\displaystyle \text{INVERSE TRIGONOMETRIC FUNCTION}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{Two men on either side of a temple }30\text{ metres high observe its top at the angles of elevation}
\displaystyle \alpha\text{ and }\beta\text{ respectively, as shown in the figure. The distance between the two men is }40\sqrt3
\displaystyle \text{metres and the distance between the first person }A\text{ and the temple is }30\sqrt3\text{ metres.}
\displaystyle \text{Based on the above information, answer the following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\angle CAB=\alpha=
\displaystyle \text{(a) }\sin^{-1}\left(\frac{2}{\sqrt3}\right)\qquad\text{(b) }\sin^{-1}\left(\frac12\right)
\displaystyle \text{(c) }\sin^{-1}(2)\qquad\text{(d) }\sin^{-1}\left(\frac{\sqrt3}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the foot of the perpendicular from }B\text{ to }AC.
\displaystyle BD=30\text{ m},\qquad AD=30\sqrt3\text{ m}.
\displaystyle \tan\alpha=\frac{BD}{AD}=\frac{30}{30\sqrt3}=\frac1{\sqrt3}.
\displaystyle \therefore \alpha=\frac{\pi}{6}.
\displaystyle \sin\alpha=\sin\frac{\pi}{6}=\frac12.
\displaystyle \therefore \alpha=\sin^{-1}\left(\frac12\right).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\angle CAB=\alpha=
\displaystyle \text{(a) }\cos^{-1}\left(\frac15\right)\qquad\text{(b) }\cos^{-1}\left(\frac25\right)
\displaystyle \text{(c) }\cos^{-1}\left(\frac{\sqrt3}{2}\right)\qquad\text{(d) }\cos^{-1}\left(\frac45\right)
\displaystyle \text{Answer:}
\displaystyle \text{From Question 1, }\alpha=\frac{\pi}{6}.
\displaystyle \cos\alpha=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}.
\displaystyle \therefore \alpha=\cos^{-1}\left(\frac{\sqrt3}{2}\right).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\angle BCA=\beta=
\displaystyle \text{(a) }\tan^{-1}\left(\frac12\right)\qquad\text{(b) }\tan^{-1}(2)
\displaystyle \text{(c) }\tan^{-1}\left(\frac1{\sqrt3}\right)\qquad\text{(d) }\tan^{-1}(\sqrt3)
\displaystyle \text{Answer:}
\displaystyle AC=40\sqrt3\text{ m}\quad\text{and}\quad AD=30\sqrt3\text{ m}.
\displaystyle \therefore DC=AC-AD=40\sqrt3-30\sqrt3=10\sqrt3\text{ m}.
\displaystyle \tan\beta=\frac{BD}{DC}=\frac{30}{10\sqrt3}=\sqrt3.
\displaystyle \therefore \beta=\tan^{-1}(\sqrt3)=\frac{\pi}{3}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\angle ABC=
\displaystyle \text{(a) }\frac{\pi}{4}\qquad\text{(b) }\frac{\pi}{6}\qquad\text{(c) }\frac{\pi}{2}\qquad\text{(d) }\frac{\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \angle CAB=\alpha=\frac{\pi}{6}\quad\text{and}\quad\angle BCA=\beta=\frac{\pi}{3}.
\displaystyle \text{Using the angle sum property of }\triangle ABC,
\displaystyle \angle ABC=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right).
\displaystyle =\pi-\frac{\pi}{2}=\frac{\pi}{2}.
\displaystyle \therefore \angle ABC=\frac{\pi}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The domain and range of }\cos^{-1}x\text{ are}
\displaystyle \text{(a) }(-1,1),\ (0,\pi)\qquad\text{(b) }[-1,1],\ (0,\pi)
\displaystyle \text{(c) }[-1,1],\ [0,\pi]\qquad\text{(d) }(-1,1),\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \text{Answer:}
\displaystyle \text{For }y=\cos^{-1}x,\text{ the principal value branch of cosine is restricted to }[0,\pi].
\displaystyle \text{On this interval, the values of }\cos y\text{ range from }-1\text{ to }1.
\displaystyle \therefore \text{Domain of }\cos^{-1}x=[-1,1].
\displaystyle \therefore \text{Range of }\cos^{-1}x=[0,\pi].
