NCERT Solutions for Class 6 Maths Chapter 3 Number Play - Ganita Prakash
Chapter 3 Number Play NCERT Solutions for Class 6 Maths is available here which will be helpful in covering the entire syllabus and solving the difficult problems given in exercise. You can also Download PDF of Class 6 Maths Chapter 3 Number Play NCERT Solutions which will prove useful guide in making a student confident. The chapter is taken from the new NCERT Mathematics textbook, Ganita Prakash.
These NCERT Solutions for Class 6 will develop you understanding of the chapter and help in gaining good marks in the examinations. We have also provided Chapter 3 Number Play Revision Notes which will help you in completing your homework on time. These NCERT Solutions will help an individual to increase concentration and you can solve questions of supplementary books easily. Students can also check Extra Questions Answer for Number Play Class 6 Maths to prepare for their examination completely.
NCERT Solutions for Chapter 3 Number Play Class 6 Maths
Page 56
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 neighbours they have.
Try answering the questions below and share your reasoning:
Question 1. Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Answer
No, the children at the ends of the line can only have one neighbour, so the maximum number they can say is '1'. It's not possible for them to say '2' because they don't have two neighbours.
Question 2. Can we arrange the children in a line so that all would say only 0s?
Answer
No, it is not possible. It is possible only when they are of equal heights.
Question 3. Can two children standing next to each other say the same number?
Answer
Yes, it is possible for two children next to each other to say the same number if both of them have the same number of taller neighbours around them. For example, two children of similar height standing between taller children both say '1'.
Question 4. 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?
Answer
No, it's not possible because if four children say '1', it means they each have one taller neighbour, which leaves the last child without any taller neighbours. Since all children are of different heights, it's impossible to arrange them in such a way that this condition is met.
Question 5. For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Answer
No, the sequence 1,1,1,1,1 is not possible because there are only five children and at least one child must have no taller neighbours, meaning they would say 'O'. It's impossible for every child to have one taller neighbour.
Question 6. Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?
Answer
Yes, the sequence 0, 1, 2, 1, 0 is possible. It represents a situation where the middle child is the tallest (and has two taller neighbours), the two children beside the middle are shorter but each has one taller neighbour and the children at the ends have no taller neighbours.
Question 7. How would you rearrange the five children so that the maximum number of children say ‘2’?
Answer
To maximize the number of children saying '2', the tallest child should be placed in the centre, with the next two tallest children directly on either side. The two shortest children should be at the ends. This way, the three middle children can say '2', as they will each have two taller neighbour, while her two children at the ends will say T.
Page 57 & 58
Question 1. Colour or mark the supercells in the table below.
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628 |
670 |
9435 |
3780 |
3708 |
7308 |
8000 |
5583 |
52 |
Answer
Question 2. Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.
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5346 |
1258 |
9635 |
Answer
Question 3. Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
|
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|
|
|
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|
|
|
|
Answer
Question 4. Out of the 9 numbers, how many supercells are there in the table above?
Answer
Out of 9 numbers, there are 5 supercells in the above table.
Question 5. 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.
Answer
If there are n odd cells then number of supercells = 
If there are n even cells then number of supercells = n/2
Yes, there is a pattern. Alternate cells can be supercells.
Method to fill a given table to get the maximum number of supercells.
- Make first cell as supercell. After that each alternate cell is to be made supercell.
- No consecutive cells can be supercell except in case of 4 cells because then first and fourth cell can be supercell.
Question 6. Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Answer
No, it is not possible to fill a supercell table without repeating numbers such that there are no supercells.
As there are two cases:
- Case I: If we fill the cells in descending order then the first cell be supercell.
- Case II: If we fill the cells in ascending order then the last cell will be supercell.
Question 7. 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?
Answer
Yes, the cell having the largest number in a table always be a supercell because if it is comer cell, then the number adjacent to it (i.e. either second cell or second last cell) will be smaller than it. If it is in between then both its adjacent number would be smaller than it. No, the cell having smallest number in a table cannot be supercell because number adjacent to it will always be larger/greater than it.
Question 8. Fill a table such that the cell having the second largest number is not a supercell.
Answer
Largest number = 850
Second largest number = 730
Question 9. 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?
Answer
Yes, it is possible. Here is one possible arrangement:
Largest number = 750
Second largest number = 640
Smallest number = 100
Second smallest number = 210
Question 10. Make other variations of this puzzle and challenge your classmates.
Answer
Fill a table such that only even numbers are supercell.
Fill a table such that all the supercells are divisible by 5.
Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.
The biggest number in the table is ______ .
The smallest even number in the table is ______.
