[New 2025-26] Chapter 3 - Number Play NCERT Solution | Worksheet | Extra Questions for Class 6 Maths Ganita Prakash.
Class 6 maths chapter 4 full chapter : Number Play.
'Number Play' is an important topic to understand in mathematics. In this chapter, we will explore some of the most basic ideas of mathematics including usages of numbers to convey information, make and discover patterns, estimate magnitudes, pose and solve puzzles, and play and win games. These ideas form the building blocks of ‘mathematics’, and will help us in understanding more advanced topics in mathematics such as the number problems and analysis of different number sequencing...
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# THIS PAGE INCLUDES :
# Notes : Number Play.
Number play class 6 notes.
# Introduction :
Numbers Play an important role in our daily life in many ways to organise our lives. We have used them in counting and in basic mathematical operation like addition, subtraction, multiplication and division to solve everyday problems.
Examples of Situations where we use Numbers :
● Reading Time : We use numbers to read time on a clock or watch.
● Counting Money : We use numbers to count money for purchase and payments.
● Measuring Quantity : We use numbers to measure quantity while purchasing fruits and vegetable and while cooking.
● Measuring Distance : We use numbers to measure the distance between two places.
● Counting Scores : We use numbers to count score while playing and watching sports.
⬇ Read more :
Number play class 6 notes cbse
# Supercells :
A supercell is a cell in which the number written in larger than the numbers written in its neighboring cells.
E.g.
Look at the grid given below with the numbers 60, 45, 90, 52 and 83.
Here, you can observe that the number 60 is larger than its neighboring number 45, So it is a supercell number. Similarly, number 90 and 83 are also greater than their neighboring numbers (45 and 52), so they are also the supercell numbers.
In case of Large grid :
let's do supercell activity with more rows.
Here, the number 7500, 9870, 8632 and 6034 are the supercell numbers as these numbers are greater than their neighboring numbers.
📌Note : Here, the neighboring cells are those that are written immediately to the left, right, top and bottom.
# Patterns of numbers on the Number Line :
We are quite familiar with number lines. So, Let’s see if we can place some numbers in their appropriate positions on the number line. Here are the numbers: 1500, 2754, 3150, 6670, 5350, 7260, 1050, 4460, 9590, 9950 and 8400.
Here :
- 1050 would be placed slightly after 1000 but before 2000.
- 2754 would be placed slightly near 3000.
- 9590 would be placed close to 10,000.
📌Note : Other given numbers can also be represented same way on the number line.
# Playing with Digits :
1, 2, 3, 4, ... and so on are called counting numbers. Digits 0 to 9 can be written in the form of one digit number, two digit number, three digit number, ... and so on.
● 1 Digit numbers : The numbers form 1 to 9 are called one digit numbers and there are 9 one-digit numbers in total.
● 2 Digit numbers : The numbers form 10 to 99 are called two digit numbers and there are 90 two-digit numbers in total.
Calculation : To find number of 2 digit numbers, we need to subtract number of one-digit numbers from the largest 2-digit number. (99 - 9 = 90)
- Largest 2-digit number = 99
- Number of 1-digit numbers = 9
- Number of 2-digit numbers = 99 - 9 = 90
● 3 Digit numbers : The numbers form 100 to 999 are called three-digit numbers and there are 900 three-digit numbers in total.
Calculation : To find number of 3 digit numbers, we need to subtract sum of one-digit numbers and two-digit numbers from the largest 3-digit number. [ 999 - (90 + 9) = 900 ]
- Largest 3-digit number = 999
- Number of 1-digit numbers = 9
- Number of 2-digit numbers = 90
- Sum of 1-digit and 2-digit numbers = 90 + 9 = 99
- Number of 3-digit numbers = 999 - 99 = 900
⏺ Digit sums of numbers :
Sometimes when we add digits of certain numbers, we get the same sum.
E.g.
- 68 : 6 + 8 = 14
- 176 : 1 + 7 + 6 = 14
- 545 : 5 + 4 + 5 = 14
All these numbers have same digit sum i.e. 14.
⏺ Digit Detectives :
How many times will the digit ‘7’ occur? Among the numbers 1–100 and between 1–1000.
● Among 1 - 100 : The number of times 7 occur in 1 to 100 is 20 times.
Calculation : Number 7 comes on 7th, 17th, 27th, 37, 47, 57, 67, 70 - 79 (11 times), 87, 97 position.
● Among 1 - 1000 : The number of times 7 occur in 1 to 1000 is 300 times.
Calculation : Number 7 comes on 7th, 17th, 27th, 37, 47, 57, 67, 70 - 79 (10 times), 87, 97 position.
- Number 7 at Unit place (1 - 100 = 10 times x 10) = 100
- Number 7 at Tens place (1 - 100 = 10 times x 10) = 100
- Number 7 at Hundreds place (700 - 799 = 100 times) = 100
- Number of times 7 comes among 1-1000 = 300
# Pretty Palindromic Patterns :
Numbers which can be read from right to left and left to right similarly are called palindromic numbers.
E.g. 454, 555, 919, etc.
⏺ Reverse and Add Palindromes :
We can get a palindromic number when we add a number to its reverse number.
E.g.
When we add 34 to its reverse number i.e. 43 we get (34 + 43) = 77 which is a palindromic number.
Similarly,
When we add 29 to its reverse number i.e. 92 we get (29 + 92) = 121 which is also a palindromic number.
But,
When we add 48 to its reverse number i.e. 84 we get (48 + 84) = 132 which is not a palindromic number. so here we need to add it again with its reverse number that is 231 to get (132 + 231) = 363 which is a palindromic number.
