case study based questions real numbers class 10

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CBSE Class 10 Maths Case Study Questions for Class 10 Maths Chapter 1 - Real Numbers (Published by CBSE)

Cbse class 10 maths cased study question bank for chapter 1 - real numbers is available here. this question bank is very useful to prepare for the class 10 maths exam 2021-2022..

The Central Board of Secondary Education has introduced the case study questions in class 10 exam pattern 2021-2022. The CBSE Class 10 questions papers of Board Exam 2022 will have questions based on case study. Therefore, students should get familiarised with these questions to do well in their board exam.

We have provided here case study questions for Class 10 Maths Chapter 1 - Real Numbers. These questions have been published by the CBSE board itself. Students must solve all these questions at the same time they finish with the chapter - Real numbers.

Case Study Questions for Class 10 Maths Chapter 1 - Real Numbers

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 32 students in section A and 36 students in section B.

case study based questions real numbers class 10

1. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Answer: c) 288

2. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is

Answer: b) 4

3. 36 can be expressed as a product of its primes as

a) 2 2 × 3 2

b) 2 1 × 3 3

c) 2 3 × 3 1

d) 2 0 × 3 0

Answer: a) 2 2 × 3 2

4. 7 × 11 × 13 × 15 + 15 is a

a) Prime number

b) Composite number

c) Neither prime nor composite

d) None of the above

Answer: b) Composite number

5. If p and q are positive integers such that p = ab 2 and q= a 2 b, where a , b are prime numbers, then the LCM (p, q) is

Answer: b) a 2 b 2

CASE STUDY 2:

A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

1. In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are

Answer: b) 12

2. What is the minimum number of rooms required during the event?

Answer: d) 21

3. The LCM of 60, 84 and 108 is

Answer: a) 3780

4. The product of HCF and LCM of 60,84 and 108 is

Answer: d) 45360

5. 108 can be expressed as a product of its primes as

a) 2 3 × 3 2

b) 2 3 × 3 3

c) 2 2 × 3 2

d) 2 2 × 3 3

Answer: d) 2 2 × 3 3

CASE STUDY 3:

A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

4. According to Fundamental Theorem of Arithmetic 13915 is a

a) Composite number

b) Prime number

d) Even number

Answer: a) Composite number

5. The prime factorisation of 13915 is

a) 5 × 11 3 × 13 2

b) 5 × 11 3 × 23 2

c) 5 × 11 2 × 23

d) 5 × 11 2 × 13 2

Answer: c) 5 × 11 2 × 23

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions at the end of his project as listed below. Answer them. (i) For what value of n, 4 n ends in 0?

(ii) If a is a positive rational number and n is a positive integer greater than 1, then for what value of n, a n is a rational number?

(iii) If x and yare two odd positive integers, then which of the following is true?

(iv) The statement 'One of every three consecutive positive integers is divisible by 3' is

(v) If n is any odd integer, then n2 - 1 is divisible by

Real numbers are extremely useful in everyday life. That is probably one of the main reasons we all learn how to count and add and subtract from a very young age. Real numbers help us to count and to measure out quantities of different items in various fields like retail, buying, catering, publishing etc. Every normal person uses real numbers in his daily life. After knowing the importance of real numbers, try and improve your knowledge about them by answering the following questions on real life based situations. (i) Three people go for a morning walk together from the same place. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance travelled when they meet at first time after starting the walk assuming that their walking speed is same?

(ii) In a school Independence Day parade, a group of 594 students need to march behind a band of 189 members. The two groups have to march in the same number of columns. What is the maximum number of columns in which they can march?

(iii) Two tankers contain 768litres and 420 litres of fuel respectively. Find the maximum capacity of the container which can measure the fuel of either tanker exactly.

(iv) The dimensions of a room are 8 m 25 cm, 6 m 75 crn and 4 m 50 cm. Find the length of the largest measuring rod which can measure the dimensions of room exactly.

(v) Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pens and notepads

In a classroom activity on real numbers, the students have to pick a number card from a pile and frame question on it if it is not a rational number for the rest of the class. The number cards picked up by first 5 students and their questions on the numbers for the rest of the class are as shown below. Answer them. (i) Suraj picked up $\sqrt{8}$ and his question was - Which of the following is true about $\sqrt{8}$ ?

(ii) Shreya picked up 'BONUS' and her question was - Which of the following is not irrational?

(iii) Ananya picked up $\sqrt{5}$ -. $\sqrt{10}$ and her question was - $\sqrt{5}$ -. $\sqrt{10}$ _________is number.

(iv) Suman picked up $\frac{1}{\sqrt{5}}$ and her question was - $\frac{1}{\sqrt{5}}$ is __________ number.

(v) Preethi picked up $\sqrt{6}$ and her question was - Which of the following is not irrational?

Decimal form of rational numbers can be classified into two types. (i) Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form $\frac{p}{\sqrt{q}}$ where p and q are co-prime and the prime faetorisation of q is of the form 2 n ·5 m , where n, mare non-negative integers and vice-versa. (ii) Let x = $\frac{p}{\sqrt{q}}$ be a rational number, such that the prime faetorisation of q is not of the form 2 n 5 m , where n and m are non-negative integers. Then x has a non-terminating repeating decimal expansion. (i) Which of the following rational numbers have a terminating decimal expansion?

(ii) 23/(2 3 x 5 2 ) =

(iii) 441/(2 2 x 5 7 x 7 2 ) is a_________decimal.

(iv) For which of the following value(s) of p, 251/(2 3 x p 2 ) is a non-terminating recurring decimal?

(v) 241/(2 5 x 5 3 ) is a _________decimal.

HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM. Based on the above information answer the following questions. (i) If two positive integers x and yare expressible in terms of primes as x = p2q3 and y = p3 q, then which of the following is true?

(ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p?

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively.

(iv) Find the least positive integer which on adding 1 is exactly divisible by 126 and 600.

(v) If A, Band C are three rational numbers such that 85C - 340A :::109, 425A + 85B = 146, then the sum of A, B and C is divisible by

*****************************************

Cbse 10th standard maths subject real number case study questions with solution 2021 answer keys.

(i) (d) : For a number to end in zero it must be divisible by 5, but 4 n = 22 n is never divisible by 5. So, 4 n never ends in zero for any value of n. (ii) (c) : We know that product of two rational numbers is also a rational number. So, a 2 = a x a = rational number a 3 = a 2 x a = rational number a 4 = a 3 x a = rational number ................................................ ............................................... a n = a n-1 x a = rational number. (iii) (d): Let x = 2m + 1 and y = 2k + 1 Then x 2 + y 2 = (2m + 1) 2 + (2k + 1) 2 = 4m 2 + 4m + 1 + 4k 2 + 4k + 1 = 4(m 2 + k 2 + m + k) + 2 So, it is even but not divisible by 4. (iv) (a): Let three consecutive positive integers be n, n + 1 and n + 2. We know that when a number is divided by 3, the remainder obtained is either 0 or 1 or 2. So, n = 3p or 3p + lor 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 Wi always divisible by 3. (v) (d): Any odd number is of the form of (2k +1), where k is any integer. So, n 2 - 1 = (2k + 1)2 -1 = 4k 2 + 4k For k = 1, 4k 2 + 4k = 8, which is divisible by 8. Similarly, for k = 2, 4k 2 + 4k = 24, which is divisible by 8. And for k = 3, 4k 2 + 4k = 48, which is also divisible by 8. So, 4k 2 + 4k is divisible by 8 for all integers k, i.e., n 2 - 1 is divisible by 8 for all odd values of n.

