CBSE Class 10 Maths 2024-25: Chapter 1 Real Numbers Important Competency-Based Questions with Answers; Download Free PDF

As the CBSE Class 10 board exams get closer, it’s important for students to understand the new exam pattern. Starting in the 2024-25 school year, CBSE will include 50% more competency-based questions. These questions will be both multiple choice and written, focusing on how to use what students have learned in real-life situations.

This article explores Chapter 1: Real Numbers. It highlights key competency-based questions and provides answers to help students succeed.

Understanding Competency-Based Questions in Chapter 1: Real Numbers

Competency-based questions are designed to see how well students can apply their knowledge in everyday life. They can come in different forms, such as case studies, true-false questions, gap-filling tasks, and long or short answer questions.

These questions are different from regular memorization. They encourage students to think critically and solve problems, helping them understand the concepts in Chapter 1: Real Numbers better. Students should connect what they know about real numbers to real-world examples.

CBSE Class 10 Maths Chapter 1: Important Competency-Based Questions

Multiple Choice Questions :

Q1. Which of these is the HCF of 1260 and 1680?

1. 210

2. 420

3. 630

4. 5040

Ans. 2. 420

Q2. Which of these is the LCM of 720 and 900?

1. 180

2. 1800

3. 3600

4. 648000

Ans. 3. 3600

Q3. 63/p has a terminating decimal expansion.

Which of these CANNOT be a factor of p ?

1. 2

2. 5

3. 13

4. 20

Ans. 3. 13

Q4. Which of the following have a terminating decimal expansion?

(Note: You need not evaluate the decimals.)

1. 1/3

2. 1/60

3. 1/90

4. 1/625

Ans. 4. 1/625

Q5. Which of the following fractions has a terminating decimal expansion?

1. 33/343

2. 19/49

3. 71/99

4. 237/625

Ans. 4. 237/625

Q6. Which of the following is an irrational number?

1. 5√4

2. √2/√8

3. 6 + √5

4. √64 - √4

Ans. 3. 6 + √5

Assertion & Reason Type Question :

Q7. Two statements are given below - one labelled Assertion (A) and the other labelled Reason (R). Read the statements carefully and choose the option that correctly describes statements (A) and (R).

Assertion (A): Product of HCF and LCM of THREE numbers is equal to the product of those numbers.

Reason (R): Product of HCF and LCM of TWO numbers is equal to the product of those numbers.

1. Both (A) and (R) are true and (R) is the correct explanation for (A).

2. Both (A) and (R) are true and (R) is not the correct explanation for (A).

3. (A) is false but (R) is true.

4. Both (A) and (R) are false.

Ans. 3. (A) is false but (R) is true.

Free Response Questions :

Q8. The prime factorisation of a natural number p is (5 × 7 × t ) where t ≠ 2, 3.

What is the prime factorisation of 42 p² ?

Ans. Writes the prime factorisation of 42 p² as (2 × 3 × 5² × 7³ × t²).

Q9. Prove that √7 is irrational.

Ans. Assumes √7 = a/b where b ≠ 0, a and b are co-primes.

Writes b √7 = a and squares both the sides to get 7 b² = a².

Concludes that a is divisible by 7 as a² is divisible by 7 because 7 is a prime number.

Writes a = 7 c and squares both the sides to get a² = 49 c².

Replaces a² with 7 b² from step 2 to get 7 b² = 49 c² and solves it to get b² = 7 c².

Concludes that b is divisible by 7 as b² is divisible by 7 because 7 is a prime number.

Mentions that 7 divides both a and b which contradicts the assumption that a and b are both co-prime and hence √7 is irrational.

Q10. The prime factorisation of a natural number k is (3 × 5 × p ) where p ≠ 2. What is the prime factorisation of 10 k² ?

Ans. Writes the prime factorisation of 10 k² as (2 × 3² × 5³ × p²).

Q11. For a positive integer n ,m is a prime factor of n . Show that m is not a factor of ( n + 1).

Ans. Uses Euclid's division lemma and writes the equation for some positive integer p as:

n = mp

=> p = n/m

Assumes that m is a factor of ( n + 1), uses Euclid's division lemma and writes the equation for some positive integer q as:

n + 1 = mq

Rearranges the above equation as:

q = n/m + 1/m

Writes that q is a positive integer, n/m is a positive integer but 1/m is not an integer and hence the above equation cannot be true.

