NCERT Solutions for Class 10 Maths

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NCERT Solutions for Class 10 Maths

Mathematics is a revered topic for engineering students pursuing professions through competitive exams such as JEE Main and JEE Advanced. It is one of those disciplines where a flawless centum can be obtained if the right preparation schedule and time are followed.

Over the last three to four years, CBSE has shifted away from regular NCERT questions and has focused more on application-based questions. This has made the paper rather tough to score if students just prepare for the questions that are expected to appear in the exam.

NCERT Solutions for Class 10 Maths Chapter-wise

Maths is a compulsory component in Class 10. Therefore, all CBSE students need to use the CBSE Class 10 Mathematics NCERT Solutions for complete understanding of the concepts to ace their preparation. These solutions cover all key mathematical ideas and assist students in developing critical problem-solving skills.

NCERT Solutions for Class 10 Maths

The solutions consist of 15 chapters, all of which are crucial for study. Students need to complete all of the exercises, even the optional ones to score well in the exam. NCERT solutions for class 10 Mathematics will assist you in thoroughly understanding each question. You can easily do well in board exams if you solve the NCERT Mathematics and practice the previous 5 years’ questions.

NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

Using Euclid’s Lemma, the Fundamental Theorem of Arithmetic. Real Numbers explains the notion of properties of real numbers in great detail. Exercises 1.1 and 1.2 require the use of division procedure to determine the divisibility of integers. Exercises 1.3 and 1.4, on the other hand, feature a question about proving the roots of any irrational number and decimal expansions of rational numbers.

NCERT Solution for Class 10 Mathematics Chapter 2 – Polynomials

This chapter covers advanced polynomial terms such as degree, coefficient, and zeroes for any type of polynomials. The chapter offers four exercises, with Exercises 2.1 and 2.2 requiring the use of a graph. Exercise 2.3 focuses on the division algorithm, and exercise 2.4 covers all subjects covered in the chapter.

NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

This chapter discusses the concept of linear equation pairs. It begins with linear equations in two variables and their solutions using the graphical approach, as well as algebraic conditions for linear equations. The chapter goes on to explain how to solve a pair of linear equations in two variables algebraically using substitution, elimination, and cross-multiplication.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

In this chapter, the learner will learn how to write quadratic equations in standard form. The chapter also covers the nature of roots, such as whether they are real and distinct, real and equal, or non-real roots. It also includes four exercises in which you must determine whether an equation is a quadratic one or not.

NCERT Solutions for Class 10 Mathematics Chapter 5 – Arithmetic Progressions

Students will learn in greater detail about arithmetic progression, which is a series in which two consecutive terms have a constant difference. The chapter serves as the foundation for other complex concepts that will be introduced in later classes. They will develop the skills to solve issues with sequences and series.

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

The Triangles chapter covers the concepts of similarity and congruence. The basics of triangles are covered in Exercise 6.1. Problems from the topic of triangular similarity and a related theorem appear in exercises 6.2, 6.3, and 6.4. Exercises 6.5 and 6.6 focus on triangular congruency and application.

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry

Students will learn how to find the coordinates of a point that splits a line segment connecting two given sites in a specified ratio. The Distance formula and Section formula are key to the chapter in 10th grade maths. Sections on the area of the triangle are also included.

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

This chapter is about the ratio of right angles to acute angles in any right-angled triangle. The questions in this chapter cover trigonometric ratios of specific angles, trigonometric identities, and trigonometric ratios of complementary angles are used. There are four exercises in this chapter: Exercise 8.1 introduces the trigonometric ratio and its calculation using Pythagoras’ theorem. Exercise 8.2 uses the trigonometric table to calculate specific angles. Exercise 8.3 is about complementary angles; and Exercise 8.4 is about trigonometric identities.

NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry

Trigonometry is the study of how to calculate the heights and distances of various things without having to measure them. This chapter explains how trigonometry is used in everyday life. Other such topics included in this chapter are the angle of deviation and elevation, and the line of sight.

NCERT Solutions for Class 10 Mathematics Chapter 10 – Circles

Students have a fundamental understanding of circles and associated terminology such as chord, segment, arc, and so on. They become familiar with the concepts of Tangent to a Circle and Number of Tangents from a Point on a Circle. The chapter discusses circle theorems which aid in the solution of many geometry problems.

NCERT Solutions for Class 10 Mathematics Chapter 11 – Constructions

This chapter teaches students how to build various geometrical figures. Methods and processes for construction are presented in simple language. This chapter also discusses Thales Theorem, often known as the Basic Proportionality Theorem. It ends with tasks that will teach pupils how to generate tangents to a circle.

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles

This chapter includes three exercises, as well as real-life difficulties and cost-related questions about the circular territory. It also introduces the concepts of circle perimeter and area. Students will be able to compute the area of a sector and a segment of a circular region.

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes

This chapter discusses calculating the areas and volumes of several solids such as the cube, cuboid, and cylinder. Questions are based on determining the volume of objects generated by merging several solid figures or converting one shape to another. NCERT Solutions for Class 10 Maths provide answers to both easy and difficult questions on the topic.

NCERT Solutions for Class 10 Maths Chapter 14 – Statistics

Students will learn how to convert individual data into grouped data and calculate the Mean, Mode, and Median. This chapter serves as the foundation for data science and analytics. The chapter teaches students how to create a graphical representation of raw, unorganised data from which tangible results can be generated.

NCERT Solutions for Class 10 Maths Chapter 15 – Probability

The final chapter is about Probability, which is the study of determining the likelihood of success or failure in a specific experiment. There are numerous examples provided to clearly explain probability and to help students understand it. Students must first solve the textbook examples before attempting to solve the exercise problems.

CBSE Class 10 Maths Exam Pattern for 2022-2023

The exam is worth 100 marks, with the theory worth 80 marks and the internal assessment worth 20 marks. The CBSE Board releases sample papers and grading schemes well in advance of the board exams to help students clear up any confusion. According to the CBSE’s revised design, 25% of theoretical questions would be objective. The number of questions has been raised, but the marks assigned to each question have been reduced. There are four parts to this paper. There are 30 questions for a total of 80 marks.

Internal Assessment Marking Scheme

Sr. No	Activity	Marks
1	Pen Paper Test	10
2	Portfolio	05
3	Lab Practical	05
TOTAL		20

Class 10 Maths Chapter-Wise Marks Weightage

Sr. No	Chapter Name	Marks
1	Number System	6
2	Algebra	20
3	Coordinate Geometry	6
4	Geometry	15
5	Trigonometry	12
6	Mensuration	10
7	Statistics and Probability	11

CBSE Class 10 Maths Question Paper Design 2022-23

Sr. No	Type of Questions	Total Marks	Weightage
1	Memory: Terms, answers, basic concepts and recalling of facts. Understanding: Compare, organise, interpret, translate, describe and stating main ideas.	43	54
2	Apply: Solving problems by applying the knowledge, facts, theorems and rules in different ways	19	24
3	Analyse: Analyse the information provided and make inferences. Evaluate: Opine about the validity of quality of work and validity of ideas. Create: Creating alternative solutions to the given problem and grouping the elements.	18	22
TOTAL		80	100

How will CBSE Class 10 Maths Solutions Help You to Score High Marks in Board Exams?

Deeper Understanding of Topics: More than knowing the answer to a problem or how to solve a specific type of question, it is critical to comprehend the ideas of algebra, geometry, statistics, and trigonometry in order to score well in Class 10 Maths. Extramarks 10th CBSE Maths solutions make it easy to understand the underlying concepts while answering questions.

Ready Reference for Practice: Students should make it a point to practice Mathematics exercises on a regular basis. As a result, anytime you intend to practice after having previously solved exercises, you may not recall everything, which is why having a ready reference for practice is vital.

Creating a Solid Foundation: While answering exercise questions is important, it is also necessary to have a good conceptual understanding. There are two reasons behind this – 1) Exam questions may be difficult to answer, but possessing conceptual understanding makes them simple to solve. 2) As you proceed to the following level, the same concepts are utilized to explain more complicated subjects, emphasizing the need for a solid foundation in Maths.

No Delay in Understanding: When students face a challenge, they may have to wait for some time with teachers to properly understand the problems and clear their questions. Class 10 Maths NCERT Solutions, on the other hand, assist students in quickly grasping key topics and avoiding delays in comprehending Mathematics.

Why Refer NCERT Maths Class 10 Solutions for Board Exams?

Class 10 is a critical time in a student’s life since the grades they obtain influence their future career route. Getting the top results on the board exams may be difficult if you do not have the correct study resources. The experts at Extramarks create NCERT solutions for class 10 Maths on a subject-by-subject basis for this purpose.

NCERT Solutions will guide you in dealing with the concepts that you find difficult to solve. These solutions use step-by-step procedures to ensure that you grasp each topic and perform well on your Class 10 board exams.

Important Tips for Scoring Well in the Board Exams

You should be familiar with the Class 10 mathematics test pattern, such as the number of very short answer questions, short answer questions, and long answer questions.
Before you begin your test preparation, familiarise yourself with the CBSE class 10 Mathematics syllabus. It will assist you in developing a better study strategy for yourself. You will also learn how many topics and chapters must be finished during the full session.
Remember the key formulas from each chapter. Keep a separate notebook for the formulas and revise it on a regular basis. This will save you time while answering questions throughout the exam.
Make a schedule before you begin studying. Keeping timetables will keep you on track and allow you to study at a consistent pace. You can modify them according to your strengths in the subject and topics.
Try solving questions from the NCERT Solutions for Class 10 Mathematics about whatever is taught in class. You can also consult other books, such as RD Sharma, for additional practice, but first complete the NCERT textbook’s exercise questions.
Take a walk or exercise for a few minutes during your study breaks. In between your studies, doing 10-20 minutes of exercise will help you keep fit and psychologically prepared for future studies.
As the exam approaches, try not to be stressed. The night before the exam, get a decent night’s sleep. Before the exam, do not study any new topics. Be calm and relaxed.
When revising, pay careful attention to the topics from which most questions are usually asked and carry more marks. Also, before the exam, answer the important questions.

Q.1

\begin{array}{l} In which of the following situations, does the list of \\ numbers involved make an arithmetic progression, \\ and why? \\ (i) The taxi fare after each km when the fare is ₹ 15 \\ for the first km and ₹ 8 for each additional km. \\ (ii) The amount of air present in a cylinder when a \\ vacuum pump removes \frac{1}{4} of the air remaining \\ in the cylinder at a time. \\ (iii) The cost of digging a well after every metre of \\ digging, when it costs ₹ 150 for the first metre \\ and rises by ₹ 50 for each subsequent metre. \\ (iv) The amount of money in the account every year, \\ when ₹ 10000 is deposited at compound interest \\ at 8 % per annum. \end{array}

Ans.

$\begin{array}{l} (i) \\ Fare for the first km = ₹ 15 \\ Fare for the second km = ₹ 15 + ₹ 8 = ₹ 23 \\ Fare for the third km = ₹ 15 + ₹ 8 + ₹ 8 = ₹ 31 \\ and so on. \\ Therefore, fare for each successive kilometre in rupees \\ are: \\ 15, 23, 31, 39, . .. \\ Difference between fares for a particular kilometre and its \\ preceding kilometre is ‘ 8. \\ Therefore, 15, 23, 31, 39, . .. forms an AP. \\ (ii) \\ Let the amount of air present in the cylinder initially is V. \\ The vaccum pump removes \frac{1}{4} of air remaining in the \\ cylinder each time. \\ Therefore, amounts of air in the cylinder after each \\ successive removal by the vacuum pump will be \\ V, \frac{3V}{4}, (\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}), \{(\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}) - (\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}) \times \frac{1}{4}\}, . .. \\ i.e., V, \frac{3V}{4}, {(\frac{3}{4})}^{2} V, {(\frac{3}{4})}^{3} V, . .. \\ Here, the differences between a term and its preceding term \\ are not equal. Therefore, list of numbers involved does \\ not make an A.P. \\ (iii) \\ According to question, cost in rupees of digging first \\ metre and thereafter each subsequent metre are \\ respectively: \\ 150, 150+50, 150+50+50, 150+50+50+50, . .. \\ 150, 200, 250, 300, . .. \\ Differences between a term and its preceding term in the \\ above list of numbers are equal. \\ Therefore, 150, 200, 250, 300, . .. \\ forms an AP. \\ (iv) \\ Amount at the end of first year=10000 {(1+ \frac{8}{100})}^{1} = ₹ 10800 \\ Amount at the end of second year=10800 {(1+ \frac{8}{100})}^{1} = ₹ 11664 \\ Amount at the end of third year=11664 {(1+ \frac{8}{100})}^{1} = ₹ 12597.12 \\ Obviously, differences between amounts in any two successive \\ years are not equal. Therefore, list of numbers involved here \\ does not make an AP. \end{array}$

Q.2 ABCD is a cyclic quadrilateral. Find its all angles.

Ans.

$\begin{array}{l} We know that sum of the measures of the opposite \\ {angles of a cyclic quadrilateral is 180}^{°} . \\ Therefore, \\ ∠ A + ∠ C = 180 \\ \Rightarrow 4 y + 20 - 4 x = 180 \\ \Rightarrow x - y = - 40 . .. (i) \\ Also, \\ ∠ B + ∠ D = 180 \\ \Rightarrow 3 y - 5 - 7 x + 5 = 180 \\ \Rightarrow - 7 x + 3 y = 180 . .. (ii) \\ Multiplying equation (i) by 3, we get \\ 3 x - 3 y = - 120 . .. (iii) \\ Adding equations (ii) and (iii), we get \\ - 7 x + 3 x = 60 \Rightarrow x = - 15 \\ Putting value of x in equation (i), we get \\ - 15 - y = - 40 \\ \Rightarrow y = 25 \\ Thus, \\ ∠ A = 4 y + 20 = 4 \times 25 + 20 = 120 ° \\ ∠ B = 3 y - 5 = 3 \times 25 - 5 = 70 ° \\ ∠ C = - 4 x = - 4 \times (- 15) = 60 ° \\ ∠ D = - 7 x + 5 = - 7 \times (- 15) + 5 = 110 ° . \end{array}$

Q.3

\begin{array}{l} Solve the following pair of linear equations: \\ (i)    px+qy = p-q                 (ii) ax+by = c \\ qx-py = p+q                        bx+ay = 1+c \\ (iii) \frac{x}{a} – \frac{y}{b} {= 0                           (iv) (a-b)x+(a+b)y = a}^{2} {-2ab-b}^{2} \\ {ax+by = a}^{2} {+b}^{2} {(a+b)(x+y) = a}^{2} {+b}^{2} \\ (v)  152x-378y = -74 \\ -378x+152y = -604 \end{array}

Ans.

$\begin{array}{l} (i) \\ The given equations can be written as: \\ px + qy - (p - q) = 0 . .. (1) \\ qx - py - (p + q) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- pq - q^{2} - (p^{2} - pq)} = \frac{y}{- pq + q^{2} + p^{2} + pq} = \frac{1}{- p^{2} - q^{2}} \\ \Rightarrow \frac{x}{- q^{2} - p^{2}} = \frac{y}{q^{2} + p^{2}} = \frac{1}{- p^{2} - q^{2}} \\ \Rightarrow x = 1 and y = - 1 \end{array}$ $\begin{array}{l} (ii) \\ The given equations can be written as: \\ ax + by - c = 0 . .. (1) \\ bx + ay - (1 + c) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- b - bc + ac} = \frac{y}{- bc + a + ac} = \frac{1}{a^{2} - b^{2}} \\ \Rightarrow \frac{x}{c (a - b) - b} = \frac{y}{c (a - b) + a} = \frac{1}{a^{2} - b^{2}} \\ \Rightarrow x = \frac{c (a - b) - b}{a^{2} - b^{2}} and y = \frac{c (a - b) + a}{a^{2} - b^{2}} \\ (iii) \\ The given equations can be written as: \\ bx - ay = 0 . .. (1) \\ ax + by - (a^{2} + b^{2}) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{a (a^{2} + b^{2})} = \frac{y}{b (a^{2} + b^{2})} = \frac{1}{a^{2} + b^{2}} \\ \Rightarrow x = a and y = b \end{array}$ $\begin{array}{l} (iv) \\ The given equations can be written as: \\ (a - b) x + (a + b) y - (a^{2} - 2 ab - b^{2}) = 0 . .. (1) \\ (a + b) x + (a + b) y - (a^{2} + b^{2}) = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- (a + b) (a^{2} + b^{2}) + (a + b) (a^{2} - 2 ab - b^{2})} \\ = \frac{y}{- (a + b) (a^{2} - 2 ab - b^{2}) + (a - b) (a^{2} + b^{2})} = \frac{1}{a^{2} - b^{2} - {(a + b)}^{2}} \\ \Rightarrow \frac{x}{(a + b) (- a^{2} - b^{2} + a^{2} - 2 ab - b^{2})} \\ = \frac{y}{- a^{3} + 2 a^{2} b + {ab}^{2} - a^{2} b + 2 {ab}^{2} + b^{3} + a^{3} + {ab}^{2} - a^{2} b - b^{3}} \\ = \frac{1}{a^{2} - b^{2} - a^{2} - b^{2} - 2 ab} \\ \Rightarrow \frac{x}{- 2 b {(a + b)}^{2}} = \frac{y}{4 {ab}^{2}} = \frac{1}{- 2 b (a + b)} \\ \Rightarrow x = a + b and y = \frac{- 2 ab}{a + b} \\ (v) \\ 152 x - 378 y = - 74 \\ - 378 x + 152 y = - 604 \\ The given equations can be written as: \\ 76 x - 189 y + 37 = 0 . .. (1) \\ - 189 x + 76 y + 302 = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{- 37 \times 76 - 189 \times 302} = \frac{y}{- 189 \times 37 - 76 \times 302} = \frac{1}{76^{2} - 189^{2}} \\ \Rightarrow x = - \frac{37 \times 76 + 189 \times 302}{76^{2} - 189^{2}} = 2 \\ and y = \frac{- 189 \times 37 - 76 \times 302}{76^{2} - 189^{2}} = 1 \end{array}$

Q.4 Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis

Ans.

The given equations are
5x – y = 5 or y = 5(x – 1) …(1)
3x – y = 3 or y = 3(x – 1) …(2)
We find the value of y when x is zero and the value of x when y is zero for both equations and write the corresponding values in tables as below.

x	0	1
y = 5(x – 1)	–5	0

(1)

x	0	1
y = 3(x – 1)	–3	0

(2)
Now, we draw graphs of the given equations as given below.

From the graph above, we find that the co-ordinates of the vertices of the triangle formed by lines and the y-axis are (1, 0), (0, –3) and (0, –5).

Q.5

$\begin{array}{l} In a ΔABC, ∠ C=3 ∠ B=2 (∠ A+ ∠ B) . Find the \\ three angles. \end{array}$

Ans.

$\begin{array}{l} Given that, \\ ∠ C = 3 ∠ B = 2 (∠ A + ∠ B) \\ \Rightarrow 3 ∠ B = 2 ∠ A + 2 ∠ \\ \Rightarrow ∠ B = 2 ∠ A \\ \Rightarrow 2 ∠ A - ∠ B = 0 . .. (1) \\ We know that the sum of measures of all angles of \\ a triangle is 180 ° . Therefore, \\ ∠ A + ∠ B + ∠ C = 180 ° \\ \Rightarrow ∠ A + ∠ B + 3 ∠ B = 180 ° \\ \Rightarrow ∠ A + 4 ∠ B = 180 ° . .. (2) \\ Multiplying equation (1) by 4, we get \\ 8 ∠ A - 4 ∠ B = 0 . .. (3) \\ Addi ng equations (2) and (3), we get \\ 9 ∠ A = 180° \Rightarrow ∠ A = 20 ° \\ From equation (2), we get \\ 20° + 4 ∠ B = 180 ° \\ \Rightarrow 4 ∠ B = 160 ° \\ \Rightarrow ∠ B = 40 ° \\ and ∠ C = 3 ∠ B \Rightarrow ∠ C = 120 ° . \\ Therefore, ∠ A, ∠ B and ∠ C are respectively \\ 20°, 40° and 120° . \end{array}$

Q.6 The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans.