(i) (b): Here 80 = 2 4 x 5, 85 = 17 x 5 and 90 = 2 x 3 2 x 5 L.C.M of 80, 85 and 90 = 2 4 x 3 x 3 x 5 x 17 = 12240 Hence, the minimum distance each should walk when they at first time is 12240 cm. (ii) (c): Here 594 = 2 x 3 3 x 11 and 189 = 3 3 x 7 HCF of 594 and 189 = 3 3 = 27 Hence, the maximum number of columns in which they can march is 27. (iii) (c) : Here 768 = 2 8 x 3 and 420 = 2 2 x 3 x 5 x 7 HCF of 768 and 420 = 2 2 x 3 = 12 So, the container which can measure fuel of either tanker exactly must be of 12litres. (iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5 HCF = 5 x 5 x 3 = 75 Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm. (v) (a): LCM of 8 and 12 is 24. $\therefore $ The least number of pack of pens = 24/8 = 3 $\therefore $ The least number of pack of note pads = 24/12 = 2

(i) (b): Here $\sqrt{8}$ = 2 $\sqrt{2}$ = product of rational and irrational numbers = irrational number (ii) (c): Here, $\sqrt{9}$ = 3 So, 2 + 2 $\sqrt{9}$ = 2 + 6 = 8 , which is not irrational. (iii) (b): Here. $\sqrt{15}$ and $\sqrt{10}$ are both irrational and difference of two irrational numbers is also irrational. (iv) (c): As $\sqrt{5}$ is irrational, so its reciprocal is also irrational. (v) (d): We know that $\sqrt{6}$ is irrational. So, 15 + 3. $\sqrt{6}$ is irrational. Similarly, $\sqrt{24}$ - 9 = 2. $\sqrt{6}$ - 9 is irrational. And 5 $\sqrt{150}$ = 5 x 5. $\sqrt{6}$ = 25 $\sqrt{6}$ is irrational.

(i) (c): Here, the simplest form of given options are 125/441 = 5 3 /(3 2 x 7 2 ), 77/210 = 11/(2 x 3 x 5), 15/1600 = 3/(2 6 x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal. (ii) (b): 23/(2 3 x 5 2 ) = 23/200 = 0.115 (iii) (a): 441/(2 2 x 5 7 x 7 2 ) = 9/(2 2 x 5 7 ), which is a terminating decimal. (iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15. (v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.

(i) (b): LCM of x and y = p 3 q 3 and HCF of x and y = p 2 q Also, LCM = pq 2 x HCF. (ii) (d): Number of marbles = 5m + 2 or 6n + 2. Thus, number of marbles, p = (multiple of 5 x 6) + 2 = 30k + 2 = 2(15k + 1) = which is an even number but not prime (iii) (d): Here, required numbers = HCF (398 - 7, 436 - 11,542 -15) = HCF (391,425,527) = 17 (iv) (b): LCMof126and600 = 2 x 3 x 21 x 100= 12600 The least positive integer which on adding 1 is exactly divisible by 126 and 600 = 12600 - 1 = 12599 (v) (a): Here 8SC - 340A = 109 and 425A + 85B = 146 On adding them, we get 85A + 85B + 85C = 255 ~ A + B + C = 3, which is divisible by 3.

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Class 10 Maths Case Study Questions of Chapter 1 Real Numbers

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Case study Questions in the Class 10 Mathematics Chapter 1 are very important to solve for your exam. Class 10 Maths Chapter 1 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 1 Real Numbers

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

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Real Numbers Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 1 Real Numbers

Case Study/Passage-Based Questions

Case Study 1: Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions at the end of his project as listed below. (i) For what value of n, 4 n ends in 0?

Answer: (d) no value of n

(ii) If a is a positive rational number and n is a positive integer greater than 1, then for what value of n, a n is a rational number?

Answer: (c) for all n > 1

(iii) If x and yare two odd positive integers, then which of the following is true?

Answer: (d) both (a) and (b)

(iv) The statement ‘One of every three consecutive positive integers is divisible by 3’ is

Answer: (a) always true

(v) If n is any odd integer, then n2 – 1 is divisible by

Answer: (d) 8

Case Study 2: HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM Based on the above information answer the following questions.

(i) If two positive integers x and y are expressible in terms of primes as x =p 2 q 3 and y=p 3 q, then which of the following is true? (a) HCF = pq 2 x LCM (b) LCM = pq 2 x HCF (c) LCM = p 2 q x HCF (d) HCF = p 2 q x LCM

Answer: (b) LCM = pq2 x HCF

ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p? (a) p is odd (b) p is even (c) p is not prime (d) both (b) and (c)

Answer: (d) both (b) and (c)

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively. (a) 3 (b) 1 (c) 34 (d) 17

Answer: (d) 17

(iv) Find the least positive integer that on adding 1 is exactly divisible by 126 and 600. (a) 12600 (b) 12599 (C) 12601 (d) 12500

Answer: (b) 12599

(v) If A, B and C are three rational numbers such that 85C – 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by (a) 3 (b) 6 (c) 7 (d) 9

Answer: (a) 3

Case Study 3: Real numbers are an essential concept in mathematics that encompasses both rational and irrational numbers. Rational numbers are those that can be expressed as fractions, where the numerator and denominator are integers and the denominator is not zero. Examples of rational numbers include integers, decimals, and fractions. On the other hand, irrational numbers are those that cannot be expressed as fractions and have non-terminating and non-repeating decimal expansions. Examples of irrational numbers include √2, π (pi), and e. Real numbers are represented on the number line, which extends infinitely in both positive and negative directions. The set of real numbers is closed under addition, subtraction, multiplication, and division, making it a fundamental number system used in various mathematical operations and calculations.

Which numbers can be classified as rational numbers? a) Fractions b) Integers c) Decimals d) All of the above Answer: d) All of the above

What are rational numbers? a) Numbers that can be expressed as fractions b) Numbers that have non-terminating decimal expansions c) Numbers that extend infinitely in both positive and negative directions d) Numbers that cannot be expressed as fractions Answer: a) Numbers that can be expressed as fractions

What are examples of irrational numbers? a) √2, π (pi), e b) Integers, decimals, fractions c) Numbers with terminating decimal expansions d) Numbers that can be expressed as fractions Answer: a) √2, π (pi), e

How are real numbers represented? a) On the number line b) In complex mathematical formulas c) In algebraic equations d) In geometric figures Answer: a) On the number line

What operations are closed under the set of real numbers? a) Addition, subtraction, multiplication b) Subtraction, multiplication, division c) Addition, multiplication, division d) Addition, subtraction, multiplication, division Answer: d) Addition, subtraction, multiplication, division

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 1 Real Numbers with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Real Numbers Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Questions for Class 10 Maths Chapter 1 Real Numbers

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Question 1:

(ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p? (a) p is odd (b) p is even (c) p is not prime (d) both (b) and (c)

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively. (a) 3 (b) 1 (c) 34 (d) 17

(iv) Find the least positive integer which on adding 1 is exactly divisible by 126 and 600. (a) 12600 (b) 12599 (C) 12601 (d) 12500

(v) If A, B and C are three rational numbers such that 85C – 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by (a) 3 (b) 6 (c) 7 (d) 9

Question 2:

(i) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B? (a) 144 (b) 128 (c) 288 (d) 272

(ii) If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is (a) 2 (b) 4 (c) 6 (d) 8

(iii) 36 can be expressed as a product of its primes as (a) (b) (c) (d)

(iv) 7 is a (a) Prime number (b) Composite number (c) Neither prime nor composite (d) None of the above

(v) If p and q are positive integers such that p = a and q= b, where a , b are prime numbers, then the LCM (p, q) is (a) ab (b) a 2 b 2 (c) a 3 b 2 (d) a 3 b 3

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CBSE Case Study Questions for Class 10 Maths Real Numbers Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Real Numbers in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Real Numbers PDF

Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.

Chapter 2: Polynomials Case Study Questions
Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
Chapter 4: Quadratic Equation Case Study Questions
Chapter 5: Arithmetic Progressions Case Study Questions

How should I study for my upcoming exams?

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Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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Case Study Class 10 Maths Questions

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

Real Numbers Case Study Question
Polynomials Case Study Question
Pair of Linear Equations in Two Variables Case Study Question
Quadratic Equations Case Study Question
Arithmetic Progressions Case Study Question
Triangles Case Study Question
Coordinate Geometry Case Study Question
Introduction to Trigonometry Case Study Question
Some Applications of Trigonometry Case Study Question
Circles Case Study Question
Area Related to Circles Case Study Question
Surface Areas and Volumes Case Study Question
Statistics Case Study Question
Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

Find the production in the 1 st year.
Find the production in the 12 th year.
Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

Find the distance between Lucknow (L) to Bhuj(B).
If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

Find the distance PA.
Find the distance PB
Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

What is the length of the line segment joining points B and F?
The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

If the first circular row has 30 seats, how many seats will be there in the 10th row?
For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

$case study based questions real numbers class 10$

Draw a neat labelled figure to show the above situation diagrammatically.

