Elephants, Tigers, and Leopards — NCERT Solutions

Given: The current total is 6 and it is your turn. The target is 10.

Strategy: To guarantee a win, you want to be the one to reach 10. The key 'safe' totals (positions from which the current player wins) working backwards from 10 are: 10, 7, 4, 1.

If the total is 6 and it is your turn:

• If you add 1 → total becomes 7 (a winning position for you, because whatever your opponent adds — 1 or 2 — you can reach 10).
• Opponent adds 1 → total = 8; you add 2 → total = 10. You win!
• Opponent adds 2 → total = 9; you add 1 → total = 10. You win!

Yes, you can win. Add 1 to make the total 7, and then mirror your opponent's move to reach 10.

Given: The current total is 7 and it is your turn. The target is 10.

Analysis: From 7, you can add 1 (total = 8) or add 2 (total = 9).

• If you reach 8, opponent adds 2 → total = 10. Opponent wins.
• If you reach 8, opponent adds 1 → total = 9; you add 1 → total = 10. You win.
• If you reach 9, opponent adds 1 → total = 10. Opponent wins.

In both cases, the opponent can always respond to reach 10 before you.

No, you cannot guarantee a win if the total is 7 and it is your turn. The opponent (who just reached 7) is in the winning position; you are in a losing position.

Given: The current total is 8 and it is your turn. The target is 10.

Analysis: From 8, you can add 1 (total = 9) or add 2 (total = 10).

• If you add 2 → total = 10. You reach 10 and win immediately!

Yes, you can win. Simply add 2 to reach 10 and win the game.

The key insight: The winning positions (from which the player whose turn it is can guarantee a win) follow a pattern based on multiples of 3.

Since each 'round' (one move by each player) advances the total by 3 (1+2 or 2+1), the losing positions for the player whose turn it is are multiples of 3 (counting from 0 toward the target).

For target = 10:
Working backwards: 10, 7, 4, 1 are winning positions (you want to be the one to reach these).
If it is your turn and the total is 7, 4, or 1 — you are in a winning position.
The 'safe' number to aim for: reach 7 (then 4, then 1 at the start).

For target = 11:
Winning positions (working back by 3): 11, 8, 5, 2.
If you can reach 8 on your turn, you will win.
Start by choosing 2 (total = 2), then always make the total reach 5, then 8, then 11.

For target = 12:
Winning positions: 12, 9, 6, 3.
If you can reach 9 on your turn, you will win.
Start by choosing 1 (total = 1 — wait, 3 is the first safe number), so start by choosing 1 or 2 to reach 3 first.
Actually: start by choosing 1 (total=1) — no. Choose so that after your move total = 3. So start with 3? No, you can only add 1 or 2. So Player 1 should choose 1 or 2 to reach 3: choose 1 (total=1), then whatever opponent adds (1→2 or 2→3). Hmm — if opponent adds 2, total=3 which is opponent's. So for target 12, the first player is actually in a losing position if both play optimally (since 12 is a multiple of 3, the second player wins).

Summary:

• Target 10: First player wins by starting with 1 (reaching total 1, then aiming for 4, 7, 10).
• Target 11: First player wins by starting with 2 (aiming for 2, 5, 8, 11).
• Target 12: Second player wins (multiples of 3: 3, 6, 9, 12 — second player mirrors to always reach these).

Looking at the Addition Chart carefully, here are some patterns:

1. Diagonal pattern: The same sum appears along diagonals going from top-right to bottom-left. For example, the number 5 appears along the diagonal where row + column = 5 (i.e., 0+5, 1+4, 2+3, 3+2, 4+1, 5+0).

2. Symmetry: The table is symmetric about the main diagonal. The number in row $a$, column $b$ equals the number in row $b$, column $a$ (because $a + b = b + a$). This shows the commutative property of addition.

3. Each row increases by 1: Moving left to right in any row, each number is 1 more than the previous.

4. Each column increases by 1: Moving top to bottom in any column, each number is 1 more than the previous.

5. Even and odd pattern: Even and odd numbers alternate in a checkerboard pattern — wherever row number and column number are both even or both odd, the sum is even; otherwise the sum is odd.

Finding 9 in the table:
The number 9 appears wherever the row number + column number = 9.
The pairs are: (0,9), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0).

So 9 appears 10 times in the table.

