Case Study / Scenario-Based MCQs for Sub-Topics of Topic 2: Algebra

Question 1. Case Study: Planning a Birthday Party Budget

Rohan is planning a birthday party for his friend. He needs to buy balloons and streamers. Each balloon costs $\textsf{₹} 15$ and each roll of streamers costs $\textsf{₹} 30$. He plans to buy $b$ balloons and $s$ rolls of streamers.

Which algebraic expression represents the total cost in $\textsf{₹}$ for the balloons and streamers?

(A) $15b + 30s$

(B) $b + s$

(C) $15s + 30b$

(D) $45(b+s)$

Question 2. Referring to the Case Study above, Rohan's father gives him a fixed amount of $\textsf{₹} 500$ for miscellaneous expenses. If Rohan decides to buy $10$ balloons and $5$ rolls of streamers, which expression helps calculate the remaining money he has?

(A) $15(10) + 30(5)$

(B) $500 - (15(10) + 30(5))$

(C) $(15+30)(10+5)$

(D) $500 + 15(10) + 30(5)$

Question 3. Referring to the Case Study above, in the expression $15b + 30s$, what does the coefficient '$15$' represent?

(A) The total number of balloons.

(B) The cost of one balloon.

(C) The cost of one roll of streamers.

(D) The total number of streamers.

Question 4. Referring to the Case Study above, Rohan decides to buy $8$ balloons and $6$ rolls of streamers. What is the value of the expression representing the total cost?

(A) $\textsf{₹} (15+8) + (30+6) = \textsf{₹} 59$

(B) $\textsf{₹} (15 \times 8) \times (30 \times 6) = \textsf{₹} 21600$

(C) $\textsf{₹} (15 \times 8) + (30 \times 6) = \textsf{₹} 120 + 180 = \textsf{₹} 300$

(D) $\textsf{₹} 15 \times 30 \times 8 \times 6 = \textsf{₹} 21600$

Question 5. Referring to the Case Study above, Rohan later realises he needs to account for a delivery fee of $\textsf{₹} 50$. The revised total cost expression becomes $15b + 30s + 50$. In this new expression, what is the constant term?

(A) $15b$

(B) $30s$

(C) $50$

(D) $15b + 30s$

Question 1. Case Study: Managing Inventory

A stationery shop maintains inventory of pens and pencils. At the beginning of the week, the number of pens is $3x + 5$ and the number of pencils is $2x - 3$. During the week, the shop receives a new stock of $x + 2$ pens and $x + 4$ pencils.

What is the total number of pens the shop has after receiving the new stock? Express your answer as a simplified algebraic expression.

(A) $4x + 7$ pens

(B) $4x + 3$ pens

(C) $3x + 7$ pens

(D) $2x + 7$ pens

Question 2. Referring to the Case Study above, the shop sells $x - 1$ pens by the end of the week. How many pens are left? Express your answer as a simplified algebraic expression.

(A) $(4x+7) + (x-1) = 5x+6$ pens

(B) $(4x+7) - (x-1) = 3x+8$ pens

(C) $(4x+7) - (x-1) = 3x+6$ pens

(D) $(4x+7) - (1-x) = 5x+6$ pens

Question 3. Referring to the Case Study above, the shop owner decides to bundle a pen and a pencil together as a set. If the number of pens is $P$ and the number of pencils is $L$, and the cost of a set is $\textsf{₹} (P+L)$, this is incorrect usage of variables. Let's rephrase. If the price of one pen is $\textsf{₹} (y+2)$ and the price of one pencil is $\textsf{₹} (y-1)$, what is the total cost of buying $5$ pens and $3$ pencils? Express your answer as a simplified algebraic expression.

(A) $5(y+2) + 3(y-1) = 5y+10 + 3y-3 = 8y+7$ $\textsf{₹}$

(B) $5(y+2) + 3(y-1) = 5y+10 + 3y+3 = 8y+13$ $\textsf{₹}$

(C) $5(y+2) \times 3(y-1) = (5y+10)(3y-3)$ $\textsf{₹}$

(D) $(y+2) + (y-1) = 2y+1$ $\textsf{₹}$

Question 4. Referring to the Case Study above, the total revenue from selling $p$ pens at $\textsf{₹} 15$ each is $15p$. If the total revenue from selling $p$ pens and $l$ pencils is $\textsf{₹} (15p + 10l)$, and the cost of $p$ pens is $\textsf{₹} (10p-5)$ and the cost of $l$ pencils is $\textsf{₹} (5l+2)$, what is the total profit? (Profit = Total Revenue - Total Cost).

(A) $(15p + 10l) - (10p - 5 + 5l + 2) = 15p + 10l - 10p + 5 - 5l - 2 = 5p + 5l + 3$ $\textsf{₹}$

(B) $(15p + 10l) + (10p - 5 + 5l + 2) = 25p + 15l - 3$ $\textsf{₹}$

(C) $(15p + 10l) - (10p - 5) - (5l + 2) = 15p + 10l - 10p + 5 - 5l - 2 = 5p + 5l + 3$ $\textsf{₹}$

(D) $(15p + 10l) - (10p + 5l) = 5p + 5l$ $\textsf{₹}$

Question 5. Referring to the Case Study above, the average profit per pen is the total profit from pens divided by the number of pens. If the total profit from selling $p$ pens is $5p+3$ (from the previous calculation, simplifying the profit expression related to pens only would be $15p - (10p-5) = 5p+5$), let's use a simpler division example. If the total cost for $x$ items is $10x^2 + 5x$ and the number of items is $5x$, what is the average cost per item? (Average cost = Total cost / Number of items).

(A) $10x^2 + 5x - 5x = 10x^2$

(B) $(10x^2 + 5x) \times 5x$

(C) $\frac{10x^2 + 5x}{5x} = \frac{10x^2}{5x} + \frac{5x}{5x} = 2x + 1$

(D) $10x^2 / 5x = 2x$

Question 1. Case Study: Modelling Physical Phenomena

A scientist is studying the motion of a small object thrown vertically upwards. The height of the object $h(t)$ in metres above the ground after $t$ seconds is given by the polynomial $h(t) = -4.9t^2 + 29.4t + 2$.

What is the degree of this polynomial, and what does it indicate about the motion?

(A) Degree 1, indicating linear motion.

(B) Degree 2, indicating parabolic motion.

(C) Degree 3, indicating cubic motion.

(D) Degree 0, indicating the object is stationary.

Question 2. Referring to the Case Study above, what is the initial height of the object when $t=0$ seconds? This can be found by evaluating the polynomial at $t=0$.

(A) $0$ metres

(B) $2$ metres

(C) $29.4$ metres

(D) $-4.9$ metres

Question 3. Referring to the Case Study above, the highest point reached by the object occurs at the vertex of the parabola. The object hits the ground when $h(t) = 0$. The zeroes of the polynomial $h(t) = -4.9t^2 + 29.4t + 2$ would give the times when the object is at height $0$. Approximately, at what time does the object reach its maximum height? (Hint: The time to reach maximum height for $at^2+bt+c$ is $-b/(2a)$).

(A) $0$ seconds

(B) $1.5$ seconds

(C) $3$ seconds

(D) $6$ seconds

Question 4. Referring to the Case Study above, if a different object's height is modelled by the polynomial $g(t) = t^3 - 6t^2 + 8t$, the zeroes of this polynomial represent times when the object is at ground level (height $0$). One zero is $t=0$. Find the other positive zeroes of $g(t)$.

(A) $t=2$ and $t=4$ seconds

(B) $t=1$ and $t=8$ seconds

(C) $t=2$ and $t=6$ seconds

(D) $t=4$ and $t=4$ seconds

Question 5. Referring to the Case Study above, the term $-4.9t^2$ in the polynomial $h(t) = -4.9t^2 + 29.4t + 2$ involves a coefficient and a variable part. What is the coefficient of the $t^2$ term, and what physical phenomenon does it relate to?

(A) Coefficient $2$; related to initial height.

(B) Coefficient $29.4$; related to initial velocity.

(C) Coefficient $-4.9$; related to gravity.

(D) Coefficient $4.9$; related to air resistance.

Question 1. Case Study: Production Batch Analysis

A company manufactures a product in batches. The total cost (in $\textsf{₹}$) to produce $x$ items in a batch is given by the polynomial $C(x) = x^3 - 6x^2 + 13x + 10$. The company wants to determine the cost when a batch size is $x-2$ units. While direct substitution is possible, they are also exploring using polynomial division concepts.