Hence, uses contradiction to conclude that m is not a factor of ( n + 1).

Q12. A rectangular arrangement of pens has rows and columns. Rohan takes away 3 rows of pens and then Sarah takes away 2 columns of pens from the remaining pens. The remaining pens are rearranged in p rows and q columns where p is a prime number.

If Rohan takes 24 pens and Sarah takes 18 pens, find all possible value(s) of p. Show your work.

Ans. Writes that in the original rectangular arrangement, there are 8 pens in each row and 12 pens in each column.

Finds the remaining number of pens as 9 rows and 6 columns or 54 pens.

Writes the prime factorisation of 54 as 2 × 3³.

Writes all the possible values of p as 2 or 3.

Q13. The number 3837425721 is divided by a number between 5621 and 5912.

State true or false for the below statements about the remainder and justify your answer.

(i) The remainder can be more than 5912.

Ans. Writes false and justifies the answer. For example, writes that Euclid's Division Lemma states that the remainder is always less than the divisor and all the divisors are less than 5912.

(ii) The remainder cannot be less than 5621.

Ans. Writes false and justifies the answer. For example, the remainder is always less than the divisor and the numbers from 0 to the divisor are all possible remainders.

(iii) The remainder is always between 5621 and 5912.

Ans. Writes false and justifies the answer. For example, writes that Euclid's Division Lemma states that the remainder always lies between 0 and the divisor

Q14. Ramesh has two rectangular fields of the same length but different widths. He wants to plant 76 trees in the smaller field and 190 trees in the larger field. In both fields, the trees will be planted in the same number of columns but in different numbers of rows. What is the most number of columns that can be planted in this arrangement? Show your work.

Ans. Identifies that the number of columns for the two fields must be HCF of 76 & 190, and applies an appropriate method to find the HCF as 38.

Q15. Use Euclid's Division Algorithm to find the HCF of 175, 225 and 465. Show your work.

Ans. Finds the HCF of 175, 225 and 465 using Euclid's Division Algorithm as follows:

225 = 175 × 1 + 50

175 = 50 × 3 + 25

50 = 25 × 2 + 0

Finds the HCF of 175 and 225 as 25.

465 = 25 × 18 + 15 1

25 = 15 × 1 + 10

15 = 10 × 1 + 5

10 = 5 × 2 + 0

Finds the HCF of 465 and 25 as 5.

Concludes that the HCF of 175, 225 and 465 is 5.

Q16. Ajay has a box of length 3.2 m, breadth 2.4 m, and height 1.6 m. What is the length of the longest ruler that can exactly measure the three dimensions of the box? Show your steps and give valid reasons.

Ans. Identifies and reasons that the length of the longest ruler should be equal to the HCF of the three lengths.

Finds the HCF of the three numbers as

Prime factorization of 32 = 2⁵

Prime factorization of 24 = 3 × 2³

Prime factorization of 16 = 2⁴

Highest Common factor, HCF = 2³

Mentions the length of the longest ruler as 80 cm or 0.8 m.

(Award 0.5 marks if the length is correct but the unit is incorrect).

Q17. m is a positive integer. HCF of m and 450 is 25. HCF of m and 490 is 35. Find the HCF of m, 450 and 490. Show your steps.

Ans. Writes that the HCF of m, 450 and 490 is nothing but the HCF of 25 and 35 and finds the same as:

35 = (25 × 1) + 10

25 = (10 × 2) + 5

10 = (5 × 2) + 0.

Concludes that HCF of m, 450 and 490 is 5.

Q18. Show that any positive even integer is of the form (8 m ) , (8 m + 2), (8 m + 4) or (8 m + 6), for some positive integer m. Show your work.

Ans. Writes Euclid's Division Lemma for a = bm + n, 0 ≤ n < b, where a is a positive 0.5 integer and substitutes b = 8 to get a = 8 m + n, 0 ≤ n < 8.

Mentions that the possible values of n for a = 8 m + n are 0, 1, 2, 3, 4, 5, 6, 7.

Writes that a can be (8 m ) , (8 m + 1 ), (8 m + 2), (8 m + 3), (8 m + 4), (8 m + 5), (8 m + 6) or (8 m + 7) where m is the quotient.

