case study questions of real numbers class 10

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

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

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

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

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

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

case study questions of real numbers class 10

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

Answer: c) 288

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

Answer: b) 4

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

a) 2 2 × 3 2

b) 2 1 × 3 3

c) 2 3 × 3 1

d) 2 0 × 3 0

Answer: a) 2 2 × 3 2

4. 7 × 11 × 13 × 15 + 15 is a

a) Prime number

b) Composite number

c) Neither prime nor composite

d) None of the above

Answer: b) Composite number

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

Answer: b) a 2 b 2

CASE STUDY 2:

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

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

Answer: b) 12

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

Answer: d) 21

3. The LCM of 60, 84 and 108 is

Answer: a) 3780

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

Answer: d) 45360

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

a) 2 3 × 3 2

b) 2 3 × 3 3

c) 2 2 × 3 2

d) 2 2 × 3 3

Answer: d) 2 2 × 3 3

CASE STUDY 3:

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

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

4. According to Fundamental Theorem of Arithmetic 13915 is a

a) Composite number

b) Prime number

d) Even number

Answer: a) Composite number

5. The prime factorisation of 13915 is

a) 5 × 11 3 × 13 2

b) 5 × 11 3 × 23 2

c) 5 × 11 2 × 23

d) 5 × 11 2 × 13 2

Answer: c) 5 × 11 2 × 23

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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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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Case Study Questions for Class 10 Maths Chapter 1 Real Numbers

Last modified on: 9 months ago
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Question 1:

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

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

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

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

Question 2:

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

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

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

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

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

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

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

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

CBSE Case Study Questions for Class 10 Maths Real Numbers PDF

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

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

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

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

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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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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024 will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

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Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

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Class 10 Maths Syllabus 2024

Chapter-1 real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2 Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3 Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4 Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5 Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6 Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7 Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8 Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9 Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10 Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11 Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12 Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13 Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14 Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15 Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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CBSE Class 10 Maths Case Study

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

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

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

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

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

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

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

CBSE Class 10th Case study Questions

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

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

CBSE Class 10th Assertion Reasoning Questions

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

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

CBSE Class 10 Maths Case Based MCQ

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

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

Class 10th Mathematics Multiple Choice Questions

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

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

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

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

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

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Class 10 Maths 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.

Two tankers contain 850 litres & 680 litres of petrol respectively. Maximum capacity of a container which can measure the petrol of either tanker in exact number of times.

Maximum capacity of container means HCF of 850 & 680, applying Euclid’s algorithm we get the HCF of two numbers is 170. Clearly HCF of 850 & 680 is 170, hence capacity of the container must be 170litres.

Three farmers have 490kg, 588kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags.

According to Euclid’s Division Lemma the HCF of 490, 588 and 882 is 98.

The number in the form of 4p +3, where p is a whole number, will always be

Because 4p will be always an even number. Sum of an even number and an odd number will be always an odd number.

When a number is divided by 7, its remainder is always

According to Euclid’s Division Lemma, If a = 7q + r, then r should be less than 7. In other words, remainder should always be less than divisor.

(6 + 5 √3) – (4 – 3 √3 ) is

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

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.

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Chapter 1 Class 10 Real Numbers

Click on any of the links below to start learning from Teachoo ...

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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Class 10th Maths - Real Number Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Real Number, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Real number case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

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

(ii) Find the LCM of 60, 84 and 108.

(iii) Find the HCF of 60, 84 and 108.

(iv) Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

(v) Based on the above (iv) conditions, find the minimum number of rooms required for all the participants and officials.

(b) Find the total number of stacks formed.

(d) If the thickness of each English book is 3 cm, then the height of each stack of English books is

(e) If each Hindi book weighs 1.5 kg, then find the weight of books in a stack of Hindi books.

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

Real number case study questions with answer key 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.