Using the Remainder Theorem, what is the remainder when $C(x)$ is divided by $x-2$? This represents the cost when considering $x=2$ (though interpretation in this cost context is theoretical for polynomial division).

(A) $C(2) = 2^3 - 6(2)^2 + 13(2) + 10 = 8 - 24 + 26 + 10 = 20$

(B) $C(-2) = (-2)^3 - 6(-2)^2 + 13(-2) + 10 = -8 - 24 - 26 + 10 = -48$

(C) $C(0) = 10$

(D) $C(1) = 1 - 6 + 13 + 10 = 18$

Question 2. Referring to the Case Study above, the company wants to know if producing exactly $5$ items ($x=5$) in a batch results in a 'zero cost' condition (theoretically speaking for factor analysis). According to the Factor Theorem, if $C(5)=0$, then $(x-5)$ is a factor of $C(x)$. Evaluate $C(5)$ to check if $(x-5)$ is a factor.

(A) $C(5) = 5^3 - 6(5)^2 + 13(5) + 10 = 125 - 150 + 65 + 10 = 50 \neq 0$. So, $(x-5)$ is not a factor.

(B) $C(5) = 50$. So, $(x-5)$ is a factor.

(C) $C(5) = 0$. So, $(x-5)$ is a factor.

(D) Evaluating $C(5)$ doesn't determine if $(x-5)$ is a factor.

Question 3. Referring to the Case Study above, the company has a total quantity of raw material for a run given by the polynomial $R(y) = y^3 + 5y^2 + 5y - 2$. This material needs to be processed by machines that handle $y+2$ units at a time. Use polynomial division to find the quotient when $R(y)$ is divided by $y+2$.

(A) $y^2 + 3y - 1$

(B) $y^2 - 3y + 1$

(C) $y^2 + 7y + 19$

(D) $y^2 - 7y + 19$

Question 4. Referring to the Case Study above, for the raw material polynomial $R(y) = y^3 + 5y^2 + 5y - 2$, use the Remainder Theorem to find the remainder when $R(y)$ is divided by $y+2$. This should match the remainder from polynomial division.

(A) $R(2) = 2^3 + 5(2)^2 + 5(2) - 2 = 8 + 20 + 10 - 2 = 36$

(B) $R(-2) = (-2)^3 + 5(-2)^2 + 5(-2) - 2 = -8 + 20 - 10 - 2 = 0$

(C) $R(-2) = -8 - 20 - 10 - 2 = -40$

(D) The remainder is constant, regardless of $y$.

Question 5. Referring to the Case Study above, since $R(-2) = 0$, what can be concluded about $(y+2)$ and $R(y)$ based on the Factor Theorem?

(A) $(y-2)$ is a factor of $R(y)$.

(B) $(y+2)$ is a factor of $R(y)$.

(C) $y=0$ is a zero of $R(y)$.

(D) $R(y)$ has no factors.

Question 1. Case Study: Designing a Community Park

A municipal corporation is designing a square community park. Due to some constraints, the side length of the park is planned to be exactly $(100 - 2)$ metres. The engineers want to calculate the area of the park efficiently using algebraic identities.

Which identity can be used to calculate the area $(100-2)^2$ metres$^2$ efficiently?

(A) $(a+b)^2 = a^2 + 2ab + b^2$

(B) $(a-b)^2 = a^2 - 2ab + b^2$

(C) $a^2 - b^2 = (a-b)(a+b)$

(D) $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$

Question 2. Referring to the Case Study above, calculate the area of the park using the appropriate identity.

(A) $(100-2)^2 = 100^2 - 2^2 = 10000 - 4 = 9996$ m$^2$

(B) $(100-2)^2 = 100^2 + 2^2 = 10000 + 4 = 10004$ m$^2$

(C) $(100-2)^2 = 100^2 - 2(100)(2) + 2^2 = 10000 - 400 + 4 = 9604$ m$^2$

(D) $(100-2)^2 = (100-2)(100+2) = 98 \times 102 = 9996$ m$^2$ (without identity)

Question 3. Referring to the Case Study above, a rectangular walking track is added around the park. The outer dimensions are $(100+2)$ metres by $(100-2)$ metres. The area of this outer rectangle is $(100+2)(100-2)$ m$^2$. Which identity can be used to calculate this area?

(A) $(a+b)^2 = a^2 + 2ab + b^2$

(B) $(a-b)^2 = a^2 - 2ab + b^2$

(C) $a^2 - b^2 = (a-b)(a+b)$

(D) $(a+b)(c+d) = ac+ad+bc+bd$

Question 4. Referring to the Case Study above, calculate the area of the outer rectangular walking track using the appropriate identity from the previous question.

(A) $(100+2)(100-2) = 100^2 + 2^2 = 10000 + 4 = 10004$ m$^2$

(B) $(100+2)(100-2) = 100^2 - 2^2 = 10000 - 4 = 9996$ m$^2$

(C) $(100+2)(100-2) = (102)(98) = 9996$ m$^2$ (without identity)

(D) $(100+2)(100-2) = 100^2 - 2(100)(2) + 2^2 = 9604$ m$^2$

Question 5. Referring to the Case Study above, consider the volume of a cubical storage unit for gardening tools, with side length $(x+5)$ metres. Use the identity for $(a+b)^3$ to find the expression for its volume.

(A) $x^3 + 5^3$

(B) $(x+5)(x^2-5x+25)$

(C) $x^3 + 15x^2 + 75x + 125$

(D) $x^3 + 5x^2 + 25x + 125$

Question 1. Case Study: Manufacturing and Packaging

A factory produces metallic sheets with an area given by the expression $6x^2y + 9xy^2$. The packaging department needs to determine the possible linear dimensions if the sheets are rectangular. This involves factorising the expression.

Factorise the expression $6x^2y + 9xy^2$ by taking out the greatest common factor.

(A) $3xy(2x + 3y)$

(B) $3(2x^2y + 3xy^2)$

(C) $xy(6x + 9y)$

(D) $3xy(2x \times 3y)$

Question 2. Referring to the Case Study above, another product's area is given by the quadratic expression $m^2 - 10m + 24$. Factorise this trinomial to find its possible linear dimensions.

(A) $(m-4)(m-6)$

(B) $(m+4)(m+6)$

(C) $(m-2)(m-12)$

(D) $(m+2)(m-12)$

Question 3. Referring to the Case Study above, a complex design uses material whose quantity is described by the expression $p^2q + pr^2 - pqr - r^3$. Factorise this expression by grouping terms.

(A) $(p^2 - r^2)(q - r)$

(B) $(p^2 + r^2)(q + r)$

(C) $(p+r)(pq-r^2)$

(D) $(p^2 - r^2)(q + r)$

Question 4. Referring to the Case Study above, the area of a circular component removed from a square plate results in a remaining area represented by $49k^2 - 25$. Factorise this expression using an algebraic identity.

(A) $(7k - 5)^2$

(B) $(7k + 5)^2$

(C) $(7k - 5)(7k + 5)$

(D) $(49k - 25)(k + 1)$

Question 5. Referring to the Case Study above, a mold has a volume described by the cubic polynomial $a^3 - 6a^2 + 11a - 6$. If it is known that $(a-1)$ is a factor of this polynomial (meaning $a=1$ is a zero), factorise the cubic polynomial completely. (Hint: Use polynomial division or synthetic division after finding one factor).

(A) $(a-1)(a-2)(a-3)$

(B) $(a-1)(a+2)(a+3)$

(C) $(a-1)(a-5)(a-6)$

(D) $(a-1)(a^2 - 5a + 6)$ only

Question 1. Case Study: Planning a School Trip

A school is planning a trip for its students. The total cost of the trip is $\textsf{₹} 25000$. The school received a grant of $\textsf{₹} 5000$, and the remaining cost is to be shared equally among the students. If each student has to pay $\textsf{₹} 400$, find the number of students going on the trip.

Let the number of students be $x$. Set up and solve the linear equation.

(A) $400x = 25000$; $x = 62.5$ students (impossible)

(B) $25000 - 5000 = 400x$; $20000 = 400x$; $x = 50$ students

(C) $400x + 5000 = 25000$; $400x = 20000$; $x = 50$ students

(D) $25000 + 5000 = 400x$; $30000 = 400x$; $x = 75$ students

Question 2. Referring to the Case Study above, suppose the school decides to reduce the per-student cost by $\textsf{₹} 50$, making it $\textsf{₹} 350$ per student, assuming the same number of students (50) and the same grant. What would the total cost of the trip be in this case?

