case study class 10 maths linear equations

Class 10 Maths Case Study Questions Chapter 3 Pair of Linear Equations in Two Variables

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Case study Questions in the Class 10 Mathematics Chapter 3 are very important to solve for your exam. Class 10 Maths Chapter 3 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 3 Pair of Linear Equations in Two Variables

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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.

Pair of Linear Equations in Two Variables Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Case Study/Passage-Based Questions

case study class 10 maths linear equations

(i) 1 st situation can be represented algebraically as

Answer: (d) 2x+3y=46

(ii) 2 nd situation can be represented algebraically as

Answer: (c) 3x + 5y = 74

(iii), Fare from Ben~aluru to Malleswaram is

Answer: (b) Rs 8

(iv) Fare from Bengaluru to Yeswanthpur is

Answer: (a) Rs 10

(v) The system oflinear equations represented by both situations has

Answer: (c) unique solution

Case Study 2: The scissors which are so common in our daily life use, its blades represent the graph of linear equations.

Let the blades of a scissor are represented by the system of linear equations:

x + 3y = 6 and 2x – 3y = 12

(i) The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is (a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

Answer: (b) (6, 0)

(ii) The points at which linear equations x + 3y = 6 and 2x – 3y = 12 intersect y – axis respectively are (a) (0, 2) and (0, 6) (b) (0, 2) and (6, 0) (c) (0, 2) and (0, –4) (d) (2, 0) and (0, –4)

Answer: (c) (0, 2) and (0, –4)

(iii) The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is (a) 0 (b) 1 (c) 2 (d) infinitely many

Answer: (d) infinitely many

(iv) If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12, then values of a and b are respectively (a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

Answer: (a) 1, 5

(v) If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

Answer: (d) intersecting or coincident

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Pair of Linear Equations in Two Variables 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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CBSE Class 10 Study Material

CBSE Class 10 Maths Case Study Questions for Chapter 3 - Pair of Linear Equations in Two Variables (Published by CBSE)

Cbse's question bank on case study for class 10 maths chapter 3 is available here. these questions will be very helpful to prepare for the cbse class 10 maths exam 2022..

Case study questions are going to be new for CBSE Class 10 students. These are the competency-based questions that are completely new to class 10 students. To help students understand the format of the questions, CBSE has released a question bank on case study for class 10 Maths. Students must practice with these questions to get familiarised with the concepts and logic used in the case study and understand how to answers them correctly. You may check below the case study questions for CBSE Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables. You can also check the right answer at the end of each question.

Check Case Study Questions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

CASE STUDY-1:

1. If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?

2. How many questions did he guess?

3. If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?

4. If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?

Let the no of questions whose answer is known to the student x and questions attempted by cheating be y

x – 1/4y =90

solving these two

x = 96 and y = 24

1. He answered 96 questions correctly.

2. He attempted 24 questions by guessing.

3. Marks = 80- ¼ 0f 40 =70

4. x – 1/4 of (120 – x) = 95

5x = 500, x = 100

CASE STUDY-2:

Amit is planning to buy a house and the layout is given below. The design and the measurement has been made such that areas of two bedrooms and kitchen together is 95 sq.m.

Based on the above information, answer the following questions:

1. Form the pair of linear equations in two variables from this situation.

2. Find the length of the outer boundary of the layout.

3. Find the area of each bedroom and kitchen in the layout.

4. Find the area of living room in the layout.

5. Find the cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m.

1. Area of two bedrooms= 10x sq.m

Area of kitchen = 5y sq.m

10x + 5y = 95

Also, x + 2+ y = 15

2. Length of outer boundary = 12 + 15 + 12 + 15 = 54m

3. On solving two equation part(i)

x = 6m and y = 7m

area of bedroom = 5 x 6 = 30m

area of kitchen = 5 x 7 = 35m

4. Area of living room = (15 x 7) – 30 = 105 – 30 = 75 sq.m

5. Total cost of laying tiles in the kitchen = Rs50 x 35 = Rs1750

Case study-3 :

It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus.

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too.

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement?

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

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What is Case Study Question / Paragraph Based Question? A case study is a scenario in a particular academic / professional context which students are expected to analyse and respond to, guided by specific questions posed concerning the situation. In many cases, the scenario or case study involves a number of issues or problems that must be dealt with in a academic / professional workplace.

Why Case Study Questions are included in academics? Case study assignments usually require students to identify problems and issues in a scenario, to demonstrate their developing knowledge of theories and academic / professional policies and to make decisions and recommendations based on these to either prevent or solve some of the issues in that scenario.

How to solve Case Study Questions? There are several steps to writing an answer to a case study assignment:

STEP 1: Read the case study and questions carefully.

Read the case and associated questions carefully.
Highlight the main points of the case and any issues that you can identify.
Read the questions closely and analyse what they are requiring you to do.
Read the case again, linking the information that is relevant to each question you have been asked.

STEP 2: Identify the issues in the case study. Case studies describe a situation which may arise in a particular profession or social context. They often involve a number of people in a complex situation. They will often describe a situation which is problematic, possibly in how it is dealt with, or in its complexity. An important part of your answer is to analyse the situation and to identify the issues/actions described in the case which may be problematic. The following questions may help you to do this:

What actions were taken in the case?
Were these actions the most appropriate and why?
Were there any consequences of the actions taken?
Was anything omitted or not considered?
Were actions/procedures in line with existing codes of practice, policy or theories?

STEP 3: Link theory to practice. Use your knowledge of existing codes of practice, theories and/or other academic / professional documents and behaviours to decide what was done appropriately and what was not.

STEP 4: Plan your answer. It can be useful to use the questions you have been set as headings and to answer each part in turn, reducing the chance of omitting set questions. You can always take out the headings before you submit if you wish. Lecturers usually set questions in a logical order, so answer in the order they are written in your question.

STEP 5: Start writing your case study answer (for theory only) Like any assignment, you will need an introduction, body sections in which you answer the questions put to you regarding the case study, and a conclusion.

STEP 6: Edit and proofread. Read through your paper yourself to detect and correct other errors and omissions. Check you have answered all questions and backed up your answer with relevant passage.

Types of Case Study Questions / Paragraph Based Questions Case Study Questions / Paragraph Based Questions can be broadly classified into two types:

MCQs type: In this type, student has to tick the correct option from various options.
Theory type: In this type, student has to write proper solution / answer in cotext to the case study.

Case Study/ Passage Based Questions Chapter 3 Pair of Linear Equations in Two Variables Mathematics

Type 1: MCQ type

Case Study Question 01

Read the following and answer any four questions from 1 to 5 given below:

From a shop, Sudhir bought 2 books of Mathematics and 3 books of Physics of class X for ₹ 850 and Suman bought 3 books of Mathematics and 2 books of Physics of class X for ₹ 900. Consider the price of one Mathematics book and that of one Physics book be ₹ x and ₹ y respectively. Based on the above information, answer the following questions.

Question.1. Represent the situation faced by Sudhir, algebraically.

(a) 2x+3y=850 (b) 3x+2y=850 (c) 2x-3y=850 (d) 3x-2y=850

Question.2. Represent the situation faced by Suman, algebraically.

(a) 2x+3y=90 (b) 3x+2y=900 (c) 2x-3y=900 (d) 3x-2y=900

Question.3. The price of one Physics book is

(a) ₹ 80 (b) ₹ 100 (c) ₹ 150 (d) ₹ 200

Question.4. The price of one Mathematics book is

Question.5. The system of linear equations represented by above situation, has

(a) unique solution (b) no solution (c) infinitely many solutions (d) none of these

Ans.1. (a) Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850 Ans.2. (b) Situation faced by Suman can be represented algebraically as 3x + 2y = 900 Ans.3. (c) We have 2x + 3y = 850 …(i) and 3x + 2y = 900 …(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get 5y=750 ⇒ y= 150 Thus, price of one Physics book is ₹ 150. Ans.4. (d) From equation (i) we have, 2x+3 \times 150=850 ⇒ 2x=850-450=400 ⇒ x=200 Hence, cost of one Mathematics book = ₹ 200 Ans.5. (a) From above, we have a_{1}=2 , b_{1}=3 , c_{1}=-850 and a_{2}=3 , b_{2}=2 , c_{2}=-900 ∴ \frac{a_{1}}{a_{2}}=\frac{2}{3} , \frac{b_{1}}{b_{2}}=\frac{3}{3} , \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} ⇒ \frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}} Thus system of linear equations has unique solution.