Writes that out of the above expressions only (8 m ) , (8 m + 2), (8 m + 4) and (8 m +  6) are even and concludes that any positive even integer is of the form (8 m ) , (8 m + 2), (8 m + 4) or (8 m + 6).

Case Study Based Question :

Q19. Answer the questions based on the given information.

For the screening of an informational documentary, three schools were selected by the district administration.

♦ During the screening, multiple rooms are used simultaneously, and each room can accommodate an equal number of students.

♦ All students in a particular room belong to the same school.

♦ As a token of appreciation, the district administration has provided an equal number of chocolates to each school.

♦ When distributing these chocolates, each school distributes chocolates equally among its students, ensuring fairness and consistency.

1. Find the maximum number of students that can be seated in one room. Show your work.

Ans. Identifies that to find the required number, HCF of 78, 117, and 130 is needed and 2 finds the HCF of 78, 117, and 130 as:

Prime factorization of 78, 117, and 130-

78 = 2¹ × 3¹ × 13¹

117 = 3² × 13¹

130 = 2¹ × 5¹ × 13¹

Concludes that the maximum number of students to be seated in a room = HCF(78, 117, 130) = 13.

2. What is the minimum number of rooms required? Show your work.

Ans. Finds the total number of students as 78 + 117 + 130 = 325.

Divides the total number of students by 13 to obtain the minimum number of rooms 1 required as 25.

3. What is the minimum number of chocolates provided to each school? Show your work.

Ans. Identifies that LCM of 78, 117, and 130 is the minimum number of chocolates 1 received by each school and uses the prime factorization used earlier to find the LCM of 78, 117, and 130 as:

LCM = 2 × 3 × 3 × 5 × 13 = 1170.

(Note: Award full marks if the student performs prime factorization.)

Q20. Answer the following questions based on the information given.

Sakshara International School organised a combined exhibition for grade 10 students of its three branches. 345 students from Mumbai branch, 405 students from Pune branch and 270 students from Nagpur branch participated in it. The following were planned for the exhibition:

1. Group projects: The students of each of the three branches were divided in groups for making various group projects such that each group had equal number of students and the number of
groups was minimum.

2. Individual project: Each of total 1020 students had to submit an individual project. A fixed number of topics were allotted such that each topic had equal number of students.

3. Inter-state model making competition: A few equal number of students were selected from each branch to participate in the competition. Each branch was supposed to submit between 3 to 7 models.

1. Use Euclid's Division Algorithm to find the number of different groups for the group projects. Show your work.

Ans. Finds HCF of 345, 405 and 270 using Euclid's Division Algorithm as follows:

405 = 345 × 1 + 60

345 = 60 × 5 + 45

60 = 45 × 1 + 15

45 = 15 × 3 + 0

Finds HCF of 405 and 345 as 15.

Finds HCF of 270 and 15 as follows: 0.5

270 = 15 × 18 + 0

Concludes that HCF of 345, 405 and 240 is 15, hence there were 15 students in each group.

Finds the number of different groups for the group projects as:

345/15 + 405/15 + 270/15 = 23 + 27 + 18 = 68

2. A maths teacher asked his students to solve the below puzzle regarding the individual project.

The number of students who got the same topic can be represented as (2ⁿ × 5) where n is a positive integer having the maximum possible value. Find n and the number of topics allotted. Show your work.

Ans. Factorises 1020 as: 2² × 3 × 5 × 17.

From the above factorisation concludes that 2ⁿ × 5 = 2² × 5 and hence finds the maximum possible value of n as 2.

Finds the number of topics allotted as the remaining factors of 1020 as:

3 × 17 = 51

3. Mumbai branch divided the students selected for inter-state model making competition into the groups of 12 students, Pune into the groups of 10 students and Nagpur into the groups of 15 students.

(i). How many students were selected from each branch?

Ans. Finds LCM of 12, 10 and 15 using prime factorisation as: 

12 = 2 × 2 × 3

10 = 2 × 5

15 = 3 × 5

LCM = 2² × 3 × 5 = 60

Concludes that 60 students were selected from each branch for the inter-state model making competition.

(ii) How many models were submitted by individual branches and all the branches together? Show your work.

Ans. Finds the number of models submitted by each branch as:

Mumbai = 60/12 = 5

Pune = 60/10 = 6

Nagpur = 60/15 = 4

Finds the total number of models submitted as 15.

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