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RD Sharma Solutions for maths Class 10

Students who follow RD Sharma Solutions for maths Class 10 understand each and every concept thoroughly as the answers are prepared in an interactive manner, based on the current board guidelines. While practising these solutions, students understand the logic in solving the questions without any difficulties.

Our aim is to make education free able experience as well as prepare our students for any type of questions. Our team of highly experienced and dedicated experts have vast experience and are alumni of some of the leading colleges in the country. They know the requirement of every student and adjust their teaching style as per the ability of every individual. Students who are trying in grasping concepts of core subjects such as Maths, Physics, Physics or Chemistry can download our solutions and kick start their preparation. Our platform Studentecare.com is perfect even for those students who want to do revision before the examinations. Our experts also give regular students updates on their study progress so that they are also aware of how their study is faring. Student Can download anytype books from Studentecare.com

As Mathematics is an intricate subject for Class 10 Students, these solutions will change every student’s approach towards Mathematics and will definitely make them realize how interesting and easy the subject is. RD Sharma Solutions are focused on learning various Mathematics tricks and shortcuts for quick and easy calculations. Download the RD Sharma Solutions for Class 10 Maths now and practice all the questions. Solving these questions will ensure that students have a good practice of all types of questions that can be framed in the examination.

RD Sharma Solutions for Class 10 PDF are provided here to help the students in their effective preparation for the exams. Expert faculty at STUDENT eCARE prepares these solutions for the exercise-wise problems after acquiring in-depth knowledge across all concepts. Students can now solve any problem from the RD Sharma textbook by referring it. The main purpose to create such a resource for students is to help them self-analyze their areas of weaknesses and also to help them solve problems of higher difficulty level in an efficient way.

The most excellent technique in which the students can prepare for their forthcoming examinations is to make sure that they are thorough with the concepts, and they have the benefit of accurate study materials. Students no longer have to waste their time and effort in finding the solutions of RD Sharma as we have provided the solutions PDF for free underneath. The RD Sharma Book for Class 10 is, undoubtedly, packed with all the essential summaries and explanations that the students need to tackle. The chapters for Class 10 are based on the lessons present in the textbook, ensuring that the students study only relevant topics and concepts with the problems related to the same. These RD Sharma solutions contain questions with varying levels of difficulty and the questions are in addition to those present in the** **RD Sharma Solutions for Class 10 Maths. Practicing these would help students gain confidence in solving the different types of questions that might appear in the examination.

RD Sharma Solutions for class 10 maths is prepared by STUDENT eCARE Expert to score good marks in class 10. Class 10 maths contains 16 chapters and 16 Excercises which are very important to score good in class 10,

**Name of chapters are mentioned below:**

RD Sharma Class 10 Maths Chapter 1 Real Numbers Solutions

RD Sharma Class 10 Maths Chapter 2 Polynomials Solutions

RD Sharma Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Solutions

RD Sharma Class 10 Maths Chapter 4 Triangles Solutions

RD Sharma Class 10 Maths Chapter 5 Trigonometric Ratios Solutions

RD Sharma Class 10 Maths Chapter 6 Trigonometric Identities Solutions

RD Sharma Class 10 Maths Chapter 7 Statistics Solutions

RD Sharma Class 10 Maths Chapter 8 Quadratic Equations Solutions

RD Sharma Class 10 Maths Chapter 9 Arithmetic Progressions Solutions

RD Sharma Class 10 Maths Chapter 10 Circles Solutions

RD Sharma Class 10 Maths Chapter 11 Constructions Solutions

RD Sharma Class 10 Maths Chapter 12 Some Applications of Trigonometry Solutions

RD Sharma Class 10 Maths Chapter 13 Probability Solutions

RD Sharma Class 10 Maths Chapter 14 Co-ordinate Geometry Solutions

RD Sharma Class 10 Maths Chapter 15 Areas Related To Circles Solutions

RD Sharma Class 10 Maths Chapter 16 Surface Areas and Volumes Solutions

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The PDF material of the solutions can be used as prime reference material by the students to get a strong grip over the concepts under each chapter of the textbook. Further, students solving the problems of RD Sharma textbook using solutions mainly improve their problem-solving skills, which are very essential from an exam point of view.

The Chapter 1 of RD Sharma textbook helps students comprehend the divisibility of integers and some important properties of integers like Euclid’s division lemma, Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. The solutions help students to discover the right method and short-cuts in solving the problem following the latest CBSE guidelines. PDF format of solutions can be downloaded by the students and practiced on a regular basis for good score in the board exams.