Let the number of rows be x and the number of students in a row be y.
Total students in the class
= (Number of rows) × (Number of students in a row)
= xy
Given that if 3 students are extra in a row, then there would be 1 row less.
Therefore,
Total number of students = (x – 1)(y + 3)
or xy =(x-1)(y+3)=xy – y + 3x – 3
or 3x – y – 3 = 0 …(1)
It is also given that if 3 students are less in a row, then there would be 2 rows more.
Therefore,
Total number of students = (x+2)(y– 3)
or xy = (x+2)(y– 3) = xy + 2y – 3x – 6
or 3x – 2y + 6 = 0 …(2)
Subtracting equation (2) from equation (1),
– y + 2y = 3 + 6
or y= 9
By using equation (1)
3x – 9 = 3
or 3x = 12
or x = 4
Number of rows = x = 4
Number of students in each row = y = 9
Hence, number of students in the class = 9 × 4 = 36.

Q.7

$\begin{array}{l} A train covered a certain distance at a uniform \\ speed. If the train would have been 10 km/h faster, \\ it would have taken 2 hours less than the scheduled \\ time. And, if the train were slower by 10 km/h; it \\ would have taken 3 hours more than the scheduled \\ time. Find the distance covered by the train. \end{array}$

Ans.

$\begin{array}{l} Let the distance covered by the train is x km and the \\ uniform speed of the train is y km/h. Then time taken \\ to travel this distance is \frac{x}{y} hour. \\ According to question, \\ \frac{x}{y} - 2 = \frac{x}{y + 10} \\ \Rightarrow (x - 2 y) (y + 10) = xy \\ \Rightarrow xy + 10 x - 2 y^{2} - 20 y = xy \\ \Rightarrow 10 x - 2 y^{2} - 20 y = 0 . .. (1) \\ Also, \\ \frac{x}{y} + 3 = \frac{x}{y - 10} \\ \Rightarrow (x + 3 y) (y - 10) = xy \\ \Rightarrow xy - 10 x + 3 y^{2} - 30 y = xy \\ \Rightarrow - 10 x + 3 y^{2} - 30 y = 0 . .. (2) \\ Adding equations (1) and (2), we get \\ 10 x - 2 y^{2} - 20 y = 0 \\ \underline{- 10 x + 3 y^{2} - 30 y = 0} \\ y^{2} - 50 y = 0 \\ \Rightarrow y (y - 50) = 0 \\ \Rightarrow y = 0 or y = 50 \end{array}$ $\begin{array}{l} Here, speed can not be taken as zero unit as then distance \\ travelled would be zero unit. \\ So, y = 50 \\ Putting this value of y in equation (1), we get \\ 10 x - 2 {(50)}^{2} - 20 \times 50 = 0 \\ \Rightarrow 10 x - 5000 - 1000 = 0 \\ \Rightarrow x = 600 \\ Hence, distance covered by the train is 600 km. \end{array}$

Q.8

$\begin{array}{l} One says, “Give me a hundred, friend! I shall then \\ become twice as rich as you”. The other replies, \\ “If you give me ten, I shall be six times as rich as \\ you”. Tell me what is the amount of their \\ (respective) capital? \\ [From the Bijaganita of Bhaskara II] \\ [Hint: x + 100 = 2 (y - 100), y + 10 = 6 (x - 10)] \end{array}$

Ans.

$\begin{array}{l} Let their respective capitals are ₹ x and ₹ y . \\ According to question, \\ x + 100 = 2 (y - 100) \\ \Rightarrow x - 2 y = - 300 . .. (1) \\ Also, \\ y + 10 = 6 (x - 10) \\ \Rightarrow y - 6 x = - 70 \\ \Rightarrow y = 6 x - 70 . .. (2) \\ Putting this value of y in equation (1), we get \\ x - 2 (6 x - 70) = - 300 \\ \Rightarrow x - 12 x = - 300 - 140 \\ \Rightarrow x = \frac{- 440}{- 11} = 40 \\ Putting this value of x in equation (2), we get \\ y = 6 \times 40 - 70 = 240 - 70 = 170 \\ Hence, their respective capitals are ₹ 40 and ₹ 170 . \end{array}$

Q.9

$\begin{array}{l} The ages of two friends Ani and Biju differ by 3 years . \\ Ani ‘ s father Dharam is twice as old as Ani and Biju is \\ twice as old as his sister Cathy . The ages of Cathy and \\ Dharam differ by 30 years . Find the ages of Ani and Biju . \end{array}$

Ans.

$\begin{array}{l} Let the ages of Ani and Biju be x and y years respectively . \\ If Ani is older than Biju then \\ x - y = 3 . .. (1) \\ Ani ‘ s father Dharam is twice as old as Ani . So, \\ Dharam ‘ s age = 2 x \\ The ages of Cathy and Dharam differ by 30 years . \\ Therefore, \\ Cathy ‘ s age = 2 x - 30 \\ Biju is twice as old as Cathy . Therefore, \\ y = 2 (2 x - 30) . .. (2) \\ Putting the value of y from equation (2) in equation (1), \\ we get \\ x - 2 (2 x - 30) = 3 \\ \Rightarrow x - 4 x + 60 = 3 \\ \Rightarrow - 3 x = 3 - 60 = - 57 \\ \Rightarrow x = 19 \\ From equation (1), we get \\ y = 19 - 3 = 16 \\ Hence, Ani is 19 years old and Biju is 16 years old . \\ Again, if Biju is older than Ani then \\ y - x = 3 . .. (3) \\ Again, putting the value of y from (2) in (3), we get \\ 2 (2 x - 30) - x = 3 \\ \Rightarrow 4 x - 60 - x = 3 \\ \Rightarrow 3 x = 3 + 60 = 63 \\ \Rightarrow x = 21 \\ Putting this value of x in equation (3), we get \\ y = 21 + 3 = 24 \\ Hence, in this case Ani is of 21 years and Biju is of 24 years . \end{array}$

Q.10 Formulate the following problem as a pair of linear equations and hence find their solution:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Ans.

(i)
Let speed of Ritu in still water be x km/h and the speed of stream be y km/h.
Speed of Ritu while rowing upstream = (x– y) km/h
Speed of Ritu while rowing downstream = (x + y) km/h
According to question,
2(x + y) = 20
or x + y = 10 …(1)
Also,
2(x – y) = 4
or x – y = 2 …(2)
Adding equations (1) and (2), we get
2x = 12
or x = 6
Putting this in equation (1), we get
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of current is 4 km/h.

(ii)

$\begin{array}{l} Let the number of days taken by a woman to finish the work \\ is x and the number of days taken by a man to finish the \\ work is y . \\ Therefore, \\ Part of the work finished by a woman in 1 day = \frac{1}{x} \\ and \\ part of the work finished by a man in 1 day = \frac{1}{y} \\ Given that \\ the . \\ Therefore, \\ 4 (\frac{2}{x} + \frac{5}{y}) = 1 \\ \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \end{array}$ $\begin{array}{l} It is also given that \\ the work . \\ Therefore, \\ 3 (\frac{3}{x} + \frac{6}{y}) = 1 \\ \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3} \\ P utting \frac{1}{x} = p and \frac{1}{y} = q in the above equations, we get \\ 2p + 5q = \frac{1}{4} \Rightarrow 8 p + 20 q = 1 \\ 3p + 6q = \frac{1}{3} \Rightarrow 9 p + 18 q = 1 \\ By cross-multiplication method, we get \\ \frac{p}{- 20 - (- 18)} = \frac{q}{- 9 - (- 8)} = \frac{1}{144 - 180} \\ \Rightarrow \frac{p}{- 2} = \frac{q}{- 1} = \frac{1}{- 36} \\ \Rightarrow \frac{p}{- 2} = \frac{1}{- 36} and \frac{q}{- 1} = \frac{1}{- 36} \\ \Rightarrow p = \frac{1}{18} and q = \frac{1}{36} \\ \Rightarrow p = \frac{1}{x} = \frac{1}{18} and q = \frac{1}{y} = \frac{1}{36} \\ \Rightarrow x = 18 and y = 36 \\ Hence, number of days taken by a woman = 18 days \\ and number of days taken by a man = 36 days. \end{array}$ $\begin{array}{l} (iii) \\ Let the speeds of train and bus be x km/h and y km/h respectively. \\ According to question, \\ \frac{60}{x} + \frac{240}{y} = 4 . .. (1) \\ and \\ \frac{100}{x} + \frac{200}{y} = \frac{25}{6} . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in these equations, we get \\ 60 p + 240 q = 4 . .. (3) \\ and 100 p + 200 q = \frac{25}{6} \\ \Rightarrow 600 p + 1200 q = 25 . .. (4) \\ Multiplying equation (3) by 10, we get \\ 600 p + 2400 q = 40 . .. (5) \\ Subtracting equation (4) from (5), we get \\ 1200q = 15 \Rightarrow q = \frac{1}{80} \\ Substituting in equation (3), we get \\ 60p = 1 \Rightarrow p = \frac{1}{60} \\ Now, p = \frac{1}{x} = \frac{1}{60} and q = \frac{1}{y} = \frac{1}{80} \end{array}$

$\begin{array}{l} \Rightarrow x = 60 km/h and y = 80 km/h \\ Hence, speed of train = 60 km/h \\ and speed of bus = 80 km/h \end{array}$

Q.11

$\begin{array}{l} Solve the following pair of equations by reducing them to a pair of linear equations: \\ (i) \frac{1}{2x} + \frac{1}{3y} =2 (ii) \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} =–1 \\ (iii) \frac{4}{x} + 3y = 14 (iv) \frac{5}{x – 1} + \frac{1}{y – 2} = 2 \\ \frac{3}{x} - 4y = 23 \frac{6}{x – 1} - \frac{3}{y – 2} = 1 \\ (v) \frac{7x – 2y}{xy} = 5 (vi) 6x + 3y = 6xy \\ \frac{8x + 7y}{xy} = 15 2x + 4y = 5xy \\ (vii) \frac{10}{x + y} + \frac{2}{x – y} = 4 (viii) \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \\ \frac{15}{x + y} - \frac{5}{x – y} = –2 \frac{1}{2(3x + y)} - \frac{1}{2(3x – y)} = \frac{–1}{8} \end{array}$

Ans.

$\begin{array}{l} (i) \\ Let \frac{1}{x} = p and \frac{1}{y} = q \\ then given equations can be written as: \\ \frac{p}{2} + \frac{q}{3} = 2 \Rightarrow 3 p + 2 q - 12 = 0 . .. (1) \\ \frac{p}{3} + \frac{q}{2} = \frac{13}{6} \Rightarrow 2 p + 3 q - 13 = 0 . .. (2) \\ Using cross-multiplication method, we get \\ \frac{p}{- 26 - (- 36)} = \frac{q}{- 24 - (- 39)} = \frac{1}{9 - 4} \\ \Rightarrow \frac{p}{10} = \frac{q}{15} = \frac{1}{5} \\ \Rightarrow \frac{p}{10} = \frac{1}{5} and \frac{q}{15} = \frac{1}{5} \\ \Rightarrow p = 2 and q=3 \\ \Rightarrow \frac{1}{x} = 2 and \frac{1}{y} = 3 \\ \Rightarrow x = \frac{1}{2} and y= \frac{1}{3} \\ (ii) \\ Given that \\ \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ and \\ \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1 \\ Let \frac{1}{\sqrt{x}} = p and \frac{1}{\sqrt{y}} = q, then we get \\ 2 p + 3 q = 2 . .. (1) \\ 4 p - 9 q = - 1 . .. (2) \\ Multiplying equation (1) by 3, we get \\ 6p + 9q = 6 \\ Adding this to equation (2), we get \\ 10p = 5 \\ \Rightarrow p= \frac{1}{2} \\ Putting value of p in equation (1), we get \\ 2 \times \frac{1}{2} + 3 q = 2 \\ \Rightarrow 3 q = 1 \\ \Rightarrow q = \frac{1}{3} \end{array}$ $\begin{array}{l} Now, \\ p = \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4 \\ q = \frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9 \\ Hence, x = 4 and y = 9 . \\ (iii) \\ Given that \\ \frac{4}{x} + 3 y = 14 \\ \frac{3}{x} - 4 y = 23 \\ Substituting \frac{1}{x} = p in above equations, we get \\ 4p + 3 y - 14 = 0 . .. (1) \\ 3 p - 4 y - 23 = 0 . .. (2) \\ By cross-multiplication method, we get \\ \frac{p}{- 69 - 56} = \frac{y}{- 42 - (- 92)} = \frac{1}{- 16 - 9} \\ \Rightarrow \frac{p}{- 125} = \frac{y}{50} = \frac{1}{- 25} \\ \Rightarrow p = 5 and y = - 2 \\ \Rightarrow p = \frac{1}{x} = 5 \Rightarrow x = \frac{1}{5} and y = - 2 \end{array}$

$\begin{array}{l} (iv) \\ Given that \\ \frac{5}{x - 1} + \frac{1}{y - 2} = 2 \\ and \\ \frac{6}{x - 1} - \frac{3}{y - 2} = 1 \\ Putting \frac{1}{x - 1} = p and \frac{1}{y - 2} = q, we get \\ 5p + q = 2 . .. (1) \\ 6p - 3 q = 1 . .. (2) \\ multiplying equation (1) by 3, we get \\ 15p + 3 q = 6 . .. (3) \\ Adding (2) and (3), we get \\ 21 p = 7 \Rightarrow p = \frac{1}{3} \\ Putting value of p in equation (1), we get \\ 5 \times \frac{1}{3} + q = 2 \Rightarrow q = 2 - \frac{5}{3} = \frac{1}{3} \\ Now, \\ p = \frac{1}{x - 1} = \frac{1}{3} \Rightarrow x - 1 = 3 \Rightarrow x = 4 \\ and q = \frac{1}{y - 2} = \frac{1}{3} \Rightarrow y - 2 = 3 \Rightarrow y = 5 \\ ∴ x = 4 and y = 5 . \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (v) \\ We have \\ \frac{7 x - 2 y}{xy} = 5 \Rightarrow \frac{7}{y} - \frac{2}{x} = 5 . .. (1) \\ \frac{8 x + 7 y}{xy} = 15 \Rightarrow \frac{8}{y} + \frac{7}{x} = 15 . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in equation (1) and (2), we get \\ - 2 p + 7 q - 5 = 0 . .. (3) \\ 7p + 8 q - 15 = 0 . .. (4) \\ By cross-multiplication method, we get \\ \frac{p}{- 105 - (- 40)} = \frac{q}{- 35 - 30} = \frac{1}{- 16 - 49} \\ \Rightarrow \frac{p}{- 65} = \frac{q}{- 65} = \frac{1}{- 65} \\ \Rightarrow \frac{p}{- 65} = \frac{1}{- 65} and \frac{q}{- 65} = \frac{1}{- 65} \\ \Rightarrow p = 1 and q = 1 \\ \Rightarrow p = \frac{1}{x} = 1 and q = \frac{1}{y} = 1 \\ \Rightarrow x = 1 and y = 1 \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (vi) \\ We have \\ 6 x + 3 y = 6 xy \Rightarrow \frac{6}{y} + \frac{3}{x} = 6 . .. (1) \\ and \\ 2 x + 4 y = 5 xy \Rightarrow \frac{2}{y} + \frac{4}{x} = 5 . .. (2) \\ Putting \frac{1}{x} = p and \frac{1}{y} = q in above equations, we get \\ 3p + 6 q - 6 = 0 . .. (3) \\ 4p + 2 q - 5 = 0 . .. (4) \\ By cross-multiplication method, we have \\ \frac{p}{- 30 - (- 12)} = \frac{q}{- 24 - (- 15)} = \frac{1}{6 - 24} \\ \Rightarrow \frac{p}{- 18} = \frac{q}{- 9} = \frac{1}{- 18} \\ \Rightarrow \frac{p}{- 18} = \frac{1}{- 18} and \frac{q}{- 9} = \frac{1}{- 18} \\ \Rightarrow p = 1 and q = \frac{1}{2} \\ \Rightarrow p = \frac{1}{x} = 1 and q = \frac{1}{y} = \frac{1}{2} \\ \Rightarrow x = 1 and y = 2 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (vii) We have \\ \frac{10}{x + y} + \frac{2}{x - y} = 4 \\ and \frac{15}{x + y} - \frac{5}{x - y} = - 2 \\ Putting \frac{1}{x + y} = p and \frac{1}{x - y} = q in above equations, we get \\ 10p + 2 q - 4 = 0 . .. (3) \\ and \\ 15p - 5 q + 2 = 0 . .. (4) \\ By cross-multiplication method, we have \\ \frac{p}{4 - 20} = \frac{q}{- 60 - 20} = \frac{1}{- 50 - 30} \\ \Rightarrow \frac{p}{- 16} = \frac{q}{- 80} = \frac{1}{- 80} \\ \Rightarrow \frac{p}{- 16} = \frac{1}{- 80} and \frac{q}{- 80} = \frac{1}{- 80} \\ \Rightarrow p = \frac{1}{5} and q = 1 \\ \Rightarrow p = \frac{1}{x + y} = \frac{1}{5} and q = \frac{1}{x - y} = 1 \\ Thus we have \\ x + y = 5 . .. (5) \\ and x - y = 1 . .. (6) \\ Adding equations (5) and (6), we get \\ 2x = 6 \Rightarrow x = 3 \end{array}$ $\begin{array}{l} Subtracting equation (6) from equation (5), we get \\ y = 2 \\ Hence, x = 3 and y = 2 . \\ (viii) \\ We have \\ \frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4} \\ and \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x + y)} = \frac{- 1}{8} \\ Putting \frac{1}{3 x + y} = p and \frac{1}{3 x - y} = q in above equations, we get \\ p + q = \frac{3}{4} . .. (3) \\ and \frac{p}{2} - \frac{q}{2} = \frac{- 1}{8} \\ \Rightarrow p - q = \frac{- 1}{4} . .. (4) \\ Adding equations (3) and (4), we get \\ 2p = \frac{3}{4} - \frac{1}{4} \Rightarrow p = \frac{1}{4} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Now, \\ p = \frac{1}{3 x + y} = \frac{1}{4} \\ \Rightarrow 3 x + y = 4 . .. (5) \\ Also, \\ q = \frac{1}{3 x - y} = \frac{1}{2} \\ \Rightarrow 3 x - y = 2 . .. (6) \\ Adding equations (5) and (6), we get \\ 6x = 6 \Rightarrow x = 1 \\ Substitute x = 1 in equation (5), we get \\ 3 + y = 4 \Rightarrow y = 1 \\ Hence, x = 1 and y = 1 . \end{array}$

Q.12

$\begin{array}{l} Form the pair of linear equations in the following \\ problems and find their solutions (if they exist) \\ by any algebraic method: \\ (i) A part of monthly hostel charges is fixed and the \\ remaining depends on the number of days one \\ has taken food in the mess. When a student A \\ takes food for 20 days she has to pay ₹ 1000 \\ as hostel charges whereas a student B, who \\ takes food for 26 days, pays ₹ 1180 as hostel \\ charges. Find the fixed charges and the cost of \\ food per day. \\ (ii) A fraction becomes \frac{1}{3} when 1 is subtracted from \\ the numerator and it becomes \frac{1}{4} when 8 is added \\ to its denominator. Find the fraction. \\ (iii) Yash scored 40 same time. If the cars travel in the same direction at \\ different speeds, they meet in 5 hours. If they \\ travel towards each other, they meet in 1 hour. \\ What are the speeds of the two cars? \\ (v) The area of a rectangle gets reduced by 9 square \\ units, if its length is reduced by 5 units and \\ breadth is increased by 3 units. If we increase \\ the length by 3 units and the breadth by 2 units, \\ the area increases by 67 square units . Find the \\ dimensions of the rectangle. \end{array}$

Ans.

