NCERT Solutions Class 6 Maths Chapter 5 Prime Time Exercise 5.1

Exercise 5.1 of chapter 5 - Prime Time is about prime numbers, composite numbers and factors. It helps you know how numbers are created and how they can be broken down into smaller components. This is a very critical chapter since these topics come up often in maths problems, not only in Class 6, but subsequently as well. 

The questions in this exercise will help you determine what numbers are prime and composite and how to check for common factors. This helps us to develop a better foundation in number systems and facilitates stronger problem-solving skills. 

We want you to be provided with NCERT solutions for all the questions in Exercise 5.1 using the latest NCERT syllabus. These solutions are trustworthy, clear and have systematic steps to follow throughout. These solutions will help you study effectively for your exam or quickly revise before a test.

1.0Download NCERT Solutions Class 6 Maths Chapter 5 Prime Time Exercise 5.1: Free PDF

The solutions for Exercise 5.1 use the latest NCERT syllabus and provide a clear and simple step-by-step method. Download the free PDF from below:

2.0NCERT Solutions Class 6 Chapter 5 Prime Time: All Exercises

3.0NCERT Class 6 Maths Chapter 5 Prime Time Exercise 5.1: Detailed Solutions

• At what number is 'idli-vada' said for the $10^{\text{th }}$ time? Sol. To determine the 10th occurrence of "idli- vada"; we need to identify the numbers that are multiples of both 3 and 5. The numbers for which "idli-vada" is said are the multiples of 15. This sequence is: $15,30,45,60,75,90,105,120,135,150,\dots$ Thus, the $10^{\text{th }}$ time for which players should say "idli-vada" is 150.
• If the game is played for the numbers from 1 till 90, find out: (a) How many times would the children say 'idli' (including the times they say 'idli-vada')? (b) How many times would the children say 'vada' (including the times they say 'idli-vada')? (c) How many times would the children say 'idli-vada'? Sol. (a) Idli is said for multiples of 3 . Between 1 and 90 , the multiples of 3 are $3,6,9,12,15,18,\dots 90$. There are 30 such numbers. Hence the children would say idli 30 times. (b) Vada is said for multiples of 5 . Between 1 and 90 , the multiples of 5 are $5,10,15,20,25,\dots$ There are 18 such numbers. (c) Idli-Vada is said for multiples of both 3 and 5, which is multiple of 15. Between 1 and 90, multiples of 15 are $15,30,45,60,75,90$. So, there are 6 such numbers.
• What if the game was played till 900 ? How would your answers change? Sol. There are 300 multiples of 3 between 1 and 900 and there are 180 multiples of 5 between 1 and 900. There are 60 multiples of 15 between 1 and 900. (a) "idli" is said: 300 times (including the times "idli-vada" is said). (b) "vada" is said: 180 times (including the times "idli-vada" is said). (c) "idli-vada" is said: 60 times.
• Is this figure somehow related to the 'idli- vada' game? (Hint: Imagine playing the game till 30. Draw the figure if the game is played till 60.).
  
  Sol. Yes, this figure is related to the 'idli-vada' game. Figure below for game played till 60.
  

• Let us now play the 'idli-vada' game with different pair of numbers: (a) 2 and 5 Sol.
  
  (b) 3 and 7 Sol.
  
  (c) 4 and 6 Sol.
  
• What jump size can reach both 15 and 30? There are multiple jump sizes possible. Try to find them all. Sol. Factors of 15: 15 can be factored into: $15=3\times 5$ The factors of 15 are: $1,3,5,15$ Factors of 30: 30 can be factored into: $30=2\times 3\times 5$ The factors of 30 are: $1,2,3,5,6,10,15,30$ The common factors between these two lists are: $1,3,5,15$. So, the jump sizes that will allow Jumpy to land on both 15 and 30 are the common factors of 15 and 30. Look at the table below. What do you notice?

