CBSE Class 6 Mathematics Number Play Worksheet Set 01

Long Answer Questions

Question 1. Answer all the parts about writing and making numbers.
(i) Write the smallest and the largest 3-digit, 4-digit and 5-digit numbers.
(ii) Using each of the digits 5, 0, 8 and 3 exactly once, make the largest 4-digit number and the smallest 4-digit number. Explain why 0 cannot be placed first.
(iii) Find the difference and the sum of the two numbers you made in part (ii).
(iv) Write the place value of the digit 8 in each of the two numbers you made.
Answer:
(i) 3-digit: smallest 100, largest 999. 4-digit: smallest 1000, largest 9999. 5-digit: smallest 10,000, largest 99,999.
(ii) Largest: put the digits from biggest to smallest, 8530.
Smallest: go from smallest to biggest, but 0 cannot come first, because 0583 is really the 3-digit number 583. So the next smallest digit 3 goes first and 0 comes next: 3058.
(iii) Difference = 8530 − 3058 = 5472. Sum = 8530 + 3058 = 11,588.
(iv) In 8530 the 8 is in the thousands place, so its place value is 8000. In 3058 the 8 is in the ones place, so its place value is 8.
In simple words: Order the digits for the largest and smallest numbers, never starting with 0.

Teacher's Note:
1. A leading 0 reduces the number of digits.
2. Place value = digit × value of its place.

Question 2. A teacher writes nine 4-digit numbers in a 3 by 3 grid. A cell is a supercell if the number in it is greater than the numbers in all the cells next to it (left, right, above and below).
(i) Write all the supercells of this grid.
(ii) How many supercells are there, and which is the largest number in the grid?
(iii) Is the largest number of the grid a supercell? Give a reason.
(iv) Can the smallest number of a grid ever be a supercell? Give a reason.
Answer:
(i) Checking each cell against its left, right, upper and lower neighbours, the supercells are 6390, 8104, 7430 and 4927.
For example, 6390 is greater than 2415, 1872 and 3256; 7430 is greater than 1872, 3256 and 2083; and 4927 is greater than 3256, 1569 and 2083.
(ii) There are 4 supercells. The largest number in the grid is 8104.
(iii) Yes. Its neighbours 2415, 3256 and 1569 are all smaller. In any grid, the largest number is bigger than every other number, so it is bigger than its neighbours and must be a supercell.
(iv) No. The smallest number is smaller than every other number. Each cell has at least one neighbour, and that neighbour is bigger, so the smallest number can never be a supercell.
In simple words: 6390, 8104, 7430 and 4927 are supercells; the largest always is one and the smallest never is.

Teacher's Note:
1. Diagonal cells are not neighbours here.
2. Tick each supercell on the grid as you find it.

Question 3. The marks on the number line below are at equal gaps. Only the first two marks have been labelled.
(i) What is the gap between two neighbouring marks? Write the labels of all the remaining marks.
(ii) Which marked number is the smallest and which is the largest?
(iii) Which marked number is nearest to 26,000? Show your working.
(iv) Between which two marks would the number 25,900 lie?
Answer:
(i) Gap = 24,400 − 23,400 = 1000. The eight marks are 23,400; 24,400; 25,400; 26,400; 27,400; 28,400; 29,400 and 30,400.
(ii) Smallest: 23,400 (first mark). Largest: 30,400 (last mark).
(iii) 26,000 − 25,400 = 600 and 26,400 − 26,000 = 400. Since 400 is less than 600, 26,400 is nearest to 26,000.
(iv) 25,900 is more than 25,400 and less than 26,400, so it lies between 25,400 and 26,400. As 25,900 − 25,400 = 500 and 26,400 − 25,900 = 500, it is exactly halfway between them.
In simple words: The marks go up by 1000; 26,400 is nearest to 26,000, and 25,900 sits midway between 25,400 and 26,400.