- Euclid’s division lemma: If we have two positive integers a and b, there should be whole numbers q and r that satisfy the equation – a = bq + r, where 0 ≤ r < b where a is the dividend and b is the divisor.
- Euclid’s division algorithm: It is the technique used to determine the HCF of two given positive integers.
- Fundamental Theorem of Arithmetic: Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Provided here are the solutions to the exercise of this chapter.

- Exercise 1.1
- Exercise 1.2
- Exercise 1.3
- Exercise 1.4
- Exercise 1.5
- Exercise 1.6

Polynomials are algebraic expressions which have both variables and coefficients. This chapter has problems based on the concepts like the geometrical representation of linear and quadratic polynomials, the geometric meaning of their zeros and relationship between the zeros and coefficients of a polynomial. Our faculty at STUDENT eCARE has explained all these concepts in an interactive manner to help students perform well in the exam. The step wise explanations for each problem help students to gain a better conceptual knowledge which is important from the exam perspective.

- Linear polynomial: A polynomial of degree one i.e. the highest exponent of the variable is one.
- Quadratic polynomial: A polynomial of degree two i.e. the highest exponent of the variable is two.
- Geometrical representation of linear and quadratic polynomial: The representation of linear and quadratic polynomial in a geometric form.

This chapter contains three exercises and their solutions links are provided by us.

- Exercise 2.1
- Exercise 2.2
- Exercise 2.3

The 3rd chapter of RD Sharma Solutions contains problems based on systems of linear equations in two variables, solution of a system of linear equations in two variables, graphical and algebraic methods of solving a system of linear equations in two variables as well as their applications. All these solutions are well explained in a simple language which matches the understanding capabilities of students to enhance their exam preparations. The solutions PDF helps students to understand the method of solving complex problems effortlessly.

- Linear equations in two variables: If a, b, and r are real numbers (a and b are not equal to 0) then ax+by = r is called a linear equation in two variables.
- Graphical and algebraic methods of solving a system of linear equations in two variables: Solving the given linear equation using graphical and algebraic methods.
- Applications of linear equations in two variables: The application of linear equations in two variables in our daily life is covered in this chapter.

Chapter 3 has 11 exercises and solutions links are provided.

- Exercise 3.1
- Exercise 3.2
- Exercise 3.3
- Exercise 3.4
- Exercise 3.5
- Exercise 3.6
- Exercise 3.7
- Exercise 3.8
- Exercise 3.9
- Exercise 3.10
- Exercise 3.11

The exercise wise problems of chapter 4 are solved by subject experts at STUDENT eCARE to help students face the exams with confidence. The important topics which are covered in this chapter are the similarity of geometric figures, similar triangles and their properties, some basic results on proportionality and some important theorems and criteria of proportionality. Students who find it difficult to solve the exercise wise problems can make use of chapter wise PDF to understand the topics with ease.

- Similarity of geometric figures: Two geometric figures are said to be similar if they have the same sides, angles or mirror of the figures.
- Similar triangles: If the two pairs of corresponding angles in two triangles are congruent, then both the triangles are said to be similar.
- Proportionality: The equality between two ratios i.e., a/b = c/d where a and b are in the same proportion as c and d.

This chapter contains 7 exercises and their solution links are given.

- Exercise 4.1
- Exercise 4.2
- Exercise 4.3
- Exercise 4.4
- Exercise 4.5
- Exercise 4.6
- Exercise 4.7

Trigonometry is an important branch of Mathematics. This chapter throws light on the basic concepts of trigonometry like trigonometric ratios and relations between them, trigonometric ratios of some specific angles and trigonometric ratios of complementary angles. RD Sharma solutions created by subject experts at STUDENT eCARE can be utilized by students to solve the exercise wise problems in this chapter. Students are advised to refer to the solutions PDF while solving the problems to gain a grip on the concepts which are important from the exam perspective.

- Trigonometry: It is a branch of Mathematics which deals with the relationship between side lengths and angles of triangles.
- Trigonometric ratios: The six trigonometric ratios are, cosine, sine, cotangent, tangent, secant and cosecant.
- Trigonometric ratios of complementary angles: If the sum of two angles is 90 degrees, then one angle is said to be the complementary of another angle.

Chapter 5 contains 3 exercises for which the solutions are given in links.

- Exercise 5.1
- Exercise 5.2
- Exercise 5.3

In this chapter students will prove some trigonometric identities, and use them to prove other useful trigonometric identities. STUDENT eCARE has made solutions for the exercise problems after conducting wide research on each concept. Further students can use these solutions when solving the chapter wise problems. The solutions contain explanations in understandable language to help students irrespective of their intelligence quotient.