Lets try again with number 76 :
When we add 76 to its reverse number i.e. 67 we get (76 + 67) = 143 which is not a palindromic number. so here we need to add it again with its reverse number that is 341 to get (143 + 341) = 484 which is a palindromic number.
⏺ Who am I ?
I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘u’ digit.
My ‘h’ digit is double of my ‘t’ digit.
Solution :
The number must be odd so we can choose only 1, 3, 5, 7 and 9 at unit place.
Choosing 1 at unit place :
- Unit place = 1
- Tens place is double of unit place = 2
- Hundreds place is double of tens place = 4
- To make palindrome we need to reverse the unit and tens place.
- Therefore,
- The number is 12421.
# The Magic Number of Kaprekar : 6174
When we subtract any 4-digit smallest number from 4-digit largest number (made up of same digits) we will always end up getting the number 6174. It is a magical phenomenon discovered by a mathematic teacher named D.R. Kaprekar in 1949.
Method :
E.g.
Let's take the number 5683.
Round 1 : The given number is 5683.
Largest number = 8653
Smallest number = 3568
Difference = 8653 - 3568 = 5085.
Round 2 : The given number is 5085.
Largest number = 8550
Smallest number = 0558
Difference = 8550 - 0558 = 7992.
Round 3 : The given number is 7992.
Largest number = 9972
Smallest number = 2799
Difference = 9972 - 2799 = 7173.
Round 4 : The given number is 7173.
Largest number = 7713
Smallest number = 1377
Difference = 7713 - 1377 = 6354.
Round 5 : The given number is 6354.
Largest number = 6543
Smallest number = 3456
Difference = 6543 - 3456 = 3087.
Round 6 : The given number is 3087.
Largest number = 8703
Smallest number = 0378
Difference = 8703 - 0378 = 8352.
Round 7 : The given number is 8352.
Largest number = 8532
Smallest number = 2358
Difference = 8532 - 2358 = 6174.
Therefore, the number 5683 takes 7 round to reach 6174 which is the Kaprekar constant.
📌Note : The number 6174 is called the 'Kaprekar constant'.
# Clock and Calendar Numbers :
Clock and Calendar are not just a tool to tell time or date, They also have interesting patterns hidden in their numbers.
⏺ Clock Patterns :
A 12 hour clock presents interesting opportunities to find patterns in time.
E.g.
- 11:11 - Here, digits are identical.
- 10:10 - Here, digits repeat in mirrored fashion.
- 12:21 - Here, digits make palindrome (reads the same forward and backward)
⏺ Calendar Patterns :
Certain dates stand out because their digits repeat or follow a particular sequence.
E.g.
- 20/12/2012 - Digits repeat in order.
- 11/02/2011 - Digits make palindrome (reads the same forward and backward).
# Mental Math :
The numbers in the middle column are added in different ways to get the numbers on the sides.
E.g.
- 3400 = 1500 + 1500 + 400
- 38,800 = 25,000 + 400 x 2 + 13000
- 3400 = 1500 + 1500 + 400.
Adding and Subtracting :
Using the numbers given in the boxes, make a number by adding and subtracting these numbers.
E.g.
- 39,800 = 40,000 - 800 + 300 + 300.
- 45,000 = 40,000 + 12,000 - 7000.
- 5,900 = 7000 - 800 - 300.
- 17,500 = 12,000 + 7000 - 1500.
- 21,400 = 12,000 + 12,000 - 1500 - 800 - 300.
Number play chapter 3 class 6 notes
# Playing with Number Patterns :
When a set of numbers are arranged in a pattern, there are often quicker ways to sum them rather than adding each number individually. Recognizing pattern can help simplify and speed up calculations.
From the Book (NCERT)
Here are some numbers arranged in some patterns. Find out the sum of the numbers in each of the below figures.
Pattern 1 :
Numbers Involved : 40 and 50
Quick Calculation : We can notice that there are 3 horizontal strips of number 40 and in each strips 40 is written 4 times. Similarly there are 2 horizontal strips of number 50 and in each strips 50 is written 5 times. So, by counting the total number of 40 and 50 we can easily get the sum of these numbers.
Solution :
Number of horizontal strips of 40 = 3
Number of times 40 written in each strip = 4
Number of times 40 written in all strips = 4 x 3 = 12
Therefore,
Sum of all 40's written in the figure pattern = 12 x 40 = 480
Similarly :
Number of horizontal strips of 50 = 2
Number of times 50 written in each strip = 5
Number of times 50 written in all strips = 5 x 2 = 10
Therefore,
Sum of all 50's written in the figure pattern = 10 x 50 = 500
Now :
Sum of all the numbers given in the figure is : 480 + 500 = 980
Pattern 2 :
Numbers Involved : Squares filled with dots.
Quick Calculation : We can notice that there are small square which are filled with dots. So, by counting the total number of square we can easily get the sum of dots.
Solution :
Step 1 :
Number of small square in a horizontal strips = 8
Number of small square in a Vertical strips = 8
Total number of small squares in the given large Square (8 x 8) = 64
Step 2 :
Number of small square having 5 dots (5 x 4) = 20
Total Number of dots in small square having 5 dots (20 x 5) = 100
Step 3 :
Number of small square having 1 dots (64 - 20) = 44
Therefore :
Total Number of dots in all small square (100 + 44) = 144
Pattern 3 :
Numbers Involved : 32 and 64.
Quick Calculation : We can notice that there are some square which are filled with 32 and 64. So, by counting the total number of square we can easily get the sum of numbers.
