Decoration for Festival (Addition and Subtraction) — NCERT Solutions

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.