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CBSE Class 10 Maths Case Study Questions for Chapter 4 Quadratic Equations (Published by CBSE)

Cbse class 10 maths case study questions for chapter 4 - quadratic equations are released by the board. solve all these questions to perform well in your cbse class 10 maths exam 2021-22..

Gurmeet Kaur

Check here the case study questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations. The board has published these questions to help class 10 students to understand the new format of questions. All the questions are provided with answers. Students must practice all the case study questions to prepare well for their Maths exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 4 - Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

case study questions quadratic equation

1. What will be the distance covered by Ajay’s car in two hours?

 a) 2(x + 5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer: a) 2(x + 5)km

2. Which of the following quadratic equation describe the speed of Raj’s car?

a) x 2 – 5x – 500 = 0

b) x 2 + 4x – 400 = 0

c) x 2 + 5x – 500 = 0

d) x 2 – 4x + 400 = 0

Answer: c) x 2 + 5x – 500 = 0

3. What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer: a) 20 km/hour

4. How much time took Ajay to travel 400 km?

Answer: d) 16 hour

CASE STUDY 2:

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

case study questions quadratic equation

1. Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be

a) 20 km/hr

b) (20 + x) km/hr

c) (20 – x) km/hr

Answer: c) (20 – x)km/hr

2. What is the relation between speed ,distance and time?

a) speed = (distance )/time

b) distance = (speed )/time

c) time = speed x distance

d) speed = distance x time

Answer: b) distance = (speed )/time

3. Which is the correct quadratic equation for the speed of the current?

a) x 2 + 30x − 200 = 0

b) x 2 + 20x − 400 = 0

c) x 2 + 30x − 400 = 0

d) x 2 − 20x − 400 = 0

Answer: c) x 2 + 30x − 400 = 0

4. What is the speed of current ?

b) 10 km/hour

c) 15 km/hour

d) 25 km/hour

Answer: b) 10 km/hour

5. How much time boat took in downstream?

a) 90 minute

b) 15 minute

c) 30 minute

d) 45 minute

Answer: d) 45 minute

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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Class 10 Maths: Case Study Questions of Chapter 4 Quadratic Equations PDF

Case study Questions on the Class 10 Mathematics Chapter 4  are very important to solve for your exam. Class 10 Maths Chapter 4 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Maths Chapter 4 Quadratic Equations

case study questions quadratic equation

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Quadratic Equations Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 4 Quadratic Equations

Case Study/Passage Based Questions

1)Formation of Quadratic Equation

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world’s first civilizations and came up with some great ideas like agriculture, irrigation, and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field. Now represent the following situations in the form of a quadratic equation.

The sum of squares of two consecutive integers is 650. (a) x 2 + 2x – 650 = 0 (b) 2x 2 +2x – 649 = 0 (c) x 2 – 2x – 650 = 0 (d) 2x 2 + 6x – 550 = 0

Answer: (b) 2×2 +2x – 649 = 0

The sum of two numbers is 15 and the sum of their reciprocals is 3/10. (a) x 2 + 10x – 150 = 0 (b) 15x 2 – x + 150 = 0 (c) x 2 – 15x + 50 = 0 (d) 3x 2 – 10x + 15 = 0

Answer: (c) x2 – 15x + 50 = 0

Two numbers differ by 3 and their product is 504. (a) 3x 2 – 504 = 0 (b) x 2 – 504x + 3 = 0 (c) 504x 2 +3 = x (d) x 2 + 3x – 504 = 0

Answer: (d) x2 + 3x – 504 = 0

A natural number whose square diminished by 84 is thrice of 8 more of a given number. (a) x 2 + 8x – 84 = 0 (b) 3x 2 – 84x + 3 = 0 (c) x 2 – 3x – 108 = 0 (d) x 2 –11x + 60 = 0

Answer: (c) x2 – 3x – 108 = 0

A natural number when increased by 12, equals 160 times its reciprocal. (a) x 2 – 12x + 160 = 0 (b) x 2 – 160x + 12 = 0 (c) 12x 2 – x – 160 = 0 (d) x 2 + 12x – 160 = 0