The smallest number greater than 50,000 in the table is ______.
Once you have filled the table above, put commas appropriately after the thousands digit.
Answer
- The biggest number in the table is 96, 310.
- The smallest even number in the table is 10,936.
- The smallest number greater than 50,000 in the table is 60,193.
Page 59
Figure it Out
Question 1. 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.
Answer
Page 60
Question 1. 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?
Answer
(a) Some numbers whose digits add up to 14 are:
59, 68, 77, 86, 95, 149, 158, 167, 176, 185, 194, 239, 248, 257, 266, 275, 281, 293
(b) The smallest number whose digit sum is 14 = 59.
(c) The largest 5 digit number containing 0 whose digit sum is 14 = 95,000.
The largest 5 digit number not containing 0 whose digit sum is 14 = 92,111.
(d) A very big number having the digit sum 14 can be made. e.g. 95000000000000.
Yes, we can make even bigger number e.g. 9500000000000000000000.
Question 2. Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Answer
Question 3. Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue?
Answer
If we take numbers in reverse order, sum of digits will remain same.
Yes, we observe a pattern.
i.e. (first number + 1)×3 = digit sum.
Page 61
Question 1. Among the numbers 1-100, how many times will the digit ‘7’occur?
Answer
The total count of 7 that we get is 20.
Question 2. Among the numbers 1-1000, how many times will the digit ‘7’occur?
Answer
The number of times 7 will be written when listing the numbers from 1 to 1000 is 300.
Question 3. Write all possible 3-digit palindromes using these digits.
Answer
- 111
- 121
- 131
- 212
- 222
- 232
- 313
- 323
- 333
Page 62
Q1: Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out.
Answer
All two-digit numbers eventually become palindromes after repeated reversal and addition. About 80% of all numbers under 10,000 resolves into a palindrome ip four or fewer steps; about 90% of those resolve in seven steps or fewer.
Example 1: Number 12
- Initial Number: 12
- Reverse: 21
- Add: 12 + 21 = 33
- Palindrome
- Check: 33 is a palindrome.
- Result: 33 is a palindrome.
Example 2: Number 89
- Initial Number: 89
- Reverse: 98
- Add: 89 + 98 = 187
- Palindrome Check: 187 is not a palindrome.
- Reverse 187: 781
- Add: 187 + 781 =968
- Palindrome Check: 968 is not a palindrome.
Puzzle time
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.
Who am I?______
Answer
12421
Page 63
Explore
Take different 4-digit numbers and try carrying out these steps. Find out what happens.
Answer
Selected a 4-digit number 1234
- Starting number: 1234
- Descending order: 4321
- Ascending order: 1234
- Subtract: 4321 – 1234 = 3087
Repeat:
- Descending order of 3087: 8730
- Ascending order of 3087: 0378
- Subtract: 8730 – 0378 = 8352
Repeat:
- Descending order of 8352: 8532
- Ascending order of 8352: 2358
- Subtract: 8532 – 2358 = 6174
Result: 6174 (Kaprekar constant)
Carry out these same steps with a few 3-digit numbers. What number will start repeating?
Answer
We will do this with help of two examples:
Number 123- Starting number: 123
- Descending order: 321
- Ascending order: 123
- Subtract: 321 – 123 = 198
Repeat:
- Descending order of 198: 981
- Ascending order of 198: 189
- Subtract: 981 – 189 = 792
Repeat:
- Descending order of 792: 972
- Ascending order of 792: 279
- Subtract: 972 – 279 = 693
Repeat:
- Descending order of 693: 963
- Ascending order of 693: 369
- Subtract: 963 – 369 = 594′
Repeat:
- Descending order of 594: 954
- Ascending order of 594: 459
- Subtract: 954 – 459 = 495
Repeat:
- Descending order of 495: 954
- Ascending order of 495: 459
- Subtract: 954 – 459 = 495
Result: The number 495 starts repeating.
- Starting number: 317
- Descending order: 731
- Ascending order: 137
- Subtract: 731 – 137 = 594
Repeat:
- Descending order of 594: 954
- Ascending order of 594: 459
- Subtract: 954 – 459 = 495 ,
Repeat:
- Descending order of 495: 954
- Ascending order of 495: 459
- Subtract: 954 – 459 = 495
Result: The number 495 starts repeating. When applying Kaprekar’s routine to 3- digit numbers, the number 495 is often reached and starts repeating. This number is known as the Kaprekar constant for 3-digit numbers.
Page 64 & 65
Question 1. Try and find out all possible times on a 12-hour clock of each of these types. For example, 4:44, 10:10, 12:21.
