NCERT Solutions for Class 10 Maths
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NCERT Solutions for Class 10 Maths
Mathematics is a revered topic for engineering students pursuing professions through competitive exams such as JEE Main and JEE Advanced. It is one of those disciplines where a flawless centum can be obtained if the right preparation schedule and time are followed.
Over the last three to four years, CBSE has shifted away from regular NCERT questions and has focused more on application-based questions. This has made the paper rather tough to score if students just prepare for the questions that are expected to appear in the exam.
NCERT Solutions for Class 10 Maths Chapter-wise
Maths is a compulsory component in Class 10. Therefore, all CBSE students need to use the CBSE Class 10 Mathematics NCERT Solutions for complete understanding of the concepts to ace their preparation. These solutions cover all key mathematical ideas and assist students in developing critical problem-solving skills.
NCERT Solutions for Class 10 Maths
The solutions consist of 15 chapters, all of which are crucial for study. Students need to complete all of the exercises, even the optional ones to score well in the exam. NCERT solutions for class 10 Mathematics will assist you in thoroughly understanding each question. You can easily do well in board exams if you solve the NCERT Mathematics and practice the previous 5 years’ questions.
NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers
Using Euclid’s Lemma, the Fundamental Theorem of Arithmetic. Real Numbers explains the notion of properties of real numbers in great detail. Exercises 1.1 and 1.2 require the use of division procedure to determine the divisibility of integers. Exercises 1.3 and 1.4, on the other hand, feature a question about proving the roots of any irrational number and decimal expansions of rational numbers.
NCERT Solution for Class 10 Mathematics Chapter 2 – Polynomials
This chapter covers advanced polynomial terms such as degree, coefficient, and zeroes for any type of polynomials. The chapter offers four exercises, with Exercises 2.1 and 2.2 requiring the use of a graph. Exercise 2.3 focuses on the division algorithm, and exercise 2.4 covers all subjects covered in the chapter.
NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
This chapter discusses the concept of linear equation pairs. It begins with linear equations in two variables and their solutions using the graphical approach, as well as algebraic conditions for linear equations. The chapter goes on to explain how to solve a pair of linear equations in two variables algebraically using substitution, elimination, and cross-multiplication.
NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations
In this chapter, the learner will learn how to write quadratic equations in standard form. The chapter also covers the nature of roots, such as whether they are real and distinct, real and equal, or non-real roots. It also includes four exercises in which you must determine whether an equation is a quadratic one or not.
NCERT Solutions for Class 10 Mathematics Chapter 5 – Arithmetic Progressions
Students will learn in greater detail about arithmetic progression, which is a series in which two consecutive terms have a constant difference. The chapter serves as the foundation for other complex concepts that will be introduced in later classes. They will develop the skills to solve issues with sequences and series.
NCERT Solutions for Class 10 Maths Chapter 6 – Triangles
The Triangles chapter covers the concepts of similarity and congruence. The basics of triangles are covered in Exercise 6.1. Problems from the topic of triangular similarity and a related theorem appear in exercises 6.2, 6.3, and 6.4. Exercises 6.5 and 6.6 focus on triangular congruency and application.
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry
Students will learn how to find the coordinates of a point that splits a line segment connecting two given sites in a specified ratio. The Distance formula and Section formula are key to the chapter in 10th grade maths. Sections on the area of the triangle are also included.
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry
This chapter is about the ratio of right angles to acute angles in any right-angled triangle. The questions in this chapter cover trigonometric ratios of specific angles, trigonometric identities, and trigonometric ratios of complementary angles are used. There are four exercises in this chapter: Exercise 8.1 introduces the trigonometric ratio and its calculation using Pythagoras’ theorem. Exercise 8.2 uses the trigonometric table to calculate specific angles. Exercise 8.3 is about complementary angles; and Exercise 8.4 is about trigonometric identities.
NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry
Trigonometry is the study of how to calculate the heights and distances of various things without having to measure them. This chapter explains how trigonometry is used in everyday life. Other such topics included in this chapter are the angle of deviation and elevation, and the line of sight.
NCERT Solutions for Class 10 Mathematics Chapter 10 – Circles
Students have a fundamental understanding of circles and associated terminology such as chord, segment, arc, and so on. They become familiar with the concepts of Tangent to a Circle and Number of Tangents from a Point on a Circle. The chapter discusses circle theorems which aid in the solution of many geometry problems.