(i) (d): Total number of participants = 60 + 84 + 108 = 252 (ii) (d): 60 = 22 x 3 x 5 84 = 22 x 3 x 7 108 = 22 x 33 LCM(60, 84, 108) = 22 x 33 x 5 x 7 = 3780 (iii) (a): 60 = 22 x 3 x 5 84 = 22 x 3 x 7 108 = 22 x 33 HCF(60, 84, 108) = 22 x 3 = 12 (iv) (c): Minimum number of rooms required for all the participants = 252/12 = 21 (v) (d): Minimum number of rooms required for all = 21 + 1 = 22

(a) (ii) 96 = 2 5 x 3 240 = 2 4 x3 x5 (b) (iii) Total number of books = 96 +240+336=672 Number of books in each stack = 48 \(\therefore\) Number of stacks formed -= \(\frac{672}{48}=14\) (c) (i) Number of mathmatics books = 336 Number of stacks of mathematics books formed = \(\frac{336}{48}\) = 7 (d) (iv) Number of books in each stack of english books = 48 Thickness of each english book = 3 cm \(\therefore\) Height of each stack of english books = (48X3) cm = 144cm (e) (iii) Number of books in a stack of hindi books = 48 Weight of each hindi book = 1.5kg \(\therefore\) The weight of books in a stack of hindi books = (48X1.5)kg = 72kg

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Important Questions for Class 10 Maths Chapter 1 Real Numbers

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

Important Questions for Chapter 1 Real Numbers Class 10 Maths

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

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

2 × 7 2

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

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

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

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

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

500 = 25 × 20

= 52 × 5 × 4

= 53 × 22

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

Now decimal expansion,

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

The required answer will be HCF of 144 and 90.

144 = 2 4 × 3 2

90 = 2 × 3 2 × 5

HCF (144, 90) = 2 × 3 2 = 18

Thus each stack would have 18 cartons.

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

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

Squaring on both sides, we get

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

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

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

pn 2 = p 2 q 2

or, n 2 = pq 2

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

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

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

Hence, √p is irrational.

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

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

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

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

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

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

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

Real Numbers Class 10 Maths Case Based Questions

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

Given, p = ab 2 = a × b × b

q = a 2 b = a × a × b

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

Prime factor 32 is 2×2×2×2×2

Prime factor 36 is 2×2×3×3

HCF is 2×2 = 4

HCF of 32 and 36 is 4.

3. 7 × 11 × 13 × 15 + 15 is a

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

B. Composite number

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

Take 15 common

15 ( 7 × 11 × 13 + 1)

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

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

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

We have to find the LCM of 32 and 36.

LCM(32, 36) = 2 5 × 9 = 288

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

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

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

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

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

Total number of plants = 135 + 225 = 360

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

= 23 × 32 × 51

∴ Sum of exponents = 3 + 2 + 1 = 6.

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

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

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

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

Total number of rows = 3 + 5 = 8

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

We have proved that the maximum number of columns = 45

So, prime factors of 45

= 3 × 3 × 5 = 32 × 51

∴ Sum of exponents = 2 + 1 = 3.

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Real Numbers For Class 10

Real Numbers Class 10 Notes: Chapter 1

Cbse real numbers class 10 notes:- download pdf here, class 10 maths chapter 1 real number 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. Numbers can be found all around us. Natural numbers are being used to count objects, integers are used to measure temperature, rational numbers are used to represent fractions, irrational numbers are used to calculate the square root of a number, etc. These various types of numbers form a collection of real numbers. Here, we are going to learn what a real number is, Euclid’s division algorithm, the fundamental theorem of arithmetic, methods of finding LCM and HCF and the complete explanation of rational and irrational numbers with examples.

Real Numbers

Positive integers, negative integers, irrational numbers, and fractions are all examples of real numbers. In other words, we can say that any number is a real number, except for complex numbers. Examples of real numbers include -1, ½, 1.75, √2, and so on. In general,

Real numbers constitute the union of all rational and irrational numbers.
Any real number can be plotted on the number line.

To know more about real numbers, visit here .

Students can refer to the short notes and MCQ questions along with a separate solution pdf of this chapter for quick revision from the links below.