Solution :
Step 1 :
Number of square in a horizontal strips of 32 = 8
Number of square in a Vertical strips of 32 = 4
Total number of squares with number 32 (8 x 2) = 32
Sum of number 32 = 32 x 32 = 1024
Step 2 :
Number of square with 64 (3 x 4 + 4) = 16
Sum of number 64 = 16 x 64 = 1024
Therefore :
Sum of number 32 and 64 in the given figure pattern = 1024 + 1024 = 2048
Pattern 4 :
Numbers Involved : Squares filled with 3 dots and 4 dots.
Quick Calculation : We can notice that there are small square which are filled with dots. So, by counting the total number of square we can easily get the sum of dots.
Solution :
Step 1 :
Number of square in horizontal line = 5
Number of square in vertical line = 7
Total number of squares = 35
Step 2 :
Number of square filled with 3 dots = 17
Number of dots in 3 dots squares (17 x 3) = 51
Step 3 :
Number of square filled with 4 dots (35 - 17) = 18
Number of dots in 4 dots squares (18 x 4) = 72
Therefore :
Total Number of dots in all square (51 + 72) = 123
Pattern 5 :
Numbers Involved : 15, 25 and 35.
Quick Calculation : We can notice that these numbers are not in any pattern. So, here we need to count the number of times each number appear in the given figure pattern to find the sum of all the numbers.
Solution :
Step 1 :
Number of times 15 appear = 22
Sum of number 15 = 15 x 22 = 330
Step 2 :
Number of times 25 appear = 22
Sum of number 25 = 25 x 22 = 550
Step 3 :
Number of times 35 appear = 22
Sum of number 35 = 35 x 22 = 770
Therefore :
Sum of number 15, 25 and 35 in the given figure pattern = 330 + 550 + 770 = 1650.
Pattern 6 :
Numbers Involved : 125, 250, 500 and 1000.
Quick Calculation : We can notice that these numbers are not in any pattern. So, here we need to count the number of times each number appear in the given figure pattern to find the sum of all the numbers.
Solution :
Step 1 :
Number of times 125 appear = 18
Sum of number 125 = 18 x 125 = 2250
Step 2 :
Number of times 250 appear = 8
Sum of number 250 = 250 x 8 = 2000
Step 3 :
Number of times 500 appear = 4
Sum of number 500 = 500 x 4 = 2000
Step 4 :
Number of times 1000 appear = 1
Sum of number 1000 = 1000 x 1 = 1000
Therefore :
Sum of number 125, 250, 500 and 1000 in the given figure pattern = 2250 + 2000 + 2000 + 1000 = 7250.
# The Collatz Conjecture :
The Collatz Conjecture is a mathematical sequence that remains unsolved.
Here is how the sequence works :
Sequence :
- 12, 6, 3, 10, 5, 16, 8, 4, 2, 1.
- 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
- 21, 64, 32, 16, 8, 4, 2, 1.
- 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Rule Applied :
- Even number : Divide by 2.
- Odd number : Multiply by 3 and add 1.
📌 NOTE : Each sequence, no matter what number you start with, eventually reaches 1. This observation is called Collatz's conjecture.
# Simple Estimation :
Estimation helps when you don't need the exact number and just want quick idea. For example, if you are estimating how many books are their in school library, you might guess based on what you see rather than counting each one.
⏺ Estimating the number of books in the school library :
You first need to count exact number of books in one shelf. Then, count the number of shelfs in the library. After that just multiply the number of books in one shelf with the number of shelfs in the library. Here you will get the estimated number of books in school library.
📌Note : Exact number of books in the library will be different as the number of books in each shelf might be different.
⏺ Estimating the number of students in your school :
You first need to count exact number of students in one section of your class. Then, count the number of sections of your class. After that just multiply the number of students in one section with the number of section of your class. Here you will get the estimated number of students in one class. Now you just need to multiply this number with the number of class your school have and you will get the estimated number of students in your school.
📌Note : Exact number of students in the school will be different as the number of students in each class might be different.
----- The End -----
# Key Points :
● Numbers can be used for many different purposes, including to convey information, make and discover patterns, estimate magnitudes, pose and solve puzzles, and play and win games.
● Thinking about and formulating set procedures to use numbers for these purposes is a useful skill and capacity (called “computational thinking”).
● Many problems about numbers can be very easy to pose, but very difficult to solve. Indeed, numerous such problems are still unsolved (e.g. Collatz’s Conjecture).
# NCERT Solutions :
Ganita Prakash class 6 maths chapter 3 number play Try answering ncert page 56.
Ncert class 6 maths chapter 3 solutions.
Did you figure out what these numbers represent?
A child says ‘1’ if there is only one taller child standing next to them. A child says ‘2’ if both the children standing next to them are taller. A child says ‘0’, if neither of the children standing next to them are taller. That is each person says the number of taller neighbors they have.
Try answering the questions below and share your reasoning:
Q1.
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution :
No, the children standing at the end can not say 2 because they can have only one neighbour.
Q2.
Can we arrange the children in a line so that all would say only 0s?
Solution :
Yes, if all the children in the line are of the same height, then they would say '0' because none of them would have a taller neighbour.
Q3.
Can two children standing next to each other say the same number?
Solution :
Yes, two children standing next to each other can say the same number if they have the same number of taller neighbors. For example, if both have one taller neighbor, they will both say '1'.
Q4.
There are 5 children in a group, all of different heights. Can they stand such that four of them say ‘1’ and the last one says ‘0’? Why or why not?
Solution :
Yes, it is possible only if the 5 children are standing in ascending order of their heights.
Q5.
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Solution :
No, this is not possible because the last child doesn't have a taller neighbour and thus can never say '1'.
