2\right)$ and $B' \left(-4, -4\right)$.

$ABA'B'$ is a rhombus.

Linear Inequations

Find the values of $x$ which satisfies the inequation $\;$ $-2 \leq \dfrac{1}{2} - \dfrac{2x}{3} \leq 1 \dfrac{5}{6}, \; x \in N$
Graph the solution set on a number line.

Consider $\;$ $-2 \leq \dfrac{1}{2} - \dfrac{2x}{3}$

i.e. $\;$ $-2 \leq \dfrac{3 - 4x}{6}$

i.e. $\;$ $-12 \leq 3 - 4x$

i.e. $\;$ $4x \leq 15$

i.e. $\;$ $x \leq \dfrac{15}{4}$

i.e. $\;$ $x \leq 3.75$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{1}{2} - \dfrac{2x}{3} \leq 1 \dfrac{5}{6}$

i.e. $\;$ $\dfrac{1}{2} - \dfrac{2x}{3} \leq \dfrac{11}{6}$

i.e. $\;$ $\dfrac{3 - 4x}{6} \leq \dfrac{11}{6}$

i.e. $\;$ $3 - 4x \leq 11$

i.e. $\;$ $-4x \leq 8$

i.e. $\;$ $-x \leq 2$

i.e. $\;$ $-2 \leq x$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, the values of $x$ which satisfy the given inequation as:

$\left\{x \; \big | -2 \leq x \leq 3.75 \right\}$

$\because \;$ $x \in N$, $\;$ the required values of $x$ are: $\;$ $\left\{1, \; 2, \; 3 \right\}$

Commercial Mathematics - Banking

A person has a recurring deposit account of ₹ $400$ per month at $10 \%$ per annum. If the person gets ₹ $260$ as interest at the time of maturity, find the total time for which the account was held.

Let the account be held for $n$ months.

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

$\therefore \;$ Total money deposited $= $ ₹ $\left(400 \times n\right)$

Rate of interest $= r = 10\%$

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

i.e. $\;$ $I = $ ₹ $\left[400 \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{10}{100}\right]$

i.e. $\;$ $I = $ ₹ $\left[\dfrac{5 n \left(n + 1\right)}{3}\right]$

But, interest earned $= $ ₹ $260$

$\therefore \;$ $\dfrac{5n \left(n + 1\right)}{3} = 260$

i.e. $\;$ $5n^2 + 5n = 780$

i.e. $\;$ $n^2 + n = 156$

i.e. $\;$ $n^2 + n - 156 = 0$

i.e. $\;$ $n^2 + 13 n - 12n - 156 = 0$

i.e. $\;$ $n \left(n + 13\right) - 12 \left(n + 13\right) = 0$

i.e. $\;$ $\left(n + 13\right) \left(n - 12\right) = 0$

i.e. $\;$ $n + 13 = 0$ $\;$ or $\;$ $n - 12 = 0$

i.e. $\;$ $n = -13$ $\;$ or $\;$ $n = 12$

Since time cannot be negative, the time for which the account was held $= n = 12$ months $= 1$ year

Statistics

Determine the mean of the following frequency distribution by short-cut (assumed mean) method:

Class Interval	$10 - 16$	$16 - 22$	$22 - 28$	$28 - 34$	$34 - 40$
Frequency	$1$	$10$	$5$	$3$	$6$

Class Interval	Frequency $f_i$	Mid value $x_i$	Assumed Mean $A = 25; \;\; d_i = x_i - A$	$f_i d_i$
$10 - 16$	$1$	$13$	$-12$	$-12$
$16 - 22$	$10$	$19$	$-6$	$-60$
$22 - 28$	$5$	$25$	$0$	$0$
$28 - 34$	$3$	$31$	$6$	$18$
$34 - 40$	$6$	$37$	$12$	$72$

$\Sigma f_i = N = 25$, $\;\;$ $\Sigma f_i d_i = 18$

Mean $= A + \dfrac{\Sigma f_i d_i}{\Sigma f_i} = 25 + \dfrac{18}{25} = 25 + 0.72 = 25.72$

Statistics

Find the mean and the median for the following frequency distribution:

$x$	$8$	$9$	$10$	$11$	$12$
$f$	$5$	$4$	$2$	$6$	$3$

$x_i$	$f_i$	$f_i x_i$	Cumulative frequency
$8$	$5$	$40$	$5$
$9$	$4$	$36$	$9$
$10$	$2$	$20$	$11$
$11$	$6$	$66$	$17$
$12$	$3$	$36$	$20$