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{The Government of India is planning to fix a hoarding board on the face of a building on the}
\displaystyle \text{road of a busy market for awareness on the COVID-19 protocol. Ram, Robert and Rahim are the}
\displaystyle \text{three engineers working on this project. }A\text{ is considered to be a person viewing the hoarding}
\displaystyle \text{board from }20\text{ metres away from the building, standing at the edge of a nearby pathway.}
\displaystyle \text{Ram, Robert and Rahim suggested placing the hoarding board at three different locations }C,D
\displaystyle \text{and }E.\ C\text{ is at a height of }10\text{ metres from the ground level.}
\displaystyle \text{For viewer }A,\text{ the angle of elevation of }D\text{ is double the angle of elevation of }C.
\displaystyle \text{The angle of elevation of }E\text{ is triple the angle of elevation of }C\text{ for the same viewer.}
\displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Measure of }\angle CAB=
\displaystyle \text{(a) }\tan^{-1}(2)\qquad\text{(b) }\tan^{-1}\left(\frac12\right)
\displaystyle \text{(c) }\tan^{-1}(1)\qquad\text{(d) }\tan^{-1}(3)
\displaystyle \text{Answer:}
\displaystyle AB=20\text{ m}\quad\text{and}\quad BC=10\text{ m}.
\displaystyle \text{Let }\angle CAB=\alpha.
\displaystyle \tan\alpha=\frac{BC}{AB}=\frac{10}{20}=\frac12.
\displaystyle \therefore \alpha=\tan^{-1}\left(\frac12\right).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Measure of }\angle DAB=
\displaystyle \text{(a) }\tan^{-1}\left(\frac34\right)\qquad\text{(b) }\tan^{-1}(3)
\displaystyle \text{(c) }\tan^{-1}\left(\frac43\right)\qquad\text{(d) }\tan^{-1}(4)
\displaystyle \text{Answer:}
\displaystyle \text{From Question 1, }\tan\alpha=\frac12.
\displaystyle \text{The angle of elevation of }D\text{ is }2\alpha.
\displaystyle \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}.
\displaystyle =\frac{2\left(\frac12\right)}{1-\left(\frac12\right)^2} =\frac{1}{1-\frac14}=\frac43.
\displaystyle \therefore 2\alpha=\tan^{-1}\left(\frac43\right).
\displaystyle \therefore \angle DAB=\tan^{-1}\left(\frac43\right).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Measure of }\angle EAB=
\displaystyle \text{(a) }\tan^{-1}(11)\qquad\text{(b) }\tan^{-1}(3)
\displaystyle \text{(c) }\tan^{-1}\left(\frac{2}{11}\right)\qquad\text{(d) }\tan^{-1}\left(\frac{11}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{The angle of elevation of }E\text{ is }3\alpha,\text{ where }\tan\alpha=\frac12.
\displaystyle \tan3\alpha=\frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}.
\displaystyle =\frac{3\left(\frac12\right)-\left(\frac12\right)^3} {1-3\left(\frac12\right)^2}.
\displaystyle =\frac{\frac32-\frac18}{1-\frac34} =\frac{\frac{11}{8}}{\frac14}=\frac{11}{2}.
\displaystyle \therefore 3\alpha=\tan^{-1}\left(\frac{11}{2}\right).
\displaystyle \therefore \angle EAB=\tan^{-1}\left(\frac{11}{2}\right).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }A'\text{ is another viewer standing on the same line of observation across the road. If}
\displaystyle \text{the width of the road is }5\text{ metres, then the difference between }\angle CAB\text{ and }\angle CA'B\text{ is}
\displaystyle \text{(a) }\tan^{-1}\left(\frac{1}{12}\right)\qquad\text{(b) }\tan^{-1}\left(\frac18\right)
\displaystyle \text{(c) }\tan^{-1}\left(\frac25\right)\qquad\text{(d) }\tan^{-1}\left(\frac{11}{21}\right)
\displaystyle \text{Answer:}
\displaystyle AB=20\text{ m}\quad\text{and}\quad AA'=5\text{ m}.
\displaystyle \therefore A'B=20+5=25\text{ m}.
\displaystyle \text{Let }\angle CAB=\alpha\quad\text{and}\quad\angle CA'B=\theta.