Q6.
Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
Solution :
Yes, it is possible if the shortest child is in the middle, with the height increasing symmetrically as you move outwards from the middle (centre).
Q7.
How would you rearrange the five children so that the maximum number of children say ‘2’?
Solution :
Yes, it is possible if the tall and short children arrange themselves alternately.
Ganita Prakash class 6 maths chapter 3 number play figure it out ncert page 57.
Class 6th math chapter 3 number play.
Q1.
Colour or mark the supercells in the table below.
Solution :
Q2.
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.
Solution :
Q3.
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
Solution :
Q4.
Out of the 9 numbers, how many supercells are there in the table above? 5
Q5.
Find out how many supercells are possible for different numbers of cells.
Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy.
Solution :
For 'n' cells :
● If 'n' is even, the maximum number of supercells is given by n/2 .
● If 'n' is odd, the maximum number of supercells is given by (n+1)/2 .
Page 58 : Figure it out.
Q6.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution :
No, it is not possible because there will always be at least one number that is greater than its neighboring cell unless all the numbers are identical.
Q7.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution :
The cell with the largest number will always be a supercell because it is larger than its neighboring cells. On the other hand, the cell with the smallest number can never be a supercell because it is always smaller than its neighboring cells.
Q8.
Fill a table such that the cell having the second largest number is not a supercell.
Solution :
● Largest number = 850
● Second largest number = 730
Q9.
Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
Solution :
Yes, it is possible. here is one possible arrangement :
● Largest number = 750
● Second largest number = 640
● Smallest number = 100
● Second smallest number = 210
Q10.
Make other variations of this puzzle and challenge your classmates.
Solution :
Fill a table such that the cell having the largest number, the second largest number and the second smallest number is a supercell.
Ganita Prakash class 6 maths chapter 3 number play Figure it out ncert page 59.
Number play class 6 solutions new book cbse.
Q1.
Identify the numbers marked on the number lines below, and label the remaining positions.
Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Solution :
Ganita Prakash class 6 maths chapter 3 number play Figure it out ncert page 60.
Number play chapter 3 class 6 solutions.
Q1.
Digit sum 14
- a. Write other numbers whose digits add up to 14.
- b. What is the smallest number whose digit sum is 14?
- c. What is the largest 5-digit whose digit sum is 14?
- d. How big a number can you form having the digit sum 14? Can you make an even bigger number?
Solution :
a.
- 59 : 5 + 9 = 14
- 77 : 7 + 7 = 14
- 86 : 8 + 6 = 14
- 95 : 9 + 5 = 14
- 149 : 1 + 4 + 9 = 14
- 257 : 2 + 5 + 7 = 14
- 653 : 6 + 5 + 3 = 14
b. 59 is the smallest number whose sum is 14. (5 + 9 = 14)
c. 95000 is the largest 5 digit number whose digit sum is 14. (9 + 5 + 0 + 0 + 0 = 14)
d. We can form an infinitely big number with digit sum 14 by adding zeros to the number 95.
Q2.
Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Solution :
| 41 - 50 | 51 - 60 | 61 - 70 |
|---|---|---|
| 4 + 1 = 05 | 5 + 1 = 06 | 6 + 1 = 07 |
| 4 + 2 = 06 | 5 + 2 = 07 | 6 + 2 = 08 |
| 4 + 3 = 07 | 5 + 3 = 08 | 6 + 3 = 09 |
| 4 + 4 = 08 | 5 + 4 = 09 | 6 + 4 = 10 |
| 4 + 5 = 09 | 5 + 5 = 10 | 6 + 5 = 11 |
| 4 + 6 = 10 | 5 + 6 = 11 | 6 + 6 = 12 |
| 4 + 7 = 11 | 5 + 7 = 12 | 6 + 7 = 13 |
| 4 + 8 = 12 | 5 + 8 = 13 | 6 + 8 = 14 |
| 4 + 9 = 13 | 5 + 9 = 14 | 6 + 9 = 15 |
| 4 + 10 = 14 | 5 + 10 = 15 | 6 + 10 = 16 |
Observation : The digit sum of number is increased by 1.
Q3.
Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue?
Solution :
123 : 1 + 2 + 3 = 6
234 : 2 + 3 + 4 = 9
345 : 3 + 4 + 5 = 12
456 : 4 + 5 + 6 = 15
567 : 5 + 6 + 7 = 18
678 : 6 + 7 + 8 = 21
789 : 7 + 8 + 9 = 24
The digit sum are 6, 9, 12, 15, 18, 21, 24. Yes, there is a pattern. The digit sum of three consecutive numbers are multiple of 3 or Each successive number is formed by adding 3 to the predecessor. Yes, this pattern will continue for valid 3 digit consecutive numbers.
Ganita Prakash class 6 maths chapter 3 number play Figure it out ncert page 64.
Number play class 6 with answers.
Q1.
Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4-digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4–digits to make:
- a. the difference between the largest and smallest numbers greater than 5085.
- b. the difference between the largest and smallest numbers less than 5085.
- c. the sum of the largest and smallest numbers greater than 9779.
- d. the sum of the largest and smallest numbers less than 9779.
Solution :
a. Digits : 9, 6, 5, 3.
Largest number = 9653
Smallest number =3569
Difference = 9653-3569 = 6084.
b. Digits : 8, 7, 5, 4.
Largest number = 8754
Smallest number = 4578
Difference = 8754 - 4578 = 4176.
c. Digits : 9, 8, 7, 1.
Largest number = 9871
Smallest number = 1789
Sum = 9871 + 1789 = 11660.
d. Digits : 8, 3, 2, 1.