$\Sigma f_i = N = 20$ (Even), $\;$ $\Sigma f_i x_i = 198$

Mean $= \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{198}{20} = 9.9$

$\because$ $N$ is even, median $= \dfrac{\left(\dfrac{N}{2}\right)^{th} \text{term} + \left(\dfrac{N}{2} + 1\right)^{th} \text{term}}{2}$

i.e. $\;$ Median $= \dfrac{\left(\dfrac{20}{2}\right)^{th} \text{term} + \left(\dfrac{20}{2} + 1\right)^{th} \text{term}}{2}$

i.e. $\;$ Median $= \dfrac{10^{th} \; \text{term} + 11^{th} \; \text{term}}{2}$

i.e. $\;$ Median $= \dfrac{10 + 10}{2} = \dfrac{20}{2} = 10$

Probability

If two digit numbers are made with the digits $2, \; 3, \; 5$, what is the probability (when repetition of digits is not allowed) that the number is

1. greater than $35$;
2. a multiple of $2$;
3. a prime number.

Sample space $= S = \left\{23, \; 25, \; 32, \; 35, \; 52, \; 53 \right\}$

Number of elements in sample space $= n \left(S\right) = 6$

Let $A =$ event of having a two digit number greater than $35$

Then $A = \left\{52, \; 53 \right\}$

Number of elements in $A = n \left(A\right) = 2$

Probability of $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{2}{6} = \dfrac{1}{3}$

Let $B =$ event of having a two digit number which is a multiple of $2$

Then $B = \left\{32, \; 52 \right\}$

Number of elements in $B = n \left(B\right) = 2$

Probability of $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{2}{6} = \dfrac{1}{3}$

Let $C =$ event of having a two digit number which is a prime number

Then $C = \left\{23, \; 53 \right\}$

Number of elements in $C = n \left(C\right) = 2$

Probability of $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{2}{6} = \dfrac{1}{3}$

Mensuration

A circus tent in the form of a cylinder is surmounted by a cone. The height of the tent is $13 \; m$ and the height of the cylinder is $8 \; m$. If the diameter of its base is $24 \; m$, calculate the total surface area of the tent to the nearest square meter. Take $\pi = \dfrac{22}{7}$

Height of tent $= 13 \; m$ (given)

Height of cylinder $= H = 8 \; m$ (given)

$\therefore \;$ Height of cone $= h = 13 - 8 = 5 \; m$

Diameter of base of tent $= 24 \; m$ (given)

$\therefore \;$ Radius of cylinder $= $ Radius of cone $= R = 12 \; m$

Slant height of cone $= \ell = \sqrt{h^2 + R^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \; m$

Surface area of the conical portion of the tent $= s = \pi R \left(\ell + R\right)$

i.e. $\;$ $s = \pi \times 12 \left(13 + 12\right) = \pi \times 12 \times 25 = 300 \pi \; m^2$

Surface area of the cylindrical portion of the tent $= S = 2 \pi R \left(H + R\right)$

i.e. $\;$ $S = 2 \times \pi \times 12 \left(8 + 12\right) = \pi \times 24 \times 20 = 480 \pi \; m^2$

$\therefore \;$ Total surface area of the tent $= A = s + S$

i.e. $\;$ $A = 300 \pi + 480 \pi = 780 \times \dfrac{22}{7} = 2451.4 \; m^2$

i.e. $\;$ Total surface area of the tent $= 2451 \; m^2$ (to the nearest square meter)

Quadratic Equations

Some students planned a picnic. The budget for the food was ₹ $2400$. Since $8$ of them failed to come, the cost of food for each member increased by ₹ $50$. Find the number of students who went for the picnic.