\displaystyle \tan\alpha=\frac{10}{20}=\frac12\quad\text{and}\quad \tan\theta=\frac{10}{25}=\frac25.
\displaystyle \tan(\alpha-\theta)=\frac{\tan\alpha-\tan\theta} {1+\tan\alpha\tan\theta}.
\displaystyle =\frac{\frac12-\frac25}{1+\left(\frac12\right)\left(\frac25\right)}.
\displaystyle =\frac{\frac{1}{10}}{\frac65}=\frac{1}{12}.
\displaystyle \therefore \alpha-\theta=\tan^{-1}\left(\frac{1}{12}\right).
\displaystyle \therefore \text{The required difference is }\tan^{-1}\left(\frac{1}{12}\right).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The domain and range of }\tan^{-1}x\text{ are}
\displaystyle \text{(a) }\mathbb{R}^{+},\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\displaystyle \text{(b) }\mathbb{R}^{-},\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\displaystyle \text{(c) }\mathbb{R},\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
\displaystyle \text{(d) }\mathbb{R},\left(0,\frac{\pi}{2}\right)
\displaystyle \text{Answer:}
\displaystyle \text{The function }\tan x\text{ is one-one and onto }\mathbb{R}\text{ when restricted to}
\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right).
\displaystyle \therefore \text{Domain of }\tan^{-1}x=\mathbb{R}.
\displaystyle \therefore \text{Range of }\tan^{-1}x=\left(-\frac{\pi}{2},\frac{\pi}{2}\right).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

 

\displaystyle \text{MATRICES}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A manufacturer produces three stationery products: pencils, erasers and sharpeners, which are}
\displaystyle \text{sold in two markets }A\text{ and }B.\text{ The annual sales of these products are given below.} \displaystyle \begin{array}{c|ccc} \text{Market}&\text{Pencil}&\text{Eraser}&\text{Sharpener}\\ \hline A&10000&2000&18000\\ B&6000&20000&8000 \end{array}
\displaystyle \text{The unit selling prices of a pencil, eraser and sharpener are Rs. }2.50,\text{ Rs. }1.50\text{ and}
\displaystyle \text{Rs. }1.00\text{ respectively. Their respective unit costs are Rs. }2.00,\text{ Rs. }1.00\text{ and Rs. }0.50.
\displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The total revenue from market }A\text{ is}
\displaystyle \text{(a) Rs. }64,000\qquad\text{(b) Rs. }60,400\qquad\text{(c) Rs. }46,000\qquad\text{(d) Rs. }40,600
\displaystyle \text{Answer:}
\displaystyle \text{Sales matrix }S=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}.
\displaystyle \text{Selling price matrix }P=\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}.
\displaystyle SP=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix} \begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}.
\displaystyle =\begin{bmatrix}25000+3000+18000\\15000+30000+8000\end{bmatrix} =\begin{bmatrix}46000\\53000\end{bmatrix}.
\displaystyle \therefore \text{Total revenue from market }A=\text{Rs. }46,000.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The total revenue from market }B\text{ is}
\displaystyle \text{(a) Rs. }35,000\qquad\text{(b) Rs. }53,000\qquad\text{(c) Rs. }50,300\qquad\text{(d) Rs. }30,500
\displaystyle \text{Answer:}
\displaystyle \text{From }SP=\begin{bmatrix}46000\\53000\end{bmatrix},\text{ the second entry gives the revenue from market }B.
\displaystyle \therefore \text{Total revenue from market }B=\text{Rs. }53,000.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The cost incurred in market }A\text{ is}
\displaystyle \text{(a) Rs. }13,000\qquad\text{(b) Rs. }30,100\qquad\text{(c) Rs. }10,300\qquad\text{(d) Rs. }31,000
\displaystyle \text{Answer:}
\displaystyle \text{Unit cost matrix }C=\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}.
\displaystyle SC=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix} \begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}.
\displaystyle =\begin{bmatrix}20000+2000+9000\\12000+20000+4000\end{bmatrix} =\begin{bmatrix}31000\\36000\end{bmatrix}.