Largest number = 8321
Smallest number = 1238
Sum = 8321 + 1238 = 9559.
Page 65 : Figure it out.
Q2.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution :
The largest 5 digit palindrome = 99999
The smallest 5 digit palindrome = 10001
Sum = 99999 + 10001 = 110000.
Difference = 99999 - 10001 = 89998.
Q3.
The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?
Solution :
The next palindromic time is 11:11 which is 70 minutes after 10:01. The one after that is 12:21 which is again 70 minutes after 11:11.
Q4.
How many rounds does the number 5683 take to reach the Kaprekar constant?
Solution :
Round 1 : The given number is 5683.
Largest number = 8653
Smallest number = 3568
Difference = 8653 - 3568 = 5085.
Round 2 : The given number is 5085.
Largest number = 8550
Smallest number = 5058
Difference = 8550 - 5058 = 3492.
Round 3 : The given number is 3492.
Largest number = 9432
Smallest number = 2349
Difference = 9432 - 2349 = 7083.
Round 4 : The given number is 7083.
Largest number = 8730
Smallest number = 3078
Difference = 8730 - 3078 = 5652.
Round 5 : The given number is 5652.
Largest number = 6552
Smallest number = 2556
Difference = 6552 - 2556 = 3996.
Round 6 : The given number is 3996.
Largest number = 9963
Smallest number = 3699
Difference = 9963 - 3699 = 6264.
Round 7 : The given number is 6264.
Largest number = 6642
Smallest number = 2466
Difference = 6642 - 2466 = 4176.
Round 8 : The given number is 4176.
Largest number = 7641
Smallest number = 1467
Difference = 7641 - 1467 = 6174.
Therefore, the number 5683 takes 8 round to reach 6174 which is the Kaprekar constant.
Ganita Prakash class 6 maths chapter 3 number play Figure it out ncert page 66.
Number play class 6 solutions new book
Q1.
Write an example for each of the below scenarios whenever possible.
Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Solution :
a. 5 digit + 5 digit to give a 5 digit sum more than 90,250.
1st 5 digit number = 60,000.
2nd 5 digit number = 31,000.
Sum = 60,000 + 31,000 = 91,000.
Here, 91,000 is greater than 90,250.
b. 5 digit + 3 digit to give a 6 digit sum.
5 digit number = 99,900.
3 digit number = 900.
Sum = 99,900 + 900 = 1,00,800.
c. 4 digit + 4 digit to give a 6 digit sum.
It is not possible to get a 6 digit number by adding two 4 digit numbers because the maximum possible sum of two 4 digit numbers would be 9,999 + 9,999 = 19,998.
d. 5 digit + 5 digit to give a 6 digit sum.
1st 5 digit number = 90,000.
2nd 5 digit number = 11,000.
Sum = 90,000 + 11,000 = 1,01,000.
e. 5 digit + 5 digit to give 18,500.
It is not possible to get a 18,500 by adding two 5 digit numbers because the smallest sum of two 5 digit numbers would be 10,000 + 10,000 = 20,000 which is greater then 18,500.
f. 5 digit - 5 digit to give a difference less than 56,503.
1st 5 digit number = 80,000.
2nd 5 digit number = 40,000.
Difference = 80,000 - 40,000 = 40,000.
Here, 40,000 is less than 56,503.
g. 5 digit - 3 digit to give a 4 digit difference.
5 digit number = 10,600.
3 digit number = 900.
Difference = 10,600 - 700 = 9,900.
h. 5 digit - 4 digit to give a 4 digit difference.
5 digit number = 15,000.
4 digit number = 7,000.
Difference = 15,000 - 7,000 = 8,000.
i. 5 digit - 5 digit to give a 3 digit difference.
1st 5 digit number = 25,700.
2nd 5 digit number = 25,000.
Sum = 25,700 + 25,000 = 700.
j. 5 digit - 5 digit to give 91,500.
It is not possible to get 91,500 by subtracting two 5 digit numbers because if we subtract smallest 5 digit number (10,000) from largest 5 digit number (99,999) we get 89,999 which is not equals to 91,500.
Page 67 : Figure it out.
Q2.
Always, Sometimes, Never? Below are some statements. Think, explore and find out if each of the statement is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning; discuss this with the class.
- a. 5-digit number + 5-digit number gives a 5-digit number
- b. 4-digit number + 2-digit number gives a 4-digit number
- c. 4-digit number + 2-digit number gives a 6-digit number
- d. 5-digit number – 5-digit number gives a 5-digit number
- e. 5-digit number – 2-digit number gives a 3-digit number
Solution :
a.
The given statement is only sometimes true because when we add two 5 digit numbers (50,000 + 40,000) it gives 90,000 which is a 5 digit number. But, if we add two 5 digit numbers (60,000 + 40,000) it give 1,00,000 which is not a 5 digit number.
b.
The given statement is only sometimes true because when we add 4 digit numbers with a 2 digit number (9,000 + 40) it gives 9,040 which is a 4 digit number. But, if we add 4 digit number with 2 digit number (9,990 + 40) it give 10,030 which is not a 4 digit number.
c.
The given statement can never true because when we add a 4 digit number with 2 digit number (9,999 + 99) it gives 10,098. which is not a 6 digit number.
d.
The given statement is only sometimes true because when we subtract two 5 digit numbers (50,000 - 40,000) it gives 10,000 which is a 5 digit number. But, if we subtract two 5 digit numbers (50,000 - 49,000) it give 1000 which is not a 5 digit number.
e.
The given statement can never true because when we subtract 5 digit number with 2 digit number (10,000 - 99) it gives 9,901. which is not a 3 digit number.
