NCERT Solutions for Class 6 Maths Chapter 5 Prime Time Ex 5.5

Exercise 5.5

1. 2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. 

a. From the year you were born till now, which years were leap years? 

b. From the year 2024 till 2099, how many leap years are there? 

Ans:

a. From the year you were born till now, which years were leap years?
(Assuming you were born in 1990)
The leap years from 1990 to 2024 are:
1992, 1996, 2000, 2004, 2008, 2012, 2016, 2020, 2024.

b. From the year 2024 till 2099, how many leap years are there?
Leap years occur every 4 years. The leap years between 2024 and 2099 are:
2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096.
There are 19 leap years from 2024 to 2099.

2. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes. 

Ans:

• A palindrome is a number that reads the same forward and backward (e.g., 1221, 1441).

• A number is divisible by 4 if the last two digits form a number divisible by 4.

• Smallest 4-digit palindrome divisible by 4:
  The smallest 4-digit palindrome divisible by 4 is 1221 (last two digits "21" is divisible by 4).

• Largest 4-digit palindrome divisible by 4:
  The largest 4-digit palindrome divisible by 4 is 9229 (last two digits "29" is divisible by 4).

3. Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning.

a. Sum of two even numbers gives a multiple of 4. 

b. Sum of two odd numbers gives a multiple of 4. 

Ans:

a. Sum of two even numbers gives a multiple of 4.

• Sometimes true.
  For example, 2 + 6 = 8 (multiple of 4), but 2 + 4 = 6 (not a multiple of 4).

b. Sum of two odd numbers gives a multiple of 4.

• Sometimes true.
  For example, 3 + 5 = 8 (multiple of 4), but 1 + 3 = 4 (not a multiple of 4).

4. Find the remainders obtained when each of the following numbers are divided by a) 10, b) 5, c) 2. 

78, 99, 173, 572, 980, 1111, 2345 

Ans:

78:

• a) 10: remainder 8

• b) 5: remainder 3

• c) 2: remainder 0

99:

• a) 10: remainder 9

• b) 5: remainder 4

• c) 2: remainder 1

173:

• a) 10: remainder 3

• b) 5: remainder 3

• c) 2: remainder 1

572:

• a) 10: remainder 2

• b) 5: remainder 2

• c) 2: remainder 0

980:

• a) 10: remainder 0

• b) 5: remainder 0

• c) 2: remainder 0

1111:

• a) 10: remainder 1

• b) 5: remainder 1

• c) 2: remainder 1

2345:

• a) 10: remainder 5

• b) 5: remainder 0

• c) 2: remainder 1

5. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be? 

Ans: To check divisibility, Guna only needs to check 5 and 8:

• Divisibility by 5: The number ends in 0, so it's divisible by 5.

• Divisibility by 8: The last three digits (560) are divisible by 8.

Since 14560 is divisible by both 5 and 8, it is divisible by all of 2, 4, 5, 8, and 10.

6. Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160. 

Ans:

572: Divisible by 2 and 4, but not by 5, 8, or 10.

2352: Divisible by 2, 4, and 8, but not by 5 or 10.

5600: Divisible by 2, 4, 5, 8, and 10.

6000: Divisible by 2, 4, 5, 8, and 10.

77622160: Divisible by 2, 4, 5, 8, and 10.

7. Write two numbers whose product is 10000. The two numbers should not have 0 as the unit digit.

Ans: 625 × 16 = 10000.

Benefits of NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.5 Prime Time

• Clear Understanding of Prime Numbers: The solutions explain concepts like prime numbers, composite numbers, and divisibility rules in simple steps, making it easy for students to understand.

• Step-by-Step Problem Solving: Each problem is solved in a step-by-step manner, helping students learn how to approach different types of questions related to prime numbers.

• Practice for Exams: By using these solutions, students can practice key questions that are often asked in exams, which helps build confidence.

• Improved Problem-Solving Skills: The exercises challenge students to apply their knowledge and think critically, enhancing their overall problem-solving abilities.

• Aligned with CBSE Syllabus: These solutions follow the CBSE curriculum closely, ensuring that students are learning what is required for their exams.

• Helps with Homework: The solutions provide clear answers to each exercise, making it easier for students to complete their homework on time.

Class 6 Maths Chapter 5: Exercises Breakdown

Important Study Material Links for Class 6 Maths Chapter 5 - Prime Time

Conclusion

The NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.5, Prime Time, offer an excellent resource for students to master the concepts of prime factorisation and the highest common factor (HCF). With clear, step-by-step explanations, these solutions simplify complex problems, making it easier for students to learn and apply mathematical concepts. Aligned with the CBSE syllabus and available for FREE PDF download, these solutions ensure that students are well-prepared for their exams and gain a strong foundation in number theory, essential for future mathematical studies.

Chapter-Specific NCERT Solutions for Class 6 Maths

The chapter-wise NCERT Solutions for Class 6 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Related Important Links for Maths Class 6

Along with this, students can also download additional study materials provided by Vedantu for Maths Class 6.