\displaystyle \therefore \text{Cost incurred in market }A=\text{Rs. }31,000.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The profits in markets }A\text{ and }B\text{ respectively are}
\displaystyle \text{(a) (Rs. }15,000,\text{ Rs. }17,000)\qquad\text{(b) (Rs. }17,000,\text{ Rs. }15,000)
\displaystyle \text{(c) (Rs. }51,000,\text{ Rs. }71,000)\qquad\text{(d) (Rs. }10,000,\text{ Rs. }20,000)
\displaystyle \text{Answer:}
\displaystyle \text{Profit matrix}=\text{Revenue matrix}-\text{Cost matrix}.
\displaystyle =\begin{bmatrix}46000\\53000\end{bmatrix} -\begin{bmatrix}31000\\36000\end{bmatrix} =\begin{bmatrix}15000\\17000\end{bmatrix}.
\displaystyle \therefore \text{Profit in market }A=\text{Rs. }15,000\text{ and profit in market }B=\text{Rs. }17,000.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The gross profit in both markets is}
\displaystyle \text{(a) Rs. }23,000\qquad\text{(b) Rs. }20,300\qquad\text{(c) Rs. }32,000\qquad\text{(d) Rs. }30,200
\displaystyle \text{Answer:}
\displaystyle \text{Profit in market }A=\text{Rs. }15,000.
\displaystyle \text{Profit in market }B=\text{Rs. }17,000.
\displaystyle \therefore \text{Gross profit}=15,000+17,000=\text{Rs. }32,000.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Amit, Biraj and Chirag were given the task of creating a square matrix of order }2.
\displaystyle \text{The matrices created by Amit, Biraj and Chirag are }A,\ B\text{ and }C\text{ respectively, where}
\displaystyle A=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\qquad B=\begin{bmatrix}4&0\\1&5\end{bmatrix},\qquad C=\begin{bmatrix}2&0\\1&-2\end{bmatrix}.
\displaystyle \text{If }a=4\text{ and }b=-2,\text{ based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The sum of the matrices }A,\ B\text{ and }C,\ A+(B+C),\text{ is}
\displaystyle \text{(a) }\begin{bmatrix}1&6\\2&7\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}6&1\\7&2\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}7&2\\1&6\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}2&1\\7&6\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle B+C=\begin{bmatrix}4&0\\1&5\end{bmatrix}+\begin{bmatrix}2&0\\1&-2\end{bmatrix}
\displaystyle =\begin{bmatrix}6&0\\2&3\end{bmatrix}.
\displaystyle A+(B+C)=\begin{bmatrix}1&2\\-1&3\end{bmatrix}+\begin{bmatrix}6&0\\2&3\end{bmatrix}
\displaystyle =\begin{bmatrix}7&2\\1&6\end{bmatrix}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }(A^T)^T\text{ is equal to}
\displaystyle \text{(a) }\begin{bmatrix}1&2\\-1&3\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}2&1\\3&-1\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}1&-1\\2&3\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}2&3\\-1&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle A^T=\begin{bmatrix}1&-1\\2&3\end{bmatrix}.
\displaystyle (A^T)^T=\begin{bmatrix}1&2\\-1&3\end{bmatrix}=A.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }(bA)^T\text{ is equal to}
\displaystyle \text{(a) }\begin{bmatrix}-2&-4\\2&-6\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}-2&2\\-6&-4\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}-6&-2\\2&4\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle b=-2.
\displaystyle bA=-2\begin{bmatrix}1&2\\-1&3\end{bmatrix} =\begin{bmatrix}-2&-4\\2&-6\end{bmatrix}.
\displaystyle (bA)^T=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }AC-BC\text{ is equal to}
\displaystyle \text{(a) }\begin{bmatrix}-4&-6\\-4&4\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}-4&-4\\4&-6\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}-6&4\\-4&-4\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle AC-BC=(A-B)C.
\displaystyle A-B=\begin{bmatrix}1&2\\-1&3\end{bmatrix}-\begin{bmatrix}4&0\\1&5\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&2\\-2&-2\end{bmatrix}.
\displaystyle (A-B)C=\begin{bmatrix}-3&2\\-2&-2\end{bmatrix} \begin{bmatrix}2&0\\1&-2\end{bmatrix}.
\displaystyle =\begin{bmatrix}-6+2&0-4\\-4-2&0+4\end{bmatrix} =\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}.