Teacher's Note:
1. Find the gap first from the two labels.
2. Compare distances to decide 'nearest'.

Question 4. Answer all the parts about palindromic numbers.
(i) What is a palindromic number? Write two examples, one of three digits and one of four digits.
(ii) Write all the 3-digit palindromes that use only the digits 2 and 5.
(iii) Start with 39 and use the reverse-and-add process until you reach a palindrome. Show every addition and state the number of steps taken.
(iv) Write the smallest 5-digit palindrome and the largest 5-digit palindrome.
Answer:
(i) A palindromic number reads the same from left to right as from right to left. Examples: 353 and 7227.
(ii) The first and last digits must be equal. With 2 at both ends: 222 and 252. With 5 at both ends: 525 and 555. So the palindromes are 222, 252, 525 and 555.
(iii) 39 + 93 = 132 (not a palindrome); 132 + 231 = 363 (a palindrome).
The palindrome 363 is reached in 2 steps.
(iv) Smallest 5-digit palindrome: 10,001. Largest: 99,999.
In simple words: Palindromes read the same both ways; 39 reaches 363 in two steps.

Teacher's Note:
1. For (ii), choose the end digit, then the middle digit.
2. Check every sum before stopping.

Question 5. Kaprekar’s process is applied to the number 6152.
(i) Write the rule of the process in your own words.
(ii) Carry out the process round by round, showing A, B and C = A − B each time, until the number stops changing.
(iii) How many rounds were needed, and what is this final number called?
(iv) Why must the starting number have at least two different digits?
Answer:
(i) Take a 4-digit number with at least two different digits. Arrange its digits to make the largest number A and the smallest number B. Find C = A − B. Repeat the same steps with the digits of C.
(ii) Round 1: A = 6521, B = 1256, C = 6521 − 1256 = 5265
Round 2: A = 6552, B = 2556, C = 6552 − 2556 = 3996
Round 3: A = 9963, B = 3699, C = 9963 − 3699 = 6264
Round 4: A = 6642, B = 2466, C = 6642 − 2466 = 4176
Round 5: A = 7641, B = 1467, C = 7641 − 1467 = 6174
(iii) It takes 5 rounds. 6174 is called the Kaprekar constant, found by D. R. Kaprekar. Once reached, it repeats for ever, since 7641 − 1467 = 6174 again.
(iv) If all four digits are the same, as in 3333, then A and B are equal and C = 3333 − 3333 = 0. The process stops at once instead of reaching 6174.
In simple words: 6152 reaches 6174 in 5 rounds; equal digits would just give 0.

Teacher's Note:
1. Write A, B and C in a neat table.
2. 6174 gives itself back, so the process stops changing.

Question 6. The Collatz rule says: if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; then repeat.
(i) Write the full Collatz chain starting from 7.
(ii) How many steps does the chain take to reach 1?
(iii) Explain why the chain starting from 64 reaches 1 very quickly.
(iv) Why is this called the Collatz conjecture and not a Collatz rule that has been proved?
Answer:
(i) 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
7 is odd, so 3 × 7 + 1 = 22; 22 is even, so half is 11; 11 is odd, so 3 × 11 + 1 = 34; and so on.
(ii) There are 17 numbers in the chain, so it takes 16 steps.
(iii) 64 = 2 × 2 × 2 × 2 × 2 × 2 is a power of 2. Every number is even and is simply halved: 64, 32, 16, 8, 4, 2, 1. No odd number other than 1 appears, so it reaches 1 in only 6 steps.
(iv) A conjecture is a statement believed to be true but not yet proved. Collatz said in 1937 that the chain always reaches 1. It has been checked for a huge number of starting numbers, but no one has proved it for every number, so it is still an unsolved problem.
In simple words: 7 takes 16 steps, 64 only 6, and nobody has proved the rule for all numbers.

Teacher's Note:
1. Steps = numbers in the chain − 1.
2. Checking many cases is not a proof.

