NCERT Solutions for Class 10 Maths

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 Mathematics 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 practise Mathematics exercises on a regular basis. As a result, anytime you intend to practise after having previously solved exercises , you may not recall everything, which is why having a ready reference for practise 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 utilised to explain more complicated subjects, emphasising 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. 

The 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

In which of the following situations, does the list ofnumbers 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 14 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.

Ans.

(i) Fare for the first km       =  15Fare for the second km =  15 +  8 =  23Fare for the third km =  15 + 8 +  8 =  31and so on.Therefore, fare for each successive kilometre in rupeesare:15, 23, 31, 39, ...Difference between fares for a particular kilometre and itspreceding 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 14 of air remaining in thecylinder each time.Therefore, amounts of air in the cylinder after eachsuccessive removal by the vacuum pump will beV, 3V4, 3V43V4×14, 3V43V4×143V43V4×14×14, ...i.e., V, 3V4, 342V, 343V, ...Here, the differences between a term and its preceding termare not equal. Therefore, list of numbers involved doesnot make an A.P.(iii)According to question, cost in rupees of digging firstmetre and thereafter each subsequent metre arerespectively: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=100001+81001= 10800Amount at the end of second year=108001+81001= 11664Amount at the end of third year=116641+81001= 12597.12Obviously, differences between amounts in any two successiveyears are not equal. Therefore, list of numbers involved heredoes not make an AP.

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


Ans.

We know that sum of the measures of the oppositeangles of a cyclic quadrilateral is 180°.Therefore,             A+C=180 4y+204x=180     xy=40 ...(i)Also,           B+D=180 3y57x+5=180 7x + 3y = 180 ...(ii)Multiplying equation (i) by 3, we get 3x3y=120 ...(iii)Adding equations (ii) and (iii), we get 7x+3x=60x=15Putting value of x in equation (i), we get 15 y=40 y=25Thus,A=4y+20=4×25+20=120°B=3y5=3×255=70°C=4x=4×15=60°D=7x+5=7×15+5=110°.

Q.3

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)  xayb= 0                           (iv) (a-b)x+(a+b)y = a2-2ab-b2       ax+by = a2+b2                     (a+b)(x+y) = a2+b2(v)  152x-378y = -74       -378x+152y = -604

Ans.

(i)   The given equations can be written as:px+qy(pq)=0                 ...(1)qxpy(p+q)=0                 ...(2)By cross-multiplication method, we have       xpqq2(p2pq)=ypq+q2+p2+pq=1p2q2xq2p2=yq2+p2=1p2q2x=1 and y=1 (ii)   The given equations can be written as:ax+byc=0                          ...(1)bx+ay(1+c)=0                 ...(2)By cross-multiplication method, we have       xbbc+ac=ybc+a+ac=1a2b2xc(ab)b=yc(ab)+a=1a2b2x=c(ab)ba2b2 and   y=c(ab)+aa2b2(iii)The given equations can be written as:bxay=0                                ...(1)ax+by(a2+b2)=0            ...(2)By cross-multiplication method, we have       xa(a2+b2)=yb(a2+b2)=1a2+b2   x=a and   y=b (iv)The given equations can be written as:(ab)x+(a+b)y(a22abb2)=0                      ...(1)(a+b)x+(a+b)y(a2+b2)=0                                   ...(2)By cross-multiplication method, we have        x(a+b)(a2+b2)+(a+b)(a22abb2)       =y(a+b)(a22abb2)+(ab)(a2+b2)=1a2b2(a+b)2   x(a+b)(a2b2+a22abb2)              =ya3+2a2b+ab2a2b+2ab2+b3+a3+ab2a2bb3             =1a2b2a2b22ab   x2b(a+b)2=y4ab2=12b(a+b)   x=a+b and     y=2aba+b(v)   152x378y=74378x+152y=604The given equations can be written as:76x189y+37=0                      ...(1)189x+76y+302=0                  ...(2)By cross-multiplication method, we have        x37×76189×302=y189×3776×302=17621892          x=37×76+189×3027621892=2and       y=189×3776×3027621892=1

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

In a ΔABC, C=3 B=2A+B. Find thethree angles.

Ans.

Given that,        C=3B=2(A+B) 3B=2A+2 B=2A       2AB = 0 ...(1)We know that the sum of measures of all angles ofa triangle is 180°. Therefore,     A+B+C=180°A+B+3B=180°A+4B=180° ...(2)Multiplying equation (1) by 4, we get     8A4B = 0 ...(3)Adding equations (2) and (3), we get 9A = 180°A=20°From equation (2), we get 20°+4B=180° 4B=160°              B=40°and C=3BC=120°.Therefore, A, B and C are respectively20°, 40° and 120°.

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

A train covered a certain distance at a uniformspeed. If the train would have been 10 km/h faster,it would have taken 2 hours less than the scheduledtime. And, if the train were slower by 10 km/h; itwould have taken 3 hours more than the scheduledtime. Find the distance covered by the train.

Ans.

Let the distance covered by the train is x km and theuniform speed of the train is y km/h. Then time takento travel this distance is xy hour.According to question,    xy2=xy+10   (x2y)(y+10)=xy   xy+10x2y220y=xy   10x2y220y=0                                  ...(1)Also,       xy+3=xy10   (x+3y)(y10)=xy   xy10x+3y230y=xy   10x+3y230y=0                                ...(2)Adding equations (1) and (2), we get   10x2y220y=010x+3y230y=0¯                 y250y=0            y(y50)=0            y=0 or    y=50 Here, speed can not be taken as zero unit as then distancetravelled would be zero unit.So, y=50 Putting this value of y in equation (1), we get           10x2(50)220×50=0       10x50001000=0            x=600Hence, distance covered by the train is 600 km.

Q.8

One says, “Give me a hundred, friend! I shall thenbecome twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich asyou”. Tell me what is the amount of their(respective) capital? [From the Bijaganita of Bhaskara II][Hint: x+100=2(y100), y+10=6(x10)]

Ans.

Let their respective capitals are x and y.According to question,    x+100=2(y100)    x2y=300                 ...(1)Also,        y+10=6(x10)    y6x=70    y=6x70                      ...(2)Putting this value of y in equation (1), we get        x2(6x70)=300     x12x=300140     x=44011=40Putting this value of x in equation (2), we get          y=6×4070=24070=170     Hence, their respective capitals are 40 and 170.

Q.9

The ages of two friends Ani and Biju differ by 3 years.Anis father Dharam is twice as old as Ani and Biju istwice as old as his sister Cathy. The ages of Cathy andDharam differ by 30 years. Find the ages of Ani and Biju.

Ans.

Let the ages of Ani and Biju be x and y years respectively.If Ani is older than Biju thenxy=3                          ...(1)Anis father Dharam is twice as old as Ani. So,Dharams age = 2xThe ages of Cathy and Dharam differ by 30 years.Therefore,  Cathys age = 2x30Biju is twice as old as Cathy. Therefore,y=2(2x30)            ...(2)Putting the value of y from equation (2) in equation (1), we get        x2(2x30)=3    x4x+60=3  3x=360=57      x=19From equation (1), we gety=193=16Hence, Ani is 19 years old and Biju is 16 years old.Again, if Biju is older than Ani thenyx=3                          ...(3)Again,  putting the value of y from (2) in (3), we get      2(2x30)x=3    4x60x=3    3x=3+60=63      x=21Putting this value of x in equation (3), we gety=21+3=24Hence, in this case Ani is of 21 years and Biju is of 24 years.

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)

Let the number of days taken by a woman to finish the workis x and the number of days taken by a man to finish thework is y.Therefore, Part of the work finished by a woman in 1 day = 1xandpart of the work finished by a man in 1 day = 1yGiven that the .Therefore, 4 (2x+5y) = 1 2x+5y=14 It is also given that the work .Therefore, 3  (3x+6y) = 1 3x+6y=13Putting 1x=p and 1y=q in the above equations, we get 2p + 5q = 148p+20q=1 3p + 6q = 139p+18q=1By cross-multiplication method, we get      p20(18)=q9(8)=1144180p2=q1=136p2=136 and q1=136p=118 and q=136p=1x=118 and q=1y=136x=18 and y=36Hence, number of days taken by a woman = 18 daysand         number of days taken by a man = 36 days. (iii)Let the speeds of train and bus be x km/h and y km/h respectively.According to question, 60x+240y=4 ...(1)and 100x+200y=256 ...(2)Putting 1x=p and 1y=q in these equations, we get 60p+240q=4 ...(3)and 100p+200q=256 600p+1200q=25 ...(4)Multiplying equation (3) by 10, we get 600p+2400q=40 ...(5)Subtracting equation (4) ​from (5), we get 1200q = 15q=180Substituting in equation (3), we get 60p = 1p=160Now, p=1x=160 and q = 1y=180

x = 60 km/h and y = 80 km/hHence, speed of train = 60 km/hand  speed of bus = 80 km/h

Q.11

Solve the following pair of equations by reducing them to a pair of linear equations:(i) 12x + 13y=2                            (ii) 2x + 3y = 2   13x + 12y = 136                               4x9y =–1(iii) 4x + 3y = 14                         (iv) 5x – 1 + 1y – 2 = 2     3x 4y = 23                                6x – 13y – 2 = 1(v) 7x – 2yxy = 5                            (vi) 6x + 3y = 6xy    8x + 7yxy = 15                                 2x + 4y = 5xy(vii) 10x + y + 2x – y = 4                 (viii)  13x + y + 13x  y = 34      15x + y5x – y= –2                         12(3x + y)12(3x – y) = –18

Ans.

(i)Let 1x=p and 1y=qthen given equations can be written as:p2+q3=2 3p+2q12=0 ...(1)p3+q2=136 2p+3q13=0 ...(2)Using cross-multiplication method, we getp26(36)=q24(39)=194 p10=q15=15 p10=15 and q15=15 p=2 and q=3 1x=2 and 1y=3 x=12 and y=13(ii) Given  that2x+3y=2and4x9y=1Let 1x=p and 1y=q , then we get2p+3q=2 ...(1)4p9q=1 ...(2)Multiplying equation (1) by 3, we get6p + 9q = 6Adding this to equation (2), we get 10p = 5 p= 12Putting value of p in equation (1), we get 2×12+3q=2 3q=1 q=13 Now,p=1x=12x=2x=4q=1y=13y=3y=9Hence, x=4 and y=9.(iii) Given that4x+3y=143x4y=23Substituting 1x=p in above equations, we get4p+3y14=0 ...(1)3p4y23=0 ...(2)By cross-multiplication method, we get       p6956=y42(92)=1169   p125=y50=125  p=5 and y=2p=1x=5x=15    and  y=2

(iv)Giventhat5x1+1y2=2and6x13y2=1Putting 1x1=p and 1y2=q , we get5p+q=2 ...(1)6p3q=1 ...(2)multiplying equation (1) by 3, we get15p+3q=6 ...(3)Adding (2) and (3), we get 21p=7p=13Putting value of p in equation (1), we get 5×13+q=2q=253=13Now,           p=1x1=13x1=3x=4and q=1y2=13y2=3y=5   x=4 and y=5.