Case Study Question 02

A boat in the river Ganga near Rishikesh covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km downstream in 6 \frac{1}{2} hours. Consider speed of the boat in still water be x km/hr and speed of the stream be y km/hr and answer the following questions.

Question.1. Represent the 1^{st} situation algebraically.

(a) \frac{24}{x-y}+\frac{36}{x+y}=6 (b) \frac{24}{x+y}+\frac{36}{x-y}=6 (c) 24x+36y=6 (d) 24x-36y=6

Question.2. Represent the 2^{nd} situation algebraically.

(a) \frac{36}{x+y}+\frac{24}{x-y}=\frac{13}{2} (b) \frac{36}{x-y}+\frac{24}{x+y}=\frac{13}{2} (c) 36x-24y= \frac{13}{2} (d) 36x+24y= \frac{13}{2}

Question.3. If u= \frac{1}{x-y} and v= \frac{1}{x+y} , then u=

(a) \frac{1}{4} (b) \frac{1}{12} (c) \frac{1}{8} (d) \frac{1}{6}

Question.4. Speed of boat in still water is

(a) 4 km/hr (b) 6 km/hr (c) 8 km/hr (d) 10 km/hr

Question.5. Speed of stream is

(a) 3 km/hr (b) 4 km/hr (c) 2 km/hr (d) 5 km/hr

Speed of boat in upstream = (x-y) km/hr and speed of boat in downstream = (x+y) km/hr. Ans.1. (a) 1^{st} situation can be represented algebraically as \frac{24}{x-y}+\frac{36}{x+y}=6 Ans.2. (b) 2^{nd} situation can be represented algebraically as \frac{36}{x-y}+\frac{24}{x+y}=\frac{13}{2} Ans.3. (c) Putting \frac{1}{x-y}=u and \frac{1}{x+y}=v , we get 24u+36v=6 and 36u+24v= \frac{13}{2} Solving the above equations, we get u=\frac{1}{8} , v= \frac{1}{12} Ans.4. (d) ∵ u= \frac{1}{8}= \frac{1}{x-y} ⇒ x-y=8 …(i) and v= \frac{1}{12}= \frac{1}{x+y} ⇒ x+y=12 …(ii) Adding equations (i) from (ii), we get 2x=20 ⇒ x=10 Speed of boat in still water = 10 km/hr Ans.5. (c) From equation (i), 10-y=8 ⇒ y=2 ∴ Speed of stream = 2 km/hr

Case Study Question 03

Piyush sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum of ₹ 1008. If he had sold the saree at 10% profit and the sweater at 8% discount, he would have got ₹ 1028. Denote the cost price of the saree and the list price (price before discount) of the sweater by ₹ x and ₹ y respectively and answer the following questions.

Question.1. The 1^{st} situation can be represented algebraically as

(a) 2.08x+1.9y=2008 (b) 1.08x+0.9y=1008 (c) 10x+8y=1008 (d) 8x+10y=1008

Question.2. The 2^{nd} situation can be represented algebraically as

(a) 10x+8y=1028 (b) 2.1x+1.92y=1028 (c) 1.1x+0.92y=1028 (d) 8x+10y=1028

Question.3. Linear equation represented by 1^{st} situation intersect the x -axis at

(a) (2800, 0) (b) (2500, 0) (c) \left(\frac{2500}{3}, 0\right) (d) \left(\frac{2800}{3}, 0\right)

Question.4. Linear equation represented by 2^{nd} situation intersect the y -axis at

(a) \left(0, \frac{25700}{23}\right) (b) (0, 25700) (c) \left(0, \frac{25800}{23}\right) (d) (0, 26800)

Question.5. Both linear equations represented by situation 1^{st} and 2^{nd} intersect each other at

(a) (400, 600) (b) (600, 400) (c) (200, 200) (d) (800, 600)

Ans.1. (b) Piyush sells a saree at 8% profit + sells a sweater at 10% discount = ₹ 1008 ⇒ (100+8)% of x + (100-10)% of y = 1008 ⇒ 108% of x + 90% of y = 1008 …(i) Ans.2. (c) Piyush sold the saree at 10% profit + sold the sweater at 8% discount = ₹ 1028 ⇒ (100+10) % of x + (100-8) % of y = 1028 ⇒ 110% of x + 92% of x = 1028 ⇒ 1.1x+0.92y= 1028 …(ii) Ans.3. (d) At x -axis, y=0 1.08x=1008 ⇒ x= \frac{1008}{1.08}=\frac{2800}{3} Ans.4. (a) At y -axis, x=0 ⇒ 0.92y=1028 ⇒ y= \frac{1028}{0.92}=\frac{25700}{23} Ans.5. (b) Solving equations (i) and (ii), we get x = 600 and y = 400 Hence both linear equations intersect at (600, 400).

Case Study Question 04

Puneet went for shopping in the evening by metro with his father who is an expert in mathematics. He told Puneet that path of metro A is given by the equation 2x+4y=8 and path of metro B is given by the equation 3x+6y=18 . His father put some questions to Puneet. Help Puneet to solve the questions.

Question.1. Equation 2x+4y=8 intersects the x -axis and y -axis respectively at

(a) (4, 0), (0, 2) (b) (0, 4), (2, 0) (c) (4, 0), (2, 0) (d) (0, 4), (0, 2)

Question.2. Equation 3x+6y=18 intersects the x -axis and y -axis respectively at

(a) (6, 0), (0, 8) (b) (0, 6), (0, 8) (c) (6, 0), (0, 3) (d) (0, 6), (0, 3)

Question.3. Coordinates of point of intersection of two given equations are

(a) (1, 2) (b) (2, 4) (c) (3, 7) (d) does not exist

Question.4. Represent the equations, 2x+4y=8 and 3x+6y=18 graphically.

Question.5. System of linear equations represented by two given lines is

(a) inconsistent (b) having infinitely many solutions (c) consistent (d) overlapping each other

Ans.1. (a) At x -axis, y=0 ∴ 2x+4y=8 ⇒ x=4 At y -axis, x=0 ∴ 2x+4y=8 ⇒ y=2 ∴ Required coordinates are (4, 0), (0, 2). Ans.2. (c) At x -axis, y=0 ∴ 3x+6y=18 ⇒ 3x=18 ⇒ x=6 At y -axis, x=0 ∴ 3x+6y=18 ⇒ 6y=18 ⇒ y=3 ∴ Required coordinates are (6, 0), (0, 3). Ans.3. (d) Since, lines are parallel. So, point of intersection of these lines does not exist. Ans.4. (a) Ans.5. (a) Since the lines are parallel. ∴ These equations have no solution i.e., the given system of linear equations is inconsistent.

Case Study Question 05

Raman usually go to a dry fruit shop with his mother. He observes the following two situations. On 1^{st} day: The cost of 2 kg of almonds and 1 kg of cashew was ₹ 1600. On 2^{nd} day: The cost of 4 kg of almonds and 2 kg of cashew was ₹ 3000. Denoting the cost of 1 kg almonds by ₹ x and cost of 1 kg cashew by ₹ y , answer the following questions.

Question.1. Represent algebraically the situation of day-I.