Question 7. The four numbers 30,000, 6,000, 500 and 200 are written in boxes. Any box may be used as many times as needed, and both addition and subtraction are allowed.
(i) Make 36,700 using addition only.
(ii) Make 42,000 and show your steps.
(iii) Make 29,800 using subtraction somewhere in your working.
(iv) Explain clearly why the number 50 can never be made from these boxes.
Answer:
(i) 36,700 = 30,000 + 6,000 + 500 + 200. Each box is used once.
(ii) 42,000 = 30,000 + 6,000 + 6,000. The box 6,000 is used twice: 30,000 + 12,000 = 42,000.
(iii) 29,800 = 30,000 − 200. Going 200 below 30,000 is much quicker than adding many small numbers.
(iv) Every box number (30,000, 6,000, 500 and 200) is a multiple of 100. Adding or subtracting multiples of 100 always gives a multiple of 100. 50 is not a multiple of 100, so it can never be made.
In simple words: Combine the boxes cleverly; 50 is impossible because every box is a multiple of 100.

Teacher's Note:
1. Subtraction can be a short cut.
2. Look for what all the boxes have in common.

Question 8. Find the sum of each of these designs in the quickest way, showing the grouping you used.
(i) A design has 40 written in 10 circles and 50 written in 6 squares.
(ii) A design is a grid of 5 rows and 8 columns, with 20 written in every cell.
(iii) A design has 1000 written once, 500 written twice, 250 written four times and 125 written eight times.
(iv) Write the short rule you followed to add such designs quickly.
Answer:
(i) 10 × 40 = 400 and 6 × 50 = 300. Sum = 400 + 300 = 700.
(ii) Cells = 5 × 8 = 40, each with 20. Sum = 40 × 20 = 800.
(iii) 1 × 1000 = 1000, 2 × 500 = 1000, 4 × 250 = 1000 and 8 × 125 = 1000. Each group gives 1000, so the sum = 4 × 1000 = 4000.
(iv) Do not add one by one. Group the equal numbers, multiply each number by how many times it appears, and add the products. When the number is halved but the count is doubled, every group gives the same total.
In simple words: Group, multiply, then add; halving with doubling keeps each group the same.

Teacher's Note:
1. Spot groups that make round numbers.
2. In (iii) each group is worth 1000.

Question 9. In the game “Reach 48”, the first player says any number from 1 to 7. After that the players take turns adding a number from 1 to 7 to the number just said. The player who says 48 wins.
(i) Working backwards from 48, write the list of numbers that a player must say in order to be sure of winning.
(ii) Which player can always win if they play correctly? Explain why.
(iii) Write the winning strategy as a short rule a classmate could follow.
(iv) In a game where a player may add any number from 1 to k and the target is N, write the general rule for the safe numbers.
Answer:
(i) To be sure of saying 48, a player must say 40: from 40 the opponent can reach only 41 to 47, and from any of these the player jumps to 48. Going back in steps of 8, the safe numbers are 8, 16, 24, 32, 40 and 48.
(ii) The second player. The first player can say only 1 to 7, never 8. The second player then adds what is needed to make 8. After that, if the first player adds a, the second adds 8 − a, so the total goes 16, 24, 32, 40 and 48.
(iii) Always say a multiple of 8. If the opponent has just added a, you add 8 − a.
(iv) The safe numbers are N, N − (k + 1), N − 2(k + 1), and so on, going down in steps of k + 1.
If N is a multiple of k + 1, the second player wins. Otherwise the first player wins by first saying the remainder when N is divided by k + 1.
In simple words: Jump in steps of 8 (one more than 7); the second player wins Reach 48.

Teacher's Note:
1. Step size is always one more than the biggest allowed number.
2. Check who can say the first safe number.

