By using Ganita Prakash Book Class 6 Solutions and Chapter 3 Number Play Class 6 NCERT Solutions Question Answer, students can improve their problem-solving skills.
Class 6 Maths Chapter 3 Number Play Solutions
Number Play Class 6 Solutions Questions and Answers
3.1 Numbers can Tell us Things 3.2 Supercells Figure it Out (Page No. 56)
Question 1.
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution:
No, the children at the ends can’t say’ll because there can only be one child taller than them, on one side.
To say ‘2’, a child must have one taller child on each side, which isn’t possible at the ends of the line.
Question 2.
Can we arrange the children in a line so that all would say only 0 s?
Solution:
No. We cannot arrange the children in a line so that all would say only 0, because a child says 0 only if neither of the children standing next to them are taller.
Question 3.
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 i.e. 1 because there is one taller child standing next to them.
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?
Solution:
Yes, because a child can say 0 only if neither of the children standing next to them are taller.
Question 5.
For this group of 5 children, is the sequence 1,1,1,1,1 possible?
Solution:
Yes, the sequence 1,1,1,1,1 is possible because a child can say 1 only if there is one1 taller child standing next to them.
Question 6.
Is the sequence 0, 1, 2,1, 0 possible? Why or why not?
Solution:
Yes, the sequence 0,1,2,1,0 is possible because each person says the number of taller neighbours they have.
Question 7.
How would you rearrange the five children so that the maximum number of children say 2?
Solution:
We can rearrange the five children so that the maximum number of children say 2 as 0, 2, 0, 2, 0.
Thus, maximum number of children who can say 2 is 2.
3.1 Numbers can Tell us Things 3.2 Supercells Figure it Out (Page No. 57-58)
Question 1.
Colour or mark the supercells in the table below.
Solution:
Question 2.
Fill the table below with only 4-digit numbers such that the supercells are exactly the c cells.
Solution:
Question 3.
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.
Solution:
Question 4.
Out of the 9 numbers, how many supercells are there in the table above? _____________
Solution:
5 supercells
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.
Solution:
If there are n odd cells then number of supercells = \(\frac{n+1}{2}\)
If there are n even cells then number of supercells = \(\frac{n}{2}\)
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?
Solution:
It is not possible to fill a supercell table without repeating numbers and have no supercells. A supercell is defined as a number that is greater than both of its adjacent numbers. If you place numbers in a sequence without repeating any number, some numbers will inevitably be larger than their neighbors.
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?
Solution:
Yes, the cell having the largest number in a table always be a supercell and the cell having the smallest number in a table will never be a supercell, as its adjacent cell has bigger number.
Question 8.
Fill a table such that the cell having the second largest number is not a supercell.
Solution:
The following is a table having the second largest number not being a supercell.
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?
Solution:
Yes, this is possible if we take an even number of cells in our sequence.
The second largest number 123 is not a supercell but the second smallest number 115 is a supercell. However, this arrangement is not possible with an odd number of cells.
Question 10.
Make other variations of this puzzle and challenge your classmates.
Solution:
Fill a table such that only even numbers are supercell.
Fill a table such that all the supercells are divisible by 5.
3.3 Patterns of Numbers on the Number Line Figure it Out (Page No. 59)
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.
Solution:
3.4 Playing with Digits (Page No. 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?
(b) How big a number can you form having the digit sum 14? Can you make an even bigger number?
Solution:
(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.
Solution:
40 = 4 + 0 = 4
41 = 4 + 1 = 5
42 = 4 + 2 = 6
43 = 4 + 3 = 7
44 = 4 + 4 = 8
45 = 4 + 5 = 9
46 = 4 + 6 = 10
47 = 4 + 7 = 11
48 = 4 + 8 = 12
49 = 4 + 9 = 13
50 = 5 + 0 = 5
69 = 6 + 9 = 15
70 = 7 + 0 = 7,
Do it yourself
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?
Solution:
The digit sum of 3-digit numbers whose digits are consecutive:
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
Pattern: All the digit sums are multiples of 3.
This pattern will continue for all 3-digit numbers whose digits are consecutive.
3.5 Pretty Palindromic Patterns 3.6 The Magic Number of Kaprekar 3.7 Clock and Calendar Numbers Figure it Out (Page No. 64-65)
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 is less than 5085.
(c) the sum of the largest and smallest numbers greater than 9779.
(d) the sum of the largest and smallest numbers is less than 9779.