Let the original number of students to go on a picnic be $= x$

Original cost of food for each student $= $ ₹ $\left(\dfrac{2400}{x}\right)$

New number of students who went for the picnic $= x - 8$

New cost of food per student $= $ ₹ $\left(\dfrac{2400}{x - 8}\right)$

As per question,

New cost of food per student $= $ Old cost of food per student $+ $ ₹ $50$

i.e. $\;$ $\dfrac{2400}{x - 8} = \dfrac{2400}{x} + 50$

i.e. $\;$ $2400 \left[\dfrac{1}{x - 8} - \dfrac{1}{x}\right] = 50$

i.e. $\;$ $48 \left[\dfrac{x - x + 8}{x \left(x - 8\right)}\right] = 1$

i.e. $\;$ $48 \times 8 = x^2 - 8x$

i.e. $\;$ $x^2 - 8x - 384 = 0$

i.e. $\;$ $x^2 - 24x + 16x - 384 = 0$

i.e. $\;$ $x \left(x - 24\right) + 16 \left(x - 24\right) = 0$

i.e. $\;$ $\left(x - 24\right) \left(x + 16\right) = 0$

i.e. $\;$ $x = 24$ $\;$ or $\;$ $x = -16$

$\therefore \;$ Original number of students $= 24$

$\therefore \;$ Number of students who went for a picnic $= 24 - 8 = 16$

Matrices

If $M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}$, find the order of matrix $M$ and the matrix $M$.

$M$ is a $1 \times 2$ matrix.

Let $M = \begin{bmatrix} a & b \end{bmatrix}$

Then, $\;$ $M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}$ $\implies$ $\begin{bmatrix} a & b \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} a \times 1 + b \times 0 & a \times 1 + b \times 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} a & a + 2b \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}$

When two matrices are equal, their corresponding elements are equal. Therefore, we have

$a = 1$

and $\;$ $a + 2b = 2$ $\implies$ $1 + 2b = 2$ $\implies$ $b = \dfrac{1}{2}$

$\therefore \;$ $M = \begin{bmatrix} 1 & \dfrac{1}{2} \end{bmatrix}$

Factor and Remainder Theorems

If $x^3 + ax^2 -x + b$ 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 $f \left(x\right) = x^3 + ax^2 -x + b$

$\because \;$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$,

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

Now, $\;$ $f \left(2\right) = 2^3 + a \times 2^2 - 2 + b = 8 + 4a -2 + b = 6 + 4a + b$

$\therefore \;$ $f \left(2\right) = 0$ $\implies$ $6 + 4a + b = 0$

i.e. $\;$ $4a + b = -6$ $\;\;\; \cdots \; (1)$

$\because \;$ $f \left(x\right)$ when divided by $\left(x - 3\right)$ leaves a remainder of $3$,

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

Now, $\;$ $f \left(3\right) = 3^3 + a \times 3^2 -3 + b = 27 + 9a -3 + b = 24 + 9a + b$

$\because \;$ $f \left(3\right) = 3$ $\implies$ $24 + 9a + b = 3$

i.e. $\;$ $9a + b = -21$ $\;\;\; \cdots \; (2)$

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

$5a = -15$ $\implies$ $a = -3$

and $\;$ $b = -6 - 4a = -6 -4 \times \left(-3\right) = 6$

$\therefore \;$ $a = -3, \; b = 6$

Ratio and Proportion

$6$ is the mean proportion between two numbers $x$ and $y$, and $48$ is the third proportional to $x$ and $y$. Find the numbers.

$6$ is the mean proportion between $x$ and $y$.

$\implies$ $x$, $6$ and $y$ are in continued proportion.

$\implies$ $x : 6 = 6 : y$

$\implies$ $6^2 = x \times y$ $\implies$ $36 = xy$ $\;\;\; \cdots \; (1)$

$48$ is the third proportional to $x$ and $y$.

$\implies$ $x$, $y$ and $48$ are in continued proportion.

$\implies$ $x : y = y : 48$

$\implies$ $y^2 = 48 x$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $y^2 = 48 \times \dfrac{36}{y}$ $\;\;\;$ [substituting the value of $x$ from equation $(1)$]

i.e. $\;$ $y^3 = 1728 = 12^3$ $\implies$ $y = 12$

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

$x = \dfrac{36}{y} = \dfrac{36}{12} = 3$

Ratio and Proportion

Given that $\dfrac{a^3 + 3 a b^2}{b^3 + 3 a^2 b} = \dfrac{62}{63}$, find $a : b$ using componendo-dividendo.