NCERT Solutions for Class 10 Mathematics Chapter 11 – Constructions
This chapter teaches students how to build various geometrical figures. Methods and processes for construction are presented in simple language. This chapter also discusses Thales Theorem, often known as the Basic Proportionality Theorem. It ends with tasks that will teach pupils how to generate tangents to a circle.
NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles
This chapter includes three exercises, as well as real-life difficulties and cost-related questions about the circular territory. It also introduces the concepts of circle perimeter and area. Students will be able to compute the area of a sector and a segment of a circular region.
NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes
This chapter discusses calculating the areas and volumes of several solids such as the cube, cuboid, and cylinder. Questions are based on determining the volume of objects generated by merging several solid figures or converting one shape to another. NCERT Solutions for Class 10 Maths provide answers to both easy and difficult questions on the topic.
NCERT Solutions for Class 10 Maths Chapter 14 – Statistics
Students will learn how to convert individual data into grouped data and calculate the Mean, Mode, and Median. This chapter serves as the foundation for data science and analytics. The chapter teaches students how to create a graphical representation of raw, unorganised data from which tangible results can be generated.
NCERT Solutions for Class 10 Maths Chapter 15 – Probability
The final chapter is about Probability, which is the study of determining the likelihood of success or failure in a specific experiment. There are numerous examples provided to clearly explain probability and to help students understand it. Students must first solve the textbook examples before attempting to solve the exercise problems.
CBSE Class 10 Maths Exam Pattern for 2022-2023
The exam is worth 100 marks, with the theory worth 80 marks and the internal assessment worth 20 marks. The CBSE Board releases sample papers and grading schemes well in advance of the board exams to help students clear up any confusion. According to the CBSE’s revised design, 25% of theoretical questions would be objective. The number of questions has been raised, but the marks assigned to each question have been reduced. There are four parts to this paper. There are 30 questions for a total of 80 marks.
Internal Assessment Marking Scheme
Sr. No | Activity | Marks |
1 | Pen Paper Test | 10 |
2 | Portfolio | 05 |
3 | Lab Practical | 05 |
TOTAL | 20 |
Class 10 Maths Chapter-Wise Marks Weightage
Sr. No | Chapter Name | Marks |
1 | Number System | 6 |
2 | Algebra | 20 |
3 | Coordinate Geometry | 6 |
4 | Geometry | 15 |
5 | Trigonometry | 12 |
6 | Mensuration | 10 |
7 | Statistics and Probability | 11 |
CBSE Class 10 Maths Question Paper Design 2022-23
Sr. No | Type of Questions | Total Marks | Weightage |
1 | Memory: Terms, answers, basic concepts and recalling of facts.
Understanding: Compare, organise, interpret, translate, describe and stating main ideas. |
43 | 54 |
2 | Apply: Solving problems by applying the knowledge, facts, theorems and rules in different ways | 19 | 24 |
3 | Analyse: Analyse the information provided and make inferences.
Evaluate: Opine about the validity of quality of work and validity of ideas. Create: Creating alternative solutions to the given problem and grouping the elements. |
18 | 22 |
TOTAL | 80 | 100 |
How will CBSE Class 10 Maths Solutions Help You to Score High Marks in Board Exams?
Deeper Understanding of Topics: More than knowing the answer to a problem or how to solve a specific type of question, it is critical to comprehend the ideas of algebra, geometry, statistics, and trigonometry in order to score well in Class 10 Maths. Extramarks 10th CBSE Maths solutions make it easy to understand the underlying concepts while answering questions.
Ready Reference for Practice: Students should make it a point to practice Mathematics exercises on a regular basis. As a result, anytime you intend to practice after having previously solved exercises, you may not recall everything, which is why having a ready reference for practice is vital.
Creating a Solid Foundation: While answering exercise questions is important, it is also necessary to have a good conceptual understanding. There are two reasons behind this – 1) Exam questions may be difficult to answer, but possessing conceptual understanding makes them simple to solve. 2) As you proceed to the following level, the same concepts are utilized to explain more complicated subjects, emphasizing the need for a solid foundation in Maths.
No Delay in Understanding: When students face a challenge, they may have to wait for some time with teachers to properly understand the problems and clear their questions. Class 10 Maths NCERT Solutions, on the other hand, assist students in quickly grasping key topics and avoiding delays in comprehending Mathematics.