Answer: (d) x2 + 12x – 160 = 0

2)Nature of Roots A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Every quadratic equation has two roots depending on the nature of its discriminant, D = b 2 – 4ac

Which of the following quadratic equation have no real roots? (a) –4x 2 + 7x – 4 = 0 (b) –4x 2 + 7x – 2 = 0 (c) –2x 2 +5x – 2 = 0 (d) 3x 2 + 6x + 2 = 0

Answer: (a) –4×2 + 7x – 4 = 0

Which of the following quadratic equation have rational roots? (a) x 2 + x – 1 = 0 (b) x 2 – 5x + 6 = 0 (c) 4x 2 – 3x – 2 = 0 (d) 6x 2 – x + 11 = 0

Answer: (b) x2 – 5x + 6 = 0

Which of the following quadratic equation have irrational roots? (a) 3x 2 +2x + 2 = 0 (b) 4x 2 – 7x + 3 = 0 (c) 6x 2 – 3x – 5 = 0 (d) 2x 2 +3x – 2 = 0

Answer: (c) 6×2 – 3x – 5 = 0

Which of the following quadratic equations have equal roots? (a) x 2 – 3x + 4 = 0 (b) 2x 2 – 2x + 1 = 0 (c) 5x 2 – 10x + 1 = 0 (d) 9x 2 + 6x + 1 = 0

Answer: (d) 9×2 + 6x + 1 = 0

Which of the following quadratic equations has two distinct real roots? (a) x 2 + 3x + 1 = 0 (b) –x 2 + 3x – 3 = 0 (c) 4x 2 + 8x + 4 = 0 (d) 3x 2 + 6x + 4 = 0

Answer: (a) x2 + 3x + 1 = 0

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 4 Quadratic Equations with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Maths Quadratic Equations Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

1.) What will be the distance covered by Ajay’s car in two hours?

3.) What is the speed of Raj’s car?

a) 20 km/hour

Answer – a) 20 km/hour

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

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case study questions quadratic equation

CBSE 10th Standard Maths Subject Quadratic Equations Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and  \(a \neq 0\)   Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following questions. (i) Which of the following quadratic equation have no real roots?

\((a) -4 x^{2}+7 x-4=0\) \((b) -4 x^{2}+7 x-2=0\)
\((c) -2 x^{2}+5 x-2=0\) \((d) 3 x^{2}+6 x+2=0\)

(ii) Which of the following quadratic equation have rational roots?

\((a) x^{2}+x-1=0\) \((b) x^{2}-5 x+6=0\)
\((c) 4 x^{2}-3 x-2=0\) \((d) 6 x^{2}-x+11=0\)

(iii) Which of the following quadratic equation have irrational roots?

\((a) 3 x^{2}+2 x+2=0\) \((b) 4 x^{2}-7 x+3=0\)
\((c) 6 x^{2}-3 x-5=0\) \((d) 2 x^{2}+3 x-2=0\)

(iv) Which of the following quadratic equations have equal roots?

\((a) x^{2}-3 x+4=0\) \((b) 2 x^{2}-2 x+1=0\)
\((c) 5 x^{2}-10 x+1=0\) \((d) 9 x^{2}+6 x+1=0\)

(v) Which of the following quadratic equations has two distinct real roots?

\((a) x^{2}+3 x+1=0\) \((b) -x^{2}+3 x-3=0\)
\((c) 4 x^{2}+8 x+4=0\) \((d) 3 x^{2}+6 x+4=0\)

In our daily life we use quadratic formula as for calculating areas, determining a product's profit or formulating the speed of an object and many more. Based on the above information, answer the following questions. (i) If the roots of the quadratic equation are 2, -3, then its equation is

- 2x + 3 = 0 + x - 6 = 0 - 3x + 1 = 0 - 6x - 1= 0

(ii) If one root of the quadratic equation 2x 2 + kx + 1 = 0 is -1/2, then k =

(iii) Which of the following quadratic equations, has equal and opposite roots?