$\begin{array}{l} (i) \\ Let the fixed charge be ₹ x and the cost of food per day be ₹ y . \\ According to question, \\ x + 20 y = 1000 . .. (1) \\ and \\ x + 26 y = 1180 . .. (2) \\ Subtracting equation (1) from equation (2), we get \\ \underline{\begin{array}{l} x + 26 y = 1180 \\ x + 20 y = 1000 \\ - - - \end{array}} \\ 6 y = 180 \\ or y = 30 \\ Putting this value of y in equation (1), we get \\ x + 20 \times 30 = 1000 \\ or x = 1000 - 600 = 400 \\ Hence, fixed charge is ₹ 400 and the cost of food per \\ day is ₹ 30 . \end{array}$ $\begin{array}{l} (ii) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x - 1}{y} = \frac{1}{3} \\ \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 x - y = 3 . .. (1) \\ Also, \\ \frac{x}{y + 8} = \frac{1}{4} \\ \Rightarrow x = \frac{y + 8}{4} \\ \Rightarrow 4 x = y + 8 \\ \Rightarrow 4 x - y = 8 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 5 \\ Substituting this value in equation (1), we get \\ 3 \times 5 - y = 3 \\ or - y = 3 - 15 \\ or y = 12 \\ Therefore, \\ \frac{x}{y} = \frac{5}{12} \\ Hence, the fraction is \frac{5}{12} . \end{array}$ $\begin{array}{l} (iii) \\ Let the number of right and the number of wrong \\ answers be x and y respectively. \\ According to questions, \\ 3 x - y = 40 . .. (1) \\ and \\ 4 x - 2 y = 50 \\ \Rightarrow 2 (2 x - y) = 2 \times 25 \\ \Rightarrow 2 x - y = 25 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 15 \\ Putting this value of x in equation (2), we get \\ 2 \times 15 - y = 25 \\ \Rightarrow - y = 25 - 30 = - 5 \\ \Rightarrow y = 5 \\ Thus, \\ Number of right answers = x = 15 \\ Number of wrong answers = y = 5 \\ and \\ Number of questions = x + y = 15 + 5 = 20 \end{array}$ $\begin{array}{l} (iv) \\ {Let the speeds of I}^{st} {and 2}^{nd} cars be x km/h and y km/h \\ respectively. \\ According to question, \\ x + y = 100 . .. (1) \\ and \\ 5 x - 100 = 5 y \\ \Rightarrow 5 (x - 20) = 5 y \\ \Rightarrow x - 20 = y \\ \Rightarrow x - y = 20 . .. (2) \\ Adding equations (1) and (2), we get \\ 2 x = 120 \Rightarrow x = 60 \\ Putting this value of x in equation (1), we get \\ 60 + y = 100 \\ \Rightarrow y = 100 - 60 = 40 \\ Thus, \\ {Speed of I}^{st} car = x km/h = 60 km/h \\ {Speed of 2}^{nd} car = y km/h = 40 km/h \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (v) \\ Let length and breadth of the rectangle be x units and y units \\ respectively. \\ According to question, \\ (x - 5) (y + 3) = xy - 9 \\ \Rightarrow xy - 5 y + 3 x - 15 = xy - 9 \\ \Rightarrow - 5 y + 3 x = 15 - 9 = 6 \\ \Rightarrow 3 x - 5 y - 6 = 0 . .. (1) \\ Also, \\ (x + 3) (y + 2) = xy + 67 \\ \Rightarrow xy + 3 y + 2 x + 6 = xy + 67 \\ \Rightarrow 3 y + 2 x = 67 - 6 = 61 \\ \Rightarrow 2 x + 3 y - 61 = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{305 - (- 18)} = \frac{y}{- 12 - (- 183)} = \frac{1}{9 - (- 10)} \\ \Rightarrow x = \frac{323}{19} = 17 and y = \frac{171}{19} = 9 \\ Hence, length and breadth of the rectangle are 17 units and \\ 9 units respectively. \end{array}$

Q.13

$\begin{array}{l} Solve the following pair of linear equations by the \\ substitution and cross-multiplication methods: \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \end{array}$

Ans.

$\begin{array}{l} 8 x + 5 y = 9 . .. (1) \\ 3 x + 2 y = 4 . .. (2) \\ By substitution mwthod: \\ From equation (2), we get \\ y = \frac{4 - 3 x}{2} \\ Substituting this value of y in equation (1), we get \\ 8 x + 5 y = 9 \\ or 8 x + 5 \times \frac{4 - 3 x}{2} = 9 \\ or 16 x + 20 - 15 x = 18 \\ or x = 18 - 20 = - 2 \\ Putting this value of x in equation (2), we get \\ 3 \times (- 2) + 2 y = 4 \\ or - 6 + 2 y = 4 \\ or 2y = 4 + 6 = 10 \\ or y = 5 \end{array}$ $\begin{array}{l} By cross-multiplication method : \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \\ Or \\ 8 x + 5 y - 9 = 0 . .. (1) \\ 3 x + 2 y - 4 = 0 . .. (2) \\ {Here, a}_{1} = 8, b_{1} {= 5, c}_{1} = - 9 \\ a_{2} = 3, b_{2} {= 2, c}_{2} = - 4 \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 20 + 18} = \frac{y}{- 27 + 32} = \frac{1}{16 - 15} \\ \Rightarrow \frac{x}{- 2} = \frac{y}{5} = \frac{1}{1} \\ \Rightarrow x = - 2 and y = 5 \end{array}$

Q.14

$\begin{array}{l} (i) For which values of a and b does the following pair of linear equations have an \\ infinite number of solutions? \\ 2x + 3y =7 \\ (a – b)x + (a + b)y = 3a + b - 2 \\ (ii) For which value of k will the following pair of linear equations have no solution? \\ 3x + y = 1 \\ (2k – 1)x + (k – 1) y = 2k + 1 \end{array}$

Ans.

$\begin{array}{l} (i) \\ Given pair of linear equations are: \\ 2 x + 3 y = 7 \\ (a - b) x + (a + b) y = 3 a + b - 2 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{a - b}, \frac{b_{1}}{b_{2}} = \frac{3}{a + b}, \frac{c_{1}}{c_{2}} = \frac{7}{3 a + b - 2} \\ For a pair of linear equations to have infinitely many solutions: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ So, we need \\ \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3 a + b - 2} \\ or \frac{2}{a - b} = \frac{7}{3 a + b - 2} and \frac{2}{a - b} = \frac{3}{a + b} \\ or 2 (3 a + b - 2) = 7 (a - b) and 2 (a + b) = 3 (a - b) \\ or 6 a + 2 b - 4 = 7 a - 7 b and 2 a - 3 a = - 3 b - 2 b \\ or - a + 9 b - 4 = 0 and a = 5 b \\ Now, we put a = 5 b in - a + 9 b - 4 = 0 . \\ Thus, \\ - 5 b + 9 b = 4 \\ or 4 b = 4 \\ or b = 1 \\ Also, \\ a = 5 b = 5 \times 1 = 5 \end{array}$ $\begin{array}{l} (ii) \\ Given pair of linear equations are: \\ 3 x + y = 1 \\ (2 k - 1) x + (k - 1) y = 2 k + 1 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{2 k - 1}, \frac{b_{1}}{b_{2}} = \frac{1}{k - 1}, \frac{c_{1}}{c_{2}} = \frac{1}{2 k + 1} \\ For a pair of linear equations to have no solution: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ So, the given pair of linear equations have no solution when \\ \frac{3}{2 k - 1} = \frac{1}{k - 1} \\ or 3 k - 3 = 2 k - 1 \\ or 3 k - 2 k = - 1 + 3 \\ or k = 2 \end{array}$

Q.15

$\begin{array}{l} Which of the following pairs of linear equations has unique solution, \\ no solution, or infinitely many solutions. In case there is a unique \\ solution, find it by using cross multiplication method. \\ (i) x - 3y - 3 = 0 (ii) 2x + y = 5 \\ 3x - 9y - 2 = 0 3x + 2y = 8 \\ (iii) 3x - 5y = 20 (iv) x - 3y - 7 = 0 \\ 6x - 10y = 40 3x - 3y - 15 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) x - 3 y - 3 = 0 \\ 3 x - 9 y - 2 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 9} = \frac{1}{3}, \frac{c_{1}}{c_{2}} = \frac{- 3}{- 2} = \frac{3}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have no solution. \\ (ii) 2 x + y = 5 i.e., 2 x + y - 5 = 0 \\ 3 x + 2 y = 8 i.e., 3 x + 2 y - 8 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{5}{8} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 8 - (- 10)} = \frac{y}{- 15 - (- 16)} = \frac{1}{4 - 3} \\ \Rightarrow \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \\ \Rightarrow x = 2 and y = 1 \end{array}$ $\begin{array}{l} (iii) 3 x - 5 y = 20 i.e., 3 x - 5 y - 20 = 0 \\ 6 x - 10 y = 40 i.e., 6 x - 10 y - 40 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 5}{- 10} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 20}{- 40} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have infinitely \\ many solutions. \\ (iv) x - 3 y - 7 = 0 \\ 3 x - 3 y - 15 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 3} = 1, \frac{c_{1}}{c_{2}} = \frac{- 7}{- 15} = \frac{7}{15} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{45 - 21} = \frac{y}{- 21 - (- 15)} = \frac{1}{- 3 - (- 9)} \\ \Rightarrow \frac{x}{24} = \frac{y}{- 6} = \frac{1}{6} \\ \Rightarrow x = 4 and y = - 1 \end{array}$

Q.16 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans.

$\begin{array}{l} (i) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x + 1}{y - 1} = 1 \\ \Rightarrow x + 1 = y - 1 \\ \Rightarrow x - y = - 2 . .. (1) \\ Also, \\ \frac{x}{y + 1} = \frac{1}{2} \\ \Rightarrow x = \frac{y + 1}{2} \\ \Rightarrow 2 x = y + 1 \\ \Rightarrow 2 x - y = 1 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 3 \\ Substituting this value in equation (1), we get \\ y = 5 \\ Hence, the fraction is \frac{3}{5} . \end{array}$ $\begin{array}{l} (ii) \\ Let the present age of Nuri and Sonu be x years \\ and y years respectively . \\ So, five years ago, \\ Nuri ‘ s age = (x - 5) years \\ and \\ Sonu’s age = (y - 5) years \\ According to question, \\ x - 5 = 3 (y - 5) \\ \Rightarrow x - 5 = 3 y - 15 \\ \Rightarrow x - 3 y = - 10 . .. (1) \\ Again, \\ Ten years hence, \\ Nuri ‘ s age = (x + 10) years \\ and \\ Sonu’s age = (y + 10) years \\ According to the question, \\ x + 10 = 2 (y + 10) \\ or x - 2 y = 10 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ y = 20 \\ Putting this value in (2), we get \\ x = 50 \\ Hence, Nuri and Sonu are 50 years \\ and 20 years old respectively . \end{array}$ $\begin{array}{l} (iii) \\ Let the two digit number be 10 x + y . \\ According to question, \\ x + y = 9 . .. (1) \\ and \\ 9 (10 x + y) = 2 (10 y + x) \\ \Rightarrow 90x + 9 y = 20 y + 2 x \\ \Rightarrow 90 x - 2 x + 9 y - 20 y = 0 \\ \Rightarrow 88 x - 11 y = 0 \\ \Rightarrow 8 x - y = 0 . .. (2) \\ Adding equations (1) and (2), we get \\ x = 1 \\ Putting this value of x in equation (2), we get \\ y = 8 \\ Hence, \\ Number = 10 x + y = 10 + 8 = 18 \end{array}$ $\begin{array}{l} (iv) \\ Let number of notes of ₹ 50 and ₹ 100 bexandyrespectively . \\ According to question, \\ x + y = 25 . .. (1) \\ and \\ 50 x + 100 y = 2000 \\ \Rightarrow 50 (x + 2 y) = 2000 \\ \Rightarrow x + 2 y = 40 . .. (2) \\ Subtracting equation (1) from equation(2), we get \\ x + 2 y = 40 \\ \underline{\begin{array}{l} x + y = 25 \\ - - - \end{array}} \\ y = 15 \\ Putting this value in equation (1), we get \\ x = 10 \\ Hence, \\ Meena received 10 notes of ‘ 50 and 15 notes of ‘ 100 . \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (v) \\ Let the fixed charge for first three day be ₹ x and additional \\ charge for each day thereafter be ₹ y . \\ According to question, \\ x + 4 y = 27 . .. (1) \\ and \\ x + 2 y = 21 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x + 4 y = 27 \\ \underline{\begin{array}{l} x + 2 y = 21 \\ - - - \end{array}} \\ 2 y = 6 \\ or y = 3 \\ Putting this value in equation (1), we get \\ x = 27 - 12 = 15 \\ Hence, fixed charge is ₹ 15 and additional charge for \\ each extra day is ₹ 3 . \end{array}$

Q.17

$\begin{array}{l} Solve the following pair of linear equations by the \\ elimination method and the substitution method: \\ (i) x + y = 5 and 2x – 3y = 4 \\ (ii) 3x + 4y = 10 and 2x – 2y = 2 \\ (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 \\ (iv) \frac{x}{2} + \frac{2y}{3} = –1 and x – \frac{y}{3} = 3 \end{array}$

Ans.

$\begin{array}{l} (i) \\ By elimination method : \\ Given that \\ x + y = 5 . .. (1) \\ 2 x - 3 y = 4 . .. (2) \\ We multiply equation (1) by 3 and then add it to \\ equation (2) to eliminate the variable y and get \\ the value of x as follows. \\ 3 x + 3 y = 15 \\ \underline{2 x - 3 y = 4} \\ 5 x = 19 \\ or x = \frac{19}{5} \\ Using this value of x in equation (1), we get \\ \frac{19}{5} + y = 5 \\ or y = 5 - \frac{19}{5} = \frac{6}{5} \\ By substitution method : \\ From equation (1), we get \\ y = 5 - x \\ Substituting this in equation (2), we get \\ 2 x - 3 (5 - x) = 4 \\ or 5 x = 4 + 15 = 19 \\ or x = \frac{19}{5} \\ Substituting this value of x in equation (1), we get \\ \frac{19}{5} + y = 5 \\ or y = 5 - \frac{19}{5} = \frac{6}{5} \end{array}$ $\begin{array}{l} (ii) \\ By elimination method : \\ Given that \\ 3 x + 4 y = 10 . .. (1) \\ 2 x - 2 y = 2 . .. (2) \\ We multiply equation (2) by 2 and then add it to \\ equation (1) to eliminate the variable y and get \\ the value of x as follows. \\ 3 x + 4 y = 10 \\ \underline{4 x - 4 y = 4} \\ 7 x = 14 \\ or x = 2 \\ Using this value of x in equation (1), we get \\ 6 + 4 y = 10 \\ or y = \frac{10 - 6}{4} = 1 \\ By substitution method : \\ From equation (2), we get \\ x = 1 + y \\ Substituting this in equation (1), we get \\ 3 (1 + y) + 4 y = 10 \\ or 7 y = 10 - 3 = 7 \\ or y = 1 \\ Substituting this value of y in equation (2), we get \\ 2 x - 2 = 2 \\ or x = \frac{2 + 2}{2} = 2 \end{array}$ $\begin{array}{l} (iii) \\ By elimination method : \\ Given that \\ 3 x - 5 y - 4 = 0 . .. (1) \\ 9 x = 2 y + 7 . .. (2) \\ We multiply equation (1) by 3 and then subtract \\ equation (2) from it to eliminate the variable x and get \\ the value of y as follows. \\ 9 x - 15 y - 12 = 0 \\ \underline{\begin{array}{l} 9 x = 2 y + 7 \\ - - - \end{array}} \\ - 15 y - 12 = - 2 y - 7 \\ or - 15 y + 2 y = - 7 + 12 = 5 \\ or y = - \frac{5}{13} \\ Using this value of y in equation (2), we get \\ 9 x = - \frac{5}{13} \times 2 + 7 \\ or x = \frac{- 10}{117} + \frac{7}{9} = \frac{- 10 + 91}{117} = \frac{81}{117} = \frac{9}{13} \\ By substitution method : \\ From equation (2), we get \\ x = \frac{2 y + 7}{9} \\ Substituting this in equation (1), we get \\ 3 \times \frac{2 y + 7}{9} - 5 y - 4 = 0 \\ or \frac{2 y + 7}{3} - 5 y = 4 \\ or 2 y + 7 - 15 y = 12 \\ or - 13 y = 12 - 7 \\ or y = - \frac{5}{13} \\ Substituting this value of y in equation (2), we get \\ 9 x = - 2 \times \frac{5}{13} + 7 = \frac{- 10 + 91}{13} \\ or x = \frac{81}{13 \times 9} = \frac{9}{13} \end{array}$ $\begin{array}{l} (iv) \\ By elimination method : \\ Given that \\ \frac{x}{2} + \frac{2 y}{3} = - 1 . .. (1) \\ x - \frac{y}{3} = 3 . .. (2) \\ We multiply equation (2) by 2 and then add it to \\ equation (1) to eliminate the variable y and get \\ the value of x as follows. \\ \frac{x}{2} + \frac{2 y}{3} = - 1 \\ \underline{2 x - \frac{2 y}{3} = 6} \\ \frac{x}{2} + 2 x = 5 \\ or \frac{5 x}{2} = 5 \\ or x = 2 \\ Using this value of x in equation (2), we get \\ 2 - \frac{y}{3} = 3 \\ or - y = 3 (3 - 2) = 3 \\ or y = - 3 \\ By substitution method : \\ From equation (2), we get \\ x = 3 + \frac{y}{3} \\ Substituting this in equation (1), we get \\ \frac{1}{2} (3 + \frac{y}{3}) + \frac{2 y}{3} = - 1 \\ or \frac{9 + y}{6} + \frac{2 y}{3} = - 1 \\ or \frac{9 + y + 4 y}{6} = - 1 \\ or 9 + y + 4 y = - 6 \\ or y = \frac{- 15}{5} = - 3 \\ Substituting this value of y in equation (2), we get \\ x = 3 + \frac{- 3}{3} = 2 \end{array}$

Q.18

$\begin{array}{l} Form the pair of linear equations for the following \\ problems and find their solution by substitution method. \\ (i) The difference betrween two numbers is 26 and one \\ number is three times the other. Find them. \\ (ii) The larger of two supplementary angles exceeds \\ the smaller by 18 degrees. Find them. \\ (iii) The coach of a cricket team buys a 7 bats and 6 balls \\ for ₹ 3800. Later she buys the 3 bats and 5 balls \\ for ₹ 1750. Find the cost of each bat and each ball. \\ (iv) The taxi charges in a city consist of a fixed charge \\ together with the charge for the distance covered. \\ For a distance of 10 km, the charge paid is ₹ 105 and \\ for a journey of 15 km, the charge paid is 155. What \\ are the fixed charges and the charge per km? How \\ much does a person have to pay for travelling a \\ distance of 25 km? \\ (v) A fraction becomes \frac{9}{11}, if 2 is added to both the \\ numerator and the denominator. If, 3 is added to \\ both the numerator and the denominator it \\ becomes \frac{5}{6} . Find the fraction. \\ (vi) Five years hence, the age of Jacob will be three times \\ that of his son. Five years ago, Jacob’s age was seven \\ times that of his son. What are their present ages? \end{array}$

Ans.

(i)
Let the larger number is y and the smaller number is x.
According to question,
y = 3x …(1)
and
y – x = 26 …(2)
We substitute the value of y from equation (1) in equation (2) and get
3x – x = 26
or 2x = 26
or x = 13
Putting this value of x in equation (1), we get
y = 39
Hence, the numbers are 13 and 39.

(ii)
Let the larger angle is y and the smaller angle is x.
According to question,
y – x = 18° …(1)
and
y + x = 180° …(2)
We write y in terms of x from (1) to get
y = 18° + x
Putting this value of y in equation (2), we get
18° + x + x = 180°
or 2x = 180° – 18° = 162°
or x = 81°
Putting this value of x in equation (1), we get
y = 18° + 81° = 99°
Hence, the two supplementary angles are 81° and 99°.

(iii)
Let the cost of one bat and one ball be ₹ x and ₹ y respectively.
According to question,
7x + 6y = ₹ 3800 …(1)
and
3x + 5y = ₹ 1750 …(2)
We write y in terms of x from (1) to get
y= (3800 – 7x)/6
Putting this value of y in equation (2), we get
3x + 5(3800 – 7x)/6 = 1750
or 18x – 35x = 1750 × 6 – 19000 = –8500
or x = 8500/17 = 500
Putting this value of x in equation (1), we get
7×500 + 6y = 3800
or y = (3800 – 3500)/6 = 50
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

(iv)
Let the fixed charge be ₹ x and charge per km be ₹ y.
According to question,
x + 10y = ₹ 105 …(1)
and
x + 15y = ₹ 155 …(2)
We write x in terms of y from (1) to get
x = 105 – 10y
Putting this value of x in equation (2), we get
105 – 10y + 15y = 155
or 5y = 155 – 105 = 50
or y = 10
Putting this value of y in equation (1), we get
x + 10×10 = 105
or x = 105 – 100 = 5
Hence, the fixed charge is ₹ 5 and charge per km is ₹ 10.
Charge for 25 km = 5 + 25 ×10 = ₹ 255

$\begin{array}{l} (v) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x + 2}{y + 2} = \frac{9}{11} \Rightarrow 11 x - 9 y = - 4 . .. (1) \\ and \\ \frac{x + 3}{y + 3} = \frac{5}{6} \Rightarrow 6 x - 5 y = - 3 . .. (2) \\ From equation (2), we get \\ x = \frac{5 y - 3}{6} \\ We put this value of x in equation (1) and get \\ 11 \frac{5 y - 3}{6} - 9 y = - 4 \\ \Rightarrow 55 y - 33 - 54 y = - 24 \\ \Rightarrow y = - 24 + 33 = 9 \\ Putting this value of y in equation (1), we get \\ 11 x - 9 \times 9 = - 4 \\ \Rightarrow 11 x = 81 - 4 = 77 \\ \Rightarrow x = 7 \\ Hence the fraction is \frac{7}{9} . \end{array}$

(vi)
Let the age of Jacob be x and the age of his son be y.
According to question,
x + 5 = 3(y + 5)
or x – 3y = 10 …(1)
Also,
x – 5 = 7(y – 5)
or x – 7y = –30 …(2)
From equation (1), we find
x = 10 + 3y
We substitute this value of x in equation (2) and get
10 + 3y – 7y = –30
or –4y = –40
or y = 10
Putting this value of y in (1), we get
x – 3×10 = 10
or x = 40
Hence the present age of Jacob is 40 years and that of his son is 10 years.

