Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This exercise covers real-life applications of quadratic equations – population growth, profit maximization, motion problems, and optimization. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
Class 10 Math Unit 2 Exercise 2.7 – Quadratic Applications: Complete Guide
Exercise 2.7 of Unit 2 applies quadratic equations to real-life problems. Students learn to model scenarios using quadratic functions and solve them using equations and inequalities. This exercise bridges theoretical algebra with practical applications.
What You Will Learn
By working through Exercise 2.7, students master: modeling population growth with quadratics, finding maximum profit using the vertex formula, solving motion problems (distance and height), solving quadratic inequalities in context, and interpreting solutions in real-world settings.
Topics Covered in This Exercise
- Population Modeling – using \(P(t) = at^2 + bt + c\) to predict when population reaches a certain level.
- Profit Maximization – finding the price that maximizes profit using \(x = -b/(2a)\).
- Motion (Distance) – solving for time when distance is given in \(d = t^2 - 0.5t\).
- Projectile Height – finding time intervals when height is at least a certain value.
- Maximum Height – using equations of motion to find maximum height.
- Earnings Optimization – finding hours to maximize earnings.
Why Exercise 2.7 Is Important
Exercise 2.7 develops critical thinking and problem-solving skills. Board exam papers frequently feature application-based questions. This exercise helps students understand how mathematics models the world around us.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. Quadratic applications are a frequently tested topic. Practising every question by hand is highly recommended.
Exam Tips for Quadratic Applications
- Identify the type of problem – optimization, motion, or inequality.
- For optimization, use the vertex formula \(x = -b/(2a)\).
- For inequalities, set up the inequality, solve, and express as an interval.
- Always check that your answer makes sense in the real-world context.
📖 Exercise 2.7 – Solved Problems
1
A town's population is modeled by \(P(t) = -2t^2 + 40t + 800\), where \(t\) is years since 2020. Find the years when the population will be at least 1000.
Given modeled equation is
\[P(t) = -2t^2 + 40t + 800\]We have to find year when population will be at least 1000. So,
\[\begin{aligned} P(t) &\geq 1000 \\ -2t^2 + 40t + 800 &\geq 1000 \\ -2t^2 + 40t + 800 - 1000 &\geq 0 \\ -2t^2 + 40t - 200 &\geq 0 \\ -2(t^2 - 20t + 100) &\geq 0 \\ \Rightarrow t^2 - 20t + 100 &\leq 0 \end{aligned}\]Associated equation
\[t^2 - 20t + 100 = 0\]Here \(a = 1\), \(b = -20\), \(c = 100\)
\[\begin{aligned} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6pt] t &= \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(100)}}{2(1)} \\[6pt] t &= \frac{20 \pm \sqrt{400 - 400}}{2} \\[6pt] t &= \frac{20 \pm \sqrt{0}}{2} \\[6pt] t &= \frac{20 \pm 0}{2} \\[6pt] t &= \frac{20}{2} \\[4pt] t &= 10 \end{aligned}\]Since \(t = 10\) represents the number of years since 2020:
\[Year = 2020 + 10 = 2030\]The population will be at least 1000 in the year 2030.
2
A company models its profit \(P\) in thousands of rupees by \(P(x) = -5x^2 + 150x - 1000\), where \(x\) is the price per item in rupees. Find the price that gives maximum profit.
\[P(x) = -5x^2 + 150x - 1000\]
Here \(a = -5\), \(b = 150\), \(c = -1000\)
Since \(a = -5 < 0\), the parabola opens downward and the vertex gives the maximum point.
\[\begin{aligned} x &= -\frac{b}{2a} \\[6pt] x &= -\frac{150}{2(-5)} \\[6pt] x &= -\frac{150}{-10} \\[4pt] \boldsymbol{x} &= \boldsymbol{15} \end{aligned}\]The price per item that gives maximum profit is Rs. 15.
3
A toy car rolls down an incline and covers a distance given by \(d = t^2 - 0.5t\) metres, where \(t\) is the time in seconds. Find the time when the car has travelled a distance 12.5 metres.
\[\begin{aligned}
d &= t^2 - 0.5t \\
12.5 &= t^2 - 0.5t \quad \because d = 12.5 \\
0 &= t^2 - 0.5t - 12.5 \\
t^2 - 0.5t - 12.5 &= 0
\end{aligned}\]
Here \(a = 1\), \(b = -0.5\), \(c = -12.5\)
\[\begin{aligned} t &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6pt] t &= \frac{-(-0.5) \pm \sqrt{(-0.5)^2 - 4(1)(-12.5)}}{2(1)} \\[6pt] t &= \frac{0.5 \pm \sqrt{0.25 + 50}}{2} \\[6pt] t &= \frac{0.5 \pm \sqrt{50.25}}{2} \\[6pt] t &= \frac{0.5 \pm 7.0887}{2} \end{aligned}\]either
\[\begin{aligned}
t &= \frac{0.5 + 7.0887}{2} \\[4pt]
t &= \frac{7.5887}{2} \\[4pt]
t &= 3.79 \text{ s}
\end{aligned}\]
or
\[\begin{aligned}
t &= \frac{0.5 - 7.0887}{2} \\[4pt]
t &= \frac{-6.5887}{2} \\[4pt]
t &= -3.29 \text{ s}
\end{aligned}\]
Rejected, time cannot be negative
The car will travel a distance of 12.5 metres after approximately 3.79 seconds.