(a) x+2y=1000 (b) 2x+y=1600 (c) x-2y=1000 (d) 2x-y=1000

Question.2. Represent algebraically the situation of day-II.

(a) 2x+y=1500 (b) 2x-y=1500 (c) x+2y=1500 (d) 2x+y=750

Question.3. The linear equation represented by day-I, intersect the x -axis at

(a) (0, 800) (b) (0, -800) (c) (800, 0) (d) (-800, 0)

Question.4. The linear equation represented by day-II, intersect the y -axis at

(a) (1500, 0) (b) (0, -1500) (c) (-1500, 0) (d) (0, 1500)

Question.5. Linear equations represented by day-I and day -II situations, are

(a) non parallel (b) parallel (c) intersect at one point (d) overlapping each other.

Ans.1. (b) Algebraic representation of situation of day-I is 2x+y=1600 Ans.2. (a) Algebraic representation of situation of day-II is 4x+2y=3000 ⇒ 2x+y=1500 Ans.3. (c) At x -axis, y=0 ∴ At y=0 , 2x+y= 1600 becomes 2x = 1600 ⇒ x=800 ∴ Linear equation represented by day-I intersect the x -axis at (800, 0). Ans.4. (d) At y -axis, x=0 ∴ 2x+y=1500 ⇒ y=1500 Linear equation represented by day-II intersect the y -axis at (0, 1500). Ans.5. (b) We have, 2x+y=1600 and 2x+y=1500 Since \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}} i.e., \frac{1}{1}=\frac{1}{1}≠\frac{16}{15} ∴ System of equations have no solution. ∴ Lines are parallel.

Case Study Question 06

A part of monthly hostel charges in a college is fixed and the remaining depends on the number of days one has taken food in the mess. When a student Anu takes food for 25 days, she has to pay ₹ 4500 as hostel charges, whereas another student Bindu who takes food for 30 days, has to pay ₹ 5200 as hostel charges. Considering the fixed charges per month be ₹ x and the cost of food per day be ₹ y , then answer the following questions.

Question.1. Represent algebraically the situation faced by both Anu and Bindu.

(a) x+25y=4500 , x+30y=5200 (b) 25x+y=4500 , 30x+y=5200 (c) x-25y=4500 , x-30y=5200 (d) 25x-y=4500 , 30x-y=5200

Question.2. The system of linear equations, represented by above situations has

(a) No solution (b) Unique solution (c) Infinitely many solutions (d) None of these

Question.3. The cost of food per day is

(a) ₹ 120 (b) ₹ 130 (c) ₹ 140 (d) ₹ 1300

Question.4. The fixed charges per month for the hostel is

(a) ₹ 1500 (b) ₹ 1200 (c) ₹ 1000 (d) ₹ 1300

Question.5. If Bindu takes food for 20 days, then what amount she has to pay?

(a) ₹ 4000 (b) ₹ 3500 (c) ₹ 3600 (d) ₹ 3800

Ans.1. (a) For student Anu: Fixed charge + cost of food for 25 days = ₹ 4500 i.e., x+25y=4500 For student Bindu: Fixed charges + cost of food for 30 days = ₹ 5200 i.e., x+30y=5200 Ans.2. (b) From above, we have a_{1}=1 , b_{1}=25 , c_{1}=-4500 and a_{2}=1 , b_{2}=30 , c_{2}=-5200 ∴ \frac{a_{1}}{a_{2}}=1 , \frac{b_{1}}{b_{2}}=\frac{25}{30}=\frac{5}{6} , \frac{c_{1}}{c_{2}}=\frac{-4500}{-5200}=\frac{45}{52} ⇒ \frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}} Thus, system of linear equations has unique solution. Ans.3. (c) We have, x+25y=4500 …(i) and x+30y=5200 …(ii) Subtracting (i) from (ii), we get 5y=700 ⇒ y= 140 ∴ Cost of food per day is ₹ 140 Ans.4. (c) We have, x+25y=4500 ⇒ x=4500-25 \times 140 ⇒ x=4500-3500=1000 ∴ Fixed charges per month for the hostel is ₹ 1000 Ans.5. (d) We have, x=1000 , y=140 and Bindu takes food for 20 days. ∴ Amount that Bindu has to pay = (1000+ 20 \times 140) = ₹ 3800

Case Study Question 07

From Bengaluru bus stand, if Riddhima buys 2 tickets to Malleswaram and 3 tickets to Yeswanthpur, then total cost is ₹ 46; but if she buys 3 tickets to Malleswaram and 5 tickets to Yeswanthpur, then total cost is ₹ 74. Consider the fares from Bengaluru to Malleswaram and that to Yeswanthpur as ₹ x and ₹ y respectively and answer the following questions.

Question.1. 1^{st} situation can be represented algebraically as

(a) 3x-5y=74 (b) 2x+5y=74 (c) 2x-3y=46 (d) 2x+3y=46

Question.2. 2^{nd} situation can be represented algebraically as

(a) 5x+3y=74 (b) 5x-3y=74 (c) 3x+5y=74 (d) 3x-5y=74

Question.3. Fare from Bengaluru to Malleswaram is

(a) ₹ 6 (b) ₹ 8 (c) ₹ 10 (d) ₹ 2

Question.4. Fare from Bengaluru to Yeswanthpur is

(a) ₹ 10 (b) ₹ 12 (c) ₹ 14 (d) ₹ 16

Question.5. The system of linear equations represented by both situations has

(a) infinitely many solutions (b) no solution (c) unique solution (d) none of these

Ans.1. (d) 1^{st} situation can be represented algebraically as 2x+3y=46 Ans.2. (c) 2^{nd} situation can be represented algebraically as 3x+5y=74 Ans.3. (b) We have, 2x+3y=46 … (i) 3x+5y=74 … (ii) Multiplying (i) by 5 and (ii) by 3 and then subtracting, we get 10x-9x=230-222 ⇒ x=8 ∴ Fare from Bengaluru to Malleswaram is ₹ 8. Ans.4. (a) Putting the value of x in equation (i), we get 3y=46-2 \times 8 = 30 ⇒ y=10 ∴ Fare from Bengaluru to Yeswanthpur is ₹ 10. Ans.5. (c) We have, a_{1}=2 , b_{1}=3 , c_{1}=-46 and a_{2}=3 , b_{2}=5 , c_{2}=-74 ∴ \frac{a_{1}}{a_{2}}=\frac{2}{3} , \frac{b_{1}}{b_{2}}=\frac{3}{5} , \frac{c_{1}}{c_{2}}=\frac{-46}{-74}=\frac{23}{37} ⇒ \frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}} Thus system of linear equations has unique solution.

Case Study Question 08

Points A and B representing Chandigarh and Kurukshetra respectively are almost 90 km apart from each other on the highway. A car starts from Chandigarh and another from Kurukshetra at the same time. If these cars go in the same direction, they meet in 9 hours and if these cars go in opposite direction they meet in 9/7 hours. Let X and y be two cars starting from points A and B respectively and their speed be x km/hr and y km/hr respectively. Then, answer the following questions.

Question.1. When both cars move in the same direction, then the situation can be represented algebraically as

(a) x-y=10 (b) x+y=10 (c) x+y=9 (d) x-y=9

Question.2. When both cars move in opposite direction, then the situation can be represented algebraically as

(a) x-y=70 (b) x+y=90 (c) x+y=70 (d) x+y=10

Question.3. Speed of car X is

(a) 30 km/hr (b) 40 km/hr (c) 50 km/hr (d) 60 km/hr

Question.4. Speed of car Y is

(a) 50 km//hr (b) 40 km/hr (c) 30 km/hr (d) 60 km/hr

Question.5. If speed of car X and car Y, each is increased by 10 km/hr, and cars are moving in opposite direction, then after how much time they will meet?