$case study based questions real numbers class 10$

What is the speed of the plane in km/hr.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Class 10 Maths Chapter 1 MCQ

Class 10 Maths Chapter 1 Real Numbers MCQ (Multiple Choice Objective Questions) with answers and complete explanation case study type questions for the first term examination 2024-25. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily. This page of Class 10 Maths MCQ contains the questions released by CBSE as well as extra questions for practice.

Case Study – 1

What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Factors of 32 = 2 х 2 х 2 х 2 х 2 = 2⁵ Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² LCM of 32 and 36 = 2⁵ х 3² = 32 х 9 = 288 Hence, the correct option is (C).

View Answer

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32, 36) is

Factors of 32 = 2 х 2 х 2 х 2 х 2 = 2⁵ Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² LCM of 32 and 36 = 2⁵ х 3² = 32 х 9 = 288 HCF (32, 36) = (32 х 36) / LCM = (32 х 36) / 288 = 4 Hence, the correct option is (B).

36 can be expressed as a product of its primes as

Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² Hence, the correct option is (A).

7 х 11 х 13 х 15 + 15 is a

7 х 11 х 13 х 15 + 15 = 15 х (7 х 11 х 13 + 1) = 15 х (Integer) It has more than two factor. So, it is a composite number. Hence, the correct option is (B).

If p and q are positive integers such that p = ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is

p = ab² q = a²b LCM = highest powers of common factors of ab² and a²b = a²b² Hence, the correct option is (B).

Case Study – 2

A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English, and Mathematics are 60, 84, and 108 respectively.

Class 10 Maths Chapter 1 Real Numbers MCQ

In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 Hence, the correct option is (B).

What is the minimum number of rooms required during the event?

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 Number of room required for Hindi participants = 60/12 = 5 Number of room required for English participants = 84/12 = 7 Number of room required for Mathematics participants = 108/12 = 9 Total number of room required = 5 + 7 + 9 = 21 Hence, the correct option is (D).

The LCM of 60, 84, and 108 is

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ LCM of 60, 84, and 108 = 2² х 3³ х 5 х 7 = 4 х 27 х 5 х 7 = 3780 Hence, the correct option is (A).

The product of HCF and LCM of 60, 84, and 108 is

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 LCM of 60, 84, and 108 = 2² х 3³ х 5 х 7 = 4 х 27 х 5 х 7 = 3780 Product of HCF and LCM of 60, 84, and 108 = 12 х 3780 = 45360 Hence, the correct option is (D).

108 can be expressed as a product of its primes as

Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ Hence, the correct option is (D).

Case Study – 3

Rohit Singh is a worker in a petrol pump. He along with the other co-workers, use to transfer petrol from tanker to storage. On Monday, there were two tankers containing 850 litres and 680 litres of petrol respectively.

Case study based MCQ for 10th Maths chapter 1

What is the maximum capacity of a container which can measure the petrol of either tanker in exact number of time?

The maximum capacity of the container is the HCF of 850 and 680. Factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 HCF of 850 and 680 = 2 х 5 х 17 = 170 Hence, the correct option is (C).

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the LCM (850, 680) is

Factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 HCF of 850 and 680 = 2 х 5 х 17 = 170 LCM (850, 680) = (850 х 680) / HCF = (850 х 680) / 170 = 3400 Hence, the correct option is (D).

680 can be expressed as a product of its primes as

Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 Hence, the correct option is (C).

2 х 3 х 5 х 11 х 17 + 11 is a

2 х 3 х 5 х 11 х 17 + 11 = 11 х (2 х 3 х 5 х 17 + 1) = 11 х (Integer) It has more than two factor. So, it is a composite number. Hence, the correct option is (B).

If p and q are positive integers such that p = a³b² and q = a²b³, where a, b are prime numbers, then the LCM (p, q) is

p = a³b² q = a²b³ LCM = highest powers of common factors of a³b² and a²b³ = a³b³ Hence, the correct option is (B).

Case Study – 4

Class 10 Maths Case study MCQ Factor tree

What will be the value of x?

X = 5 х 2783 = 13915 Hence, the correct option is (B).

What will be the value of y?

Y = 2783/253 = 11 Hence, the correct option is (C).

What will be the value of z?

Z = 253/11 = 23 Hence, the correct option is (B).

According to Fundamental Theorem of Arithmetic 13915 is a

Because 13915 can be written into the product of primes. 13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23 Hence, the correct option is (A).

The prime factorisation of 13915 is

13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23 Hence, the correct option is (C).

Case Study – 5

We all know that morning walk is good for health. In a morning walk, three friends Anjali, Sofia, and Angelina step of together. There steps measure 80 cm, 85 cm, and 90 cm. respectively.

What is the minimum distance each should walk so that they can cover the distance in complete steps?

The minimum distance covered by each in complete steps must be the LCM of 80 cm, 85 cm, and 90 cm. Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 Now, 12240 cm = 122 m 40 cm Hence, the correct option is (B).

What is the minimum number of steps taken by any of the three friends, when they meet again?

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 The step size of Angelina is maximum among these three. So, she will take minimum number of steps to cover the same distance. Number of steps = 12240/90 = 136 Hence, the correct option is (D).

The HCF of 80, 85, and 90 is

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 HCF of 80, 85, and 90 = 5 Hence, the correct option is (A).

The product of HCF and LCM of 80, 85, and 90 is

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 HCF of 80, 85, and 90 = 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 Therefore, the product of HCF and LCM of 80, 85, and 90 = 12240 х 5 = 61200 Hence, the correct option is (C).

90 can be expressed as a product of its primes as

Factors of 108 = 2 х 3 х 3 х 5 = 2 х 3² х 5 Hence, the correct option is (D).

Class 10 Maths Chapter 1 MCQ are given below. There are total of 5 questions with four choices. Only one option is correct and the explanation of correct answer is given below the questions. Every time the students will get a new set of five questions with different levels of questions. For any further discussion, please join the Discussion Forum.

The decimal form of 120 /(2^2 × 5^7 × 7^5 )is

Because the denominator contains the factors of 7.

HCF of 8, 9, 25 is

We can write 8, 9, 25 as 8 × 1 + 0, 8 × 1 + 1, 8 × 3 + 1. Because there is no common divisor for all three numbers where remainder will be zero Hence HCF will be 1.

The decimal expansion of the rational number 14587/250 will terminate after

The factors of denominator 250 = 2 × 5 × 5 × 5. There are three factors of 5, that is why it will terminate after three decimal places. [Always more number of factors will be considered, whether it is 5 or 2, does not matter.] By dividing actually 14587 by 250 we get the quotient as 58.348.

If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is

We know that: HCF × LCM = 16 × y So, 8 × 48 = 16 × y y = 8 × 48/16 = 24

The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates after one place of decimal, is

1/3 × 3/10 = 1/10 = 0.1

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What are the important topics in Class 10 Maths Chapter 1 MCQ?

Euclid’ division lemma and the Fundamental Theorem of Arithmetic are the two main topics in 10th Maths chapter 1 Real Numbers. Now questions are designed on the basis of case study. So practice MCQ questions based on daily life events which will be more helpful in CBSE exams.

In which of the four exercise of 10th Maths Chapter 1, are Case Study MCQ asked?

There are questions from each exercise of Chapter 1 of 10th Maths, but most of the MCQs can be formed from Exercise 1.4. Now CBSE introduces the questions based on CASE STUDY which may be asked from any section of class 10 Maths chapter 1.

How many MCQ are required to be perfect in Chapter 1 of Class 10 Maths?

If your concepts are clear, the MCQs provide more confidence in that section. More practice means more to retain and better understanding with the concepts of topics.

How many questions from Chapter 1 of Class 10 Maths asked in CBSE Board?

There is no classification of number of questions from the different chapters. There may be one, more than one or none from Chapter 1 Real Numbers of Class 10 Maths.

We are adding more questions frequently, so that students can have a good practice of Class 10 Maths Chapters. If you have suggestion or feedback about this page or website improvement, you are welcome. Important questions with solutions and answers will be added very soon for each chapter of class 10 Maths.