(A) $350 \times 50 + 5000 = 17500 + 5000 = \textsf{₹} 22500$

(B) $350 \times 50 - 5000 = 17500 - 5000 = \textsf{₹} 12500$

(C) $350 \times 50 = \textsf{₹} 17500$

(D) $400 \times 50 = \textsf{₹} 20000$

Question 3. Referring to the Case Study above, another class collected $\textsf{₹} 15000$ for their portion of the trip cost. If the total amount they needed to collect was $\textsf{₹} x$, and they still need $\textsf{₹} 5000$ more, which equation represents this situation?

(A) $x + 15000 = 5000$

(B) $x - 15000 = 5000$

(C) $15000 - x = 5000$

(D) $x + 5000 = 15000$

Question 4. Referring to the Case Study above, suppose the total number of students is $60$, and the total cost is $\textsf{₹} 30000$. If the grant remains $\textsf{₹} 5000$, what is the per-student cost? Let the per-student cost be $p$ $\textsf{₹}$. Set up an equation and solve for $p$.

(A) $60p + 5000 = 30000$; $60p = 25000$; $p \approx \textsf{₹} 416.67$

(B) $60p - 5000 = 30000$; $60p = 35000$; $p \approx \textsf{₹} 583.33$

(C) $p + 5000 = 30000/60$; $p = 500 - 5000 = -\textsf{₹} 4500$ (impossible)

(D) $60p = 30000 + 5000$; $60p = 35000$; $p \approx \textsf{₹} 583.33$

Question 5. Referring to the Case Study above, the trip duration is planned for $d$ days. The cost per day is $\textsf{₹} 1500$ per student (after accounting for grant, etc.). If the total per-student cost for the trip is $\textsf{₹} 4500$, how many days is the trip planned for? Set up and solve the equation.

(A) $d + 1500 = 4500$; $d=3000$ days (impossible)

(B) $1500d = 4500$; $d=3$ days

(C) $4500d = 1500$; $d = 1/3$ days (impossible)

(D) $d/1500 = 4500$; $d = 6750000$ days (impossible)

Question 1. Case Study: Purchasing Fruits

Anjali goes to a fruit vendor to buy oranges and mangoes. The price of one orange is $\textsf{₹} 8$ and the price of one mango is $\textsf{₹} 12$. She wants to spend exactly $\textsf{₹} 240$. Let $x$ be the number of oranges and $y$ be the number of mangoes she buys.

Which linear equation in two variables represents the total cost of the fruits?

(A) $x + y = 240$

(B) $8x + 12y = 240$

(C) $12x + 8y = 240$

(D) $20(x+y) = 240$

Question 2. Referring to the Case Study above, one possible combination of fruits Anjali could buy is $15$ oranges and $10$ mangoes. Do these values satisfy the equation derived in the previous question?

(A) Yes, because $8(15) + 12(10) = 120 + 120 = 240$.

(B) No, because $15 + 10 \neq 240$.

(C) Yes, because $8+12 = 20$ and $15+10 = 25$.

(D) No, because $8 \times 15 \neq 240$ individually.

Question 3. Referring to the Case Study above, if Anjali decides to buy only oranges, how many oranges can she buy? This corresponds to setting $y=0$ in the equation and solving for $x$.

(A) $8x = 240 \implies x = 30$ oranges

(B) $12y = 240 \implies y = 20$ oranges (incorrect variable)

(C) $x = 240/8 = 30$ oranges

(D) $x + 0 = 240 \implies x=240$ oranges (incorrect equation)

Question 4. Referring to the Case Study above, if the linear equation $8x + 12y = 240$ is plotted on a graph where $x$ is on the horizontal axis and $y$ is on the vertical axis, what does the point $(0, 20)$ represent in the context of this problem?

(A) Buying $0$ oranges and $20$ mangoes for $\textsf{₹} 240$.

(B) Buying $20$ oranges and $0$ mangoes for $\textsf{₹} 240$.

(C) The price of one orange and one mango.

(D) A combination that exceeds the budget.

Question 5. Referring to the Case Study above, suppose another friend, Bhavesh, buys fruits such that the equation $8x + 12y = 300$ represents his total spending. If Anjali's spending is represented by $8x + 12y = 240$, what can be said about the graphs of these two equations? (Note: Assume $x$ and $y$ can be any non-negative real numbers for the graph).

(A) The graphs are identical lines.

(B) The graphs are parallel lines.

(C) The graphs intersect at one point.

(D) The graphs are perpendicular lines.

Question 1. Case Study: Solving a Pricing Puzzle

A shopkeeper sells two types of sweets, Laddus and Jalebis. The cost of $500$ grams of Laddus and $700$ grams of Jalebis is $\textsf{₹} 210$. The cost of $700$ grams of Laddus and $500$ grams of Jalebis is $\textsf{₹} 190$. Let the cost per gram of Laddus be $\textsf{₹} l$ and the cost per gram of Jalebis be $\textsf{₹} j$.

Which pair of linear equations represents this situation?

(A) $500l + 700j = 210$, $700l + 500j = 190$

(B) $500l + 700j = 190$, $700l + 500j = 210$

(C) $l + j = 210$, $l + j = 190$

(D) $500l + 700j = 21000$, $700l + 500j = 19000$ (if prices were per kg and amounts in grams)

Question 2. Referring to the Case Study above, solve the system of equations (using elimination or substitution) to find the cost per gram of Laddus ($l$) and Jalebis ($j$).

(A) $l = \textsf{₹} 0.20$, $j = \textsf{₹} 0.30$

(B) $l = \textsf{₹} 0.30$, $j = \textsf{₹} 0.20$

(C) $l = \textsf{₹} 2$, $j = \textsf{₹} 3$

(D) $l = \textsf{₹} 3$, $j = \textsf{₹} 2$

Question 3. Referring to the Case Study above, suppose another shop has prices modelled by the system: $x + 2y = 5$ and $3x + 6y = 15$. What type of system is this, and how many solutions does it have? (Consider the ratios of coefficients).

(A) Consistent, unique solution

(B) Inconsistent, no solution

(C) Consistent, infinitely many solutions (Dependent system)

(D) Consistent, exactly two solutions

Question 4. Referring to the Case Study above, consider the system $x+y=10$ and $x-y=2$. If we solve this graphically, we would plot both lines and find their intersection point. What is the solution $(x, y)$ to this system?

(A) $(6, 4)$

(B) $(4, 6)$

(C) $(5, 5)$

(D) $(12, -2)$

Question 5. Referring to the Case Study above, a classic problem involves a boat in a river. A boat travels $30$ km upstream and $40$ km downstream in $10$ hours. In $13$ hours, it travels $40$ km upstream and $50$ km downstream. If $u$ is the speed upstream and $v$ is the speed downstream, the equations are $\frac{30}{u} + \frac{40}{v} = 10$ and $\frac{40}{u} + \frac{50}{v} = 13$. These can be reduced to linear equations by substituting $x = 1/u$ and $y = 1/v$. What is the resulting pair of linear equations in terms of $x$ and $y$?

(A) $30x + 40y = 10$, $40x + 50y = 13$ (already linear in $x, y$)

(B) $30/x + 40/y = 10$, $40/x + 50/y = 13$

(C) $30x + 40y = 10xy$, $40x + 50y = 13xy$ (multiplying by $uv$)

(D) $u+v=10$, $u-v=13$

Question 1. Case Study: Designing a Garden Pathway

A rectangular garden is $12$ metres long and $8$ metres wide. A pathway of uniform width is to be built around it. The total area of the garden with the pathway is $220$ square metres. Let the width of the pathway be $x$ metres.

Which quadratic equation represents the total area of the garden with the pathway?

(A) $(12+x)(8+x) = 220$

(B) $(12+2x)(8+2x) = 220$

(C) $(12-x)(8-x) = 220$

(D) $(12-2x)(8-2x) = 220$

Question 2. Referring to the Case Study above, expand and simplify the quadratic equation from the previous question to the standard form $ax^2 + bx + c = 0$.

(A) $x^2 + 20x + 96 = 220 \implies x^2 + 20x - 124 = 0$

(B) $4x^2 + 40x + 96 = 220 \implies 4x^2 + 40x - 124 = 0 \implies x^2 + 10x - 31 = 0$

(C) $4x^2 + 40x + 96 = 220 \implies 4x^2 + 40x + 316 = 0 \implies x^2 + 10x + 79 = 0$

(D) $x^2 + 20x + 96 = 220 \implies x^2 + 20x + 316 = 0$

Question 3. Referring to the Case Study above, the equation for the pathway width is $x^2 + 10x - 31 = 0$. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the possible value(s) for the pathway width $x$. (Note: Width must be positive).