Ganita Prakash class 6 maths chapter 3 number play figure it out ncert page 69.
Number play class 6 question answer
We shall do some simple estimates. It is a fun exercise, and you may find it amusing to know the various numbers around us. Remember, we are not interested in the exact numbers for the following questions.
Share your methods of estimation with the class.
Q1.
Steps you would take to walk :
- a. From the place you are sitting to the classroom door.
- b. Across the school ground from start to end.
- c. From your classroom door to the school gate.
- d. From your school to your home.
Solution :
a. 25 steps.
b. 500 steps.
c. 800 steps.
d. 10,000 steps.
Q2.
Number of times you blink your eyes or number of breaths you take:
- a. In a minute
- b. In an hour
- c. In a day
Solution :
a. I blink my eyes about 20 times and I breadth about 15 times in a minute.
b. I blink my eyes about 20 x 60 = 1200 times and I breadth about 15 x 60 = 900 times in an hour.
c. I blink my eyes about 1200 x 16 = 19,200 times and I breadth about 900 x 24 = 21,600 times in an hour.
📌Note : Here, we take 8 hours sleep so we blink our eyes only 16 hours in a day.
Q3.
Name some objects around you that are :
- a. A few thousand in number
- b. More than ten thousand in number
Solution :
a. Trees, animals, people, etc.
b. Leaves. hairs, grains, etc.
Ganita Prakash class 6 maths chapter 3 number play figure it out ncert page 70.
Number play chapter 3 class 6
Try to guess within 30 seconds. Check your guess with your friends.
Q1.
Number of words in your maths textbook:
- a. More than 5000
- b. Less than 5000
Solution :
a. More than 5000.
Q2.
Number of students in your school who travel to school by bus:
- a. More than 200
- b. Less than 200
Solution :
a. More than 200.
Q3.
Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹ 100. Do you agree with him? Why or why not?
Solution :
No, I don't agree with him because the estimated cost of making custard for 5 people is likely to be more than ₹100.
Estimation :
Cost of 1 litre of milk = ₹70
Cost of 3 types of fruits ₹50 x 3 = ₹150
Total estimated cost of food items = ₹150 + ₹70 = ₹220.
Q4.
Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland).
[Hint: Look at the map of India to locate these cities.]
Solution :
The Estimated distance between Gandhinagar to Kohima is about 3000 kms.
Page 71 : Figure it out.
Q5.
Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date. Do you agree with her? Why or why not?
Solution :
No, I don't agree with her because she is in grade 6 and I think she has spent less than 13,000 hours in school till date.
Estimation :
No. of years Sheetal has spent in school = 8
No. of school days in a year = 225
No. of school hours in a day = 6
Total estimated hours she spent in her school = 8 x 225 x 6 = 10,800 hours
Q6.
Earlier, people used to walk long distances as they had no other means of transport. Suppose you walk at your normal pace. Approximately how long would it take you to go from :
- a. Your current location to one of your favourite places nearby.
- b. Your current location to any neighboring state’s capital city.
- c. The southernmost point in India to the northernmost point in India.
Solution :
a. Around 3 hours.
b. About 15 days.
c. Approx.. 2 year.
Q7.
Make some estimation questions and challenge your classmates!
Solution :
Class Activity to be done in class.
Hint : Ask your classmate to estimate the distance between his/her home and your home.
Ganita Prakash class 6 maths chapter 3 number play figure it out ncert page 72.
Number play class 6 new book
Q1.
There is only one supercell (number greater than all its neighbors) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.
Solution :
If you swap '6' of the supercell 62871 with '1', then there will be 4 supercells.
Q2.
How many rounds does your year of birth take to reach the Kaprekar constant?
Solution :
Year of birth of a 6th standard student : 2012
Round 1 : The given number is 2012.
Largest number = 2210
Smallest number = 0122
Difference = 2210 - 0122 = 2088.
Round 2 : The given number is 2088.
Largest number = 8820
Smallest number = 0288
Difference = 8820 - 0288 = 8532.
Round 3 : The given number is 8532.
Largest number = 8532
Smallest number = 2358
Difference = 8532 - 2358 = 6174.
Therefore, the number 2012 takes 3 round to reach 6174 which is the Kaprekar constant.
Q3.
We are the group of 5-digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution :
● Largest number in the group = 73,999
● Smallest number in the group = 35,111
● Number closest to 50,000 = 51,111
Q4.
Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then try to get an exact number and see how close your estimate is.
Solution :
The Estimated number of holidays we get in a year including weekends, festivals and vacation are 150 days.
Estimation :
Holidays of Sundays in a year = 52
Holidays of Saturday (2 holidays per month) = 24
Summer vacations = 45 days
Winter vacations = 15 days
Public holidays = 15 days
Actual number of holidays in a year = 52 + 24 + 45 + 15 + 15 = 151.
Q5.
Estimate the number of liters a mug, a bucket and an overhead tank can hold.
Solution :
Estimation :
● Capacity of a mug = 250 ml
● Capacity of a bucket = 20 L
● Capacity of a overhead tank = 500 L
Q6.
Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
Solution :
5 digit number = 17,200
3 digit number = 800
3 digit number = 670
Sum = 17,200 + 800 + 670 = 18,670.
Q7.
Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number.
Solution :
Chosen number is 360.
● 30 x 12 = 360
● 90 x 4 = 360
Page 73 : Figure it out.
Q8.
Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution :
The sequence for power of 2 is 2, 4, 8, 16, 32, 64, ...