Given: $\;$ $\dfrac{a^3 + 3 a b^2}{b^3 + 3 a^2 b} = \dfrac{62}{63}$

i.e. $\;$ $\dfrac{a^3 + 3 ab^2 + b^3 + 3 a^2 b}{a^3 + 3ab^2 - b^3 - 3a^2 b} = \dfrac{62 + 63}{62 - 63}$ $\;\;\;$ [by componendo-dividendo]

i.e. $\;$ $\dfrac{\left(a + b\right)^3}{\left(a - b\right)^3} = \dfrac{125}{-1}$

i.e. $\;$ $\dfrac{\left(a + b\right)^3}{\left(a - b\right)^3} = \dfrac{5^3}{-1^3}$

i.e. $\;$ $\dfrac{a + b}{a - b} = \dfrac{5}{-1}$

i.e. $\;$ $\dfrac{a + b + a - b}{a + b - a + b} = \dfrac{5 + \left(-1\right)}{5 - \left(-1\right)}$ $\;\;\;$ [by componendo-dividendo]

i.e. $\;$ $\dfrac{2a}{2b} = \dfrac{4}{6}$

i.e. $\;$ $\dfrac{a}{b} = \dfrac{2}{3}$

Constructions

Construct a triangle $ABC$ given that $AB = 5 \; cm$, $BC = 6 \; cm$ and $\angle ABC = 120^\circ$. Construct the incircle of the triangle. Measure and record the radius of the incircle.

$\triangle ABC$ is constructed with $AB = 5 \; cm$, $BC = 6 \; cm$ and $\angle ABC = 120^\circ$ (blue and orange construction arcs).

The angle bisectors of $\angle A$ and $\angle C$ are drawn. The two angle bisectors intersect at point $I$ (violet construction arcs and lines).

From the point $I$ draw $IP$ perpendicular to side $AC$ of the triangle (red construction arcs and lines).

With $I$ as center and $IP$ as radius, draw a circle which will touch all the sides of the triangle drawn.

This circle is required inscribing circle of $\triangle ABC$.

Inradius $= IP = 1.2 \; cm$ (by measurement).

Coordinate Geometry: Section Formula, Equation of a Line

Find the equation of a line whose slope is $\dfrac{3}{2}$ and passes through the point $P$ which divides the line segment joining $A \left(-2, 6\right)$ and $B \left(3, -4\right)$ in the ratio $2 : 3$.

Point $P \left(x,y\right)$ divides the line segment $A \left(-2, 6\right)$ and $B \left(3, -4\right)$ in the ratio $2 : 3$.

$\therefore \;$ By section formula

$x = \dfrac{2 \times 3 + 3 \times \left(-2\right)}{2 + 3} = 0$; $\;\;\;$ $y = \dfrac{2 \times \left(-4\right) + 3 \times 6}{2 + 3} = 2$

$\therefore \;$ $P \left(x, y\right) = P \left(0, 2\right)$

Equation of line passing through the point $P$ and slope $= \dfrac{3}{2}$ is

$y - 2 = \dfrac{3}{2} \left(x - 0\right)$

i.e. $\;$ $3x - 2y + 4 = 0$

Matrices

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find $A^2 - 5A + 7I$.

$\begin{aligned} A^2 = A \times A & = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 3 \times 3 + 1 \times \left(-1\right) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times \left(-1\right) & -1 \times 1+ 2 \times 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned}$

$\begin{aligned} 5 A & = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \end{aligned}$

$\begin{aligned} 7 I & = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \end{aligned}$

$\begin{aligned} \therefore \; A^2 - 5A + 7 I & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\\\ & = \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 +5 + 0 & 3 -10 + 7 \end{bmatrix} \\\\ & = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{aligned}$

Mensuration

The surface area of a solid metallic sphere is $2464 \; cm^2$. It is melted and recast into solid right circular cones of radius $3.5 \; cm$ and height $7 \; cm$. Calculate the radius of the sphere and the number of cones recast. Take $\pi = \dfrac{22}{7}$.

Let the radius of the sphere $= R$

Surface area of sphere $= 4 \pi R^2 = 2464$

i.e. $\;$ $R^2 = \dfrac{2464}{4 \pi} = \dfrac{2464 \times 7}{4 \times 22} = 196$ $\implies$ $R = \sqrt{196} = 14 \; cm$

$\therefore \;$ Radius of sphere $= R = 14 \; cm$

Radius of cone $= r = 3.5 \; cm$

Height of cone $= h = 7 \; cm$

Let the number of cones recast $= n$

Since the sphere is melted and recast into $n$ number of cones,

Volume of sphere $= n \times$ volume of cone

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

i.e. $\;$ $n = \dfrac{4 R^3}{r^2 h} = \dfrac{4 \times 14^3}{3.5^2 \times 7} = 128$

$\therefore \;$ Number of cones recast $= 128$

Quadratic Equations

Without solving the equation, find the value of $p$ for which the equation $4x^2 + \left(p - 2\right)x + 1 = 0$ has equal roots.

