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CBSE Class 10 Study Material

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.

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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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

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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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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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CBSE Class 10 Maths: Case Study Questions of Chapter 1 Real Numbers PDF Download

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 case study-based questions for Class 10 Maths Chapter 1 Real Numbers

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

Question 1:

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

Question 2:

Answer: (b) LCM = pq2 x HCF

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

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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

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.

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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Case Study Class 10 Maths Questions and Answers (Download PDF)

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study.

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them.

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers.

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

Features Of Class 10 Maths Case Study Questions And Answers Pdf

Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-

Accurate answers of all the Case-based questions given in the PDF.
Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.

Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers

Since, CBSE Class 10 Maths Case Study Questions and Answers are prepared by our maths experts referring to the CBSE Class 10 Syllabus, it provided benefits in various way:-

Case study class 10 maths helps in exam preparation since, CBSE Class 10 Question Papers contain case-based questions.
It allows students to utilise their learning to solve real life problems.
Solving case study questions class 10 maths helps students in developing their observation skills.
Those students who solve Case Study Class 10 Maths on a regular basis become extremely good at answering normal formula based maths questions.
By using class 10 Maths Case Study questions and answers pdf, students focus more on Selfstudys instead of wasting their valuable time.
With the help of given solutions students learn to solve all Case Study questions class 10 Maths CBSE chapter wise pdf regardless of its difficulty level.

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Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

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

$case study ch 1 class 10 maths$

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

$case study ch 1 class 10 maths$

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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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?

First, learn to sit for at least 2 hours at a stretch

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

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Chapter 1 Class 10 Real Numbers

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Updated for NCERT 2023-2024 Book.

Answers to all exercise questions and examples are solved for Chapter 1 Class 10 Real numbers. Solutions of all these NCERT Questions are explained in a step-by-step easy to understand manner

In this chapter, we will study

What is a Real Number
What is Euclid's Division Lemma , and
How to find HCF (Highest Common Factor) using Euclid's Division Algorithm
Then, we study Fundamental Theorem of Arithmetic, which is basically Prime Factorisation
And find HCF and LCM using Prime Factorisation
We also use the formula of HCF and LCM of two numbers a and b HCF × LCM = a × b
Then, we see what is an Irrational Number
and Prove numbers irrational (Like Prove √ 2, √ 3 irrational)
We revise our concepts about Decimal Expansion (Terminating, Non-Terminating Repeating, Non Terminating Non Repeating)
And find out Decimal Expansion of numbers without performing long division

Click on an NCERT Exercise below to get started.

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Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths

Ncert solutions for class 10 maths chapter 1 real numbers.

Exercise 1.1

1. Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(i) 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that, 225= 135 × 1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that, 135= 90 × 1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as 90 = 2 × 45+ 0

As there are no remainder so divisor 45 is our HCF.

(ii) 38220 > 196 we always divide greater number with smaller one.

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as 38220 = 196 × 195 + 0

As there is no remainder so divisor 196 is our HCF.

(iii) 867 > 255 we always divide greater number with smaller one.

Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as 867 = 255 × 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as 255 = 102 × 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder so we can write it as 102 = 51 × 2 + 0

As there is no remainder so divisor 51 is our HCF.

2. Show that any positive odd integer is of the form 6 q + 1, or 6 q + 3, or 6 q + 5, where q is some integer.

Let take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get a = 6 q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So, total possible forms will 6 q + 0 , 6 q + 1 , 6 q + 2,6 q + 3, 6 q + 4, 6 q + 5

6 q + 0 6 is divisible by 2 so it is a even number

6 q + 1 6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6 q + 2 6 is divisible by 2 and 2 is also divisible by 2 so it is a even number

6 q + 3 6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6 q + 4 6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6 q + 5 6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So, odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid's algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.

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

[Hint: Let x be any positive integer then it is of the form 3 q , 3 q + 1 or 3 q + 2. Now square each of these and show that they can be rewritten in the form 3 m or 3 m + 1.]

Let a be any positive integer and b = 3.

Then a = 3 q + r for some integer q ≥ 0 And r = 0, 1, 2 because 0 ≤ r < 3 Therefore, a = 3 q or 3 q + 1 or 3 q + 2 Or, a 2 = (3 q ) 2 or (3 q + 1) 2 or (3 q + 2) 2 a 2 = (9 q ) 2 or 9 q 2 + 6 q + 1 or 9 q 2 + 12 q + 4 = 3 × (3 q 2 ) or 3(3 q 2 + 2 q ) + 1 or 3(3 q 2 + 4 q + 1) + 1 = 3 k 1 or 3 k 2 + 1 or 3 k 3 + 1

Where k 1 , k 2 , and k 3 are some positive integers Hence, it can be said that the square of any positive integer is either of the form 3 m or 3 m + 1.

