Decoration for Festival (Addition and Subtraction)

Given: Each garland has 10 flowers.

We write the number of tens (garlands) and ones (extra flowers) and find the total.

| Name | Tens (garlands) | Ones (extra flowers) | Total |
|---|---|---|---|
| Rohan | 1 | 2 | $10 + 2 = 12$ |
| Simran | 3 | 7 | $30 + 7 = 37$ |
| Akash | 4 | 8 | $40 + 8 = 48$ |
| Suwali | 5 | 3 | $50 + 3 = 53$ |
| Javed | 4 | 3 | $40 + 3 = 43$ |
| Zoha | 2 | 0 | $20 + 0 = 20$ |

(Note: Javed's and Zoha's values are read from the picture described in the chapter. Javed = 43 flowers, Zoha = 20 flowers.)

Given:
- Rohan's flowers = $12$
- Suwali's flowers = $53$

We add using the column method (Tens and Ones):

$$\begin{array}{r} \text{T} \quad \text{O} \\ 1 \quad 2 \\ +\; 5 \quad 3 \\ \hline 6 \quad 5 \end{array}$$

Step 1: Add Ones → $2 + 3 = 5$
Step 2: Add Tens → $1 + 5 = 6$

$$12 + 53 = \boxed{65}$$

Rohan and Suwali together have 65 flowers.

Given:
- Zoha's flowers = $20$
- Javed's flowers = $43$

Using column method:

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 0 \\ +\; 4 \quad 3 \\ \hline 6 \quad 3 \end{array}$$

Step 1: Add Ones → $0 + 3 = 3$
Step 2: Add Tens → $2 + 4 = 6$

$$20 + 43 = \boxed{63}$$

Zoha and Javed together have 63 flowers.

Given:
- Zoha's flowers = $20$ (Tens = 2, Ones = 0)
- Akash's flowers = $48$ (Tens = 4, Ones = 8)

Using column method:

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 0 \\ +\; 4 \quad 8 \\ \hline 6 \quad 8 \end{array}$$

Step 1: Add Ones → $0 + 8 = 8$
Step 2: Add Tens → $2 + 4 = 6$

$$20 + 48 = \boxed{68}$$

Zoha and Akash together have 68 flowers.

| | Tens | Ones | Total |
|---|---|---|---|
| Zoha brings | 2 | 0 | $20 + 0 = 20$ |
| Akash brings | 4 | 8 | $40 + 8 = 48$ |
| Together | 6 | 8 | $60 + 8 = 68$ |

Given:
- Rohan's flowers = $12$ (Tens = 1, Ones = 2)
- Simran's flowers = $37$ (Tens = 3, Ones = 7)

Using column method:

$$\begin{array}{r} \text{T} \quad \text{O} \\ 1 \quad 2 \\ +\; 3 \quad 7 \\ \hline 4 \quad 9 \end{array}$$

Step 1: Add Ones → $2 + 7 = 9$
Step 2: Add Tens → $1 + 3 = 4$

$$12 + 37 = \boxed{49}$$

Rohan and Simran together have 49 flowers.

| | Tens | Ones | Total |
|---|---|---|---|
| Rohan brings | 1 | 2 | $10 + 2 = 12$ |
| Simran brings | 3 | 7 | $30 + 7 = 37$ |
| Together | 4 | 9 | $40 + 9 = 49$ |

Given: $25 + 12$

Method: Start at 25. First jump 10 steps forward to reach 35. Then jump 2 more steps forward to reach 37.

$$25 + 10 = 35$$
$$35 + 2 = 37$$

$$25 + 12 = \boxed{37}$$

Given: $57 + 34$

Method (breaking 34 into tens and ones):
$$57 + 34 = 57 + 30 + 4$$

Step 1: $57 + 30 = 87$
Step 2: $87 + 4 = 91$

$$57 + 34 = \boxed{91}$$

On the number line: Count 34 ahead of 57.

Given: $24 + 43$

Method (breaking 43 into tens and ones):
$$24 + 43 = 24 + 40 + 3$$

Step 1: $24 + 40 = 64$
Step 2: $64 + 3 = 67$

$$24 + 43 = \boxed{67}$$

Given: $23 + 14$

Start at 23 on the number line. Jump 10 forward → reach 33. Jump 4 more forward → reach 37.

$$23 + 10 = 33, \quad 33 + 4 = 37$$

$$23 + 14 = \boxed{37}$$

Given: $24 + 16$

Start at 24. Jump 10 forward → reach 34. Jump 6 more forward → reach 40.

$$24 + 10 = 34, \quad 34 + 6 = 40$$

$$24 + 16 = \boxed{40}$$

Given: $11 + 22$

Start at 11. Jump 20 forward → reach 31. Jump 2 more forward → reach 33.

$$11 + 20 = 31, \quad 31 + 2 = 33$$

$$11 + 22 = \boxed{33}$$

Given: 5 tens and 11 ones.