(A) $x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-31)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 124}}{2} = \frac{-10 \pm \sqrt{224}}{2}$. Only the positive root is valid.

(B) $x = \frac{-10 \pm \sqrt{10^2 - 4(1)(31)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 124}}{2} = \frac{-10 \pm \sqrt{-24}}{2}$ (complex roots)

(C) $x = \frac{10 \pm \sqrt{100 - 124}}{2}$ (complex roots)

(D) $x = \frac{-10 \pm \sqrt{100 + 124}}{2} = \frac{-10 \pm \sqrt{224}}{2}$. Both roots are valid.

Question 4. Referring to the Case Study above, the discriminant $\Delta = b^2 - 4ac$ of the quadratic equation $ax^2 + bx + c = 0$ determines the nature of the roots. For the equation $x^2 + 10x - 31 = 0$, calculate the discriminant and determine the nature of the roots.

(A) $\Delta = 10^2 - 4(1)(-31) = 100 + 124 = 224 > 0$. Roots are real and distinct.

(B) $\Delta = 10^2 - 4(1)(31) = 100 - 124 = -24 < 0$. Roots are complex conjugates.

(C) $\Delta = (-10)^2 - 4(1)(-31) = 100 + 124 = 224 > 0$. Roots are real and distinct.

(D) $\Delta = 10^2 - 4(10)(-31) = 100 + 1240 = 1340 > 0$. Roots are real and distinct.

Question 5. Referring to the Case Study above, consider a simpler garden problem: A square garden's side length is increased by $2$ metres, resulting in a new square garden whose area is $64$ square metres. Let the original side length be $s$ metres. Which quadratic equation represents this? And what was the original side length?

(A) $s^2 + 2 = 64$; $s^2 = 62$; $s = \sqrt{62}$

(B) $(s+2)^2 = 64$; $s+2 = \pm 8$. Since $s>0$, $s+2=8 \implies s=6$ metres.

(C) $s^2 + 4s + 4 = 64$; $s^2 + 4s - 60 = 0$; $(s+10)(s-6)=0$; $s=6$ or $s=-10$. Since $s>0$, $s=6$ metres.

(D) Both (B) and (C) lead to the correct solution.

Question 1. Case Study: Electrical Circuit Analysis

In alternating current (AC) circuits, impedance ($Z$) is a measure of the opposition to current flow, represented by a complex number $Z = R + iX$, where $R$ is resistance and $X$ is reactance. Engineers use $j$ instead of $i$ to avoid confusion with current $I$. Assume we use $i$ as the imaginary unit here.

If the impedance of a resistor is $Z_R = 5$ ohms and the impedance of an inductor is $Z_L = 3i$ ohms, and these are connected in series, the total impedance is $Z_{total} = Z_R + Z_L$. What is $Z_{total}$?

(A) $5i + 3$ ohms

(B) $5 + 3i$ ohms

(C) $8i$ ohms

(D) $8$ ohms

Question 2. Referring to the Case Study above, if another circuit component has an impedance of $Z_C = 2 - 4i$ ohms, and it is connected in series with a component of impedance $Z_D = 1 + i$ ohms, what is the total impedance $Z_C + Z_D$?

(A) $3 - 3i$ ohms

(B) $3 + 3i$ ohms

(C) $1 - 5i$ ohms

(D) $1 + 5i$ ohms

Question 3. Referring to the Case Study above, when components are connected in parallel, the reciprocal of the total impedance is the sum of the reciprocals of individual impedances. For two impedances $Z_1$ and $Z_2$ in parallel, $\frac{1}{Z_{total}} = \frac{1}{Z_1} + \frac{1}{Z_2}$. This requires complex number division. If $Z_1 = 1+i$ and $Z_2 = 1-i$, calculate $\frac{1}{Z_1}$.

(A) $\frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1-i}{1^2 - i^2} = \frac{1-i}{1 - (-1)} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i$

(B) $1-i$

(C) $1+i$

(D) $\frac{1}{2} + \frac{1}{2}i$

Question 4. Referring to the Case Study above, the square of the current $I$ (a complex number) times the impedance $Z$ gives power in some analyses. If $I = 2+i$ amperes and $Z = 1+3i$ ohms, the complex power $S = I^2 Z$ is relevant. Calculate $I^2 = (2+i)^2$ using algebraic identities.

(A) $2^2 + i^2 = 4 - 1 = 3$

(B) $2^2 + 2(2)(i) + i^2 = 4 + 4i - 1 = 3 + 4i$

(C) $4 + i^2 = 4 - 1 = 3$

(D) $2^2 - i^2 = 4 - (-1) = 5$

Question 5. Referring to the Case Study above, simplify the power of the imaginary unit $i^3$.

(A) 1

(B) -1

(C) $i$

(D) $-i$

Question 1. Case Study: Representing Signals

In electrical engineering and signal processing, signals are often represented as complex numbers. The complex number $Z = a + bi$ represents a signal with a real component $a$ and an imaginary component $b$. This can be visualized as a point $(a, b)$ on the Argand plane.

A signal is represented by the complex number $Z = 3 + 2i$. Plot this signal on the Argand plane. What are the coordinates corresponding to this complex number?

(A) $(3, 2)$

(B) $(2, 3)$

(C) $(-3, -2)$

(D) $(3i, 2)$

Question 2. Referring to the Case Study above, the strength or amplitude of a signal represented by a complex number $Z = a+bi$ is given by its modulus, $|Z| = \sqrt{a^2 + b^2}$. If a signal is represented by $Z = -5 + 12i$, what is its amplitude?

(A) $|Z| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

(B) $|Z| = -5 + 12 = 7$

(C) $|Z| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = 13$

(D) $|Z| = -5^2 + 12^2 = -25 + 144 = 119$

Question 3. Referring to the Case Study above, sometimes the "phase-reversed" version of a signal represented by $Z = a+bi$ is needed, which corresponds to its conjugate $\bar{Z} = a - bi$. If a signal is $Z = 7 - 3i$, what is the phase-reversed signal?

(A) $7 + 3i$

(B) $-7 + 3i$

(C) $-7 - 3i$

(D) $7 - 3i$ (itself)

Question 4. Referring to the Case Study above, signals are often represented in polar form $r(\cos\theta + i\sin\theta)$ because multiplication of complex numbers in polar form corresponds to multiplying amplitudes and adding phases. Convert the signal $Z = -1 + i$ into polar form.

(A) $\sqrt{2}(\cos(3\pi/4) + i\sin(3\pi/4))$

(B) $\sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$

(C) $1(\cos(3\pi/4) + i\sin(3\pi/4))$

(D) $\sqrt{2}(\cos(-\pi/4) + i\sin(-\pi/4))$

Question 5. Referring to the Case Study above, the square root of a complex number might represent possible signal components. Which of the following is a square root of the complex number $-8 - 6i$?

(A) $1 - 3i$

(B) $-1 + 3i$

(C) $1 + 3i$

(D) $-1 - 3i$

Question 1. Case Study: Analysing Mechanical Vibrations

In mechanical engineering, the behaviour of vibrating systems is often described by differential equations. The characteristic equation of such a system can be a quadratic equation with real coefficients. The roots of this equation determine whether the system oscillates and decays, grows, or remains stable.

If the characteristic equation of a system is $r^2 + 4r + 13 = 0$, find the roots of this equation using the quadratic formula. These roots are crucial for determining the system's response.

(A) $r = -2 \pm 3i$

(B) $r = -2 \pm \sqrt{17}$

(C) $r = 2 \pm 3i$

(D) $r = -4 \pm \sqrt{36}$

Question 2. Referring to the Case Study above, the damping ratio and natural frequency of the mechanical system are related to the real and imaginary parts of the complex roots. For the equation $r^2 + 4r + 13 = 0$, the roots are $r = -2 \pm 3i$. The real part is $-2$ and the imaginary part is $\pm 3$. This indicates a specific type of system behaviour. If the roots of a quadratic equation with real coefficients are complex, what can be said about them?

(A) The roots are always real and distinct.

(B) The roots are always real and equal.

(C) The roots are always complex conjugates of each other.

(D) The roots can be any complex numbers.

Question 3. Referring to the Case Study above, consider another mechanical system whose characteristic equation is $s^2 - 2s + 2 = 0$. Find the roots of this equation.

(A) $s = 1 \pm i$

(B) $s = -1 \pm i$

(C) $s = 1 \pm \sqrt{3}$

(D) $s = -1 \pm \sqrt{3}$

Question 4. Referring to the Case Study above, if a system has characteristic roots that are real and negative, it is typically overdamped (returns to equilibrium without oscillation). If the roots are complex with a negative real part, it is typically underdamped (oscillates while returning to equilibrium). Based on the roots of $r^2 + 4r + 13 = 0$ ($r = -2 \pm 3i$), what type of behaviour does this system likely exhibit?

