CBSE Class 6 Mathematics Number Play Worksheet Set 01

Short Answer Questions

Question 1. Four children of different heights stand in a line in increasing order of height, the shortest at the left end and the tallest at the right end. Each child says the number of taller children standing next to them.
(i) Write the four numbers that are said, from left to right.
(ii) Give a reason for the number said by the child at the right end.
Answer:
Each child tells how many of their neighbours are taller.
(i) Every child except the last has a taller child on the right and (if any) a shorter one on the left. The numbers are 1, 1, 1, 0.
(ii) The child at the right end is the tallest of all and has only one neighbour, who is shorter. So this child has no taller neighbour and says 0.
In simple words: In increasing order everyone sees one taller neighbour, except the tallest, who says 0.

Teacher's Note:
1. Look at the left and right neighbours of each child.
2. The tallest child always says 0.

Teacher's Note:
1. The end cards 408 and 309 have only one neighbour each.
2. Check both sides of every card.

Question 3. Answer the following about how many numbers there are of each size.
(i) How many 2-digit numbers and how many 3-digit numbers are there?
(ii) Write the smallest and the largest 5-digit number, and say how many 5-digit numbers there are in all.
Answer:
(i) 2-digit numbers: 10 to 99, so 90. 3-digit numbers: 100 to 999, so 900.
(ii) Smallest 5-digit number = 10,000; largest = 99,999.
Number of 5-digit numbers = 99,999 − 10,000 + 1 = 90,000.
In simple words: Each extra digit gives ten times as many numbers: 90, 900, …, 90,000.

Teacher's Note:
1. Largest − smallest + 1 gives the count.
2. Notice the pattern 9, 90, 900, 9,000, 90,000.

Question 4. Look at these numbers: 707, 1230, 6776, 8008, 9190.
(i) Which of them are palindromes? Give a reason for leaving out the others.
(ii) Write the smallest and the largest 3-digit palindrome.
Answer:
(i) The palindromes are 707, 6776 and 8008, because each reads the same from both ends.
1230 backwards is 0321 and 9190 backwards is 0919, so these two are not palindromes.
(ii) Smallest 3-digit palindrome = 101; largest = 999.
In simple words: A palindrome reads the same both ways; 101 and 999 are the smallest and largest 3-digit ones.

Teacher's Note:
1. A number ending in 0 cannot be a palindrome (it cannot start with 0).
2. Read each number backwards slowly.

Question 5. Start with the number 7261 and carry out one round of Kaprekar’s process. Show the largest number A, the smallest number B and the difference C = A − B.
Answer:
The digits of 7261 are 7, 2, 6 and 1.
Largest number A = 7621
Smallest number B = 1267
C = A − B = 7621 − 1267 = 6354
So one round of Kaprekar’s process turns 7261 into 6354.
In simple words: Arrange the digits biggest-first and smallest-first, then subtract: 6354.

Teacher's Note:
1. Write A and B clearly before subtracting.
2. Check: 6354 + 1267 = 7621.

Question 6. Apply the Collatz rule (halve an even number, and for an odd number take 3n + 1) starting from 40.
(i) Write the full chain up to 1.
(ii) How many steps does it take to reach 1?
Answer:
(i) 40, 20 and 10 are even, so they are halved. 5 is odd, so 3 × 5 + 1 = 16. Then 16, 8, 4 and 2 are halved down to 1.
Chain: 40, 20, 10, 5, 16, 8, 4, 2, 1.
(ii) The chain has 9 numbers, so it takes 8 steps to reach 1.
In simple words: 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 takes 8 steps.

Teacher's Note:
1. Steps = numbers in the chain − 1.
2. Check even or odd before each step.

Question 7. A palindromic time on a 12-hour digital clock reads the same forwards and backwards, for example 3:03.
(i) Write all the palindromic times of the form “h:0h” from 1:00 to 9:59.
(ii) How many such times are there?
Answer:
(i) 1:01, 2:02, 3:03, 4:04, 5:05, 6:06, 7:07, 8:08 and 9:09.
(ii) There are 9 such times, one for each hour from 1 to 9.
In simple words: One time per hour, 1:01 to 9:09, gives 9 times.

Teacher's Note:
1. The minutes are 0 followed by the hour digit.
2. One for each hour from 1 to 9.

