case study ch 3 class 10 maths

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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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Situation 1: In city A, for a journey of 10 km, the charge paid is Rs 75 and for a journey of 15 km, the charge paid is Rs 110.

Situation 2: In a city B, for a journey of 8km, the charge paid is Rs91 and for a journey of 14km, the charge paid is Rs 145.

Refer situation 1

1. 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

Answer: b) x + 10y = 75, x + 15y = 110

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

Answer: c) Rs.355

Refer situation 2

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

Answer: b) Rs.289

4. The graph of lines representing the conditions are: (situation 2)

Answer: (iii)

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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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

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

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

(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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Case Study on Pair of Equations in Two Variables Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Pair of Equations in Two Variables Class 10 Maths can use this page to download the PDF file. 

The case study questions on Pair of Equations in Two Variables are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Pair of Equations in Two Variables case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Pair of Equations in Two Variables Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Pair of Equations in Two Variables, therefore, they prepared a set of solutions along with the case study questions on Pair of Equations in Two Variables.

The case study on Pair of Equations in Two Variables Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Pair of Equations in Two Variables case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Pair of Equations in Two Variables Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Pair of Equations in Two Variables 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 Pair of Equations in Two Variables Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Pair of Equations in Two Variables 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 Pair of Equations in Two Variables Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Pair of Equations in Two Variables as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Pair of Equations in Two Variables?

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

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Pair of Equations in Two Variables questions quickly.
• 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 Pair of Equations in Two Variables 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 Pair of Equations in Two Variables?

 A few essential things to know to solve Case Study Questions on Class 10 Pair of Equations in Two Variables are -

• Basic Formulas of Pair of Equations in Two Variables: One of the most important things to know to solve Case Study Questions on Class 10 Pair of Equations in Two Variables is to learn about the basic formulas or revise them before solving the case-based questions on Pair of Equations in Two Variables.
• 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 Pair of Equations in Two Variables 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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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
• Reading Time: 5 Minutes

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)

(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)

(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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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

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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?

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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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