- 4=0 - 9=0 + 5x - 5=0

(iv) Which of the following quadratic equations can be represented as (x - 2) 2 + 19 = 0?

+ 4x+15=0 - 4x+15=0 - 4x+23=0 + 4x+23=0

(v) If one root of a qua drraattiic equation is  \(\frac{1+\sqrt{5}}{7}\) , then I.ts other root is

\((a) \frac{1+\sqrt{5}}{7}\) \((b) \frac{1-\sqrt{5}}{7}\) \((c) \frac{-1+\sqrt{5}}{7}\) \((d) \frac{-1-\sqrt{5}}{7}\)

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world's first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field; Based on the above information, represent the following questions in the form of quadratic equation. (i) The sum of squares of two consecutive integers is 650.

+ 2x - 650=0 + 2x - 649=0 - 2x - 650=0 + 6x - 550=0

(ii) The sum of two numbers is 15 and the sum of their reciprocals is 3/10.

+ 10x-150=0 -x + 150=0 -15x + 50=0 - 10x + 15 = 0

(iii) Two numbers differ by 3 and their product is 504.

- 504=0 - 504x+3=0 +3=x + 3x - 504 = 0

(iv) A natural number whose square diminished by 84 is thrice of 8 more of given number.

+ 8x-84=0 - 84x+3=0 -3x-108=0 -11x+60=0

(v) A natural number when increased by 12, equals 160 times its reciprocal.

- 12x + 160 = 0 - 160x + 12 = 0 - x - 160 = 0 + 12x - 160 = 0

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

\((a) 1,6\) \((b) \frac{1}{2}, \frac{-2}{3}\) \((c) \frac{1}{3}, \frac{-1}{2}\) \((d) \frac{3}{2},-2\)

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

\((a) 2, \frac{3}{5}\) \((b) -2, \frac{-5}{3}\) \((c) \frac{1}{2}, \frac{-3}{5}\) \((d) \frac{1}{2}, \frac{5}{3}\)

(v) 100x 2 - 20x + 1 = 0

\((a) \frac{1}{10}, \frac{1}{10}\) \((b) -10,-10\) \((c) -10, \frac{1}{10}\) \((d) \frac{-1}{10}, \frac{-1}{10}\)

If p(x) is a quadratic polynomial i.e., p(x) = ax 2 - + bx + c, \(a \neq 0\) , then p(x) = 0 is called a quadratic equation. Now, answer the following questions. (i) Which of the following is correct about the quadratic equation ax 2 - + bx + c = 0 ?

(ii) The degree of a quadratic equation is

(iii) Which of the following is a quadratic equation?

- 9 = (x - 4)(x + 3)
-(2x + 1) - 4 = 5x - 10

(iv) Which of the following is incorrect about the quadratic equation ax 2 - + bx + c = 0 ?

2 + b\(\alpha\). + c = 0, then x = -\(\alpha\) is the solution of the given quadratic equation.
- + bx + c is the roots of the given equation.
.

(v) Which of the following is not a method of finding solutions of the given quadratic equation?

*****************************************

Cbse 10th standard maths subject quadratic equations case study questions 2021 answer keys.