Question 8. Rahul has to add 243 + 456 + 181 quickly.
(i) Round each number to the nearest hundred and write the estimated sum.
(ii) Find the exact sum and write how far the estimate is from it.
Answer:
(i) 243 → 200, 456 → 500, 181 → 200. Estimated sum = 200 + 500 + 200 = 900.
(ii) Exact sum = 243 + 456 + 181 = 880. The estimate is 20 more than the exact answer, which is close enough for a quick check.
In simple words: Rounding gives 900, only 20 more than the exact 880.

Teacher's Note:
1. Round using the tens digit.
2. An estimate helps you check your exact answer.

Question 9. On a number line the marks are placed at equal gaps. Two neighbouring marks are labelled 2,497 and 2,498.
(i) Write the next four marks to the right and the two marks to the left.
(ii) Among these eight marks, which one has a digit sum of 7?
Answer:
Since 2,498 comes just after 2,497, the marks go up in steps of 1.
(i) To the right: 2,499; 2,500; 2,501; 2,502. To the left: 2,496 and 2,495.
(ii) Digit sums from 2,495 to 2,502 are 20, 21, 22, 23, 24, 7, 8 and 9. So 2,500 is the only one, since 2 + 5 + 0 + 0 = 7.
In simple words: The marks go up by 1, and 2,500 is the one with digit sum 7.

Teacher's Note:
1. Crossing 2,500 changes the digit sum sharply.
2. List all eight marks before checking.

Question 10. In any supercell table, explain why the cell holding the largest number is always a supercell, while the cell holding the smallest number can never be a supercell.
Answer:
Largest number: it is bigger than every other number in the table. Its neighbours are some of those numbers, so it is bigger than all of them. Hence it is always a supercell.
Smallest number: it is smaller than every other number. It has at least one neighbour, and that neighbour is bigger. So it can never be greater than all its neighbours, and it is never a supercell.
In simple words: The biggest number beats all neighbours; the smallest loses to all of them.

Teacher's Note:
1. Bigger than all numbers means bigger than the neighbours too.
2. Every cell has at least one neighbour.

Question 11. Komal is looking for numbers whose digits add up to 20.
(i) Find the largest 5-digit number whose digits add up to 20 and explain your choice.
(ii) Find the smallest 4-digit number whose digits add up to 20 and explain your choice.
Answer:
(i) For the largest number, make the left digits as big as possible. Put 9 and 9 in the first two places; they use 18, leaving 2. Put 2 next and 0 in the rest.
The number is 99,200 (9 + 9 + 2 + 0 + 0 = 20).
(ii) For the smallest number, keep the left digits small and push the large digits to the right. The last two digits can give at most 9 + 9 = 18, so the first two must add up to 2; the first digit cannot be 0, so take 1 and 1.
The number is 1199 (1 + 1 + 9 + 9 = 20).
In simple words: Big digits go left for the largest number and right for the smallest.

Teacher's Note:
1. The first digit of a number is never 0.
2. Always check the digit sum at the end.

Question 12. Begin with 68 and keep reversing and adding until a palindrome appears. Show each addition and state how many additions were needed.
Answer:
68 + 86 = 154
154 + 451 = 605
605 + 506 = 1111
The palindrome 1111 is reached after 3 additions.
In simple words: 68 → 154 → 605 → 1111 in three additions.

Teacher's Note:
1. Check for a palindrome after every addition.
2. Reverse all the digits, not just the last two.

Question 13. How many rounds does the number 2451 take to reach the Kaprekar constant? Show every round.
Answer:
The digits of 2451 are 2, 4, 5 and 1.
Round 1: A = 5421, B = 1245, so 5421 − 1245 = 4176
Round 2: A = 7641, B = 1467, so 7641 − 1467 = 6174
The Kaprekar constant 6174 is reached in 2 rounds.
In simple words: 2451 → 4176 → 6174 takes 2 rounds.

Teacher's Note:
1. Each round starts from the digits of the last answer.
2. Stop when 6174 appears.

Question 14. Say whether each statement is always true, only sometimes true or never true, and give a reason.
(i) A 4-digit number added to a 4-digit number gives a 6-digit sum.
(ii) A 5-digit number minus a 3-digit number gives a 2-digit difference.
Answer:
(i) Never true. The biggest sum is 9,999 + 9,999 = 19,998, which has only 5 digits.
(ii) Never true. The smallest difference is 10,000 − 999 = 9,001, which has 4 digits.
In simple words: Use the extreme numbers: sums stay below 6 digits and differences stay at 4 digits or more.

Teacher's Note:
1. Largest + largest gives the biggest sum.
2. Smallest − largest gives the smallest difference.