5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9 m , 9 m + 1 or 9 m + 8.

Let a be any positive integer and b = 3

a = 3 q + r , where q ≥ 0 and 0 ≤ r < 3

∴ a = 3q or 3 q + 1 or 3 q + 2

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3 q , a 3 = (3 q ) 3 = 27 q 3 = 9(3 q ) 3 = 9 m , where, m is an integer such that m = 3 q 3

Case 2: When a = 3q + 1, a 3 = (3 q +1) 3 a 3 = 27 q 3 + 27 q 2 + 9 q + 1 a 3 = 9(3 q 3 + 3 q 2 + q ) + 1 a 3 = 9 m + 1 where, m is an integer such that m = (3 q 3 + 3 q 2 + q )

Case 3: When a = 3 q + 2, a 3 = (3 q +2) 3 a 3 = 27 q 3 + 54 q 2 + 36 q + 8 a 3 = 9(3 q 3 + 6 q 2 + 4q) + 8 a 3 = 9 m + 8 where m is an integer such that m = (3 q 3 + 6 q 2 + 4 q )

Therefore, the cube of any positive integer is of the form 9 m , 9 m + 1, or 9 m + 8.

Page No: 11

Exercise 1.2

1. Express each number as product of its prime factors:

(i) 140 = 2 × 2 × 5 × 7 = 2 2 × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2 2 × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3 2 × 5 2 × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

(i) 26 = 2 × 13

91 =7 × 13

LCM =2 × 7 × 13 =182

Product of two numbers 26 × 91 = 2366

Product of HCF and LCM 13 × 182 = 2366

Hence, product of two numbers = product of HCF × LCM

(ii) 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 × 92 = 46920

Product of HCF and LCM 2 × 23460 = 46920

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024

Product of two numbers 336 × 54 =18144

Product of HCF and LCM 6 × 3024 = 18144

Hence, product of two numbers = product of HCF × LCM.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(i) 12 = 2 × 2 × 3

15 = 3 × 5

21 =3 × 7

LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 =1 × 2 × 2 × 2

9 =1 × 3 × 3

25 =1 × 5 × 5

LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

We have the formula that,

Product of LCM and HCF = product of number

LCM × 9 = 306 × 657

Divide both side by 9 we get,

⇒ LCM = 34 × 657 = 22338

5. Check whether 6 n can end with the digit 0 for any natural number n .

If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.

So, value 6 n should be divisible by 2 and 5 both.

6 n is divisible by 2 but not divisible by 5 as the prime factors of 6 are 2 and 3.

So, it can not end with 0.

6. Explain why 7×11×13 + 13 and 7×6×5×4×3×2×1 + 5 are composite numbers.

7 × 11 × 13 + 13

Taking 13 common, we get

13 (7×11 +1 )

⇒ 13(77 + 1 )

⇒ 13 (78)

It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 common, we get

5(7 × 6 × 4 × 3 × 2 × 1 +1)

⇒ 5(1008 + 1)

⇒ 5(1009)

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point.

They will be meet again after LCM of both values at the starting point.

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM = 2 × 2 × 3 × 3 = 36

Therefore, they will meet together at the starting point after 36 minutes.

Page No: 14

Exercise 1.3

1. Prove that √5 is irrational.

Let us take √5 as rational number

If a and b are two co prime number and b is not equal to 0.

We can write √5 = a/b

Multiply by b both side we get,

b√5 = a

To remove root, Squaring on both sides, we get

5 b 2 = a 2 … (i)

Therefore, 5 divides a 2 and according to theorem of rational number, for any prime number p which is divides a 2 then it will divide a also.

That means 5 will divide a . So we can write,

Putting value of a in equation (i) we get

5 b 2 = (5 c ) 2

⇒ 5 b 2 = 25 c 2

Divide by 25 we get,

b 2 /5 = c 2

Similarly, we get that b will divide by 5 and we have already get that a is divide by 5 but a and b are co prime number. So it contradicts.

Hence, √5 is not a rational number, it is irrational.

2. Prove that 3 + 2√5 is irrational.

Let take that 3 + 2√5 is a rational number.

So we can write this number as,

3 + 2√5 = a / b

Here, a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = a/b – 3

⇒ 2√5 = (a-3b)/b

Now, divide by 2, we get

√5 = (a-3b)/2b

Here, a and b are integer so ( a -3 b ) / 2 b is a rational number so √5 should be a rational number. But √5 is a irrational number, so it contradicts.

