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 cannot rearrange themselves so that the children standing at the ends say two because a child says 2 if both the children standing next to them are taller.
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 coloured 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:
Question 6.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, we can’not fill’a supercell table without supercell numbers because at least one number will always be larger than its adjacent cell unless repetition is allowed.
Here, if the last number is greater than 21 it is a supercell and if less than 21 then 21, itself is a supercell number.
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 will always be a supercell, because largest number will always be largest among the other numbers.
But, the cell having smallest number in a table be never be a supercell, because it will always be smallest.
Question 8.
Fill a table such that the cell having the second largest number is not a supercell.
Solution:
Yes, it is possible.
In this table, the second highest number is 888 which is not a supercell number.
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.
In this table, the second highest number is 696, it is not a supercell number and second smallest number is 245, it is a supercell number.
Question 10.
Make other variations of this puzzle and challenge your classmates.
Solution:
Fill a table such that the cell having the second smallest number is not a supercell.
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.
Answer:
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?
(d) How big a number can you form having the digit sum 14? Can you make an even biggermumber?
Solution:
(a) 590, 770, 248, 680 etc.
(b) 59, (∵ 5 + 9 = 14)
(c) 95000,(∵ 9 + 5 + 0 + 0 + 0 = 14)
(d) Infinite numbers can be formed having the digit sum 14.
Yes, we can make an bigger even number.
Question 2.
Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Solution:
Digit sum of 40 = (4 + 0 = 4)
Digit sum of 41 = (4 +1 = 5)
Digit sum of 42 = (4 + 2 = 6)
Digit sum of 43 = (4 + 3 = 7)
Digit sum of 44 = (4 + 4 = 8)
Digit sum of 45 = (4 + 5 = 9)
Digit sum ,of 46 = (4 + 6 = 10)
Digit sum of 47 = (4 + 7 = 11)
Digit sum of 48 = (4 + 8 = 12)
Digit sum of 49 = (4 +9 = 13)
Digit sum of 50 = (5 + 0 = 5)
Digit sum of 51 = (5 + 1 = 6)
Digit sum of 52 = (5 + 2 = 7)
Digit sum of 53 = (5 + 3 = 8)
Digit sum of 54 = (5 + 4 = 9)
Digit sum of 55 = (5 + 5 = 10)
Digit sum of 56 = (5 + 6 = 11)
Digit sum of 57 = (5 + 7 = 12)
Digit sum of 58 = (5 + 8 = 13)
Digit sum of 59 = (5 + 9 = 14)
Digit sum of 60 = (6 + 0 = 6)
Digit sum of 61 = (6 + 1 = 7)
Digit sum of 62 = (6 + 2 = 8)
Digit sum of 63 = (6 + 3 = 9)
Digit sum of 64 = (6 + 4 = 10)
Digit sum of 65 = (6 + 5 = 11)
Digit sum of 66 = (6 + 6 = 12)
Digit sum of 67 = (6 + 7 = 13)
Digit sum of 68 = (6 + 8 = 14)
Digit sum of 69 = (6 + 9 = 15)
Digit sum of 70 = (7 + 0 = 7)
Observations
For numbers 40 to 49, we get
a sequence 4, 5, 6, 7, 8 … 13
For numbers 50 to 59, we get
a sequence 5,6, 7, 8,… 14
For numbers 60 to 69, we get
a sequence 6, 7, 8, 9, 10,,… 15
Here, the sum of the digits of all the numbers from 40 to 70 is present in the numbers from 4 to 15.
Question 3.
Calculate the digit sums of three-digit numbers whose digits are consecutive (e.g. 345). Do you see a pattern? Will this pattern continue?
Solution:
Sum of consecutive three-digit numbers.
Digit sum of 123 = (1 + 2 + 3 = 6)
Digit sum of 234 = (2 + 3 + 4 = 9)
Digit sum of 345 = (3 + 4 + 5 = 12)
Digit sum of 456 = (4 + 5 + 6 = 15)
Digit sum of 567 = (5 + 6 + 7 = 18)
Digit sum of 678 = (6 + 7 + 8 = 21)
Digit sum of 789 = (7 + 8 + 9 = 24)
Here, we get a pattern of numbers with a difference of 3.