Concept: 10 ones = 1 ten (regrouping)

$$11 \text{ ones} = 1 \text{ ten} + 1 \text{ one}$$

So: $5 \text{ tens} + 1 \text{ ten} = 6 \text{ tens}$, and $1 \text{ one}$ remains.

$$\text{Total} = 6 \text{ tens } 1 \text{ one} = \boxed{61}$$

Given:
- First group: 3 tens 6 ones = 36
- Second group: 4 tens 7 ones = 47

Step 1: Add ones → $6 + 7 = 13$ ones = 1 ten + 3 ones
Step 2: Add tens → $3 + 4 + 1 = 8$ tens

$$36 + 47 = \boxed{83}$$

Verification: $80 + 3 = 83$ ✓

Given:
- First: 4 tens 5 ones = 45
- Second: 4 tens 9 ones = 49

Step 1: Add ones → $5 + 9 = 14$ ones = 1 ten + 4 ones
Step 2: Add tens → $4 + 4 + 1 = 9$ tens

$$45 + 49 = \boxed{94}$$

Given:
- First: 5 tens 8 ones = 58
- Second: 3 tens 2 ones = 32

Step 1: Add ones → $8 + 2 = 10$ ones = 1 ten + 0 ones
Step 2: Add tens → $5 + 3 + 1 = 9$ tens

$$58 + 32 = \boxed{90}$$

Given:
- First: 8 tens 4 ones = 84
- Second: 1 ten 0 ones = 10

Step 1: Add ones → $4 + 0 = 4$
Step 2: Add tens → $8 + 1 = 9$

$$84 + 10 = \boxed{94}$$

Given:
- First: 2 tens 6 ones = 26
- Second: 4 tens 8 ones = 48

Step 1: Add ones → $6 + 8 = 14$ ones = 1 ten + 4 ones
Step 2: Add tens → $2 + 4 + 1 = 7$ tens

$$26 + 48 = \boxed{74}$$

Given:
- Runs in first match = 43
- Runs in second match = 58

We need to find the total runs.

$$\begin{array}{r} \text{T} \quad \text{O} \\ 4 \quad 3 \\ +\; 5 \quad 8 \\ \hline \end{array}$$

Step 1: Add Ones → $3 + 8 = 11$ → write 1, carry 1
Step 2: Add Tens → $4 + 5 + 1 = 10$ → write 10

$$43 + 58 = \boxed{101}$$

Rahim made 101 runs in the two matches together.

Given:
- Pencils Simarpreet had = 12
- Pencils gifted by mother = 36

$$\begin{array}{r} \text{T} \quad \text{O} \\ 1 \quad 2 \\ +\; 3 \quad 6 \\ \hline 4 \quad 8 \end{array}$$

Step 1: Add Ones → $2 + 6 = 8$
Step 2: Add Tens → $1 + 3 = 4$

$$12 + 36 = \boxed{48}$$

Simarpreet now has 48 colour pencils.

Given:
- Red marbles = 34
- Blue marbles = 57

$$\begin{array}{r} \text{T} \quad \text{O} \\ 3 \quad 4 \\ +\; 5 \quad 7 \\ \hline \end{array}$$

Step 1: Add Ones → $4 + 7 = 11$ → write 1, carry 1
Step 2: Add Tens → $3 + 5 + 1 = 9$

$$34 + 57 = \boxed{91}$$

Heena has 91 marbles in total.

Given:
- Sarika spent = ₹56
- Manish spent = ₹35

$$\begin{array}{r} \text{T} \quad \text{O} \\ 5 \quad 6 \\ +\; 3 \quad 5 \\ \hline \end{array}$$

Step 1: Add Ones → $6 + 5 = 11$ → write 1, carry 1
Step 2: Add Tens → $5 + 3 + 1 = 9$

$$56 + 35 = \boxed{91}$$

They both spent ₹91 in the fair together.

Given:
- Number of men = 36
- Number of women = 47

$$\begin{array}{r} \text{T} \quad \text{O} \\ 3 \quad 6 \\ +\; 4 \quad 7 \\ \hline \end{array}$$

Step 1: Add Ones → $6 + 7 = 13$ → write 3, carry 1
Step 2: Add Tens → $3 + 4 + 1 = 8$

$$36 + 47 = \boxed{83}$$

There are 83 people in the bus.