Power of 2 sequence contain all even number. On dividing an even number by 2, we will get an even number but by repeatedly dividing an even number by 2, we will eventually reach 1.
Calculation :
64 ÷ 2 = 32
32 ÷ 2 = 16
16 ÷ 2 = 08
08 ÷ 2 = 04
04 ÷ 2 = 02
02 ÷ 2 = 01
Q9.
Check if the Collatz Conjecture holds for the starting number 100.
Solution :
Yes, Collatz conjecture holds for the starting number 100.
Collatz conjecture : If the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat.
Calculation :
| Number | Calculation | = |
|---|---|---|
| Even | 100 ÷ 2 | 50 |
| Even | 50 ÷ 2 | 25 |
| Odd | 25 x 3 + 1 | 76 |
| Even | 76 ÷ 2 | 38 |
| Even | 38 ÷ 2 | 19 |
| Odd | 19 x 3 + 1 | 58 |
| Even | 58 ÷ 2 | 29 |
| Odd | 29 x 3 + 1 | 88 |
| Even | 88 ÷ 2 | 44 |
| Even | 44 ÷ 2 | 22 |
| Even | 22÷ 2 | 11 |
| Odd | 11 x 3 + 1 | 34 |
| Even | 34 ÷ 2 | 17 |
| Odd | 17 x 3 + 1 | 52 |
| Even | 52 ÷ 2 | 26 |
| Even | 26 ÷ 2 | 13 |
| Odd | 13 x 3 + 1 | 40 |
| Even | 40 ÷ 2 | 20 |
| Even | 20 ÷ 2 | 10 |
| Even | 20 ÷ 2 | 05 |
| Odd | 05 x 3 + 1 | 16 |
| Even | 16 ÷ 2 | 08 |
| Even | 08 ÷ 2 | 04 |
| Even | 04 ÷ 2 | 02 |
| Even | 02 ÷ 2 | 01 |
Q10.
Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?
Solution :
Start by adding 3 on your first turn to reach 3. Then, always move in such a way that you leave the opponent on a multiple of 4. Following this strategy will ensure that you reach 22 first, securing the win.
# Worksheet - Extra Questions :
Class 6 maths chapter 3 number play worksheet.
Number play class 6 worksheet.
# Fill in the blanks :
1. The number of digits in 11111 is _____.
2. When placing the number 2300 on a number line, it would be placed just after _____.
3. In a grid, a supercell is a number that is _____ than its neighbors directly above, below, left and right.
4. The digit '7' appears _____ times in the tens place from 1 to 100.
5. on a 12 hour clock, the time 11:11 is interesting because its forms a _____ pattern.
# True - False :
1. The number 7245 would be placed between 2000 and 3000 on a number line.
2. The number 121 is a palindrome.
3. On a number line, 3330 would be placed between 3000 and 4000.
4. The digit '6' appears 100 times in the tens place between 1 to 1000.
5. If you reverse the number 1234 and add it to the original number, you will always get a palindrome.
# Multiple Choice Questions :
Q1. What is the next number in the Collatz conjecture sequence after 8?
(a) 4
(b) 3
(c) 2
(d) 1
Q2. Which is the largest 5 digit palindromic number?
(a) 99999
(b) 10001
(c) 10000
(d) 99990
Q3. Which pattern of time can not be found on a clock?
(a) 11:11
(b) 10:10
(c) 13:13
(d) 24:24
Q4. How many two digits number are their between 1 to 100?
(a) 90
(b) 91
(c) 89
(d) 99
Q5. Which number is called Kaprekar constant?
(a) 6471
(b) 6173
(c) 6174
(d) 7164
# Numerical Question :
Q1.
Arrange the following numbers (3030, 6400, 2170, 5100) in ascending order and place them on number line.
Q2.
How many rounds does the number 5344 take to reach the Kaprekar constant?
Q3.
Make a 5 digit palindromic number whose sum is 14.
Q4.
How many '5' are their between 1 to 1000?
Q5.
How you will reach Collatz's conjecture if you starts with 26.
Number play chapter 3 class 6 worksheet.
Number play class 6 extra questions.
# Fill in the Blanks :
1. 5
2. 2299
3. Greater
4. 20
5. Palindromic
# True - False :
1. False
2. True
3. True
4. True
5. True
# MCQ's :
1. 4 (a)
2. 99999 (a)
3. 24:24 (d)
4. 90 (a)
5. 6174 (c)
# Numerical Questions :
Solution Q1 :
Ascending order : 2170, 3030, 5100, 6400.
Number Line :
Solution Q2 :
Round 1 : The given number is 5344.
Largest number = 5443
Smallest number = 3445
Difference = 5443 - 3445 = 1998.
Round 2 : The given number is 1998.
Largest number = 9981
Smallest number = 1899
Difference = 9981 - 1998 = 7983.
Round 3 : The given number is 7983.
Largest number = 9873
Smallest number = 3789
Difference = 9873 - 3789 = 6084.
Round 4 : The given number is 6084.
Largest number = 8460
Smallest number = 4068
Difference = 8460 - 4068 = 4392.
Round 5 : The given number is 4392.
Largest number = 9432
Smallest number = 2349
Difference = 9432 - 2349 = 7083.
Round 6 : The given number is 7083.
Largest number = 8703
Smallest number = 0378
Difference = 8703 - 0378 = 8352.
Round 7 : The given number is 8352.
Largest number = 8532
Smallest number = 2358
Difference = 8532 - 2358 = 6174.
Therefore, the number 5344 takes 7 round to reach 6174 which is the Kaprekar constant.
Solution Q3 :
Example 1 :
5 digit palindromic number = 12321
Sum of all digits = 9
Here, the sum of the digits of the palindromic number 12321 is not 14.