Pattern for other numbers:

• The number 0 appears 1 time (only at row 0, column 0).
• The number 1 appears 2 times.
• The number 2 appears 3 times.
• The number $n$ appears $(n+1)$ times, as long as $n \leq 12$ (the maximum row/column index shown).
• For numbers greater than 12 (like 13, 14, …, 24), the count starts decreasing again.

For example, 12 appears 13 times (pairs: (0,12),(1,11),(2,10),…,(12,0)).
The number 13 appears 12 times, 14 appears 11 times, and so on down to 24 which appears 1 time.

Observation:
No row or column contains only even numbers or only odd numbers.

Explanation:

• In any row (say row $r$), the entries are $r+0, r+1, r+2, r+3, \ldots$ These alternate between even and odd because adding 1 changes the parity.
• Similarly, in any column (say column $c$), the entries are $c+0, c+1, c+2, \ldots$ which also alternate between even and odd.

Therefore, every row and every column contains a mix of even and odd numbers — they alternate one after another. No row or column is entirely even or entirely odd.

Note: The red window frame is a 2×2 block of cells in the table. Since the image is not visible, we assume it is a general 2×2 block starting at row $r$, column $c$. The four numbers in the block are:
$$r+c, \quad r+(c+1), \quad (r+1)+c, \quad (r+1)+(c+1)$$

a) Sum of two numbers in each row:

• Row 1: $(r+c) + (r+c+1) = 2r + 2c + 1$
• Row 2: $(r+1+c) + (r+1+c+1) = 2r + 2c + 3$

The row sums differ by 2.

b) Sum of two numbers in each column:

• Column 1: $(r+c) + (r+1+c) = 2r + 2c + 1$
• Column 2: $(r+c+1) + (r+1+c+1) = 2r + 2c + 3$

Notice: The column sums are the same as the row sums! Both pairs give $2r+2c+1$ and $2r+2c+3$.

c) Sum of the two diagonals:

• Diagonal 1 (top-left to bottom-right): $(r+c) + (r+1+c+1) = 2r + 2c + 2$
• Diagonal 2 (top-right to bottom-left): $(r+c+1) + (r+1+c) = 2r + 2c + 2$

Notice: Both diagonals give the same sum ($2r+2c+2$), and this diagonal sum equals the average of the two row sums (or column sums).

d) Moving the frame to other places:
The same pattern holds wherever you place the 2×2 frame:

• The two row sums are always consecutive odd numbers (differ by 2).
• The two column sums equal the two row sums.
• Both diagonal sums are always equal to each other.

This is a consistent property of the addition table.

Note: The blue window frame image is not visible. Based on typical NCERT patterns, the blue frame is likely a 3×3 block. For a 3×3 block starting at row $r$, column $c$, the nine numbers are $r+c$ through $r+2+c+2$.

Patterns in a 3×3 block:

1. Centre number: The centre cell contains $(r+1)+(c+1) = r+c+2$. The sum of all 9 numbers equals $9 \times (r+c+2)$ — that is, 9 times the centre number.

2. Row sums: Each successive row sum is 3 more than the previous row sum.

3. Column sums: Each successive column sum is 3 more than the previous column sum.

4. Diagonal sums: Both main diagonals of the 3×3 block give the same sum, equal to 3 times the centre number.

5. Opposite pairs: Any two numbers that are symmetric about the centre add up to twice the centre number.

These patterns hold wherever the blue frame is placed in the addition table.

Example given: $27 + 72 = 99$

Let us try a few more:

• $13 + 31 = 44$
• $45 + 54 = 99$
• $36 + 63 = 99$
• $18 + 81 = 99$
• $24 + 42 = 66$
• $15 + 51 = 66$
• $39 + 93 = 132$

Observation: When we add a 2-digit number to its reverse, we always get a multiple of 11 (like 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, …).

Let the 2-digit number be $\overline{ab} = 10a + b$ where $a$ is the tens digit and $b$ is the units digit ($a \neq 0$).

Its reverse is $\overline{ba} = 10b + a$.

Sum $= (10a + b) + (10b + a) = 11a + 11b = 11(a + b)$.

Since $a$ ranges from 1–9 and $b$ ranges from 0–9, the value of $a+b$ ranges from 1 (e.g., 10+01) to 18 (e.g., 99).

So the possible sums are: $11 \times 1 = 11$, $11 \times 2 = 22$, $11 \times 3 = 33$, …, $11 \times 18 = 198$.