Given: $18 - 9$

On the number line: Start at 18, move 9 steps backward.

$$18 - 9 = \boxed{9}$$

Given: $12 - 8$

On the number line: Start at 12, move 8 steps backward.

$$12 - 8 = \boxed{4}$$

Given: $30 - 18$

On the number line: Start at 30, jump back 10 steps → reach 20, then jump back 8 more steps → reach 12.

$$30 - 10 = 20, \quad 20 - 8 = 12$$

$$30 - 18 = \boxed{12}$$

Given: $23 - 14$

On the number line: Start at 23, jump back 10 → reach 13, then jump back 4 → reach 9.

$$23 - 10 = 13, \quad 13 - 4 = 9$$

$$23 - 14 = \boxed{9}$$

Given: $26 - 17$

On the number line: Start at 26, jump back 10 → reach 16, then jump back 7 → reach 9.

$$26 - 10 = 16, \quad 16 - 7 = 9$$

$$26 - 17 = \boxed{9}$$

Given: Starting number is 26. From the number line, we move back 13 steps to reach 13.

$$26 - 13 = \boxed{13}$$

Given: $39 - 28$

Method: $39 - 20 = 19$, then $19 - 8 = 11$

$$39 - 28 = \boxed{11}$$

Note: This question depends on the number line image. Based on typical values in this chapter:

A common example: $45 - 23 = 22$

$$45 - 23 = \boxed{22}$$

(Students should read their specific number line and fill accordingly.)

Given: $38 - 11$

Step 1: Subtract ones → $8 - 1 = 7$
Step 2: Subtract tens → $3 - 1 = 2$

$$38 - 11 = \boxed{27}$$

Given: $22 - 22$

Concept: Any number subtracted from itself equals zero.

$$22 - 22 = \boxed{0}$$

Given: $14 - 0$

Concept: Subtracting zero from any number gives the same number.

$$14 - 0 = \boxed{14}$$

Given: $35 - 12$

Step 1: Subtract ones → $5 - 2 = 3$
Step 2: Subtract tens → $3 - 1 = 2$

$$35 - 12 = \boxed{23}$$

Given: $24 - 13$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 4 \\ -\; 1 \quad 3 \\ \hline 1 \quad 1 \end{array}$$

Step 1: Subtract ones → $4 - 3 = 1$
Step 2: Subtract tens → $2 - 1 = 1$

$$24 - 13 = \boxed{11}$$

Given: $54 - 31$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 5 \quad 4 \\ -\; 3 \quad 1 \\ \hline 2 \quad 3 \end{array}$$

Step 1: Subtract ones → $4 - 1 = 3$
Step 2: Subtract tens → $5 - 3 = 2$

$$54 - 31 = \boxed{23}$$

Given: $32 - 12$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 3 \quad 2 \\ -\; 1 \quad 2 \\ \hline 2 \quad 0 \end{array}$$

Step 1: Subtract ones → $2 - 2 = 0$
Step 2: Subtract tens → $3 - 1 = 2$

$$32 - 12 = \boxed{20}$$

Given: $46 - 34$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 4 \quad 6 \\ -\; 3 \quad 4 \\ \hline 1 \quad 2 \end{array}$$

Step 1: Subtract ones → $6 - 4 = 2$
Step 2: Subtract tens → $4 - 3 = 1$

$$46 - 34 = \boxed{12}$$

Given:
- Money Shikha has = ₹82
- Cost of pencils = ₹22

$$\begin{array}{r} \text{T} \quad \text{O} \\ 8 \quad 2 \\ -\; 2 \quad 2 \\ \hline 6 \quad 0 \end{array}$$

Step 1: Subtract ones → $2 - 2 = 0$
Step 2: Subtract tens → $8 - 2 = 6$

$$82 - 22 = \boxed{60}$$

Shikha is left with ₹60.

Given:
- Money Ruby has = ₹60
- Cost of notebook = ₹20

$$\begin{array}{r} \text{T} \quad \text{O} \\ 6 \quad 0 \\ -\; 2 \quad 0 \\ \hline 4 \quad 0 \end{array}$$

Step 1: Subtract ones → $0 - 0 = 0$
Step 2: Subtract tens → $6 - 2 = 4$

$$60 - 20 = \boxed{40}$$

Ruby is left with ₹40.