Q.19 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Ans.

$\begin{array}{l} The given linear equations are: \\ 2 x + 3 y = 11 . .. (1) \\ 2 x - 4 y = - 24 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = \frac{11 - 3 y}{2} \\ We substitute this value of x in equation (2) to get \\ 11 - 3 y - 4 y = - 24 \\ i.e., 11 - 7 y = - 24 \\ i.e., 7 y = 24 + 11 = 35 \\ i.e., y = 5 \\ Putting this value of y in equation (1), we get \\ 2 x + 3 \times 5 = 11 \\ i.e., 2 x = 11 - 15 = - 4 \\ i.e., x = - 2 \\ Now, \\ y = mx + 3 \\ or 5 = - 2 m + 3 \\ or m = - 1 \end{array}$

Q.20

$\begin{array}{l} Solve the following pair of linear equations by the \\ substitution method. \\ (i) x + y = 14 (ii) s - t = 3 \\ x - y = 4 \frac{s}{3} + \frac{t}{2} = 6 \\ (iii) 3 x - y = 3 (iv) 0 .2 x + 0 .3 y = 1 .3 \\ 9 x - 3 y = 9 0 .4 x + 0 .5 y = 2 .3 \\ (v) \sqrt{2} x + \sqrt{3} y = 0 (vi) \frac{3 x}{2} - \frac{5 y}{3} = - 2 \\ \sqrt{3} x - \sqrt{8} y = 0 \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \end{array}$

Ans.

$\begin{array}{l} (i) The given pair of linear equations are: \\ x + y = 14 . .. (1) \\ x - y = 4 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = 14 - y \\ We substitute this value of x in equation (2) to get \\ 14 - y - y = 4 \\ i.e., - 2 y = 4 - 14 = - 10 \\ i.e., y = 5 \\ Putting this value of y in equation (2), we get \\ x - 5 = 4 \\ i.e., x = 9 \\ (ii) The given pair of linear equations are: \\ s - t = 3 . .. (1) \\ \frac{s}{3} + \frac{t}{2} = 6 . .. (2) \\ We express s in terms of t from equation (1) to get \\ s = t + 3 \\ We substitute this value of s in equation (2) to get \\ \frac{t + 3}{3} + \frac{t}{2} = 6 \\ i.e., \frac{5 t + 6}{6} = 6 \\ i.e., 5 t + 6 = 36 \\ i.e., t = 6 \\ Putting this value of t in equation (1), we get \\ s - 6 = 3 \\ i.e., s = 9 \end{array}$ $\begin{array}{l} (iii) The given pair of linear equations are: \\ 3 x - y = 3 . .. (1) \\ 9 x - 3 y = 9 . .. (2) \\ We express y in terms of x from equation (1) to get \\ y = 3 x - 3 \\ We substitute this value of y in equation (2) to get \\ 9 x - 9 x + 9 = 9 \\ i.e., 9 = 9 \\ This statement is true for all values of x and so we can not \\ obtain a specific value of x. We observe that both the given \\ equations are the same as one is derived from another. \\ Therefore, given equations have infinitely many solutions. \\ (iv) The given pair of linear equations are: \\ 0 .2 x + 0 .3 y = 1 .3 . .. (1) \\ 0 .4 x + 0 .5 y = 2 .3 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = (1 .3 - 0 .3 y) / (0 .2) = (13 - 3 y) / 2 \\ We substitute this value of x in equation (2) to get \\ 0.2 (13 - 3 y) + 0 .5 y = 2 .3 \\ i.e., 2 .6 - 0 .6 y + 0 .5 y = 2 .3 \\ i.e., - 0.1 y = 2 .3 - 2 .6 = - 0 .3 \\ i.e., y = 3 \\ Putting this value of y in equation (2), we get \\ 0 .4 x + 0 .5 \times 3 = 2 .3 \\ i.e., x = (2 .3 - 1 .5) / 0 .4 = 8 / 4 = 2 \end{array}$ $\begin{array}{l} (v) The given pair of linear equations are : \\ \sqrt{2} x + \sqrt{3} y = 0 . .. (1) \\ \sqrt{3} x - \sqrt{8} y = 0 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = \frac{\sqrt{3}}{\sqrt{2}} y \\ We substitute this value of x in equation (2) to get \\ \sqrt{3} \frac{\sqrt{3}}{\sqrt{2}} y - \sqrt{8} y = 0 \\ i . e ., \frac{3 - 4}{\sqrt{2}} y = 0 \\ i . e ., y = 0 \\ Putting this value of y in equation (2), we get \\ \sqrt{3} x = 0 \\ i . e ., x = 0 \\ (vi) The given pair of linear equations are : \\ \frac{3 x}{2} - \frac{5 y}{3} = - 2 \\ or 9 x - 10 y = - 12 . .. (1) \\ and \\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \\ or 2 x + 3 y = 13 . .. (2) \\ We express x in terms of y from equation (1) to get \\ x = (- 12 + 10 y) / 9 \\ We substitute this value of x in equation (2) to get \\ 2(\frac{- 12 + 10 y}{9})+ 3 y = 13 \\ i . e ., - 24 + 20 y + 27 y = 117 \\ i . e ., 47 y = 117 + 24 = 141 \\ i . e ., y = \frac{141}{47} = 3 \\ Putting this value of y in equation (2), we get \\ 2 x + 9 = 13 \\ i . e ., x = \frac{4}{2} = 2 \end{array}$

Q.21 Draw the graphs of the equations
x – y + 1 = 0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Ans.

We have,
x – y + 1 = 0
or y = x + 1
We have the following table.

x	0	–1
y = x + 1	1	0

Again,
3x + 2y – 12 = 0
or y = (12–3x)/2
and so we have the following table.

x	0	4
y = (12–3x)/2	6	0

Graphs of the given equations are drawn below and from there we find that coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0); (4, 0) and (2, 3).

Q.22 Given the linear equation 2x+ 3y– 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Ans.

$\begin{array}{l} (i) Intersecting lines: \\ A line which will intersect the given line 2 x + 3 y - 8 = 0 \\ is 3 x + 2 y - 8 = 0 as \\ \frac{a_{1}}{a_{2}} = \frac{2}{3} and \frac{b_{1}}{b_{2}} = \frac{3}{2} impllies that \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} . \\ (ii) P lines: \\ A line which is paralel to the given line 2 x + 3 y - 8 = 0 \\ is 4 x + 6 y - 7 = 0 as \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2} and \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} impllies that \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} . \\ (iii) : \\ A line which is to the given line 2 x + 3 y - 8 = 0 \\ is 4 x + 6 y - 16 = 0 because \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{1}{2} . \end{array}$

Q.23 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans.

Let the width of the garden be x and length be y.
According to the question, length is 4 m more than its width. Therefore,
y − x = 4 …(1)
Also, half the perimeter of a rectangular garden is 36 m.
Therefore,
x + y = 36 …(2)
Now,
y − x = 4 …(1)
or y = x + 4

x	–4	0
y = x + 4	0	4

Similarly,
x + y = 36 …(2)
or y = 36 – x

x	28	20
y = 36 – x	8	16

Graphs of the equations (1) and (2) are drawn below and from there we observe that they intersect at (16, 20).
Therefore, x = 16 and y = 20. Hence, width of the garden is 16 m and length of the garden is 20 m.

Q.24

$\begin{array}{l} Which of the following pairs of linear equations are consistent/inconsistent? \\ If consistent, obtain the solution graphically: \\ (i) x + y = 5, 2x + 2y = 10 \\ (ii) x – y = 8, 3x – 3y = 16 \\ (iii) 2x + y - 6 = 0, 4x – 2y - 4 = 0 \\ (iv) 2x – 2y – 2 = 0, 4x – 4y - 5 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) x + y = 5, 2 x + 2 y = 10 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 1, b_{1} = 1, c_{1} = - 5 \\ a_{2} = 2, b_{2} = 2, c_{2} = - 10 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 5}{- 10} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they are consistent. \\ Graphs of the two equations coincide and hence each and \\ every point on this graph is a solution of these equations. \end{array}$

$\begin{array}{l} (ii) x - y = 8, 3 x - 3 y = 16 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 1, b_{1} = - 1, c_{1} = - 8 \\ a_{2} = 3, b_{2} = - 3, c_{2} = - 16 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{3}, \frac{c_{1}}{c_{2}} = \frac{- 8}{- 16} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel \\ and hence they are inconsistent. \end{array}$ $\begin{array}{l} (iii) 2 x + y - 6 = 0, 4 x - 2 y - 4 = 0 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = 1, c_{1} = - 6 \\ a_{2} = 4, b_{2} = - 2, c_{2} = - 4 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = - \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 6}{- 4} = \frac{3}{2} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given linear equations are consistent. \\ Now, \\ 2 x + y - 6 = 0 \Rightarrow y = 6 - 2 x \\ Some points which satisfy this equation are written \\ in the following table. \\ \begin{matrix} x & 0 & 2 & 3 \\ y = 6 - 2 x & 6 & 2 & 0 \end{matrix} \\ Again, \\ 4 x - 2 y - 4 = 0 \Rightarrow y = 2 x - 2 \\ Some points which satisfy this equation are written \\ in the following table. \\ \begin{matrix} x & 0 & 2 & 3 \\ y = 2 x - 2 & - 2 & 2 & 4 \end{matrix} \\ We get the following graphs of the given equations and \\ find that they intersect at (2, 2). Hence, x = 2 and y = 2. \end{array}$

$\begin{array}{l} (iv) 2 x - 2 y - 2 = 0, 4 x - 4 y - 5 = 0 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = - 2, c_{1} = - 2 \\ a_{2} = 4, b_{2} = - 4, c_{2} = - 5 \\ Now, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{1}{3} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 2}{- 5} = \frac{2}{5} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel \\ and hence they are inconsistent. \end{array}$

Q.25

$\begin{array}{l} On comparing the ratios \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}, \frac{c_{1}}{c_{2}}, find out whether \\ the following pair of linear equations are consistent, \\ or inconsistent. \\ (i) 3 x + 2 y = 5; 2 x - 3 y = 7 \\ (ii) 2 x - 3 y = 8; 4 x - 6 y = 9 \\ (iii) \frac{3}{2} x + \frac{5}{3} y = 7; 9 x - 10 y = 14 \\ (iv) 5 x - 3 y = 11; - 10 x + 6 y = - 22 \\ (v) \frac{4}{3} x + 2 y = 8; 2 x + 3 y = 12 \end{array}$

Ans.

$\begin{array}{l} (i) 3 x + 2 y = 5; 2 x - 3 y = 7 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 3, b_{1} = 2, c_{1} = - 5 \\ a_{2} = 2, b_{2} = - 3, c_{2} = - 7 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{3}{2}, \frac{b_{1}}{b_{2}} = - \frac{2}{3}, \frac{c_{1}}{c_{2}} = \frac{5}{7} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pair of linear equations has a unique \\ solution and hence the two linear equations are consistent. \end{array}$ $\begin{array}{l} (ii) 2 x - 3 y = 8; 4 x - 6 y = 9 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 2, b_{1} = - 3, c_{1} = - 8 \\ a_{2} = 4, b_{2} = - 6, c_{2} = - 9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{8}{9} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations are parallel and \\ hence the two linear equations are inconsistent . \end{array}$ $\begin{array}{l} (iii) \frac{3}{2} x + \frac{5}{3} y = 7; 9 x - 10 y = 14 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = \frac{3}{2}, b_{1} = \frac{5}{3}, c_{1} = - 7 \\ a_{2} = 9, b_{2} = - 10, c_{2} = - 14 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \frac{b_{1}}{b_{2}} = \frac{\frac{5}{3}}{- 10} = - \frac{1}{6}, \frac{c_{1}}{c_{2}} = \frac{- 7}{- 14} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pair of linear equations has a unique \\ solution and hence the two linear equations are consistent. \end{array}$ $\begin{array}{l} (iv) 5 x - 3 y = 11; - 10 x + 6 y = - 22 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = 5, b_{1} = - 3, c_{1} = - 11 \\ a_{2} = - 10, b_{2} = 6, c_{2} = 22 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{5}{- 10} = - \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{6} = - \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 11}{22} = - \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they consistent. \end{array}$ $\begin{array}{l} (v) \frac{4}{3} x + 2 y = 8; 2 x + 3 y = 12 \\ Comparing these equations with a_{1} x + b_{1} y + c_{1} = 0 \\ and a_{2} x + b_{2} y + c_{2} = 0, we get \\ a_{1} = \frac{4}{3}, b_{1} = 2, c_{1} = - 8 \\ a_{2} = 2, b_{2} = 3, c_{2} = - 12 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{2}{3}, \frac{c_{1}}{c_{2}} = \frac{- 8}{- 12} = \frac{2}{3} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given linear equations have infinitely many \\ solutions and hence they are consistent. \end{array}$

Q.26

$\begin{array}{l} On c omparing the ratios \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}, \frac{c_{1}}{c_{2}}, find out whether the lines representing the \\ following pairs of linear equations intersect at a point, are parallel or coincident: \\ (i) 5x – 4y + 8 = 0 \\ 7x + 6y – 9 = 0 \\ (ii) 9x + 3y + 12 = 0 \\ 18x + 6y + 24 = 0 \\ (iii) 6x – 3y + 10 = 0 \\ 2x – y + 9 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) 5x-4y+8 = 0 \\ 7x+6y-9 = 0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} = 0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {=5, b}_{1} {=-4, c}_{1} =8 \\ a_{2} {=7, b}_{2} {=6, c}_{2} =-9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{5}{7}, \frac{b_{1}}{b_{2}} = \frac{-4}{6} = \frac{-2}{3} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Therefore, the given pairs of linear equations intersect at a point. \end{array}$ $\begin{array}{l} (ii) 9x+3y+12=0 \\ 18x+6y+24=0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} = 0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {=9, b}_{1} {=3, c}_{1} =12 \\ a_{2} {=18, b}_{2} {=6, c}_{2} =24 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given pairs of linear equations are coincident. \end{array}$ $\begin{array}{l} (iii) 6x-3y+10 = 0 \\ 2x-y+9 = 0 \\ {Comparing these equations with a}_{1} {x+b}_{1} {y+c}_{1} =0 \\ {and a}_{2} {x+b}_{2} {y+c}_{2} = 0, we get \\ a_{1} {= 6, b}_{1} {= -3, c}_{1} = 10 \\ a_{2} {= 2, b}_{2} {= -1, c}_{2} = 9 \\ Also, \\ \frac{a_{1}}{a_{2}} = \frac{6}{2} = 3, \frac{b_{1}}{b_{2}} = \frac{-3}{-1} = 3, \frac{c_{1}}{c_{2}} = \frac{10}{9} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given pairs of linear equations are parallel. \end{array}$

Q.27 Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans.

(i) Let the number of girls and the number of boys be x and y respectively.
According to question,
x + y = 10
x – y = 4
For x + y = 10,
y = 10 – x

x	4	6	8
y = 10 – x	6	4	2

For x – y = 4,
y = x – 4

x	4	6	8
y = x – 4	0	2	4

The graphs of equations are drawn below which shows that the two lines intersect at (7, 3).
Hence the number of boys and the number of girls are 7 and 3 respectively.

(ii)
Let the cost of 1 pencil and the cost of 1 pen be ₹ x and ₹ y respectively.
According to the question,
5x + 7y = 50
and
7x + 5y = 46
For 5x + 7y = 50,
y = (50 – 5x)/7

x	3	-4	10
y = (50 – 5x)/7	5	10	0

For 7x + 5y = 46,
y = (46 – 7x)/5

x	8	3	–2
y = (46 – 7x)/5	–2	5	12

The graphs of equations are drawn below which shows that the two lines intersect at (3, 5).
Hence the cost of 1 pencil and the cost of 1 pen are ₹ 3 and ₹ 5 respectively.

Q.28 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.

Ans.

Let the price of 1 kg of apple and 1 kg of grapes be ₹ x and ₹ y respectively.
According to the question,
2x + y = 160 …(1)
And
4x + 2y = 300
or 2x + y = 150 …(2)
Equations (1) and (2) represent the given situation algebraically.
To represent the given situation graphically, we need at least two solutions for each equation. We write these solutions in table.
2x + y = 160 …(1)
or y = 160 – 2x

x	60	40
y = 160 – 2x	40	80

(i)
Also,
2x + y = 150 …(2)
or y = 150 – 2x

x	60	40
y = 150 – 2x	30	70

(ii)
The graphical representation of the situation is given below. The two lines never intersect each other, i.e., the two lines are parallel.

Q.29 The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

Ans.

Let the price of a bat and a ball be ₹ x and ₹ y respectively.
According to the question,
3x + 6y = 3900 …(1)
or 3 (x + 2y) = 3900
or x + 2y = 3900/3 =1300 …(2)
Equation (1) represents the total price of 3 bats and 6 balls whereas equation (2) represents the total price of one bat and 2 balls.
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
3x + 6y = 3900
or y = (3900 – 3x)/6

x	100	300
y	600	500

(i)
Also,
x + 2y =1300
or y = (1300 – x)/2

x	500	900
y	400	200

(ii)
The below given graphical representation shows that
graphs of both the equations coincide.

Q.30 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Ans.

Let the present age of Aftab and his daughter be x years and y years respectively.
So, seven years ago,
Aftab’s age = (x–7) years and his daughter’s age = (y–7) years
According to the question,
x–7 = 7(y–7)
or x – 7y + 42 = 0
Three years hence,
Age of Aftab = (x + 3) years
Age of daughter = (y + 3) years
According to the question,
x + 3 = 3(y + 3)
or x – 3y – 6 = 0
Hence, the given information is represented algebraically by the two equations below.
x – 7y + 42 = 0
x – 3y – 6 = 0
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
x – 7y + 42 = 0
or x = 7y – 42

x	0	–7	7
y	6	5	7

Also,
x – 3y – 6 = 0
or x = 3y + 6

x	0	–3	3
y	–2	–3	–1

The graphical representation is given below.

Q.31 The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield

(in kg/ha)

50-55

55-60

60-65

65-70

70-75

75-80

Number of farms

Change the distribution to a more than type distribution, and draw its graph.

Ans.

$Following is the cumulative frequency distributions of more than type of the given data..$

$Production yield (in kg/ha)$	$\begin{array}{l} Number of farms \\ (Cumulative frequency) \end{array}$
More than or equal to 50	100
More than or equal to 55	100 – 2 = 98
More than or equal to 60	98 – 8 = 90
More than or equal to 65	90 – 12 = 78
More than or equal to 70	78 – 24 = 54
More than or equal to 75	54 – 38 = 16

Now, draw the graph by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

Q.32 During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)

Number of students

Less than 38

Less than 40

Less than 42

Less than 44

Less than 46

Less than 48

Less than 50

Less than 52

Draw a less than type graph for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Ans.

$Following is the given cumulative frequency distributions of less than type$

$\begin{array}{l} Weight (in kg) \\ upper class limits \end{array}$	$\begin{array}{l} Number of students \\ (Cumulative frequency) \end{array}$
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Now locate 17.5 (half of 35) on y-axis and draw a line parallel to x-axis to cut the graph at a point.
Draw a perpendicular from this point to x-axis. The point where this perpendicular meets the x-axis
determines the median of the data. From the graph we see the median is 46.5.