Solution:
(a) The digits such that the difference between the largest and smallest numbers is greater than 5085 are 9, 8, 2, and 1.
Largest number: 9821
Smallest number: 1289
Difference: 9821 – 1289 = 8532
(b) The digits such that the difference between the largest and smallest numbers is less than 5085 are 5, 6, 4, and 3.
Largest number: 6543
Smallest number: 3456
Difference: 6543 – 3456 = 3087
(c) The digits such that the sum of the largest and smallest numbers is greater than 9779 are 9, 8, 7, and 6.
Largest number: 9876
Smallest number: 6789
Sum; 9876 + 6789 = 16665
(d) The digits such that the sum of the largest and smallest numbers is less than 9779 are 5, 4, 2, and 1.
Largest number: 5421
Smallest number: 1245
Sum: 5421 + 1245 = 6666
Question 2.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Largest 5-digit palindromic number = 99999
Smallest 5-digit palindromic number = 10001
Sum = 99999 + 10001 = 110000
Difference = 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?
Solution:
Time now – 10 : 01
Now next palindrome time is 11 : 11
Hence, 11:11 – 10:01 = 70 minutes.
Question 4.
How many rounds does the number 5683 take to reach the Kaprekar constant?
3.8 Mental Math Figure it Out (Page No. 66 – 67)
Question 1.
Write an example for each of the below scenarios whenever possible.
Solution:
(a) 5-digit + 5-digit to give a 5-digit sum more than 90,250: 58673 + 35987 = 94660
(b) 5-digit + 3-digit to give a 6-digit sum: 99876 + 456 = 100332
(c) 4-digit + 4-digit to give a 6-digit sum: Not Possible
(d) 5-digit + 5-digit to give a 6-digit sum: 56789 + 45678 = 102467
(e) 5-digit + 5-digit to give 18500: Not possible
(f) 5-digit – 5-digit to give a difference less than 56503: 87654 – 45678 = 41976
(g) 5-digit – 3-digit to give a 4-digit difference: 10234 – 876 = 9358
(h) 5-digit – 4-digit to give a 4-digit difference: 12345 – 8567 = 3778
(i) 5-digit – 5-digit to give a 3-digit difference: 54321 – 53789 = 532
(j) 5-digit – 5-digit to give 91500: Not Possible
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
Solution:
3.9 Playing with Number Patterns Figure Out(Pages.no 69 – 70)
Question 1.
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.
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
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
Solution:
Do yourself.
(a) 20 times (approx.)
(b) Let number of times eyes blink in a minute = 20
∵ Number of minutes in an hour = 60
Number of times eyes blink in an hour = 20 × 60 = 1200
(c) Number of times eyes blink in an hour = 1200
Number of hours in a day = 24
Let number of hours we sleep in a day = 7
Number of time eyes blink in a day = (24 – 7) × 1200
= 17 × 1200 = 20400
Question 3.
Name some objects around you that are
(a) a few thousand in number
(b) more than ten thousand in number
Solution:
(a) Leaves of trees, grain of rice, seeds, dry fruits.
(b) Grains of sand, words in a book, strands of hair, threads in a fabric, pixels on a TV screen.
3.9 Playing with Number Patterns Figure Out(Pages.no 69 – 70)
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
Solution:
(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
Solution:
(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?
Solution:
Let estimated cost of milk = ₹ 60
Estimated cost of 3 fruits = ₹ 150 (let ₹ 50 for each fruit)
Estimated cost of custard powder = ₹ 50
Total estimated cost of making custard
= 60 + 50 + 150 = ₹ 260
Thus, no, I do not 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.
Solution:
3000 km
Question 5.
Sheetal is in Grade 6 and says she has spent around 13000 hours in school till date. Do you agree with her? Why or why not?
Solution:
Let 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 ye;yrs in school by Grade 6 = 8 years
Therefore, total hours spent in school in 8 years
= 1260 × 8 = 10080 hours
Thus, no, I do not agree with Sheetal’s estimate that she has spent around 13000 hours in school till date because I think the more resonable estimate would be around 10000 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) Yqur 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.
Solution:
(a) Approximately 1 hour.
(b) Around 7-8 days
(c) Around 100 days
Question 7.
Make some estimation questions and challenge your classmates.
Solution:
Do yourself.
3.10 An Unsolved Mystery – the Collatz Conjecture! 3.11 Simple Estimation 3.12 Games and Winning Strategies Figure it Out (Page No. 72 – 73)
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.