\displaystyle \therefore AC-BC=\begin{bmatrix}-4&-4\\-6&4\end{bmatrix}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }(a+b)B\text{ is equal to}
\displaystyle \text{(a) }\begin{bmatrix}0&8\\10&2\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}2&10\\8&0\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}8&0\\2&10\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}2&0\\8&10\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle a+b=4+(-2)=2.
\displaystyle (a+b)B=2\begin{bmatrix}4&0\\1&5\end{bmatrix}.
\displaystyle =\begin{bmatrix}8&0\\2&10\end{bmatrix}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{Three schools, DPS, CVC and KVS, decided to organize a fair to collect money for helping flood}
\displaystyle \text{victims. They sold handmade fans, mats and plates made from recycled material at prices of}
\displaystyle \text{Rs. }25,\text{ Rs. }100\text{ and Rs. }50\text{ each respectively. The numbers of articles sold are given below.} \displaystyle \begin{array}{c|ccc} \text{School/Article}&\text{DPS}&\text{CVC}&\text{KVS}\\ \hline \text{Handmade fans}&40&25&35\\ \text{Mats}&50&40&50\\ \text{Plates}&20&30&40 \end{array}
\displaystyle \text{Based on the information given above, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{What is the total money collected by the school DPS?}
\displaystyle \text{(a) Rs. }700\qquad\text{(b) Rs. }7,000\qquad\text{(c) Rs. }6,125\qquad\text{(d) Rs. }7,875
\displaystyle \text{Answer:}
\displaystyle \text{For DPS, the numbers of fans, mats and plates sold are }40,\ 50\text{ and }20\text{ respectively.}
\displaystyle \text{Total money collected}=40(25)+50(100)+20(50).
\displaystyle =1000+5000+1000=\text{Rs. }7,000.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the total amount of money collected by schools CVC and KVS?}
\displaystyle \text{(a) Rs. }14,000\qquad\text{(b) Rs. }15,725\qquad\text{(c) Rs. }21,000\qquad\text{(d) Rs. }13,125
\displaystyle \text{Answer:}
\displaystyle \text{Money collected by CVC}=25(25)+40(100)+30(50).
\displaystyle =625+4000+1500=\text{Rs. }6,125.
\displaystyle \text{Money collected by KVS}=35(25)+50(100)+40(50).
\displaystyle =875+5000+2000=\text{Rs. }7,875.
\displaystyle \therefore \text{Total amount collected by CVC and KVS}=6125+7875=\text{Rs. }14,000.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the total amount of money collected by all three schools, DPS, CVC}
\displaystyle \text{and KVS?}
\displaystyle \text{(a) Rs. }15,775\qquad\text{(b) Rs. }14,000\qquad\text{(c) Rs. }21,000\qquad\text{(d) Rs. }17,125
\displaystyle \text{Answer:}
\displaystyle \text{Amount collected by DPS}=\text{Rs. }7,000.
\displaystyle \text{Amount collected by CVC}=\text{Rs. }6,125.
\displaystyle \text{Amount collected by KVS}=\text{Rs. }7,875.
\displaystyle \therefore \text{Total amount collected}=7000+6125+7875=\text{Rs. }21,000.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the numbers of handmade fans and plates are interchanged for all the schools,}
\displaystyle \text{then what is the total money collected by all the schools?}
\displaystyle \text{(a) Rs. }18,000\qquad\text{(b) Rs. }6,750\qquad\text{(c) Rs. }5,000\qquad\text{(d) Rs. }21,250
\displaystyle \text{Answer:}
\displaystyle \text{Total number of fans}=40+25+35=100.
\displaystyle \text{Total number of mats}=50+40+50=140.
\displaystyle \text{Total number of plates}=20+30+40=90.
\displaystyle \text{After interchanging the numbers of fans and plates, the numbers become }90,\ 140\text{ and }100.
\displaystyle \text{Total money collected}=90(25)+140(100)+100(50).
\displaystyle =2250+14000+5000=\text{Rs. }21,250.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How many articles in total are sold by the three schools?}
\displaystyle \text{(a) }230\qquad\text{(b) }130\qquad\text{(c) }430\qquad\text{(d) }330
\displaystyle \text{Answer:}
\displaystyle \text{Total number of fans}=40+25+35=100.
\displaystyle \text{Total number of mats}=50+40+50=140.
\displaystyle \text{Total number of plates}=20+30+40=90.