$\begin{array}{l} Now, we mark the point A(17.5, 46.5). So median of the given data is 46.5. \\ We find class intervals with their respective frequencies as the following. \end{array}$

Weight (in kg)	Frequency (f)	Cumulative Frequency
Less than 38	0	0
38 – 40	3 – 0 = 3	3
40 – 42	5 – 3 = 2	5
42 – 44	9 – 5 = 4	9
44 – 46	14 – 9 = 5	14
46 – 48	28 – 14 = 14	28
48 – 50	32 – 28 = 4	32
50 – 52	35 – 32 = 3	35
Total (n)	35

$\begin{array}{l} \frac{n}{2} = \frac{35}{2} = 17 .5 \\ Cumulative frequency just greater that 17.5 is 28 which belongs \\ to class interval 46 - 48 . \\ ∴ Median class = 46 - 48 \\ Now, we have \\ l = 46, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 14, \\ Frequency of median class (f) = 14, \\ Class size (h) = 2 \end{array}$ $\begin{array}{l} We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 46 + (\frac{\frac{35}{2} - 14}{14}) \times 2 \\ or Median = 46 .5 \\ So median of the given data is 46 .5. \\ Hence, value of the median is verified. \end{array}$

Q.33 The following distribution gives the daily income of 50 workers of a factory.

Daily income

(in ₹)

100-120

120-140

140-160

160-180

180-200

Number of workers

Convert the distribution above to a less than type cumulative frequency distribution, and draw its graph.

Ans.

$We find less than type cumulative frequency table as below.$

Daily income (in ₹)	Cumulative Frequency
Less than 120	12
Less than 140	12 + 14 = 26
Less than 160	26 + 8 = 34
Less than 180	34 + 6 = 40
Less than 200	40 + 10 = 50

$\begin{array}{l} Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw its \\ ogive as following. \end{array}$

Q.34 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number of students	2	3	8	6	6	3	2

Ans.

$\begin{array}{l} We find the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Weight (in kg)	Number of students (f_i)	Cumulative frequency (cf )
40 – 45	2	2
45 – 50	3	5
50 – 55	8	13
55 – 60	6	19
60 – 65	6	25
65 – 70	3	28
70 – 75	2	30
Total (n)	30

$\begin{array}{l} We have, n = 30 . \\ ∴ \frac{n}{2} = \frac{30}{2} = 15 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 15 is 19 \\ which belongs to interval 55 - 60 . \\ ∴ Median class = 55 - 60 \\ Now, we have \\ l = 55, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 13, \\ Frequency of median class (f) = 6, \\ Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 55 + (\frac{\frac{30}{2} - 13}{6}) \times 5 \\ or Median = 56 .67 \\ Hence, median weight of students is 56 .67 kg. \end{array}$

Q.35 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Ans.

$\begin{array}{l} We find the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Number of letters	Number of surnames (f_i)	Cumulative frequency (cf)
1 – 4	6	6
4 – 7	30	36
7 – 10	40	76
10 – 13	16	92
13 – 16	4	96
16 – 19	4	100

$\begin{array}{l} Total(n) = 100 \\ We have, n = 100 . \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 50 is 76 \\ which belongs to interval 7 - 10 . \\ ∴ Median class = 7 - 10 \end{array}$ $\begin{array}{l} Now, we have \\ l = 7, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 36, \\ Frequency of median class (f) = 40, \\ Class size (h) = 3 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 7 + (\frac{\frac{100}{2} - 36}{40}) \times 3 \\ or Median = 8 .05 \\ ∴ Median number of letters in the surnames is 8.05. \\ Now, we find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 11 .5 \end{array}$

We proceed as below to find d_i, u_i, f_iu_i .

Number of letters	Number of surnames (f_i)	x_i	d_i = x_i – a	$u_{i} = \frac{x_{i} - a}{h}$	f_iu_i
1 – 4	6	2.5	-9	-3	-18
4 – 7	30	5.5	-6	-2	-60
7 – 10	40	8.5	-3	-1	-40
10 – 13	16	11.5	0	0	0
13 – 16	4	14.5	3	1	4
16 – 19	4	17.5	6	2	8
Total (n)	100				-106

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 11 .5 + \frac{- 106}{100} \times 3 \\ = 8 .32 \\ We know that \\ 3 Median = Mode + 2 Mean \\ 3 \times 8 .05 = Mode + 2 \times 8 .32 \\ or Mode = 7 .51 \\ Hence, mean number of letters in the surnames is 8 .32 \\ and modal size of the surnames is 7 .51. \end{array}$

Q.36 The following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	14 56 60 86 74 62 48

Find the median life time of a lamp.

Ans.

$\begin{array}{l} We write the cumulative frequency (cf) as below in the \\ following table on the basis of the given information. \end{array}$

Life time (in hours)	Number of lamps (f_i)	Cumulative frequency (cf)
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400
Total (n)	400

$\begin{array}{l} We have, n = 400 . \\ ∴ \frac{n}{2} = \frac{400}{2} = 200 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 200 is 216 \\ which belongs to interval 3000 - 3500 . \\ ∴ Median class = 3000 - 3500 \\ Now, we have \\ l = 3000, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 130, \\ Frequency of median class (f) = 86, \\ Class size (h) = 500 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 3000 + (\frac{\frac{400}{2} - 130}{86}) \times 500 \\ or Median = 3406 .98 \\ Hence, median life time of a lamp is 3406 .98 hours. \end{array}$

Q.37 The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm)

Number of leaves

118 – 126

127– 135

136– 144

145–153

154–162

163–171

172–180

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180)

Ans.

$\begin{array}{l} The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from each \\ of lower boundary and adding 0.5 in each of upper boundary. \\ We write the class interval, number of leaves (f_{i}), and cumulative frequency (cf) as below in the following table \\ on the basis of the given information. \end{array}$

Class interval	Number of leaves (f_i)	Cumulative frequency (cf)
117.5 – 126.5	3	3
126.5 – 135.5	5	8
13.5 – 144.5	9	17
144.5 – 153.5	12	29
153.5 – 162.5	5	34
162.5 – 171.5	4	38
171.5 – 180.5	2	40
Total	40

$\begin{array}{l} We have, n = 40 . \\ ∴ \frac{n}{2} = \frac{40}{2} = 20 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 20 is 29 \\ which belongs to interval 144 .5 - 153 .5. \\ ∴ Median class = 144 .5 - 153 .5 \\ Now, we have \\ l = 144 .5, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 17, \\ Frequency of median class (f) = 12, \\ Class size (h) = 9 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 144 .5 + (\frac{\frac{40}{2} - 17}{12}) \times 9 \\ or Median = 146 .75 \\ Hence, median length of leaves is 146 .75 mm. \end{array}$

Q.38 A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Class interval

Frequency

Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60

100

Ans.

$\begin{array}{l} We write the class interval, number of policy holders (f_{i}), and cumulative frequency (cf) as below in the following table \\ on the basis of the given information. \end{array}$

Class interval	Number of policy holders (f_i)	Cumulative frequency (cf)
18 – 20	2	2
20 – 25	6 – 2 = 4	6
25 – 30	24 – 6 = 18	24
30 – 35	45 – 24 = 21	45
35 – 40	78 – 45 = 33	78
40 – 45	89 – 78 = 11	89
45 – 50	92 – 89 = 3	92
50 – 55	98 – 92 = 6	98
55 – 60	100 – 98 = 2	100
Total (n)	100

$\begin{array}{l} We have, n = 100 . \\ ∴ \frac{n}{2} = \frac{100}{2} = 50 \\ Cumulative frequency (cf) just greater than \frac{n}{2} = 50 is 78 \\ which belongs to interval 35 - 40 . \\ ∴ Median class = 35 - 40 \\ Now, we have \\ l = 35, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 45, \\ Frequency of median class (f) = 33, \end{array}$ $\begin{array}{l} Class size (h) = 5 \\ We know that \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ ∴ Median = 35 + (\frac{\frac{100}{2} - 45}{33}) \times 5 \\ or x = 35 .76 \\ Hence, median age is 35 .76 years. \end{array}$

Q.39 If the median of the distribution given below is 28.5, find the values of x and y.

Class interval

Frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60

Total

Ans.

We find cumulative frequency of the given data as below.

Class interval

Frequency

Cumulative frequency

0 – 10

10– 20

20– 30

30–40

40–50

50–60

5 + x

25 + x

40 + x

40 + x + y

45 + x + y

Total

Here,
n = 60
or 45 + x + y = 60
or x + y = 15 …(1)
Median of data is given as 28.5 that lies in the interval 20 – 30.
So, median class = 20 – 30

$\begin{array}{l} From the given data, we have \\ l = 20, \\ Cumulative frequency (cf) of the class preceding the \\ median class = 5 + x, \\ Frequency of median class (f) = 20, \\ Class size (h) = 10 \\ Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h \\ or 28 .5 = 20 + (\frac{\frac{60}{2} - 5 - x}{20}) \times 10 \\ or x = 8 \\ On putting this value of x in equation (1), we get \\ x + y = 15 \\ or 8 + y = 15 \\ or y = 15 - 8 = 7 \\ Hence, values of x and y are 8 and 7 respectively. \end{array}$

Q.40 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)

Number of consumers

65 – 85

85– 105

105– 125

125–145

145–165

165–185

185–205

Ans.

$\begin{array}{l} We find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 135 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Monthly consumption (in units)	Number of consumers (f_i)	x_i	d_i = x_i – 135	$u_{i} = \frac{x_{i} - 135}{h}$	f_iu_i
65 – 85	4	75	-60	-3	-12
85 – 105	5	95	-40	-2	-10
105 – 125	13	115	-20	-1	-13
125 – 145	20	135	0	0	0
145 – 165	14	155	20	1	14
165 – 185	8	175	40	2	16
185 – 205	4	195	60	3	12
Total	68				7

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 135 + \frac{7}{68} \times 20 \\ = 137 .058 \\ Modal class = 125 - 145 \\ From the given data, we have \\ l = 125, f_{1} = 20, f_{0} = 13, f_{2} = 14, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 125 + (\frac{20 - 13}{2 \times 20 - 13 - 14}) \times 20 \\ = 135 .76 \\ We know that \\ 3 Median = Mode + 2 Mean \\ or Median = \frac{135 .76 + 2 \times 137 .058}{3} = 136 .625 \end{array}$

We observe all these three measures are approximately same.

Q.41 A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Number

of cars

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Frequency

Ans.

$\begin{array}{l} From the given data, we have \\ l = 40, f_{1} = 20, f_{0} = 12, f_{2} = 11, h = 10 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 40 + (\frac{20 - 12}{2 \times 20 - 12 - 11}) \times 10 \\ = 44 .7 \\ ∴ Mode of the given data is 44 .7. \end{array}$

Q.42 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored

Number of batsmen

3000 – 4000

4000– 5000

5000– 6000

6000–7000

7000–8000

8000–9000

9000–10000

10000–11000

Find the mode of the data.

Ans.

$\begin{array}{l} From the given data, we have \\ l = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 9, h = 1000 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 4000 + (\frac{18 - 4}{2 \times 18 - 4 - 9}) \times 1000 \\ = 4608 .695 \\ Mode of the given data is 4608 .7 (approx .) \end{array}$

Q.43 The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher

Number of states / U .T.

15– 20

20– 25

25–30

30–35

35–40

40–45

45–50

50–55

Ans.

$\begin{array}{l} From the given data, we have \\ l = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3, h = 5 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 30 + (\frac{10 - 9}{2 \times 10 - 9 - 3}) \times 5 \\ = 30 .625 \approx 30 .6 \\ Interpretation: \\ Most of the states/UT have a teacher student ratio as 30.6. \\ Now, we find class mark for each interval by using the \\ following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 32 .5 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Number of students per teacher	Number of states/UT (f_i)	x_i	d_i = x_i – 32.5	$u_{i} = \frac{x_{i} - 32 .5}{h}$ b	f_iu_i
15 – 20	3	17.5	-15	-3	-9
20 – 25	8	22.5	-10	-2	-16
25 – 30	9	27.5	-5	-1	-9
30 – 35	10	32.5	0	0	0
35 – 40	3	37.5	5	1	3
40 – 45	0	42.5	10	2	0
45 – 50	0	47.5	15	3	0
50 – 55	2	52.5	20	4	8
Total	35				-23

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 32 .5 + \frac{- 23}{35} \times 5 \\ = 29 .2 \\ Interpretation: \\ Average teacher student ratio of the states/UT is 29 .2. \end{array}$

Q.44 The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Expenditure (in ₹)	Number of families
1000 – 1500 1500– 2000 2000– 2500 2500–3000 3000–3500 3500–4000 4000–4500 4500–5000	24 40 33 28 30 22 16 7

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 2750 \\ We proceed as below to find d_{i}, u_{i}, f_{i} u_{i} . \end{array}$

Expenditure (in ₹)	Number of families (f_i)	x_i	d_i = x_i – 2750	$u_{i} = \frac{x_{i} - 2750}{h}$	f_iu_i
1000 – 1500	24	1250	-1500	-3	-72
1500 – 2000	40	1750	-1000	-2	-80
2000 – 2500	33	2250	-500	-1	-33
2500 – 3000	28	2750	0	0	0
3000 – 3500	30	3250	500	1	30
3500 – 4000	22	3750	1000	2	44
4000 – 4500	16	4250	1500	3	48
4500 – 5000	7	4750	2000	4	28
Total	200				-35

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} u_{_{i}}}{\sum f_{i}} \times h \\ = 2750 + \frac{- 35}{200} \times 500 \\ = 2662 .5 \\ Therefore, mean monthly expenditure is ₹ 2662 .5. \\ From the given data, we have \\ l = 1500, f_{1} = 40, f_{0} = 24, f_{2} = 33, h = 500 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 1500 + (\frac{40 - 24}{2 \times 40 - 24 - 33}) \times 500 \\ = 1847 .826 \approx 1847 .83 \\ So, modal monthly expenditure is ₹ 1847 .83. \end{array}$

Q.45 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components.

Lifetimes (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Ans.

$\begin{array}{l} From the given data, we have \\ l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38, h = 20 \\ Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 60 + (\frac{61 - 52}{2 \times 61 - 52 - 38}) \times 20 \\ = 65 .625 \\ So, modal lifetime of electrical components is 65.625 hours. \end{array}$

Q.46 The following table shows the ages of the patients admitted in a hospital during a year.

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 30 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Age (in years)	Number of patients (f_i)	Class mark (x_i)	d_i = x_i – 30	f_id_i
5 -15	6	10	-20	-120
15 – 25	11	20	-10	-110
25 – 35	21	30	0	0
35 – 45	23	40	10	230
45 – 55	14	50	20	280
55 – 65	5	60	30	150

$\begin{array}{l} Total = 80 Total = 430 \\ Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 30 + \frac{430}{80} \\ = 35 .375 \approx 35 .38 \\ Therefore, mean of the given data is 35 .38 years . \\ Modal class is 35 - 45 . \\ l = 35, f_{1} = 23, f_{0} = 21, f_{2} = 14, h = 10 \end{array}$ $\begin{array}{l} Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h \\ = 35 + (\frac{23 - 21}{2 \times 23 - 21 - 14}) \times 10 \\ = 36 .8 \\ Mode is 36.8 which represents that maximum number \\ of patients admitted in hospital are of 36.8 years . \end{array}$

$\begin{array}{l} While on average the age of a patient admitted to the \\ hospital is 35 .38 years \end{array}$

Q.47 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 10 \\ Here, we take assumed mean(a) = 70 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Literacy rate (in %)	Number of cities f_i	x_i	d_i = x_i – 70	$u_{i} = \frac{x_{i} - 70}{h}$	f_iu_i
45 – 55	3	50	-20	-2	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	35				-2

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 70 + (\frac{- 2}{35}) \times 10 = 69 .43 \\ Therefore, mean literacy rate is 69 .43. \end{array}$

Q.48 A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Assumed mean(a) = 17 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Number of days	Number of students f_i	x_i	d_i = x_i – 17	f_id_i
0 – 6	11	3	-14	-154
6 – 10	10	8	-9	-90
10 – 14	7	12	-5	-35
14 – 20	4	17	0	0
20 – 28	4	24	7	28
28 – 38	3	33	16	48
38 – 40	1	39	22	22
Total	40			-181

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ = 17 + \frac{- 181}{40} \\ = 12 .475 \approx 12 .48 \\ Therefore, mean number of days for which a student was absent is 12 .48. \end{array}$

Q.49 To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below.

Concentration of SO₂ (in ppm)

Frequency

0.00 – 0.04

0.04– 0.08

0.08 – 0.12

0.12–0.16

0.16–0.20

0.20–0.24

Find the mean concentration of SO₂ in the air.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 0 .04 \\ Here, we take assumed mean(a) = 0 .14 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Concentration of SO₂ (in ppm)	Frequency (f_i)	x_i	d_i = x_i – 0.14	$u_{i} = \frac{x_{i} - 0 .14}{h}$ b	f_iu_i
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 0 .14 + (\frac{- 31}{30}) \times 0 .04 \\ \approx 0 .099 ppm \\ {Therefore, mean concentration of SO}_{2} in the air is 0 .099 ppm . \end{array}$

Q.50 The table below shows the daily expenditure on food of 25 households in a locality.

Daily Expenditure (in ₹)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 50 \\ Here, we take assumed mean(a) = 225 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Class interval	f_i	x_i	d_i = x_i – 225	$u_{i} = \frac{x_{i} - 57}{h}$	f_iu_i
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
Total					-7

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 225 + (\frac{- 7}{25}) \times 50 = 211 \\ Therefore, mean daily expenditure is ₹ 211 . \end{array}$

Q.51 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ans.

$\begin{array}{l} The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limit \\ of each interval. We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 57 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Class interval	f_i	x_i	d_i = x_i – 57	$u_{i} = \frac{x_{i} - 57}{h}$	$f_{i} u_{i}$
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	400				25

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 57 + (\frac{25}{400}) \times 3 = 57 .1875 \approx 57 .19 \\ Therefore, mean number of mangoes is 57 .19. \\ We have chosen step deviation method as values of f_{i}, d_{i} are big \\ and there is a common multiple between all d_{i} . \end{array}$

Q.52 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 3 \\ Here, we take assumed mean(a) = 75 .5 and proceed as below \\ to find d_{i}, u_{i} and f_{i} u_{i} . \end{array}$

Number of heartbeats per minute	Number of women (f_i)	x_i	d_i = x_i – 75.5	$u_{i} = \frac{x_{i} - 75 .5}{h}$	f_iu_i
65-68	2	66.5	-9	-3	-6
68-71	4	69.5	-6	-2	-8
71-74	3	72.5	-3	-1	-3
74-77	8	75.5	0	0	0
77-80	7	78.5	3	1	7
80-83	4	81.5	6	2	8
83-86	2	84.5	9	3	6
Total	30				4

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 75 .5 + (\frac{4}{30}) \times 3 = 75 .9 \\ Therefore, mean heartbeats per minute is 75 .9. \end{array}$

Q.53 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Daily pocket allowance

(in ₹)

11-13

13-15

15-17

17-19

18-21

21-23

23-25

Number of children

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 2 \\ Given mean(a) = 18 \\ We proceed as below to find d_{i}, f_{i} d_{i} . \end{array}$

Daily pocket allowance (in ₹)	Number of children (f_i)	x_i	d_i = x_i – 18	f_id_i
11 – 13	7	12	-6	-42
13 – 15	6	14	-4	-24
15 – 17	9	16	-2	-18
17 – 19	13	18	0	0
19 – 21	f	20	2	2f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total	$\sum f_{i} = 44 + f$			2f – 40

$\begin{array}{l} Mean = \bar{x} = a + \frac{\sum f_{i} d_{_{i}}}{\sum f_{i}} \\ or 18 = 18 + \frac{2 f - 40}{44 + f} \\ or f = 20 \\ Therefore, missing frequency f is 20 . \end{array}$

Q.54 Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹)	500-120	520-140	540-160	560-180	580-200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans.

$\begin{array}{l} We find class mark for each interval by using the following \\ relation. \\ x_{i} = \frac{Upper class limit + Lower class limit}{2} \\ Class size(h) of the given data = 20 \\ Here, we take assumed mean(a) = 550 and proceed as below. \end{array}$

Daily wages (in ₹)	Number of workers (f_i)	x_i	d_i = x_i – 550	$u_{i} = \frac{x_{i} - 550}{h}$	f_iu_i
500 – 520	12	510	-40	-2	-24
520 – 540	14	530	-20	-1	-14
540 – 560	8	550	0	0	0
560 – 580	6	570	20	1	6
580 – 600	10	590	40	2	20
Total	50				-12

$\begin{array}{l} Mean = \bar{x} = a + (\frac{\sum f_{i} u_{_{i}}}{\sum f_{i}}) h = 550 + (\frac{- 12}{50}) \times 20 = 545 .2 \end{array}$

$\begin{array}{l} Therefore, mean daily wages of the workers is ₹ 545 .20 \end{array}$

Q.55 A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Ans.