Answer
01:10, 02:20, 03:30, 04:40, 05:50, 10:01, 11:11, 12:21
Question 2. Find some other dates of this form from the past like 20/12/2012 where the digits ‘2’, ‘0’, ‘1 ’, and ‘2 ’ repeat in that order.
Answer
11/02/2011, 22/02/2022, 01/10/2010, 10/01/2010, 02/02/2020
Figure it Out
Question 1. 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.
Answer
(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, 5, 2, 1
Largest number = 8321
Smallest number = 1238
Sum = 8321 + 1238 = 9559
Question 2. What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Answer
The smallest 5-digit palindrome = 10001
The largest 5-digit palindrome = 99999
Sum = 10001 + 99999 = 110000
Difference = 99999 - 10001 = 89998
Question 3. The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?
Answer
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.
Question 4. How many rounds does the number 5683 take to reach the Kaprekar constant?
Answer
The given number is 5683
1st round
Largest number = 8653
Smallest Number = 3568
Subtract = 8653 - 3568
= 5085
2nd round
Largest number = 8550
Smallest number = 0558
Subtract = 8550 - 0558
= 7992
3rd round
Largest number = 9972
Smallest number = 2799
Subtract= 9972 - 2799 = 7173
4th round
Largest number = 7731
Smallest number = 1377
Subtract = 7731 - 1377 = 6354
5th round
Largest number = 6543
Smallest number = 3456
Subtract = 6543 - 3456 = 3087
6th round
Largest number = 8730
Smallest number = 0378
Difference = 8730 - 0378
= 8352
7th round
Largest number = 8532
Smallest number = 2358
Difference = 8532-2358
= 6174
Therefore, the number 5683 takes 7 rounds to reach 6174 which is the Kaprekar constant.
Page 66 & 67
Question 1. 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.
Answer
Question 2. 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
Answer
Question 3. Find out the sum of the numbers in each of the below figures. Should we add them one by one or can we use a quicker way? Share and discuss in class the different methods each of you used to solve these questions.
Answer
(a) In figure (a), number 40 is repeated 12 times and number 50 is repeated 10 times
Hence sum of all numbers = 40×12 + 50 × 10
= 480 + 500 = 980
(b) In figure (b), 1 dot (•) is 44 times and 5 dots (•) are 20 times
Hence sum of all dots = 1×44 + 5×20
= 44 + 100
= 144
(c) In figure (c), number 32 is 32 times and number 64 is 16 times
Hence sum of all numbers = 32×32 + 64×16
= 1024 + 1024
= 2048
(d) In figure (d), 3 dots (•) are 17 times and 4 dots (•) are 18 times
Hence sum of all dots = 17×3 + 18×4
= 51 + 72
= 123
(e) In figure (e), number 15 is 22 times, number 25 is 22 times and number 35 is 22 times
Hence sum of all numbers = 15×22 + 25×22 + 35×22
= 330 + 550 + 770
= 1650
(f) In figure (f), number 125 is 18 times, number 250 is 8 times and number 500 is 4 times and number 1000 is one time.
Hence sum of all numbers = 125×18 + 250×8 + 500×4 + 1000
= 2250 + 2000 + 2000 +1000
= 7250
Page 69 & 70
Figure it Out
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.
Question 1. 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
Answer
(a) 35 steps.
(b) 600 steps.
(c) 400 steps.
(d) 10,000 steps.
Question 2. 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
Answer
(a) 20 times
(b) No. of times eyes blink in a minute = 20
No. of minutes in a hour = 60 minutes
No. of times eyes blink in an hour = 20×60 = 1200.
(c) No. of times eyes blink in an hour = 1200
No. of hours in a day = 24 hours
No. of hours we sleep in a day = 7 hours
No of times eyes blink in a day = (24 - 7)×1200
= 17×1200
= 20,400
Question 3. Name some objects around you that are:
(a) a few thousand in number
(b) more than ten thousand in number
Answer
(a) Leaves of tree, grains of rice, seeds, dry fruits.
(b) Grains of sand, words in a book, strands of hair, threads in a fabric, pixels on a T.V screen.
Try to guess within 30 seconds. Check your guess with your friends.
Question 1. Number of words in your maths textbook:
(a) More than 5000
(b) Less than 5000
Answer
(a) More than 5000
Question 2. Number of students in your school who travel to school by bus:
(a) More than 200
(b) Less than 200
Answer
(a) More than 200
Question 3. 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?
Answer
Estimated cost of milk = 60
Estimated cost of 3 fruits = 150 (₹50 for each fruit)
Estimated cost of custard powder = 50
Total estimated cost of making custard = 60 + 50 + 150 = 260.