Real Numbers Short Notes
Real Numbers MCQ Practice Questions
Real Numbers MCQ Practice Solutions

Euclid’s Division Lemma

Euclid’s Division Lemma states that given two integers a and b , there exists a unique pair of integers q and r such that a = b × q + r a n d 0 ≤ r < b .
This lemma is essentially equivalent to : dividend = divisor × quotient + remainder
In other words, for a given pair of dividend and divisor, the quotient and remainder obtained are going to be unique.

For more information on Euclid’s Division Lemma, watch the below video

To know more about Euclid’s Division Lemma, visit here .

Euclid’s Division Algorithm

Euclid’s Division Algorithm is a method used to find the H.C.F of two numbers, say a and b where a> b.
We apply Euclid’s Division Lemma to find two integers q and r such that a = b × q + r a n d 0 ≤ r < b .
If r = 0, the H.C.F is b; else, we apply Euclid’s division Lemma to b (the divisor) and r (the remainder) to get another pair of quotient and remainder.
The above method is repeated until a remainder of zero is obtained. The divisor in that step is the H.C.F. of the given set of numbers.

For more information on Euclid’s Division Algorithm, watch the below video

The Fundamental Theorem of Arithmetic

Prime Factorisation

Prime Factorisation is the method of expressing a natural number as a product of prime numbers.
Example: 36 = 2 × 2 × 3 × 3 is the prime factorisation of 36.

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.
Example: 36 = 2 × 2 × 3 × 3 OR, 36 = 2 × 3 × 2 × 3
Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.

To know more about the Fundamental Theorem of Arithmetic, visit here .

Method of Finding LCM

As we know, the smallest of the common multiples of two or more numbers is called the lowest common multiple (LCM). Example: To find the Least Common Multiple ( L.C.M ) of 36 and 56,

36 = 2 × 2 × 3 × 3 56 = 2 × 2 × 2 × 7
The common prime factors are 2 × 2
The uncommon prime factors are 3 × 3 for 36 and 2 × 7 for 56.
LCM of 36 and 56 = 2 × 2 × 3 × 3 × 2 × 7 which is 504

To know more about LCM, visit here .

Method of Finding HCF

We know that the greatest number that divides each of the given numbers without leaving any remainder is the highest common factor (HCF) of two or more given numbers. H.C.F can be found using two methods – Prime factorisation and Euclid’s division algorithm.

Given two numbers, we express both of them as products of their respective prime factors. Then, we select the prime factors that are common to both the numbers
Example – To find the H.C.F of 20 and 24 20 = 2 × 2 × 5 and 24 = 2 × 2 × 2 × 3
The factor common to 20 and 24 is 2 × 2 , which is 4, which in turn is the H.C.F of 20 and 24.
It is the repeated use of Euclid’s division lemma to find the H.C.F of two numbers.

The required HCF is 6 .

To know more about HCF, visit here .

For more information on HCF and LCM, watch the below video

To know more about the Properties of HCF and LCM, visit here .

Product of Two Numbers = HCF X LCM of the Two Numbers

For any two positive integers a and b, a × b = H . C . F × L . C . M .
Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504 36 × 56 = 2016 4 × 504 = 2016 Thus, 36 × 56 = 4 × 504
Let us consider another example: For 5 and 6, the H.C.F is 1 and the L.C.M is 30 5 × 6 = 30 1 × 30 =30 Thus, 5 × 6 = 1 × 30
The above relationship, however, doesn’t hold true for 3 or more numbers

To know the Relationship between LCM and HCF, visit here .

Applications of HCF & LCM in Real-World Problems

L.C.M can be used to find the points of common occurrence. For example,ringing of bells that ring with different frequencies, the time at which two persons running at different speeds meet, and so on.

For more information on Applications Of LCM, watch the below video

Revisiting Irrational Numbers

Irrational numbers.

Any number that cannot be expressed in the form of p/q (where p and q are integers and q ≠ 0 .) is an irrational number. Examples √2 , π , e and so on.

To know more about Irrational Numbers, visit here .