Consider the sum of consecutive four-digit numbers. Digit sum of 1234 = (1 + 2 + 3 + 4 = 10)
Digit sum of 2345 = (2 + 3 + 4 + 5 = 14)
Digit sum of 3456 = (3 + 4 + 5 +6 = 18)
Digit sum of 4567 = (4 + 5 + 6 + 7 = 22)
Digit sum of 5678 = (5 + 6 + 7 + 8 = 26)
Digit sum of 6789 = (6 + 7 + 8 + 9 = 30)
Here, we get a pattern of numbers with a difference of 4. Thus, this pattern will continue
i. e. for 3-digit numbers, difference will be 3.
For 4-digit numbers, difference will be 4.
For 5-digit numbers, difference will be 5 and so on.
Yes, it is a pattern and this pattern will continue in every number whose number is 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 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) Let the 4-digits be 1, 3,6, and 9
Largest number = 9631 Smallest number 1369 Difference = 9631 -1369 = 8262
(b) Let the 4-digits be 1, 2, 3 and 5.
Largest number = 5321 Smallest number 1235 Difference = 5321-1235 = 4086
(c) Let the 4-digits be 1, 2, 3, and 9.
Largest number = 9321 Smallest number 1239
Sum = 9321+1239 = 10560
(d) Let the 4-digits be 1, 2, 5 and 7 Largest number = 7521 Smallest number = 1257
Sum = 7521 +1257 = 8778
Question 2.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
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?
Solution:
Prime time = 10:01
Next palindromic time after 10:01 = 11:11
Difference = 11:11-10:01 = 70 minutes
Thus, the next palindromic time shows after 70 minutes.
Next palindromic time after 11 : 11=12:21
Difference = 12:21 – 11:11 = 70 minutes
= 1 hour 10 minutes
Thus, the next palindromic time shows after 1 hour 10 minutes and next one is 12 : 21 which comes total 2 hours 20 minutes later.
Question 4.
How many rounds does the number 5683 take to reach the Kaprekar constant?
Solution:
1st round,
A = 8653
B = 3568
C =8653 – 3568 = 5085
2nd round,
A = 8550
B = 0558
C = 8550 – 0558 = 7992
3rd round,
A = 9972
B = 2799
C = 9972 – 2799 = 7173
4th round,
A = 7731
B = 1377
C= 7731 – 1377 = 6354
5th round,
A = 6543
B = 3456
C= 6543 – 3456 = 3087
6th round,
A = 8730
B = 0378
C = 8730 – 0378 = 8352
7th round,
A = 8532
B = 2358
C = 8532 – 2358 = 6174
Thus, the number 5683 reaches the Kaprekar’s constant 6174 after 7 rounds.
3.8 Mental Math Figure it Out (Page No. 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.
Solution:
5-digit +5-digit sum to give a 5-digit sum more than 90250
⇒ 45000 + 50000 = 95000
5-digit +3-digit to give a 6-digit sum
⇒ 99999 +900 = 100899
4-digit + 4-digit to give a 6-digit sum. Since, the maximum sum for two 4-digit numbers (9999 + 9999) is 19998. Thus, it is impossible to get a 6-digit number from adding two 4-digit numbers.
5-digit + 5-digit to give a 6-digit sum
⇒ 50000 + 61000 = 111000
5-digit + 5-digit to give 18,500. The minimum sum of two 5-digit numbers (10000 +10000) is 20000, which is greater than 18500. Therefore, it is not possible to achieve a sum of 18500 with two 5-digit numbers.
5-digit -5-digit to give a difference less than 56503
⇒ 70000-15000 = 55000
5-digit -3-digit to give a 4-digit difference
⇒ 10000-999=9001
5-digit -4-digit to give a 4-digit difference.
⇒ 12000-8000=4000
5-digit -5-digit to give a 3-digit difference
⇒ 20000-19500 = 500
5-digit -5-digit to give 91500. The difference between the largest 5-digit number and smallest 5-digit number (99999 – 10000) is 89999 which is less than 91500. Therefore, it not possible to achieve the difference of 91500 between two 5-digit numbers.
Question 2.
Always, Sometimes, Never?
Below are some statements. Think, explore and find out if each of the statements is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning and 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) Only sometimes true.