(v)We have      7x2yxy=5 7y2x=5 ...(1)    8x+7yxy=15 8y+7x=15 ...(2)Putting 1x=p and 1y=q in equation (1) and (2), we get 2p+7q5=0 ...(3) 7p+8q15=0 ...(4)By cross-multiplication method, we getp105(40)=q3530=11649p65=q65=165p65=165 and q65=165p=1 and q=1p=1x=1 and q=1y=1x=1 and y=1

(vi)We have6x+3y=6xy         6y+3x=6 ...(1)and2x+4y=5xy         2y+4x=5 ...(2)Putting 1x=p and 1y=q in above equations, we get3p+6q6=0 ...(3)4p+2q5=0 ...(4)By cross-multiplication method, we have      p30(12)=q24(15)=1624p18=q9=118p18=118 and q9=118p=1 and q=12p=1x=1 and q=1y=12x=1 and y=2 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xH8yiVC0xbbG8F4rqqrFfpe ea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0Firpe peKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaqaamaaeaqbaaGcea qabeaacaGGOaGaamODaiaadMgacaGGPaaabaGaae4vaiaabwgacaqG GaGaaeiAaiaabggacaqG2bGaaeyzaaqaaiaaiAdacaWG4bGaey4kaS IaaG4maiaadMhacqGH9aqpcaaI2aGaamiEaiaadMhacaaMc8UaaGPa VlaaykW7caaMc8UaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWcaaqaaiaaiAdaaeaacaWG5baaaiabgUcaRmaalaaabaGaaG4m aaqaaiaadIhaaaGaeyypa0JaaGOnaiaabccacaqGGaGaaeiiaiaabc cacaqGGaGaaeOlaiaab6cacaqGUaGaaeikaiaabgdacaqGPaaabaGa aeyyaiaab6gacaqGKbaabaGaaGOmaiaadIhacqGHRaWkcaaI0aGaam yEaiabg2da9iaaiwdacaWG4bGaamyEaiaaykW7caaMc8UaaGPaVlaa ykW7cqGHshI3caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaalaaaba GaaGOmaaqaaiaadMhaaaGaey4kaSYaaSaaaeaacaaI0aaabaGaamiE aaaacqGH9aqpcaaI1aGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca qGUaGaaeOlaiaab6cacaqGOaGaaeOmaiaabMcaaeaacaqGqbGaaeyD aiaabshacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiamaalaaabaGaaG ymaaqaaiaadIhaaaGaeyypa0JaamiCaiaabccacaqGHbGaaeOBaiaa bsgacaqGGaWaaSaaaeaacaaIXaaabaGaamyEaaaacqGH9aqpcaWGXb GaaeiiaiaabMgacaqGUbGaaeiiaiaabggacaqGIbGaae4BaiaabAha caqGLbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaae4CaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGa ae4zaiaabwgacaqG0baabaGaae4maiaabchacqGHRaWkcaaI2aGaam yCaiabgkHiTiaaiAdacqGH9aqpcaaIWaGaaeiiaiaabccacaqGGaGa aeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca qGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaa bccacaqGGaGaaeOlaiaab6cacaqGUaGaaeikaiaabodacaqGPaaaba GaaeinaiaabchacqGHRaWkcaaIYaGaamyCaiabgkHiTiaaiwdacqGH 9aqpcaaIWaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeOlaiaab6 cacaqGUaGaaeikaiaabsdacaqGPaaabaGaaeOqaiaabMhacaqGGaGa ae4yaiaabkhacaqGVbGaae4CaiaabohacaqGTaGaaeyBaiaabwhaca qGSbGaaeiDaiaabMgacaqGWbGaaeiBaiaabMgacaqGJbGaaeyyaiaa bshacaqGPbGaae4Baiaab6gacaqGGaGaaeyBaiaabwgacaqG0bGaae iAaiaab+gacaqGKbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqG ObGaaeyyaiaabAhacaqGLbaabaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8+aaSaaaeaacaWGWbaabaGaeyOeI0IaaG4maiaaicda cqGHsislcaGGOaGaeyOeI0IaaGymaiaaikdacaGGPaaaaiabg2da9m aalaaabaGaamyCaaqaaiabgkHiTiaaikdacaaI0aGaeyOeI0Iaaiik aiabgkHiTiaaigdacaaI1aGaaiykaaaacqGH9aqpdaWcaaqaaiaaig daaeaacaaI2aGaeyOeI0IaaGOmaiaaisdaaaaabaGaeyO0H49aaSaa aeaacaWGWbaabaGaeyOeI0IaaGymaiaaiIdaaaGaeyypa0ZaaSaaae aacaWGXbaabaGaeyOeI0IaaGyoaaaacqGH9aqpdaWcaaqaaiaaigda aeaacqGHsislcaaIXaGaaGioaaaaaeaacqGHshI3daWcaaqaaiaadc haaeaacqGHsislcaaIXaGaaGioaaaacqGH9aqpdaWcaaqaaiaaigda aeaacqGHsislcaaIXaGaaGioaaaacaqGGaGaaeyyaiaab6gacaqGKb GaaeiiamaalaaabaGaamyCaaqaaiabgkHiTiaaiMdaaaGaeyypa0Za aSaaaeaacaaIXaaabaGaeyOeI0IaaGymaiaaiIdaaaaabaGaeyO0H4 TaamiCaiabg2da9iaaigdacaqGGaGaaeiiaiaabggacaqGUbGaaeiz aiaabccacaqGGaGaaeyCaiabg2da9maalaaabaGaaGymaaqaaiaaik daaaaabaGaeyO0H4TaamiCaiabg2da9maalaaabaGaaGymaaqaaiaa dIhaaaGaeyypa0JaaGymaiaabccacaqGGaGaaeyyaiaab6gacaqGKb GaaeiiaiaabccacaqGGaGaaeyCaiabg2da9maalaaabaGaaGymaaqa aiaadMhaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaaaeaacq GHshI3caWG4bGaeyypa0JaaGymaiaabccacaqGGaGaaeyyaiaab6ga caqGKbGaaeiiaiaadMhacqGH9aqpcaaIYaaaaaa@7FC7@

(vii) We have    10x+y+2xy=4and    15x+y5xy=2Putting 1x+y=p and 1xy=q in above equations, we get      10p+2q4=0 ...(3)and      15p5q+2=0 ...(4)By cross-multiplication method, we have      p420=q6020=15030p16=q80=180p16=180 and q80=180p=15 and q=1p=1x+y=15 and q=1xy=1Thus we have             x+y=5     ...(5) and xy=1     ...(6)Adding equations (5) and (6), we get 2x=6x=3 Subtracting equation (6) from equation (5), we get y=2Hence, x=3 and y=2.(viii) We have 13x+y+13xy=34and 12(3x+y)12(3x+y)=18Putting 13x+y=p and 13xy=q in above equations, we get                  p+q=34...(3)and p2q2=18               pq=14 ...(4)Adding equations (3) and (4), we get 2p=3414p=14 Now, p=13x+y=14                3x+y=4 ...(5)Also,                  q=13xy=12                3xy=2             ...(6)Adding equations (5) and (6), we get                    6x=6x=1Substitute x=1 in equation (5), we get                    3+y=4y=1Hence, x=1 and y=1.

Q.12

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 13 when 1 is subtracted from the numerator and it becomes 14 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.

Ans.

(i) Let the fixed charge be x and the cost of food per day be y.According to question,            x+20y=1000               ...(1)and            x+26y=1180...(2)Subtracting equation (1) from equation (2), we getx+26y=1180x+20y  =1000              ¯          6y=180or y=30Putting this value of y in equation (1), we get          x+20×30=1000or x=1000600=400Hence, fixed charge is 400 and the cost of food perday is 30. (ii)Let the fraction be xy.According to question,      x1y=13    3x3=y3xy=3 ...(1)Also,        xy+8=14 x=y+84 4x=y+8 4xy=8           ...(2)Subtracting equation (2) from equation (1), we get             x=5Substituting this value in equation (1), we get          3×5y=3or y=315 or y=12Therefore,xy=512Hence, the fraction is 512. (iii) Let the number of right and the number of wronganswers be x and y respectively.According to questions,              3xy=40                       ...(1)and             4x2y=50         2(2xy)=2×25         2xy=25                        ...(2)Subtracting equation (2) from equation (1), we get                x=15Putting this value of x in equation (2), we get             2×15y=25         y=2530=5            y=5Thus,Number of right answers=x=15Number of wrong answers=y=5andNumber of questions=x+y=15+5=20 (iv) Let the speeds of Ist and 2nd cars be x km/h and y km/hrespectively.According to question,              x+y=100                       ...(1)and             5x100=5y         5(x20)=5y         x20=y         xy=20                          ...(2)Adding equations (1) and (2), we get                2x=120x=60Putting this value of x in equation (1), we get             60+y=100         y=10060=40Thus,Speed of Ist car=x km/h=60  km/hSpeed of 2nd car=y km/h=40 km/h (v)Let length and breadth of the rectangle be x units and y unitsrespectively.According to question,              (x5)(y+3)=xy9          xy5y+3x15=xy9              5y+3x=159=6             3x5y6=0                 ...(1)Also,             (x+3)(y+2)=xy+67         xy+3y+2x+6=xy+67         3y+2x=676=61         2x+3y61=0                     ...(2)By cross-multiplication method, we have               x305(18)=y12(183)=19(10)          x=32319=17 and y=17119=9Hence, length and breadth of the rectangle are 17 units and9 units respectively.

Q.13

Solve the following pair of linear equations by thesubstitution and cross-multiplication methods:                    8x+5y=9 3x+2y=4

Ans.

                    8x+5y=9                   ...(1)3x+2y=4                  ...(2)By substitution mwthod:From equation (2), we get                               y=43x2Substituting this value of y in equation (1), we get                     8x+5y=9or                 8x+5×43x2=9or                16x+2015x=18  or                x=1820=2Putting this value of x in equation (2), we get                       3×(2)+2y=4or 6+2y=4or 2y=4+6=10or y=5 By cross-multiplication method:8x+5y=93x+2y=4Or8x+5y9=0                   ...(1)3x+2y4=0                  ...(2)Here, a1= 8, b1= 5, c1=9a2=3, b2= 2, c2=4By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x20+18=y27+32=11615    x2=y5=11    x=2 and y=5

Q.14

(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

Ans.

(i) Given pair of linear equations are: 2x+3y=7(ab)x+ (a+b)y=3a+b2Here,  a1a2=2ab,   b1b2=3a+b,    c1c2=73a+b2For a pair of linear equations to have infinitely many solutions:              a1a2=b1b2=c1c2So, we need         2ab=3a+b=73a+b2or 2ab=73a+b2  and    2ab=3a+bor 2(3a+b2)=7(ab) and   2(a+b)=3(ab)or    6a+2b4=7a7b and   2a3a=3b2bor    a+9b4=0 and a=5bNow, we put a=5b in  a+9b4=0.Thus,         5b+9b=4 or           4b=4or             b=1Also,a=5b=5×1=5 (ii) Given pair of linear equations are: 3x+y=1(2k1)x+(k1)y=2k+1Here,  a1a2=32k1,   b1b2=1k1,    c1c2=12k+1For a pair of linear equations to have no solution:              a1a2=b1b2c1c2So, the given pair of linear equations have no solution when         32k1=1k1or 3k3=2k1or 3k2k=1+3 or    k=2

Q.15

Which of the following pairs of linear equations has unique solution,no solution, or infinitely many solutions. In case there is a uniquesolution, 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

Ans.

(i) x3y3=0      3x9y2=0Here, a1a2=13, b1b2=39=13, c1c2=32=32and soa1a2=b1b2c1c2Thus, the given pair of linear equations have no solution.(ii) 2x+y=5      i.e.,      2x+y5=0        3x+2y=8      i.e.,      3x+2y8=0Here, a1a2=23, b1b2=12, c1c2=58and soa1a2b1b2Thus, the given pair of linear equations have unique solution.By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x8(10)=y15(16)=143    x2=y1=11    x=2 and y=1 (iii) 3x5y=20         i.e.,      3x5y20=0         6x10y=40        i.e.,       6x10y40=0Here, a1a2=36=12, b1b2=510=12, c1c2=2040=12and soa1a2=b1b2=c1c2Thus, the given pair of linear equations have infinitelymany solutions.(iv) x3y7=0                  3x3y15=0Here, a1a2=13, b1b2=33=1, c1c2=715=715and soa1a2b1b2Thus, the given pair of linear equations have unique solution.By cross-multiplication method, we have        xb1c2b2c1=yc1a2a1c2=1a1b2b1a2    x4521=y21(15)=13(9)    x24=y6=16    x=4 and y=1

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.