Number theory: Interesting results

If a number p (a prime number) divides a 2 , then p divides a. Example: 3 divides 6 2 i.e 36, which implies that 3 divides 6.
The sum or difference of a rational and an irrational number is irrational
The product and quotient of a non-zero rational and irrational number are irrational.
√p is irrational when ‘p’ is a prime. For example, 7 is a prime number, and √7 is irrational. The above statement can be proved by the method of “Proof by contradiction”.

To know more about Number theory, visit here .

Proof by Contradiction

In the method of contradiction, to check whether a statement is TRUE (i) We assume that the given statement is TRUE. (ii) We arrive at some result which contradicts our assumption, thereby proving the contrary. Eg: Prove that √7 is irrational. Assumption: √7 is rational. Since it is rational √7 can be expressed as √7 = a/b , where a and b are co-prime Integers, b ≠ 0. On squaring, a 2 /b 2 = 7 ⇒ a 2 = 7 b 2 . Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a 2 = 49 c 2 . Hence, 7 b 2 = 49 c 2 or b 2 = 7 c 2 . Hence 7 divides b. Since 7 is a common factor for both a and b, it contradicts our assumption that a and b are coprime integers. Hence, our initial assumption that √7 is rational is wrong. Therefore, √7 is irrational.

Revisiting Rational Numbers and Their Decimal Expansions

Rational numbers.

Rational numbers are numbers that can be written in the form p/q, where p and q are integers and q ≠ 0 . Examples -1/2 , 4/5, 1 , 0 , − 3 and so on.

To know more about Rational Numbers, visit here .

Terminating and Non-Terminating Decimals

Terminating decimals are decimals that end at a certain point. Example: 0.2, 2.56 and so on. Non-terminating decimals are decimals where the digits after the decimal point don’t terminate. Example: 0.333333….., 0.13135235343… Non-terminating decimals can be : a) Recurring – a part of the decimal repeats indefinitely (0 . 142857 142857 …. ) b) Non-recurring – no part of the decimal repeats indefinitely. Example: π = 3.1415926535…

To know more about terminating and non-terminating decimals, click here .

Check if a given rational number is terminating or not

If a/b is a rational number, then its decimal expansion would terminate if both of the following conditions are satisfied : a) The H.C.F of a and b is 1. b) b can be expressed as a prime factorisation of 2 and 5 i.e b = 2 m × 5 n where either m or n, or both can = 0. If the prime factorisation of b contains any number other than 2 or 5, then the decimal expansion of that number will be recurring

1/40 = 0.025 is a terminating decimal, as the H.C.F of 1 and 40 is 1, and the denominator (40) can be expressed as 2 3 × 5 1 .

3/7 = 0.428571 is a recurring decimal as the H.C.F of 3 and 7 is 1 and the denominator (7) is equal to 7 1

Real Numbers for Class 10 Solved Examples

Find the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively.

First, subtract the remainder from the number.

(i.e) 70-5 = 65

125-8 = 117.

Thus, we need to find the largest number that divides 65 and 117 and leaves the remainder 0.

To find the largest number, take the HCF of 65 and 117.

Finding HCF of 65 and 117.

65 = 5×13

117 = 3×3×13.

Hence, HCF (65, 117) = 13.

Therefore, the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively is 13.

Find the LCM of 306 and 657, given that HCF (306, 657) = 9.

Given that, HCF (306, 657) = 9.

We know that HCF × LCM = Product of Numbers

Hence, 9×LCM = 306×657

9×LCM = 201042

LCM = 201042/9

LCM = 22338.

Therefore, LCM of 306 and 657 is 22338.

Prove that 1/√2 is an irrational number.

To prove 1/√2 is an irrational number.

Now, let us take the opposite assumption.

(i.e) Take 1/√2 is a rational number.

We know that rational numbers are the numbers that can be written in the form of p/q, where q is not equal to 0. (p and q are two co-prime numbers)

Hence, 1/√2 = p/q.

Now, simplify the above equation by multiplying √2 on both sides.

1 = (p√2)/q

q = p√2

Hence, we get q/p = √2.

Here, p and q are integers, and hence q/p is a rational number.

But, √2 is an irrational number.

Hence, our assumption is wrong.