Download NCERT Books and Offline Apps 2024-25 based on new CBSE Syllabus. Ask your doubts related to NIOS or CBSE Board and share your knowledge with your friends and other users through Discussion Forum.

Class 10 Maths Chapter 1 Case Based Questions - Real Numbers

Case study – 1.

Q1: What is the minimum distance each should walk so that they can cover the distance in complete steps? (a) 120 m 40 cm (b) 122 m 40 cm (c) 12 m 4 cm (d) None of these Ans: (b) Explanation: The process of solving this problem involves finding the least common multiple (LCM) of the three given measurements. The LCM of a set of numbers is the smallest number that is a multiple of each of the numbers in the set. Here are the steps to find the LCM of 80 cm, 85 cm, and 90 cm: Step 1: Prime factorization The first step is to find the prime factors of each number.

For 80, the prime factors are 2, 2, 2, 2, and 5 (or 2⁴ х 5)
For 85, the prime factors are 5 and 17
For 90, the prime factors are 2, 3, 3, and 5 (or 2 х 3² х 5)

Step 2: Find the LCM Now, we find the LCM by taking the highest power of each prime factor from all the numbers.

The highest power of 2 is 2⁴ from 80
The highest power of 3 is 3² from 90
The highest power of 5 is 5 from 80 or 90
The highest power of 17 is 17 from 85

So, the LCM is 2⁴ х 3² х 5 х 17 = 12240 Step 3: Convert cm to m Finally, we convert the LCM from centimeters to meters. Since 1 meter is 100 cm, we divide 12240 by 100 to get 122.4 meters, which can also be written as 122 meters and 40 cm. Therefore, the minimum distance each should walk so that they can cover the distance in complete steps is 122 meters and 40 cm. This corresponds to option (b). Q2: What is the minimum number of steps taken by any of the three friends, when they meet again? (a) 120 (b) 125 (c) 130 (d) 136 Ans: (d) Explanation: To solve this problem, we need to find the least common multiple (LCM) of the step sizes of the three friends: 80 cm, 85 cm, and 90 cm. The LCM will give us the smallest distance that all three friends can walk together, taking whole steps. Here is the step-by-step process: Step 1: Prime Factorization We start by breaking down each number into its prime factors.

80 = 2 x 2 x 2 x 2 x 5 = 2⁴ x 5
85 = 5 x 17
90 = 2 x 3 x 3 x 5 = 2 x 3² x 5

Step 2: Find the LCM The LCM is found by taking the highest power of all the prime numbers that appear in the prime factorization of any of the numbers. LCM = 2⁴ x 3² x 5 x 17 = 12240 cm This means that the three friends will meet again after walking a distance of 12240 cm. Step 3: Determine the Minimum Steps Among the three friends, Angelina has the longest step size (90 cm). Therefore, she will take the smallest number of steps to cover the distance of 12240 cm. Number of steps taken by Angelina = Total distance / Step size = 12240 cm / 90 cm = 136 steps Hence, the correct answer is (d) 136 steps. Q3: The HCF of 80, 85, and 90 is (a) 5 (b) 10 (c) 12 (d) 18 Ans: (a) Explanation: The Highest Common Factor (HCF) of a set of numbers is the largest number that divides evenly into all the numbers in the set. In this case, we are looking for the HCF of 80, 85, and 90. The first step is to determine the prime factors of each of the numbers. Prime factors are the factors of a number that are prime numbers. 1. For 80, the prime factors are 2 and 5. We obtain this by dividing 80 by the smallest prime number (2) as many times as possible until we are left with a prime number. This gives us 2 x 2 x 2 x 2 x 5 = 2⁴ x 5. 2. For 85, the prime factors are 5 and 17. We obtain this by dividing 85 by the smallest prime number (2) as many times as possible until we are left with a prime number. This gives us 5 x 17. 3. For 90, the prime factors are 2, 3, and 5. We obtain this by dividing 90 by the smallest prime number (2) as many times as possible, then doing the same with the next smallest prime number (3) until we are left with a prime number. This gives us 2 x 3 x 3 x 5 = 2 x 3² x 5. Now that we have the prime factors of each number, we can determine the HCF by finding the largest number that is a factor of all three numbers. In this case, the only common factor among 80, 85, and 90 is 5. Therefore, the HCF of 80, 85, and 90 is 5, which corresponds to answer choice (a). Q4: The product of HCF and LCM of 80, 85, and 90 is (a) 60400 (b) 61000 (c) 61200 (d) 65500 Ans: (c) Explanation: The problem requires us to find the product of the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the numbers 80, 85, and 90. Step 1: To find the HCF and LCM, we first need to find the prime factors of the three numbers. For 80, the prime factors are 2 x 2 x 2 x 2 x 5 (or 2 4 x 5). For 85, the prime factors are 5 x 17. For 90, the prime factors are 2 x 3 x 3 x 5 (or 2 x 3 2 x 5). Step 2: To find the HCF, we look for common prime factors. The only common factor among all three numbers is 5. So, HCF = 5. Step 3: For the LCM, we take the highest power of all the prime numbers in the factorization of each number. So, LCM = 2 2 x 3 2 x 5 x 17 = 12240. Step 4: Finally, we need to find the product of the HCF and LCM. This is done by multiplying the HCF (5) with the LCM (12240), which gives us 61200. So, the product of the HCF and LCM of 80, 85, and 90 is 61200. Therefore, the correct answer is option (C). Q5: 90 can be expressed as a product of its primes as (a) 2 х 3² х 5² (b) 2 х 3³ х 5 (c) 2² х 3² х 5 (d) 2 х 3² х 5 Ans: (d) Explanation: The question asks us to express 90 as a product of its prime factors. Prime factors are the factors of a number that are prime numbers. A prime number is a number that only has two factors: 1 and itself. Here are the steps to find the prime factors of 90: Step 1: Start by dividing the number 90 with the smallest prime number, which is 2. 90 is divisible by 2. So, divide 90 by 2. You get 45. Step 2: Now, try dividing 45 by 2. It can't be divided evenly. So, we move to the next prime number, which is 3. 45 divided by 3 gives 15. Step 3: Try dividing 15 by 3. It can't be divided evenly. So, we move to the next prime number, which is 5. 15 divided by 5 gives 3. Step 4: Now, we are left with 3. 3 is a prime number itself, so we stop here. So, the prime factors of 90 are 2, 3, 3, and 5. We can write this as 2 x 3² x 5, which matches option (d). Therefore, the correct answer is (d).

Case Study – 2

Q1: What will be the value of x? (a) 15005 (b) 13915 (c) 56920 (d) 17429 Ans: (b) Explanation: The factor tree is a method used to break down any given number into its prime factors. In this case, we don't have the factor tree visually, but the question suggests that 'x' can be obtained by multiplying the numbers 5 and 2783. Step-by-step process: Step 1: Identify the numbers given. Here, we have 5 and 2783. Step 2: Multiply the given numbers. In this case, x = 5 * 2783 Step 3: Perform the multiplication. 5 * 2783 = 13915 So, by using these steps, we find that the value of 'x' is 13915. Therefore, the correct option is (b) 13915. Q2: What will be the value of y? (a) 23 (b) 22 (c) 11 (d) 19 Ans: (c) Explanation: The given factor tree shows how a number is broken down into its prime factors. The number at the top of the tree is the original number and the numbers at the bottom are all prime factors. In the question, we are not given the specific factor tree, but we are asked to find the value of 'y' given that Y = 2783/253. To solve this, we need to perform the division operation: 2783 divided by 253 equals to 11. Hence, the correct answer is option (c), i.e., y = 11. Q3: What will be the value of z? (a) 22 (b) 23 (c) 17 (d) 19 Ans: (b) Explanation: The given factor tree is not explicitly provided here, but from the available solution, we can assume that the number 253 is divided by 11 on the factor tree to obtain the value of z. The process for solving the problem is as follows: Step 1: Identify the numbers given in the factor tree. Here, it's 253 divided by 11 to get 'z'. Step 2: Divide the larger number (253) by the smaller number (11). 253 ÷ 11 = 23 So, z = 23. Therefore, the correct answer is (b) 23. In conclusion, a factor tree is a tool that breaks down any number into its prime factors. In this case, it helped to find the value of z by dividing 253 by 11. Q4: According to Fundamental Theorem of Arithmetic 13915 is a (a) Composite number (b) Prime number (c) Neither prime nor composite (d) Even number Ans: (a) Explanation: The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number, or can be represented as a unique product of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For example, the first six prime numbers are 2, 3, 5, 7, 11, and 13. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, a composite number is any positive integer greater than one that is not a prime number. Now, let's consider the number 13915. We are given that 13915 can be written as the product of primes: 13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23. Here, we can see that 13915 has more divisors than just 1 and itself (which are 5, 11 and 23). This means that 13915 is not a prime number. Also, as 13915 can be expressed as a product of prime numbers, it is not a number that falls into the category of 'neither prime nor composite'. As for being an even number, we know that an even number is any integer that can be divided by 2. In the case of 13915, it is not divisible by 2, so it is not an even number. Therefore, by process of elimination and based on the definitions, we can conclude that 13915 is a composite number (option a). Q5: The prime factorisation of 13915 is (a) 5 х 11³ х 13² (b) 5 х 11³ х 23² (c) 5 х 11² х 23 (d) 5 х 11² х 13² Ans: (c) Explanation: The prime factorisation of a number is the representation of that number as the product of its prime factors. Here's how you would calculate the prime factorisation of 13915 step-by-step:

First, find the smallest prime number that divides 13915. This will be 5, because 13915 is not divisible by 2 (it's not an even number), nor by 3 (the sum of its digits is not divisible by 3). So, you can start with 5.
Divide 13915 by 5, which gives you 2783.
Now, repeat the process with 2783. The smallest prime number that divides 2783 is 11. Divide 2783 by 11 to get 253.
Repeat the process with 253. It's not divisible by 2, 3, 5, or 7, but it is divisible by 11. Dividing by 11 gives you 23.
23 is a prime number itself, so that's the end of the process.

Therefore, the prime factorisation of 13915 is 5 x 11 x 11 x 23, or 5 x 11² x 23, which matches option (c).

Case Study – 3

Q1: What is the maximum capacity of a container which can measure the petrol of either tanker in exact number of time? (a) 150 litres (b) 160 litres (c) 170 litres (d) 180 litres Ans: (c) Explanation: The question is asking for the highest common factor (HCF) of 850 and 680. The HCF is the largest number that can evenly divide both numbers. Step 1: Find the prime factors of both numbers. Prime factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Prime factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 Step 2: Identify the common prime factors. The common prime factors of 850 and 680 are 2, 5, and 17. Step 3: Multiply the common prime factors to get the HCF. HCF of 850 and 680 = 2 х 5 х 17 = 170 Therefore, the maximum capacity of a container that can measure the petrol of either tanker an exact number of times is 170 litres, which corresponds to option (c). Q2: If the product of two positive integers is equal to the product of their HCF and LCM is true then, the LCM (850, 680) is (a) 3100 (b) 3200 (c) 3300 (d) 3400 Ans: (d) Explanation: The question is asking for the least common multiple (LCM) of 850 and 680. The LCM is the smallest number that is a multiple of both numbers. Step 1: We already have the prime factors of both numbers from the previous question, and the HCF. Prime factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Prime factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 HCF of 850 and 680 = 2 х 5 х 17 = 170 Step 2: Use the formula for finding the LCM when the HCF is known. LCM (850, 680) = (850 х 680) / HCF Step 3: Substitute the values into the formula. LCM (850, 680) = (850 х 680) / 170 = 3400 Therefore, the LCM of 850 and 680 is 3400, which corresponds to option (d). Q3: 680 can be expressed as a product of its primes as (a) 2² х 5 х 17 (b) 2¹ х 5 х 17 (c) 2³ х 5 х 17 (d) 2³ х 5 х 17⁰ Ans: (c) Explanation: To solve this problem, you need to understand what prime factorization is. Prime factorization is the process of breaking down a number into its smallest prime factors. Let's try to factorize the number 680. First, we need to find a prime number that can divide 680. The smallest prime number is 2, and it can divide 680, so we use it as our first factor. 680 ÷ 2 = 340 Now we continue the process with 340. Again, it can be divided by 2, so we use 2 as our next factor. 340 ÷ 2 = 170 We repeat the process with 170. It can be divided by 2, so we use 2 as our next factor. 170 ÷ 2 = 85 Now, 85 cannot be divided by 2, so we move to the next prime number, which is 3. However, 85 cannot be divided by 3 either. We continue this process until we find a prime number that can divide 85, which is 5. 85 ÷ 5 = 17 Finally, we have 17, which is a prime number itself, so our factorization process stops here. Therefore, the prime factorization of 680 is 2 × 2 × 2 × 5 × 17, or in the exponential form, it is 2³ × 5 × 17. Hence, option (c) is correct. Q4: 2 х 3 х 5 х 11 х 17 + 11 is a (a) Prime number (b) Composite number (d) Neither prime nor composite (d) None of the above Ans: (b) Explanation: The provided answer appears to be incorrect. The number 11 is indeed a prime number, not a composite number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In other words, if a number is prime, it can only be divided without a remainder by 1 and itself. The number 11 meets this criteria, as it can only be divided evenly by 1 and 11. On the other hand, a composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, it has more than two distinct divisors. Here, the number 11 does not have more than two distinct divisors. Thus, the number 11 is a prime number. Therefore, the correct answer is (a) Prime number. Q5: If p and q are positive integers such that p = a³b² and q = a²b³, where a, b are prime numbers, then the LCM (p, q) is (a) ab (b) a³b³ (c) a³b⁵ (d) a⁵b³ Ans: (b) Explanation: To find the least common multiple (LCM) of two numbers, we need to consider the highest powers of all the factors in the numbers. In this case, we are given that p = a³b² and q = a²b³, where a and b are prime numbers. The factors of p are a and b, with a having a power of 3 and b having a power of 2. The factors of q are also a and b, but here a has a power of 2 and b has a power of 3. When finding the LCM, we need to take the highest powers of these common factors. So, we take a to the power of 3 (since 3 is higher than 2) and b to the power of 3 (since 3 is higher than 2). Hence, the LCM of p and q is a³b³. Therefore, the correct option is (b) a³b³.

Case Study – 4

Q1: In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are (a) 14 (b) 12 (c) 16 (d) 18 Ans: (b) Explanation: In order to find the maximum number of participants that can be accommodated in each room, we need to find the Highest Common Factor (HCF) of the number of participants in each subject. The HCF of a set of numbers is the largest number that divides each of them without leaving a remainder. It can be found by listing all the factors of each number and finding the largest one that they have in common. Here are the factors of each number:

Factors of 60: 2 x 2 x 3 x 5 = 2² x 3 x 5
Factors of 84: 2 x 2 x 3 x 7 = 2² x 3 x 7
Factors of 108: 2 x 2 x 3 x 3 x 3 = 2² x 3³

The HCF of 60, 84, and 108 is 2² x 3 = 12. Therefore, the maximum number of participants that can be accommodated in each room is 12, which corresponds to the option (b). Q2: What is the minimum number of rooms required during the event? (a) 11 (b) 31 (c) 41 (d) 21 Ans: (d) Explanation: The question requires us to calculate the minimum number of rooms required for the seminar. This can be done by finding the highest common factor (HCF) of the number of participants in each subject. The HCF tells us the maximum number of participants that can be accommodated in each room such that all rooms have the same number of participants. Let's start by finding the prime factorization of the numbers. For 60, the prime factors are 2, 2, 3, and 5 (2² х 3 х 5). For 84, the prime factors are 2, 2, 3, and 7 (2² х 3 х 7). For 108, the prime factors are 2, 2, 3, 3, and 3 (2² х 3³). Now, the HCF is found by multiplying the lowest power of the common prime factors. In this case, the common prime factors are 2 and 3. The lowest power of 2 is 2 (as in 2²), and the lowest power of 3 is 1 (as in 3). So, the HCF is 2² х 3 = 12. Now, to find the number of rooms required for each subject, we divide the number of participants by the HCF. For Hindi, we need 60/12 = 5 rooms. For English, we need 84/12 = 7 rooms. For Mathematics, we need 108/12 = 9 rooms. Adding these together, the total number of rooms required is 5 + 7 + 9 = 21 rooms. Therefore, the answer is (d) 21. Q3: The LCM of 60, 84, and 108 is (a) 3780 (b) 3680 (c) 4780 (d) 4680 Ans: (a) Explanation: The problem revolves around finding the Least Common Multiple (LCM) of three numbers: 60, 84, and 108. To find the LCM of these numbers, we first need to find their prime factors. Here's how:

The prime factors of 60 are 2, 2, 3, and 5 (since 2*2*3*5 = 60). We can write it as 2² * 3 * 5.
The prime factors of 84 are 2, 2, 3, and 7 (since 2*2*3*7 = 84). We can write it as 2² * 3 * 7.
The prime factors of 108 are 2, 2, 3, 3, and 3 (since 2*2*3*3*3 = 108). We can write it as 2² * 3³.