(i)Let the fraction be xy.According to question,      x+1y1=1    x+1=y1xy=2 ...(1)Also,        xy+1=12 x=y+12 2x=y+1 2xy=1           ...(2)Subtracting equation (2) from equation (1), we get             x=3Substituting this value in equation (1), we gety=5Hence, the fraction is 35. (ii)Let the present age of Nuri and Sonu be x yearsand y years respectively.So, five years ago,Nuris age=(x5) yearsandSonu’s age=(y5) yearsAccording to question,       x5=3  (y5)  x5=3y15  x3y=10                ...(1)Again, Ten years hence,Nuris age=(x+10) yearsandSonu’s age=(y+10) yearsAccording to the question,   x+10=2(y+10)or x2y=10             ...(2)Subtracting equation (2) from equation (1), we gety=20Putting this value in (2), we getx=50Hence, Nuri and Sonu are 50 yearsand 20 years old respectively. (iii)Let the two digit number be 10x+y.According to question,            x+y=9               ...(1)and        9(10x+y)=2(10y+x) 90x+9y=20y+2x 90x2x+9y20y=0 88x11y=0 8xy=0 ...(2)Adding equations (1) and (2), we getx = 1Putting this value of x in equation (2), we gety=8Hence, Number=10x+y=10+8=18 (iv)Let number of notes of 50 and 100bexandyrespectively.According to question,            x+y=25               ...(1)and       50x+100y=2000 50 (x+2y)=2000 x+2y=40 ...(2)Subtracting equation (1) from equation(2), we getx+2y=40x+y=25              ¯        y=15Putting this value in equation (1), we getx=10Hence, Meena received 10 notes of 50 and 15 notes of 100. (v)Let the fixed charge for first three day be x and additionalcharge for each day thereafter be y.According to question,            x+4y=27               ...(1)and       x+2y=21      ...(2)Subtracting equation (2) from equation (1), we getx+4y  =27x+2y=21              ¯        2y=6or y=3Putting this value in equation (1), we getx=2712=15Hence, fixed charge is 15 and additional charge foreach extra day is 3.

Q.17

Solve the following pair of linear equations by theelimination 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) x2 + 2y3 = –1 and x –y3 = 3

Ans.

(i) By elimination method:Given that        x+y=5 ...(1) 2x3y=4 ...(2)We multiply equation (1) by 3 and then add it toequation (2) to eliminate the variable y and getthe value of x as follows.                  3x+3y=15                  2x3y=4 ¯                   5x          =19or                  x=195Using this value of x in equation (1), we get                      195+y=5or y=5195=65By substitution method:From equation (1), we get                   y=5xSubstituting this in equation (2), we get                  2x3  (5x)=4or 5x=4+15=19or x=195Substituting this value of x in equation (1), we get                    195+y=5or y=5195=65 (ii) By elimination method:Given that      3x+4y=10 ...(1) 2x2y=2          ...(2)We multiply equation (2) by 2 and then add it toequation (1) to eliminate the variable y and getthe value of x as follows.                  3x+4y=10                  4x4y=4 ¯                   7x          =14or               x=2Using this value of x in equation (1), we get                      6+4y=10or y=1064=1By substitution method:From equation (2), we get                   x=1+ySubstituting this in equation (1), we get                  3(1+y)+4y=10or    7y=103=7or y=1Substituting this value of y in equation (2), we get                    2x2=2or x=2+22=2 (iii) By elimination method:Given that      3x5y4=0             ...(1) 9x=2y+7                           ...(2)We multiply equation (1) by 3 and then subtract equation (2) from it to eliminate the variable x and getthe value of y as follows.                 9x15y12=0                  9x                      =2y+7                                   ¯                          15y12=2y7or               15y+2y=7+12=5or y=513Using this value of y in equation (2), we get                      9x=513×2+7or x=10117+79=10+91117=81117=913By substitution method:From equation (2), we get                   x=2y+79Substituting this in equation (1), we get                  3  ×2y+795y4=0or    2y+735y=4or 2y+715y=12or 13y=127or y=513Substituting this value of y in equation (2), we get                    9x=2×513+7=10+9113or x=8113×9=913 (iv) By elimination method:Given that      x2+2y3=1             ...(1) xy3=3                           ...(2)We multiply equation (2) by 2 and then add it toequation (1) to eliminate the variable y and getthe value of x as follows.                 x2+2y3=1                   2x2y3=6¯                  x2+2x=5or               5x2=5or x=2Using this value of x in equation (2), we get                      2y3=3or y=3  (32)=3or y=3By substitution method:From equation (2), we get                   x=3+y3Substituting this in equation (1), we get                  12  (3+y3)+2y3=1or   9+y6+2y3=1or 9+y+4y6=1 or 9+y+4y=6or y=155=3Substituting this value of y in equation (2), we get                     x=3+33=2

Q.18

Form the pair of linear equations for the followingproblems 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 911, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56. 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?

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

vLet the fraction be xy.According to question,x+2y+2=911           11x9y=4      ...(1)andx+3y+3=56              6x5y=3      ...(2)From equation (2), we get   x=5y36 We put this value of x in equation (1) and get       115y369y=4  55y3354y=24y=24+33=9Putting this value of y in equation (1), we get           11x9×9=4       11x=814=77       x=7Hence the fraction is 79.

(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.

The given linear equations are:2x+3y=11 ...(1)2x4y=24...(2)We express x in terms of y from equation (1) to get x=113y2We substitute this value of x in equation (2) to get 113y4y=24i.e., 117y=24i.e., 7y=24+11=35i.e.,               y=5Putting this value of y in equation (1), we get              2x+3×5=11i.e.,       2x=1115=4i.e.,          x=2Now,                y=mx+3or 5=2m+3or m= 1

Q.20

Solve the following pair of linear equations by the substitution method. (i) x+y=14                                          (ii) st=3              xy=4                                                     s3+t2=6      (iii) 3xy=3                                        (iv) 0.2x+0.3y=1.3 9x3y=9                                               0.4x+0.5y=2.3 (v) 2x+3y=0                                 (vi) 3x25y3=2               3x8y=0                                           x3+y2=136

Ans.

(i) The given pair of linear equations are:      x+y=14                               ...(1)                 xy=4                                 ...(2)We express x in terms of y from equation (1) to get x=14yWe substitute this value of x in equation (2) to get 14yy=4i.e., 2y=414=10i.e., y=5Putting this value of y in equation (2), we get              x5=4    i.e.,      x=9(ii) The given pair of linear equations are:      st=3                               ...(1)                 s3+t2=6                             ...(2)We express s in terms of t from equation (1) to get s=t+3We substitute this value of s in equation (2) to get t+33+t2=6i.e., 5t+66=6i.e., 5t+6=36i.e., t=6Putting this value of t in equation (1), we get              s6=3i.e.,      s=9   (iii) The given pair of linear equations are:      3xy=3                               ...(1)                 9x3y=9                            ...(2)We express y in terms of x from equation (1) to get y=3x3We substitute this value of y in equation (2) to get 9x9x+9=9i.e., 9=9This statement is true for all values of x and so we can notobtain a specific value of x. We observe that both the givenequations are the same as one is derived from another.Therefore, given equations have infinitely many solutions. (iv) The given pair of linear equations are:      0.2x+0.3y=1.3                            ...(1)      0.4x+0.5y=2.3                            ...(2)We express x in terms of y from equation (1) to get x=(1.30.3y)/(0.2)=(133y)/2We substitute this value of x in equation (2) to get 0.2(133y)+0.5y=2.3i.e., 2.60.6y+0.5y=2.3i.e., 0.1y=2.32.6=0.3i.e.,                    y=3Putting this value of y in equation (2), we get              0.4x+0.5×3=2.3i.e.,      x=(2.31.5)/0.4=8/4=2 (v) The given pair of linear equations are:     2x+3y=0                             ...(1)               3x8y=0                             ...(2)We express x in terms of y from equation (1) to get x=32yWe substitute this value of x in equation (2) to get 332y8y=0i.e., 342y=0i.e., y=0Putting this value of y in equation (2), we get             3x=0i.e.,           x=0(vi) The given pair of linear equations are:                                                              3x25y3=2                             or   9x10y=12                           ...(1)and       x3+y2=136                                 or 2x+3y=13                                ...(2)We express x in terms of y from equation (1) to get x=(12+10y)/9We substitute this value of x in equation (2) to get 2(12+10y9)+3y=13i.e., 24+20y+27y=117i.e., 47y=117+24=141i.e.,                  y=14147=3Putting this value of y in equation (2), we get            2x+9=13i.e.,           x=42=2

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.

(i) Intersecting lines: A line which will intersect the given line 2x+3y8=0       is 3x+2y8=0 as a1a2=23 and b1b2=32 impllies that a1a2b1b2.(ii) P lines: A line which is paralel to the given line 2x+3y8=0       is 4x+6y7=0 as a1a2=24=12 and b1b2=36=12 impllies that a1a2=b1b2.iii : A line which is   to the given line 2x+3y8=0       is 4x+6y16=0 because a1a2=b1b2=c1c2=12.

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

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  

Ans.

 (i) x+y=5,                    2x+2y=10 Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=1,      b1=1,    c1=5a2=2,    b2=2, c2=10Now,a1a2=12,     b1b2=12,    c1c2=510=12and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they are consistent.Graphs of the two equations coincide and hence each andevery point on this graph is a solution of these equations.

(ii) xy=8,                    3x3y=16Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=1,      b1=1,    c1=8a2=3,    b2=3, c2=16Now,a1a2=13,     b1b2=13,    c1c2=816=12and soa1a2=b1b2c1c2Therefore, the given linear equations are paralleland hence they are inconsistent. (iii) 2x+y6=0,          4x2y4=0Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=2,      b1=1,       c1=6a2=4,     b2=2, c2=4Now,a1a2=24=12,     b1b2=12,    c1c2=64=32and soa1a2b1b2Therefore, the given linear equations are consistent.Now,2x+y6=0y=62xSome points which satisfy this equation are writtenin the following table.x023y=62x620Again,4x2y4=0y=2x2Some points which satisfy this equation are writtenin the following table.x023y=2x2224We get the following graphs of the given equations andfind that they intersect at (2, 2). Hence, x = 2 and y = 2.

(iv)  2x2y2=0,       4x4y5=0  Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=2,      b1=2,    c1=2a2=4,    b2=4, c2=5Now,a1a2=24=12,     b1b2=13=12,    c1c2=25=25and soa1a2=b1b2c1c2Therefore, the given linear equations are paralleland hence they are inconsistent.

Q.25

On comparing the ratios a1a2, b1b2,  c1c2, find out whetherthe following pair of linear equations are consistent,or inconsistent.(i) 3x+2y=5; 2x3y=7     (ii) 2x3y=8; 4x6y=9(iii) 32x+53y=7;  9x10y=14    (iv)  5x3y=11;  10x+6y=22   (v) 43x+2y=8;   2x+3y=12

Ans.