(A) Overdamped

(B) Critically damped

(C) Underdamped

(D) Unstable (growing oscillations)

Question 5. Referring to the Case Study above, sometimes complex roots are written in polar form. For the equation $s^2 - 2s + 2 = 0$ with roots $s = 1 \pm i$, what is the polar form of the root $1+i$?

(A) $\sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$

(B) $\sqrt{2}(\cos(\pi/2) + i\sin(\pi/2))$

(C) $2(\cos(\pi/4) + i\sin(\pi/4))$

(D) $1(\cos(\pi/4) + i\sin(\pi/4))$

Question 1. Case Study: Planning Daily Diet

A nutritionist advises a person to consume at least $1500$ calories per day but not more than $2000$ calories. The person's calorie intake from breakfast is $400$ calories. The remaining calories must come from lunch, snacks, and dinner. Let $C$ be the total daily calorie intake.

Which inequality represents the allowed range for the total daily calorie intake?

(A) $1500 < C < 2000$

(B) $1500 \leq C \leq 2000$

(C) $C \geq 1500$

(D) $C \leq 2000$

Question 2. Referring to the Case Study above, let $R$ be the remaining calories the person needs to consume from lunch, snacks, and dinner. Which inequality represents the allowed range for $R$?

(A) $1500 - 400 \leq R \leq 2000 - 400 \implies 1100 \leq R \leq 1600$

(B) $1500 \leq R \leq 2000$

(C) $1500 + 400 \leq R \leq 2000 + 400 \implies 1900 \leq R \leq 2400$

(D) $R \geq 1500$

Question 3. Referring to the Case Study above, suppose lunch typically provides $500$ calories. Let $s$ be the calories from snacks and $d$ be the calories from dinner. The inequality $500 + s + d \leq 1600$ represents a constraint on snacks and dinner calories to stay within the upper limit. If this is plotted on a graph with $s$ on the x-axis and $d$ on the y-axis, which region represents the valid combinations of calories from snacks and dinner?

(A) The region above the line $s+d = 1100$.

(B) The region below or on the line $s+d = 1100$.

(C) The region above or on the line $s+d = 1100$.

(D) The region below the line $s+d = 1100$.

Question 4. Referring to the Case Study above, the person also needs to consume at least $50$ grams of protein per day. Breakfast provides $10$ grams. Let $p_L$ be protein from lunch, $p_S$ from snacks, and $p_D$ from dinner. The constraint is $10 + p_L + p_S + p_D \geq 50$. Which inequality represents the minimum protein needed from lunch, snacks, and dinner combined?

(A) $p_L + p_S + p_D > 40$

(B) $p_L + p_S + p_D \geq 40$

(C) $p_L + p_S + p_D < 40$

(D) $p_L + p_S + p_D \leq 40$

Question 5. Referring to the Case Study above, the nutritionist recommends that the fat intake should be between $50$ grams and $70$ grams, inclusive. Let $F$ be the total fat intake. Which pair of inequalities represents this recommendation?

(A) $F > 50$ and $F < 70$

(B) $F \geq 50$ and $F \leq 70$

(C) $F \leq 50$ or $F \geq 70$

(D) $F < 50$ or $F > 70$

Question 1. Case Study: Saving for a Goal

Deepak starts a savings plan. In the first month, he saves $\textsf{₹} 2000$. In each subsequent month, he increases his savings by $\textsf{₹} 150$. This forms an Arithmetic Progression (AP).

How much money will Deepak save in the 10th month?

(A) $\textsf{₹} 2000 + 10 \times 150 = \textsf{₹} 3500$

(B) $\textsf{₹} 2000 + 9 \times 150 = \textsf{₹} 2000 + 1350 = \textsf{₹} 3350$

(C) $\textsf{₹} 2000 + (10-1) \times 150 = \textsf{₹} 2000 + 9 \times 150 = \textsf{₹} 3350$

(D) $\textsf{₹} 2000 \times 10 + 150 = \textsf{₹} 20150$

Question 2. Referring to the Case Study above, what is the total amount of money Deepak will have saved at the end of $12$ months? (Sum of the first $12$ terms of the AP).

(A) $S_{12} = \frac{12}{2} (2 \times 2000 + (12-1) \times 150) = 6 (4000 + 11 \times 150) = 6 (4000 + 1650) = 6 \times 5650 = \textsf{₹} 33900$

(B) $S_{12} = \frac{12}{2} (2000 + 150 \times 12) = 6 (2000 + 1800) = 6 \times 3800 = \textsf{₹} 22800$

(C) $S_{12} = 12 \times 3350 = \textsf{₹} 40200$ (assuming he saves the 10th month amount every month)

(D) $S_{12} = \frac{12}{2} (2000 + (2000 + 11 \times 150)) = 6 (2000 + 3650) = 6 \times 5650 = \textsf{₹} 33900$

Question 3. Referring to the Case Study above, if another person, Alia, starts saving $\textsf{₹} 1000$ in the first month and doubles her savings each month, this forms a Geometric Progression (GP). How much will Alia save in the 5th month?

(A) $\textsf{₹} 1000 \times 2^5 = \textsf{₹} 32000$

(B) $\textsf{₹} 1000 \times 2^{5-1} = \textsf{₹} 1000 \times 2^4 = \textsf{₹} 1000 \times 16 = \textsf{₹} 16000$

(C) $\textsf{₹} 1000 + 4 \times 2 = \textsf{₹} 1008$

(D) $\textsf{₹} 1000 \times 5 \times 2 = \textsf{₹} 10000$

Question 4. Referring to the Case Study above, what is the total amount Alia will have saved at the end of $6$ months? (Sum of the first $6$ terms of the GP).

(A) $S_6 = \frac{1000(2^6 - 1)}{2-1} = 1000(64 - 1) = 1000 \times 63 = \textsf{₹} 63000$

(B) $S_6 = \frac{1000(1 - 2^6)}{1-2} = \frac{1000(-63)}{-1} = \textsf{₹} 63000$

(C) $S_6 = \frac{1000(2^6 - 1)}{2+1} = \frac{1000 \times 63}{3} = \textsf{₹} 21000$

(D) $S_6 = 6 \times 16000 = \textsf{₹} 96000$ (incorrect assumption)

Question 5. Referring to the Case Study above, the annual percentage growth of Deepak's savings (in the amount saved each month, not total) is constant, while Alia's monthly savings amount grows by a constant *factor*. This relates to the definitions of AP and GP. Which statement is correct?

(A) Deepak's monthly savings form an AP because the difference between consecutive terms is constant.

(B) Alia's monthly savings form an AP because the ratio between consecutive terms is constant.

(C) Deepak's monthly savings form a GP because the difference between consecutive terms is constant.

(D) Alia's monthly savings form neither an AP nor a GP.

Question 1. Case Study: Proving Mathematical Statements

A mathematician is teaching students about proving statements involving positive integers using the Principle of Mathematical Induction (PMI). They want to prove the statement: The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$, i.e., $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$ for all positive integers $n$.

What is the first step in the proof by PMI for this statement?

(A) Assume the statement is true for $n=k$, where $k$ is a positive integer.

(B) Prove the statement is true for $n=1$ (the base case).

(C) Prove the statement is true for $n=k+1$, assuming it is true for $n=k$.

(D) Assume the statement is false for all $n$.

Question 2. Referring to the Case Study above, perform the base case step for the statement $1 + 2 + \dots + n = \frac{n(n+1)}{2}$.

(A) For $n=1$, LHS = $1$. RHS = $\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1$. LHS = RHS, so the statement is true for $n=1$.

(B) For $n=1$, LHS = $1+2=3$. RHS = $\frac{1(1+1)}{2} = 1$. LHS $\neq$ RHS.

(C) Assume $1+2+\dots+k = \frac{k(k+1)}{2}$.

(D) Consider the formula $\frac{n(n+1)}{2}$.

Question 3. Referring to the Case Study above, the inductive step in PMI requires assuming the statement is true for some positive integer $k$ (the inductive hypothesis) and then proving it is true for $k+1$. For the statement $1 + 2 + \dots + n = \frac{n(n+1)}{2}$, what does the inductive hypothesis state?

(A) $1 + 2 + \dots + k + (k+1) = \frac{(k+1)((k+1)+1)}{2}$

(B) $1 + 2 + \dots + k = \frac{k(k+1)}{2}$

(C) The statement is true for all positive integers $n \leq k$.