Now, to find the LCM, we take the highest power of all the prime factors obtained from these numbers. If a prime factor is not present in one number but is present in another, we take the factor from the number where it is present.

We have the factor 2 in all three numbers, and the highest power is 2². So, we take 2².
We have the factor 3 in all three numbers, and the highest power is 3³. So, we take 3³.
We have the factor 5 only in 60. So, we take 5.
We have the factor 7 only in 84. So, we take 7.

Q4: The product of HCF and LCM of 60, 84, and 108 is (a) 55360 (b) 35360 (c) 45500 (d) 45360 Ans: (d) Explanation: The first step to solving this problem is understanding what HCF (Highest Common Factor) and LCM (Least Common Multiple) are. The HCF is the highest number that can divide two or more numbers without leaving a remainder. The LCM is the smallest number that is a multiple of two or more numbers. To find the HCF and LCM, we first need to find the prime factors of each number. For 60, the prime factors are 2, 2, 3, and 5 (or 2², 3, 5). For 84, the prime factors are 2, 2, 3, and 7 (or 2², 3, 7). For 108, the prime factors are 2, 2, 3, 3, and 3 (or 2², 3³). The HCF of these three numbers is found by taking the highest common factor of all three numbers, which is 2² (or 4) and 3. Multiplying these together gives us an HCF of 12. The LCM is found by taking the highest power of all the prime factors present in the numbers. This gives us 2², 3³, 5, and 7. Multiplying these together gives us an LCM of 3780. Finally, to find the product of the HCF and LCM, we multiply 12 and 3780 together, which gives us 45360. Hence, the correct answer is (d) 45360. Q5: 108 can be expressed as a product of its primes as (a) 2³ х 3² (b) 2³ х 3³ (c) 2² х 3² (d) 2² х 3³ Ans: (d) Explanation: The process of finding the answer is called prime factorization. Step 1: Start with the smallest prime number, which is 2. Check if 108 is divisible by 2. If it is, then write down 2 as a factor and divide 108 by 2. Step 2: You get 54 as the quotient. Now, repeat the process with 54. Is it divisible by 2? Yes, it is. So, write down 2 as a factor again and divide 54 by 2. Step 3: You now have a quotient of 27. Repeat the process. Is 27 divisible by 2? No, it's not. So, move on to the next prime number, which is 3. Step 4: Is 27 divisible by 3? Yes, it is. So, write down 3 as a factor and divide 27 by 3. Step 5: You get a quotient of 9. Repeat the process. Is 9 divisible by 3? Yes, it is. So, write down 3 as a factor again and divide 9 by 3. Step 6: You now have a quotient of 3. Repeat the process. Is 3 divisible by 3? Yes, it is. So, write down 3 as a factor again and divide 3 by 3. Step 7: You now have a quotient of 1. When you reach 1, you can stop the process. Step 8: Now, count the number of times each prime number appears in your list of factors. You have two 2s and three 3s. Step 9: Write down your answer as the product of the prime numbers, each raised to the power of its count. So, 108 = 2² х 3³. That's how you get the answer (d) 2² х 3³.

Case Study – 5

Q1: What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B? (a)144 (b) 128 (c) 288 (d) 272 Ans: (c) Explanation: The question requires finding the minimum number of books that can be equally distributed among the students of either section A or section B. This is essentially finding the least common multiple (LCM) of the number of students in both sections. Step 1: Find the prime factors of both 32 and 36. Prime factors of 32 are 2 x 2 x 2 x 2 x 2 = 2^5 Prime factors of 36 are 2 x 2 x 3 x 3 = 2^2 x 3^2 Step 2: Find the LCM of 32 and 36. For finding the LCM, we take the highest power of each prime factor that appears in the factorization of either 32 or 36. Here, for 2, the highest power is 5 (from 32) and for 3, it is 2 (from 36). So, LCM of 32 and 36 = 2^5 x 3^2 = 32 x 9 = 288. Hence, you would need a minimum of 288 books so that they can be equally distributed among the students of either section A or section B. So, the correct option is (C). Q2: If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32, 36) is (a) 2 (b) 4 (c) 6 (d) 8 Ans: (b) Explanation: To arrive at the solution, we need to understand two key concepts - Highest Common Factor (HCF) and Lowest Common Multiple (LCM). The HCF of two numbers is the highest number that can divide both of them without leaving a remainder. On the other hand, LCM of two numbers is the smallest number that can be divided by both of them without leaving a remainder. Let's break down the problem into steps: Step 1: Find the factors of the given numbers 32 and 36. Factors of 32: 2*2*2*2*2 = 2⁵ Factors of 36: 2*2*3*3 = 2²*3² Step 2: Find the LCM of 32 and 36. The LCM is found by multiplying the highest power of all the factors that appear in either number. Here we have 2⁵ from 32 and 2²*3² from 36. The higher power of 2 is 2⁵ from 32 and the higher power of 3 is 3² from 36. So, LCM = 2⁵*3² = 32*9 = 288 Step 3: Find the HCF of 32 and 36. The product of two integers (32 and 36) is equal to the product of their HCF and LCM. So, we can find the HCF by dividing the product of the two numbers by their LCM. HCF = (32*36) / LCM = (32*36) / 288 = 4 Therefore, the HCF of 32 and 36 is 4. Hence, the correct option is (b) 4. Q3: 36 can be expressed as a product of its primes as (a) 2² х 3² (b) 2¹ х 3³ (c) 2³ х 3¹ (d) 2⁰ х 3⁰ Ans: (a) Explanation: Prime factorization is the process of breaking down a number into its smallest prime factors. A prime number is a number that has only two distinct positive divisors: 1 and itself. For example, the first six prime numbers are 2, 3, 5, 7, 11, and 13. To express 36 as a product of its primes, we follow these steps: Step 1: Begin by dividing the number 36 with the smallest prime number, i.e., 2. 36 divided by 2 is 18. Step 2: Now divide 18 by 2 to get 9. Step 3: As 9 cannot be divided by 2, we move to the next prime number, which is 3. 9 divided by 3 is 3. Step 4: Finally, divide 3 by 3 to get 1. Now we stop because we have reached 1. The prime factors of 36 are therefore 2, 2, 3, and 3. We write this as 2² х 3². So, the answer is (a) 2² х 3². Q4: 7 х 11 х 13 х 15 + 15 is a (a) Prime number (b) Composite number (d) Neither prime nor composite (d) None of the above Ans: (b) Explanation: The question is asking to determine the type of number 15 is, when it's part of the given multiplication equation. Step 1: Let's understand the definitions first. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are 2, 3, 5, 7, 11, and 13. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, a composite number is any positive integer greater than one that is not a prime number. Step 2: Now, consider the number 15. We can see that 15 is not a prime number because it has more than two factors, which are 1, 3, 5, and 15. Step 3: Therefore, 15 is a composite number. Hence, the correct option is (B). Q5: If p and q are positive integers such that p = ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is (a) ab (b) a²b² (c) a³b² (d) a³b³ Ans: (b) Explanation: Let's break down the solution to understand it better. Firstly, we are given two positive integers, p and q which are represented as p = ab² and q = a²b, where a and b are prime numbers. The LCM (Least Common Multiple) is the smallest number that is a multiple of both numbers. In other words, it is the smallest number that both numbers can divide into evenly. When finding the LCM of two numbers represented as the product of prime numbers raised to some powers (as in this case), the LCM is simply the product of these primes each raised to the highest power that appears in either number. In this case, the prime number 'a' is raised to the first power in p and to the second power in q. Thus, in the LCM, 'a' is raised to the highest power of these, which is 2. Similarly, the prime number 'b' is raised to the second power in p and to the first power in q. In the LCM, 'b' is raised to the highest power of these, which is 2. Hence, the LCM of p and q is a²b² which corresponds to option (b). So, the correct answer is option (b) a²b².