Question 10. Estimation is used when an exact count is not needed.
(i) A school works for about 200 days in a year and has 6 school hours each day. Estimate the number of school hours a student spends in school in one year.
(ii) Tanvi is 11 years old and says she has slept for about 50,000 hours so far. Taking 8 hours of sleep each day and 365 days in a year, check whether her claim is sensible.
(iii) Write the steps you followed in part (ii) as a general method of checking such a claim.
(iv) Give two everyday situations in which an estimate is more useful than an exact count.
Answer:
(i) Hours in a year = 6 × 200 = 1200 hours.
(ii) Sleep in one year = 8 × 365 = 2920 hours, about 3000 hours.
Sleep in 11 years = 2920 × 11 = 32,120 hours, about 32,000 hours.
50,000 hours is far more than this, so the claim is not sensible. Even 10 hours a day would give only about 40,000 hours.
(iii) Choose a sensible amount for one day, multiply by 365 for one year, then multiply by the number of years. Round the answer and compare it with the claim. If the claim is much bigger or smaller, it is not sensible.
(iv) Checking whether the money in your pocket is enough for the items in a shopping basket, and guessing the size of a crowd at a fair or a match. An exact count is hard to make and is not needed in these cases.
In simple words: Multiply day → year → years and compare; 50,000 hours of sleep is too many.

Teacher's Note:
1. Round only at the end for a fair check.
2. Any two sensible daily-life examples are fine.

Case Study Based Questions

Question 11. The morning assembly line
At the morning assembly of Class 6, six children stand in a row in the order Anu, Biju, Chetan, Divya, Esha and Farid. Their heights in centimetres are 132, 145, 128, 150, 138 and 141 in the same order. Their teacher plays a game. Each child looks only at the children standing immediately next to them and says how many of those neighbours are taller than themselves. A child at an end of the row has only one neighbour. The children enjoy the game and want to play it again after changing places.
(i) What number does Chetan say?
(ii) Write the numbers said by all six children, in order from Anu to Farid.
(iii) Name the children who say 0, and explain what saying 0 tells us about them.
(iv) Can a child standing at an end of the row ever say 2? Give a reason.
Answer:
(i) Chetan is 128 cm. His neighbours Biju (145 cm) and Divya (150 cm) are both taller, so Chetan says 2.
(ii) 1, 0, 2, 0, 2, 0. Anu says 1 (Biju is taller); Biju says 0 (Anu and Chetan are shorter); Chetan says 2; Divya says 0 (Chetan and Esha are shorter); Esha says 2 (Divya and Farid are taller); Farid says 0 (Esha is shorter).
(iii) Biju, Divya and Farid say 0. It means no child standing next to them is taller than they are.
(iv) No. An end child has only one neighbour, so at most one taller child stands next to them. The most such a child can say is 1.
In simple words: Compare each child only with neighbours: 1, 0, 2, 0, 2, 0.

Teacher's Note:
1. Write the heights under the names first.
2. End children have one neighbour only.

Teacher's Note:
1. End houses need to beat only one neighbour.
2. Supercells cannot stand side by side.

Question 13. Page numbers of a storybook
Meera buys a storybook whose pages are numbered from 1 to 150. While turning the pages she notices that some page numbers have one digit, some have two digits and some have three digits. She also notices that a few page numbers read the same from left to right as from right to left, such as page 22 and page 99.
(i) How many pages of the book carry a 1-digit number?
(ii) How many pages carry a 2-digit number?
(iii) How many pages carry a 3-digit number?
(iv) Write all the page numbers between 100 and 150 that read the same both ways.
Answer:
(i) Pages 1 to 9: 9 pages.
(ii) Pages 10 to 99: 90 pages.
(iii) Pages 100 to 150: 150 − 100 + 1 = 51 pages.
(iv) 101, 111, 121, 131 and 141, five pages in all.
In simple words: 9 + 90 + 51 = 150 pages; five palindromic page numbers lie between 100 and 150.

Teacher's Note:
1. Check: 9 + 90 + 51 = 150.
2. A 3-digit palindrome from 100 to 150 must start and end with 1.