Question 15. Find the sum of each design in a quick way and describe the short cut you used.
(i) The number 25 is written in every small square of a design that has 4 rows and 6 columns.
(ii) A design has 125 written in eight circles and 500 written in two squares.
Answer:
(i) Number of squares = 4 × 6 = 24, each with 25. Sum = 24 × 25 = 600.
Short cut: four 25s make 100, and there are 6 such groups, so 6 × 100 = 600.
(ii) 8 × 125 = 1000 and 2 × 500 = 1000. Sum = 1000 + 1000 = 2000.
Short cut: group the equal numbers, multiply, then add.
In simple words: Group equal numbers and multiply instead of adding one by one.

Teacher's Note:
1. Look for groups that make 100 or 1000.
2. Count the cells first.

Question 16. In the game of 21, the first player says 1, 2 or 3; after that the players take turns adding 1, 2 or 3 to the number just said, and whoever says 21 wins.
(i) Which player can always win if they play correctly?
(ii) Write the numbers that the winning player should say.
Answer:
Going back from 21 in steps of 4 gives the safe numbers 21, 17, 13, 9, 5, 1.
(i) The first player can always win, because 1 can be said in the first turn.
(ii) The winner should say 1, 5, 9, 13, 17 and 21. If the other player adds 1, 2 or 3, add 3, 2 or 1, so that each jump is 4.
In simple words: Say 1 first and keep jumping by 4 to 5, 9, 13, 17 and 21.

Teacher's Note:
1. Step size = 3 + 1 = 4.
2. Work backwards from the target.

Question 17. Dinesh writes all the numbers from 1 to 200 in his notebook. How many times does he write the digit ‘9’? Explain how you counted without writing all the numbers.
Answer:
Ones place: 9, 19, 29, …, 99 and 109, 119, …, 199. That is 10 numbers in each hundred, so 10 + 10 = 20 times.
Tens place: 90 to 99 and 190 to 199, again 10 + 10 = 20 times.
Total = 20 + 20 = 40 times. (In 99 and 199 the digit 9 is written twice, and both are counted.)
In simple words: Count 9s in the ones place and the tens place separately: 20 + 20 = 40.

Teacher's Note:
1. Count place by place.
2. The hundreds digit is never 9 here.

Question 18. A date written as dd/mm/yyyy is called palindromic if the eight digits read the same forwards and backwards.
(i) Show that 12/02/2021 is a palindromic date.
(ii) Find the palindromic date in the year 2022 and explain how you found it.
Answer:
(i) The eight digits are 1, 2, 0, 2, 2, 0, 2, 1. Read backwards they are 1, 2, 0, 2, 2, 0, 2, 1, exactly the same. So it is a palindromic date.
(ii) The day must be the last two digits of the year reversed, and the month must be the first two digits of the year reversed.
2022 has digits 2, 0, 2, 2, so the day is 22 and the month is 02. The date is 22/02/2022, which exists because February has more than 22 days.
In simple words: Reverse the year to get the day and month: 22/02/2022.

Teacher's Note:
1. The year decides the whole date.
2. Check that the date really exists.

Question 19. Meera claims that she takes about ten lakh breaths in one day. Taking about 20 breaths in a minute, estimate the number of breaths in a day and decide whether her claim is sensible.
Answer:
Breaths in one minute = 20
Breaths in one hour = 20 × 60 = 1200
Breaths in one day = 1200 × 24 = 28,800, which is about 30,000.
Ten lakh = 10,00,000, which is more than 30 times this. So Meera’s claim is not sensible.
In simple words: About 30,000 breaths a day is far less than ten lakh, so the claim is wrong.

Teacher's Note:
1. Build up minute → hour → day.
2. Compare by how many times bigger.

Question 20. Answer both parts about sums of multi-digit numbers.
(i) Write one 5-digit number and two 3-digit numbers whose sum is exactly 16,480.
(ii) Explain why no two 5-digit numbers can ever have the sum 18,500.
Answer:
(i) One answer: 15,000 + 980 + 500 = 16,480. (Another: 15,280 + 700 + 500 = 16,480.)
(ii) The smallest 5-digit number is 10,000, so two 5-digit numbers add up to at least 10,000 + 10,000 = 20,000. Since 18,500 is less than 20,000, it is impossible.
In simple words: Many answers work for (i); two 5-digit numbers always add to 20,000 or more.

Teacher's Note:
1. Check the digit count of every number you use.
2. Use the smallest numbers to find the least sum.