• Is there anything common among the shaded numbers? Sol. The shaded numbers in the table are: $33,36,39,42,45,48,51,54,57,60,63,66,69$. These are all multiples of 3 . So, the numbers in shaded boxes are multiples of 3 .
• Is there anything common among the circled numbers? Sol. The circled numbers in the table are: $32,36,40,44,48,52,56,60,64,68$. These are all multiples of 4 . So, the numbers in circles are multiples of 4.
• Which numbers are both shaded and circled? What are these numbers called? Sol. The numbers that are both shaded and circled are: $36,48,60$. These numbers are called common multiples of 12 (3 and 4 both].
• Find all multiples of 40 that lie between 310 and 410 . Sol. Here, multiples of 40 are 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440. Hence, multiples of 40 that lie between 310 and 410 are 320, 360 and 400.
• Who am I? (a) I am a number less than 40 . One of my factors is 7 . The sum of my digits is 8 . (b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other. Sol. (a) The number is 35 [since $3+5=8$ and 35 is divisible by 7]. (b) Common multiples of 3 and 5 are 15, 30, 45, 60, 75, 90, (which are less than 100). And there is one number in which one of digit is 1 more than the other that is 45 . So, I am 45.
• A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are $1,2,4,7,14$ and 28 . Their sum is 56 which is twice 28 . Find a perfect number between 1 and 10 . Sol. The only perfect number between 1 and 10 is 6 . Factors of 6 are 1, 2, 3, 6 Sum of factors/divisors : $1+2+3+6=12$ Since 12 is twice of 6 , hence 6 is a perfect number.
• Find the common factors of: (a) 20 and 28 (b) 35 and 50 (c) 4, 8 and 12 (d) 5, 15 and 25 Sol. (a) Factors of 20 are $1,2,4,5,10,20$ Factors of 28 are 1, 2, 4, 7, 14, 28 Common factors are 1, 2, 4. (b) Factors of 35 are 1, 5, 7, 35 Factors of 50 are 1, 2, 5, 10, 25, 50 Common factors are 1, 5. (c) Factors of 4 are 1, 2, 4 Factors of 8 are 1, 2, 4, 8 Factors of 12 are 1, 2, 3, 4, 6, 12 Common factors are 1, 2, 4. (d) Factors of 5 are 1, 5 Factors of 15 are 1, 3, 5, 15 Factors of 25 are 1,5, 25 Common factors are 1, 5.
• Find any three numbers that are multiples of 25 but not multiples of 50 . Sol. Numbers that are multiples of 25 are $25,50,75,100,125,150,175,\dots$ Numbers that are multiples of 50 are $50,100,150,200,250,300,\dots$ Hence, the numbers that are multiples of 25 but not multiples of 50 are $25,75,125,175,\dots$. So, three numbers are $25,75,125$.
• Anshu and his friends play the 'idli-vada' game with two numbers, which are both smaller than 10. The first time anybody says 'idli-vada' is after the number 50. What could the two numbers be which are assigned 'idli'and 'vada'? Sol. The next number after 50 is 51 since, 51 is multiple of 3 and 17 . But 17 is greater than 10 . The next is 52. 52 is multiple of $2,4,13$ but 13 is greater than 10 and first common multiple of 2 and 4 is 4 . 54 is multiple of $2,3,6,9,18,27,54$. Here, $18,27,54$ are greater than 10 . Out of $2,3,6,9$. First common multiple of $6,9=54$.
• In the treasure hunting game, Grumpy has kept treasures on 28 and 70 . What jump sizes will land on both the numbers? Sol. Factors of $28=1,2,4,7,14,28$ Factors of $70=1,2,5,7,10,14,35,70$ Common factors are 1, 2, 7 and 14. Hence jump sizes which will land at both 28 and 70 are 1, 2, 7 and 14.
• In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.
  
  Sol.
  
• Find the smallest number that is a multiple of all the numbers from 1 to 10 except for 7. Sol. Numbers: 1, 2, 3, 4, 5, 6, 8, 9, 10 Here, we have to find the smallest number which is multiple of above numbers. (i) Number divisible by 8 will be divisible by 2,4 also. (ii) Number divisible by 9 will be divisible by 3 also. (iii) Number divisible by 2 and 3 will be divisible by 6 also (which we have taken into consideration above two points). (iv) Number divisible by 10 will be divisible by 5 and 2 (divisible by 2 is taken into consideration in point (i)). So, required smallest number will be $\Rightarrow 8\times 9\times 5=360$.
• Find the smallest number that is a multiple of all the numbers from 1 to 10 . Sol. Given, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 (i) Number divisible by 8 will be divisible by 2,4 also. (ii) Number divisible by 9 will divisible by 3 also. (iii) Number divisible by 2 and 3 will be divisible by 6 also (which we have taken into consideration above two points). (iv) Number divisible by 10 will be divisible by 5 and 2 (divisible by 2 is taken into consideration in point (i)). (v) Number should also be divisible by 7 . So, the required smallest number should be divisible by 8, 9, 5, 7 $\Rightarrow 8\times 9\times 5\times 7=2520$

4.0Key Features and benefits for Class 6 Maths Chapter 5 Exercise 5.1

• The exercise focuses on concepts such as prime numbers, composite numbers and common factors. 
• All the questions come directly from the revised NCERT syllabus for Class 6 learners. 
• Solving NCERT solutions builds a good foundation for the CBSE exam, and the various forms of end of unit and school based assessment. 
• These solutions also assist in developing one's thinking skills and preparation for maths olympiads.
• Regular practice of these solutions helps develop confidence in the number system and problem solving in general.