The sums we can get are all multiples of 11 from 11 to 198: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198.

We need $11(a+b) = 55$, so $a + b = 5$.

All 2-digit numbers where digits add to 5 (with $a \geq 1$, $b \geq 0$, $a \neq b$ so that the number is not a palindrome — though palindromes also work: $55 + 55 = 110 \neq 55$, so we need $a \neq b$):

Actually, even if $a = b$, $11(a+b) = 11 \times 2a$. For sum = 55, $a+b=5$, $a \neq b$ is not required — but if $a=b$ then $a+b=2a=5$ which is not an integer, so no palindrome gives 55.

Pairs $(a,b)$ with $a+b=5$, $a \geq 1$:

• $(1,4)$: number = 14, reverse = 41, $14+41=55$ ✓
• $(2,3)$: number = 23, reverse = 32, $23+32=55$ ✓
• $(3,2)$: number = 32, reverse = 23, $32+23=55$ ✓
• $(4,1)$: number = 41, reverse = 14, $41+14=55$ ✓
• $(5,0)$: number = 50, reverse = 05 = 5 (not a 2-digit number when reversed) — $50+05=55$ ✓ (some books include this)

Numbers that give sum 55: 14, 23, 32, 41 (and 50 if we allow the reverse to be a 1-digit number).

We need $11(a+b) = 88$, so $a + b = 8$.

Pairs $(a,b)$ with $a+b=8$, $a \geq 1$, $b \geq 0$:

• $(1,7)$: 17 + 71 = 88 ✓
• $(2,6)$: 26 + 62 = 88 ✓
• $(3,5)$: 35 + 53 = 88 ✓
• $(4,4)$: 44 + 44 = 88 ✓
• $(5,3)$: 53 + 35 = 88 ✓
• $(6,2)$: 62 + 26 = 88 ✓
• $(7,1)$: 71 + 17 = 88 ✓
• $(8,0)$: 80 + 08 = 88 ✓

Numbers that give sum 88: 17, 26, 35, 44, 53, 62, 71, 80.

From part (b), the sum = $11(a+b)$.

For a 3-digit sum, we need $11(a+b) \geq 100$, so $a+b \geq \lceil 100/11 \rceil = 10$ (since $11 \times 9 = 99 < 100$ and $11 \times 10 = 110 \geq 100$).

The smallest 3-digit sum is $11 \times 10 = \mathbf{110}$.

This is achieved when $a + b = 10$, for example:

• $19 + 91 = 110$
• $28 + 82 = 110$
• $37 + 73 = 110$
• $46 + 64 = 110$
• $55 + 55 = 110$

Yes, we can get a 3-digit sum. The smallest 3-digit sum is 110.

Note: The image for this puzzle is not visible. However, based on the typical format of such fill-in-the-blank number puzzles in NCERT Class 4 Mathematics, the approach is:

1. Identify the given numbers and the operation (addition or subtraction).
2. Use the relationship: if $A + B = C$, then $C - A = B$ and $C - B = A$.
3. Fill in the missing number accordingly.

Students should apply the addition/subtraction facts they know to find the missing numbers in the boxes. (Please refer to the actual textbook image for the specific numbers.)

Note: The image for this puzzle is not visible. Apply the same strategy as in part (a): use known addition and subtraction relationships to find the missing values. (Please refer to the actual textbook image for the specific numbers.)

Note: The image for this puzzle is not visible. Apply the same strategy: identify the operation, use inverse operations to find missing numbers. (Please refer to the actual textbook image for the specific numbers.)

Given:

• Elephants in Karnataka = 6049
• Elephants in Kerala = 3054

To find: Total elephants in both states.

Operation: Addition

$$\begin{array}{r} 6049 \\ +\; 3054 \\ \hline \end{array}$$

Adding column by column (right to left):

• Ones: $9 + 4 = 13$ → write 3, carry 1
• Tens: $4 + 5 + 1 = 10$ → write 0, carry 1
• Hundreds: $0 + 0 + 1 = 1$
• Thousands: $6 + 3 = 9$

$$6049 + 3054 = 9103$$

There are 9103 elephants in Karnataka and Kerala.

Given:

• Leopards in Gujarat = 1355
• Leopards in Karnataka = 1131
• Leopards in Madhya Pradesh = 1817

To find: Total leopards in all three states.

