NCERT Solutions for Class 5th Maths Chapter 1 Pastime Mathematics — Pastime Mathematics
Book page 11–13 Updated on2026-09-19
Q1.
1. Mira poses the river crossing puzzle to Sanju. A boatman wants to cross a river in a boat. He has to take a lion, a sheep, and a bundle of grass with him. He can take one of them at a time. If the sheep and grass are left on the shore, the sheep will eat the grass. And, if the sheep and lion are left on the shore, the lion will eat the sheep. How can the boatman take the lion, sheep, and grass across the river? Help him so that he can ferry the lion, sheep, and grass across the river safely, and in the minimum number of trips.
The river crossing puzzle. The grass, the lion and the sheep are all on this bank; the boatman must take them across, one at a time.
Answer
It can be done in 7 trips. The trick is that the boatman is allowed to bring something back.
Trip
What the boatman does
Left behind on this side
Waiting on the far side
1
Takes the sheep across
lion, grass
sheep
2
Comes back alone
lion, grass
sheep
3
Takes the grass across
lion
sheep, grass
4
Brings the sheep back
lion, sheep
grass
5
Takes the lion across
sheep
lion, grass
6
Comes back alone
sheep
lion, grass
7
Takes the sheep across
—
lion, sheep, grass
Check every stage. The lion and the sheep are never left alone together. The sheep and the grass are never left alone together. So nothing gets eaten.
Why the sheep goes first: The sheep is the troublemaker — it is in danger from the lion and it is a danger to the grass. The lion and the grass, however, can safely sit together, because a lion does not eat grass. So the sheep must be moved out of the way first, and moved back later to keep the lion and grass apart.
Try This: Play it with three paper slips on your desk. Then try starting with the grass instead of the sheep and see how quickly it goes wrong.
Q2.
2. Sanju introduces a game called pile of pebbles to Mira. There are two piles of pebbles. Each pile contains 7 pebbles. Each player can pick as many pebbles they want from either of the piles. The player who picks the last pebble wins. Try this game with your friends. Now, how do you play so that you win? To find a winning strategy, try playing with 1 pebble in each pile, two in each, three in each, and so on.
Answer
Play second, and always make the two piles equal again. Do that every turn and you will win.
Step 1 — try the smallest game: 1 pebble in each pile. Whatever the first player takes, one pebble is left in the other pile, and you take it and win. So the second player wins.
Step 2 — try 2 in each pile. Say the first player takes 1 from a pile, leaving 1 and 2. You take 1 from the other pile, making it 1 and 1 — back to the game you already know you win. If instead they take both from a pile, you take both from the other. Second player wins again.
Step 3 — try 3 in each. Same thing. Copy their move in the other pile, so the piles stay equal.
Step 4 — the rule for 7 and 7.
Let the other player go first. Whatever they take from one pile, take exactly the same number from the other pile. The piles are equal after every one of your turns.
An example game:
Move
Player
Takes
Piles now
Start
—
—
7 · 7
1
Friend
3 from pile A
4 · 7
2
You
3 from pile B
4 · 4
3
Friend
4 from pile B
4 · 0
4
You
4 from pile A
0 · 0 — you took the last pebble, you win
Why copying works: As long as the piles are equal, there is always something for you to copy — the other pile still has at least as many pebbles as your friend just removed. So you can never get stuck. And you always leave equal piles behind, which means your friend can never leave the table empty. The person who empties it is always you.
Try This: Start with unequal piles, say 5 and 3. Now the first player wins: take 2 from the pile of 5 to make it 3 and 3, and then just copy.
Q3.
3. Now, it's Mira's turn. She gives a fun puzzle to Sanju: (a) Take any two different digits. (b) Make two 2-digit numbers using them. (c) Subtract the smaller number from the bigger number. Now, use the two digits in the difference and repeat steps (b) and (c). Continue until you get a 1-digit number. Mira exclaimed, No matter which two numbers you choose, you will get 9 in the end. How did Mira know what the 1-digit number in the end would be?
Answer
Mira knew because every difference you can ever get is a number in the 9 times table, and the chain of 9s always ends at 9 itself.
Step 1 — see what one subtraction really does. Take the digits 3 and 7.
Difference of the numbers = 9 × (difference of the two digits) Here: 7 – 3 = 4, and 9 × 4 = 36 ✓
Step 3 — so every answer is a multiple of 9. The only 2-digit multiples of 9 are 18, 27, 36, 45, 54, 63, 72, 81. Whichever one you land on, its two digits get subtracted again — and that gives another multiple of 9.
Why it always stops at 9: The only 1-digit multiple of 9 is 9 itself. So once the chain shrinks to one digit, that digit has to be 9. There is nowhere else for it to land.
Q4.
(1) Observe the differences you get in each step above. Do you notice anything in common?
Answer
The differences in the book's example are 36, 27, 45, 9. Two things are true of every one of them.
Each one is in the 9 times table.
The digits of each one add up to 9.
Difference
Is it 9 × something?