Given:
- Flowers Jyoti has = 28
- Flowers given away = 5

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 8 \\ -\; 0 \quad 5 \\ \hline 2 \quad 3 \end{array}$$

Step 1: Subtract ones → $8 - 5 = 3$
Step 2: Subtract tens → $2 - 0 = 2$

$$28 - 5 = \boxed{23}$$

Jyoti is left with 23 flowers.

Given:
- Flowers Jyoti has = 54
- Flowers given to mother = 28

Since ones digit: 4 < 8, we need to regroup (borrow).

Step 1: Borrow 1 ten from tens place. Tens become $5 - 1 = 4$. Ones become $14$.
Step 2: Subtract ones → $14 - 8 = 6$
Step 3: Subtract tens → $4 - 2 = 2$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 4 \quad 14 \\ -\; 2 \quad 8 \\ \hline 2 \quad 6 \end{array}$$

$$54 - 28 = \boxed{26}$$

Jyoti is left with 26 flowers.

Given:
- Flowers Jyoti has = 73
- Flowers given to mother = 47

Since ones digit: 3 < 7, we need to regroup.

Step 1: Borrow 1 ten from tens place. Tens become $7 - 1 = 6$. Ones become $13$.
Step 2: Subtract ones → $13 - 7 = 6$
Step 3: Subtract tens → $6 - 4 = 2$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 6 \quad 13 \\ -\; 4 \quad 7 \\ \hline 2 \quad 6 \end{array}$$

$$73 - 47 = \boxed{26}$$

Jyoti is left with 26 flowers.

Given: $45 - 18$

Since 5 < 8, regroup: borrow 1 ten.
Tens: $4 - 1 = 3$, Ones: $15$

Step 1: $15 - 8 = 7$
Step 2: $3 - 1 = 2$

$$45 - 18 = \boxed{27}$$

Given: $84 - 37$

Since 4 < 7, regroup: borrow 1 ten.
Tens: $8 - 1 = 7$, Ones: $14$

Step 1: $14 - 7 = 7$
Step 2: $7 - 3 = 4$

$$84 - 37 = \boxed{47}$$

Given: $65 - 45$

Step 1: Subtract ones → $5 - 5 = 0$
Step 2: Subtract tens → $6 - 4 = 2$

$$65 - 45 = \boxed{20}$$

Given: $27 - 27$

Concept: Any number subtracted from itself = 0.

$$27 - 27 = \boxed{0}$$

Given: $93 - 48$

Since 3 < 8, regroup: borrow 1 ten.
Tens: $9 - 1 = 8$, Ones: $13$

Step 1: $13 - 8 = 5$
Step 2: $8 - 4 = 4$

$$93 - 48 = \boxed{45}$$

Given:
- Total skips needed = 20
- Skips done = 14

We find: $20 - 14$

$$\begin{array}{r} 2 \quad 0 \\ -\; 1 \quad 4 \\ \hline 0 \quad 6 \end{array}$$

Step 1: Since 0 < 4, regroup. Borrow 1 ten. Ones = 10, Tens = 1.
Step 2: $10 - 4 = 6$
Step 3: $1 - 1 = 0$

$$20 - 14 = \boxed{6}$$

Also: $14 + \boxed{6} = 20$

Aman needs 6 more skips.

Given:
- Total skips needed = 20
- Skips done = 12

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 0 \\ -\; 1 \quad 2 \\ \hline 0 \quad 8 \end{array}$$

Step 1: Since 0 < 2, regroup. Borrow 1 ten. Ones = 10, Tens = 1.
Step 2: $10 - 2 = 8$
Step 3: $1 - 1 = 0$

$$20 - 12 = \boxed{8}$$

Also: $12 + \boxed{8} = 20$

Avni needs 8 more skips.

Given:
- Shells collected = 63
- Shells given away = 26

Since 3 < 6, regroup.

$$\begin{array}{r} \text{T} \quad \text{O} \\ 5 \quad 13 \\ -\; 2 \quad 6 \\ \hline 3 \quad 7 \end{array}$$

Step 1: $13 - 6 = 7$
Step 2: $5 - 2 = 3$

$$63 - 26 = \boxed{37}$$

Anushka is left with 37 shells.

Given:
- Passengers in bus = 54
- Passengers who got down = 16

Since 4 < 6, regroup.

$$\begin{array}{r} \text{T} \quad \text{O} \\ 4 \quad 14 \\ -\; 1 \quad 6 \\ \hline 3 \quad 8 \end{array}$$

Step 1: $14 - 6 = 8$
Step 2: $4 - 1 = 3$

$$54 - 16 = \boxed{38}$$

There are 38 passengers in the bus now.