Hence, 3 + 2√ 5 is a irrational number.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

(i) Let take that 1/√2 is a rational number.

So, we can write this number as

Here, a and b are two co prime number and b is not equal to 0

Multiply by √2 both sides we get,

1 = ( a √ 2 )/ b

Now, multiply by b

b = a √ 2

Divide by a we get

Here, a and b are integer, so b/a is a rational number. So, √2 should be a rational number

But √2 is a irrational number so it contradicts.

Hence, 1/√2 is a irrational number

So we can write this number as

7√5 = a / b

Divide by 7 we get

√5 = a /(7 b )

Here, a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.

Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.

6 + √2 = a / b

Subtract 6 both side we get

√2 = a / b – 6

⇒ √2 = ( a - 6 b )/ b

Here, a and b are integer so (a-6 b )/ b is a rational number, So, √2 should be a rational number.

Hence, 6 + √ 2 is a irrational number.

Page No: 17

Exercise 1.4

1. 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

(v) 29/343

(vi) 23/2 3 × 5 2

(vii) 129/2 2 × 5 7 × 7 5

(viii) 6/15

(ix) 35/50

Factorize the denominator we get

3125 =5 × 5 × 5 × 5 × 5 = 5 5

So, denominator is in form of 5 m so it is terminating.

Factorize the denominator we get,

8 =2 × 2 × 2 = 2 3

So, denominator is in form of 2 m so it is terminating.

(iii) 64 / 455

455 = 5 × 7 × 13

There are 7 and 13 also in denominator so denominator is not in form of 2 m × 5 n . So, it is not terminating.

(iv) 15 / 1600

1600 = 2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 2 6 × 5 2

So, denominator is in form of 2 m × 5 n

Hence, it is terminating.

(v) 29 / 343

343 = 7 × 7 × 7 = 7 3

There are 7 also in denominator so denominator is not in form of 2 m × 5 n

Hence, it is non-terminating.

(vi) 23 / (2 3 × 5 2 )

Denominator is in form of 2 m × 5 n

Hence, it is terminating.

(vii) 129 / (2 2 × 5 7 × 7 5 )

Denominator has 7 in denominator so denominator is not in form of 2 m × 5 n

Hence, it is none terminating.

(viii) 6 / 15

Divide nominator and denominator both by 3 we get 2 / 5

Denominator is in form of 5 m so it is terminating.

(ix) 35 / 50 divide denominator and nominator both by 5 we get 7 / 10

10 = 2 × 5

So, denominator is in form of 2 m × 5 n so it is terminating.

(x) 77 / 210

Simplify it by dividing nominator and denominator both by 7 we get, 11 / 30

30 = 2 × 3 × 5

Denominator has 3 also in denominator so denominator is not in form of 2 m × 5 n

Page No: 18

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) 13 / 3125

(ii) 17 / 8

(vi) 23 / 2 3 5 2

Dividing numerator and denominator by 3.

(ix) 35 / 50

Dividing numerator and denominator by 5.

3. 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 you say about the prime factors of q ?

(i) 43.123456789

(ii) 0.120120012000120000...

(iii) 43. 123456789

(i) Since this number has a terminating decimal expansion, it is a rational number of the form p / q, and q is of the form 2 m × 5 n .

(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p / q, and q is not of the form 2 m × 5 n .

Chapterwise NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

How many exercises in Chapter 1 Real Numbers

What is lemma, what do you mean by fundamental theorem of arithmetic., what is algorithm, contact form.

CASE STUDY BASED QUESTIONS CLASS 10 MATHS CHAPTER-1

Table of Contents

CHAPTER – 1 REAL NUMBERS

CASE STUDY BASED QUESTIONS CLASS 10 MATHS

CASE STUDY 1.

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?

A) 144 b) 128 c) 288 d) 272, 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 , a) 2 b) 4 c) 6 d) 8, 3. 36 can be expressed as a product of its primes as, a) b) c) d) , 4. 7x11x13x15+15 is a .

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

5. 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) a²b² c) a³b² d) a³b³.

ANSWERS: 1.c 2. b 3. a 4. b 5. b

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 a)14 b)12 c) 16 d) 18
What is the minimum number of rooms required during the event? a)11 b) 31 c) 41 d) 21
The LCM of 60, 84 and 108 is a)3780 b) 3680 c) 4780 d) 4680
The product of HCF and LCM of 60,84 and 108 is a)55360 b) 35360 c) 45500 c) 45360
108 can be expressed as a product of its primes as a) 2³x3² b)2²x3² c) 2³x3³ d)2²x3³

ANSWERS: 1.b 2. d 3. a 4. d 5. d

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

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.

Largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is

This means 60 – 8 = 52 and 75 – 10 = 65, are completely divided by a integer. Therefore, we have to find the HCF of 52 and 65.. According to Euclid’s Division Lemma, 65 = 52 × 1 +13 and 52 = 13 × 4 + 0, Hence 13 is HCF.

The least number that is divisible by all the numbers from 1 to 8 is

The LCM of numbers 1 to 8 is 840. We can write the factors of the numbers from 1 to 8 as follows: 1, 2, 3, 2 × 2, 5, 2 × 3, 7, 2 × 2 × 2. Therefore, LCM = 1 × 2 × 2 × 2 × 3 × 5 × 7 = 840

(6 + 5 √3) – (4 – 3 √3 ) is

After simplifying the expression: 6 +5 √3 – 4 + 3 √3 = 2 + 8 √3 is a irrational number

The number π is

We know the value of π = 3.14285714….. Which is non-repeating non-terminating. So this is an irrational.

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.

NCERT Solutions
NCERT Solutions for Class 10
NCERT Solutions for Class 10 Maths
Chapter 1: Real Numbers
Exercise 1.1

NCERT Solutions for Class 10 Maths Chapter 1- Real Number Exercise 1.1

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 are provided here. These solutions are prepared by our subject expert faculty to help students in their Class 10 exam preparations. These experts review these NCERT Maths Solutions chapter-wise to help students to solve problems easily while using them as a reference. They also focus on preparing the NCERT Solutions for these exercises in such a way that it is easy to understand for the students.

The first exercise in Real Numbers- Exercise 1.1 explains the divisibility of integers using Euclid’s Division Algorithm. They provide a detailed and step-wise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 10. The NCERT Class 10 Maths solutions are always prepared by following NCERT guidelines so students can cover the whole syllabus accordingly. These are very helpful in scoring well in examinations.

Chapter 1 Real Numbers
Chapter 2 Polynomials
Chapter 3 Pair of Linear Equations in Two Variables
Chapter 4 Quadratic Equations
Chapter 5 Arithmetic Progressions
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Some Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Areas Related to Circles
Chapter 13 Surface Areas and Volumes
Chapter 14 Statistics
Chapter 15 Probability
Exercise 1.1 Chapter 1 Real Numbers
Exercise 1.2 Chapter 1 Real Numbers
Exercise 1.3 Chapter 1 Real Numbers
Exercise 1.4 Chapter 1 Real Numbers

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Access answers of Maths NCERT Class 10 Chapter 1 – Real Number Exercise 1.1 page number 7

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

As you can see from the question, 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have

225 = 135 × 1 + 90

Now, the remainder 90 ≠ 0; thus, again, using the division lemma for 90, we get

135 = 90 × 1 + 45

Again, 45 ≠ 0; repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45; therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45

Hence, the HCF of 225 and 135 is 45.

In the given question, 38220>196; therefore, by applying Euclid’s division algorithm and taking 38220 as the divisor, we get

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196

Hence, the HCF of 196 and 38220 is 196.

As we know, 867 is greater than 255. Let us now apply Euclid’s division algorithm on 867 to get

867 = 255 × 3 + 102

Remainder 102 ≠ 0; therefore, taking 255 as the divisor and applying the division lemma method, we get

255 = 102 × 2 + 51

Again, 51 ≠ 0. Now, 102 is the new divisor, so by repeating the same step, we get

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. In the last step, the divisor is 51; therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51

Hence, the HCF of 867 and 255 is 51.

2. Show that any positive odd integer is of form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now, substituting the value of r, we get

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

Number of army contingent members=616

Number of army band members = 32

If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF (616, 32) gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm to find their HCF, we get

Since 616>32, therefore,

616 = 32 × 19 + 8

Since 8 ≠ 0, therefore, taking 32 as the new divisor, we have

32 = 8 × 4 + 0

Now, we have the remainder as 0; therefore, HCF (616, 32) = 8

Hence, the maximum number of columns in which they can march is 8.