Solution:
Swap the digits 6 and 1 in the number 62871.
Question 2.
How many rounds does your year of birth take to reach the Kaprekar constant?
Solution:
Birth year 1994.
Let largest number and smallest number be denoted by A and B, respectively.
1st round,
A = 9941
B = 1499
C = 9941 – 1499 = 8442
2nd round,
A = 8442
B = 2448
C = 8442 – 2448 = 5994
3rd round,
A = 9954
B = 4599
C = 9954 – 4599 = 5355
4th round,
A = 5553
B = 3555
C = 5553 – 3555 =1998
5th round,
A = 9981
B = 1899
C = 9981 -1899 = 8082
6th round,
A = 8820
B = 0288
C = 8820 – 0288 = 8532
7th round,
A = 8532
B = 2358
C = 8532 – 2358 = 6174
Thus, it takes 7 rounds for the nurilber 1994 to reach Kaprekar constant 6174.
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?
Solution:
The largest number with all odd digits (different) = 73951
The largest number with all odd digits (repetitive) = 73999
The smallest number (non repetitive) = 35,179
The smallest number (repetitive) = 57111
Closest to 50,000 (in case of non-repetition) = 49751
Closest to 50,000 (in case of repetition) = 49999
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.
Solution:
Will be done by students.
Question 5.
Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Solution:
Will be done by students
Question 6.
Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
Solution:
Let the 5-digit number = 18,000
And the 1st 3-digit number = 345
The 2nd 3-digit number = 325
So, their sum = 18,000 + 345 + 325 = 18,670 (Answer may vary)
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.
Solution:
Let the number be 350.
Corresponding number pattern:
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?
Solution:
The sequence of powers of 2 is as follows: 1,2, 4, 8, 16, 32, 64,…
Since,
- if the sequence starts with 2, an even number then half of it is 1.
- if the sequence starts with 4, an even number then half of it is 2 and further its half is 1.
- if the sequence starts with 8, an even number then half of it is 4, further its half is 2 and further 1.
- In the same way if the sequence is starts with 64, then half of it is 32, and further the sequence is as follows: 64, 32, 16, 8, 4, 2, 1.
Hence, the Collatz Conjecture rule is correct for all the starting numbers in this sequence.
Question 9.
Check if the Collatz Conjecture holds for the starting number 100.
Solution:
Collatz Process for 100:
1. Starting number: 100 (even)
Apply: 100 ÷ 2 = 50
2. Next number: 50 (even)
Apply: 50 ÷ 2 = 25
3. Next number: 25 (odd)
Apply: 25 × 3 + 1 = 76
4. Next number: 76 (even)
Apply: 76 ÷ 2 = 38
5. Next number: 38 (even)
Apply: 38 ÷ 2 = 19
6. Next number: 19 (odd)
Apply: 19 × 3 + 1 = 58
7. Next number: 58 (even)
Apply: 58 ÷ 2 = 29
8. Next number: 29 (odd)
Apply: 29 × 3 + 1 = 88
9. Next number: 88 (even)
Apply: 88 ÷ 2 = 44
10. Next number: 44 (even)
Apply: 44 ÷ 2 = 22
11. Next number: 22 (even)
Apply: 22 ÷ 2 = 11
12. Next number: 11 (odd)
Apply: 11 × 3 + 1 = 34
13. Next number: 34 (even)
Apply: 34-2=17
14. Next number: 17 (odd)
Apply: 17 × 3 + 1 = 52
15. Next number: 52 (even)
Apply: 52 ÷ 2 = 26
16. Next number: 26 (even)
Apply: 26 ÷ 2 = 13
17. Next number: 13 (odd)
Apply: 13 × 3 + 1 = 40
18. Next number: 40 (even)
Apply: 40 ÷ 2 = 20
19. Next number: 20 (even)
Apply: 20 ÷ 2 = 10
20. Next number: 10 (even)
Apply: 10 ÷ 2 = 5
21. Next number: 5 (odd)
Apply: 5 × 3 + 1 = 16
22. Next number: 16 (even)
Apply: 16 ÷ 2 = 8
23. Next number: 8 (even)
Apply: 8 ÷ 2 = 4
24. Next number: 4 (even)
Apply: 4 ÷ 2 = 2
25. Next number: 2 (even)
Apply: 2 ÷ 2 = 1
Starting with the number 100, the Collatz process eventually reaches 1 after several steps. This confirms that the Collatz conjecture holds for the starting number 100.
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?
Solution:
Look following steps
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