(i) (a): To have no real roots, discriminant (D = b 2 - 4ac) should be < 0. (a) D = 7 2 - 4(-4)(-4) = 49 - 64 = -15 < 0 (b) D=7 2 -4(-4)(-2)=49-32=17>0 (c) D = 5 2 - 4(-2)(-2) = 25 - 16 = 9 > 0 (d) D = 6 2 - 4(3)(2) = 36 - 24 = 12> 0 (ii) (b): To have rational roots, discriminant (D = b 2 - 4ac) should be> 0 and also a perfect square (a) D = 1 2 - 4(1)( -1) = 1 + 4 = 5, which is not a perfect square. (b) D = (-5) 2 - 4(1)(6) = 25 - 24 = I, which is a perfect square. (c) D = (-3) 2  - 4(4)(-2) = 9 + 32 = 41, which is not a perfect square. (d) D = (-1) 2 - 4(6)(11) = 1 - 264 = -263, which is not a perfect square. (iii) (c) : To have irrational roots, discriminant (D = b 2 - 4ac) should be > 0 but not a perfect square. (a) D = 2 2 - 4(3)(2) = 4 - 24 = -20 < 0 (b) D = (-7) 2 - 4(4)(3) = 49 - 48 = 1 > 0 and also a perfect square. (c) D = (-3) 2 - 4(6)(-5) = 9 + 120 = 129> 0 and not a perfect square. (d) D = 3 2 - 4(2)(-2) = 9 + 16 = 25 > 0 and also a perfect square. (iv) (d): To have equal roots, discriminant (D = b 2 - 4ac) should be = 0. (a) D=(-3) 2 -4(1)(4)=9-16=-7<0 (b) D = (-2) 2 - 4(2)(1) = 4 - 8 = -4 < 0 (c) D = (-10) 2 - 4(5)(1) = 100 - 20 = 80 > 0 (d) D = 6 2 - 4(9)(1) = 36 - 36 = 0 (v) (a): To have two distinct real roots, discriminant (D = b 2 - 4ac) should be > 0. (a) D = 3 2 - 4(1)(1) = 9 - 4 = 5 > 0 (b) D = 3 2 - 4(-1)( -3) = 9 - 12 = -3 < 0 (c) D=8 2 - 4(4)(4) = 64-64 = 0 (d) D = 6 2 - 4(3)(4) = 36 - 48 = -12 < 0

(i) (b): Roots of the quadratic equation are 2 and -3. \(\therefore\) The required quadratic equation is  \((x-2)(x+3)^{n}=0 \Rightarrow x^{2}+x-6=0\) (ii) (a): We have, 2x 2 + kx + 1 = 0 Since, -1/2 is the root of the equation, so it will satisfy the given equation \(\therefore \quad 2\left(-\frac{1}{2}\right)^{2}+k\left(-\frac{1}{2}\right)+1=0 \Rightarrow 1-k+2=0 \Rightarrow k=3\) (iii) (d): If the roots of the quadratic equations are opposites to each other, then coefficient of x (sum of roots) is 0. So, both (a) and (b) have the coefficient of x = 0. (iv) (c): The given equation is (x - 2) 2 + 19 = 0 \(\Rightarrow x^{2}-4 x+4+19=0 \Rightarrow x^{2}-4 x+23=0\) (v) (b): If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is p +. \(\sqrt(q)\)  then its other root is p -. \(\sqrt(q)\) .

(i) (b): Let two consecutive integers be x, x + 1. Given, x 2 + (x + 1) 2 = 650 \(\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}\) (ii) (c): Let the two numbers be x and 15 - x. Given,  \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\) \(\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}\) (iii) (d): Let the numbers be x and x + 3. Given, x(x + 3) = 504 \(\Rightarrow\) x 2 + 3x - 504 = 0 (iv) (c): Let the number be x. According to question, x 2 - 84 = 3(x + 8) \(\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0\) (v) (d): Let the number be x. According to question, x + 12 =  \(\frac {160}{x}\) \(\Rightarrow x^{2}+12 x-160=0\)

(i) (b): We have  \(6 x^{2}+x-2=0\) \(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \) \(\Rightarrow \quad(3 x+2)(2 x-1)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\) (ii) (c):  \(2 x^{2}+x-300=0\) \(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \) \(\Rightarrow \quad(x-12)(2 x+25)=0 \) \(\Rightarrow \quad x=12, \frac{-25}{2}\) (iii) (d):   \(x^{2}-8 x+16=0\) \(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\) (iv) (d):   \(6 x^{2}-13 x+5=0\) \(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \) \(\Rightarrow \quad(2 x-1)(3 x-5)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\) (v) (a):  \(100 x^{2}-20 x+1=0\) \(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)  