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus.

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too.

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement?

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

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Class 10 Maths
Important Questions for Class 10 Maths Chapter 1- Real Numbers

Important Questions Class 10 Maths Chapter 1 Real Numbers

Important Class 10 maths questions for Chapter 1 real numbers are provided here to help the students practise better and score well in their CBSE Class 10 maths exam 2022-23. Additional questions of real numbers given here are as per the NCERT book. It covers the complete syllabus and will help the students to develop confidence and problem-solving skills for their exams.

Students can practise solving these questions after they have covered the syllabus for this chapter. Solving different types of questions will help them to clear their doubts. Let us see some questions here with their solutions. Get the important questions of Maths Class 10 for all the chapters here. Solving these important questions will boost the confidence level of students. They will also understand the difficulty level of the questions expected to be asked from this chapter.

Students can also get access to Class 10 Maths Chapter 1 Real Numbers MCQs here.

Before we move ahead with important questions let us learn about real numbers.

What are real numbers?

Real numbers are a combination of rational and irrational numbers. They are the numbers upon which we easily perform mathematical operations. All the numbers which are not imaginary are real numbers. For example, 22, -11, 7.99, 3/2, π(3.14), √2, etc.

Also Check:

Important 2 Marks Questions for CBSE 10th Maths
Important 3 Marks Questions for CBSE 10th Maths
Important 4 Marks Questions for CBSE 10th Maths

Class 10 Real Numbers Important Questions and Answers

Q.1: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Let x be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

x = 3q, 3q + 1 and 3q + 2

As per the given question, if we take the square on both the sides, we get;

x 2 = (3q) 2 = 9q 2 = 3.3q 2

Let 3q 2 = m

x 2 = 3m ………………….(1)

x 2 = (3q + 1) 2

= (3q) 2 + 1 2 + 2 × 3q × 1

= 9q 2 + 1 + 6q

= 3(3q 2 + 2q) + 1

Substituting 3q 2 +2q = m we get,

x 2 = 3m + 1 ……………………………. (2)

x 2 = (3q + 2) 2

= (3q) 2 + 2 2 + 2 × 3q × 2

= 9q 2 + 4 + 12q

= 3(3q 2 + 4q + 1) + 1

Again, substituting 3q 2 + 4q + 1 = m, we get,

x 2 = 3m + 1…………………………… (3)

Hence, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Q.2: Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution: (i) 140 Using the division of a number by prime numbers method, we can get the product of prime factors of 140. Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2 2 × 5 × 7

(ii) 156 Using the division of a number by prime numbers method, we can get the product of prime factors of 156.

Hence, 156 = 2 × 2 × 13 × 3 = 2 2 × 13 × 3

(iii) 3825 Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 3 2 × 5 2 × 17

(iv) 5005 Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.

Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429 Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.

Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Q.3: Given that HCF (306, 657) = 9, find LCM (306, 657).

As we know that,

HCF × LCM = Product of the two given numbers

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Therefore, LCM(306,657) = 22338

Q.4: Prove that 3 + 2√5 is irrational.

Let 3 + 2 √ 5 be a rational number.

Then the co-primes x and y of the given rational number where (y ≠ 0) is such that:

3 + 2 √ 5 = x/y

Rearranging, we get,

2 √ 5 = (x/y) – 3

Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.

Therefore, √ 5 is also a rational number. But this confronts the fact that √ 5 is irrational.

Thus, our assumption that 3 + 2 √ 5 is a rational number is wrong.

Hence, 3 + 2 √ 5 is irrational.

Q.5: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600

Note: If the denominator has only factors of 2 and 5 or in the form of 2 m × 5 n then it has a terminating decimal expansion. If the denominator has factors other than 2 and 5 then it has a non-terminating repeating decimal expansion.

(i) 13/3125

Factoring the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 5 5

= 2 0 × 5 5

Since the denominator is of the form 2 m × 5 n then, 13/3125 has a terminating decimal expansion.

8 = 2× 2 × 2 = 2 3

= = 2 3 × 5 0

Since the denominator is of the form 2 m × 5 n then, 17/8 has a terminating decimal expansion.

(iii) 64/455

455 = 5 × 7 × 13

Since the denominator is not in the form of 2 m × 5 n , therefore 64/455 has a non-terminating repeating decimal expansion.

(iv) 15/1600

1600 = 2 6 × 5 2

Since the denominator is in the form of 2 m × 5 n , 15/1600 has a terminating decimal expansion.

Q.6: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p/q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

(i) 43.123456789 Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

(ii) 0.120120012000120000. . . Since it has a non-terminating and non-repeating decimal expansion, it is an irrational number.

Q.7: Check whether 6 n can end with the digit 0 for any natural number n.

If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.

Prime factorization of 6 n = (2 × 3) n

Therefore, the prime factorization of 6 n doesn’t contain the prime number 5.

Hence, it is clear that for any natural number n, 6 n is not divisible by 5 and thus it proves that 6 n cannot end with the digit 0 for any natural number n.

Q.8: What is the HCF of the smallest prime number and the smallest composite number?

The smallest prime number = 2

The smallest composite number = 4

Prime factorisation of 2 = 2

Prime factorisation of 4 = 2 × 2

HCF(2, 4) = 2

Therefore, the HCF of the smallest prime number and the smallest composite number is 2.

Q.9: Using Euclid’s Algorithm, find the HCF of 2048 and 960.

2048 > 960

Using Euclid’s division algorithm,

2048 = 960 × 2 + 128

960 = 128 × 7 + 64

128 = 64 × 2 + 0

Therefore, the HCF of 2048 and 960 is 64.

Q.10: Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

Prime factorisation of 404 = 2 × 2 × 101

Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2 5 × 3

HCF = 2 × 2 = 4

LCM = 2 5 × 3 × 101 = 9696

HCF × LCM = 4 × 9696 = 38784

Product of the given two numbers = 404 × 96 = 38784

Hence, verified that LCM × HCF = Product of the given two numbers.

Video Lesson on Numbers

Extra questions for Class 10 Maths Chapter 1

Q.1: Find three rational numbers lying between 0 and 0.1. Find twenty rational numbers between 0 and 0.1. Give a method to determine any number of rational numbers between 0 and 0.1.

Q.2: Which of the following rational numbers have the terminating decimal representation?

(v) 133/125

Q.3: Write the following rational numbers in decimal form:

(v) 327/500

(viii) 11/17

Q.4: If a is a positive rational number and n is a positive integer greater than 1, prove that a n is a rational number.

Q.5: Show that 3 √6 and 3 √3 are not rational numbers.

Q.6: Show that 2 + √2 is not a rational number.

Q.7: Give an example to show that the product of a rational number and an irrational number may be a rational number.

Q.8: Prove that √3 – √2 and √3 + √5 are irrational.

Q.9: Express 7/64, 12/125 and 451/13 in decimal form.

Q.10: Find two irrational numbers lying between √2 and √3.

Q.11: Mention whether the following numbers are rational or irrational:

(i) (√2 + 2)

(ii) (2 – √2) x (2 + √2)

(iii) (√2 + √3) 2

Q.12: Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.

Q.13: Write the smallest number which is divisible by both 360 and 657.

Q.14: Show that the square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q.

Q.15: Prove that one of every three consecutive positive integers is divisible by 3.

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Unit 1: Real numbers

Fundamental theorem of arithmetic.

Intro to Euclid's division algorithm (Opens a modal)
Example: Euclid's division algorithm (Opens a modal)
The fundamental theorem of arithmetic (Opens a modal)
Prime factorization exercise (Opens a modal)
Prime factorization of larger numbers Get 3 of 4 questions to level up!