(a) 5 hrs (b) 4 hrs (c) 2 hrs (d) 1 hr

Ans.1. (a) Suppose two cars meet at point Q. Then, Distance travelled by car X = AQ, Distance travelled by car Y = BQ. It is given that two cars meet in 9 hours. ∴ Distance travelled by car X in 9 hours = 9 x km ⇒ AQ = 9x Distance travelled by car Y in 9 hours = 9 y km ⇒ BQ = 9y Clearly, AQ – BQ = AB ⇒ 9x-9y=90 ⇒ x-y=10 Ans.2. (c) Suppose two cars meet at point P. Then Distance travelled by car X = AP and Distance travelled by car Y = BP. In this case, two cars meet in 9/7 hours. ∴ Distance travelled by car X in 9/7 hours = \frac{9}{7}x km ⇒ AP = \frac{9}{7}x Distance travelled by car Y in 9/7 hours = \frac{9}{7}y km ⇒ BP = \frac{9}{7}y Clearly, AP + BP = AB ⇒ \frac{9}{7}x + \frac{9}{7}y = 90 ⇒ \frac{9}{7}(x+y)=90 ⇒ x+y=70 Ans.3. (b) We have x-y=10 …(i) ⇒ x+y=70 …(ii) Adding equations (i) and (ii), we get 2x=80 ⇒ x=40 Hence, speed of car X is 40 km/hr. Ans.4. (c) We have x-y=10 ⇒ 40-y=10 ⇒ y=30 Hence, speed of car Y is 30 km/hr Ans.5. (d)

Case Study Question 09

Read the following and answer any four questions from 1 to 4 given below:

Mr Manoj Jindal arranged a lunch party for some of his friends. The expense of the lunch are partly constant and partly proportional to the number of guests. The expenses amount to ₹ 650 for 7 guests and ₹ 970 for 11 guests. Denote the constant expense by ₹ x and proportional expense per person by ₹ y and answer the following questions.

Question.1. Represent both the situations algebraically.

(a) x+7y=650 , x+11y=970 (b) x-7y=650 , x-11y=970 (c) x+11y=650 , x+7y=970 (d) 11x+7y=650 , 11x-7y=970

Question.2. Proportional expense for each person is

(a) ₹ 50 (b) ₹ 80 (c) ₹ 90 (d) ₹ 100

Question.3. The fixed (or constant) expense for the party is

Question.4. If there would be 15 guests at the lunch party, then what amount Mr Jindal has to pay?

(a) ₹ 1500 (b) ₹ 1300 (c) ₹ 1200 (d) ₹ 1290

Question.5. The system of linear equations representing both the situations will have

Ans.1. (a) 1st situation can be represented as x+7y=650 …(i) and 2nd situation can be represented as x+11y=970 …(ii) Ans.2. (b) Subtracting equations (i) from (ii), we get 4y=320 ⇒ y=80 ∴ Proportional expense for each person is ₹ 80. Ans.3. (c) Putting y = 80 in equation (i), we get x+7 \times 80 = 650 ⇒ x= 650-560 = 90 .. Fixed expense for the party is ₹ 90 Ans.4. (d) If there will be 15 guests, then amount that Mr Jindal has to pay = ₹(90 + 15 \times 80) = ₹ 1290 Ans.5. (a) We have, a_{1}=1 , b_{1}=7 , c_{1}=-650 and a_{2}=1 , b_{2}=11 , c_{2}=-970 ∴ \frac{a_{1}}{a_{2}}=1 , \frac{b_{1}}{b_{2}}=\frac{7}{11} , \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97} ⇒ \frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}} Thus system of linear equations has unique solution.

Case Study Question 10

In a office, 8 men and 12 women together can finish a piece of work in 10 days, while 6 men and 8 women together can finish it in 14 days. Let one day’s work of a man be \frac{1}{x} and one day’s work of a woman be \frac{1}{y} . Based on the above information, answer the following questions.

(a) \frac{80}{x}-\frac{120}{y}=1 (b) \frac{120}{x}-\frac{80}{y}=1 (c) \frac{120}{x}+\frac{80}{y}=1 (d) \frac{80}{x}+\frac{120}{y}=1

(a) \frac{112}{x}-\frac{84}{y}=1 (b) \frac{84}{x}-\frac{112}{y}=1 (c) \frac{84}{x}+\frac{112}{y}=1 (d) \frac{112}{x}+\frac{84}{y}=1

Question.3. One woman alone can finish the work in

(a) 220 days (b) 140 days (c) 280 days (d) 160 days

Question.4. One man alone can finish the work in

(a) 140 days (b) 220 days (c) 160 days (d) 280 days

Question.5. If 14 men and 28 women work together, then in what time, the work will be completed?

(a) 2 days (b) 3 days (c) 4 days (d) 5 days

Ans.1. (d) Since 8 men and 12 women can finish the work in 10 days. ∴ \left(\frac{8}{x}+\frac{12}{y}\right)=\frac{1}{10} ⇒ \left(\frac{80}{x}+\frac{120}{y}\right)=1 Ans.2. (c) Since 6 men and 8 women can finish a piece of work in 14 days. ∴ \left(\frac{6}{x}+\frac{8}{y}\right)=\frac{1}{14} ⇒ \left(\frac{84}{x}+\frac{112}{y}\right)=1 Ans.3. (c) Let \frac{1}{x}=u , \frac{1}{y}=v Thus, we have 80u + 120v = 1 and 84u + 112y = 1 Solving above two equations, we get v=\frac{1}{280} ⇒ \frac{1}{y}=\frac{1}{280} ⇒ y=280 Thus one woman alone can finish the work in 280 days. Ans.4. (a) We have ⇒ \left(\frac{80}{x}+\frac{120}{y}\right)=1 ⇒ \left(\frac{80}{x}+\frac{120}{280}\right)=1 ⇒ \frac{80}{x}=1-\frac{3}{7} ⇒ \frac{80}{x}=\frac{4}{7} ⇒ x=140 Thus one man alone can finish the work in 140 days. Ans.5. (d) We have, x=140 and y=280 One day’s work of 14 men and 28 women = \frac{14}{140}+\frac{28}{280} = \frac{1}{10}+\frac{1}{10} = \frac{2}{10} ⇒ \frac{1}{5} Thus, work will be finished in 5 days.

Case Study Question 11

Question.1. If answers to all questions that he attempted by guessing were wrong, how many questions did he answer correctly?

(a) 96 (b) 86 (c) 76 (d) 106

Question.2. How many questions did he guess?

(a) 12 (b) 18 (c) 24 (d) 26

Question.3. If answers to all the questions he attempted by guessing were wrong and answered 80 correctly, then how many marks did he get?

(a) 50 (b) 70 (c) 80 (d) 20

Question.4. If answers to all the questions he attempted by guessing were wrong, then how many questions did answer correctly to score 95 marks?

(a) 100 (b) 200 (c) 250 (d) 150

Ans.1. (a) 96 Let the no. of questions whose answers are known to the student x and questions attempted by guessing be y . x+y=120 …(i) x- \frac{1}{4}=90 …(ii) Solving (i) and (ii), we get x = 96 and y = 24 ∴ No. of questions whose answer are known = 96 Ans.2. (c) 24 Ans.3. (b) 70 Total no. of questions = 96 + 24 = 120 Marks = 80 – \frac{1}{4} of 40 = 70 Ans.4. (a) 100 x- \frac{1}{4} of (120 – x ) = 95 ⇒ 5x = 380 + 120 ⇒ 5x = 500 ⇒ x = 100

Case Study Question 12

Question.1. Which is the correct equation in two variables from this situation.

(a) x+y=13 (b) x+y=15 (c) x+y=17 (d) x+y=19

Question.2. Find the length of the outer boundary of the layout.

(a) 86 m (b) 45 m (c) 34 m (d) 54 m

Question.3. Find the area of each bedroom and kitchen in the layout.