Therefore, 1/√2 is an irrational number.

Hence, proved.

Frequently Asked Questions on Class 10 Real Numbers

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Extra questions for class 10 maths real numbers, euclid's division lemma | hcf and lcm | irrational numbers.

NCERT Class 10 Maths ➤ Real Numbers Extra Questions

Two core concepts tested in this chapter are LCM and HCF. Watch these three videos to understand the two concepts and when to use LCM and when to use HCF ?

Chapter 1 of CBSE NCERT Class 10 Math covers Real Numbers. Concepts covered in chapter 1 include Euclid's Division Lemma, Fundamental Theorem of Arithmetic and Prime Factorization, Highest Common Factor(HCF), Lowest Common Multiple(LCM), Product of HCF and LCM, Prime Factorisation, Irrational Numbers, and Rational Numbers. The extra questions given below include questions akin to HOTS (Higher Order Thinking Skills) questions and exemplar questions of NCERT.

Here is a quick recap of the key concepts that are covered in this chapter in the CBSE NCERT Class 10 Math text book. Already, know these concepts? Jump to extra questions directly.

What is Euclid’s Division Lemma

If we have two positive integers n and d, then there would be whole numbers q and r that satisfy the equation: n= qd + r , where 0 ≤ r < d. n is the dividend. q is the quotient. d is the divisor.

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that, "Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur."

What is HCF? How to find HCF?

Highest Common Factor (HCF) of two or more numbers is the largest number that divides all of those numbers without leaving a remainder.

Highest Common Factor (HCF) is the product of the lowest power of common prime factors in the numbers.

How to use Euclid’s Division Lemma to compute HCF?

Euclid's Division Algorithm can be used to compute the Highest Common Factor (HCF) of the two or more positive integers. Let the given positive integers be a and b such that a > b Step 1: Apply Euclid's Division Lemma to find q and r where a = bq + r, 0 ≤ r < b. Step 2: If r = 0, then the HCF is b. If r ≠ 0, apply Euclid's Lemma to b and r. Step 3: Continue the process till the remainder is zero. The divisor at the step in which the remainder is zero is the HCF(a, b). Also HCF(a, b) = HCF(b, r).

What is LCM? How to find LCM?

Lowest Common Multiple (LCM) is the product of the greatest power of all the prime factors found in the numbers.

What are irrational numbers?

A number ‘s’ is irrational if it cannot be written in the form \\frac{\text{p}}{\text{q}}), where p and q are integers and q ≠ 0. The decimal expansion of an irrational number is non-terminating and non-recurring.

Extra Questions for Class 10 Maths - Real Numbers

Karan has 180 blue marbles and 150 red marbles. He wants to pack them into packets containing equal number of marbles of the same colour. What is the maximum number of marbles that each packet can hold?

Find the largest number that will divide 382 and 710 and leaves a remainder 13 in each case.

What is the largest number that divides 437, 732, and 1263 leaving remainder of 24 in each case?

What is the largest number that divides 967 and 1767 leaving remainders of 71 and 103 respectively?

What is the largest number that divides 170, 220, and 420 leaving remainder 8, 4 and 15 respectively?

Find the LCM and HCF of the following : (i) 2 5 × 5 4 × 7 2 × 13 6 and 2 3 × 5 6 × 7 × 17 3 (ii) a 5 × b 2 × c 2 × d 5 and a 7 × b 3 × e × f 3 where a, b, c, d, e, and f are prime.

The Muscle Gym has bought 63 treadmills and 108 elliptical machines. The gym divides them into several identical sets of treadmills and elliptical machines for its branches located throughout the city, with no exercise equipment left over. What is the greatest number of branches the gym can have in the city?

Katya has 49 paintings and 35 medals. She wants to display them in groups throughout her house, each with the same combination of paintings and medals, with none left over. What is the greatest number of groups Katya can display?

Anish goes fishing every 5th day and Balaji goes fishing every 7th day. If Anish and Balaji both went fishing today, how many days until they will go fishing on the same day again?