The given statement is ‘5-digit number + 5-digit number gives a 5-digit number’. It is only sometimes true.
e.g. 10000 +10000 = 20000 i.e. 5-digit number
and 99999 + 99999 = 199998 i.e. 6-digit number
(b) The given statement is ‘4-digit number + 2-digit number gives a 4-digit number’. It is only sometimes true.
e.g. 1000 +10 = 1010 i.e. 4-digit number
and 9999 +10 = 10009 i.e. 5-digit number
(c) The given statement is ‘4-digit number + 2-digit number gives a 6-digit number’. It is never true.
e.g. 9999 + 99 = 10098 i.e. 5-digit number
(d) The given statement is ‘5-digit number – 5-digit number gives a 5-digit number’.
It is only sometimes true.
e.g. 99999 -10000 = 89999 i.e. 5-digit number
and 98765 – 94321 = 4444 i.e. 4-digit number
(e) The given statement is ‘5-digit number – 2-digit number gives a 3-digit number’. It is never true,
e.g. 10000 – 99 = 9901 i.e. 4-digit number.
3.9 Playing with Number Patterns Figure Out(Pages.no 69 – 70)
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
Solution:
Do yourself.
(a) 35 steps
(b) 600 steps
(c) 400 steps
(d) 10000 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
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 35000 and 75000 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 50000?
Solution:
The odd numbers between 35000 and 75000 are
35001, 35003, 35005, ……..74999.
Therefore, largest number = 74999,
smallest number = 35001
and closest to 50000 = 49999 or 50001
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:
Number of holidays = 170 [estimate]
Number of holidays = 166 [exact]
Question 5.
Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Solution:
Mug = 0.35 litres
Bucket = 20 litres
Overhead tank = 2000 litres.
Question 6.
Write one 5-digit number and two 3-digit numbers such that their sum is 18670.
Solution:
5-digit number = 18000
3-digit number = 670
5-digit + 3-digit = 18000 + 670 = 18670
Question 7.
Choose a number between 210 and 390. Create a number pattern sirpilar to those shown in section 3.9 that will sum up to this number.
Solution:
Let the number be 320.
i. e. 40 + 120 + 40 + 120 = 320
or 20 × 4 + 40 × 6 = 320
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:
Powers of 2 → 2,4, 8,16,
For 2 (even) → 2 divide by 2 = 1
For 4 (even) → 4 divide by 2 = 2 (even)
→ 2 divide by 2 = 1
For 8 (even) → 8 divide by 2 = 4 (even)
→ 4 divide by 2 = 2 (even)
→ 2 divide by 2 = 1
Thus, for starting numbers that are powers of 2, the Collatz conjecture holds true because the sequence of operations simply involves a series of divisions by 2, which eventual leads to 1.
Question 9.
Check if the Collatz conjecture holds for the starting number 100.
Solution:
Collatz conjecture for 100
100 (even ) → \(\frac{100}{2}\) = 50
50 (even) → \(\frac{50}{2}\) = 25
25 (odd) → 25 × 3 + 1 =76 7 ft
76 (even) → \(\frac{76}{2}\) = 38
38 (even) → \(\frac{38}{2}\) = 19
19 (odd) → 19 × 3 + 1 = 58
58 (even) → \(\frac{58}{2}\) = 29
29 (odd) → 29 × 3 + 1 =88 88
88 (even) → \(\frac{88}{2}\) =44
44 (even) → \(\frac{44}{2}\) = 22
22(even) → \(\frac{22}{2}\) = 11
11 (odd) → 11 × 3 + 1 = 34
34 (even) → \(\frac{34}{2}\) = 17
17 (odd) → 17 × 3 + 1 = 52
52 (even) → \(\frac{52}{2}\) = 26
26 (even) → \(\frac{26}{2}\) = 13
13 (odd) → 13 × 3 + 1 =40 40
20 (even) → \(\frac{40}{2}\) = 10
10 (even) → \(\frac{10}{2}\) = 5
5 (odd) → 5 × 3 + 1 = 16
16 (even) → \(\frac{16}{2}\) = 8
8 (even) → \(\frac{8}{2}\) = 4;
4 (even) → \(\frac{4}{2}\) = 2
2 (even) → \(\frac{2}{2}\) = 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?
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