(a) Bedroom 20 m^{2} ; Kitchen 25 m^{2} (b) Bedroom 50 m^{2} ; Kitchen 55 m^{2} (c) Bedroom 30 m^{2} ; Kitchen 35 m^{2} (d) Bedroom 40 m^{2} ; Kitchen 45 m^{2}

Question.4. Find the area of living room in the layout.

(a) 85 m^{2} (b) 65 m^{2} (c) 45 m^{2} (d) 75 m^{2}

Question.5. Find the cost of laying tiles in kitchen at the rate of ₹ 50 per sq.m.

(a) ₹ 1850 (b) ₹ 1750 (c) ₹ 1950 (d) ₹ 1650

Ans.1. (a) ⇒ x+2+y=15 ∴ x+y=13 Ans.2. (d) 54 m Length = 12 + 15 + 12 + 15 = 54 m Ans.3. (c) Bedroom 30 m^{2} ; Kitchen 35 m^{2} Area of two bedrooms = 5x + 5x = 10x m^{2} Area of Kitchen = 5y m^{2} According to Question, 10x + 5y = 95 …(i) x + y = 13 …[From point (i)] Solving the above, we get x = 6 and y = 7 ∴ Area of each Bedroom = 5x = 30 m^{2} ∴ Area of the Kitchen = 5y = 35 m^{2} Ans.4. (d) 75 m^{2} Total Area = 15 \times 12 = 180 m^{2} Area of Bathroom = 2 \times 5 = 10 m^{2} ∴ Area of living room = 180 – (30 \times 2) – 10 – 35 = 75 m^{2} Ans.5. (b) ₹ 1750 Total cost = 7 m × 5 m × ₹ 50 = ₹ 1750

Case Study Question 13

Situation 1: Question.1. If the fixed charges of auto rickshaw be ₹ x and the running charges be ₹ y per km, the pair of linear equations representing the situation is

(a) x + 10y = 110 , x + 15y = 75 (b) x + 10y = 75 , x + 15y = 110 (c) 10x + y = 110 , 15x + y = 75 (d) 10x + y = 75 , 15 x + y =110

Question.2. A person travels a distance of 50 km. The amount he has to pay is

(a) ₹ 155 (b) ₹ 255 (c) ₹ 355 (d) ₹ 455

Situation 2: Question.3. What will a person have to pay for travelling a distance of 30 km?

(a) ₹ 185 (b) ₹ 289 (c) ₹ 275 (d) ₹ 305

Question.4. The graph of lines representing the conditions of both the equations are depicted below. Which one is correct.

Question.5. Out of both the city, which one has cheaper fare?

(a) City A (b) City A (c) Both are same (d) cannot decided

Case Study Question 14

Read the following and answer the questions given below:

A man is trying to choose between two phone plans. The first plan of company A, cost ₹ 20 per month, with calls costing an additional 25 paise per minute. The second plan of company B charges ₹ 40 per month, but calls cost 8 paise per minute. These two situations are shown below which represent linear equations. The total cost for the two company’s are given by y = 0.25x + 20 and y = 0.08x + 40 where x is the minutes used and y is the total cost per month.

Question.1. If a person takes first plan and calls for 90 minutes in a month then how much amount he will have to pay whose cost for a month is given by y = 0.25x + 20 ?

(a) ₹ 20 (b) ₹ 40 (c) ₹ 42.50 (d) ₹ 45

Question.2. Another person takes second plan and also calls for 90 minutes in a month, then the amount which he has to pay when total cost is given by y = 0.08x + 40 , is

(a) ₹ 45 (b) ₹ 47 (c) ₹ 45.20 (d) ₹ 47.20

Question.3. Solution of system of linear equations x + 2y = – 1 and 2x – 3y = 12 is

(a) (–3, 2) (b) (–3, –2) (c) (3, – 2) (d) (3, 2)

Question.4. If the system of pair of linear equations kx + 2y = 5 , 3x + y = 1 has a unique solution, then

(a) k ≠ \frac{3}{2} (b) k = 6 (c) k ≠ 6 (d) k ≠ \frac{2}{3}

Question.5. Given system of linear equations x + 2y – 4 = 0 , 2x + 4y – 12 = 0 represents

(a) parallel lines (b) intersecting lines (c) coincident lines (d) can’t say

Ans.1. (c) Ans.2. (d) Ans.3. (c) Ans.4. (c) Ans.5. (a)

Case Study Question 15

A cricket bat manufacturer’s revenue is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R = xp , where x = quantity and p = price. The revenue function is shown in the figure. The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs such as utilities. The cost function is shown in blue colour in figure. The x -axis represents quantity (in hundreds) of units, and the y -axis represents either cost or revenue (in thousand of rupees). The profit function is the difference of revenue function and the cost function, written as P(x) = R(x) – C(x) Now, let C(x) = 100x + 30000 and R(x) = 150x If we replace function by y , we get Linear equations are y = 100x + 30000 and y = 150x

Question.1. The number of bat manufactured ( x ) so that there is no profit or loss for the manufacturer is

(a) x = 200 (b) x = 300 (c) x = 400 (d) x = 600

Question.2. If the cost and revenue are given by the linear equations y = 0.85x + 35000 and y = 1.55x respectively, and the break-even point is the point at which the two lines intersect, then its break even point is

(a) (50000, 77500) (b) (50000, 60000) (c) (45000, 50000) (d) (45000, 60000)

Question.3. The system of linear equations 3x + 2y = 12 , 5x – 2y = 4 represents

Question.4. Solution of the system of linear equations x + y = 1 , 2x – 3y = 7 , is

(a) (1, 2) (b) (2, 1) (c) (2, –1) (d) (–2, –1)

Question.5. The value of k for which the system of linear equations 4x + 5y = 3 and kx + 15y = 9 has infinitely many solutions is

(a) k = 3 (b) k = 4 (c) k = 12 (d) k = 8

Ans.1. (d) Ans.2. (a) Ans.3. (b) Ans.4. (c) Ans.5. (c)

Case Study Question 16

The scissors which is so common in our daily life use, its blades represent the graph of linear equations. Let the blades of a scissor are represented by the system of linear equations: x + 3y = 6 and 2x – 3y = 12

Question.1. The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is

(a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

Question.2. The points at which linear equations x + 3y = 6 and 2x – 3y = 12 intersect y –axis respectively are

(a) (0, 2) and (0, 6) (b) (0, 2) and (6, 0) (c) (0, 2) and (0, –4) (d) (2, 0) and (0, –4)

Question.3. The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is

(a) 0 (b) 1 (c) 2 (d) infinitely many

Question.4. If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12 , then values of a and b are respectively

(a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

Question.5. If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

Ans.1. (b) Ans.2. (c) Ans.3. (d) Ans.4. (a) Ans.5. (d)

Case Study Question 17

A pen stand with a pen is represented by the system of linear equations y = 0 and 3x + 2y = 6 .

Question.1. The system of linear equations y = 0 and 3x + 2y = 6 represents the pen stand and a pen respectively then their point of contact (intersection) is

(a) (0, 3) (b) (2, 0) (c) (3, 2) (d) (3, 0)

Question.2. The pair of linear equation y = x and x + y = 6 intersect each other at point

(a) (4, 2) (b) (4, 4) (c) (3, 3) (d) (3, 2)

Question.3. The system of linear equations 3x + 6y = 3900 and x + 3y = 1300 represent the lines which are

(a) parallel (b) intersecting (c) coincident (d) can’t say

Question.4. The value of k for which the system of equations kx – 5y = 2 and 6x + 2y = 7 has no solution, is

(a) 30 (b) –30 (c) 15 (d) –15

Question.5. The linear equation 3x + 2y = 6 intersects the y –axis at the point

(a) (0, 3) (b) (0, 2) (c) (0, –3) (d) (2, 3)

Ans.1. (b) Ans.2. (c) Ans.3. (b) Ans.4. (d) Ans.5. (a)

Case Study Question 18

A boy enjoying the pizza with his friends and share with them by slicing it. During slicing the pizza, he noticed that the pair of linear equations formed. Let these pair of linear equations be y – 2x = 1 and 5y – x = 14 .