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Tamanna is arranging black marbles in groups of 13 and purple marbles in groups of 25. If she has the same number of black and purple marbles, what is the smallest number of marbles of each colour that she could have?

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What the data says about gun deaths in the U.S.

More Americans died of gun-related injuries in 2021 than in any other year on record, according to the latest available statistics from the Centers for Disease Control and Prevention (CDC). That included record numbers of both gun murders and gun suicides. Despite the increase in such fatalities, the rate of gun deaths – a statistic that accounts for the nation’s growing population – remained below the levels of earlier decades.

Here’s a closer look at gun deaths in the United States, based on a Pew Research Center analysis of data from the CDC, the FBI and other sources. You can also read key public opinion findings about U.S. gun violence and gun policy .

This Pew Research Center analysis examines the changing number and rate of gun deaths in the United States. It is based primarily on data from the Centers for Disease Control and Prevention (CDC) and the Federal Bureau of Investigation (FBI). The CDC’s statistics are based on information contained in official death certificates, while the FBI’s figures are based on information voluntarily submitted by thousands of police departments around the country.

For the number and rate of gun deaths over time, we relied on mortality statistics in the CDC’s WONDER database covering four distinct time periods: 1968 to 1978 , 1979 to 1998 , 1999 to 2020 , and 2021 . While these statistics are mostly comparable for the full 1968-2021 period, gun murders and suicides between 1968 and 1978 are classified by the CDC as involving firearms and explosives; those between 1979 and 2021 are classified as involving firearms only. Similarly, gun deaths involving law enforcement between 1968 and 1978 exclude those caused by “operations of war”; those between 1979 and 2021 include that category, which refers to gun deaths among military personnel or civilians due to war or civil insurrection in the U.S . All CDC gun death estimates in this analysis are adjusted to account for age differences over time and across states.

The FBI’s statistics about the types of firearms used in gun murders in 2020 come from the bureau’s Crime Data Explorer website . Specifically, they are drawn from the expanded homicide tables of the agency’s 2020 Crime in the United States report . The FBI’s statistics include murders and non-negligent manslaughters involving firearms.

How many people die from gun-related injuries in the U.S. each year?

In 2021, the most recent year for which complete data is available, 48,830 people died from gun-related injuries in the U.S., according to the CDC. That figure includes gun murders and gun suicides, along with three less common types of gun-related deaths tracked by the CDC: those that were accidental, those that involved law enforcement and those whose circumstances could not be determined. The total excludes deaths in which gunshot injuries played a contributing, but not principal, role. (CDC fatality statistics are based on information contained in official death certificates, which identify a single cause of death.)

A pie chart showing that suicides accounted for more than half of U.S. gun deaths in 2021.

What share of U.S. gun deaths are murders and what share are suicides?

Though they tend to get less public attention than gun-related murders, suicides have long accounted for the majority of U.S. gun deaths . In 2021, 54% of all gun-related deaths in the U.S. were suicides (26,328), while 43% were murders (20,958), according to the CDC. The remaining gun deaths that year were accidental (549), involved law enforcement (537) or had undetermined circumstances (458).

What share of all murders and suicides in the U.S. involve a gun?

About eight-in-ten U.S. murders in 2021 – 20,958 out of 26,031, or 81% – involved a firearm. That marked the highest percentage since at least 1968, the earliest year for which the CDC has online records. More than half of all suicides in 2021 – 26,328 out of 48,183, or 55% – also involved a gun, the highest percentage since 2001.

A line chart showing that the U.S. saw a record number of gun suicides and gun murders in 2021.

How has the number of U.S. gun deaths changed over time?

The record 48,830 total gun deaths in 2021 reflect a 23% increase since 2019, before the onset of the coronavirus pandemic .

Gun murders, in particular, have climbed sharply during the pandemic, increasing 45% between 2019 and 2021, while the number of gun suicides rose 10% during that span.

The overall increase in U.S. gun deaths since the beginning of the pandemic includes an especially stark rise in such fatalities among children and teens under the age of 18. Gun deaths among children and teens rose 50% in just two years , from 1,732 in 2019 to 2,590 in 2021.