Given quadratic equation: $\;$ $4x^2 + \left(p -2\right)x + 1 = 0$

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

$a = 4, \; b = p - 2, \; c = 1$

The given quadratic equation has equal roots when discriminant $= \Delta = b^2 - 4ac = 0$

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

$\therefore \;$ We have for equal roots,

$\left(p - 2\right)^2 - 16 = 0$

i.e. $\;$ $p - 2 = \sqrt{16} = \pm 4$

i.e. $\;$ $p = 6$ $\;$ or $\;$ $p = -2$

Coordinate Geometry

Find the equation of a line which has Y intercept $4$ and is parallel to the line $2x - 3y = 7$. Find the coordinates of the point where it cuts the X-axis.

Y intercept $= c = 4$

Given line: $2x - 3y = 7$

i.e. $\;$ $y = \dfrac{2}{3}x - \dfrac{7}{3}$

Slope of the given line $= m = \dfrac{2}{3}$

Since the required line is parallel to the given line, slope of required line $= m = \dfrac{2}{3}$

Let the equation of the required line be $\;$ $y = mx + c$

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

i.e. $\;$ $2x - 3y = -12$

When the line cuts the X-axis, its Y coordinate is $0$.

Then we have, $\;$ $2x = -12$ $\implies$ $x = -6$

The required line cuts the X-axis at the point $\left(-6,0\right)$.

Arithmetic Progression

Find: $\;$ $21 + 18 + 15 + \cdots - 81$

For the given arithmetic progression (A.P) $\;$ $21 + 18 + 15 + \cdots - 81$

first term $= a = 21$

common difference $= d = -3$

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

$n^{th}$ term $= t_n = -81$

$t_n = a + \left(n - 1\right)d$

i.e. $\;$ $-81 = 21 + \left(n - 1\right) \times \left(-3\right)$

i.e. $\;$ $n - 1 = \dfrac{21 + 81}{3} = 34$

$\implies$ $n = 35$

For the given A.P, sum to $n$ terms is

$\begin{aligned} S_n & = 21 + 18 + 15 + \cdots - 81 \\\\ & = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right] \\\\ & = \dfrac{35}{2} \left[2 \times 21 + \left(35 - 1\right) \times \left(-3\right)\right] \\\\ & = \dfrac{35}{2} \left[42 - 102\right] \\\\ & = \dfrac{35}{2} \times \left(-60\right) = -1050 \end{aligned}$

Geometric Progression

How many terms of the geometric progression (G.P) $\;$ $3, \; \dfrac{3}{2}, \; \dfrac{3}{4}, \cdots$ $\;$ are needed to give the sum $\dfrac{186}{32}$?

First term of G.P $= a = 3$

Common ratio of G.P $= r = \dfrac{3 / 2}{3} = \dfrac{1}{2} < 1$

Let, number of terms of G.P needed to get the required sum $= n$

Sum of G.P $= S_n = \dfrac{186}{32}$

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

i.e. $\;$ $\dfrac{186}{32} = \dfrac{3 \left[1 - \left(\dfrac{1}{2}\right)^n\right]}{1 - \dfrac{1}{2}}$

i.e. $\;$ $\dfrac{62}{32} = \dfrac{1 - \left(\dfrac{1}{2}\right)^n}{\dfrac{1}{2}}$

i.e. $\;$ $\dfrac{31}{16} \times \dfrac{1}{2} = 1 - \left(\dfrac{1}{2}\right)^n$

i.e. $\;$ $\left(\dfrac{1}{2}\right)^n = 1 - \dfrac{31}{32} = \dfrac{1}{32} = \left(\dfrac{1}{2}\right)^5$

$\implies$ $n = 5$

Trigonometry

Prove the identity: $\;$ $\dfrac{\tan^2 \theta}{\tan^2 \theta - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} = \dfrac{1}{1 - 2 \cos^2 \theta}$

$\begin{aligned} LHS & = \dfrac{\tan^2 \theta}{\tan^2 \theta - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\dfrac{\sin^2 \theta}{\cos^2 \theta}}{\dfrac{\sin^2 \theta}{\cos^2 \theta} - 1} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\sin^2 \theta}{\sin^2 \theta - \cos^2 \theta} + \dfrac{cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{\sin^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{1 - \cos^2 \theta - \cos^2 \theta} \\\\ & = \dfrac{1}{1 - 2 \cos^2 \theta} = RHS \end{aligned}$

Hence proved.