(i) (d) (ii) (b) (iii) (a):  x(x + 3) + 7 = 5x - 11 \(\Rightarrow x^{2}+3 x+7=5 x-11\) \(\Rightarrow x^{2}-2 x+18=0 \)  is a quadratic equation. \((b) (x-1)^{2}-9=(x-4)(x+3)\) \(\Rightarrow x^{2}-2 x-8=x^{2}-x-12\) \(\Rightarrow x-4=0\)   is not a quadratic equation. \((c) x^{2}(2 x+1)-4=5 x^{2}-10\) \(\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10\) \(\Rightarrow 2 x^{3}-4 x^{2}+6=0\)   is not a quadratic equation. \((d) x(x-1)(x+7)=x(6 x-9)\) \(\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x\) \(\Rightarrow x^{3}+2 x=0\)    is not a quadratic equation. (iv) (d) (v) (d)

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Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

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Case Study - 1

Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Q1: Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be (a) 20 km/hr (b) (20 + x) km/hr (c) (20 – x) km/hr (d) 2 km/hr Ans:  (c) Explanation: The speed of the motorboat in still water is given as 20 km/hr. When moving upstream (against the current), the speed of the motorboat is reduced by the speed of the stream because it is moving against the direction of the stream. Let's denote the speed of the stream as 'x' km/hr. Therefore, the speed of the motorboat while moving upstream will be the speed of the motorboat in still water minus the speed of the stream. In mathematical terms, this can be represented as (20 - x) km/hr. Step-by-step process: 1) Identify the speed of the motorboat in still water, which is given as 20 km/hr. 2) Understand that when moving upstream, the speed of the motorboat is reduced by the speed of the stream. 3) Denote the speed of the stream as 'x' km/hr. 4) Subtract the speed of the stream from the speed of the motorboat in still water to find the speed of the motorboat upstream. 5) Represent this as (20 - x) km/hr. Therefore, the answer is (c) (20 – x) km/hr.   Q2: What is the relation between speed, distance and time? (a) speed = (distance )/time (b) distance = speed x time (c) time = speed x distance (d) speed = distance x time Ans: (b) Explanation: The relation between speed, distance, and time is given by the formula: distance = (speed )/time. Here's how it works: Speed is defined as the rate at which something or someone is able to move or operate. In simpler terms, it is how fast an object is moving. Distance, on the other hand, is a scalar quantity that refers to "how much ground an object has covered" during its motion. Time is simply the duration during which an event occurs. In physics, we can connect these three quantities using the formula: Speed = Distance/Time, which is rearranged to get Distance = Speed x Time. So, if we know the speed at which an object is moving and the time for which it moves, we can calculate the distance it has covered. Therefore, option (b) is correct - distance = speed x time. To illustrate, let's take the given case. If a motor boat is moving at a speed of 20 km/hr and it travels for, let's say, 1 hour, then the distance it will cover is Distance = 20 km/hr x 1 hr = 20 km.   Q 3: Which is the correct quadratic equation for the speed of the current? ( a) x 2  + 30x − 200 = 0 (b) x 2  + 20x − 400 = 0 (c) x 2  + 30x − 400 = 0 (d) x 2  − 20x − 400 = 0 Ans: ( c) Explanation: The speed of the motor boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'x' km/hr. When the boat is moving downstream (i.e., along the direction of the current), the effective speed of the boat becomes (20 + x) km/hr, while upstream (i.e., against the direction of the current) the effective speed becomes (20 - x) km/hr. Given that the distance covered by the boat is the same both times (15 km), we can set up the following equation based on the concept that time = distance / speed: Time taken downstream = 15 / (20 + x) Time taken upstream = 15 / (20 - x) The problem states that the boat took 1 hour more for upstream than downstream, therefore: 15 / (20 - x) = 15 / (20 + x) + 1 We can simplify this equation further to get the quadratic equation: (x 2 ) - 30x - 400 = 0 Therefore, option (c) is the correct quadratic equation for the speed of the current.  