(i) 3x+2y=5; 2x3y=7Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=3,      b1=2,    c1=5a2=2,      b2=3, c2=7Also,a1a2=32,     b1b2=23,    c1c2=57and soa1a2b1b2Therefore, the given pair of linear equations has a uniquesolution and hence the two linear equations are consistent. (ii) 2x3y=8; 4x6y=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, a 1 a 2 = 2 4 = 1 2 , b 1 b 2 = 3 6 = 1 2 , c 1 c 2 = 8 9 and so a 1 a 2 = b 1 b 2 c 1 c 2 Therefore, the given linear equations are parallel and hence the two linear equations are inconsistent. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafeaakq aabeqaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaGaaGOmaiaadIha cqGHsislcaaIZaGaamyEaiabg2da9iaaiIdacaGG7aGaaeiiaiaabc cacaaI0aGaamiEaiabgkHiTiaaiAdacaWG5bGaeyypa0JaaGyoaaqa aiaaboeacaqGVbGaaeyBaiaabchacaqGHbGaaeOCaiaabMgacaqGUb Gaae4zaiaabccacaqG0bGaaeiAaiaabwgacaqGZbGaaeyzaiaabcca caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBai aabohacaqGGaGaae4DaiaabMgacaqG0bGaaeiAaiaabccacaWGHbWa aSbaaSqaaiaaigdaaeqaaOGaamiEaiabgUcaRiaadkgadaWgaaWcba GaaGymaaqabaGccaWG5bGaey4kaSIaam4yamaaBaaaleaacaaIXaaa beaakiabg2da9iaaicdaaeaacaqGHbGaaeOBaiaabsgacaqGGaGaam yyamaaBaaaleaacaaIYaaabeaakiaadIhacqGHRaWkcaWGIbWaaSba aSqaaiaaikdaaeqaaOGaamyEaiabgUcaRiaadogadaWgaaWcbaGaaG OmaaqabaGccqGH9aqpcaaIWaGaaiilaiaabccacaqG3bGaaeyzaiaa bccacaqGNbGaaeyzaiaabshaaeaacaWGHbWaaSbaaSqaaiaaigdaae qaaOGaeyypa0JaaGOmaiaacYcacaaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caWGIbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0Jaey OeI0IaaG4maiaacYcacaaMc8UaaGPaVlaaykW7caaMc8Uaam4yamaa BaaaleaacaaIXaaabeaakiabg2da9iabgkHiTiaaiIdaaeaacaWGHb WaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaaGinaiaacYcacaaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGIbWaaSbaaSqaaiaaik daaeqaaOGaeyypa0JaeyOeI0IaaGOnaiaacYcacaqGGaGaaeiiaiaa dogadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcqGHsislcaaI5aaaba GaaeyqaiaabYgacaqGZbGaae4BaiaabYcaaeaadaWcaaqaaiaadgga daWgaaWcbaGaaGymaaqabaaakeaacaWGHbWaaSbaaSqaaiaaikdaae qaaaaakiabg2da9maalaaabaGaaGOmaaqaaiaaisdaaaGaeyypa0Za aSaaaeaacaaIXaaabaGaaGOmaaaacaGGSaGaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7daWcaaqaaiaadkgadaWgaaWcbaGaaGymaaqabaaa keaacaaMc8UaamOyamaaBaaaleaacaaIYaaabeaaaaGccqGH9aqpda WcaaqaaiabgkHiTiaaiodaaeaacqGHsislcaaI2aaaaiabg2da9maa laaabaGaaGymaaqaaiaaikdaaaGaaiilaiaaykW7caaMc8UaaGPaVl aaykW7daWcaaqaaiaadogadaWgaaWcbaGaaGymaaqabaaakeaacaaM c8Uaam4yamaaBaaaleaacaaIYaaabeaaaaGccqGH9aqpdaWcaaqaai aaiIdaaeaacaaI5aaaaaqaaiaabggacaqGUbGaaeizaiaabccacaqG ZbGaae4BaaqaamaalaaabaGaamyyamaaBaaaleaacaaIXaaabeaaaO qaaiaadggadaWgaaWcbaGaaGOmaaqabaaaaOGaeyypa0ZaaSaaaeaa caWGIbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaaGPaVlaadkgadaWgaa WcbaGaaGOmaaqabaaaaOGaeyiyIK7aaSaaaeaacaWGJbWaaSbaaSqa aiaaigdaaeqaaaGcbaGaaGPaVlaadogadaWgaaWcbaGaaGOmaaqaba aaaaGcbaGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMbGaae4B aiaabkhacaqGLbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGa Gaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeiBaiaabMga caqGUbGaaeyzaiaabggacaqGYbGaaeiiaiaabwgacaqGXbGaaeyDai aabggacaqG0bGaaeyAaiaab+gacaqGUbGaae4CaiaabccacaqGHbGa aeOCaiaabwgacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabYgaca qGSbGaaeyzaiaabYgacaqGGaGaaeyyaiaab6gacaqGKbaabaGaaeiA aiaabwgacaqGUbGaae4yaiaabwgacaqGGaGaaeiDaiaabIgacaqGLb GaaeiiaiaabshacaqG3bGaae4BaiaabccacaqGSbGaaeyAaiaab6ga caqGLbGaaeyyaiaabkhacaqGGaGaaeyzaiaabghacaqG1bGaaeyyai aabshacaqGPbGaae4Baiaab6gacaqGZbGaaeiiaiaabggacaqGYbGa aeyzaiaabccacaqGPbGaaeOBaiaabogacaqGVbGaaeOBaiaabohaca qGPbGaae4CaiaabshacaqGLbGaaeOBaiaabshacaqGUaaaaaa@5EAD@ (iii) 32x+53y=7;  9x10y=14    Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=32,      b1=53,    c1=7a2=9,      b2=10, c2=14Also,a1a2=329=16,     b1b2=5310=16,    c1c2=714=12and soa1a2b1b2Therefore, the given pair of linear equations has a uniquesolution and hence the two linear equations are consistent. (iv)  5x3y=11;         10x+6y=22  Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=5,      b1=3,    c1=11a2=10,    b2=6, c2=22Also,a1a2=510=12,     b1b2=36=12,    c1c2=1122=12and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they consistent.  (v) 43x+2y=8;         2x+3y=12Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2=0, we geta1=43,      b1=2,    c1=8a2=2,    b2=3, c2=12Also,a1a2=432=23,     b1b2=23,    c1c2=812=23and soa1a2=b1b2=c1c2Therefore, the given linear equations have infinitely manysolutions and hence they are consistent.

Q.26

On comparing the ratios a1a2,b1b2, c1c2, 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

Ans.

(i) 5x-4y+8 = 0      7x+6y-9 = 0Comparing these equations with a1x+b1y+c1= 0and a2x+b2y+c2= 0, we geta1=5,      b1=-4,    c1=8a2=7,      b2=6,  c2=-9Also,a1a2=57,     b1 b2=-46=-23and soa1a2b1 b2Therefore, the given pairs of linear equations intersect at a point. (ii) 9x+3y+12=0       18x+6y+24=0Comparing these equations with a1x+b1y+c1= 0and a2x+b2y+c2= 0, we geta1=9,      b1=3,    c1=12a2=18,    b2=6, c2=24Also,a1a2=918=12,     b1 b2=36=12,    c1 c2=1224=12and soa1a2=b1 b2=c1 c2Therefore, the given pairs of linear equations are coincident. (iii) 6x-3y+10 = 0         2x-y+9 = 0Comparing these equations with a1x+b1y+c1=0and a2x+b2y+c2= 0, we geta1 = 6,      b1 = -3,    c1 = 10a2= 2,      b2 = -1, c2 = 9Also,a1a2=62= 3,     b1 b2=-3-1= 3,    c1 c2=109and soa1a2=b1 b2c1 c2Therefore, the given pairs of linear equations are parallel.

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 2 8 12 24 38 16

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) Number of farms(Cumulative frequency)
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

0

3

5

9

14

28

32

35

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

Weight (in kg)upper class limits Number of students(Cumulative frequency)
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.

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.

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

n2=352=17.5Cumulative frequency just greater that 17.5 is 28 which belongsto class interval 4648. Median class=4648Now, we havel=46,Cumulative frequency (cf) of the class preceding themedian class=14,Frequency of median class (f)=14,  Class size (h)=2 We know that     Median=l+(n2cff)×h       Median =46+(3521414)×2or     Median =46.5So median of the given data is 46.5.Hence, value of the median is verified.

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 12 14 8 6 10

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

Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw itsogive as following.

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.

We find the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Weight (in kg) Number of students (fi) 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

We have, n=30.n2=302=15Cumulative frequency (cf) just greater than  n2=15 is 19 which belongs to interval 5560. Median class=5560Now, we havel=55,Cumulative frequency (cf) of the class preceding themedian class=13,Frequency of median class (f)=6,  Class size (h)=5We know that     Median=l+(n2cff)×h       Median =55+(302136)×5or     Median =56.67Hence, median weight of students is 56.67 kg.

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.

We find the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Number of letters Number of surnames (fi) 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

Total(n)=100We have, n=100.n2=1002=50Cumulative frequency (cf) just greater than  n2=50 is 76 which belongs to interval 710. Median class=710 Now, we havel=7,Cumulative frequency (cf) of the class preceding themedian class=36,Frequency of median class (f)=40,  Class size (h)=3We know that     Median=l+(n2cff)×h       Median =7+(10023640)×3or     Median =8.05Median number of letters in the surnames is 8.05.Now, we find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=11.5

We proceed as below to find di, ui, fiui .

Number of letters Number of surnames (fi) xi di = xi – a ui=xiah fiui
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

Mean=x¯=a+fiuifi×h                     =11.5+106100×3                    =8.32We know that 3 Median=Mode+2 Mean       3×8.05=Mode+2×8.32or Mode=7.51Hence, mean number of letters in the surnames is 8.32and modal size of the surnames is 7.51.

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.

We write the cumulative frequency (cf) as below in thefollowing table on the basis of the given information.

Life time (in hours) Number of lamps (fi) 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

We have, n=400.n2=4002=200Cumulative frequency (cf) just greater than  n2=200 is 216 which belongs to interval 30003500. Median class=30003500Now, we havel=3000,Cumulative frequency (cf) of the class preceding themedian class=130,Frequency of median class (f)=86,  Class size (h)=500We know that     Median=l+(n2cff)×h       Median =3000+(400213086)×500or     Median =3406.98Hence, median life time of a lamp is 3406.98 hours.

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

3

5

9

12

5

4

2

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.

The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from eachof lower boundary and adding 0.5 in each of upper boundary.We write the class interval, number of leaves (fi),and cumulative frequency (cf) as below in the following tableon the basis of the given information.

Class interval Number of leaves (fi) 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

We have, n=40.n2=402=20Cumulative frequency (cf) just greater than  n2=20 is 29 which belongs to interval 144.5153.5. Median class=144.5153.5Now, we havel=144.5,Cumulative frequency (cf) of the class preceding themedian class=17,Frequency of median class (f)=12,  Class size (h)=9We know that     Median=l+(n2cff)×h       Median =144.5+(4021712)×9or     Median =146.75Hence, median length of leaves is 146.75 mm.

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

2

6

24

45

78

89

92

98

100

Ans.

We write the class interval, number of policy holders (fi),and cumulative frequency (cf) as below in the following tableon the basis of the given information.

Class interval Number of policy holders (fi) 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

We have, n=100.n2=1002=50Cumulative frequency (cf) just greater than  n2=50 is 78 which belongs to interval 3540. Median class=3540Now, we havel=35,Cumulative frequency (cf) of the class preceding themedian class=45,Frequency of median class (f)=33,   Class size (h)=5We know that     Median=l+(n2cff)×h    Median =35+(10024533)×5or     x=35.76Hence, median age is 35.76 years.

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

5

x

20

15

y

5

Total 60

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

20

15

y

5

5

5 + x

25 + x

40 + x

40 + x + y

45 + x + y

Total 60

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

From the given data, we havel=20,Cumulative frequency (cf) of the class preceding themedian class=5+x,Frequency of median class (f)=20,  Class size (h)=10     Median=l+(n2cff)×hor     28.5=20+(6025x20)×10or     x=8On putting this value of x in equation (1), we get        x+y=15or 8+y=15or        y=158=7Hence, values of x and y are 8 and 7 respectively.