HCF and LCM

HCF and LCM product (Opens a modal)
Least common multiple of three numbers (Opens a modal)
HCF & LCM product Get 3 of 4 questions to level up!
HCF & LCM of three numbers Get 3 of 4 questions to level up!

Revisiting Irrational Numbers

Proof: √2 is irrational (Opens a modal)
Proof: square roots of prime numbers are irrational (Opens a modal)

Important Questions for Class 10 Maths Chapter 1 Real Numbers

Chapter 1 real numbers important questions for cbse class 10 maths board exams.

Important Questions for Chapter 1 Real Numbers Class 10 Maths

Real numbers class 10 maths important questions very short answer (1 mark).

2. The decimal expansion of the rational number 43/2 4 5 3 will terminate after how many places of decimals?

2 × 7 2

Real Numbers Class 10 Maths Important Questions Short Answer-I (2 Marks)

18. Check whether 4 n can end with the digit 0 for any natural number n.

If the number 4 n for any n, were to end with the digit zero, then it would be divisible by 5 and 2.

That is, the prime factorization of 4 n would contain the prime 5 and 2. This is not possible because the only prime in the factorization of 4 n = 2 2n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4 n . So, there is no natural number n for which 4 n ends with the digit zero. Hence 4 n cannot end with the digit zero.

19. Write the denominator of the rational number 257/500 in the form 2 m × 5 n , where m and n are non-negative integers. Hence write its decimal expansion without actual division.

500 = 25 × 20

= 52 × 5 × 4

= 53 × 22

Here, denominator is 500 which can be written as 22 × 53.

Now decimal expansion,

Real Numbers Class 10 Maths Important Questions Short Answer-II (3 Marks)

The required answer will be HCF of 144 and 90.

144 = 2 4 × 3 2

90 = 2 × 3 2 × 5

HCF (144, 90) = 2 × 3 2 = 18

Thus each stack would have 18 cartons.

Let p be a prime number and if possible, let √p be rational

where m and n are co-primes and n ≠0 .

Squaring on both sides, we get

or, pn 2 = m 2 ...(1)

Here p divides pn 2 . Thus p divides m 2 and in result p also divides m.

Let m = pq for some integer q and putting m = pq in eq. (1), we have

pn 2 = p 2 q 2

or, n 2 = pq 2

Here, p divides pq 2 . Thus p divides n 2 and in result p also divides n.

[∵ p is prime and p divides n 2 ⇒ p divides n]

Thus p is a common factor of m and n but this contradicts the fact that m and n are primes. The contradiction arises by assuming that √p is rational.

Hence, √p is irrational.

30. Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together? (2013)

To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20. 10 = 2 × 5 16 = 24 20 = 22 × 5 LCM = 24 × 5 = 16 × 5 = 80 minutes They will start preparing a new card together after 80 minutes.

31. If two positive integers x and y are expressible in terms of primes as x = p 2 q 3 and y = p 3 q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain. (2014)

x = p 2 q 3 and y = p 3 q LCM = p 3 q 3 HCF = p 2 q …(i) Now, LCM = p 3 q 3 ⇒ LCM = pq 2 (p 2 q) ⇒ LCM = pq 2 (HCF) Yes, LCM is a multiple of HCF. Explanation: Let a = 12 = 2 2 × 3 b = 18 = 2 × 3 2 HCF = 2 × 3 = 6 …(ii) LCM = 2 2 × 3 2 = 36 LCM = 6 × 6 LCM = 6 (HCF) [From (ii)] Here LCM is 6 times HCF.

32. Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer. Solution Let a be a positive odd integer By Euclid’s Division algorithm: a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4] a = 4q or 4q + 1 or 4q + 2 or 4q + 3 But 4q and 4q + 2 are both even a is of the form 4q + 1 or 4q + 3.

Real Numbers Class 10 Maths Important Questions Long Answer (4 Marks)

We have n 2 - n = n(n-1) Thus n 2 - n is product of two consecutive positive integers. Any positive integer is of the form 2q or 2q + 1, for some integer q. Case 1: n = 2q If n = 2q we have n(n-1) = 2q(2q-1) = 2m where m = q(2q-1) which is divisible by 2. Case 2: n = 2q+1 If n = 2q+1, we have n(n-1) = (2q+1) (2q+1-1) = 2q(2q+1) = 2m where m = q(2q+1) which is divisible by 2. Hence, n 2 - n is divisible by 2 for every positive integer n.

Real Numbers Class 10 Maths Case Based Questions

1. If p and q are positive integers such that p = ab2 and q = a2b, where a, b are prime numbers, then the LCM (p, q) is

Given, p = ab 2 = a × b × b

q = a 2 b = a × a × b

LCM of (p, q) = a 2 b 2

Prime factor 32 is 2×2×2×2×2

Prime factor 36 is 2×2×3×3

HCF is 2×2 = 4

HCF of 32 and 36 is 4.

3. 7 × 11 × 13 × 15 + 15 is a

A. Prime number B. Composite number C. Neither prime nor composite D. None of the above

B. Composite number

7 × 11 × 13 × 15 + 15 is composite number.

Take 15 common

15 ( 7 × 11 × 13 + 1)

So, given number has factor other than 1 and itself so it is a composite number.

4. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

A. 144 B. 128 C. 288 D. 272

We have to find the LCM of 32 and 36.

LCM(32, 36) = 2 5 × 9 = 288

Hence, the minimum number of books required to distribute equally among students of section A and section B are 288.

B. Read the following text and answer the following questions on the basis of the same:

A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

1. Find the sum of exponents of the prime factors of total number of plants.

A. 2 B. 3 C. 5 D. 6

Total number of plants = 135 + 225 = 360

The prime factors of 360 = 2 × 2 × 2 × 3 × 3 × 5

= 23 × 32 × 51

∴ Sum of exponents = 3 + 2 + 1 = 6.

2. What is total numbers of row in which they can be planted

A. 3 B. 5 C. 8 D. 15

Number of rows of Rose plants = 135/45 = 3

Number of rows of marigold plants = 225/45 = 5

Total number of rows = 3 + 5 = 8

3. Find the sum of exponents of the prime factors of the maximum number of columns in which they can be planted.

We have proved that the maximum number of columns = 45

So, prime factors of 45

= 3 × 3 × 5 = 32 × 51

∴ Sum of exponents = 2 + 1 = 3.

More Study Materials for Class 10

Important Questions Class 10
NCERT Solutions Class 10 Maths
Important Questions Class 10 Maths
Study Materials for Class 10 Maths
Maths Formula Class 10
Solved MCQ Questions Class 10

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Class 10 Maths Case Study Questions of Chapter 1 Real Numbers

Real Numbers Case Study Questions With Answers

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Class 10 Maths Chapter 1 MCQ

Case Study – 1

What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32, 36) is

36 can be expressed as a product of its primes as

7 х 11 х 13 х 15 + 15 is a

If p and q are positive integers such that p = ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is

Case Study – 2

In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are

What is the minimum number of rooms required during the event?

The LCM of 60, 84, and 108 is

The product of HCF and LCM of 60, 84, and 108 is

108 can be expressed as a product of its primes as

Case Study – 3

What is the maximum capacity of a container which can measure the petrol of either tanker in exact number of time?

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the LCM (850, 680) is

680 can be expressed as a product of its primes as

2 х 3 х 5 х 11 х 17 + 11 is a

If p and q are positive integers such that p = a³b² and q = a²b³, where a, b are prime numbers, then the LCM (p, q) is

Case Study – 4

What will be the value of x?

What will be the value of y?

What will be the value of z?

According to Fundamental Theorem of Arithmetic 13915 is a

The prime factorisation of 13915 is

Case Study – 5

What is the minimum distance each should walk so that they can cover the distance in complete steps?

What is the minimum number of steps taken by any of the three friends, when they meet again?

The HCF of 80, 85, and 90 is

The product of HCF and LCM of 80, 85, and 90 is

90 can be expressed as a product of its primes as

The decimal form of 120 /(2^2 × 5^7 × 7^5 )is

HCF of 8, 9, 25 is

The decimal expansion of the rational number 14587/250 will terminate after

If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is

The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates after one place of decimal, is