Example 2 :
5 digit palindromic number = 23432
Sum of all digits = 14
Here, the sum of the digits of the palindromic number 23432 is 14.
Solution Q4 :
Number of times digit '5' comes in between 1 to 1000 :
● Among 1 - 1000 : The number of times 5 occur in 1 to 1000 is 300 times.
Calculation : Number 5 comes on 5th, 15th, 25th, 35, 45, 50 - 59 (11 times), 65, 75, 85, 95th position.
- Number 5 at Unit place (1 - 100 = 10 times x 10) = 100
- Number 5 at Tens place (1 - 100 = 10 times x 10) = 100
- Number 5 at Hundreds place (500 - 599 = 100 times) = 100
- Number of times 5 comes among 1-1000 = 300
Solution Q5 :
Yes, Collatz conjecture holds for the starting number 26.
Collatz conjecture : If the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat.
Calculation :
| Number | Calculation | = |
|---|---|---|
| Even | 26 ÷ 2 | 13 |
| Odd | 13 x 3 + 1 | 40 |
| Even | 40 ÷ 2 | 20 |
| Even | 20 ÷ 2 | 10 |
| Even | 20 ÷ 2 | 05 |
| Odd | 05 x 3 + 1 | 16 |
| Even | 16 ÷ 2 | 08 |
| Even | 08 ÷ 2 | 04 |
| Even | 04 ÷ 2 | 02 |
| Even | 02 ÷ 2 | 01 |
# MCQ Test :
# Still have some Doubts?
--Feel FREE to Ask you Doubts Here--
# Learn More :
CBSE Class 6th Mathematics teaches students about the basics, which have a wide range of application in their higher studies. All the chapters given below includes Solutions for all the question available in Class 6th Mathematics NCERT Textbook Ganita Prakash. It also includes some Important Extra Questions related to the Chapters and we have also provided Free Quiz Based Test which is consist of Objective Type Questions which help the students to test their Understanding about the given Chapters. This material is available for Free For the Students So that they can prepare & score good marks in their upcoming exams.
| Chapters | Class 6 Maths Syllabus |
|---|---|
| Chapter 01 : | Patterns in Mathematics |
| Chapter 02 : | Lines and Angles |
| Chapter 03 : | Number Play |
| Chapter 04 : | Data Handling and Presentation |
| Chapter 05 : | Prime Time |
| Chapter 06 : | Perimeter and Area |
| Chapter 07 : | Fractions |
| Chapter 08 : | Playing with Constructions |
| Chapter 09 : | Symmetry |
| Chapter 10 : | The Other Side of Zero |
Curiosity, Textbook of Science for Grade 6, comprises twelve chapters. As the name of the textbook suggests, there are numerous opportunities for the learners to explore the world of science and its nature. Through the chapters, learners will embark on a journey that will connect them to the world around and spark curiosity for further exploration. The hands-on activities embedded within each chapter engages the learners and provide them an opportunity to reflect on learning. The primary aim of Curiosity is to prepare the children for becoming the responsible members of the society, and therefore efforts have been made to raise awareness about various issues, such as gender, region, environment, health and hygiene, water scarcity and energy conservation.
| Chapters | Class 6 Science Syllabus |
|---|---|
| Chapter 01 : | The Wonderful World of Science |
| Chapter 02 : | Diversity in the Living World |
| Chapter 03 : | Mindful Eating: A Path to a Healthy Body |
| Chapter 04 : | Exploring Magnets |
| Chapter 05 : | Measurement of Length and Motion |
| Chapter 06 : | Materials Around Us |
| Chapter 07 : | Temperature and its Measurement |
| Chapter 08 : | A Journey through States of Water |
| Chapter 09 : | Methods of Separation in Everyday Life |
| Chapter 10 : | Living Creatures: Exploring their Characteristics |
| Chapter 11 : | Nature’s Treasures |
| Chapter 12 : | Beyond Earth |
Class 6th English NCERT Poorvi has five thematic units that comprise stories, poems, conversation, narrative and descriptive pieces. Themes such as friendship, wellness, sports, nature, art and culture, etc. have been included. Cross-cutting themes, such as Indian Knowledge Systems, values, heritage, gender sensitivity and inclusion have been integrated in all the units. Each unit has three literary pieces― story or conversation, poem and non-fiction. There are intext questions, ‘Let us discuss’ to assess comprehension of the text. The end-of-the-text questions given in ‘Let us think and reflect’ are designed to encourage critical thinking, reasoning, responding, analyzing, etc. These literary pieces are not only entertaining but also instill valuable life lessons, fostering personal growth and helping children navigate social situations with confidence. The selected pieces will resonate with children’s daily experiences and encourage positive values like resilience, empathy and emotional intelligence that can have a profound impact on their development.
| No. | Class 6 English Syllabus |
|---|---|
| Unit 1 : | Fables and Folk Tales |
| A Bottle of Dew | |
| The Raven and The Fox | |
| Rama to the Rescue | |
| Unit 2 : | Friendship |
| The Unlikely Best Friends | |
| A Friend's Prayer | |
| The Chair | |
| Unit 3 : | Nurturing Nature |
| Neem Baba | |
| What a Bird Thought | |
| Spices that Heal Us | |
| Unit 4 : | Sports and Wellness |
| Change of Heart | |
| The Winner | |
| Yoga - A Way of Life | |
| Unit 5 : | Culture and Tradition |
| Hamara Bharat - Incredible India! | |
| The Kites | |
| Ila Sachani : Embroidering Dreams with her Feet | |
| National War Memorial |