- Trigonometric identities: These are equalities which involve trigonometric functions and are true for every value of occurring variables.

Students can access the exercise-wise solutions of Chapter 6 from the links provided.

- Exercise 6.1
- Exercise 6.2

The Chapter 7 of RD Sharma explains the techniques for finding mean, median and mode of grouped data as well as about cumulative frequency graphs of a frequency distribution. The solutions have explanations in a simple language to enhance conceptual knowledge. Also, the students can access the solutions PDF at ease to improve their speed in solving problems. The objective of providing solutions is to help the students to understand topics which are important from the exam point of view.

- Statistics: The collection, organization, analysis, interpretation and presentation of numerical data are called statistics.
- Mean: It is done by adding all the numbers in the data set and then dividing by the number of values.
- Median: The middle value when a data set is ordered from least to greatest.
- Mode: The number which occurs most frequently in a data set.
- Frequency distribution: The overview of all different values in some variable and the number of times they occur.

The solutions of all 6 exercises present in this chapter created by experts at STUDENT eCARE are given here.

- Exercise 7.1
- Exercise 7.2
- Exercise 7.3
- Exercise 7.4
- Exercise 7.5
- Exercise 7.6

The exercise problems in this chapter mainly deal with various ways of finding their zeros or roots and some applications of quadratic equations in daily life situations. The PDF format of solutions is made available in order to boost the exam preparation of students. The solutions cover all the problems from the prescribed textbook according to the latest CBSE syllabus. The problems contain step wise explanations based on the exam pattern and marks weightage allotted for each step.

- Quadratic equations: The quadratic polynomials when equated to zero are the quadratic equations.
- Standard form of quadratic equation: ax2 + bx + c = 0.

There are about 13 exercises in Chapter 8 and students can access the solutions from the links provided.

- Exercise 8.1
- Exercise 8.2
- Exercise 8.3
- Exercise 8.4
- Exercise 8.5
- Exercise 8.6
- Exercise 8.7
- Exercise 8.8
- Exercise 8.9
- Exercise 8.10
- Exercise 8.11
- Exercise 8.12
- Exercise 8.13

A sequence is an arrangement of numbers in a definite order according to some rule. Arithmetic Progressions is a sequence of numbers in which the difference of any two successive numbers is always a constant. The Chapter 9 contains important concepts like general, nth, selection, the sum of terms of an A.P. Students can make use of the solutions PDF prepared by experts at STUDENT eCARE as a reference to perform well in their exams. The PDF is available in both chapter wise and exercise wise format based on the requirements of students.

- Arithmetic Progression: It is a sequence of numbers where the difference between successive numbers is constant.
- General term of AP: Formula for the nth term of the A.P. with a as first term and d as common difference is an = a + (n-1) d. Here the nth term (an) is called the general term of AP.
- Sum of terms of AP: It is the sum of first n terms of the arithmetic sequence.

The exercise solutions links for the concepts covered in this chapter are given here.

- Exercise 9.1
- Exercise 9.2
- Exercise 9.3
- Exercise 9.4
- Exercise 9.5
- Exercise 9.6

Circle is nothing but the locus of all points which are equidistant from the centre. The 10th chapter of RD Sharma gives students the concepts of secant of a circle and tangent to a circle along with their properties. Additionally, problems including circle, triangle and quadrilateral are tested in the exercise problems. The students referring to the solutions will get an idea about the steps involved in solving the problems. The solutions are curated by highly experienced faculty at STUDENT eCARE based on the latest syllabus of CBSE board.

- Circle: Locus of points equidistant from the centre.
- Secant of a circle: It will intersect the circle at exactly two points.
- Tangent to a circle: A straight line which touches the circle at a single point.

This chapter has 2 exercises such as.

Exercise 10.1

Exercise 10.2

Construction is one of the important concepts in geometry. Chapter 11 will help students gain knowledge about constructing a triangle similar to a given triangle and tangents to a circle. The steps involved in the construction of geometric figures are explained in an interactive way to help students understand the method easily. The solutions PDF can be downloaded for free by the students who wish to master the steps of the constructions. By regularly practicing the exercise wise problems, students will be able to solve complex problems which would arise in the board exams.

- Constructions: Drawing lines and shapes using various geometrical instruments.
- Angle: The amount of turn between two lines from their common point.
- Length: The size of a line segment is called the length.