4. 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, then

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

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

Now, as per the question given, by squaring both sides, we get

x 2 = (3q) 2 = 9q 2 = 3 × 3q 2

Let 3q 2 = m

Therefore, 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

Substitute 3q 2 +2q = m to 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, substitute 3q 2 +4q+1 = m to get

x 2 = 3m + 1…………………………… (3)

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

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get

Now, by taking the cube of all the three above expressions, we get

Case (i): When r = 0, then,

x 2 = (3q) 3 = 27q 3 = 9(3q 3 )= 9m; where m = 3q 3

Case (ii): When r = 1, then,

x 3 = (3q+1) 3 = (3q) 3 +1 3 +3×3q×1(3q+1) = 27q 3 +1+27q 2 +9q

Taking 9 as a common factor, we get

x 3 = 9(3q 3 +3q 2 +q)+1

Putting = m, we get

Putting (3q 3 +3q 2+ q) = m, we get

Case (iii): When r = 2, then,

x 3 = (3q+2) 3 = (3q) 3 +2 3 +3×3q×2(3q+2) = 27q 3 +54q 2 +36q+8

x 3 =9(3q 3 +6q 2 +4q)+8

Putting (3q 3 +6q 2 +4q) = m, we get

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Exercise 1.1 of NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers is the first exercise of Chapter 1 of Class 10 Maths. Real Numbers are introduced in Class 9 and are discussed further in detail in Class 10 by studying Euclid’s division Algorithm. The exercise discusses the divisibility of integers. The divisibility of integers using Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b, such that the remainder will be which is smaller than b.

Euclid’s Division Algorithm – It includes 5 questions based on Theorem 1.1 – Euclid’s Division Lemma.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1 – Real Number Exercise 1.1 Page number 7

These NCERT Solutions for Class 10 help students solve and revise all questions of Exercise 1.1.
After going through the step-wise solutions given by our subject expert teachers, learners will be able to score more marks.
They help in scoring well in Maths exams.
They follow NCERT guidelines which help in preparing the students accordingly.
They contain all the important questions from the examination point of view.

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Case Study Class 10 Maths Questions
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.
CBSE Case Study Questions for Class 10 Maths Real Numbers Free PDF
Level 1. First, learn to sit for at least 2 hours at a stretch. Level 2. Solve every question of NCERT by hand, without looking at the solution. Level 3. Solve NCERT Exemplar (if available) Level 4. Sit through chapter wise FULLY INVIGILATED TESTS. Level 5.
Chapter 1 Class 10 Real Numbers
Updated for NCERT 2023-2024 Book. Answers to all exercise questions and examples are solved for Chapter 1 Class 10 Real numbers. Solutions of all these NCERT Questions are explained in a step-by-step easy to understand manner. In this chapter, we will study. Click on an NCERT Exercise below to get started.
Case Study Based Questions Class 10 Chapter 1 Real Numbers ...
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Class 10 Maths NCERT Solutions Chapter 1 Real Numbers
These NCERT Solutions for Class 10 of Maths subject includes detailed answers of all the questions in Chapter 1 - Class 10 Real Numbers provided in NCERT Book which is prescribed for class 10 in schools. Class 10 Real Numbers Ex 1.4 - 3 Questions Based in which you have to expand fractions into decimals and write decimals in their fraction ...
Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Here, we have providing Class 10 Maths NCERT Solutions for Chapter 1 Real Numbers which will be beneficial for students.These solutions are updated according to 2020-21 syllabus. As NCERT Solutions are prepared by Studyrankers experts, we have taken of every steps so you can understand the concepts without any difficulty.
Case Study Based Questions Class 10 Maths Chapter-1
CASE STUDY BASED QUESTIONS CLASS 10 MATHS. 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. What is the minimum number of rooms required during the event? a)11 b) 31 c) 41 d) 21.
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.
Important Questions Class 10 Maths Chapter 1 Real Numbers
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.
Real Numbers Class 10 Maths Chapter 1 Notes
CBSE Class 10 Maths Chapter 1 Real Numbers Notes are provided here in detail. As we all know, any number, excluding complex numbers, is a real number. Positive and negative integers, irrational numbers, and fractions are all examples of real numbers. To put it another way, any number found in the real world is a real number.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1
Access answers of Maths NCERT Class 10 Chapter 1 - Real Number Exercise 1.1 page number 7. 1. Use Euclid's division algorithm to find the HCF of: i. 135 and 225. ii. 196 and 38220. iii. 867 and 255. Solutions: i. 135 and 225. As you can see from the question, 225 is greater than 135.

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Class 10 Maths Chapter 1 Case Based Questions - Real Numbers

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Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths

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CASE STUDY BASED QUESTIONS CLASS 10 MATHS CHAPTER-1

CHAPTER – 1 REAL NUMBERS

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?

5. 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

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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