How has the rate of U.S. gun deaths changed over time?

While 2021 saw the highest total number of gun deaths in the U.S., this statistic does not take into account the nation’s growing population. On a per capita basis, there were 14.6 gun deaths per 100,000 people in 2021 – the highest rate since the early 1990s, but still well below the peak of 16.3 gun deaths per 100,000 people in 1974.

A line chart that shows the U.S. gun suicide and gun murder rates reached near-record highs in 2021.

The gun murder rate in the U.S. remains below its peak level despite rising sharply during the pandemic. There were 6.7 gun murders per 100,000 people in 2021, below the 7.2 recorded in 1974.

The gun suicide rate, on the other hand, is now on par with its historical peak. There were 7.5 gun suicides per 100,000 people in 2021, statistically similar to the 7.7 measured in 1977. (One caveat when considering the 1970s figures: In the CDC’s database, gun murders and gun suicides between 1968 and 1978 are classified as those caused by firearms and explosives. In subsequent years, they are classified as deaths involving firearms only.)

Which states have the highest and lowest gun death rates in the U.S.?

The rate of gun fatalities varies widely from state to state. In 2021, the states with the highest total rates of gun-related deaths – counting murders, suicides and all other categories tracked by the CDC – included Mississippi (33.9 per 100,000 people), Louisiana (29.1), New Mexico (27.8), Alabama (26.4) and Wyoming (26.1). The states with the lowest total rates included Massachusetts (3.4), Hawaii (4.8), New Jersey (5.2), New York (5.4) and Rhode Island (5.6).

A map showing that U.S. gun death rates varied widely by state in 2021.

The results are somewhat different when looking at gun murder and gun suicide rates separately. The places with the highest gun murder rates in 2021 included the District of Columbia (22.3 per 100,000 people), Mississippi (21.2), Louisiana (18.4), Alabama (13.9) and New Mexico (11.7). Those with the lowest gun murder rates included Massachusetts (1.5), Idaho (1.5), Hawaii (1.6), Utah (2.1) and Iowa (2.2). Rate estimates are not available for Maine, New Hampshire, Vermont or Wyoming.

The states with the highest gun suicide rates in 2021 included Wyoming (22.8 per 100,000 people), Montana (21.1), Alaska (19.9), New Mexico (13.9) and Oklahoma (13.7). The states with the lowest gun suicide rates were Massachusetts (1.7), New Jersey (1.9), New York (2.0), Hawaii (2.8) and Connecticut (2.9). Rate estimates are not available for the District of Columbia.

How does the gun death rate in the U.S. compare with other countries?

The gun death rate in the U.S. is much higher than in most other nations, particularly developed nations. But it is still far below the rates in several Latin American countries, according to a 2018 study of 195 countries and territories by researchers at the Institute for Health Metrics and Evaluation at the University of Washington.

The U.S. gun death rate was 10.6 per 100,000 people in 2016, the most recent year in the study, which used a somewhat different methodology from the CDC. That was far higher than in countries such as Canada (2.1 per 100,000) and Australia (1.0), as well as European nations such as France (2.7), Germany (0.9) and Spain (0.6). But the rate in the U.S. was much lower than in El Salvador (39.2 per 100,000 people), Venezuela (38.7), Guatemala (32.3), Colombia (25.9) and Honduras (22.5), the study found. Overall, the U.S. ranked 20th in its gun fatality rate that year .

How many people are killed in mass shootings in the U.S. every year?

This is a difficult question to answer because there is no single, agreed-upon definition of the term “mass shooting.” Definitions can vary depending on factors including the number of victims and the circumstances of the shooting.

The FBI collects data on “active shooter incidents,” which it defines as “one or more individuals actively engaged in killing or attempting to kill people in a populated area.” Using the FBI’s definition, 103 people – excluding the shooters – died in such incidents in 2021 .

The Gun Violence Archive, an online database of gun violence incidents in the U.S., defines mass shootings as incidents in which four or more people are shot, even if no one was killed (again excluding the shooters). Using this definition, 706 people died in these incidents in 2021 .