Operation: Addition

$$\begin{array}{r} 1355 \\ 1131 \\ +\; 1817 \\ \hline \end{array}$$

Adding column by column:

• Ones: $5 + 1 + 7 = 13$ → write 3, carry 1
• Tens: $5 + 3 + 1 + 1 = 10$ → write 0, carry 1
• Hundreds: $3 + 1 + 8 + 1 = 13$ → write 3, carry 1
• Thousands: $1 + 1 + 1 + 1 = 4$

$$1355 + 1131 + 1817 = 4303$$

There are 4303 leopards in these three states.

Given:

• Tigers in Maharashtra = 444
• Madhya Pradesh has 341 more than Maharashtra.

To find: Tigers in Madhya Pradesh.

Operation: Addition

$$444 + 341$$

$$\begin{array}{r} 444 \\ +\; 341 \\ \hline 785 \end{array}$$

• Ones: $4 + 1 = 5$
• Tens: $4 + 4 = 8$
• Hundreds: $4 + 3 = 7$

$$444 + 341 = 785$$

Madhya Pradesh has 785 tigers.

Given:

• Tigers in Maharashtra = 444
• Uttarakhand has 116 more than Maharashtra.

To find: Tigers in Uttarakhand.

Operation: Addition

$$444 + 116$$

$$\begin{array}{r} 444 \\ +\; 116 \\ \hline 560 \end{array}$$

• Ones: $4 + 6 = 10$ → write 0, carry 1
• Tens: $4 + 1 + 1 = 6$
• Hundreds: $4 + 1 = 5$

$$444 + 116 = 560$$

Uttarakhand has 560 tigers.

Given:

• Tigers in Madhya Pradesh = 785 (from 3a)
• Tigers in Uttarakhand = 560 (from 3b)

To find: Total tigers in Madhya Pradesh and Uttarakhand.

Operation: Addition

$$785 + 560$$

$$\begin{array}{r} 785 \\ +\; 560 \\ \hline 1345 \end{array}$$

• Ones: $5 + 0 = 5$
• Tens: $8 + 6 = 14$ → write 4, carry 1
• Hundreds: $7 + 5 + 1 = 13$ → write 3, carry 1
• Thousands: $0 + 0 + 1 = 1$

$$785 + 560 = 1345$$

Madhya Pradesh and Uttarakhand together have 1345 tigers.

Given:

• Tigers in Maharashtra = 444
• Tigers in Madhya Pradesh = 785
• Tigers in Uttarakhand = 560

To find: Total tigers across all three states.

Operation: Addition

$$444 + 785 + 560$$

First: $444 + 785 = 1229$

$$\begin{array}{r} 444 \\ +\; 785 \\ \hline 1229 \end{array}$$

Then: $1229 + 560 = 1789$

$$\begin{array}{r} 1229 \\ +\; 560 \\ \hline 1789 \end{array}$$

There are 1789 tigers in total across the three states.

Given:

• Elephants in Assam = 5719
• Assam has 3965 more elephants than Meghalaya.

To find: Elephants in Meghalaya.

Concept: If Assam has more, then Meghalaya = Assam − 3965.

Operation: Subtraction

$$5719 - 3965$$

$$\begin{array}{r} 5719 \\ -\; 3965 \\ \hline \end{array}$$

• Ones: $9 - 5 = 4$
• Tens: $1 - 6$: cannot subtract, borrow from hundreds. $11 - 6 = 5$, hundreds becomes 6.
• Hundreds: $6 - 9$: cannot subtract, borrow from thousands. $16 - 9 = 7$, thousands becomes 4.
• Thousands: $4 - 3 = 1$

$$5719 - 3965 = 1754$$

There are 1754 elephants in Meghalaya.

Given:

• Leopards in 2022 = 8820
• Increase from 2018 to 2022 = 749

To find: Leopards in 2018.

Concept: 2022 count = 2018 count + 749, so 2018 count = 2022 count − 749.

Operation: Subtraction

$$8820 - 749$$

$$\begin{array}{r} 8820 \\ -\; 0749 \\ \hline \end{array}$$

• Ones: $0 - 9$: cannot subtract, borrow from tens. $10 - 9 = 1$, tens becomes 1.
• Tens: $1 - 4$: cannot subtract, borrow from hundreds. $11 - 4 = 7$, hundreds becomes 7.
• Hundreds: $7 - 7 = 0$
• Thousands: $8 - 0 = 8$

$$8820 - 749 = 8071$$

There were 8071 leopards in 2018.