No, I don't agree with Roshan's estimate because I think cost of making custard for 5 people is likely to be around 260.
Question 4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland). [Hint: Look at the map of India to locate these cities.]
Answer
3000 km.
Page 71
Question 5. 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?
Answer
Total school days per year = 210
Total hours spent in school per day = 6 hours
Total hours spent in school per year = 210 × 6
= 1260 hours
Total number of years in school by Grade 6 = 8 years
Therefore, total hours spent in school in 8 years = 1260×8
= 10,080 hours.
No, I don't agree with Sheetal's estimate that she has spent around 13,000 hours in school till date because I think the more reasonable estimate would be around 10,000 hours.
Question 6. 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 neighbouring state's capital city.
(c) The southernmost point in India to the northernmost point in India.
Answer
(a) Approximately 1 hour.
(b) Around 7-8 days.
(c) Around 100 days.
Question 7. Make some estimation questions and challenge your classmates!
Answer
Students should this themselves.
Page 72
Figure it Out
Question 1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.
Answer
If you swap '6' of the supercell 62,871 with '1', then there will be 4 supercells.
Question 2. How many rounds does your year of birth take to reach the Kaprekar constant?
Answer
Year of birth: 2012
1st round
Largest number = 2210
Smallest Number = 0122
Subtract = 2210 - 0122
= 2088
2nd round
Largest number = 8820
Smallest number = 0288
Subtract = 8820 - 0288
3rd round
Largest number = 8532
Smallest number =2358
Subtract = 8532-2358 = 6174
Therefore, the number 2012 takes 3 rounds to reach 6174 which is the Kaprekar constant.
Question 3. 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?
Answer
Largest number in the group = 73999
Smallest number in the group = 35,111
Number closest to 50,000 = 51,111.
Question 4. 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.
Answer
Estimated number of holidays in a year including weekends, festivals and vacations = 120 days
Number of weeks in a year = 52 weeks
Holidays on Sundays = 52 days
Holidays on Saturdays (2 holidays per month) = 24
Public holidays = 15 days
Summer vacations = 45 days
Winter vacations = 15 days
Actual number of holidays in a year = 52 + 24 + 15 + 45 + 15 = 151 days.
Question 5. Estimate the number of liters a mug, a bucket and an overhead tank can hold.
Answer
Capacity of a mug = 0.3 liters
Capacity of a bucket = 25 liters
Capacity of an overhead tank = 1000 liters.
Question 6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
Answer
5-digit number = 16,945
3-digit number = 825
3-digit number = 900
Sum 16,945 + 825 +900 = 18,670.
Question 7. 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.
Answer
Chosen number = 320
In case of 10: 4 × 8 × 10 = 320
In case of 20: 4 × 4 × 20 = 320
Page 73
Question 8. 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?
Answer
Power of 2 sequence: 1, 2, 4, 8, 16, 32, 64,...
Power of 2 sequence contains all even numbers.
When 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.
This is why Collatz conjecture is correct for all the starting numbers in this sequence.
Example: 16 (power of 2)
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1
Question 9. Check if the Collatz Conjecture holds for the starting number 100.
Answer
Collatz conjecture: If the number is even, take half of it;if the number is odd, multiply it by 3 and add 1. If you continue doing this you will always reach number 1, regardless of the number you start with.
100 (even) : 100/2 = 50
50 (even): 50/2 = 25
25 (odd): 25×3 + 1 = 76
76 (even): 76/2 = 38
38 (even): 38/2= 19
19 (even): 19×3 + 1 = 58
58 (even): 58/2 = 29
29 (odd): 29×3 + 1 = 88
88 (even): 88/2 = 44
44 (even): 44/2 = 22
22 (even): 22/2 = 11
11 (odd): 11 x 3 + 1 = 34
34 (even): 34/2 = 17
17 (even): 17×3 + 1 = 52
52 (even): 52/2 = 26
26 (even): 26/2 = 13
13 (even): 13×3 + 1 = 40
40 (even): 40/2 = 20
20 (even): 20/2= 10
10 (even): 10/2 = 5
5 (even): 5×3 + 1 = 16
16 (even): 16/2 = 8
8 (even): 8/2 = 4
4 (even): 4/2 = 2
2 (even): 2/2 = 1
Question 10. Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?
Answer
(i) Start by adding 3 on your first turn to reach 3.
(ii) Then, always move in such a way that you leave the opponent on a multiple of 4.
(iii) Following this strategy will ensure that you reach 22 first, securing the win.