4
A ball's height (in metres) after \(t\) seconds is \(h(t) = -4t^2 + 24t\). For what time interval is the ball at least 20 m above the ground?
Given
\[h(t) = -4t^2 + 24t\]According to given condition
\[\begin{aligned} h(t) &\geq 20 \\ -4t^2 + 24t &\geq 20 \\ -4t^2 + 24t - 20 &\geq 0 \\ -4(t^2 - 6t + 5) &\geq 0 \\ \Rightarrow t^2 - 6t + 5 &\leq 0 \end{aligned}\]Associated equation
\[\begin{aligned} t^2 - 6t + 5 &= 0 \\ t^2 - 5t - t + 5 &= 0 \\ t(t-5) - 1(t-5) &= 0 \\ (t-5)(t-1) &= 0 \end{aligned}\]either
\[\begin{aligned} t - 5 &= 0 \\ t &= 5 \end{aligned}\]
or
\[\begin{aligned} t - 1 &= 0 \\ t &= 1 \end{aligned}\]
Since \((t-5)(t-1) \leq 0\)
The ball is at least 20 m above the ground when \(\boldsymbol{1 \leq t \leq 5}\).
5
A ball is thrown upward with an initial velocity of \(40 \text{ ms}^{-1}\). Calculate the maximum height it reaches above ground level.
\(Initial\ velocity = v_i = 40\ \text{ms}^{-1}\)
\(Final\ Velocity = v_f = 0\ \text{ms}^{-1}\)
\(Gravitational\ acceleration = g = -10\ \text{ms}^{-2}\)
\(Height = h = ?\)
By using third equation of motion
\[\begin{aligned} 2aS &= v_f^2 - v_i^2 \\ 2gh &= v_f^2 - v_i^2 \\ 2(-10)(h) &= (0)^2 - (40)^2 \\ -20h &= 0 - 1600 \\ h &= \frac{-1600}{-20} \\[4pt] \boldsymbol{h} &= \boldsymbol{80 \text{ m}} \end{aligned}\]The maximum height reached by the ball is 80 metres.
6
A freelancer's earnings follow the model \(E(h) = -2h^2 + 40h\), where \(E\) is earning in rupees and \(h\) is hours worked per week. What is the maximum number of hours he should work to maximize earnings?
\[E(h) = -2h^2 + 40h + 0\]
Here \(a = -2\), \(b = 40\), \(c = 0\)
Since \(a = -2 < 0\), the parabola opens downward and the vertex gives the maximum point.
\[\begin{aligned} x &= -\frac{b}{2a} \\[6pt] x &= -\frac{40}{2(-2)} \\[6pt] x &= -\frac{40}{-4} \\[4pt] \boldsymbol{x} &= \boldsymbol{10} \end{aligned}\]He should work a maximum of 10 hours to maximize earnings.
📈 Key Concepts – Applications of Quadratics
- Optimization: For \(ax^2+bx+c\), vertex \(x = -\frac{b}{2a}\) gives maximum (if \(a<0\)) or minimum (if \(a>0\)).
- Inequalities: For "at least" or "at most" problems, set up and solve quadratic inequalities.
- Motion: Use equations of motion (\(v_f^2 = v_i^2 + 2as\)) for projectile problems.
- Real-world: Quadratic models appear in population, profit, distance, height, and optimization scenarios.
❓ Frequently Asked Questions
What is taught in Exercise 2.7 of Class 10 Math Unit 2?
Exercise 2.7 covers real-life applications of quadratic equations. It includes problems on population growth, profit maximization, projectile motion (distance and height), and optimization. Students learn to model real-world scenarios using quadratic functions and solve them using quadratic equations and inequalities.
How many questions are there in Unit 2 Exercise 2.7?
Exercise 2.7 has 6 questions covering various applications of quadratic equations including population modeling, profit maximization, motion problems, and height-time problems.
What is the vertex formula used for in quadratic applications?
The vertex formula \(x = -b/(2a)\) is used to find the maximum or minimum value of a quadratic function. When \(a < 0\), the vertex gives the maximum point, which is used in optimization problems like maximizing profit or earnings.
How do you solve quadratic inequalities in real-life problems?
For 'at least' or 'at most' problems, set up the inequality, bring all terms to one side, factor or use the quadratic formula to find critical points, then test intervals to determine the solution set. The answer is usually expressed as an interval.
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Exercise 2.7 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Exercise 2.7?
Yes, a complete solved PDF for Exercise 2.7 is embedded on this page and available to download for free.
Can I download the Unit 2 Exercise 2.7 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Exercise 2.7 notes to your device.
Is Exercise 2.7 important for Class 10 board exams?
Yes, applications of quadratic equations are frequently tested in board exams. Exercise 2.7 helps students understand how quadratics are used to solve real-world problems, which is an important skill.
Who prepared these Class 10 Math Unit 2 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.