Digits added
36
9 × 4
3 + 6 = 9
27
9 × 3
2 + 7 = 9
45
9 × 5
4 + 5 = 9
9
9 × 1
9
Tip: Adding the digits is a quick test for the 9 times table. If the digits of a number add up to 9, the number is a multiple of 9.
Q5.
(2) Try the puzzle using any other pair of digits. What is common to these differences? What do you get in the end? The steps of Mira’s puzzle, with the book’s worked example beside them.
What is common: every difference is a multiple of 9 — 63, 27, 45, 9. And you always end at 9, however many steps it takes.
Why: One subtraction always gives 9 × (difference of the digits), so the answer can only be 9, 18, 27, 36, 45, 54, 63, 72 or 81. Each of these feeds back into the machine and gives another multiple of 9. The chain shrinks until only the single-digit multiple of 9 is left — and that is 9.
Q6.
(3) What digits can you choose so that you get a 1-digit number in the first step itself? Give some examples. Describe the pattern in the digits. The steps of Mira’s puzzle, with the book’s worked example beside them.
Answer
Choose two digits that are next-door neighbours — digits that differ by 1.
Examples: 2 and 3 → 32 – 23 = 9 4 and 5 → 54 – 45 = 9 6 and 7 → 76 – 67 = 9 8 and 9 → 98 – 89 = 9
The full list of such pairs: (1, 2) · (2, 3) · (3, 4) · (4, 5) · (5, 6) · (6, 7) · (7, 8) · (8, 9).
The pattern: the two digits are consecutive — one is exactly 1 more than the other.
Why it works: The difference is always 9 × (difference of the digits). To get a 1-digit answer, that difference must be 9 itself. So 9 × (difference of digits) = 9, which means the two digits must differ by exactly 1.
Careful: The pair (0, 1) does not work, because the two numbers would be 10 and 01 — and 01 is not a proper 2-digit number.
Q7.
(4) Now, find different digits such that the difference between the numbers is 27.
Answer
Choose two digits that are 3 apart.
Why 3? Because difference of numbers = 9 × (difference of digits) 27 = 9 × 3, so the digits must differ by 3.
All the pairs that work:
Digits
The two numbers
Subtraction
1, 4
41 and 14
41 – 14 = 27
2, 5
52 and 25
52 – 25 = 27
3, 6
63 and 36
63 – 36 = 27
4, 7
74 and 47
74 – 47 = 27
5, 8
85 and 58
85 – 58 = 27
6, 9
96 and 69
96 – 69 = 27
Check it yourself: Pick any row and do the subtraction on paper. Every single one gives 27.
Q8.
(5) Mira found an interesting relationship between the two digits and the difference obtained. Can you see it in the table that Mira made?
Digits
Differences in digits
Difference in numbers formed by the digits
3, 7
7 – 3 = 4
73 – 37 = 36
1, 9
9 – 1 = 8
91 – 19 = 72
2, 8
8 – 2 = 6
82 – 28 = 54
4, 5
5 – 4 = 1
54 – 45 = 9
The table Mira made, printed on page 13.
Answer
Yes. The relationship is:
Difference in the numbers = 9 × Difference in the digits
Check it on every row of Mira's table.
Digits
Difference in digits
9 × that
Difference in numbers
3, 7
4
9 × 4 = 36
73 – 37 = 36 ✓
1, 9
8
9 × 8 = 72
91 – 19 = 72 ✓
2, 8
6
9 × 6 = 54
82 – 28 = 54 ✓
4, 5
1
9 × 1 = 9
54 – 45 = 9 ✓
Why the 9 appears: Say the bigger digit is written first. The bigger number has that digit in the tens place; the smaller number has it in the ones place. So the tens go up by the digit difference and the ones go down by the same digit difference. That is 10 times it minus 1 times it — which is 9 times it.
Q9.
Extend this table by choosing appropriate digits so that the resulting differences are 2, 3, 5, and 7 respectively.
Digits
Differences in digits
Difference in numbers formed by the digits
3, 7
7 – 3 = 4
73 – 37 = 36
1, 9
9 – 1 = 8
91 – 19 = 72
2, 8
8 – 2 = 6
82 – 28 = 54
4, 5
5 – 4 = 1
54 – 45 = 9
The table Mira made, printed on page 13, with four empty rows for you to add.
Answer
Mira's table already covers digit differences 4, 8, 6 and 1. The four missing ones are 2, 3, 5 and 7. Here they are.
Digits
Difference in digits
Difference in the numbers formed
3, 5
5 – 3 = 2
53 – 35 = 18
1, 4
4 – 1 = 3
41 – 14 = 27
2, 7
7 – 2 = 5
72 – 27 = 45
1, 8
8 – 1 = 7
81 – 18 = 63
Check each one against the rule (difference in numbers = 9 × difference in digits):
Why 18 takes the longest: 18 sends you to 63, and 63 still needs three more rounds (27, then 45, then 9). Every other starting point joins the chain further along. The book's own example, 3 and 7, starts at 36 and takes 4 subtractions — one short of the record.