Question.1. The point of intersection of the lines given by the equations y – 2x = 1 and 5y – x = 14 is

(a) (–2, 3) (b) (–4, 2) (c) (6, 4) (d) (1, 3)

Question.2. The linear equation y – 2x = 1 intersect the y –axis at point

(a) \left(-\frac{1}{2},0\right) (b) (0, 1) (c) (0, –14) (d) \left(0,\frac{14}{5}\right)

Question.3. The system of linear equations 2x – 3y + 6 = 0 and 2x + 3y – 18 = 0

(a) has a unique solution (b) has no solution (c) has infinitely many solution (d) may or may not have a solution

Question.4. The value(s) of k for which the system of linear equations 2x – ky + 3 = 0 and 3x + 2y – 1 = 0 has no solution, is

(a) \frac{4}{3} (b) -\frac{4}{3} (c) 6 (d) –6

Question.5. If a pair of linear equations in two variables is inconsistent, then the lines represented by two equations are

(a) intersecting (b) parallel (c) always coincident (d) Intersecting or coincident

Ans.1. (d) Ans.2. (b) Ans.3. (a) Ans.4. (b) Ans.5. (b)

Case Study Question 19

Special offers are short-term pricing strategies that businesses, especially shops will adopt to encourage customers to buy from them. During winter season, a shopkeeper sells a jacket at 8% profit and a sweater at 10 % discount thereby getting a sum of ₹1008. If she had sold the jacket at 10 % profit and the sweater at 8 % discount, she would have got ₹ 1028. Denoting the cost price of one jacket by ₹ x and the list price of one sweater by ₹ y , answer the following situations.

Question.1. Represent the first situation algebraically.

(a) 12x+10y=11200 (b) 10x+12y=11200 (c) 12x-10y=11200 (d) 10x-12y=1120

Question.2. Represent the second situation algebraically

(a) 46x+55y=51400 (b) 55x+46y=51400 (c) 55x-46y=51400 (d) 46x-55y=51400

Question.3. The system of linear equations representing both the situations will have.

(a) Infinite number of solutions (b) Unique solution (c) No Solutions (d) Exactly two solutions

Question.4. The graph of the system of linear equations representing both the situations will be

(a) Parallel lines (b) Coincident lines (c) Intersecting lines (d) None of these

Let the cost price of one jacket be ₹ x and the list price of one sweater be ₹ y Ans.1. (a) According to the first condition, x+8100x+y-10100 y=1008 108 \times 100+90100y=1008 108x+90y=100800 Dividing through out by 9, we get 12x+10y=11200 Ans.2. (b) According to the second condition x+10100x+y-8100 y=1028 110×100+92100y=1028 110x+92y=102800 Dividing through out by 2, we get 55x+46y=51400 Ans.3. (b) Ans.4. (c)

Case Study Question 20

An alumni association is an association of former students. These associations often organize social events, publish newsletters or magazines and raise funds for the organisation. The alumni meet of two batches of a college- batch A & batch B were held on the same day in the same hotel in two separate halls “Rose” and “Jasmine”. The rents were the same for both the halls. The expense for each hall is equal to the fixed rent of each hall and proportional to the number of persons attending each meet. 50 persons attended the meet in “Rose” hall, and the organisers had to pay ₹ 10000 towards the hotel charges. 25 guests attended the meet in “Jasmine” hall and the organisers had to pay ₹ 7500 towards the hotel charges. Denote the fixed rent by ₹ x and proportional expense per person by ₹ y .

Question.1. Represent algebraically the situation in hall “Rose”.

(a) 50x+y=10000 (b) 50x-y=10000 (c) x+50y=10000 (d) x-50y=10000

Question.2. Represent algebraically the situation in hall “Jasmine”

(a) x+25y=7500 (b) x-25y=7500 (c) 25x+y=7500 (d) 25x-y=7500

Question.3. What is the fixed rent of the halls?

(a) ₹ 2500 (b) ₹ 3300 (c) ₹ 4000 (d) ₹ 5000

Question.4. Find the amount the hotel charged per person.

(a) ₹ 150 (b) ₹ 190 (c) ₹ 130 (d) ₹ 100

Let us denote the fixed rent by ₹ x and proportional expense per person by ₹ y . Ans.1. (c) Algebraic representation of the situation in “Rose” hall x+50y=10000 Ans.2. (a) Algebraic representation of the situation in “Jasmine” hall x+25y=7500 Subtracting the equations represented by (i) and (ii) x+50y-x+25y=10000-7500 ⇒ 25y=2500 ⇒ y=100 Substituting y=100 in x+50y=10000 , we get x+50 \times 100=10000 ⇒ x+5000=10000 ⇒ x=5000 Ans.3. (d) Ans.4. (d)

Case Study Question 21

Question.1. What can you say about the pair of linear equations?

(a) Consistent (b) Inconsistent (c) Dependent (d) Cannot be determined

Question.2. From the graph, find the coordinates of the point, where the line AB intersects the x -axis

(a) (5,0) (b) (-2,0) (c) (0,2) (d) (0,0)

Question.3. From the graph, find the solution of the pair of linear equations

(a) (4,2) (b) (2,4) (c) (-2,0) (d) (5,0)

Question.4. What is the area of the shaded region?

(a) 11 sq. units (b) 12 sq. units (c) 13 sq. units (d) 14 sq.units

Ans.1. (a) Consistent Ans.2. (b) (-2,0) Ans.3. (b) (2,4) Ans.4. (d) Area of the shaded region = 12 \times 7 \times 4 = 14 sq. units

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CBSE Case Study Questions for Class 10 Maths Linear Equations in Two Variables Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Linear Equations in Two Variables in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Linear Equations in Two Variables PDF

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Class 10 Maths Chapter 3 Case Based Questions - Pair of Linear Equations in Two Variables

Case study - 1.

Refer situation 1 : Q1: If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y km/hr, the pair of linear equations representing the situation is (a) x + 10y =110, x + 15y = 75 (b) x + 10y = 75, x + 15y = 110 (c) 10x + y = 110, 15x + y = 75 (d) 10x + y = 75, 15x + y = 110 Ans: (b) Explanation: The cost of the auto rickshaw ride consists of two components - a fixed charge (x) and a variable charge based on the distance of the journey (y per km). The total cost of the ride would therefore be the sum of these two components. In the first situation, the auto rickshaw ride in city A costs Rs 75 for a 10 km journey and Rs 110 for a 15 km journey. We can express these two situations as two linear equations: For the 10 km journey: x + 10y = 75 (equation 1) For the 15 km journey: x + 15y = 110 (equation 2) Where x is the fixed charge and y is the cost per km. Therefore, the pair of linear equations representing the situation is x + 10y = 75, x + 15y = 110 (option b). Q2: A person travels a distance of 50km. The amount he has to pay is (a) Rs.155 (b) Rs.255 (c) Rs.355 (d) Rs.455 Ans: (c) Explanation: To solve this case-based problem, we first need to understand the relationship between distance traveled and the charge paid. This can be represented as a linear equation in the form of y = mx + c, where y is the charge paid, m is the rate charged per km, x is the distance traveled, and c is the fixed charge. From Situation 1, we have two equations based on the given data: 1) 75 = 10m + c 2) 110 = 15m + c Subtracting equation 1 from equation 2, we get: 35 = 5m So, m = 35/5 = 7. This means the rate charged per km in city A is Rs. 7. Substituting m = 7 in equation 1, we get: 75 = 10*7 + c So, c = 75 - 70 = 5. This means the fixed charge in city A is Rs. 5. So, the total charge for traveling a distance of x km in city A is given by the equation: Charge = 7x + 5 Now, let's calculate the charge for traveling a distance of 50km in city A: Charge = 7*50 + 5 = Rs. 355 Therefore, the answer is (c) Rs. 355. Refer situation 2: Q3: What will a person have to pay for travelling a distance of 30km? (a) Rs.185 (b) Rs.289 (c) Rs.275 (d) Rs.305 Ans: (b) Explanation: To solve this problem, we first need to find the rate of the fare per kilometer and the fixed charge in city B. We can solve this problem by making two equations from the given situations and then solve them simultaneously. Let's take the fixed charge as F (in Rs.) and the rate per kilometer as R (in Rs. per km). From the given situation 2, we know that: For a journey of 8km, the charge paid is Rs 91. For a journey of 14km, the charge paid is Rs 145. So, we can formulate two equations: 1) F + 8R = 91 2) F + 14R = 145 Subtract equation 1 from equation 2 and we get 6R = 54, therefore R = 54/6 = 9 Rs per km. Substitute the value of R in equation 1, we get F = 91 - 8*9 = 91 - 72 = 19 Rs. So, the fixed charge is Rs 19 and the fare per kilometer is Rs 9 in city B. Now, to find out what a person will have to pay for traveling a distance of 30km, we add the fixed charge to the product of the fare per kilometer and the distance. Therefore, the fare for 30km = F + 30R = 19 + 30*9 = 19 + 270 = Rs 289. So, the correct answer is (b) Rs.289.