$\begin{array}{l} Here, we observe that class Number of plants Number of houses – 21112 – 42364 – 61556 – 857358 – 10695410 – 122112212 – 1431339 Total20
162 \end{array}$

$\begin{array}{l} Mean = \bar{x} = \frac{\sum f_{i} x_{_{i}}}{\sum f_{i}} = \frac{162}{20} = 8 .1 \\ Therefore, mean number of plants per house is 8.1. \end{array}$

Q.56 A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3⋅ Find the number of blue marbles in the jar.

Ans.

$\begin{array}{l} Let the number of green marbles = x \\ Total number of marbles in the jar = 24 \\ P(drawing a green marble) \\ or \frac{2}{3} = \frac{Number of favourable outcomes}{Total number of outcomes} \\ or \frac{2}{3} = \frac{x}{24} \\ or x = 16 \\ Number of blue marbles = 24 - x = 24 - 16 = 8 \end{array}$

Q.57 A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Ans.

$\begin{array}{l} Number of black balls = x \\ Total number of balls in the bag = 12 \\ P(drawing a black ball) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{x}{12} \\ Now, 6 more black balls are putted in the box. \\ ∴ Number of black balls = x + 6 \\ Total number of balls in the bag = 12 + 6 = 18 \\ P(drawing a black ball) = 2 \times \frac{x}{12} \\ or \frac{x + 6}{18} = 2 \times \frac{x}{12} \\ or x = 3 \end{array}$

Q.58 A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Ans.

$\begin{array}{l} Let number of blue balls be n . \\ Number of red balls = 5 \\ Total number of balls in the bag = 5 + n \\ P(drawing a red ball) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{5}{5 + n} \\ P(drawing a blue ball) = 2 \times P(drawing a red ball) \\ or \frac{n}{5 + n} = 2 \times \frac{5}{5 + n} \\ or n = 10 \\ ∴ Number of blue balls = 10 \end{array}$

Q.59 A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws.

What is the probability that the total score is
(i) even? (ii) 6? (iii) at least 6?

Ans.

The given table can be completed as below.

$\begin{array}{l} Total number of possible outcomes = 6 \times 6 = 36 \\ (i) \\ Number of times when the total score is even = 18 \\ P(total score being even) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{18}{36} = \frac{1}{2} \\ (ii) \\ Number of times when the total score is 6 = 4 \\ P(total score being 6) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{4}{36} = \frac{1}{9} \\ (iii) \\ Number of times when the total score is at least 6 = 15 \\ P(total score being at least 6) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{15}{36} = \frac{5}{12} \end{array}$

Q.60 Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?

Ans.

$\begin{array}{l} There are 5 days namely Tuesday, Wednesday, Thursday, \\ Friday, Saturday. \\ Shyam can go to the shop on any of the five days. \\ Also, Ekta can go to the shop on any of the five days. \\ So, total number of outcomes = 5 \times 5 = 25 \\ (i) \\ Outcomes of visiting the shop on same day are \\ (T, T), (W, W), (Th, Th), (F, F), (S, S). \\ P(both will visit the shop on the same day) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{5}{25} = \frac{1}{5} \\ (ii) \\ Outcomes of visiting the shop on consecutive days are \\ (T, W), (W, Th), (Th, F), (F, S), (W, T), (Th, W), (F, Th), (S, F). \\ Number of odd numbers = 3 \\ ∴ P(both will visit on consecutive days) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{8}{25} \\ (iii) \\ P(both will visit on different days) \\ = 1 - both will visit on the same day \\ = 1 - \frac{1}{5} = \frac{4}{5} \end{array}$

Q.61 Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes — two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes — an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

Ans.

$\begin{array}{l} (i) \\ The given statement is incorrect. There are 4 possible \\ outcomes which are (H, H), (T, T), (H, T), (T, H) . \\ P(getting two heads) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{4} \\ P(getting two tails) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{4} \\ P(getting one head and one tail) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{2}{4} = \frac{1}{2} \\ (ii) Correct because the two outcomes are equally likely. \end{array}$

Q.62 A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Ans.

$\begin{array}{l} Total number of outcomes = 6 \times 6 = 36 \\ (i) \\ Possible outcomes when 5 comes up either time are \\ (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) . \\ Number of favourable outcomes = 11 \\ P(5 will come up either time) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{11}{36} \\ ∴ P(5 will not come up either time) \\ = 1 - P(5 will come up either time) \\ = 1 - \frac{11}{36} = \frac{25}{36} \\ (ii) \\ Number of cases when 5 will come up at least once = 11 \\ P(5 will come up at least once) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{11}{36} \end{array}$

Q.63 A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans.

$\begin{array}{l} Possible outcomes are \\ {HHH, TTT, HTT, THH, HTH, THT, HHT, TTH} . \\ Number of possible outcomes = 8 \\ Number of favourable outcomes = 2 \\ P(Hanif will win the game) \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{2}{8} = \frac{1}{4} \\ ∴ P(Hanif will lose the game) \\ = 1 - P(Hanif will win the game) \\ = 1 - \frac{1}{4} = \frac{3}{4} \end{array}$

Q.64 Two dice, one blue and one grey, are thrown at the same time.
(i) Complete the following table:

Event: ‘Sum of two dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

Ans.

(i) Total possible outcomes = 36
Possible outcomes for getting the sum 3 are (1, 2) and (2, 1).
Possible outcomes for getting the sum 4 are (1, 3); (3, 1) and (2, 2).
Possible outcomes for getting the sum 5 are (1, 4); (2, 3) (3, 2) and (4,1).
Possible outcomes for getting the sum 6 are (1, 5); (2, 4); (3, 3) (4, 2) and (5, 1).
Possible outcomes for getting the sum 7 are ((1, 6); (2, 5); (3, 4) (4, 3); (5, 2) and (6, 1).
Possible outcomes for getting the sum 9 are (3, 6); (4, 5) ; (5, 4) and (6, 3).
Possible outcomes for getting the sum 10 are (4, 6); (5, 5) and (6, 4).
Possible outcomes for getting the sum 11 are (5, 6) and (6, 5).
Now the complete table is as follows:

Event: ‘Sum of two dice’	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

(ii) No. The eleven sums are not equally likely.

Q.65 A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?

Ans.

$\begin{array}{l} Number of defective pens = 20 \\ Number of good pens = 144 - 20 = 124 \\ Total number of outcomes = 144 \\ (i) \\ Probability of buying a pen = Probability of drawing a good pen \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{124}{144} = \frac{31}{36} \\ (ii) \\ Probability of not buying pen = 1 - Probability of buying a pen \\ = 1 - \frac{31}{36} = \frac{5}{36} \end{array}$

Q.66 Suppose you drop a die at random on the rectangular region shown in the following figure. What is the probability that it will land inside the circle with diameter 1m?

Ans.

$\begin{array}{l} Favourable outcomes = Area of the circle = {πr}^{2} = π {(0 .5)}^{2} \\ = 0 .25 π m^{2} \\ Total outcomes = Area of the rectangle = 2 m \times 3 m = 6 m^{2} \\ Probability of landing inside the circle \\ = \frac{Area of the circle}{Area of the rectangle} \\ = \frac{0 .25 π}{6} = \frac{π}{24} \end{array}$

Q.67 A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?

Ans.

$\begin{array}{l} (i) \\ Probability of getting A \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{2}{6} = \frac{1}{3} \\ (ii) \\ Probability of getting D \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{6} \end{array}$

Q.68 A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Ans.

$\begin{array}{l} Numbers of two-digit numbers among 1 to 90 = 81 \\ Numbers of perfect square numbers among 1 to 90 = 9 \\ Numbers of multiples of 5 which are less than or equal 90 = 18 \\ Total number of outcomes = 90 \\ (i) \\ Probability of getting a two-digit number bearing disc \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{81}{90} = \frac{9}{10} \\ (ii) \\ Probability of getting a perfect square number \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{9}{90} = \frac{1}{10} \\ (iii) \\ Probability of getting a number divisible by 5 \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{18}{90} = \frac{1}{5} \end{array}$

Q.69 (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Ans.

$\begin{array}{l} (i) \\ Number of defective balls = 4 \\ Total number of outcomes = 20 \\ Probability of getting a defective ball \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{4}{20} = \frac{1}{5} \\ (ii) \\ Number of defective balls = 4 \\ Number of good balls = 16 - 1 = 15 \\ Total number of outcomes = 15 + 4 = 19 \\ Probability of not getting a defective ball \\ = 1 - Probability of getting a defective ball \\ = 1 - \frac{4}{19} = \frac{15}{19} \end{array}$

Q.70 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Ans.

$\begin{array}{l} Number of defective pens = 12 \\ Number of good pens = 132 \\ Total number of outcomes = 132 + 12 = 144 \\ Probability of getting a good pen \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{132}{144} = \frac{11}{12} \end{array}$

Q.71 Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Ans.

$\begin{array}{l} (i) \\ Total number of outcomes = 5 \\ Probability of getting the queen \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{5} \\ (ii) \\ Queen is put aside. \\ ∴ total number of outcomes = 4 \\ (a) \\ Probability of getting an ace \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{4} \\ (b) \\ Probability of getting a queen \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{0}{4} = 0 \end{array}$

Q.72 One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamond

Ans.

$\begin{array}{l} Total number of outcomes = 52 \\ (i) \\ Number of kings of red colour = 2 \\ Probability of getting a king of red colour \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{2}{52} = \frac{1}{26} \\ (ii) \\ Number of face cards = 12 \\ Probability of getting a face card \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{12}{52} = \frac{3}{13} \\ (iii) \\ Number of red face cards = 6 \\ Probability of getting a red face card \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{6}{52} = \frac{3}{26} \\ (iv) \\ Number of the jack of hearts = 1 \\ Probability of getting the jack of hearts \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{52} \\ (v) \\ Number of spades = 13 \\ Probability of getting a spade \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{13}{52} = \frac{1}{4} \\ (vi) \\ Number of the queen of diamond = 1 \\ Probability of getting the queen of diamond \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{52} \end{array}$

Q.73 A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Ans.

$\begin{array}{l} Total number of outcomes = 6 \\ (i) \\ Prime numbers on the die are 2, 3 and 5. \\ ∴ Number of prime numbers on the die = 3 \\ Probability of getting a prime number \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{3}{6} = \frac{1}{2} \\ (ii) \\ Numbers lying between 2 and 6 on the die are 3, 4 and 5. \\ ∴ Number of prime numbers on the die = 3 \\ Probability of getting a number lying between 2 and 6 \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{3}{6} = \frac{1}{2} \\ (iii) \\ Odd numbers on the die are 1, 3 and 5. \\ ∴ Number of odd numbers on the die = 3 \\ Probability of getting an odd number \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{3}{6} = \frac{1}{2} \end{array}$

Q.74 A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the following figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Ans.

$\begin{array}{l} (i) Probability of the arrow pointing at 8 \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{1}{8} \\ (ii) Probability of the arrow pointing at an odd number \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{4}{8} = \frac{1}{2} \\ (iii) Probability of the arrow pointing at a number greater than 2 \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{6}{8} = \frac{3}{4} \\ (iv) Probability of the arrow pointing at a number less than 9 \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{8}{8} = 1 \end{array}$

Q.75 Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see the following figure). What is the probability that the fish taken out is a male fish?

Ans.

$\begin{array}{l} Number of male fishes = 5 \\ Number of female fishes = 8 \\ Total number of outcomes = 5 + 8 = 13 \\ (i) Probability of getting a male fish \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{5}{13} \end{array}$

Q.76 A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a ₹ 5 coin?

Ans.

$\begin{array}{l} Number of 50p coins = 100 \\ Number of ₹1 coins = 50 \\ Number of ₹2 coins = 20 \\ Number of ₹5 coins = 10 \\ Total number of outcomes = 100 + 50 + 20 + 10 = 180 \\ (i) Probability of getting a 50p coin \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{100}{180} = \frac{5}{9} \\ (ii) \\ Probability of not getting a ₹5 coin \\ = 1 - Probability of getting a ₹ 5 coin \\ = 1 - \frac{Number of favourable outcomes}{Total number of outcomes} \\ = 1 - \frac{10}{180} = \frac{17}{18} \end{array}$

Q.77 A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green?

Ans.

$\begin{array}{l} Number of red marbles = 5 \\ Number of white marbles = 8 \\ Number of green marbles = 4 \\ Total number of outcomes = 5 + 8 + 4 = 17 \\ (i) Probability of getting a red marble \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{5}{17} \\ (ii) Probability of getting a white marble \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{8}{17} \\ (iii) Probability of getting a marble not green \\ =1 - Probability of getting a green marble \\ = 1 - \frac{4}{17} = \frac{13}{17} \end{array}$

Q.78 A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Ans.

$\begin{array}{l} Number of red balls = 3 \\ Number of black balls = 5 \\ Total number of outcomes = 3 + 5 = 8 \\ (i) Probability of getting a red ball \\ = \frac{Number of favourable outcomes}{Total number of outcomes} \\ = \frac{3}{8} \\ (ii) Probability of not getting a red ball \\ = 1 - Probability of getting a red ball \\ = 1 - \frac{3}{8} = \frac{5}{8} \end{array}$

Q.79 It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Ans.

The probability of 2 students not having the same birthday = P(E’) = 0.992
The probability of 2 students having the same birthday= 1 – P(E’) = 1 – 0.992 = 0.008

Q.80 A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Ans.

(i) There is no orange flavoured candy in the bag. So, the probability of taking out an orange flavoured candy is zero.
(ii) The probability of taking out lemon flavoured candy is 1 as there are only lemon flavoured candies in the bag.

Q.81 If P(E) = 0.05, what is the probability of ‘not E’?

Ans.

Probability of ‘not E’ = 1 – P(E) = 1 – 0.05 = 0.95

Q.82 Which of the following cannot be the probability of an event?
(A) 2/3
(B) –1.5
(C) 15%
(D) 0.7

Ans.

The correct answer is (B).

Q.83 Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Ans.

The outcomes of a coin toss are equally likely. So, the result of a coin toss is completely unpredictable.

Q.84 Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Ans.

(i) It does not have equally likely outcomes. The car may never start for some fault.
(ii) It does not have equally likely outcomes as player may take more attempts to shoot a basketball.
(iii) It has equally likely outcomes. The answer of a true-false question is either right or wrong.
(iv) It has equally likely outcomes. The baby is either a boy or a girl.

Q.85 Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ =_____.
(ii) The probability of an event that cannot happen is_____. Such an event is called _____.
(iii) The probability of an event that is certain to happen is_____. Such an event is called______.
(iv) The sum of the probabilities of all the elementary events of an experiment is _____.
(v) The probability of an event is greater than or equal to _____ and less than or equal to_____.

Ans.

(i) Probability of an event E + Probability of the event ‘not E’ =1.
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

Q.86 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the following figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Ans.

$\begin{array}{l} Let AB be the height of the tip of fishing rod from water \\ surface. \\ Let BC be the horizontal distance of fly from the tip of \\ fishing rod. \\ AC is the length of string. \\ On applying Pythagoras theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \Rightarrow {AB}^{2} = {(1.8)}^{2} + {(2.4)}^{2} \\ \Rightarrow {AB}^{2} = 3.24 + 5.76 \\ \Rightarrow {AB}^{2} = 9.00 \\ \Rightarrow AB = \sqrt{9} = 3 \\ Therefore, length of the string is 3 m. \\ It is given that string is pulled at the rate of 5 cm per second. \\ So, string pulled in 12 seconds = 12 \times 5 = 60 cm = 0 .6 m \end{array}$

$\begin{array}{l} Let after 12 second Fly be at point D. \\ Length of string out after 12 second is AD \\ AD = AC – string pulled by Nazima in 12 seconds \\ = 3.00 – 0.6 \\ = 2.4 \\ In ΔADB, \\ {AB}^{2} {+ BD}^{2} {= AD}^{2} \\ \Rightarrow {(1.8)}^{2} {+ BD}^{2} = {(2.4)}^{2} \\ \Rightarrow {BD}^{2} = 5.76 - 3.24 = 2.52 \\ ⇒ BD = 1.587 \\ Horizontal distance of fly = BD + 1.2 \\ = 1.587 + 1.2 \\ = 2.787 \\ = 2.79 m \end{array}$

Q.87

$\begin{array}{l} In the following figure, D is a point on side BC of ΔABC such that \frac{BD}{CD} = \frac{AB}{AC} . \\ Prove that AD is the bisector of ∠ BAC . \end{array}$

Ans.

$\begin{array}{l} Let us extend BA to P such that AP = AC. Join PC. \\ It is given that \\ \frac{BD}{CD} = \frac{AB}{AC} \\ \Rightarrow \frac{BD}{CD} = \frac{AB}{AP} \\ Therefore, by converse of basic proportinality theorem, AD ∥ PC. \\ Therefore, \\ ∠ BAD = ∠ APC . .. (1) \\ ∠ DAC = ∠ ACP . .. (2) \\ Also, \\ AP = AC \\ \Rightarrow ∠ APC = ∠ ACP \\ \Rightarrow ∠ APC = ∠ ACP = ∠ DAC [From (2)] \\ \Rightarrow ∠ BAD = ∠ DAC [From (1)] \\ Therefore, AD is the bisector of angle BAC. \end{array}$

Q.88

$\begin{array}{l} In the following figure, two chords AB and CD of a circle intersect each other at the point P \\ (when produced) outside the circle. Prove that \\ (i) Δ PAC ~ Δ PDB (ii) PA \cdot PB = PC \cdot PD \end{array}$

Ans.

$\begin{array}{l} (i) \\ In Δ PAC and Δ PDB, \\ ∠ P = ∠ P (common) \\ ∠ PAC = ∠ PDB \\ (Exterior angle of a cyclic quadrilateral is \\ equal to opposite interior angle) \\ Δ PAC ~ Δ PDB \\ (ii) \\ We know that corresponding sides of similar triangles \\ are proportional. \\ ∴ Δ PAC ~ Δ PDB \\ \Rightarrow \frac{PA}{PD} = \frac{PC}{PB} \\ \Rightarrow PA \cdot PB = PC \cdot PD \end{array}$

Q.89

$\begin{array}{l} In the following figure, two chords AB and CD intersect each other at the point P. Prove that : \\ (i) Δ APC ~ Δ DPB (ii) AP.PB = CP.DP \end{array}$

Ans.

$\begin{array}{l} Let us join CB. \\ (i) In Δ APC and Δ DPB, \\ ∠ APC = ∠ DPB [Vertically opposite angles] \\ ∠ CAP = ∠ BDP [Angles in same segment for chord CB] \\ ∴ Δ APC ~ Δ DPB [By AA similarity criterion] \\ (ii) \\ We know that corresponding sides of similar triangles \\ are proportional. \\ Therefore, \\ Δ APC ~ Δ DPB \\ \Rightarrow \frac{AP}{DP} = \frac{PC}{PB} \\ \Rightarrow AP \cdot PB = PC \cdot DP \end{array}$

Q.90 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Ans.

Let ABCD be a parallelogram. We draw perpendiculars AF on CD and DE on extended side BA.