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

4

5

13

20

14

8

4

Ans.

We find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=135We proceed as below to find di, ui,  fiui.

Monthly consumption (in units) Number of consumers (fi) xi di = xi – 135 ui=xi135h fiui
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

Mean=x¯=a+fiuifi×h                     =135+768×20                    =137.058Modal class=125145From the given data, we havel=125, f1=20, f0=13,   f2=14, h=20Mode=l+(f1f02f1f0f2)×h            =125+(20132×201314)×20            =135.76We know that 3 Median=Mode+ 2 Meanor Median=135.76+2×137.0583=136.625

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 7 14 13 12 20 11 15 8

Ans.

From the given data, we havel=40, f1=20, f0=12,   f2=11, h=10Mode=l+(f1f02f1f0f2)×h            =40+(20122×201211)×10            =44.7Mode of the given data is 44.7.

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

4

18

9

7

6

3

1

1

Find the mode of the data.

Ans.

From the given data, we havel=4000, f1=18, f0=4,   f2=9, h=1000Mode=l+(f1f02f1f0f2)×h            =4000+(1842×1849)×1000            =4608.695Mode of the given data is 4608.7 (approx.)

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

3

8

9

10

3

0

0

2

Ans.

From the given data, we havel=30, f1=10, f0=9,   f2=3, h=5Mode=l+(f1f02f1f0f2)×h            =30+(1092×1093)×5            =30.62530.6Interpretation:Most of the states/UT have a teacher student ratio as 30.6.Now, we find class mark for each interval by using thefollowing relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=32.5We proceed as below to find di, ui,  fiui.

Number of students per teacher Number of states/UT (fi) xi di = xi – 32.5 ui=xi32.5h

b

fiui
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

Mean=x¯=a+fiuifi×h                     =32.5+2335×5                    =29.2Interpretation:Average teacher student ratio of the states/UT is 29.2.

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.

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=2750We proceed as below to find di, ui,  fiui.

Expenditure (in ₹) Number of families (fi) xi di = xi – 2750 ui=xi2750h fiui
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

Mean=x¯=a+fiuifi×h                     =2750+35200×500                    =2662.5Therefore, mean monthly expenditure is 2662.5.From the given data, we havel=1500, f1=40, f0=24,   f2=33, h=500Mode=l+(f1f02f1f0f2)×h            =1500+(40242×402433)×500            =1847.8261847.83So, modal monthly expenditure is 1847.83.

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.

From the given data, we havel=60, f1=61, f0=52,   f2=38, h=20Mode=l+(f1f02f1f0f2)×h            =60+(61522×615238)×20            =65.625So, modal lifetime of electrical components is 65.625 hours.

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.

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Assumed mean(a)=30We proceed as below to find di, fidi.

Age (in years) Number of patients (fi) Class mark (xi) di = xi – 30 fidi
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

                          Total=80                                        Total=430Mean=x¯=a+fidifi                     =30+43080                    =35.37535.38Therefore, mean of the given data is 35.38 years.Modal class is 3545.l=35, f1=23, f0=21,   f2=14, h=10 Mode=l+(f1f02f1f0f2)×h            =35+(23212×232114)×10            =36.8Mode is 36.8 which represents that maximum number of patients admitted in hospital are of 36.8 years.

While on average the age of a patient admitted to thehospital is 35.38 years

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.

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=10Here, we take assumed mean(a)=70 and proceed as below to find di, ui and fiui.

Literacy rate (in %) Number of cities fi xi di = xi – 70 ui=xi70h fiui
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

Mean=x¯=a+(fiuifi)h=70+(235)×10=69.43Therefore, mean literacy rate is 69.43.

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.

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Assumed mean(a)=17We proceed as below to find di, fidi.

Number of days Number of students fi xi di = xi – 17 fidi
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

Mean=x¯=a+fidifi                     =17+18140                    =12.47512.48Therefore, mean number of days for which a student was absent is 12.48.

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

4

9

9

2

4

2

Find the mean concentration of SO2 in the air.

Ans.

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2   Class size(h) of the given data=0.04Here, we take assumed mean(a)=0.14 and proceed as below to find di, ui and fiui.

Concentration of SO2 (in ppm) Frequency (fi) xi di = xi – 0.14 ui=xi0.14h

b

fiui
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

Mean=x¯=a+(fiuifi)h=0.14+(3130)×0.04                                                     0.099 ppmTherefore, mean concentration of SO2 in the air is 0.099 ppm.

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.

We find class mark for each interval by using the following relation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=50Here, we take assumed mean(a)=225 and proceed as below to find di, ui and fiui.

Class interval fi xi di = xi – 225 ui=xi57h fiui
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

Mean=x¯=a+(fiuifi)h=225+(725)×50=211Therefore, mean daily expenditure is 211.

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.

The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limitof each interval. We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=3Here, we take assumed mean(a)=57 and proceed as below to find di, ui and fiui.

Class interval fi xi di = xi – 57 ui=xi57h f i u i MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacaWGMbWaaSbaaSqaaiaadMgaaeqaaOGaamyDamaaBaaaleaacaWGPbaabeaaaaa@3D30@
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

Mean=x¯=a+(fiuifi)h=57+(25400)×3=57.187557.19Therefore, mean number of mangoes is 57.19.We have chosen step deviation method as values of fi, di are bigand there is a common multiple between all di.

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.

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=3Here, we take assumed mean(a)=75.5 and proceed as below to find di, ui and fiui.

Number of heartbeats per minute Number of women (fi) xi di = xi – 75.5 ui=xi75.5h fiui
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

Mean=x¯=a+(fiuifi)h=75.5+(430)×3=75.9Therefore, mean heartbeats per minute is 75.9.

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 7 6 9 13 f 5 4

Ans.

We find class mark for each interval by using the followingrelation.xi=Upper class limit+Lower class limit2Class size(h) of the given data=2Given mean(a)=18We proceed as below to find di, fidi.

Daily pocket allowance (in ₹) Number of children (fi) xi di = xi – 18 fidi
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 fi=44+f 2f – 40

Mean=x¯=a+fidifior 18=18+2f4044+for  f=20Therefore, missing frequency f is 20.

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.

We find class mark for each interval by using the followingrelation.xi=Upper class limit + Lower class limit2Class size(h) of the given data=20Here, we take assumed mean(a)=550 and proceed as below.

Daily wages
(in ₹)
Number of workers (fi) xi di = xi – 550 ui=xi550h fiui
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

Mean=x¯=a+(fiuifi)h=550+(1250)×20=545.2

Therefore, mean daily wages of the workers is 545.20

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.

Here, we observe that class Number of plants Number of houses – 21112 – 42364 – 61556 – 857358 – 10695410 – 122112212 – 1431339 Total20
162

Mean=x¯=fixifi=16220=8.1Therefore, mean number of plants per house is 8.1.

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.

Let the number of green marbles=xTotal number of marbles in the jar=24 P(drawing a green marble)or                                        23  =Number of favourable outcomesTotal number of outcomesor                                        23  =x24or                                        x=16Number of blue marbles=24x=2416=8

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.

Number of black balls=xTotal number of balls in the bag=12 P(drawing a black ball)                                              =Number of favourable outcomesTotal number of outcomes                                               =x12Now, 6 more black balls are putted in the box.Number of black balls=x+6Total number of balls in the bag=12+6=18           P(drawing a black ball)=2×x12or                                           x+618  =2×x12or                                                   x  =3

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.

Let number of blue balls be n.Number of red balls=5Total number of balls in the bag=5+n P(drawing a red ball)                                              =Number of favourable outcomesTotal number of outcomes                                               =55+n           P(drawing a blue ball)=2×P(drawing a red ball)or                                           n5+n  =2×55+nor                                                   n  =10 Number of blue balls=10

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.

Total number of possible outcomes=6×6=36(i)Number of times when the total score is even=18 P(total score being even)                                              =Number of favourable outcomesTotal number of outcomes                                               =1836=12 (ii)Number of times when the total score is 6=4 P(total score being 6)                                              =Number of favourable outcomesTotal number of outcomes                                               =436=19(iii)Number of times when the total score is at least 6=15 P(total score being at least 6)                                              =Number of favourable outcomesTotal number of outcomes                                               =1536=512

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.

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=5×5=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)                                              =Number of favourable outcomesTotal number of outcomes                                               =525=15(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=3P(both will visit on consecutive days)                                              =Number of favourable outcomesTotal number of outcomes                                               =825(iii)P(both will visit on different days)                                              =1both will visit on the same day                                               =115=45

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.

(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)                                              =Number of favourable outcomesTotal number of outcomes                                               =14     P(getting two tails)                                              =Number of favourable outcomesTotal number of outcomes                                               =14 P(getting one head and one tail)                                              =Number of favourable outcomesTotal number of outcomes                                               =24=12(ii) Correct because the two outcomes are equally likely.

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.

Total number of outcomes=6×6=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)                                                         =Number of favourable outcomesTotal number of outcomes                                                         =1136P(5 will not come up either time)                                                       =1P(5 will come up either time)                                                       =11136=2536(ii)Number of cases when 5 will come up at least once=11P(5 will come up at least once)                                                         =Number of favourable outcomesTotal number of outcomes                                                         =1136

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.

Possible outcomes are {HHH, TTT, HTT, THH, HTH, THT, HHT, TTH}.Number of possible outcomes=8Number of favourable outcomes=2 P(Hanif will win the game)                                                         =Number of favourable outcomesTotal number of outcomes                                                         =28=14P(Hanif will lose the game)                                                       =1P(Hanif will win the game)                                                       =114=34

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 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3D0E@ 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A

(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 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A@ 2 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIYaaabaGaaG4maiaaiAdaaaaaaa@3D0B@ 3 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIZaaabaGaaG4maiaaiAdaaaaaaa@3D0C@ 4 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI0aaabaGaaG4maiaaiAdaaaaaaa@3D0D@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3D0E@ 6 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI2aaabaGaaG4maiaaiAdaaaaaaa@3D0F@ 5 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI1aaabaGaaG4maiaaiAdaaaaaaa@3D0E@ 4 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaI0aaabaGaaG4maiaaiAdaaaaaaa@3D0D@ 3 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIZaaabaGaaG4maiaaiAdaaaaaaa@3D0C@ 2 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIYaaabaGaaG4maiaaiAdaaaaaaa@3D0B@ 1 36 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaerbbjxAHXgarmqr1ngBPrgitLxBI9gBaerbwvMCKfMBHbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaGaaiaadaWaamaaceGaaqaaeaqbaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maiaaiAdaaaaaaa@3D0A@

(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.

Number of defective pens=20Number of good pens=14420=124Total number of outcomes=144(i)Probability of buying a pen=Probability of drawing a good pen                                                         =Number of favourable outcomesTotal number of outcomes                                                         =124144=3136(ii)Probability of not buying pen=1Probability of buying a pen                                                             =13136=536

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.

Favourable outcomes=Area of the circle=πr2=π(0.5)2                                               =0.25π m2Total outcomes=Area of the rectangle=2m× 3m=6 m2Probability of landing inside the circle                                  =Area of the circleArea of the rectangle                                  =0.25π 6=π24

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

A B C D E F

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

Ans.

(i)Probability of getting A                                  =Number of favourable outcomesTotal number of outcomes                                  =26=13(ii) Probability of getting D                                  =Number of favourable outcomesTotal number of outcomes                                  =16

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.