Question 14. The station clock
A digital clock at a railway station shows the time in the 12-hour form, such as 9:45 or 11:20. Raghav is waiting for a train and passes the time by looking for times that read the same forwards and backwards when the colon is ignored. Such a time is called a palindromic time. He notices the clock showing 11:11 just as the announcement for his train is made.
(i) Is 11:11 a palindromic time? Give a reason.
(ii) Which is the next palindromic time the clock will show after 11:11?
(iii) How many minutes after 11:11 will that time appear?
(iv) Write the palindromic time that the clock shows next, after the one found in part (ii).
Answer:
(i) Yes. Without the colon the digits are 1111, which read the same from both ends.
(ii) 12:21, since 1221 reads the same both ways.
(iii) 11:11 to 12:11 is 60 minutes, and 12:11 to 12:21 is 10 more. So it appears 70 minutes later.
(iv) 1:01 (digits 101), which comes 40 minutes after 12:21.
In simple words: After 11:11 come 12:21 (70 minutes later) and then 1:01.

Teacher's Note:
1. On a 12-hour clock, 12:59 is followed by 1:00.
2. Count the minutes hour by hour.

Teacher's Note:
1. Round each amount before adding.
2. Use the exact bill for the change.

Question 16. Stones along the highway
Along a straight highway, small stones are fixed at equal gaps. Each stone shows the distance of that point from the city, measured in metres. A bus travelling away from the city passes a stone showing 48,500 and the next stone shows 49,000. The gap between two stones stays the same all along the road. The stones therefore work exactly like the marks on a number line, with each stone standing for the next number in the pattern.
(i) What is the gap between two stones, in metres?
(ii) Write the next three readings after 49,000.
(iii) How many stones are there from the one showing 48,500 up to the one showing 50,500, counting both?
(iv) Which stone reading lies exactly halfway between 48,500 and 50,500, and what is this distance in kilometres?
Answer:
(i) Gap = 49,000 − 48,500 = 500 metres.
(ii) 49,500; 50,000; 50,500.
(iii) Readings: 48,500; 49,000; 49,500; 50,000; 50,500, so 5 stones. (2000 m ÷ 500 m = 4 gaps, and 4 + 1 = 5 stones.)
(iv) 48,500 + 50,500 = 99,000 and half of it is 49,500. Since 1 km = 1000 m, this is 49 km 500 m.
In simple words: The stones go up by 500 m; there are 5 from 48,500 to 50,500, and 49,500 is the middle one.

Teacher's Note:
1. Stones = gaps + 1.
2. Halfway = half of the sum of the two ends.

Question 17. A page from the calendar
Rohit keeps a wall calendar in his room. He writes every date in the dd/mm/yyyy form, so that 5 March 2024 is written as 05/03/2024. One day he notices that his mother’s birthday, 12/02/2021, is made of the eight digits 1, 2, 0, 2, 2, 0, 2, 1, and that these read the same from left to right and from right to left. His friend Manish was born on 20/12/2012, where the four digits 2, 0, 1, 2 simply repeat in the same order. Rohit begins hunting for more such special dates.
(i) Write 2 February 2022 in the dd/mm/yyyy form.
(ii) Show that 22/02/2022 reads the same forwards and backwards.
(iii) Write one more date from the past of Manish’s type, in which the four digits of the day and month repeat as the year.
(iv) Rohit asks whether an old calendar can ever be used again. Answer with a reason.
Answer:
(i) 02/02/2022.
(ii) The eight digits are 2, 2, 0, 2, 2, 0, 2, 2. From the other end they are 2, 2, 0, 2, 2, 0, 2, 2, exactly the same. So the date reads the same both ways.
(iii) 20/11/2011: the digits 2, 0, 1, 1 of the day and month repeat as the year 2011. (20/04/2004 and 20/06/2006 also work.)
(iv) Yes. In an ordinary year each date moves on by one weekday, and in a leap year by two. When these shifts add up to whole weeks and the leap-year pattern also matches, every date falls on the same weekday again.
An ordinary year’s calendar can usually be used again after 6 or 11 years; a leap year’s calendar after 28 years.
In simple words: Palindromic dates read the same both ways, and old calendars do come back after some years.