\displaystyle \therefore \text{Total number of articles}=100+140+90=330.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 4}

\displaystyle \text{On her birthday, Seema decided to donate some money to children of an orphanage home. If there}
\displaystyle \text{were }8\text{ children less, everyone would have received Rs. }10\text{ more. However, if there were }16
\displaystyle \text{children more, everyone would have received Rs. }10\text{ less. Let the number of children be }x
\displaystyle \text{and the amount distributed by Seema to each child be Rs. }y. \displaystyle \text{Based on the information given above, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The equations in terms of }x\text{ and }y\text{ are}
\displaystyle \text{(a) }5x-4y=40,\quad 5x-8y=-80
\displaystyle \text{(b) }5x-4y=40,\quad 5x-8y=80
\displaystyle \text{(c) }5x-4y=40,\quad 5x+8y=-80
\displaystyle \text{(d) }5x+4y=40,\quad 5x-8y=-80
\displaystyle \text{Answer:}
\displaystyle \text{Total amount distributed by Seema}=xy.
\displaystyle \text{If there are }8\text{ children less, each child receives Rs. }10\text{ more.}
\displaystyle xy=(x-8)(y+10).
\displaystyle xy=xy+10x-8y-80.
\displaystyle 10x-8y=80.
\displaystyle \therefore 5x-4y=40.
\displaystyle \text{If there are }16\text{ children more, each child receives Rs. }10\text{ less.}
\displaystyle xy=(x+16)(y-10).
\displaystyle xy=xy-10x+16y-160.
\displaystyle 10x-16y=-160.
\displaystyle \therefore 5x-8y=-80.
\displaystyle \therefore \text{The equations are }5x-4y=40\text{ and }5x-8y=-80.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following matrix equations represents the information given above?}
\displaystyle \text{(1) }\begin{bmatrix}5&4\\5&8\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}
\displaystyle \text{(2) }\begin{bmatrix}5&-4\\5&-8\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\80\end{bmatrix}
\displaystyle \text{(3) }\begin{bmatrix}5&-4\\5&-8\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}
\displaystyle \text{(4) }\begin{bmatrix}5&4\\5&-8\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle 5x-4y=40,\qquad 5x-8y=-80.
\displaystyle \text{Writing these equations in matrix form,}
\displaystyle \begin{bmatrix}5&-4\\5&-8\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}.
\displaystyle \therefore \text{Option (3) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The number of children who were given money by Seema is}
\displaystyle \text{(a) }30\qquad\text{(b) }40\qquad\text{(c) }23\qquad\text{(d) }32
\displaystyle \text{Answer:}
\displaystyle 5x-4y=40\qquad\ldots(1)
\displaystyle 5x-8y=-80\qquad\ldots(2)
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle 4y=120.
\displaystyle \therefore y=30.
\displaystyle \text{Substituting }y=30\text{ in (1),}
\displaystyle 5x-4(30)=40.
\displaystyle 5x=160.
\displaystyle \therefore x=32.
\displaystyle \therefore \text{The number of children is }32.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{How much amount is given to each child by Seema?}
\displaystyle \text{(a) Rs. }32\qquad\text{(b) Rs. }30\qquad\text{(c) Rs. }62\qquad\text{(d) Rs. }26
\displaystyle \text{Answer:}
\displaystyle \text{From Question 3, }y=30.
\displaystyle \therefore \text{The amount given to each child is Rs. }30.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How much amount does Seema spend in distributing the money to all the children}
\displaystyle \text{of the orphanage?}
\displaystyle \text{(a) Rs. }609\qquad\text{(b) Rs. }960\qquad\text{(c) Rs. }906\qquad\text{(d) Rs. }690
\displaystyle \text{Answer:}
\displaystyle x=32\quad\text{and}\quad y=30.
\displaystyle \text{Total amount distributed}=xy.
\displaystyle =32\times30=\text{Rs. }960.
\displaystyle \therefore \text{Seema distributes a total amount of Rs. }960.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 5}

\displaystyle \text{Two farmers, Ramakishan and Gurucharan Singh, cultivate only three varieties of rice, namely}
\displaystyle \text{Basmati, Permal and Naura. The sales, in rupees, of these varieties of rice by both farmers in}
\displaystyle \text{the months of September and October are represented by the matrices }A\text{ and }B\text{ respectively.} \displaystyle \text{September sales:}\qquad A=\begin{bmatrix}10000&20000&30000\\50000&30000&10000\end{bmatrix}.