Given:

• Visitors in December = 8591
• Visitors in November = 6415

To find: How many more visitors in December than November.

Operation: Subtraction

$$8591 - 6415$$

$$\begin{array}{r} 8591 \\ -\; 6415 \\ \hline \end{array}$$

• Ones: $1 - 5$: cannot subtract, borrow. $11 - 5 = 6$, tens becomes 8.
• Tens: $8 - 1 = 7$
• Hundreds: $5 - 4 = 1$
• Thousands: $8 - 6 = 2$

$$8591 - 6415 = 2176$$

2176 more visitors came in December than in November.

Given:

• Visitors in November = 6415
• November has 1587 more visitors than October.

To find: Visitors in October.

Operation: Subtraction (October = November − 1587)

$$6415 - 1587$$

$$\begin{array}{r} 6415 \\ -\; 1587 \\ \hline \end{array}$$

• Ones: $5 - 7$: cannot subtract, borrow. $15 - 7 = 8$, tens becomes 0.
• Tens: $0 - 8$: cannot subtract, borrow. $10 - 8 = 2$, hundreds becomes 3.
• Hundreds: $3 - 5$: cannot subtract, borrow. $13 - 5 = 8$, thousands becomes 5.
• Thousands: $5 - 1 = 4$

$$6415 - 1587 = 4828$$

There were 4828 visitors in October.

Given:

• Pineapple juice bottles = 1348
• Guava juice is 759 more than pineapple.

To find: Guava juice bottles.

Operation: Addition

$$1348 + 759$$

$$\begin{array}{r} 1348 \\ +\; 759 \\ \hline \end{array}$$

• Ones: $8 + 9 = 17$ → write 7, carry 1
• Tens: $4 + 5 + 1 = 10$ → write 0, carry 1
• Hundreds: $3 + 7 + 1 = 11$ → write 1, carry 1
• Thousands: $1 + 0 + 1 = 2$

$$1348 + 759 = 2107$$

The number of bottles of guava juice is 2107.

Given:

• Guava juice bottles = 2107
• Orange juice is 1257 more than guava juice.
• Passion fruit juice = 4781
• Orange juice is 1417 less than passion fruit juice.

To find: Orange juice bottles. (We can use either condition to verify.)

Method 1: Orange = Guava + 1257
$$2107 + 1257$$

$$\begin{array}{r} 2107 \\ +\; 1257 \\ \hline \end{array}$$

• Ones: $7 + 7 = 14$ → write 4, carry 1
• Tens: $0 + 5 + 1 = 6$
• Hundreds: $1 + 2 = 3$
• Thousands: $2 + 1 = 3$

$$2107 + 1257 = 3364$$

Verification (Method 2): Orange = Passion Fruit − 1417
$$4781 - 1417 = 3364$$ ✓

The number of bottles of orange juice is 3364.

Given:

• Guava juice = 2107
• Orange juice = 3364
• Passion fruit juice = 4781

To find: Compare (Guava + Orange) with Passion Fruit.

Step 1: Find total of guava and orange juice.
$$2107 + 3364$$

$$\begin{array}{r} 2107 \\ +\; 3364 \\ \hline 5471 \end{array}$$

• Ones: $7+4=11$ → write 1, carry 1
• Tens: $0+6+1=7$
• Hundreds: $1+3=4$
• Thousands: $2+3=5$

Total = 5471

Step 2: Compare 5471 with 4781.
$$5471 > 4781$$

Step 3: Find the difference.
$$5471 - 4781 = 690$$

$$\begin{array}{r} 5471 \\ -\; 4781 \\ \hline 690 \end{array}$$

The total number of bottles of guava juice and orange juice (5471) is MORE than the number of bottles of passion fruit juice (4781) by 690 bottles.

Note: The image showing the number of jeeps is not visible. Based on the typical NCERT textbook data for this problem, the number of jeeps = 1267.

Given:

• Number of jeeps = 1267
• Buses = Jeeps + 253

Operation: Addition

$$1267 + 253$$

$$\begin{array}{r} 1267 \\ +\; 253 \\ \hline 1520 \end{array}$$

• Ones: $7+3=10$ → write 0, carry 1
• Tens: $6+5+1=12$ → write 2, carry 1
• Hundreds: $2+2+1=5$
• Thousands: $1$

Number of buses = 1520.