(D) The statement is true for $n=k+1$.

Question 4. Referring to the Case Study above, in the inductive step, we start with the left side for $n=k+1$, which is $1 + 2 + \dots + k + (k+1)$. Using the inductive hypothesis, we replace $1 + 2 + \dots + k$ with $\frac{k(k+1)}{2}$. What is the next step to show that this equals the right side for $n=k+1$, which is $\frac{(k+1)(k+2)}{2}$?

(A) Add $k+1$ to both sides of the inductive hypothesis.

(B) Show that $\frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2}$.

(C) Show that $1 + 2 + \dots + k + (k+1) = \frac{k(k+1)}{2}$.

(D) Replace $k+1$ with $n$ in the inductive hypothesis.

Question 5. Referring to the Case Study above, the mathematician also considers proving divisibility statements. For example, prove that $2^n - 1$ is divisible by 1 for all positive integers $n$. (This is trivially true). Let's use a better example: Prove that $5^n - 1$ is divisible by 4 for all positive integers $n$. What is the base case ($n=1$) result?

(A) $5^1 - 1 = 4$, which is divisible by 4. The base case holds.

(B) $5^1 - 1 = 4$, which is not divisible by 4.

(C) $5^k - 1$ is divisible by 4 for some $k \geq 1$.

(D) $5^{k+1} - 1$ is divisible by 4.

Question 1. Case Study: Arranging and Selecting Items

A security manager needs to set up passwords and access codes. A password must consist of $5$ distinct digits from $0$ to $9$. An access code must consist of selecting $3$ specific features out of $8$ available features.

How many different $5$-digit passwords can be created if the digits must be distinct?

(A) $C(10, 5)$

(B) $P(10, 5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$

(C) $10^5$

(D) $5!$

Question 2. Referring to the Case Study above, how many ways can $3$ features be selected from $8$ available features for the access code, if the order of selection does not matter?

(A) $P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$

(B) $C(8, 3) = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

(C) $8^3$

(D) $3!$

Question 3. Referring to the Case Study above, suppose the security manager needs to arrange $7$ security guards in a line for a photo. In how many distinct ways can they be arranged?

(A) $C(7, 7) = 1$

(B) $P(7, 7) = 7! = 5040$

(C) $7^7$

(D) $P(7, 1) = 7$

Question 4. Referring to the Case Study above, suppose there are $15$ applicants for $4$ different security positions (Supervisor, Senior Guard, Junior Guard A, Junior Guard B). In how many ways can these positions be filled if each applicant can hold only one position?

(A) $C(15, 4)$

(B) $P(15, 4) = \frac{15!}{(15-4)!} = \frac{15!}{11!} = 15 \times 14 \times 13 \times 12 = 32760$

(C) $15^4$

(D) $4!$

Question 5. Referring to the Case Study above, suppose the security team consists of $10$ members, and they need to form two groups of $5$ members each for patrolling different areas. How many ways can the team be divided into two groups of $5$?

(A) $C(10, 5) \times C(5, 5) = \frac{10!}{5!5!} \times 1 = 252$

(B) $\frac{C(10, 5) \times C(5, 5)}{2!} = \frac{252 \times 1}{2} = 126$ (since the two groups are indistinguishable)

(C) $P(10, 5) \times P(5, 5)$

(D) $10!$

Question 1. Case Study: Probability in Quality Control

A factory produces light bulbs. The probability that a randomly selected bulb is defective is $p = 0.05$. The probability that it is non-defective is $q = 1-p = 0.95$. When a sample of $n$ bulbs is tested, the number of defective bulbs follows a binomial distribution. The probability of getting exactly $k$ defective bulbs in a sample of size $n$ is given by $\binom{n}{k} p^k q^{n-k}$.

If a sample of $5$ bulbs is tested, what is the probability of finding exactly $1$ defective bulb?

(A) $\binom{5}{1} (0.05)^1 (0.95)^4$

(B) $\binom{5}{1} (0.05)^1 (0.95)^5$

(C) $\binom{5}{1} (0.05)^4 (0.95)^1$

(D) $(0.05)^1 (0.95)^4$

Question 2. Referring to the Case Study above, consider the expansion of $(q+p)^5 = (0.95 + 0.05)^5$. The terms in the expansion represent the probabilities of getting $0, 1, 2, 3, 4$, or $5$ defective bulbs. What is the sum of the coefficients in the expansion of $(a+b)^n$ for any positive integer $n$?

(A) $0$

(B) $1$

(C) $n$(D) $2^n$

Question 3. Referring to the Case Study above, the expansion of $(x + \frac{1}{x^2})^9$ is used in a theoretical calculation related to the manufacturing process. Find the term independent of $x$ in this expansion.

(A) $\binom{9}{3} x^3 (\frac{1}{x^2})^6$

(B) $\binom{9}{6} x^3 (\frac{1}{x^2})^6$

(C) $\binom{9}{3} x^6 (\frac{1}{x^2})^3$

(D) $\binom{9}{6} x^6 (\frac{1}{x^2})^3$

Question 4. Referring to the Case Study above, the total variance in a process is related to the middle term(s) in a binomial expansion. Consider the expansion of $(A+B)^8$. Find the middle term in this expansion.

(A) 4th term

(B) 5th term

(C) 4th and 5th terms

(D) $\binom{8}{4} A^4 B^4$

Question 5. Referring to the Case Study above, approximate the value of $(1.01)^{10}$ using the binomial theorem by considering the first two terms of the expansion of $(1+0.01)^{10}$.

(A) $1 + 10(0.01) = 1 + 0.1 = 1.1$

(B) $1 + 10(0.01) + \binom{10}{2}(0.01)^2 = 1 + 0.1 + 45(0.0001) = 1.1 + 0.0045 = 1.1045$

(C) $(1.01)^{10} \approx 10 \times 1.01 = 10.1$

(D) $10 \times (0.01) = 0.1$

Question 1. Case Study: Sales Data Management

A company has two stores (Store 1, Store 2) and sells three products (Product A, Product B, Product C). The sales (in thousands of $\textsf{₹}$) for a particular month are represented by a matrix $S$, where rows represent products and columns represent stores.

The sales matrix is given by $S = \begin{pmatrix} 50 & 60 \\ 70 & 80 \\ 90 & 100 \end{pmatrix}$. What are the dimensions (order) of this matrix?

(A) $2 \times 3$

(B) $3 \times 2$

(C) $3 \times 3$

(D) $2 \times 2$

Question 2. Referring to the Case Study above, in the sales matrix $S = \begin{pmatrix} 50 & 60 \\ 70 & 80 \\ 90 & 100 \end{pmatrix}$, what does the element in the second row and first column ($S_{21}$) represent?

(A) Sales of Product A in Store 2.

(B) Sales of Product B in Store 1.

(C) Sales of Product C in Store 1.

(D) Sales of Product B in Store 2.

Question 3. Referring to the Case Study above, the sales target for the next month is $10\%$ higher than the current month's sales for each product in each store. To find the target sales matrix, you need to multiply the current sales matrix $S$ by a scalar factor. What is the scalar factor and the operation used?

(A) Scalar factor $0.10$, Matrix Addition.

(B) Scalar factor $1.10$, Scalar Multiplication.

(C) Scalar factor $10$, Matrix Multiplication.

(D) Scalar factor $0.90$, Scalar Multiplication.

Question 4. Referring to the Case Study above, the prices (in $\textsf{₹}$ per thousand units) of the three products are given by a row matrix $P = \begin{pmatrix} 10 & 15 & 20 \end{pmatrix}$. To find the total revenue for each store, you need to perform matrix multiplication. The number of units sold (in thousands) is given by the matrix $U = \begin{pmatrix} 5 & 6 \\ 7 & 8 \\ 9 & 10 \end{pmatrix}$ (same as S, but representing units). The revenue matrix is $P \times U$. What are the dimensions of the resulting revenue matrix?

(A) $1 \times 2$

(B) $2 \times 1$

(C) $3 \times 3$

(D) Matrix multiplication is not possible.

Question 5. Referring to the Case Study above, suppose the company also tracks expenses. The expense matrix (in thousands of $\textsf{₹}$) for the same month and stores is $E = \begin{pmatrix} 20 & 25 \\ 30 & 35 \\ 40 & 45 \end{pmatrix}$. To find the profit matrix, you subtract the expense matrix from the sales matrix: $Profit = S - E$. What is the profit matrix?