\displaystyle \text{October sales:}\qquad B=\begin{bmatrix}5000&10000&6000\\20000&10000&10000\end{bmatrix}.
\displaystyle \text{The first and second rows represent Ramakishan and Gurucharan respectively, while the three}
\displaystyle \text{columns represent Basmati, Permal and Naura respectively.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The total sales in September and October for each farmer in each variety can be}
\displaystyle \text{represented as}
\displaystyle \text{(a) }A+B\qquad\text{(b) }A-B\qquad\text{(c) }A>B\qquad\text{(d) }A<B
\displaystyle \text{Answer:}
\displaystyle \text{Matrix }A\text{ represents September sales and matrix }B\text{ represents October sales.}
\displaystyle \therefore \text{Total sales for the two months are obtained by adding the corresponding entries.}
\displaystyle \therefore \text{Total sales matrix}=A+B.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the value of }A_{23}\text{?}
\displaystyle \text{(a) }10000\qquad\text{(b) }20000\qquad\text{(c) }30000\qquad\text{(d) }40000
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}10000&20000&30000\\50000&30000&10000\end{bmatrix}.
\displaystyle A_{23}\text{ denotes the element in the second row and third column.}
\displaystyle \therefore A_{23}=10000.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The decrease in sales from September to October is represented by}
\displaystyle \text{(a) }A+B\qquad\text{(b) }A-B\qquad\text{(c) }A>B\qquad\text{(d) }A<B
\displaystyle \text{Answer:}
\displaystyle \text{September sales are represented by }A\text{ and October sales are represented by }B.
\displaystyle \therefore \text{Decrease in sales}=\text{September sales}-\text{October sales}.
\displaystyle \therefore \text{Decrease in sales matrix}=A-B.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If Ramakishan receives }2\%\text{ profit on gross sales, compute his profit for each}
\displaystyle \text{variety sold in October.}
\displaystyle \text{(a) Rs. }100,\text{ Rs. }200\text{ and Rs. }120
\displaystyle \text{(b) Rs. }100,\text{ Rs. }200\text{ and Rs. }130
\displaystyle \text{(c) Rs. }100,\text{ Rs. }220\text{ and Rs. }120
\displaystyle \text{(d) Rs. }110,\text{ Rs. }200\text{ and Rs. }120
\displaystyle \text{Answer:}
\displaystyle \text{Ramakishan's October sales are Rs. }5000,\text{ Rs. }10000\text{ and Rs. }6000.
\displaystyle \text{Profit on Basmati}=\frac{2}{100}\times5000=\text{Rs. }100.
\displaystyle \text{Profit on Permal}=\frac{2}{100}\times10000=\text{Rs. }200.
\displaystyle \text{Profit on Naura}=\frac{2}{100}\times6000=\text{Rs. }120.
\displaystyle \therefore \text{His profits are Rs. }100,\text{ Rs. }200\text{ and Rs. }120\text{ respectively.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If Gurucharan receives }2\%\text{ profit on gross sales, compute his profit for each}
\displaystyle \text{variety sold in September.}
\displaystyle \text{(a) Rs. }100,\text{ Rs. }200,\text{ Rs. }120
\displaystyle \text{(b) Rs. }1000,\text{ Rs. }600,\text{ Rs. }200
\displaystyle \text{(c) Rs. }400,\text{ Rs. }200,\text{ Rs. }120
\displaystyle \text{(d) Rs. }1200,\text{ Rs. }200,\text{ Rs. }120
\displaystyle \text{Answer:}
\displaystyle \text{Gurucharan's September sales are Rs. }50000,\text{ Rs. }30000\text{ and Rs. }10000.
\displaystyle \text{Profit on Basmati}=\frac{2}{100}\times50000=\text{Rs. }1000.
\displaystyle \text{Profit on Permal}=\frac{2}{100}\times30000=\text{Rs. }600.
\displaystyle \text{Profit on Naura}=\frac{2}{100}\times10000=\text{Rs. }200.
\displaystyle \therefore \text{His profits are Rs. }1000,\text{ Rs. }600\text{ and Rs. }200\text{ respectively.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

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