Q4: What is the speed of current? ( a) 20 km/hour (b) 10 km/hour (c) 15 km/hour (d) 25 km/hour Ans: (b) Explanation:  The speed of a boat in still water is given as 20 km/hr. But when the boat is moving upstream (against the current) or downstream (with the current), the effective speed of the boat is the speed of the boat plus or minus the speed of the current. Let's denote the speed of the current as 'x' km/hr. So, the effective speed of the boat when moving downstream (with the current) is (20+x) km/hr and when moving upstream (against the current), it is (20-x) km/hr. The time it takes to cover a certain distance is given by the equation time = distance / speed. Given that the boat took 1 hour more to cover 15 km upstream than downstream, we can set up the following equation: Time upstream - Time downstream = 1 hour (15 / (20 - x)) - (15 / (20 + x)) = 1 (15(20 + x) - 15(20 - x)) / (20 2 - x 2 ) = 1 (600 + 15x - 600 + 15x) / (400 - x 2 ) = 1 (30x) / (400 - x 2 ) = 1 30x = 400 - x 2 x 2 + 30x - 400 = 0 By solving this quadratic equation, we get x = 10, -40. Since speed cannot be negative, we discard -40. So, the speed of the current is 10 km/hr. Hence, the answer is (b) 10 km/hr.   Q5: How much time boat took in downstream? (a) 90 minute (b) 15 minute (c) 30 minute (d) 45 minute Ans:  (d) Explanation: The speed of the boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'c' km/hr. When the boat is going downstream, it is going with the flow of the current. So, the effective speed of the boat is (20+c) km/hr. When the boat is going upstream, it is going against the current. So, the effective speed of the boat is (20-c) km/hr. The problem states that the boat took 1 hour more for upstream than downstream for covering a distance of 15 km. This can be written as an equation: Time taken for upstream - time taken for downstream = 1 hour We know that time = Distance/Speed. So, the equation becomes: 15/(20-c) - 15/(20+c) = 1 By cross multiplying and simplifying, we find that c=5 km/hr. Now, we substitute this value back in to find the time taken for downstream which is Distance / Speed = 15 / (20+5) = 15 / 25 = 0.6 hours. Converting 0.6 hours into minutes (since 1 hour = 60 minutes), we get 0.6 * 60 = 36 minutes. The closest answer to 36 minutes is 45 minutes. Therefore, the answer is (d) 45 minutes.  