This chapter has 3 exercises such as.

- Exercise 11.1
- Exercise 11.2
- Exercise 11.3

The important concepts based on trigonometric results are explained in an easy and understandable language in this chapter. The main topic on which the problems are created is regarding heights and distances. The height of certain objects and the distance from a certain point is discussed in brief under each exercise to make students familiar with the type of questions that would arise in the board exams. The students can also use the PDF to solve problems of RD Sharma textbook to score well in the board exams.

- Trigonometry: It is a branch of Mathematics which deals with the relationship between angles and side lengths of triangles.
- Height of a triangle: It is the length of a perpendicular line segment which originates on a side and intersects the opposite angle.

Chapter 12 has only one exercise such as.

- Exercise 12.1

There is always a degree of certainty or uncertainty about what has happened or going to happen. This leads to finding the probability of an occurrence of an event. The branch of Mathematics which deals with numerical descriptions of how likely an event are to occur or how likely a proposition is true is called probability. By using solutions PDF, students will get a quick overview of the concepts which are detailed under each exercise. It’s also a great help for students to practice problems on a daily basis. By solving the textbook problems, students will gain a strong hold on the fundamental concepts which would be continued in higher classes as well.

- Probability: The numerical description of how likely an event will occur.
- Event: It is an outcome or defined collection of outcomes of a random experiment.

This chapter has two exercises such as.

- Exercise 13.1
- Exercise 13.2

Chapter 14 of RD Sharma textbook comprises 5 exercises and the questions covers key concepts like understanding rectangular coordinates, the distance between two points, sections formulae and its applications, collinearity of points and area of triangle and quadrilateral. Methods used in solving problems along with shortcut tips are well highlighted to help students boost their exam preparations. The solutions PDF contains step wise explanations of each problem for a better understanding of concepts which are important for the exam.

- Coordinate Geometry: Study of geometry using coordinate systems.
- Distance between two points: The distance between two points in a geometric shape can be determined by using the distance formula which is an application of Pythagoras theorem.
- Section formula: It is used to determine the ratio in which a line segment is divided by a point externally or internally.

This chapter has 5 exercises such as.

- Exercise 14.1
- Exercise 14.2
- Exercise 14.3
- Exercise 14.4
- Exercise 14.5

Students can now access the chapter-wise solutions of RD Sharma textbook prepared according to the latest CBSE guidelines. This chapter deals with finding the area of circular regions and two special parts of the circular region known as sector and segment of a circle. The important formulas in finding the areas related to circles and steps included in them are explained in a comprehensive manner. The students can self-assess their knowledge about the entire chapter by solving problems using the solutions PDF.

- Area of circular regions: The area of a circle is pi times the radius squared.
- Sector: A part of the circle which is made of arc of the circle along with its two radii.
- Segment: It is the region bounded by the arc and the intercepted arc of the circle.

All four exercise problems of this chapter are solved by subject experts.

- Exercise 15.1
- Exercise 15.2
- Exercise 15.3
- Exercise 15.4

In this chapter, we shall discuss problems on the conversion of one of these solids into another as well as finding the surface areas and volumes of a combination of two or more such solids. The sum of the areas on a 3 dimensional figure is called the surface area. The space that a substance or shape occupies is called volume. The solutions are well-explained in an interactive manner to make the subject more understanding and easier for students. The exercise wise problems are solved by the experts at STUDENT eCARE after conducting vast research on each topic.

- Surface area: The region occupied by the surface of the object.
- Volume: The amount of space which is available in the object.

Students can hereby access the solutions of 3 exercises from the links provided.

- Exercise 16.1
- Exercise 16.2
- Exercise 16.3

Class 10 is a crucial stage for a school student – and the results of Class 10 exams may determine the academic and career path one may follow in the future. The detailed R.D. Sharma Class 10 solutions we present here will help you lay a strong foundation of the basic Mathematics concepts that can help you ace the board exam as well as other competitive exams you are appearing for.

- RD Sharma Solutions provide an algorithmic approach to solve each problem.
- These solutions give more than one way to solve, with detailed illustrations and brief summaries, which consist of concepts and formulae.
- These solutions are prepared by our STUDENT eCARE expert team, focusing completely on accuracy.
- Students get detailed study material on topics from Class 10 Maths.
- To make learning more interesting, each solution is provided with a pictorial representation to improve analyzing skills among students.

RD Sharma Solutions provide a detailed approach, which helps to reassure students and makes the exam preparation better for them. By using these reference books we can find a vast number of questions, enough to practice through the whole academic year. By exercising regularly, students will be thorough with concepts provided in the syllabus.