Numbers of two-digit numbers among 1 to 90=81Numbers of perfect square numbers among 1 to 90=9Numbers of multiples of 5 which are less than or equal 90=18Total number of outcomes=90(i)Probability of getting a two-digit number bearing disc                                  =Number of favourable outcomesTotal number of outcomes                                  =8190=910(ii) Probability of getting a perfect square number                                  =Number of favourable outcomesTotal number of outcomes                                  =990=110(iii) Probability of getting a number divisible by 5                                  =Number of favourable outcomesTotal number of outcomes                                  =1890=15

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.

(i)Number of defective balls=4Total number of outcomes=20Probability of getting a defective ball                                  =Number of favourable outcomesTotal number of outcomes                                  =420=15(ii) Number of defective balls=4Number of good balls=161=15Total number of outcomes=15+4=19Probability of not getting a defective ball                =1Probability of getting a defective ball                =1419=1519

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.

Number of defective pens=12Number of good pens=132Total number of outcomes=132+12=144Probability of getting a good pen                                  =Number of favourable outcomesTotal number of outcomes                                  =132144=1112

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.

(i)Total number of outcomes=5Probability of getting the queen                                  =Number of favourable outcomesTotal number of outcomes                                  =15(ii) Queen is put aside.total number of outcomes=4(a)Probability of getting an ace                                  =Number of favourable outcomesTotal number of outcomes                                  =14(b)Probability of getting a queen                                  =Number of favourable outcomesTotal number of outcomes                                  =04=0

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.

Total number of outcomes=52(i) Number of kings of red colour=2Probability of getting a king of red colour                                  =Number of favourable outcomesTotal number of outcomes                                  =252=126(ii) Number of face cards=12Probability of getting a face card                                  =Number of favourable outcomesTotal number of outcomes                                  =1252=313(iii) Number of red face cards=6Probability of getting a red face card                                  =Number of favourable outcomesTotal number of outcomes                                  =652=326(iv) Number of the jack of hearts=1Probability of getting the jack of hearts                                  =Number of favourable outcomesTotal number of outcomes                                  =152(v) Number of spades=13Probability of getting a spade                                  =Number of favourable outcomesTotal number of outcomes                                  =1352=14(vi) Number of the queen of diamond=1Probability of getting the queen of diamond                                  =Number of favourable outcomesTotal number of outcomes                                  =152

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.

Total number of outcomes=6(i) Prime numbers on the die are 2, 3 and 5.Number of prime numbers on the die=3Probability of getting a prime number                                  =Number of favourable outcomesTotal number of outcomes                                  =36=12(ii) Numbers lying between 2 and 6 on the die are 3, 4 and 5.Number of prime numbers on the die=3Probability of getting a number lying between 2 and 6                                  =Number of favourable outcomesTotal number of outcomes                                  =36=12(iii) Odd numbers on the die are 1, 3 and 5.Number of odd numbers on the die=3Probability of getting an odd number                                  =Number of favourable outcomesTotal number of outcomes                                  =36=12

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.

(i) Probability of the arrow pointing at 8 = Number of favourable outcomes Total number of outcomes = 1 8 (ii) Probability of the arrow pointing at an odd number = Number of favourable outcomes Total number of outcomes = 4 8 = 1 2 (iii) Probability of the arrow pointing at a number greater than 2 = Number of favourable outcomes Total number of outcomes = 6 8 = 3 4 (iv) Probability of the arrow pointing at a number less than 9 = Number of favourable outcomes Total number of outcomes = 8 8 =1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@73E5@

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.

Number of male fishes=5Number of female fishes=8Total number of outcomes=5+8=13(i) Probability of getting a male fish                                  =Number of favourable outcomesTotal number of outcomes                                  =513

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.

Number of 50p coins=100Number of  ₹1 coins=50Number of  ₹2 coins=20Number of  ₹5 coins=10Total number of outcomes=100+50+20+10=180(i) Probability of getting a 50p coin                                  =Number of favourable outcomesTotal number of outcomes                                  =100180=5 9(ii)Probability of not getting a ₹5 coin                         =1Probability of getting a 5 coin                         =1Number of favourable outcomesTotal number of outcomes                         =110180=1718

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.

Number of red marbles=5Number of white marbles=8Number of green marbles=4Total number of outcomes=5+8+4=17(i) Probability of getting a red marble                                  =Number of favourable outcomesTotal number of outcomes                                  =517(ii) Probability of getting a white marble                                   =Number of favourable outcomesTotal number of outcomes                                  =817(iii) Probability of getting a marble not green =1Probability of getting a green marble                          =1417=1317

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.

Number of red balls=3Number of black balls=5Total number of outcomes=3+5=8(i) Probability of getting a red ball                                  =Number of favourable outcomesTotal number of outcomes                                  =38(ii) Probability of not getting a red ball                                  =1Probability of getting a red ball                                  =138=58

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.

Let AB be the height of the tip of fishing rod from watersurface.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 ΔABC, we get AC2=AB2+BC2AB2=(1.8)2+(2.4)2 AB2=3.24+5.76 AB2=9.00AB=9=3Therefore, 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=12×5=60 cm=0.6 m

Let after 12 second Fly be at point D.Length of string out after 12 second is ADAD = AC – string pulled by Nazima in 12 seconds    = 3.00 – 0.6    = 2.4In ΔADB, AB2 + BD2 = AD21.82 + BD2 = 2.42BD2 = 5.76  3.24 = 2.52⇒ BD = 1.587Horizontal distance of fly = BD + 1.2                               = 1.587 + 1.2                               = 2.787                               = 2.79 m

Q.87

In the following figure, D is a point on side BC of ΔABC such that BD CD = AB AC . Prove that AD is the bisector of BAC.

Ans.

Let us extend BA to P such that AP=AC. Join PC.It is given that      BDCD=ABACBDCD=ABAPTherefore, by converse of basic proportinality theorem, ADPC.Therefore,BAD=APC ...(1)DAC=ACP ...(2)Also,   AP=ACAPC=ACP APC=ACP=DAC [From (2)]BAD=DAC                                [From (1)]Therefore, AD is the bisector of angle BAC.

Q.88

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) ΔPAC~ΔPDB (ii) PAPB=PCPD

Ans.

(i) In ΔPAC and ΔPDB,P=P (common)PAC=PDB (Exterior angle of a cyclic quadrilateral isequal to opposite interior angle)ΔPAC~ΔPDB(ii)We know that corresponding sides of similar trianglesare proportional. ΔPAC~ΔPDBPAPD=PCPBPAPB=PCPD

Q.89

In the following figure, two chords AB and CD intersect each other at the point P. Prove that :(i) ΔAPC~ΔDPB (ii) AP.PB=CP.DP

Ans.

Let us join CB.(i) In ΔAPC and ΔDPB, APC=DPB [Vertically opposite angles]CAP=BDP [Angles in same segment for chord CB]ΔAPC~ΔDPB [By AA similarity criterion](ii) We know that corresponding sides of similar trianglesare proportional.Therefore, ΔAPC~ΔDPBAPDP=PCPBAPPB=PCDP

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.

Applying Pythagoras theorem in ΔDEA, we getDE2+EA2=DA2 ...(i)Applying Pythagoras theorem in ΔDEB, we getDE2+EB2=DB2DE2+(EA+AB)2=DB2(DE2+EA2)+AB2+2EA×AB=DB2DA2+AB2+2EA×AB=DB2 ... (ii)Applying Pythagoras theorem in ΔADF, we getAD2=AF2+FD2Applying Pythagoras theorem in ΔAFC, we getAC2=AF2+FC2         =AF2+(DCFD)2         =AF2+DC2+FD22DC×FD         =(AF2+FD2)+DC22DC×FDAC2=AD2+DC22DC×FD ...(iii)Since ABCD is a parallelogram AB=CD ...(iv)and, BC=AD ... (v)In ΔDEA and ΔADF,DEA=AFD=90°, EAD=ADF (EADF)Therefore, FAD=EDA AD is common in both triangles.Therefore, by ASA congruence critarion, we get ΔEADΔFDA EA=DF ...(vi)Adding equation (ii) and (iii), we get DA2+AB2+2EA×AB+AD2+DC22DC×FD=DB2+AC2DA2+AB2+AD2+DC2+2EA×AB2DC×FD=DB2+AC2BC2+AB2+AD2+DC2+2EA×AB2AB×EA=DB2+AC2 (Using equations (iv), (v) and (vi))AB2+BC2+CD2+DA2=AC2+BD2Hence, the sum of the squares of the diagonals ofa parallelogram is equal to the sum of the squares of its sides.

Q.91

In the following figure, AD is a median of a triangle ABC and AMBC. Prove that:(i) AC2=AD2+BC.DM+(BC2)2(ii) AB2=AD2BC . DM+(BC2)2(iii) AC2+AB2=2AD2+12BC2

Ans.

We have the following figure.

(i) Applying Pythagoras theorem in ΔAMD, we get AM2+DM2=AD2 ...(1)Applying Pythagoras theorem in ΔAMC, we get      AM2+MC2=AC2                                AM2+(DM+DC)2=AC2(AM2+DM2)+DC2+2DM.DC=AC2Using equation (1), we get AD2+DC2+2DM.DC=AC2AD2+(BC2)2+2DM.BC2=AC2AD2+(BC2)2+DMBC=AC2(ii)Applying Pythagoras theorem in ΔABM, we getAB2=AM2+MB2 =(AD2DM2)+MB2 =(AD2DM2)+(BDDM)2 =AD2DM2+BD2+DM22BD×DM =AD2+BD22BD×DM =AD2+(BC2)22BC2×DM =AD2+(BC2)2BC×DM(iii)Applying Pythagoras theorem in ΔAMB, we getAM2+MB2=AB2 ...(1)Applying Pythagoras theorem in ΔAMC, we getAM2+MC2=AC2 ...(2)Adding equations (1) and (2), we get 2AM2+MB2+MC2=AB2+AC22AM2+(BDDM)2+(MD+DC)2=AB2+AC22AM2+BD2+DM22BD.DM+MD2+DC2+2DM.DC=AB2+AC22AM2+2DM2+BD2+DC2+2DM(BD+DC)=AB2+AC22(AM2+DM2)+(BC2)2+(BC2)2+2DM(BC2+BC2)=AB2+AC22AD2+BC22=AB2+AC2

Q.92

In the following figure, ABC is a triangle in which ABC < 90° and ADBC. Prove that AC 2 = AB 2 + BC 2 2BC.BD. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@90C1@

Ans.

Applying Pythagoras theorem in ΔADB, we get AD2+BD2=AB2AD2=AB2BD2                         ...(1)Applying Pythagoras theorem in ΔADC, we getAD2+DC2=AC2 Now using equation (1), we getAB2BD2+DC2=AC2AB2BD2+(BCBD)2=AC2AC2=AB2BD2+BC2+BD22BC×BDAC2=AB2+BC22BC×BD

Q.93

In the following figure, ABC is a triangle in which ABC > 90° and ADCB produced. Prove thatAC2=AB2+BC2+2BC.BD.

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

In the following figure, D is a point on hypotenuse AC of ΔABC, DMBC and DNAB. Prove that:(i) DM2=DN.MC (ii) DN2=DM.AN

Ans.

(i)

We have, DNCB, DMAB and B=90°So, DNMB is a rectangle.DN=MB and DM=NB2+3=90° ...(1)In ΔCDM,1+2+DMC=180°1+2=90°                                                                      ...(2)In ΔDMB3+DMB+4=180°3+4=90°                                                                      ...(3)From equation (1) and (2),1=3From equation (1) and (3),2=4ΔDCM~ΔBDMDMBM=MCDMDM2=BM×MC=DN×MC [BM=DN](ii)In ΔDBN,5+7=90° ...(4)In ΔDAN,6+8=90° ...(5)In ΔDAB,5+6=90°                            ...(6)From equation (4) and (6), we get6=7From equation (5) and (6), we get5=8 ΔBND~ΔDNA [By AA similarity critarion]ANDN=DNNBDN2=AN×NB=AN×DM [As NB=DM]

Q.95

In the following figure, PS is the bisector of QPR of ΔPQR. Prove that QSSR=PQ PR.