Given:
- Total balloons = 40
- Balloons burst = 13

$$\begin{array}{r} \text{T} \quad \text{O} \\ 3 \quad 10 \\ -\; 1 \quad 3 \\ \hline 2 \quad 7 \end{array}$$

Step 1: Since 0 < 3, regroup. Borrow 1 ten. Ones = 10, Tens = 3.
Step 2: $10 - 3 = 7$
Step 3: $3 - 1 = 2$

$$40 - 13 = \boxed{27}$$

27 balloons are left.

Given:
- Bangles made = 72
- Bangles sold = 36

Since 2 < 6, regroup.

$$\begin{array}{r} \text{T} \quad \text{O} \\ 6 \quad 12 \\ -\; 3 \quad 6 \\ \hline 3 \quad 6 \end{array}$$

Step 1: $12 - 6 = 6$
Step 2: $6 - 3 = 3$

$$72 - 36 = \boxed{36}$$

Kanika has 36 bangles left.

Given:
- Birds originally on tree = 56
- Birds now = 87
- Birds that came later = ?

We find: $87 - 56$

$$\begin{array}{r} \text{T} \quad \text{O} \\ 8 \quad 7 \\ -\; 5 \quad 6 \\ \hline 3 \quad 1 \end{array}$$

Step 1: $7 - 6 = 1$
Step 2: $8 - 5 = 3$

$$87 - 56 = \boxed{31}$$

31 birds came later.

Given:
- Taps done = 18
- Total taps needed = 35
- More taps needed = ?

We find: $35 - 18$

Since 5 < 8, regroup.

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 15 \\ -\; 1 \quad 8 \\ \hline 1 \quad 7 \end{array}$$

Step 1: $15 - 8 = 7$
Step 2: $2 - 1 = 1$

$$35 - 18 = \boxed{17}$$

Arman needs to dribble 17 more times.

Given numbers: 20, 30, 50

Addition facts:
$$20 + 30 = \boxed{50}$$
$$30 + \boxed{20} = 50$$

Subtraction facts:
$$50 - 30 = \boxed{20}$$
$$\boxed{50} - 20 = 30$$

Given numbers: 10, 20, 30

Addition facts:
$$10 + 20 = \boxed{30}$$
$$\boxed{20} + \boxed{10} = \boxed{30}$$

Subtraction facts:
$$30 - \boxed{20} = 10$$
$$30 - \boxed{10} = 20$$

Given numbers: 20, 80, 100

Addition facts:
$$20 + 80 = \boxed{100}$$
$$80 + 20 = \boxed{100}$$

Subtraction facts:
$$100 - 80 = \boxed{20}$$
$$100 - 20 = \boxed{80}$$

Given:
- Cost of gift wrapper = ₹24
- Cost of ribbons = ₹37

Total money spent = ₹24 + ₹37

$$\begin{array}{r} \text{T} \quad \text{O} \\ 2 \quad 4 \\ +\; 3 \quad 7 \\ \hline \end{array}$$

Step 1: Add ones → $4 + 7 = 11$ → write 1, carry 1
Step 2: Add tens → $2 + 3 + 1 = 6$

$$24 + 37 = \boxed{61}$$

They spent ₹61 in total.

Given:
- Money in purse = ₹78
- Money spent = ₹61

$$\begin{array}{r} \text{T} \quad \text{O} \\ 7 \quad 8 \\ -\; 6 \quad 1 \\ \hline 1 \quad 7 \end{array}$$

Step 1: Subtract ones → $8 - 1 = 7$
Step 2: Subtract tens → $7 - 6 = 1$

$$78 - 61 = \boxed{17}$$

₹17 is left with them.

Note: The exact sums in Daljeet's work are in the image which cannot be fully read. However, here is how to verify any answer:

Method to verify addition: If Daljeet wrote $36 + 47 = 83$, check:
$$36 + 47: \text{ Ones: } 6+7=13 \text{ (write 3, carry 1)}, \text{ Tens: } 3+4+1=8$$
$$\text{Answer} = 83 \checkmark \text{ (Correct)}$$

Method to verify subtraction: If Daljeet wrote $54 - 28 = 36$, check:
$$54 - 28: \text{ Regroup: Ones: } 14-8=6, \text{ Tens: } 4-2=2$$
$$\text{Answer} = 26 \neq 36 \text{ (Wrong — correct answer is 26)}$$

General Rule: Students should redo each sum step by step using the column method (Tens and Ones), check if regrouping is needed, and compare with Daljeet's answer. If different, write the correct answer.