RD Sharma Solutions are primarily designed for CBSE students and are based on the latest syllabus prescribed as per the CCE guidelines by CBSE Board. Most of the questions asked in the exams are from these textbooks, which in turn, help to know the marking scheme and exam pattern.

RD Sharma reference books provide very short answer questions, short answer questions, multiple-choice questions and long answer questions, which in turn, help in time management and develop critical thinking among the students.

RD Sharma Solutions for Class 10 Maths can be downloaded and viewed from the STUDENT ECARE website in free PDF. At STUDENT ECARE, students can avail the solutions both online as well as download and save them for offline reference. The solutions provided in this website have been solved very accurately by STUDENT ECARE experts in Maths.

8 tips to ace your CBSE Class 10 Maths exam with full marks!

- Understand the contents and weightage of the syllabus.
- Preparation from NCERT textbooks is the most important part.
- Strategize and THEN prepare!
- Practice sample papers and previous years' papers.
- Say goodbye to rote learning!
- Self-evaluation.

**Solution:**

From the question it can be understood that the required number when divides 285 and 1249, leaves remainder 9 and 7 respectively should be

285 – 9 = 276 and 1249 -7 = 1242 can divide them exactly.

So, if the H.C.F. of 276 and 1242 is found then that will be the required number.

Now, by applying Euclid’s division lemma, we get

1242 = 276 x 4 + 138

276 = 138 x 2 + 0. (The remainder becomes 0 here)

So, the H.C.F. = 138

∴ the required number is **138**

**Solution:**

From the question it’s understood that,

2011 – 9 = 2002 and 2623 – 5 = 2618 has to be exactly divisible by the number.

Thus, the required number should be the H.C.F. of 2002 and 2618

Applying Euclid’s division lemma, we get

2618 = 2002 x 1 + 616

2002 = 616 x 3 + 154

616 = 154 x 4 + 0. (Here the remainder becomes 0)

And hence the H.C.F. (2002, 2618) = 154

∴ the required number is **154**

**Solution:**

From the question it’s understood that,

1251 – 1 = 1250, 9377 – 2 = 9375 and 15628 – 3 = 15625 has to be exactly divisible by the number.

Thus, the required number should be the H.C.F of 1250, 9375 and 15625.

First, consider 1250 and 9375 and apply Euclid’s division lemma

9375 = 1250 x 7 + 625

1250 = 625 x 2 + 0

∴ H.C.F (1250, 9375) = 625

Next, consider 625 and the third number 15625 to apply Euclid’s division lemma

15625 = 625 x 25 + 0

We get, the HCF of 625 and 15625 to be 625.

∴ H.C.F. (1250, 9375, 15625) = 625

So, the required number is 625.

**Solution:**

From the question, it’s given that

Number of chocolates of 1st brand in a pack = 24

Number of chocolates of 2nd brand in a pack = 15.

So, the least number of both brands of chocolates need to be purchased is given by their LCM.

L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5 = 120

Hence, the number of packets of 1st brand to be bought = 120 / 24 = **5**

And, the number of packets of 2nd brand to be bought = 120 / 15 = **8**

From the question it’s given that,

Size of bathroom = 10 ft. by 8 ft.

= (10 x 12) inch by (8 x 12) inch [conversion from ft. to inch.]

= 120 inch by 96 inch

Now, the largest size of tile required will be the HCF of 120 and 96.

So, Applying Euclid’s division lemma

We get,

120 = 96 x 1 + 24

96 = 24 x 4 + 0

⇒ HCF = 24

Thus, we can conclude that the largest size of tile which required is 24 inches

And,

Number of tiles required = (area of bathroom) / (area of a tile)

= (120 x 96) / (24×24)

= 5 x 4

= 20 tiles

Therefore, **20 tiles **each of size 24inch by 24inch are required to be cut.

From the question it’s given that,

Number of pastries = 15

Number of biscuit packets = 12

So, the required number of boxes which will contain equal number of both pastries and biscuits will be the HCF of 15 and 12.

So, applying Euclid’s division lemma, we have

15 = 12 x 1 + 3

12 = 3 x 4 = 0

So, the number of boxes required = 3

∴ Each box will contain 15/3 = **5 pastries** and 12/3 = **4 biscuit packs**.

From the question it’s given that,

Number of goats = 105

Number of donkeys = 140

Number of cows = 175

So, to find the largest number of animals in one trip the HCF (105, 140 and 175) has to be calculated.