$\begin{array}{l} Applying Pythagoras theorem in Δ DEA, we get \\ {DE}^{2} + {EA}^{2} = {DA}^{2} . .. (i) \\ Applying Pythagoras theorem in ΔDEB, we get \\ {DE}^{2} + {EB}^{2} = {DB}^{2} \\ {DE}^{2} + {(EA + AB)}^{2} = {DB}^{2} \\ ({DE}^{2} + {EA}^{2}) + {AB}^{2} + 2 EA \times AB = {DB}^{2} \\ {DA}^{2} + {AB}^{2} + 2 EA \times AB = {DB}^{2} . .. (ii) \\ Applying Pythagoras theorem in ΔADF, we get \\ {AD}^{2} = {AF}^{2} + {FD}^{2} \\ Applying Pythagoras theorem in ΔAFC, we get \\ {AC}^{2} = {AF}^{2} + {FC}^{2} \\ = {AF}^{2} + {(DC - FD)}^{2} \\ = {AF}^{2} + {DC}^{2} + {FD}^{2} - 2 DC \times FD \\ = ({AF}^{2} + {FD}^{2}) + {DC}^{2} - 2 DC \times FD \\ {AC}^{2} = {AD}^{2} + {DC}^{2} - 2 DC \times FD . .. (iii) \\ Since ABCD is a parallelogram \\ AB = CD . .. (iv) \\ and, BC = AD . .. (v) \\ In Δ DEA and Δ ADF, \\ ∠ DEA = ∠ AFD = 90 °, ∠ EAD = ∠ ADF (EA ∥ DF) \\ Therefore, ∠ FAD = ∠ EDA \\ AD is common in both triangles. \\ Therefore, by ASA congruence critarion, we get \\ ΔEAD ≅ ΔFDA \\ \Rightarrow EA = DF . .. (vi) \\ Adding equation (ii) and (iii), we get \\ {DA}^{2} + {AB}^{2} + 2 EA \times AB + {AD}^{2} + {DC}^{2} - 2 DC \times FD = {DB}^{2} + {AC}^{2} \\ \Rightarrow {DA}^{2} + {AB}^{2} + {AD}^{2} + {DC}^{2} + 2 EA \times AB - 2 DC \times FD = {DB}^{2} + {AC}^{2} \\ \Rightarrow {BC}^{2} + {AB}^{2} + {AD}^{2} + {DC}^{2} + 2 EA \times AB - 2 AB \times EA = {DB}^{2} + {AC}^{2} \\ (Using equations (iv), (v) and (vi)) \\ \Rightarrow {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} \\ Hence, the sum of the squares of the diagonals of a parallelogram is \\ equal to the sum of the squares of its sides . \end{array}$

Q.91

$\begin{array}{l} In the following figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: \\ {(i) AC}^{2} = {AD}^{2} + BC.DM + {(\frac{BC}{2})}^{2} \\ {(ii) AB}^{2} = {AD}^{2} - BC . DM + {(\frac{BC}{2})}^{2} \\ {(iii) AC}^{2} + {AB}^{2} = {2AD}^{2} + \frac{1}{2} {BC}^{2} \end{array}$

Ans.

We have the following figure.

$\begin{array}{l} (i) \\ Applying Pythagoras theorem i n Δ AMD, we get \\ {AM}^{2} + {DM}^{2} = {AD}^{2} . .. (1) \\ Applying Pythagoras theorem i n Δ AMC, we get \\ {AM}^{2} + {MC}^{2} = {AC}^{2} \\ \Rightarrow {AM}^{2} + {(DM + DC)}^{2} = {AC}^{2} \\ \Rightarrow ({AM}^{2} + {DM}^{2}) + {DC}^{2} + 2 DM . DC = {AC}^{2} \\ Using equation (1), we ge t \\ {AD}^{2} + {DC}^{2} + 2 DM . DC = {AC}^{2} \\ \Rightarrow {AD}^{2} + {(\frac{BC}{2})}^{2} + 2 DM . \frac{BC}{2} = {AC}^{2} \\ \Rightarrow {AD}^{2} + {(\frac{BC}{2})}^{2} + DM \cdot BC = {AC}^{2} \\ (ii) \\ Applying Pythagoras theorem in Δ ABM, we get \\ {AB}^{2} = {AM}^{2} + {MB}^{2} \\ = ({AD}^{2} - {DM}^{2}) + {MB}^{2} \\ = ({AD}^{2} - {DM}^{2}) + {(BD - DM)}^{2} \\ = {AD}^{2} - {DM}^{2} + {BD}^{2} + {DM}^{2} - 2 BD \times DM \\ = {AD}^{2} + {BD}^{2} - 2 BD \times DM \\ = {AD}^{2} + {(\frac{BC}{2})}^{2} - 2 \frac{BC}{2} \times DM \\ = {AD}^{2} + {(\frac{BC}{2})}^{2} - BC \times DM \\ (iii) \\ Applying Pythagoras theorem in Δ AMB, we get \\ {AM}^{2} + {MB}^{2} = {AB}^{2} . .. (1) \\ Applying Pythagoras theorem in ΔAMC, we get \\ {AM}^{2} + {MC}^{2} = {AC}^{2} . .. (2) \\ Adding equations (1) and (2), we get \\ 2 {AM}^{2} + {MB}^{2} + {MC}^{2} = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + {(BD - DM)}^{2} + {(MD + DC)}^{2} = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + {BD}^{2} + {DM}^{2} - 2 BD . DM + {MD}^{2} + {DC}^{2} + 2 DM . DC = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + 2 {DM}^{2} + {BD}^{2} + {DC}^{2} + 2 DM (- BD + DC) = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 ({AM}^{2} + {DM}^{2}) + {(\frac{BC}{2})}^{2} + {(\frac{BC}{2})}^{2} + 2 DM (- \frac{BC}{2} + \frac{BC}{2}) = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AD}^{2} + \frac{{BC}^{2}}{2} = {AB}^{2} + {AC}^{2} \end{array}$

Q.92

$\begin{array}{l} In the following figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC . Prove that \\ {AC}^{2} = {AB}^{2} + {BC}^{2} - 2BC .BD . \end{array}$

Ans.

$\begin{array}{l} A pplying Pythagoras theorem i n ΔADB, we get \\ {AD}^{2} + {BD}^{2} = {AB}^{2} \\ \Rightarrow {AD}^{2} = {AB}^{2} - {BD}^{2} . .. (1) \\ A pplying Pythagoras theorem i n ΔADC, we get \\ {AD}^{2} + {DC}^{2} = {AC}^{2} \\ Now using equation (1), we get \\ {AB}^{2} - {BD}^{2} + {DC}^{2} = {AC}^{2} \\ \Rightarrow {AB}^{2} - {BD}^{2} + {(BC - BD)}^{2} = {AC}^{2} \\ \Rightarrow {AC}^{2} = {AB}^{2} - {BD}^{2} + {BC}^{2} + {BD}^{2} - 2 BC \times BD \\ \Rightarrow {AC}^{2} = {AB}^{2} + {BC}^{2} - 2 BC \times BD \end{array}$

Q.93

$\begin{array}{l} In the following figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that \\ {AC}^{2} = {AB}^{2} + {BC}^{2} + 2 BC . BD . \end{array}$

Ans.

Applying Pythagoras theorem in ΔADB, we get
AB² = AD² + DB² ….(1)
Applying Pythagoras theorem in ΔACD, we get
AC² = AD² + DC²
⇒AC² = AD² + ( BD + BC )²
⇒AC² = AD² + DB² + BC² + 2BD x BC
Now using equation ( 1 ), we get
AC² = AB² + BC² + 2BD × BC

Q.94

$\begin{array}{l} In the following figure, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that: \\ (i) {DM}^{2} = DN . MC (ii) {DN}^{2} = DM . AN \end{array}$

Ans.

(i)

$\begin{array}{l} We have, DN ∥ CB, DM ∥ AB and ∠ B = 90 ° \\ So, DNMB is a rectangle. \\ ∴ DN = MB and DM = NB \\ ∠ 2 + ∠ 3 = 90 ° . .. (1) \\ In Δ CDM, \\ ∠ 1 + ∠ 2 + ∠ DMC = 180 ° \\ ∠ 1 + ∠ 2 = 90 ° . .. (2) \\ In Δ DMB \\ ∠ 3 + ∠ DMB + ∠ 4 = 180 ° \\ ∠ 3 + ∠ 4 = 90 ° . .. (3) \\ From equation (1) and (2), \\ ∠ 1 = ∠ 3 \\ From equation (1) and (3), \\ ∠ 2 = ∠ 4 \\ ∴ Δ DCM ~ Δ BDM \\ \Rightarrow \frac{DM}{BM} = \frac{MC}{DM} \\ \Rightarrow {DM}^{2} = BM \times MC = DN \times MC [BM = DN] \\ (ii) \\ In Δ DBN, \\ ∠ 5 + ∠ 7 = 90 ° . .. (4) \\ In Δ DAN, \\ ∠ 6 + ∠ 8 = 90 ° . .. (5) \\ In Δ DAB, \\ ∠ 5 + ∠ 6 = 90 ° . .. (6) \\ From equation (4) and (6), we get \\ ∠ 6 = ∠ 7 \\ From equation (5) and (6), we get \\ ∠ 5 = ∠ 8 \\ ∴ Δ BND ~ Δ DNA [By AA similarity critarion] \\ \Rightarrow \frac{AN}{DN} = \frac{DN}{NB} \\ \Rightarrow {DN}^{2} = AN \times NB = AN \times DM [As NB = DM] \end{array}$

Q.95

$In the following figure, PS is the bisector of ∠ QPR of Δ PQR. Prove that \frac{QS}{SR} = \frac{PQ}{PR} .$

Ans.

$\begin{array}{l} We draw a line segment RT parallel to SP which \\ intersects extended line segment QP at point T. \\ It is given that PS is angle bisector of ∠ QPR. \\ Therefore, \\ ∠ QPS = ∠ SPR (1) \\ Also, \\ ∠ SPR = ∠ PRT (As PS ∥ TR) (2) \\ ∠ QPS = ∠ QTR (As PS ∥ TR) (3) \\ Using these equations we get, \\ ∠ PRT = ∠ QTR \\ So, PT = PR \\ Now in Δ QTR and Δ QPS, \\ ∠ QSP = ∠ QRT (As PS ∥ TR) \\ ∠ QPS = ∠ QTR (As PS ∥ TR) \\ ∠ Q is common \\ ∴ Δ QTR ~ Δ QPS \\ \Rightarrow \frac{QS}{SR} = \frac{QP}{PT} \\ \Rightarrow \frac{QS}{SR} = \frac{QP}{PR} [PT = PR] \end{array}$

Q.96

$\begin{array}{l} Tick the correct answer and justify: In Δ ABC, AB = 6 \sqrt{3} cm, AC = 12 cm and BC = 6 cm. \\ The angle B is : \\ (A) 120° (B) 60° \\ (C) 90° (D) 45° \end{array}$

Ans.

$\begin{array}{l} It is given that in Δ ABC, \\ AB = 6 \sqrt{3} cm, AC = 12 cm and BC = 6 cm \\ \Rightarrow {AB}^{2} = {108 cm}^{2} {, AC}^{2} = {144 cm}^{2} {and BC}^{2} = 36 {cm}^{2} \\ Now, \\ {AB}^{2} + {BC}^{2} = {108 cm}^{2} + 36 {cm}^{2} = 144 {cm}^{2} = {AC}^{2} \\ Therefore, by converse of Pythagoras theorem, we find that \\ in Δ ABC, angle B is 90°. \\ Therefore, correct answer is (C). \end{array}$

Q.97 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Ans.

$\begin{array}{l} Let ABC is an equilateral triangle and AE is its altitude. \\ Let length of each side of Δ ABC be a. \\ So, \\ BE = EC = \frac{1}{2} BC = \frac{a}{2} . \\ Using Pythagoras theorem in Δ ACE, we have \\ {AE}^{2} = {AC}^{2} - {EC}^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3}{4} a^{2} \\ \Rightarrow 3 a^{2} = 4 {AE}^{2} \\ Therefore, three times the square of one side of an \\ equilateral triangle is equal to four times the square \\ of one of its altitude. \end{array}$

Q.98

$In an equilateral triangle ABC, D is a point on side BC such that BD = \frac{1}{3} {BC. Prove that 9AD}^{2} = {7AB}^{2} .$

Ans.

$\begin{array}{l} It is given that ABC is an equilateral triangle. \\ Let length of each side of Δ ABC be a. \\ It is given that D is a point on side BC such that \\ BD = \frac{1}{3} BC = \frac{a}{3} . \\ Let AE is an altitude of equilateral triangle ABC. So, \\ BE = EC = \frac{1}{2} BC = \frac{a}{2} . \\ Also, AE = \sqrt{{AC}^{2} - {EC}^{2}} = \sqrt{a^{2} - \frac{a^{2}}{4}} = \frac{\sqrt{3}}{2} a \\ Again, DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{a}{6} \\ In Δ ADE, \\ {AD}^{2} = {AE}^{2} + {DE}^{2} [By Pythagoras Theorem] \\ \Rightarrow {AD}^{2} = {(\frac{\sqrt{3}}{2} a)}^{2} + \frac{a^{2}}{36} = \frac{3}{4} a^{2} + \frac{a^{2}}{36} = \frac{27 a^{2} + a^{2}}{36} = \frac{28}{36} a^{2} = \frac{7}{9} a^{2} \\ \Rightarrow {AD}^{2} = \frac{7}{9} a^{2} = \frac{7}{9} {AB}^{2} [AB = a] \\ \Rightarrow 9 {AD}^{2} = 7 {AB}^{2} \end{array}$

Q.99

$\begin{array}{l} The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD (see the \\ following figure). Prove that 2 {AB}^{2} = 2 {AC}^{2} + {BC}^{2} . \end{array}$

Ans.

$\begin{array}{l} In Δ ADC, \\ {AC}^{2} = {CD}^{2} + {AD}^{2} \\ \Rightarrow {AD}^{2} = {AC}^{2} - {CD}^{2} . .. (1) \\ In Δ ADB, \\ {AB}^{2} = {AD}^{2} + {DB}^{2} \\ \Rightarrow {AD}^{2} = {AB}^{2} - {DB}^{2} . .. (2) \\ From (1) and (2), we have \\ {AC}^{2} - {CD}^{2} = {AB}^{2} - {DB}^{2} . .. (3) \\ Now, it is given that DB = 3CD \\ Also, \\ BC = CD + DB \\ \Rightarrow BC = CD + 3CD = 4 CD \\ \Rightarrow CD = \frac{BC}{4} . .. (4) \\ Again, \\ BC = CD + DB \\ \Rightarrow BC = \frac{DB}{3} + DB = \frac{4 DB}{3} \\ \Rightarrow DB = \frac{3 BC}{4} . .. (5) \\ From (3), (4) and (5), we have \\ {AC}^{2} - {CD}^{2} = {AB}^{2} - {DB}^{2} \\ \Rightarrow {AC}^{2} - \frac{{BC}^{2}}{16} = {AB}^{2} - \frac{9 {BC}^{2}}{16} \\ \Rightarrow 16 {AC}^{2} - {BC}^{2} = 16 {AB}^{2} - 9 {BC}^{2} \\ \Rightarrow 16 {AB}^{2} = 16 {AC}^{2} - {BC}^{2} + 9 {BC}^{2} = 16 {AC}^{2} + 8 {BC}^{2} \\ \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + {BC}^{2} \end{array}$

Q.100

$\begin{array}{l} D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. \\ Prove that {AE}^{2} + {BD}^{2} = {AB}^{2} + {DE}^{2} . \end{array}$

Ans.

$\begin{array}{l} In Δ ACE, \\ {AE}^{2} = {AC}^{2} + {CE}^{2} . .. (1) \\ In Δ DCB, \\ {BD}^{2} = {CD}^{2} + {BC}^{2} . .. (2) \\ On adding (1) and (2), we have \\ {AE}^{2} + {BD}^{2} = {AC}^{2} + {CE}^{2} + {CD}^{2} + {BC}^{2} . .. (3) \\ In Δ ACB, \\ {AB}^{2} = {AC}^{2} + {BC}^{2} . .. (4) \\ In Δ DCE, \\ {DE}^{2} = {CD}^{2} + {CE}^{2} . .. (5) \\ On adding (4) and (5), we have \\ {AB}^{2} + {DE}^{2} = {AC}^{2} + {BC}^{2} + {CD}^{2} + {CE}^{2} . .. (6) \\ From (3) and (6), we find that \\ {AE}^{2} + {BD}^{2} = {AB}^{2} + {DE}^{2} \end{array}$

Q.101

$\begin{array}{l} Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet \\ of the poles is 12 m, find the distance between their tops. \end{array}$

Ans.

$\begin{array}{l} Let AB and CD be two poles of heights 6 m and 11m \\ respectively . Let their feet are on the ground at points B \\ and D . \\ It is given that distance between feet is 12 m . Therefore, \\ BD = 12 m . \\ Also, \\ AB = DE = 6m \\ and \\ CE = CD - DE = 11 m - 6 m = 5 m \\ Now, AEC is a right triangle . Therefore, by Pythagoras theorem, \\ AC = \sqrt{{AE}^{2} + {CE}^{2}} = \sqrt{12^{2} + 5^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 m \\ Therefore, \\ Distance between tops of the two poles = 13 m \end{array}$

Q.102

$\begin{array}{l} An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. \\ At the same time, another aeroplane leaves the same airport and flies due west at \\ a speed of 1200 km per hour. How far apart will be the two planes after 1 \frac{1}{2} hours? \end{array}$

Ans.

$\begin{array}{l} Distance travelled by the plane flying towards north in \\ 1 \frac{1}{2} hours = 1000 \times 1 \frac{1}{2} km = 1500 km \\ Distance travelled by the plane flying towards west in \\ 1 \frac{1}{2} hours = 1200 \times 1 \frac{1}{2} km = 1800 km \\ We represent thsese distances by OA and OB. \\ Also, AB represents the distance between the two planes. \\ Using Pythagoras theorem, we get \\ {AB}^{2} = {OA}^{2} + {OB}^{2} = 1500^{2} + 1800^{2} = 22, 50, 000 + 32, 40, 000 \\ \Rightarrow {AB}^{2} = 54, 90, 000 \\ \Rightarrow AB = \sqrt{54, 90, 000} = \sqrt{9 \times 6, 10, 000} = 300 \sqrt{61} \\ Therefore, distance between the two planes = 300 \sqrt{61} km \end{array}$

Q.103 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Ans.

$\begin{array}{l} Let AC is a pole and AB is a guy wire with stake at B . \\ Then, ABC is a right angle triangle. Therefore, using Pythagoras \\ theorem, we get \\ {AB}^{2} = {BC}^{2} + {AC}^{2} \\ \Rightarrow 24^{2} = {BC}^{2} + 18^{2} \\ \Rightarrow {BC}^{2} = 24^{2} - 18^{2} = 576 - 324 = 252 \\ \Rightarrow BC = \sqrt{252} = \sqrt{4 \times 9 \times 7} = 6 \sqrt{7} \\ Therefore, the distance of the foot of the ladder from the \\ base of the wall is 6 \sqrt{7} m. \end{array}$

Q.104 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Ans.

$\begin{array}{l} Let AB is a ladder and AC is wall . A represents the window \\ and BC is the distance of the foot of the ladder from the \\ base of the wall . \\ ABC is a right angle triangle. Therefore, using Pythagoras \\ theorem, we get \\ {AB}^{2} = {BC}^{2} + {AC}^{2} \\ \Rightarrow 10^{2} = {BC}^{2} + 8^{2} \\ \Rightarrow {BC}^{2} = 100 - 64 = 36 \\ \Rightarrow BC = 6 \\ Therefore, the distance of the foot of the ladder from the \\ base of the wall is 6 m. \end{array}$

Q.105

$\begin{array}{l} In the following figure, O is a point in the interior of a triangle ABC, \\ OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. \\ {(i) OA}^{2} {+OB}^{2} {+OC}^{2} - {OD}^{2} - {OE}^{2} - {OF}^{2} {= AF}^{2} {+BD}^{2} {+CE}^{2}, \\ {(ii) AF}^{2} {+BD}^{2} {+CE}^{2} {= AE}^{2} {+CD}^{2} {+BF}^{2} . \end{array}$

Ans.

$\begin{array}{l} We join O to A, B and C . \\ (i) \\ In Δ AOF, \\ {OA}^{2} = {OF}^{2} + {AF}^{2} [By Pythagoras theorem] \\ In Δ BOD, \\ {OB}^{2} = {OD}^{2} + {BD}^{2} [By Pythagoras theorem] \\ In Δ COE, \\ {OC}^{2} = {OE}^{2} + {EC}^{2} [By Pythagoras theorem] \\ Adding these equations, we get \\ {OA}^{2} + {OB}^{2} + {OC}^{2} = {OF}^{2} + {AF}^{2} + {OD}^{2} + {BD}^{2} + {OE}^{2} + {EC}^{2} \\ \Rightarrow {OA}^{2} + {OB}^{2} + {OC}^{2} - {OD}^{2} - {OE}^{2} - {OF}^{2} = {AF}^{2} + {BD}^{2} + {EC}^{2} \\ (ii) \\ We have form above result, \\ {AF}^{2} + {BD}^{2} + {EC}^{2} = {OA}^{2} + {OB}^{2} + {OC}^{2} - {OD}^{2} - {OE}^{2} - {OF}^{2} \\ = ({OA}^{2} - {OE}^{2}) + ({OB}^{2} - {OF}^{2}) + ({OC}^{2} - {OD}^{2}) \\ = {AE}^{2} + {BF}^{2} + {CD}^{2} \end{array}$

Q.106 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Ans.