Regardless of the definition being used, fatalities in mass shooting incidents in the U.S. account for a small fraction of all gun murders that occur nationwide each year.

How has the number of mass shootings in the U.S. changed over time?

A bar chart showing that active shooter incidents have become more common in the U.S. in recent years.

The same definitional issue that makes it challenging to calculate mass shooting fatalities comes into play when trying to determine the frequency of U.S. mass shootings over time. The unpredictability of these incidents also complicates matters: As Rand Corp. noted in a research brief , “Chance variability in the annual number of mass shooting incidents makes it challenging to discern a clear trend, and trend estimates will be sensitive to outliers and to the time frame chosen for analysis.”

The FBI found an increase in active shooter incidents between 2000 and 2021. There were three such incidents in 2000. By 2021, that figure had increased to 61.

Which types of firearms are most commonly used in gun murders in the U.S.?

In 2020, the most recent year for which the FBI has published data, handguns were involved in 59% of the 13,620 U.S. gun murders and non-negligent manslaughters for which data is available. Rifles – the category that includes guns sometimes referred to as “assault weapons” – were involved in 3% of firearm murders. Shotguns were involved in 1%. The remainder of gun homicides and non-negligent manslaughters (36%) involved other kinds of firearms or those classified as “type not stated.”

It’s important to note that the FBI’s statistics do not capture the details on all gun murders in the U.S. each year. The FBI’s data is based on information voluntarily submitted by police departments around the country, and not all agencies participate or provide complete information each year.

Note: This is an update of a post originally published on Aug. 16, 2019.

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Striking findings from 2023, key facts about americans and guns, for most u.s. gun owners, protection is the main reason they own a gun, gun violence widely viewed as a major – and growing – national problem, most popular.

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CBSE Class 10 Maths Case Study Questions for Class 10 Maths Chapter 1
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.
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Show Answer. (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. Show Answer. Case Study 3: Real numbers are an essential concept in mathematics that encompasses both rational and irrational numbers.
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Case Study Questions for Class 10 Maths Chapter 1 Real Numbers. Question 1: 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 ...
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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.
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Maximum capacity of a container, which can measure the petrol in exact number of times. Question 2. Find the value of: (-1) + (-1) 2n + (-l) 2n+1 + (-l) 4n+1 , where n is any positive odd integer. Solution: Question 3. Find whether decimal expansion of 13/64 is a terminating or non-terminating decimal.
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Let the given positive integers be a and b such that a > b. Step 1: Apply Euclid's Division Lemma to find q and r where a = bq + r, 0 ≤ r < b. Step 2: If r = 0, then the HCF is b. If r ≠ 0, apply Euclid's Lemma to b and r. Step 3: Continue the process till the remainder is zero. The divisor at the step in which the remainder is zero is the ...
Important Questions for Class 10 Maths Chapter 1 Real Numbers
Real Numbers Class 10 Important Questions Short Answer-I (2 Marks) Question 6. HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012) Solution: We know, 1st number × 2nd number = HCF × LCM. ⇒ 27 × 2nd number = 9 × 459. ⇒ 2nd number = 9×459 27 = 153.
What the data says about gun deaths in the U.S
The U.S. gun death rate was 10.6 per 100,000 people in 2016, the most recent year in the study, which used a somewhat different methodology from the CDC. That was far higher than in countries such as Canada (2.1 per 100,000) and Australia (1.0), as well as European nations such as France (2.7), Germany (0.9) and Spain (0.6).

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Class 10 Maths Chapter 1 MCQ

Case Study – 1

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

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

36 can be expressed as a product of its primes as

7 х 11 х 13 х 15 + 15 is a

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

Case Study – 2

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

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

The LCM of 60, 84, and 108 is

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

108 can be expressed as a product of its primes as

Case Study – 3

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

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

680 can be expressed as a product of its primes as

2 х 3 х 5 х 11 х 17 + 11 is a

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

Case Study – 4

What will be the value of x?

What will be the value of y?

What will be the value of z?