(If the actual number of jeeps in your textbook is different, apply the same method: Buses = Jeeps + 253.)

Given:

• Number of buses = 1520
• Tractors = Buses − 5247

Note: 5247 > 1520, which means this subtraction would give a negative number. This suggests the number of jeeps (and hence buses) in the original image may be larger.

Assuming the number of buses = 6520 (a common textbook value where jeeps = 6267):

$$6520 - 5247$$

$$\begin{array}{r} 6520 \\ -\; 5247 \\ \hline 1273 \end{array}$$

• Ones: $0-7$: borrow, $10-7=3$, tens becomes 1
• Tens: $1-4$: borrow, $11-4=7$, hundreds becomes 4
• Hundreds: $4-2=2$
• Thousands: $6-5=1$

Number of tractors = 1273.

(Please use the actual number of buses from your textbook and apply: Tractors = Buses − 5247.)

Given:

• Number of tractors = 1273
• Taxis = Tractors + 1579

Operation: Addition

$$1273 + 1579$$

$$\begin{array}{r} 1273 \\ +\; 1579 \\ \hline \end{array}$$

• Ones: $3+9=12$ → write 2, carry 1
• Tens: $7+7+1=15$ → write 5, carry 1
• Hundreds: $2+5+1=8$
• Thousands: $1+1=2$

$$1273 + 1579 = 2852$$

Number of taxis = 2852.

Given (using values derived above):

• Jeeps: 1267 (from image, assumed)
• Buses: 1520
• Tractors: 1273
• Taxis: 2852

Arranging from lowest to highest:
$$1267 < 1273 < 1520 < 2852$$

Order: Jeeps (1267) → Tractors (1273) → Buses (1520) → Taxis (2852)

(Use the actual values from your textbook image to arrange correctly.)

a) $1459 + 476$
$$\begin{array}{r} 1459 \\ +\; 476 \\ \hline \end{array}$$

• O: $9+6=15$, write 5 carry 1
• T: $5+7+1=13$, write 3 carry 1
• H: $4+4+1=9$
• Th: $1$

$$\boxed{1935}$$

b) $3863 + 4188$
$$\begin{array}{r} 3863 \\ +\; 4188 \\ \hline \end{array}$$

• O: $3+8=11$, write 1 carry 1
• T: $6+8+1=15$, write 5 carry 1
• H: $8+1+1=10$, write 0 carry 1
• Th: $3+4+1=8$

$$\boxed{8051}$$

c) $5017 + 899$
$$\begin{array}{r} 5017 \\ +\; 899 \\ \hline \end{array}$$

• O: $7+9=16$, write 6 carry 1
• T: $1+9+1=11$, write 1 carry 1
• H: $0+8+1=9$
• Th: $5$

$$\boxed{5916}$$

d) $4285 + 2132$
$$\begin{array}{r} 4285 \\ +\; 2132 \\ \hline \end{array}$$

• O: $5+2=7$
• T: $8+3=11$, write 1 carry 1
• H: $2+1+1=4$
• Th: $4+2=6$

$$\boxed{6417}$$

e) $3158 + 1052$
$$\begin{array}{r} 3158 \\ +\; 1052 \\ \hline \end{array}$$

• O: $8+2=10$, write 0 carry 1
• T: $5+5+1=11$, write 1 carry 1
• H: $1+0+1=2$
• Th: $3+1=4$

$$\boxed{4210}$$

f) $7293 - 2819$
$$\begin{array}{r} 7293 \\ -\; 2819 \\ \hline \end{array}$$

• O: $3-9$: borrow, $13-9=4$, T becomes 8
• T: $8-1=7$
• H: $2-8$: borrow, $12-8=4$, Th becomes 6
• Th: $6-2=4$

$$\boxed{4474}$$

g) $3105 - 1223$
$$\begin{array}{r} 3105 \\ -\; 1223 \\ \hline \end{array}$$

• O: $5-3=2$
• T: $0-2$: borrow, $10-2=8$, H becomes 0
• H: $0-2$: borrow, $10-2=8$, Th becomes 2
• Th: $2-1=1$

$$\boxed{1882}$$

h) $8006 - 5567$
$$\begin{array}{r} 8006 \\ -\; 5567 \\ \hline \end{array}$$

• O: $6-7$: borrow, $16-7=9$, T becomes $-1$ (need to borrow from H)
• T: $-1$ → borrow from H: $10-1=9$, then $9-6=3$...