(A) $\begin{pmatrix} 30 & 35 \\ 40 & 45 \\ 50 & 55 \end{pmatrix}$

(B) $\begin{pmatrix} 70 & 85 \\ 100 & 115 \\ 130 & 145 \end{pmatrix}$

(C) $\begin{pmatrix} -30 & -35 \\ -40 & -45 \\ -50 & -55 \end{pmatrix}$

(D) Matrix subtraction is not possible.

Question 1. Case Study: Analysing Data Relationships

A researcher collects data and organises it into matrices. Matrix $A$ represents the correlations between different variables. If the correlation between variable $i$ and variable $j$ is the same as the correlation between variable $j$ and variable $i$, the correlation matrix $A$ will be equal to its transpose $A'$.

If a square matrix $M$ satisfies $M' = M$, where $M'$ is the transpose of $M$, what type of matrix is $M$?

(A) Symmetric matrix

(B) Skew Symmetric matrix

(C) Diagonal matrix

(D) Identity matrix

Question 2. Referring to the Case Study above, suppose a matrix $N$ represents flows in a directed network. If the flow from node $i$ to node $j$ is the negative of the flow from node $j$ to node $i$ (implying a balanced flow in opposite directions or no net flow), the matrix $N$ might satisfy $N' = -N$. What type of matrix is $N$ if it satisfies this condition?

(A) Symmetric matrix

(B) Skew Symmetric matrix

(C) Identity matrix

(D) Zero matrix

Question 3. Referring to the Case Study above, a data matrix is being processed to solve a system of linear equations using Gaussian elimination. This process involves applying elementary row operations. Which of the following operations would change the set of solutions of the corresponding system of linear equations?

(A) Interchanging two rows.

(B) Multiplying a row by a non-zero scalar.

(C) Replacing a row by the sum of that row and a non-zero multiple of another row.

(D) Replacing a row by the sum of that row and a non-zero constant (not a multiple of another row).

Question 4. Referring to the Case Study above, a transformation matrix $T$ is used to rotate and scale data points. If the transformation is reversible, an inverse matrix $T^{-1}$ exists. For a square matrix $A$, its inverse $A^{-1}$ exists if and only if $A$ is non-singular. What does it mean for a matrix to be non-singular?

(A) Its determinant is zero.

(B) Its determinant is non-zero.

(C) It is a symmetric matrix.

(D) It is a diagonal matrix.

Question 5. Referring to the Case Study above, consider the matrix $P = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Find its transpose $P'$.

(A) $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

(B) $\begin{pmatrix} 4 & 3 \\ 2 & 1 \end{pmatrix}$

(C) $\begin{pmatrix} -1 & -2 \\ -3 & -4 \end{pmatrix}$

(D) $\begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$

Question 1. Case Study: Geometric Area Calculation

A civil engineer is working with land survey data. The coordinates of three vertices of a plot are given as A$(2, 3)$, B$(5, 7)$, and C$(-1, 4)$. The engineer needs to calculate the area of the triangular plot using determinants.

The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ using determinants is $\frac{1}{2} |\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}|$. Calculate the area of the triangle ABC.

(A) $\frac{1}{2} |2(7-4) - 3(5-(-1)) + 1(5 \times 4 - 7 \times -1)| = \frac{1}{2} |2(3) - 3(6) + 1(20+7)| = \frac{1}{2} |6 - 18 + 27| = \frac{1}{2} |15| = 7.5$ square units.

(B) $\frac{1}{2} |2(7+4) - 3(5-1) + 1(5 \times 4 - 7 \times -1)| = \frac{1}{2} |2(11) - 3(4) + 27| = \frac{1}{2} |22 - 12 + 27| = \frac{1}{2} |37| = 18.5$ square units.

(C) $|2(7-4) - 3(5-(-1)) + 1(20+7)| = |6 - 18 + 27| = |15| = 15$ square units.

(D) $\frac{1}{2} |2(7-4) + 3(5-(-1)) + 1(20+7)| = \frac{1}{2} |6 + 18 + 27| = \frac{1}{2} |51| = 25.5$ square units.

Question 2. Referring to the Case Study above, suppose the coordinates of three proposed streetlights are D$(1, 1)$, E$(2, 3)$, and F$(3, 5)$. The authorities want to know if these three points lie on the same straight line (are collinear). This can be checked by seeing if the area of the triangle formed by them is zero using the determinant method.

Calculate the determinant $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 1 \end{vmatrix}$ and determine if the points are collinear.

(A) Determinant $= 1(3-5) - 1(2-3) + 1(10-9) = -2 - 1(-1) + 1(1) = -2 + 1 + 1 = 0$. Yes, the points are collinear.

(B) Determinant $= -2$. No, the points are not collinear.

(C) Determinant $= 0$. No, the points are not collinear.

(D) Determinant $= 1$. Yes, the points are collinear.

Question 3. Referring to the Case Study above, a $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is used in a transformation. The absolute value of its determinant, $|\det(M)| = |ad-bc|$, represents the scaling factor of area under the transformation. If a design element with an area of $20$ square units is transformed by the matrix $T = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$, what is the area of the transformed element?

(A) $\det(T) = 3 \times 4 - 1 \times 2 = 12 - 2 = 10$. Transformed Area $= 20 \times |10| = 200$ square units.

(B) $\det(T) = 10$. Transformed Area $= 20 / |10| = 2$ square units.

(C) $\det(T) = 10$. Transformed Area $= 20 + |10| = 30$ square units.

(D) $\det(T) = 10$. Transformed Area $= 20$ square units (no scaling).

Question 4. Referring to the Case Study above, the adjoint of a square matrix $A$, denoted by $\text{adj}(A)$, is the transpose of the cofactor matrix. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjoint is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Find the adjoint of the matrix $A = \begin{pmatrix} 5 & 2 \\ 3 & 1 \end{pmatrix}$.

(A) $\begin{pmatrix} 1 & -2 \\ -3 & 5 \end{pmatrix}$

(B) $\begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix}$

(C) $\begin{pmatrix} 1 & 2 \\ 3 & 5 \end{pmatrix}$

(D) $\begin{pmatrix} -5 & -2 \\ -3 & -1 \end{pmatrix}$

Question 5. Referring to the Case Study above, a property of determinants states that if two rows (or two columns) of a square matrix are identical, the determinant is zero. Consider a matrix $R = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 1 & 2 & 3 \end{pmatrix}$ representing some redundancy in data. What is the determinant of matrix $R$?

(A) 1

(B) -1

(C) Non-zero

(D) 0

Question 1. Case Study: Solving a Resource Allocation Problem

A company produces two types of goods, P and Q, using two resources, R1 and R2. The requirement of resources (in units per item) is given by a matrix. The total available resources are fixed. Let $x$ be the number of items of P and $y$ be the number of items of Q produced.

Suppose the resource equations are: $3x + 2y = 120$ (Total R1 units) $4x + 5y = 210$ (Total R2 units)

This system can be written in matrix form $AX=B$, where $A = \begin{pmatrix} 3 & 2 \\ 4 & 5 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, and $B = \begin{pmatrix} 120 \\ 210 \end{pmatrix}$. To find the production quantities $x$ and $y$, we can use the matrix inverse method: $X = A^{-1}B$. First, find the inverse of matrix $A$. ($\det(A) = 3 \times 5 - 2 \times 4 = 15 - 8 = 7$. $\text{adj}(A) = \begin{pmatrix} 5 & -2 \\ -4 & 3 \end{pmatrix}$).

What is the inverse matrix $A^{-1}$?

(A) $\begin{pmatrix} 5 & -2 \\ -4 & 3 \end{pmatrix}$

(B) $\frac{1}{7} \begin{pmatrix} 5 & -2 \\ -4 & 3 \end{pmatrix}$

(C) $\begin{pmatrix} 3 & 4 \\ 2 & 5 \end{pmatrix}$

(D) $\begin{pmatrix} 5 & 2 \\ 4 & 3 \end{pmatrix}$

Question 2. Referring to the Case Study above, use the matrix inverse method ($X = A^{-1}B$) to find the number of items of P ($x$) and Q ($y$) that can be produced.

(A) $X = \frac{1}{7} \begin{pmatrix} 5 & -2 \\ -4 & 3 \end{pmatrix} \begin{pmatrix} 120 \\ 210 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 5 \times 120 + (-2) \times 210 \\ -4 \times 120 + 3 \times 210 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 600 - 420 \\ -480 + 630 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 180 \\ 150 \end{pmatrix} = \begin{pmatrix} 180/7 \\ 150/7 \end{pmatrix}$. So $x=180/7, y=150/7$ (not integer items).