Case Study - 2

Based on the above information, answer the following questions: Q1: Form the pair of linear equations in two variables from this situation. Ans: Area of two bedrooms= 10x sq.m Area of kitchen = 5y sq.m 10x + 5y = 95 2x + y =19 Also, x + 2+ y = 15 x + y = 13 Q2: Find the length of the outer boundary of the layout. Ans: Length of outer boundary = 12 + 15 + 12 + 15 = 54m Q3: Find the area of each bedroom and kitchen in the layout. Ans: On solving two equation part(i) x = 6m and y = 7m area of bedroom = 5 x 6 = 30m area of kitchen = 5 x 7 = 35m Q4: Find the area of living room in the layout. Ans: Area of living room = (15 x 7) – 30 = 105 – 30 = 75 sq.m Q5: Find the cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m. Ans: Total cost of laying tiles in the kitchen = Rs.50 x 35 = Rs1750.

Case Study - 3

A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while ¼ mark is deducted for every wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.

Q2: How many questions did he guess? Ans: Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 He attempted 24 questions by guessing. Q3: If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got? Ans: Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 Marks = 80- ¼ 0f 40 =70

Q4: If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks? Ans: Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x – 1/4y =90 solving these two x = 96 and y = 24 x – 1/4 of (120 – x) = 95 5x = 500, x = 100

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Case Based Questions: Pair of Linear Equations in Two Variables Free PDF Download

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Case Study Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Last modified on: 1 year ago
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Case Study Questions

Question 1:

The scissors which is so common in our daily life use, its blades represent the graph of linear equations.

Let the blades of a scissor are represented by the system of linear equations:

x + 3y = 6 and 2x – 3y = 12

(i) The pivot point (point of intersection) of the blades represented by the linear equation x + 3y = 6 and 2x – 3y = 12 of the scissor is (a) (2, 3) (b) (6, 0) (c) (3, 2) (d) (2, 6)

(iii) The number of solution of the system of linear equations x + 2y – 8 = 0 and 2x + 4y = 16 is (a) 0 (b) 1 (c) 2 (d) infinitely many

(iv) If (1, 2) is the solution of linear equations ax + y = 3 and 2x + by = 12, then values of a and b are respectively (a) 1, 5 (b) 2, 3 (c) –1, 5 (d) 3, 5

(v) If a pair of linear equations in two variables is consistent, then the lines represented by two equations are (a) intersecting (b) parallel (c) always coincident (d) intersecting or coincident

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CBSE 10th Standard Maths Subject Case Study Questions With Solution 2021 Part - II

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

(ii) Proportional expense for each person is

(iii) The fixed (or constant) expense for the party is

(iv) If there would be 15 guests at the lunch party, then what amount Mr Jindal has to pay?

(v) The system of linear equations representing both the situations will have

(ii) Represent the situation faced by Suman, algebraically

(iii) The price of one Physics book is

(iv) The price of one Mathematics book is

(v) The system of linear equations represented by above situation, has

(ii) Represent algebraically the situation of day- II.

(iii) The linear equation represented by day-I, intersect the x axis at

(iv) The linear equation represented by day-II, intersect the y-axis at

(v) Linear equations represented by day-I and day -II situations, are

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

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

(iii) x 2 - 8x + 16 = 0

(iv) 6x 2 - 13x + 5 = 0

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

(ii) Difference of pairs of shoes in 17 th row and 10 th row is

(iii) On next day, she arranges x pairs of shoes in 15 rows, then x =

(iv) Find the pairs of shoes in 30 th row.

(v) The total number of pairs of shoes in 5 th and 8 th row is

(ii) The number on first card is

(iii) What is the number on the 19 th card?

(iv) What is the number on the 23 rd card?

(v) The sum of numbers on the first 15 cards is

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information, answer the following questions. (i) Which of the following sequence is an A.P.?

(ii) If x, y and z are in A.P., then

(iii) If a 1 a 2 , a 3 ..... , a n are in A.P., then which of the following is true?

(iv) If the n th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(v) Which of the following is incorrect about A.P.?

(ii) Find the radius of the core.

(iii) S 2 =

(iv) What is the diameter of roll when one tissue sheet is rolled over it?

(v) Find the thickness of each tissue sheet

(ii) Distance travelled by aeroplane towards west after \(1 \frac{1}{2}\) hr is

(iii) In the given figure, \(\angle\) POQ is

(iv) Distance between aeroplanes after \(1 \frac{1}{2}\) hr is

(v) Area of \(\Delta\) POQ is

(ii) The value of x is

(iii) The value of PR is

(iv) The value of RQ is

(v) How much distance will be saved in reaching city Q after the construction of highway?

(ii) Length of BC =

(iii) Length of AD =

(iv) Length of ED =

(v) Length of AE =

(ii) The value of x + y is

(iii) Which of the following is true?

(iv) The ratio in which B divides AC is

(v) Which of the following equations is satisfied by the given points?

(ii) The value of x is equal to

(iii) If M is any point exactly in between city A and city B, then coordinates of M are

(iv) The ratio in which A divides the line segment joining the points O and M is

(v) If the person analyse the petrol at the point M(the mid point of AB), then what should be his decision?

(ii) The centre of circle is the

(iii) The radius of the circle is

(iv) The area of the circle is

(v) If \(\left(1, \frac{\sqrt{7}}{3}\right)\) is one of the ends of a diameter, then its other end is

(ii) The distance between A and Cis

(iii) If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B?

(iv) If the fare for first bus is Rs10/km, then what will be the fare for total journey by that bus?

(v) If the fare for second bus is Rs 15/km, then what will be the fare to reach to the destination by this bus?

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

Cbse 10th standard maths subject case study questions with solution 2021 part - ii answer keys.