Now, first consider the numbers 105 and 140 and apply division lemma

140 = 105 x 1 + 35

105 = 35 x 3 + 0

Thus, the HCF (105 and 140) = 35

Now, consider 35 and the third number 175 and apply Euclid’s division lemma

175 = 35 x 5 +0

Hence, the HCF (105, 140, 175) = 35.

Therefore, it can be said that 35 animals went on each trip.

Given,

Diameter of the copper wire = 1 cm

So, radius of the copper wire = 1/2 cm = 0.5 cm

Length of the copper rod = 8 cm

We know that,

Volume of the cylinder = π r2h

= π × 0.52 × 8 ……. (i)

Length of the wire = 18 m = 1800 cm

Volume of the wire = π r2h

= π r2 × 1800 ….. (ii)

On equating both the equations, we have

π × 0.52 × 8 = π r2 × 1800

r2 = 2 /1800 = 1/900

r = 1/30 cm

Therefore, the diameter of the wire is 1/15 cm i.e. 0.67 mm which is the thickness of the wire.

Given: A dice is thrown once

Required to find:

(i) Probability of getting a prime number

(ii) Probability of getting 2 or 4

(iii) Probability of getting a multiple of 2 or 3.

(iv) Probability of getting an even number

(v) Probability of getting a number greater than five.

(vi) Probability of lying between 2 and 6

Total number on a dice is 6 i.e., 1, 2, 3, 4, 5 and 6.

(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, probability of getting a prime number = 3/6 = 1/2

(ii) For getting 2 and 4, clearly the number of favourable outcomes is 2.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting 2 or 4 = 2/6 = 1/3

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6.

So, the number of favourable outcomes is 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an multiple of 2 or 3 = 4/6 = 2/3

(iv) An even prime number is 2 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even prime number = 1/6

(v) A number greater than 5 is 6 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number greater than 5 = 1/6

(vi) Total number on a dice is 6.

Numbers lying between 2 and 6 are 3, 4 and 5

So, the total number of numbers lying between 2 and 6 is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number lying between 2 and 6 = 3/6 =1/2

Given: Three coins are tossed simultaneously.

When three coins are tossed then the outcome will be anyone of these combinations.

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.

So, the total number of outcomes is 8.

(i) For exactly two heads, the favourable outcome are THH, HHT, HTH

So, the total number of favourable outcomes is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting exactly two heads is 3/8

(ii) For getting at least two heads the favourable outcomes are HHT, HTH, HHH, and THH

So, the total number of favourable outcomes is 4.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting at least two heads when three coins are tossed simultaneously = 4/8 = 1/2

(iii) For getting at least one head and one tail the cases are THT, TTH, THH, HTT, HHT, and HTH.

So, the total number of favourable outcomes i.e. at least one tail and one head is 6

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting at least one head and one tail = 6/8 = 3/4

(iv) For getting an outcome of no tail, the only possibility is HHH.

So, the total number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting no tails is 1/8.

**Solution:**

Given: Tickets are marked numbers from 1 to 50. And, one ticket is drawn at random.

Required to find: Probability of getting a prime number on the drawn ticket

Total number of tickets is 50.

Tickets which are number as prime number are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Total number of tickets marked as prime is 15.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a prime number on the ticket = 15/50 = 3/10

Let the cost of a bag and a pen be ? x and ? y, respectively.

Then, according to the question

3x + 4y = 257 … (i)

4x + 3y = 324 … (ii)

On multiplying equation (i) by 3 and (ii) by 4,

We get,

9x + 12y = 770 … (iii)

16x + 12y = 1296 … (iv)

Subtracting equation (iii) from (iv), we get

16x – 9x = 1296 – 771

7x = 525

x = 525/7 = 75

Hence, the cost of a bag = ? 75

Substituting x = 75 in equation (i),

We get,

3 x 75 + 4y = 257

225 + 4y = 257

4y = 257 – 225

4y = 32

y = 32/4 = 8

Hence, the cost of a pen = ? 8

From the question, it’s required to find the value of (x + 10y) ⇒ 75 +10(8) = 20

Therefore, the total cost of 1 bag and 10 pens = 75 + 80 = ? 155.

**Solution:**

Since, 327.7081 has a terminating decimal expansion its denominator should be of the form 2m x 5n, where m, n are non-negative integers.

Further,

327.7081 can be expressed as 3277081/10000 = p/q

⇒ q = 10000 = 23 x 53

Hence, the prime factors of q has only factors of 2 and 5.