(B) $X = \begin{pmatrix} 5 & -2 \\ -4 & 3 \end{pmatrix} \begin{pmatrix} 120 \\ 210 \end{pmatrix} = \begin{pmatrix} 600 - 420 \\ -480 + 630 \end{pmatrix} = \begin{pmatrix} 180 \\ 150 \end{pmatrix}$. So $x=180, y=150$ items.

(C) Let's recheck the system. Using elimination: $5 \times (3x+2y=120) \implies 15x+10y=600$. $2 \times (4x+5y=210) \implies 8x+10y=420$. Subtracting: $(15x+10y) - (8x+10y) = 600 - 420 \implies 7x = 180 \implies x = 180/7$. $3(180/7) + 2y = 120 \implies 540/7 + 2y = 840/7 \implies 2y = 300/7 \implies y = 150/7$. Option A is correct based on calculation, but might be impractical for real items. Assuming the question implies finding the exact solution from the matrix method.

(D) $X = \frac{1}{7} \begin{pmatrix} 180 \\ 150 \end{pmatrix}$. This is the solution matrix.

Which option correctly states the solution matrix $X$ using $A^{-1}B$?

(A) $\begin{pmatrix} 180/7 \\ 150/7 \end{pmatrix}$

(B) $\begin{pmatrix} 180 \\ 150 \end{pmatrix}$

(C) $\frac{1}{7} \begin{pmatrix} 180 \\ 150 \end{pmatrix}$ (Intermediate step)

(D) $\begin{pmatrix} 180 & 150 \end{pmatrix}$

Question 3. Referring to the Case Study above, consider a system of $3$ linear equations with $3$ variables: $x+y+z=5$, $2x-y+z=2$, $x-2y+3z=6$. This system can be solved using Cramer's Rule, which involves calculating determinants. The determinant of the coefficient matrix is needed first. What is the determinant of $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & -2 & 3 \end{pmatrix}$?

(A) $\det(A) = 1(-3 - (-2)) - 1(6 - 1) + 1(-4 - (-1)) = 1(-1) - 1(5) + 1(-3) = -1 - 5 - 3 = -9$

(B) $\det(A) = 1(-3 - 2) - 1(6 - 1) + 1(-4 - 1) = 1(-5) - 1(5) + 1(-5) = -5 - 5 - 5 = -15$

(C) $\det(A) = 1(3 - (-2)) - 1(6 - 1) + 1(-4 - 1) = 1(5) - 1(5) + 1(-5) = 5 - 5 - 5 = -5$

(D) $\det(A) = -9$. Since $\det(A) \neq 0$, a unique solution exists.

Question 4. Referring to the Case Study above, in Cramer's Rule for the system $Ax=B$, the solution for $x$ is given by $x = \frac{\det(A_x)}{\det(A)}$, where $A_x$ is the matrix formed by replacing the first column of $A$ with the column matrix $B$. For the system $x+y+z=5$, $2x-y+z=2$, $x-2y+3z=6$, and $B = \begin{pmatrix} 5 \\ 2 \\ 6 \end{pmatrix}$. What is the matrix $A_x$?

(A) $\begin{pmatrix} 5 & 1 & 1 \\ 2 & -1 & 1 \\ 6 & -2 & 3 \end{pmatrix}$

(B) $\begin{pmatrix} 1 & 5 & 1 \\ 2 & 2 & 1 \\ 1 & 6 & 3 \end{pmatrix}$

(C) $\begin{pmatrix} 1 & 1 & 5 \\ 2 & -1 & 2 \\ 1 & -2 & 6 \end{pmatrix}$

(D) $\begin{pmatrix} 120 \\ 210 \end{pmatrix}$ (from the previous system)

Question 5. Referring to the Case Study above, for a $2 \times 2$ system $a_{11}x + a_{12}y = b_1$, $a_{21}x + a_{22}y = b_2$, represented by $AX=B$, where $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, $B = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$. Cramer's rule gives $x = \frac{\det(A_x)}{\det(A)}$ and $y = \frac{\det(A_y)}{\det(A)}$. Given the system from Question 106: $3x + 2y = 120$, $4x + 5y = 210$. We found $\det(A) = 7$. Calculate $\det(A_y)$ where $A_y = \begin{pmatrix} 3 & 120 \\ 4 & 210 \end{pmatrix}$ and find the value of $y$.

(A) $\det(A_y) = 3 \times 120 - 120 \times 4 = 360 - 480 = -120$. $y = -120 / 7$.

(B) $\det(A_y) = 3 \times 210 - 120 \times 4 = 630 - 480 = 150$. $y = 150 / 7$.

(C) $\det(A_y) = 3 \times 210 - 120 \times 4 = 630 - 480 = 150$. $y = 150$.

(D) $\det(A_y) = 3 \times 120 - 210 \times 4 = 360 - 840 = -480$. $y = -480 / 7$.

Question 1. Case Study: River Boat Travel

A boat travels on a river. It takes $3$ hours to travel $15$ km upstream and $22$ km downstream. It takes $3$ hours to travel $20$ km upstream and $16$ km downstream. Assume the speed of the boat in still water ($u$) and the speed of the stream ($v$) are constant.

Which pair of equations represents the time taken for the two journeys?

(A) $\frac{15}{u-v} + \frac{22}{u+v} = 3$, $\frac{20}{u-v} + \frac{16}{u+v} = 3$

(B) $\frac{15}{u+v} + \frac{22}{u-v} = 3$, $\frac{20}{u+v} + \frac{16}{u-v} = 3$

(C) $15(u-v) + 22(u+v) = 3$, $20(u-v) + 16(u+v) = 3$

(D) $15u + 22v = 3$, $20u + 16v = 3$

Question 2. Referring to the Case Study above, the equations are $\frac{15}{u-v} + \frac{22}{u+v} = 3$ and $\frac{20}{u-v} + \frac{16}{u+v} = 3$. Let $x = \frac{1}{u-v}$ and $y = \frac{1}{u+v}$. The system becomes $15x + 22y = 3$ and $20x + 16y = 3$. Solve this linear system for $x$ and $y$.

(A) $x = 1/5$, $y = 1/11$

(B) $x = 1/11$, $y = 1/5$

(C) $x = 5$, $y = 11$

(D) $x = 11$, $y = 5$

Question 3. Referring to the Case Study above, now that you have $x = \frac{1}{u-v} = \frac{1}{5}$ and $y = \frac{1}{u+v} = \frac{1}{11}$, find the speed of the boat in still water ($u$) and the speed of the stream ($v$). (This leads to another system of linear equations: $u-v=5$ and $u+v=11$).

(A) $u = 8$ km/h, $v = 3$ km/h

(B) $u = 3$ km/h, $v = 8$ km/h

(C) $u = 5$ km/h, $v = 11$ km/h

(D) $u = 11$ km/h, $v = 5$ km/h

Question 4. Case Study: Work and Time Problem

Two taps, A and B, can fill a tank. Tap A takes $10$ minutes longer than Tap B to fill the tank separately. If both taps are opened together, they can fill the tank in $12$ minutes. Let Tap B take $x$ minutes to fill the tank alone.

Which equation represents the relationship between the time taken by Tap A, Tap B, and both together? (Rate of work = $1$ / time taken).

(A) $x + (x+10) = 12$

(B) $\frac{1}{x} + \frac{1}{x+10} = \frac{1}{12}$

(C) $x(x+10) = 12$

(D) $\frac{1}{x} + \frac{1}{10} = \frac{1}{12}$

Question 5. Referring to the Case Study above, solve the equation $\frac{1}{x} + \frac{1}{x+10} = \frac{1}{12}$ to find the time taken by Tap B ($x$) and Tap A ($x+10$).

(A) $12(x+10) + 12x = x(x+10) \implies 12x + 120 + 12x = x^2 + 10x \implies x^2 - 14x - 120 = 0 \implies (x-20)(x+6) = 0$. Since time is positive, $x=20$ minutes. Tap A takes $20+10=30$ minutes.

(B) $12(x+10+x) = x(x+10) \implies 12(2x+10) = x^2+10x \implies 24x + 120 = x^2 + 10x \implies x^2 - 14x - 120 = 0$. $x=20$ minutes.

(C) $\frac{x+10+x}{x(x+10)} = \frac{1}{12} \implies \frac{2x+10}{x^2+10x} = \frac{1}{12} \implies 12(2x+10) = x^2+10x \implies 24x+120 = x^2+10x \implies x^2 - 14x - 120 = 0$. $x=20$ minutes.

(D) All the above options correctly set up and solve the equation leading to $x=20$ minutes.