(i) (a): 1 st situation can be represented as x + 7y = 650 ...(i) and 2 nd situation can be represented as x + 11y = 970 ...(ii) (ii) (b): Subtracting equations (i) from (ii), we get \(4 y=320 \Rightarrow y=80\) \(\therefore\) Proportional expense for each person is Rs 80. (iii) (c): Puttingy = 80 in equation (i), we get x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90 \(\therefore\) Fixed expense for the party is Rs 90 (iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290 (v) (a): We have a 1 = 1, b 1 = 7, c 1 = -650 and \(a_{2}=1, b_{2}=11, c_{2}=-970 \) \(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\) \(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus, system of linear equations has unique solution.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850 (ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900 (iii) (c) : We have 2x + 3y = 850 .........(i) and 3x + 2y = 900 .........(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get 5y = 750 \(\Rightarrow\) Y = 150 Thus, price of one Physics book is Rs 150. (iv) (d): From equation (i) we have, 2x + 3 x 150 = 850 \(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200 Hence, cost of one Mathematics book = Rs 200 (v) (a): From above, we have \(a_{1} =2, b_{1}=3, c_{1}=-850 \) \(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\) \(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus system of linear equations has unique solution.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600. (ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500. (iii) (c) : At x-axis, y = 0 \(\therefore\) At y = 0, 2x + y = 1600 becomes 2x = 1600 \(\Rightarrow\) x = 800 \(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0). (iv) (d) : At y-axis, x = 0 \(\therefore\) 2x + Y = 1500 \(\Rightarrow\) y = 1500 \(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500). (v) (b): We have, 2x + y = 1600 and 2x + y = 1500 Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\) \(\therefore\) System of equations have no solution. \(\therefore\) Lines are parallel.

(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): \(100 x^{2}-20 x+1=0\) \(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)

Number of pairs of shoes in 1 st , 2 nd , 3 rd row, ... are 3,5,7, ... So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2 (i) (d): Let n be the number of rows required. \(\therefore S_{n}=120 \) \(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \) \(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\) \(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\) So, 10 rows required to put 120 pairs. (ii) (b): No. of pairs in 1ih row = t 17 = 3 + 16(2) = 35 No. of pairs in 10th row = t 10 = 3 + 9(2) = 21 \(\therefore\) Required difference = 35 - 21 = 14 (iii) (c) : Here n = 15 \(\therefore\) t 15 = 3 + 14(2) = 3 + 28 = 31 (iv) (a): No. of pairs in 30 th row = t 30 = 3 +29(2) = 61 (v) (c): No. of pairs in 5 th row = t 5 = 3 + 4(2) = 11 No. of pairs in 8 th row = t 8 = 3 + 7(2) = 17 \(\therefore\) Required sum = 11 + 17 = 28

Let the numbers on the cards be a, a + d, a + Zd, ... According to question, We have (a + 5d) + (a + 13d) = -76 \(\Rightarrow\) 2a+18d = -76 \(\Rightarrow\) a + 9d= -38 ... (1) And (a + 7d) + (a + 15d) = -96 \(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2) From (1) and (2), we get 2d= -10 \(\Rightarrow\) d= -5 From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7 (i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5 (ii) (d): Number on first card = a = 7 (iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83 (iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103 (v) (d): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c) (ii) (c) (iii) (d) (iv) (b) (v) (c)

Here S n = 0.1n 2 + 7.9n (i) (c): S n -1 = 0.1(n - 1) 2 + 7.9(n - 1) = 0.1n 2 + 7.7n - 7.8 (ii) (b): S 1 = t 1 = a = 0.1(1) 2 + 7.9(1) = 8 cm = Diameter of core So, radius of the core = 4 cm (iii) (a): S 2 = 0.1(2) 2 + 7.9(2) = 16.2 (iv) (d): Required diameter = t 2 = S 2 - S 1 = 16.2 - 8 = 8.2 cm (v) (c): As d = t 2 - t 1 = 8.2 - 8 = 0.2 cm So, thickness of tissue = 0.2 \(\div\) 2 = 0.1 cm = 1 mm

(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\) Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time = \(\frac{3}{2}\) hr. \(\therefore\) Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\) [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (b): If \(\Delta\) AED and \(\Delta\) BEC, are similar by SAS similarity rule, then their corresponding proportional sides are \(\frac{B E}{A E}=\frac{C E}{D E}\) (ii) (c): By Pythagoras theorem, we have \(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\) (iii) (a): Since \(\Delta\) ADE and \(\Delta\) BCE are similar. \(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \) \(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\) (iv) (b): \(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \) \(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\) (v) (d) : \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\) \(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \) \(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (a): We have, OA = 2 \(\sqrt{2}\) km \(\Rightarrow \sqrt{2^{2}+y^{2}}=2 \sqrt{2} \) \(\Rightarrow 4+y^{2}=8 \Rightarrow y^{2}=4 \) \(\Rightarrow y=2 \quad(\because y=-2 \text { is not possible })\) (ii) (c): We have OB = 8 \(\sqrt{2}\) \(\Rightarrow \sqrt{x^{2}+8^{2}}=8 \sqrt{2} \) \(\Rightarrow x^{2}+64=128 \Rightarrow x^{2}=64 \) \(\Rightarrow x=8 \quad(\because x=-8 \text { is not possible })\) (iii) (c) : Coordinates of A and Bare (2, 2) and (8, 8) respectively, therefore coordinates of point M are \(\left(\frac{2+8}{2}, \frac{2+8}{2}\right)\) i.e .,(5.5) (iv) (d): Let A divides OM in the ratio k: 1.Then \(2=\frac{5 k+0}{k+1} \Rightarrow 2 \mathrm{k}+2=5 k \Rightarrow 3 k=2 \Rightarrow k=\frac{2}{3}\) \(\therefore\) Required ratio = 2 : 3 (v) (b): Since M is the mid-point of A and B therefore AM = MB. Hence, he should try his luck moving towards B.

(i) (c): Required coordinates are \(\left(0, \frac{4}{3}\right)\) (ii) (c) (iii) (a): Radius = Distance between (0,0) and \(\left(\frac{4}{3}, 0\right)\) \(=\sqrt{\left(\frac{4}{3}\right)^{2}+0^{2}}=\frac{4}{3} \text { units }\) (iv) (b): Area of circle = \(\pi\) (radius) 2 \(=\pi\left(\frac{4}{3}\right)^{2}=\frac{16}{9} \pi \text { sq. units }\) (v) (d): Let the coordinates of the other end be (x,y). Then (0,0) will bethe mid-point of \(\left(1, \frac{\sqrt{7}}{3}\right)\) and (x, y). \(\therefore\left(\frac{1+x}{2}, \frac{\frac{\sqrt{7}}{3}+y}{2}\right)=(0,0) \) \(\Rightarrow \frac{1+x}{2}=0 \text { and } \frac{\frac{\sqrt{7}}{3}+y}{2}=0 \) \(\Rightarrow x=-1 \text { and } y=-\frac{\sqrt{7}}{3}\) Thus, the coordinates of other end be \(\left(-1, \frac{-\sqrt{7}}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2). (i) (d): Required distance \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\) \(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) (ii) (d): Required distance \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\) \(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\) (iii) (b): Distance between Band C \(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\) Thus, distance travelled by first bus to reach to B \(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\) and distance travelled by second bus to reach to B \(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) \(\therefore\) Distance of first bus is greater than distance of the second bus, therefore second bus should be chosen. (iv) (d): Distance travelled by first bus = 8.48 km \(\therefore\) Total fare = 8.48 x 10 = Rs 84.80 (v) (b): Distance travelled by second bus = 7. 2 km \(\therefore\) Total fare = 7.2 x 15 = Rs 108

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Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus. These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam.
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STEP 1: Read the case study and questions carefully. Read the case and associated questions carefully. Highlight the main points of the case and any issues that you can identify. Read the questions closely and analyse what they are requiring you to do.
CBSE 10th Standard Maths Pair of Linear Equation in Two ...
CBSE 10th Standard Maths Subject Pair of Linear Equation in Two Variables Case Study Questions With Solution 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 .
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These Case Study Questions for Class 10 Maths Linear Equations in Two Variables are latest, comprehensive, confidence inspiring, with easy to understand explanation.
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Full syllabus notes, lecture & questions for Class 10 Maths Chapter 3 Case Based Questions - Pair of Linear Equations in Two Variables - Class 10 | Plus excerises question with solution to help you revise complete syllabus | Best notes, free PDF download
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10th Standard CBSE Subjects. 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.