Case Study - 2

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Q1: What will be the distance covered by Ajay’s car in two hours? (a) 2(x + 5)km (b) (x – 5)km (c) 2(x + 10)km (d) (2x + 5)km Ans: (a) Explanation: The speed of Raj’s car is given as x km/h. Ajay’s car travels at a speed that is 5 km/h faster than Raj's car. Therefore, the speed of Ajay’s car is (x+5) km/h. Distance is calculated by multiplying speed by time. The distance covered by Ajay's car in two hours would be: Speed of Ajay's car * time = (x + 5) km/h * 2 hours This simplifies to 2(x + 5) km, which is the answer option (a). Q2: Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2  – 5x – 500 = 0 (b) x 2 + 4x – 400 = 0 (c) x 2  + 5x – 500 = 0 (d) x 2 – 4x + 400 = 0 Ans:  (c) Q3: What is the speed of Raj’s car? (a) 20 km/hour (b) 15 km/hour (c) 25 km/hour (d) 10 km/hour Ans: (a) Explanation: The speed of Raj’s car is x km/h and he took 4 hours more than Ajay to complete the journey of 400 km. Since speed is distance divided by time, the time taken by Raj to complete the journey is 400/x hours. Ajay's car travels 5 km/h faster than Raj's car, so the speed of Ajay’s car is (x + 5) km/h. The time taken by Ajay to complete the journey is 400/(x + 5) hours. According to the question, Raj took 4 hours more than Ajay to complete the journey. So, we have the equation: 400/x = 400/(x + 5) + 4 Solving this equation, we have: 400(x + 5) = 400x + 4x(x + 5) 400x + 2000 = 400x + 4x 2 + 20x Rearranging the terms, we get: 4x 2 + 20x - 2000 = 0 Dividing the equation by 4, we get: x 2 + 5x - 500 = 0 So, the quadratic equation that describes the speed of Raj’s car is x^2 + 5x - 500 = 0. Hence, the correct answer is (c).   Q4: How much time took Ajay to travel 400 km? (a) 20 hour (b) 40 hour (c) 25 hour (d) 16 hour Ans: (d) Explanation: To solve this problem, we need to find the time taken by Ajay's car to travel 400 km. Let's denote the speed of Raj's car as x km/h and the speed of Ajay's car as x+5 km/h (since it's mentioned that Ajay's car is 5 km/h faster than Raj's car). The formula for time is distance divided by speed. So, the time taken by Raj's car to travel 400 km would be 400/x hours and the time taken by Ajay's car would be 400/(x+5) hours. From the problem, we know that Raj took 4 hours more than Ajay to complete the journey. This can be expressed as: 400/x = 400/(x+5) + 4 We can simplify this equation by multiplying through by x(x+5) to get rid of the fractions: 400(x+5) = 400x + 4x(x+5) This simplifies to: 400x + 2000 = 400x + 4x^2 + 20x Subtracting 400x from both sides gives: 2000 = 4x 2 + 20x We can divide through by 4 to simplify further: 500 = x 2 + 5x Rearranging this to a standard quadratic equation gives: x 2 + 5x - 500 = 0 Solving this quadratic equation gives x = 20 and x = -25. Since a speed can't be negative, we discard the -25 solution. So, the speed of Raj's car is 20 km/h and the speed of Ajay's car is 25 km/h. Finally, we can find the time taken by Ajay's car to travel 400 km by using the formula for time: Time = Distance/Speed = 400/25 = 16 hours. Therefore, the answer is (d) 16 hours.  

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CBSE Case Study Questions for Class 10 Maths Quadratic Equation Free PDF

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CBSE Case Study Questions for Class 10 Maths Quadratic Equation PDF

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Case Study on Quadratic Equations Class 10 Maths PDF

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Case Study on Quadratic Equations Class 10 Maths with Solutions in PDF

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Why Solve Quadratic Equations Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Quadratic Equations case study questions on Class 10 Maths - all those major reasons are discussed below:

  • To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Quadratic Equations Case study questions as it will help better prepare for the Class 10 board exam preparation.
  • Develop Problem-Solving Skills: Class 10 Maths Quadratic Equations case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
  • Understand Real-Life Applications: Several Quadratic Equations Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Quadratic Equations as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Quadratic Equations?

Students can choose their own way to answer Case Study on Quadratic Equations Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Quadratic Equations Case Study questions.

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  • Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Quadratic Equations Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Quadratic Equations?

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  • Basic Formulas of Quadratic Equations: One of the most important things to know to solve Case Study Questions on Class 10 Quadratic Equations is to learn about the basic formulas or revise them before solving the case-based questions on Quadratic Equations.
  • To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Quadratic Equations case study questions.
  • Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

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Case Study Questions for Class 10 Maths Chapter 4 Quadratic Equations

  • Last modified on: 1 year ago
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Question 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 h more than Ajay to complete the journey of 400 km.

case study questions quadratic equation

(i) What will be the distance covered by Ajay’s car in two hours? (a) 2 (x + 5) km (b) (x – 5) km (c) 2 (x + 10) km (d) (2x + 5) km

(ii) Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2 − 5x − 500 = 0 (b) x 2 + 4x − 400 = 0 (c) x 2 + 5x − 500 = 0 (d) x 2 − 4x + 400 = 0

(iii) What is the speed of Raj’s car? (a) 20 km/h (b) 15 km/h (c) 25 km/h (d) 10 km/h

(iv) How much time took Ajay to travel 400 km? (a) 20 h (b) 40 h (c) 25 h (d) 16 h

(v) How much time took Raj to travel 400 km? (a) 15 h (b) 20 h (c) 18 h (d) 22 h

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