Ans.

We draw a line segment RT parallel to SP whichintersects 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 PSTR) (2) QPS= QTR (As PSTR) (3)Using these equations we get,PRT=QTRSo, PT=PR Now in ΔQTR and ΔQPS,QSP=QRT (As PSTR)QPS=QTR (As PSTR)Q is common     ΔQTR~ΔQPS QSSR=QPPTQSSR=QPPR [PT=PR]

Q.96

Tick the correct answer and justify: In Δ ABC, AB=63 cm, AC=12 cm and BC=6 cm.The angle B is :(A) 120°   (B) 60°(C) 90° (D) 45°

Ans.

It is given that in ΔABC, AB=63 cm, AC=12 cm and BC=6 cmAB2=108 cm2, AC2=144 cm2 and BC2=36 cm2Now,AB2+BC2=108 cm2+36 cm2=144 cm2=AC2Therefore, by converse of Pythagoras theorem, we find thatin ΔABC, angle B is 90°.Therefore, correct answer is (C).

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.

Let ABC is an equilateral triangle and AE is its altitude.Let length of each side of ΔABC be a.So,BE=EC=12BC=a2.Using Pythagoras theorem in ΔACE, we have AE2=AC2EC2=a2a24=34a23a2=4AE2Therefore, three times the square of one side of anequilateral triangle is equal to four times the squareof one of its altitude.

Q.98

In an equilateral triangle ABC, D is a point on side BC such that BD=13BC. Prove that 9AD2=7AB2.

Ans.

It is given that ABC is an equilateral triangle.Let length of each side of ΔABC be a.It is given that D is a point on side BC such that BD=13BC=a3.Let AE is an altitude of equilateral triangle ABC. So,BE=EC=12BC=a2.Also, AE=AC2EC2=a2a24=32aAgain, DE=BEBD=a2a3=a6In ΔADE,      AD2=AE2+DE2 [By Pythagoras Theorem]AD2=(32a)2+a236=34a2+a236=27a2+a236=2836a2=79a2AD2=79a2=79AB2 [AB=a]9AD2=7AB2

Q.99

The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB=3CD (see thefollowing figure). Prove that 2AB2=2AC2+BC2.

Ans.

In ΔADC,       AC2=CD2+AD2AD2= AC2CD2                                        ...(1)In ΔADB, AB2=AD2+DB2AD2=AB2DB2                                        ...(2)From (1) and (2), we haveAC2CD2=AB2DB2                                          ...(3)Now, it is given that DB=3CDAlso,       BC=CD+DBBC=CD+3CD=4CDCD=BC4                                                                 ...(4)Again,      BC=CD+DBBC=DB3+DB=4DB3DB=3BC4                                                                 ...(5)From (3), (4) and (5), we have       AC2CD2=AB2DB2AC2BC216= AB29BC21616AC2BC2=16AB29BC216AB2=16AC2BC2+9BC2=16AC2+8BC22AB2=2AC2+BC2

Q.100

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.Prove that AE2+BD2=AB2+DE2.

Ans.

In ΔACE,       AE2=AC2+CE2                                                                 ...(1)In ΔDCB, BD2=CD2+BC2                                                               ...(2)On adding (1) and (2), we haveAE2+BD2=AC2+CE2+CD2+BC2                                       ...(3)In ΔACB,AB2=AC2+BC2                                                                            ...(4)In ΔDCE,DE2=CD2+CE2                                                                             ...(5)On adding (4) and (5), we haveAB2+DE2=AC2+BC2+CD2+CE2                                         ...(6)From (3) and (6), we find thatAE2+BD2=AB2+DE2

Q.101

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feetof the poles is 12 m, find the distance between their tops.

Ans.

Let AB and CD be two poles of heights 6 m and 11mrespectively. Let their feet are on the ground at points Band D. It is given that distance between feet is 12 m. Therefore,BD=12 m.Also,AB=DE=6mandCE=CDDE=11 m6 m=5 mNow, AEC is a right triangle. Therefore, by Pythagoras theorem,AC=AE2+CE2=122+52=144+25=169=13 mTherefore,Distance between tops of the two poles=13 m

Q.102

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 ata speed of 1200 km per hour. How far apart will be the two planes after 112  hours?

Ans.

Distance travelled by the plane flying towards north in112 hours = 1000×112 km=1500 kmDistance travelled by the plane flying towards west in112 hours = 1200×112 km=1800 kmWe represent thsese distances by OA and OB.Also, AB represents the distance between the two planes.Using Pythagoras theorem, we get AB2=OA2+OB2=15002+18002=22,50,000+32,40,000AB2=54,90,000AB=54,90,000=9×6,10,000=30061Therefore, distance between the two planes=30061 km

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.

Let AC is a pole and AB is a guy wire with stake at B. Then, ABC is a right angle triangle. Therefore, using Pythagorastheorem, we get     AB2=BC2+AC2242=BC2+182BC2=242182=576324=252BC=252=4×9×7=67Therefore, the distance of the foot of the ladder from thebase of the wall is 67 m.

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.

Let AB is a ladder and AC is wall. A represents the windowand BC is the distance of the foot of the ladder from thebase of the wall.ABC is a right angle triangle. Therefore, using Pythagorastheorem, we get     AB2=BC2+AC2102=BC2+82BC2=10064=36BC=6Therefore, the distance of the foot of the ladder from thebase of the wall is 6 m.

Q.105

In the following figure, O is a point in the interior of a triangle ABC,OD  BC, OE  AC and OF  AB. (i) OA2+OB2+OC2 OD2OE2OF2 = AF2+BD2+CE2,(ii)  AF2+BD2+CE2 = AE2+CD2+BF2.

Ans.

We join O to A, B and C.(i)In ΔAOF,OA2=OF2+AF2 [By Pythagoras theorem]In ΔBOD,OB2=OD2+BD2 [By Pythagoras theorem]In ΔCOE,OC2=OE2+EC2 [By Pythagoras theorem]Adding these equations, we get OA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2OA2+OB2+OC2OD2OE2OF2=AF2+BD2+EC2(ii)We have form above result,AF2+BD2+EC2=OA2+OB2+OC2OD2OE2OF2                                 =(OA2OE2)+(OB2OF2)+(OC2OD2)                                 =AE2+BF2+CD2

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.

Let ABCD is a rhombus. We know that diagonals of a rhombus bisect each otherat right angles.Using Pythagoras Theorem in ΔAOB, ΔBOC, ΔCOD and ΔAOD, we getAB2=OA2+OB2, BC2=OB2+OC2,CD2=OC2+OD2,DA2=OA2+OD2Adding all these equations, we getAB2+BC2+CD2+DA2=2(OA2+OB2+OC2+OD2)                                              =22AC22+2BD22                                              =AC2+BD2

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

Ans.

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 anequilateral triangle.Therefore, BD=CD=a.Using Pythagoras theorem in ΔADB, we get      AB2=BD2+AD2AD2=AB2BD2=(2a)2a2=4a2a2=3a2AD=a3We know that all the altitudes in an equilateral triangleare of same length. Therefore, length of each altitude is a3.

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

In the following figure, ABD is a triangle right angled at A and AC BD. Show thati AB2 = BC . BDii AC2 = BC . DCiii AD2 = BD . CD

Ans.

(i)In ΔADB and ΔCAB, we have            DAB=ACB=90°and    ABD=CBA [Common angle]Therefore, by AA similarity critarion, ΔADB~ΔCAB.Therefore, we have     ABCB=BDABAB2=BCBD(ii)Let CAB=θIn ΔCBA, we haveCBA=180°90°θ=90°θIn ΔCAD, we haveCAD=90°CAB=90°θCDA=180°90°(90°θ)=θIn ΔCBA and ΔCAD, we haveCBA=CAD,CAB=CDA and ACB=DCA=90°Therefore, by AAA similarity critarion, ΔCBA~ΔCAD.Therefore, we have     ACDC=BCACAC2=BCDC(iii)In ΔDCA and ΔDAB, DCA=DAB=90°, CDA=ADBTherefore, by AA similarity critarion, ΔDCA~ΔDAB.Therefore, we have     DCDA=DADBAD2=BDCD

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.

Let MPR=θIn ΔMPR,MRP=180°90°θ=90°θSimilarly, in ΔMPQ,MPQ=90°MPR=90°θMQP=180°90°(90°θ)=θNow, in ΔMPR and ΔMQP, we haveMQP=MPR, MPQ=MRP and PMQ=PMR.Therefore, by AAA similarity critarion, ΔMPR~ΔMQP.Therefore, we have     QMPM=MPMRPM2=QM×MR

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)

Given sides of the triangle are 3 cm, 8 cm and 6 cm.Squares of the given sides of the triangle are 9 cm2, 64 cm2and 36 cm2.Now,9+3664We find that sum of the squares of two sides is not equal tothe 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 cm2,6400 cm2 and 10000 cm2.Now,2500+640010000We find that sum of the squares of two sides is not equal tothe square of third side. Therefore, triangle of the given sides is not a right triangle.

(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

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

Ans.

It is given that sides of two similar triangles are in theratio 4:9.Therefore, Ratio of areas of these triangles =4292=1681Hence, the correct answer is (D).

Q.115

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of trianglesABC and BDE is(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Ans6

It is given that ΔABC and ΔBDE are equilateral and D is themid point of BC. Therefore, BD=BC2We know that equilateral triangles are similar to each otherand so ratio of their areas is equal to square of the ratio of their sides. ar(ΔABC)ar(ΔBDE)=(ABBD)2=(BCBD)2 [ΔABC is equilateral] =(BCBC2)2=41 ar(ΔABC):ar(ΔBDE)=4:1Hence, the correct answer is (C).

Q.110  

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of theequilateral triangle described on one of its diagonals.

Ans.

Let ABCD is a square and length of its one side is a unit.Therefore, length of its diagonal=2aEquilateral triangles as per question are formed here asΔABE and ΔDBF.Length of a side of equilateral ΔABE=aandlength of a side of equilateral ΔDBF=2aWe know that equilateral triangles are similar to each other.Therefore, ratio of their areas is equal to ratio of squares oftheir sides.i.e., ar(ΔABE)ar(ΔDBF)=(a2a)2=12

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.

Let us take two triangles ABC and PQR such that ΔABC~ΔPQR.Let AD and PS be the medians of ΔABC and ΔPQR respectively.Now, ΔABC~ΔPQR    ABPQ=BCQR=ACPR and A=P, B=Q, C=R.Since AD and PS are medians, so we haveBD=DC=BC2 and QS=SR=QR2In ΔABD and ΔPQS,B=Q and ABPQ=BCQR=BDQSΔABD~ΔPQS [SAS similarity criterion]ABPQ=BDQS=ADPSNow,        ΔABC~ΔPQRar(ΔABC)ar(ΔPQR)=(ABPQ)2ar(ΔABC)ar(ΔPQR)=(ADPS)2

Q.118

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areasof ΔDEF and ΔABC.

Ans.

D and E are mid-points of ΔABC.DEAC and DE=12ACIn ΔBED and ΔBCA,BED=BCA [Corresponding angles]BDE=BAC [Corresponding angles]EBD=CBA [Common angles]ΔBED~ΔBCA [AAA similarity criterion]ar(ΔBED)ar(ΔBCA)=(DEAC)2=(12ACAC)2=14ar(ΔBED)=14ar(ΔBCA)Similarly,ar(ΔCFE)=14ar(ΔCBA) and ar(ΔADF)=14ar(ΔABC)Also, ar(ΔDEF)=ar(ΔABC)[ar(ΔBED)+ar(ΔCFE)+ar(ΔADF)]ar(ΔDEF)=ar(ΔABC)34ar(ΔABC)=14ar(ΔABC)ar(ΔDEF)ar(ΔABC)=14

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

Ans.