$\begin{array}{l} Let ABCD is a rhombus. \\ We know that diagonals of a rhombus bisect each other \\ at right angles. \\ Using Pythagoras Theorem in Δ AOB, Δ BOC, Δ COD and Δ AOD, \\ we get \\ {AB}^{2} = {OA}^{2} + {OB}^{2}, \\ {BC}^{2} = {OB}^{2} + {OC}^{2}, \\ {CD}^{2} = {OC}^{2} + {OD}^{2}, \\ {DA}^{2} = {OA}^{2} + {OD}^{2} \\ Adding all these equations, we get \\ {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = 2 ({OA}^{2} + {OB}^{2} + {OC}^{2} + {OD}^{2}) \\ = 2 [2 {(\frac{AC}{2})}^{2} + 2 {(\frac{BD}{2})}^{2}] \\ = {AC}^{2} + {BD}^{2} \end{array}$

Q.107 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Ans.

$\begin{array}{l} It is given that ABC is an equilateral triangle of side 2a . \\ Let AD is an altitude . \\ We know that altitude bisects opposite side in an \\ equilateral triangle . \\ Therefore, BD = CD = a . \\ Using Pythagoras theorem in Δ ADB, we get \\ {AB}^{2} = {BD}^{2} + {AD}^{2} \\ \Rightarrow {AD}^{2} = {AB}^{2} - {BD}^{2} = {(2 a)}^{2} - a^{2} = 4 a^{2} - a^{2} = 3 a^{2} \\ \Rightarrow AD = a \sqrt{3} \\ We know that all the altitudes in an equilateral triangle \\ are of same length. Therefore, length of each altitude is a \sqrt{3} . \end{array}$

Q.108 ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Ans.

Given that ABC is an isosceles triangle with AC = BC and AB² = 2AC².
Therefore,
AB² = 2AC² = AC² + BC²
Therefore, by converse of Pythagoras theorem, ABC is a right triangle.

Q.109 ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Ans.

Given that ABC is an isosceles triangle right angled at C.
Therefore, AC = BC

Using Pythagoras theorem in the given triangle,
we have
AB²= AC² + BC² = AC² + AC²= 2AC²

Q.110

$\begin{array}{l} In the following figure, ABD is a triangle right angled at A and AC ⊥ B D. \\ Show that \\ (i) {AB}^{2} = BC . BD \\ (ii) {AC}^{2} = BC . DC \\ (iii) {AD}^{2} = BD . CD \end{array}$

Ans.

$\begin{array}{l} (i) \\ In Δ ADB and Δ CAB, we have \\ ∠ DAB = ∠ ACB = 90 ° \\ and ∠ ABD = ∠ CBA [Common angle] \\ Therefore, by AA similarity critarion, Δ ADB ~ Δ CAB . \\ Therefore, we have \\ \frac{AB}{CB} = \frac{BD}{AB} \\ \Rightarrow {AB}^{2} = BC \cdot BD \\ (ii) \\ Let ∠ CAB = θ \\ In Δ CBA, we have \\ ∠ CBA = 180 ° - 90 ° - θ = 90 ° - θ \\ In Δ CAD, we have \\ ∠ CAD = 90 ° - ∠ CAB = 90 ° - θ \\ ∠ CDA = 180 ° - 90 ° - (90 ° - θ) = θ \\ In Δ CBA and Δ CAD, we have \\ ∠ CBA = ∠ CAD, ∠ CAB = ∠ CDA and ∠ ACB = ∠ DCA = 90 ° \\ Therefore, by AAA similarity critarion, Δ CBA ~ Δ CAD . \\ Therefore, we have \\ \frac{AC}{DC} = \frac{BC}{AC} \\ \Rightarrow {AC}^{2} = BC \cdot DC \\ (iii) \\ In Δ DCA and Δ DAB, \\ ∠ DCA = ∠ DAB = 90 °, ∠ CDA = ∠ ADB \\ Therefore, by AA similarity critarion, Δ DCA ~ Δ DAB . \\ Therefore, we have \\ \frac{DC}{DA} = \frac{DA}{DB} \\ \Rightarrow {AD}^{2} = BD \cdot CD \end{array}$

Q.111 PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that ( PM )²= QM . MR.

Ans.

$\begin{array}{l} Let ∠ MPR = θ \\ In Δ MPR, \\ ∠ MRP = 180 ° - 90 ° - θ = 90 ° - θ \\ Similarly, in Δ MPQ, \\ ∠ MPQ = 90 ° - ∠ MPR = 90 ° - θ \\ ∠ MQP = 180 ° - 90 ° - (90 ° - θ) = θ \\ Now, in Δ MPR and Δ MQP, we have \\ ∠ MQP = ∠ MPR, ∠ MPQ = ∠ MRP and ∠ PMQ = ∠ PMR. \\ Therefore, by AAA similarity critarion, Δ MPR ~ Δ MQP . \\ Therefore, we have \\ \frac{QM}{PM} = \frac{MP}{MR} \\ \Rightarrow {PM}^{2} = QM \times MR \end{array}$

Q.112 Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Ans.

(i) Given sides of the triangle are 7 cm, 24 cm and 25 cm.
Squares of the given sides of the triangle are 49 cm², 576 cm² and 625 cm².
Now,
49 cm²+ 576 cm² = 625 cm²
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 25 cm.

(ii)

$\begin{array}{l} Given sides of the triangle are 3 cm, 8 cm and 6 cm . \\ {Squares of the given sides of the triangle are 9 cm}^{2}, {64 cm}^{2} \\ {and 36 cm}^{2} . \\ Now, \\ 9 + 36 \neq 64 \\ We find that sum of the squares of two sides is not equal to \\ the square of third side. \\ Therefore, triangle of the given sides is not a right triangle. \\ (iii) \\ Given sides of the triangle are 50 cm, 80 cm and 100 cm . \\ {Squares of the given sides of the triangle are 2500 cm}^{2}, \\ {6400 cm}^{2} {and 10000 cm}^{2} . \\ Now, \\ 2500 + 6400 \neq 10000 \\ We find that sum of the squares of two sides is not equal to \\ the square of third side. \\ Therefore, triangle of the given sides is not a right triangle. \end{array}$

(iv) Given sides of the triangle are 13 cm, 12 cm and 5 cm.
Squares of the given sides of the triangle are 169 cm², 144 cm² and 25 cm².
Now, 144 c m²+ 25 cm² = 169 cm²
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 13 cm.

Q.113

$\begin{array}{l} Sides of two similar triangles are in the ratio 4 : 9 . Areas of these triangles are in the ratio \\ (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 \end{array}$

Ans.

$\begin{array}{l} It is given that sides of two similar triangles are in the \\ ratio 4 : 9. \\ Therefore, \\ Ratio of areas of these triangles = \frac{4^{2}}{9^{2}} = \frac{16}{81} \\ Hence, the correct answer is (D). \end{array}$

Q.115

$\begin{array}{l} ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles \\ ABC and BDE is \\ (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 \end{array}$

Ans6

$\begin{array}{l} It is given that Δ ABC and Δ BDE are equilateral and D is the \\ mid point of BC. Therefore, BD = \frac{BC}{2} \\ We know that equilateral triangles are similar to each other \\ and so ratio of their areas is equal to square of the ratio of \\ their sides. \\ ∴ \frac{ar (ΔA BC)}{ar (ΔBDE)} = {(\frac{AB}{BD})}^{2} = {(\frac{BC}{BD})}^{2} [ΔA BC is equilateral] \\ = {(\frac{BC}{\frac{BC}{2}})}^{2} = \frac{4}{1} \\ \Rightarrow ar (ΔA BC) : ar (ΔBDE) = 4 : 1 \\ Hence, the correct answer is (C). \end{array}$

Q.110

$\begin{array}{l} Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the \\ equilateral triangle described on one of its diagonals. \end{array}$

Ans.

$\begin{array}{l} Let ABCD is a square and length of its one side is a unit. \\ Therefore, length of its diagonal = \sqrt{2} a \\ Equilateral triangles as per question are formed here as \\ Δ ABE and ΔDBF . \\ Length of a s ide of equilateral Δ ABE = a \\ and \\ length of a s ide of equilateral ΔDBF = \sqrt{2} a \\ We know that equilateral triangles are similar to each other. \\ Therefore, ratio of their areas is equal to ratio of squares of \\ their sides. \\ i . e ., \frac{ar (ΔA BE)}{ar (ΔDBF)} = {(\frac{a}{\sqrt{2} a})}^{2} = \frac{1}{2} \end{array}$

Q.117 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Ans.

$\begin{array}{l} Let us take two triangles ABC and PQR such that Δ ABC ~ ΔPQR . \\ Let AD and PS be the medians of Δ ABC and ΔPQR respectively. \\ Now, \\ Δ ABC ~ ΔPQR \\ \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} and ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R . \\ Since AD and PS are medians, so we have \\ BD = DC = \frac{BC}{2} and QS = SR = \frac{QR}{2} \\ In Δ ABD and ΔPQS, \\ ∠ B = ∠ Q and \frac{AB}{PQ} = \frac{BC}{QR} = \frac{BD}{QS} \\ ∴ Δ ABD ~ ΔPQS [SAS similarity criterion] \\ \Rightarrow \frac{AB}{PQ} = \frac{BD}{QS} = \frac{AD}{PS} \\ Now, \\ Δ ABC ~ ΔPQR \\ \Rightarrow \frac{ar (ΔA BC)}{ar (ΔPQR)} = {(\frac{AB}{PQ})}^{2} \\ \Rightarrow \frac{ar (ΔA BC)}{ar (ΔPQR)} = {(\frac{AD}{PS})}^{2} \end{array}$

Q.118

$\begin{array}{l} D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas \\ of Δ DEF and Δ ABC. \end{array}$

Ans.

$\begin{array}{l} D and E are mid-points of Δ ABC. \\ ∴ DE ∥ AC and DE = \frac{1}{2} AC \\ In Δ BED and Δ BCA, \\ ∠ BED = ∠ BCA [Corresponding angles] \\ ∠ BDE = ∠ BAC [Corresponding angles] \\ ∠ EBD = ∠ CBA [Common angles] \\ ∴ Δ BED ~ Δ BCA [AAA similarity criterion] \\ \Rightarrow \frac{ar (Δ BED)}{ar (Δ BCA)} = {(\frac{DE}{AC})}^{2} = {(\frac{\frac{1}{2} AC}{AC})}^{2} = \frac{1}{4} \\ \Rightarrow ar (Δ BED) = \frac{1}{4} ar (Δ BCA) \\ Similarly, \\ ar (ΔCF E) = \frac{1}{4} ar (Δ CBA) and ar (ΔADF) = \frac{1}{4} ar (Δ ABC) \\ Also, \\ ar (ΔD EF) = ar (Δ ABC) - [ar (Δ BED) + ar (ΔCF E) + ar (ΔADF)] \\ \Rightarrow ar (ΔD EF) = ar (Δ ABC) - \frac{3}{4} ar (Δ ABC) = \frac{1}{4} ar (Δ ABC) \\ \Rightarrow \frac{ar (ΔD EF)}{ar (Δ ABC)} = \frac{1}{4} \end{array}$

Q.119 If the areas of two similar triangles are equal, prove that they are congruent.

Ans.

$\begin{array}{l} Let ΔABC ~ ΔPQR, then we have \\ \frac{ar (ΔABC)}{ar (ΔPQR)} = {(\frac{AB}{PQ})}^{2} = {(\frac{BC}{QR})}^{2} = {(\frac{AC}{PR})}^{2} \\ \Rightarrow \frac{ar (ΔABC)}{ar (ΔABC)} = {(\frac{AB}{PQ})}^{2} = {(\frac{BC}{QR})}^{2} = {(\frac{AC}{PR})}^{2} [Given ar (ΔABC) = ar (ΔPQR)] \\ \Rightarrow 1 = {(\frac{AB}{PQ})}^{2} = {(\frac{BC}{QR})}^{2} = {(\frac{AC}{PR})}^{2} \\ \Rightarrow 1 = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \\ \Rightarrow AB = PQ, BC = QR and AC = PR \\ ∴ ΔABC ≅ ΔPQR [By SSS congruence criterion] \end{array}$

Q.110 In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

Ans.

$We draw perpendiculars AP and DM on line BC.$

$\begin{array}{l} We know that area of a triangle = \frac{1}{2} \times Base \times Height \\ ∴ \frac{ar (ΔABC)}{ar (ΔDBC)} = \frac{\frac{1}{2} BC \times AP}{\frac{1}{2} BC \times DM} = \frac{AP}{DM} \\ In Δ APO and ΔDM O, \\ ∠ APO = ∠ DMO = 90 ° \\ ∠ A OP = ∠ D OM [Vertically opposite angles] \\ ∴ Δ APO ~ ΔDM O [By AA similarity critarion] \\ \Rightarrow \frac{AP}{DM} = \frac{AO}{DO} \\ \Rightarrow \frac{ar (ΔABC)}{ar (ΔDBC)} = \frac{AO}{DO} \end{array}$

Q.120 Diagonals of a trapezium ABCD with AB ║ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Ans.

$\begin{array}{l} In ΔC OD and ΔA OB, \\ ∠ ODC = ∠ OBA [Alternate interior angles as AB ∥ CD] \\ ∠ DCO = ∠ BAO [Alternate interior angles as AB ∥ CD] \\ ∠ C OD = ∠ A OB [Vertically opposite angles] \\ ∴ ΔC OD ~ ΔA OB [AAA similarity critarion] \\ \Rightarrow \frac{ar (ΔC OD)}{ar (ΔA OB)} = \frac{{CD}^{2}}{{AB}^{2}} \\ \Rightarrow \frac{ar (ΔC OD)}{ar (ΔA OB)} = \frac{{CD}^{2}}{{(2 CD)}^{2}} = \frac{1}{4} [Given AB = 2 CD] \\ \Rightarrow ar (ΔA OB) : ar (ΔC OD) = 4 : 1 \end{array}$

Q.121

$Let Δ ABC ~ Δ {DEF and their areas be, respectively, 64 cm}^{2} {and 121 cm}^{2} . If EF = 15.4 cm, find BC.$

Ans.

$\begin{array}{l} It is given that Δ ABC ~ Δ DEF. \\ ∴ \frac{ar (Δ ABC)}{ar (Δ DEF)} = \frac{{BC}^{2}}{{EF}^{2}} \\ Given that \\ EF = 15 .4 cm \\ ar (Δ ABC) = 64 {cm}^{2} \\ ar (ΔDEF) = 121 {cm}^{2} \\ ∴ \frac{ar (Δ ABC)}{ar (Δ DEF)} = \frac{{BC}^{2}}{{EF}^{2}} \\ \Rightarrow \frac{64}{121} = \frac{{BC}^{2}}{{(15 .4)}^{2}} \\ \Rightarrow \frac{8}{11} = \frac{BC}{15 .4} \\ \Rightarrow BC = \frac{8 \times 15 .4}{11} \\ \Rightarrow BC = 11 .2 \end{array}$

Q.122

$\begin{array}{l} If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that \\ \frac{AB}{PQ} = \frac{AD}{PM} . \end{array}$

Ans.

$\begin{array}{l} It is given that AD and PM are medians . Therefore, \\ \frac{1}{2} BC = BD = DC and \frac{1}{2} QR = QM = MR \\ Also, Δ ABC ~ Δ PQR \\ Therefore, \\ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} and ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R \\ In Δ ABD and Δ PQM, \\ \frac{AB}{PQ} = \frac{\frac{1}{2} BC}{\frac{1}{2} QR} = \frac{BD}{QM} and ∠ B = ∠ Q \\ ∴ Δ ABD ~ Δ PQM [By SAS similarity critarion] \\ \Rightarrow \frac{AB}{PQ} = \frac{AD}{PM} \end{array}$

Q.123

$\begin{array}{l} A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower \\ casts a shadow 28 m long. Find the height of the tower. \end{array}$

Ans.

$\begin{array}{l} Let CD is a pole and DF is its shadow . \\ Let AB is a tower and BE is its shadow . \\ At the same time in a day, the sun rays will fall on pole \\ and tower at the same angle . \\ Therefore, \\ ∠ DCF = ∠ BAE \\ Also, ∠ CDF = ∠ ABE = 90 ° \\ ∴ Δ ABE ~ Δ CDF [By AA similarity critarion] \\ Corresponding sides are proportional in similar triangles . \\ Therefore, \\ \frac{AB}{CD} = \frac{BE}{DF} \\ \Rightarrow \frac{AB}{6} = \frac{28}{4} = 7 \\ \Rightarrow AB = 42 \\ Therefore, the height of the tower is 42 m. \end{array}$

Q.124

$\begin{array}{l} Sides AB and AC and median AD of a triangle ABC \\ are respectively proportional to sides PQ and PR and \\ median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR \end{array}$

Ans.

$\begin{array}{l} It is given that, \\ \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \\ We extend AD and PM up to point E and L respectively, \\ such that AD = DE and PM = ML. We join B to E, C to E, \\ Q to L and R to L. \end{array}$

$\begin{array}{l} It is given that AD and PM are medians . Therefore, \\ \frac{1}{2} BC = BD = DC and \frac{1}{2} QR = QM = MR \\ In quadrilateral ABEC, diagonals AE and BC bisect each \\ other at point D. Therefore, quadrilateral ABEC is a \\ parallelogram. \\ ∴ AC = BE and AB = EC \\ Similarly, we can prove that quadrilateral PQLR is \\ a parallelogram and PR = QL, PQ = LR. \\ It is given that \\ \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \\ \Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{2AD}{2PM} \\ \Rightarrow \frac{AB}{PQ} = \frac{BE}{QL} = \frac{AE}{PL} \\ Therefore, by SSS similarity critarion, Δ ABE ~ Δ PQL and so \\ in Δ ABE and Δ PQL, \\ ∠ BAE = ∠ QPL . .. (1) \\ Similarly, we can show that Δ AEC ~ Δ PLR and \\ ∠ CAE = ∠ RPL . .. (2) \\ Adding (1) and (2), we get \\ ∠ BAE + ∠ CAE = ∠ QPL + ∠ RPL \\ ∠ CAB = ∠ RPQ \\ Also, we have \\ \frac{AB}{PQ} = \frac{AC}{PR} \\ Therefore, by SAS similarity critarion, Δ ABC ~ Δ PQR. \end{array}$

Q.125

$D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ {BAC. Show that CA}^{2} = CB \cdot CD.$

Ans.

$\begin{array}{l} In Δ BAC and Δ ADC, \\ ∠ ADC = ∠ BAC (Given) \\ ∠ ACD = ∠ BCA (Common angle) \\ ∴ Δ ADC ~ Δ BAC (By AA similarity critarion) \\ We know that corresponding sides of similar triangles \\ are proportional. \\ ∴ \frac{CA}{CD} = \frac{CB}{CA} \\ \Rightarrow {CA}^{2} = CB \cdot CD \end{array}$

Q.126

$\begin{array}{l} Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and \\ median PM of Δ PQR (see the following figure). Show that Δ ABC ~ Δ PQR. \end{array}$

Ans.

$\begin{array}{l} It is given that AD and PM are medians . Therefore, \\ \frac{1}{2} BC = BD = DC \\ and \\ \frac{1}{2} QR = QM = MR \\ It is given that \\ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \\ \Rightarrow \frac{AB}{PQ} = \frac{\frac{1}{2} BC}{\frac{1}{2} QR} = \frac{AD}{PM} \\ \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \\ Therefore, by SSS similarity critarion, Δ ABD ~ Δ PQM and so \\ in Δ ABC and Δ PQR, ∠ B = ∠ Q and \end{array}$

NCERT Solutions for Class 10 Maths

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NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter-wise

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

NCERT Solution for Class 10 Mathematics Chapter 2 – Polynomials

NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

NCERT Solutions for Class 10 Mathematics Chapter 5 – Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry

NCERT Solutions for Class 10 Mathematics Chapter 10 – Circles

NCERT Solutions for Class 10 Mathematics Chapter 11 – Constructions

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes

NCERT Solutions for Class 10 Maths Chapter 14 – Statistics

NCERT Solutions for Class 10 Maths Chapter 15 – Probability

CBSE Class 10 Maths Exam Pattern for 2022-2023

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