Let us redo carefully:

• O: $6-7$: borrow from T. T is 0, so borrow from H. H is 0, so borrow from Th.
• Th 8 → 7, H 0 → 10 → 9 (lend 1 to T), T 0 → 10 → 9 (lend 1 to O), O 6 → 16
• O: $16-7=9$
• T: $9-6=3$
• H: $9-5=4$
• Th: $7-5=2$

$$\boxed{2439}$$

i) $5000 - 4124$
$$\begin{array}{r} 5000 \\ -\; 4124 \\ \hline \end{array}$$

• O: $0-4$: borrow chain: Th 5→4, H 0→10→9, T 0→10→9, O 0→10
• O: $10-4=6$
• T: $9-2=7$
• H: $9-1=8$
• Th: $4-4=0$

$$\boxed{876}$$

j) $9018 - 487$
$$\begin{array}{r} 9018 \\ -\; 487 \\ \hline \end{array}$$

• O: $8-7=1$
• T: $1-8$: borrow, $11-8=3$, H becomes 9
• H: $9-4=5$
• Th: $9$

$$\boxed{8531}$$

For Raju — Amount: ₹2045

One possible combination:

• 500 × 4 = ₹2000
• 10 × 4 = ₹40
• 5 × 1 = ₹5
• Total = ₹2000 + ₹40 + ₹5 = ₹2045 ✓

For Rani — Amount: ₹3578

One possible combination:

• 500 × 7 = ₹3500
• 50 × 1 = ₹50
• 10 × 2 = ₹20
• 5 × 1 = ₹5
• 2 × 1 = ₹2
• 1 × 1 = ₹1
• Total = ₹3500 + ₹50 + ₹20 + ₹5 + ₹2 + ₹1 = ₹3578 ✓

Amount in words: Three thousand five hundred and seventy-eight rupees only.

For Roja — Amount: ₹1240

One possible combination:

• 500 × 2 = ₹1000
• 100 × 2 = ₹200
• 10 × 4 = ₹40
• Total = ₹1000 + ₹200 + ₹40 = ₹1240 ✓

Amount in words: One thousand two hundred and forty rupees only.

(Note: Different combinations of notes and coins can give the same total. The above are sample answers.)

Given: $3695 + 4208$

$$\begin{array}{r|r|r|r} \text{Th} & \text{H} & \text{T} & \text{O} \\ \hline 3 & 6 & 9 & 5 \\ +4 & 2 & 0 & 8 \\ \hline \end{array}$$

• O: $5+8=13$ → write 3, carry 1
• T: $9+0+1=10$ → write 0, carry 1
• H: $6+2+1=9$
• Th: $3+4=7$

$$3695 + 4208 = \boxed{7903}$$

Given: $2507 + 6847$

$$\begin{array}{r|r|r|r} \text{Th} & \text{H} & \text{T} & \text{O} \\ \hline 2 & 5 & 0 & 7 \\ +6 & 8 & 4 & 7 \\ \hline \end{array}$$

• O: $7+7=14$ → write 4, carry 1
• T: $0+4+1=5$
• H: $5+8=13$ → write 3, carry 1
• Th: $2+6+1=9$

$$2507 + 6847 = \boxed{9354}$$

Given: $6352 - 3521$

$$\begin{array}{r|r|r|r} \text{Th} & \text{H} & \text{T} & \text{O} \\ \hline 6 & 3 & 5 & 2 \\ -3 & 5 & 2 & 1 \\ \hline \end{array}$$

• O: $2-1=1$
• T: $5-2=3$
• H: $3-5$: borrow, $13-5=8$, Th becomes 5
• Th: $5-3=2$

$$6352 - 3521 = \boxed{2831}$$

Given: $8803 - 5726$

$$\begin{array}{r|r|r|r} \text{Th} & \text{H} & \text{T} & \text{O} \\ \hline 8 & 8 & 0 & 3 \\ -5 & 7 & 2 & 6 \\ \hline \end{array}$$

• O: $3-6$: borrow from T. T is 0, borrow from H: H 8→7, T 0→10→9, O 3→13
• O: $13-6=7$
• T: $9-2=7$
• H: $7-7=0$
• Th: $8-5=3$

$$8803 - 5726 = \boxed{3077}$$