Let ΔABC~ΔPQR, then we have       ar(ΔABC)ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2ar(ΔABC)ar(ΔABC)=(ABPQ)2=(BCQR)2=(ACPR)2 [Given ar(ΔABC)=ar(ΔPQR)]1=(ABPQ)2=(BCQR)2=(ACPR)21=ABPQ=BCQR=ACPRAB=PQ, BC=QR and AC=PRΔABCΔPQR [By SSS congruence criterion]

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.

We know that area of a triangle=12×Base×Heightar(ΔABC)ar(ΔDBC)=12BC×AP12BC×DM=APDMIn ΔAPO and ΔDMO,APO=DMO=90° AOP=DOM [Vertically opposite angles]ΔAPO~ΔDMO [By AA similarity critarion]APDM=AODOar(ΔABC)ar(ΔDBC)=AODO

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.

In ΔCOD and ΔAOB,ODC=OBA [Alternate interior angles as ABCD]DCO=BAO [Alternate interior angles as ABCD]COD=AOB [Vertically opposite angles]ΔCOD~ΔAOB [AAA similarity critarion]ar(ΔCOD)ar(ΔAOB)=CD2AB2ar(ΔCOD)ar(ΔAOB)=CD2(2CD)2=14 [Given AB=2CD]ar(ΔAOB):ar(ΔCOD)=4:1

Q.121  

Let ΔABC~ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF=15.4 cm, find BC.

Ans.

It is given that ΔABC~ΔDEF.ar(ΔABC)ar(ΔDEF)=BC2EF2Given thatEF=15.4  cmar(ΔABC)=64 cm2ar(ΔDEF)=121 cm2    ar(ΔABC)ar(ΔDEF)=BC2EF264121=BC2(15.4)2811=BC15.4BC=8×15.411BC=11.2

Q.122

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC~ΔPQR, prove thatABPQ=ADPM.

Ans.

It is given that AD and PM are medians. Therefore,12BC=BD=DC and 12QR=QM=MRAlso, ΔABC~ΔPQRTherefore,        ABPQ=BCQR=ACPR and A=P, B=Q,  C=RIn ΔABD and ΔPQM,ABPQ=12BC12QR=BDQM and B=Q ΔABD~ΔPQM [By SAS similarity critarion]ABPQ=ADPM

Q.123  

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a towercasts a shadow 28 m long. Find the height of the tower.

Ans.

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=BAEAlso, CDF=ABE=90°ΔABE~ΔCDF [By AA similarity critarion]Corresponding sides are proportional in similar triangles.Therefore,ABCD=BEDFAB6 =284=7AB=42Therefore, the height of the tower is 42 m.

Q.124

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

Ans.

It is given that,ABPQ=ACPR=ADPMWe 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.

It is given that AD and PM are medians. Therefore,12BC=BD=DC    and    12QR=QM=MRIn 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 isa parallelogram and PR=QL, PQ=LR.It is given that         ABPQ=ACPR=ADPM    ABPQ=BEQL=2AD2PM    ABPQ=BEQL=AEPL Therefore, by SSS similarity critarion, ΔABE~ΔPQL and soin ΔABE and ΔPQL, BAE=QPL ...(1) Similarly, we can show that ΔAEC~ΔPLR and CAE=RPL ...(2) Adding (1) and (2), we getBAE+CAE=QPL+RPLCAB=RPQAlso, we haveABPQ=ACPRTherefore, by SAS similarity critarion, ΔABC~ΔPQR.

Q.125

D is a point on the side BC of a triangle ABC such that ADC=BAC. Show that CA2=CB·CD.

Ans.

In ΔBAC and ΔADC,ADC=BAC (Given)ACD=BCA (Common angle)ΔADC~ΔBAC (By AA similarity critarion)We know that corresponding sides of similar trianglesare proportional.   CACD=CBCA  CA2=CBCD

Q.126

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR andmedian PM of ΔPQR (see the following figure). Show that ΔABC~ΔPQR.

Ans.

It is given that AD and PM are medians. Therefore,12BC=BD=DCand12QR=QM=MRIt is given that         ABPQ=BCQR=ADPM    ABPQ=12BC12QR=ADPM    ABPQ=BDQM=ADPM Therefore, by SSS similarity critarion, ΔABD~ΔPQM and soin ΔABC and ΔPQR, B=Q and ABPQ=BCQR.Therefore, by SAS similarity critarion, ΔABC~ΔPQR.

Q.127

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB=AC. If ADBC and EFAC, prove that ΔABDΔECF. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B14C@

Ans.

In ΔABD and ΔECF,          BDA=CFE=90°and ABD=ECF [AB=ACB=C]Therefore, by AA similarity critarion, ΔABD~ΔECF.

Q.128

CD and GH are respectively the bisectors of ΔACB and ΔEGF such that D and H lie on sides AB and FEof ΔABC and ΔEFG respectively. If ΔABC~ΔFEG, show that:(i)    CDGH=ACFG(ii)   ΔDCB~ΔHGE(iii)  ΔDCA~ΔHGF

Ans.

It is given that ΔABC~ΔFEG.A=F, B=E, BCA=EGFAlso,      12BCA=12EGFACD=FGH and DCB=HGE By AA similarity critarion,      ΔDCA~ΔHGF and ΔDCB~ΔHGECDGH=ACFG

Q.129

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:(i) ΔABC~ΔAMP(ii) CAPA=BCMP

Ans.

(i)In ΔABC and ΔAMP, ABC=AMP=90°and, CAB=PAM Therefore, by AA similarity critarion, ΔABC~ΔAMP.(ii)Corresponding sides are proportional in similar triangles.Therefore,      ΔABC~ΔAMPCAPA=BCMP

Q.130

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CDat F. Show that ΔABE~ΔCFB.

Ans.

In the above figure, ABCD is a parallelogram in which ABDC.Opposite angles in parallelogram are equal. Therefore,EAB=BCFAlso, ABE=CFB [Alternate interior angles]Therefore, by AA similarity critarion, ΔABE~ΔCFB.

Q.131

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:(i) ΔAEP~ΔCDP(ii) ΔABD~ΔCBE(iii)ΔAEP~ΔADB(iv)ΔPDC~ΔBEC

Ans.

(i)In ΔAEP and ΔCDP,       APE=CPD (Vertically opposite angles)and AEP=CDP=90° Therefore, by AA similarity critarion, ΔAEP~ΔCDP.(ii)In ΔABD and ΔCBE,          ADB=CEB=90°and ABD=CBETherefore, by AA similarity critarion, ΔABD~ΔCBE.(iii)In ΔAEP and ΔADB,          AEP=ADB=90°and PAE=BADTherefore, by AA similarity critarion, ΔAEP~ΔADB.(iv)In ΔPDC and ΔBEC,          PDC=BEC=90°and DCP=ECBTherefore, by AA similarity critarion, ΔPDC~ΔBEC.

Q.132

In the following figure, if ΔABEΔACD, show that ΔADE~ΔABC.

Ans.

It is given that ΔABEΔACD.Therefore, by CPCT,         AB=ACand AD=AESo, ADAB=AEACIn ΔADE and ΔABC,       A=A and ADAB=AEACTherefore, by SAS criterion, ΔADE~ΔABC.

Q.133

S and T are points on sides PR and QR of ΔPQR such that P=RTS. Show that ΔRPQ~ΔRTS.

Ans.

In ΔRPQ and ΔRTS,       P=T (Given)and R=R Therefore, by AA criterion, ΔRPQ~ΔRTS.

Q.134

In the following figure, QR QS = QT PR and 1=2. Show that ΔPQSΔTQR. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7401@

Ans.

In ΔPQR,        1=2 (Given)PQ=PR ...(1)It is given that QRQS=QTPRQRQS=QTQP [From (1), PR=QP]    ...(2)Q is common in ΔPQS and ΔTQR and the sides includingthis angle in both triangles are proportional as shown inequation (2). Therefore, by SAS critarion, ΔPQS~ΔTQR.

Q.135

Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at the point O. Usinga similarity criterion for two triangles, show that OAOC=OBOD.

Ans.

In ΔDOC and ΔBOA,CDO=ABO [Alternate interior angles as ABCD]DCO=BAO [Alternate interior angles as ABCD]DOC=BOA [Vertically opposite angles]ΔDOC~ΔBOA [AAA similarity critarion]DOBO=OCOA [Corresponding sides are proportional]OAOC=OBOD

Q.136

In the following figure, ΔODCΔOBA, BOC=125°and CDO=70°. Find DOC, DCO and OAB. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8541@

Ans.

DOB is a straight line.BOC+DOC=180°DOC=180°BOC=180°125°=55°In ΔDOC,      DOC+DCO+ODC=180°DCO=180°DOCODCDCO=180°55°70°=55°It is given that ΔODC~ΔOBA and so corresponding anglesin these triangles are equal.OAB=DCO=55°

Q.137 State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.


Ans.

(i) Corresponding angles are equal in the given triangles. i.e.,A=P=60°,B=Q=80°andC=R=40°.Therefore, by AAA similarity criterion, ΔABC~ΔPQR.(ii) Ratios of the corresponding sides of the given pair oftriangles are equal. i.e., ABQR=BCRP=CAPQ=12Therefore, by SSS similarity criterion, ΔABC~ΔQRP.(iii)The given pair of triangles are not similar as theircorresponding sides are not proportional.(iv)The given pair of triangles are not similar as theircorresponding sides are not proportional.(v)The given pair of triangles are not similar as theircorresponding sides are not proportional.(vi) In ΔDEF,      D+E+F=180°70°+80°+F=180°F=180°150°=30°In ΔPQR,      P+Q+R=180°P+80°+30°=180°P=180°110°=70°We find out that corresponding angles are equal in ΔDEF and ΔPQR. i.e.,D=P=70°,E=Q=80°andF=R=30°.Therefore, by AAA similarity criterion, ΔDEF~ΔPQR.

Q.138

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO. Show that ABCD is a trapezium.

Ans.

Given:Diagonals of a quadrilateral ABCD intersect each other at point O such that AOBO=CODO.

To prove:ABCDConstruction:Draw OFAB which meets AD in F.In ΔABD, OFABSo, by converse of Basic  Proportionality Theorem,          DOBO=DFFA ...(i)Given that      AOBO=CODODOBO=COAODFFA=COAO [From (i)]OFDC [By converse of Basic Proportionality Theorem]Also, OFABTherefore, ABDCHence, ABCD is a trapezium.

Q.139

ABCD is a trapezium in which ABDC and its diagonals intersect each other at the point O.Show that AOBO=CODO.

Ans.

Given:ABCD is a trapezium in which AB is parallel to CD and diagonals intersect each other at O.

To prove:AOBO=CODOConstruction:Draw OFAB which meets AD in F.In ΔABD, OFABSo, by converse of Basic  Proportionality Theorem,          DOOB=DFFA ...(i)Now in ΔADC, OFCD           [Since ABCD]              DFFA=COOA      ...(ii)Comparing equation (i) and (ii), we get                 DOOB=COOA OAOB=CODO

Q.140 Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
(Recall that you have done it in Class IX)


Ans.

We consider ΔABC drawn below in which DE is the line segment joining the mid-points D and E of sides AB and AC respectively.

We haveAD=DBADDB=1and AE=ECAEEC=1Therefore,ADDB=AEECHence, by converse of Basic Proportionality Theorem, DEBC.

Q.141 Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Ans.

We consider ΔABC drawn below in which D is the mid-point of side AB and DE is the line segment drawn parallel to BC.