Teacher's Note:
1. Write all eight digits in a row to check.
2. A leap year's calendar takes longest to return.

Teacher's Note:
1. Add the runs in easy pairs.
2. A 3-digit palindrome above 461 must end in 4.

Question 19. A number game in the classroom
Ravi and Sara play a game in class. Starting from 0, the players take turns adding any number from 1 to 5 to the number just said, and the number said each time is the new total. The player who says 25 wins the game. Ravi always goes first. After losing three games in a row, Sara decides to work backwards from 25 to find the totals that are safe to say, and she soon starts winning whenever Ravi plays carelessly.
(i) Ravi begins by saying 1 and Sara then says 6. What number should Ravi say next to stay safe?
(ii) Write all the safe totals of this game, working backwards from 25.
(iii) Which player can always win if they play correctly, and what should the first move be?
(iv) The class now changes the target to 24, keeping the same rule. Which player wins now? Give a reason.
Answer:
(i) Ravi should say 7 (add 1 to 6). The safe totals go up in steps of 6, and 7 is the next one after 1.
(ii) Going back from 25 in steps of 6: 25, 19, 13, 7 and 1.
(iii) The first player, Ravi, can always win by saying 1 first. After that, if Sara adds a, he adds 6 − a, so the total goes 7, 13, 19 and 25.
(iv) Now the safe totals are 24, 18, 12 and 6. On the first move a player can say only 1 to 5, not 6. So the second player (Sara) can always reach 6 and then keep to 12, 18 and 24.
In simple words: Safe totals go down in steps of 6; with 25 the first player wins, with 24 the second player wins.

Teacher's Note:
1. Step = 5 + 1 = 6.
2. If the smallest safe total can be said in one move, the first player wins.

Question 20. Maths club day
The maths club has made number cards. On Monday the members work with 3-digit numbers. They pick three digits, make the largest and the smallest 3-digit numbers from them, subtract the smaller from the larger, and repeat the process with the digits of the answer. On Tuesday they take a number and apply a different rule: halve it if it is even, and replace it by three times the number plus one if it is odd, repeating until the chain ends. The teacher tells them that the second rule was studied by Lothar Collatz in 1937 and that mathematicians still cannot prove what happens for every starting number.
(i) Using the digits 8, 1 and 5, write the largest and the smallest 3-digit numbers.
(ii) Carry out Monday’s process starting from 815 until a number begins to repeat. Write each round and name that number.
(iii) Write Tuesday’s chain starting from 26 and count the number of steps it takes to reach 1.
(iv) Why is Collatz’s statement called a conjecture and counted among the unsolved problems of mathematics?
Answer:
(i) Largest = 851, smallest = 158.
(ii) 851 − 158 = 693; 963 − 369 = 594; 954 − 459 = 495; 954 − 459 = 495 again.
The number 495 starts repeating. It is the Kaprekar number for 3-digit numbers, just as 6174 is for 4-digit numbers.
(iii) 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. 26 is even → 13; 13 is odd → 3 × 13 + 1 = 40; and so on. The chain has 11 numbers, so it takes 10 steps.
(iv) A conjecture is a statement people believe but nobody has proved. Collatz’s chain has reached 1 for a very large number of starting numbers, but no proof works for every whole number. So it is still an unsolved problem, a question that is easy to ask but very hard to answer.
In simple words: Monday's process ends at 495, Tuesday's chain from 26 takes 10 steps, and Collatz's idea is still unproved.

Teacher's Note:
1. Rearrange the digits of each answer before the next round.
2. Tested is not the same as proved.