Why Refer NCERT Maths Class 10 Solutions for Board Exams?
Class 10 is a critical time in a student’s life since the grades they obtain influence their future career route. Getting the top results on the board exams may be difficult if you do not have the correct study resources. The experts at Extramarks create NCERT solutions for class 10 Maths on a subject-by-subject basis for this purpose.
NCERT Solutions will guide you in dealing with the concepts that you find difficult to solve. These solutions use step-by-step procedures to ensure that you grasp each topic and perform well on your Class 10 board exams.
Important Tips for Scoring Well in the Board Exams
- You should be familiar with the Class 10 mathematics test pattern, such as the number of very short answer questions, short answer questions, and long answer questions.
- Before you begin your test preparation, familiarise yourself with the CBSE class 10 Mathematics syllabus. It will assist you in developing a better study strategy for yourself. You will also learn how many topics and chapters must be finished during the full session.
- Remember the key formulas from each chapter. Keep a separate notebook for the formulas and revise it on a regular basis. This will save you time while answering questions throughout the exam.
- Make a schedule before you begin studying. Keeping timetables will keep you on track and allow you to study at a consistent pace. You can modify them according to your strengths in the subject and topics.
- Try solving questions from the NCERT Solutions for Class 10 Mathematics about whatever is taught in class. You can also consult other books, such as RD Sharma, for additional practice, but first complete the NCERT textbook’s exercise questions.
- Take a walk or exercise for a few minutes during your study breaks. In between your studies, doing 10-20 minutes of exercise will help you keep fit and psychologically prepared for future studies.
- As the exam approaches, try not to be stressed. The night before the exam, get a decent night’s sleep. Before the exam, do not study any new topics. Be calm and relaxed.
- When revising, pay careful attention to the topics from which most questions are usually asked and carry more marks. Also, before the exam, answer the important questions.
Q.1
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Q.2 ABCD is a cyclic quadrilateral. Find its all angles.
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Q.3
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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
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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
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Q.8
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Q.9
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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)
Q.11
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Q.12
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Q.13
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Q.14
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Q.15
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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.
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Q.17
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Q.18
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(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
(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.
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Q.20
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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
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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.
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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
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Q.25
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Q.26
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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.
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.
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.
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 |
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.
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 |
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.
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 |
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.
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 |
We proceed as below to find di, ui, fiui .
Number of letters | Number of surnames (fi) | xi | di = xi – a | 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 |
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.
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 |
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.
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 |
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.
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 |
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
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.
Monthly consumption (in units) | Number of consumers (fi) | xi | di = xi – 135 | 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 |
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.
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.
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.
Number of students per teacher | Number of states/UT (fi) | xi | di = xi – 32.5 |
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 |
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.
Expenditure (in ₹) | Number of families (fi) | xi | di = xi – 2750 | 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 |
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.
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.
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 |
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.
Literacy rate (in %) | Number of cities fi | xi | di = xi – 70 | 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 |
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.
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 |
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.
Concentration of SO2 (in ppm) | Frequency (fi) | xi | di = xi – 0.14 |
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 |
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.
Class interval | fi | xi | di = xi – 225 | 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 |
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.
Class interval | fi | xi | di = xi – 57 | ||
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 |
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.
Number of heartbeats per minute | Number of women (fi) | xi | di = xi – 75.5 | 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 |
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.
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 | 2f – 40 |
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.
Daily wages (in ₹) |
Number of workers (fi) | xi | di = xi – 550 | 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 |
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.
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.
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.
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.
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.
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.
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.
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.
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.
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 |
(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 |
(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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
Q.87
Ans.
Q.88
Ans.
Q.89
Ans.
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.
Q.91
Ans.
We have the following figure.
Q.92
Ans.
Q.93
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
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(i)
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Q.96
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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.
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Q.98
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Q.99
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Q.100
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Q.101
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Q.102
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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?
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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.
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Q.105
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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.
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Q.107 ABC is an equilateral triangle of side 2a. Find each of its altitudes.
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Q.108 ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
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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.
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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
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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 =
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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
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(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)
(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.
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Q.115
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Q.110
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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.
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Q.118
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Q.119 If the areas of two similar triangles are equal, prove that they are congruent.
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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.
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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.
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Q.121
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Q.122
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Q.126
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