Arithmetic Expressions

Concept: Find the value of one side and work backwards to find the missing number.

(a) $13 + 4 = \_ + 6$

LHS $= 13 + 4 = 17$

So $\_ + 6 = 17 \Rightarrow \_ = 17 - 6 = 11$

$$\boxed{13 + 4 = 11 + 6}$$

(b) $22 + \_ = 6 \times 5$

RHS $= 6 \times 5 = 30$

So $22 + \_ = 30 \Rightarrow \_ = 30 - 22 = 8$

$$\boxed{22 + 8 = 6 \times 5}$$

(c) $8 \times \_ = 64 \div 2$

RHS $= 64 \div 2 = 32$

So $8 \times \_ = 32 \Rightarrow \_ = 32 \div 8 = 4$

$$\boxed{8 \times 4 = 64 \div 2}$$

(d) $34 - \_ = 25$

$\_ = 34 - 25 = 9$

$$\boxed{34 - 9 = 25}$$

Step 1: Evaluate each expression.

(a) $67 - 19 = 48$

(b) $67 - 20 = 47$

(c) $35 + 25 = 60$

(d) $5 \times 11 = 55$

(e) $120 \div 3 = 40$

Step 2: Arrange in ascending order: $47 < 48 < 40$...

Re-listing: $40, 47, 48, 55, 60$

Corresponding expressions:

$$\underbrace{120 \div 3}_{40} < \underbrace{67-20}_{47} < \underbrace{67-19}_{48} < \underbrace{5 \times 11}_{55} < \underbrace{35+25}_{60}$$

Ascending order: $(e),\ (b),\ (a),\ (d),\ (c)$

Concept: Follow BODMAS/DMAS — Division and Multiplication before Addition and Subtraction.

(a) $28 - 7 + 8$

Terms: $+28,\ -7,\ +8$

$$= 28 - 7 + 8 = 21 + 8 = \boxed{29}$$

(b) $39 - 2 \times 6 + 11$

First evaluate $2 \times 6 = 12$.

Terms: $+39,\ -12,\ +11$

$$= 39 - 12 + 11 = 27 + 11 = \boxed{38}$$

(c) $40 - 10 + 10 + 10$

Terms: $+40,\ -10,\ +10,\ +10$

$$= 40 - 10 + 10 + 10 = 30 + 20 = \boxed{50}$$

(d) $48 - 10 \times 2 + 16 \div 2$

First: $10 \times 2 = 20$ and $16 \div 2 = 8$.

Terms: $+48,\ -20,\ +8$

$$= 48 - 20 + 8 = 28 + 8 = \boxed{36}$$

(e) $6 \times 3 - 4 \times 8 \times 5$

First: $6 \times 3 = 18$ and $4 \times 8 \times 5 = 160$.

Terms: $+18,\ -160$

$$= 18 - 160 = \boxed{-142}$$

(a) $89 + 21 - 10$

Story: A shopkeeper had 89 apples. He received 21 more apples in the morning but sold 10 apples by noon. How many apples does he have now?

Value:
$$89 + 21 - 10 = 110 - 10 = \boxed{100}$$

(b) $5 \times 12 - 6$

Story: A box contains 12 chocolates. Riya bought 5 such boxes and then ate 6 chocolates. How many chocolates are left?

Value:
$$5 \times 12 - 6 = 60 - 6 = \boxed{54}$$

(c) $4 \times 9 + 2 \times 6$

Story: In a class, 4 boys each have 9 pencils and 2 girls each have 6 pencils. What is the total number of pencils?

Value:
$$4 \times 9 + 2 \times 6 = 36 + 12 = \boxed{48}$$

(a)

Given: Elsa starts with 100 coins and doubles them → $2 \times 100$.
Anna starts with 100 coins and has half left → $100 \div 2$.

Expression:
$$2 \times 100 + 100 \div 2$$

Terms: $2 \times 100 = 200$ and $100 \div 2 = 50$

Value:
$$200 + 50 = \boxed{250 \text{ gold coins}}$$

(b)(i) Four adults and three children:

Expression: $4 \times 40 + 3 \times 20$

Terms: $4 \times 40 = 160$ and $3 \times 20 = 60$

Value:
$$160 + 60 = \boxed{₹220}$$

(b)(ii) Two groups of three adults each:

Expression: $2 \times 3 \times 40$ or $2 \times (3 \times 40)$

Value:
$$2 \times 120 = \boxed{₹240}$$

(c) (Note: The figure is not visible. A typical window problem involves a top panel, a middle glass pane, and a bottom panel.)

Assumption: If the window has a top frame of height $a$, a glass pane of height $b$, and a bottom frame of height $c$, then:

Expression: $a + b + c$

Students should substitute the values shown in their textbook figure and add the terms to get the total height.

Concept: When removing brackets preceded by $+$, signs inside remain the same. When removing brackets preceded by $-$, signs inside flip.

(a) $24 + (6 - 4) = 24 + 6\ \square\ 4$

Removing brackets (preceded by $+$): signs stay the same.
$$24 + 6 - 4$$
So the box is $-$.
$$\boxed{24 + (6-4) = 24 + 6 - 4}$$

(b) $38 + (\underline{\quad}) = 38 + 9 - 4$

RHS $= 38 + 9 - 4$. The bracket must give $9 - 4$ when opened with $+$.
$$\boxed{38 + (9 - 4) = 38 + 9 - 4}$$

(c) $24 - (6 + 4) = 24\ \square\ 6\ \square\ 4$

Removing brackets preceded by $-$: signs flip.
$$24 - 6 - 4$$
Both boxes are $-$.
$$\boxed{24 - (6+4) = 24 - 6 - 4}$$

(d) $24 - 6 - 4 = 24 - (6\ \square\ 4)$

For $24 - (6\ \square\ 4)$ to equal $24 - 6 - 4$, we need the bracket to give $+4$ when the minus sign flips it to $-4$. So inside the bracket: $6 + 4$.
$$\boxed{24 - 6 - 4 = 24 - (6 + 4)}$$

(e) $27 - (8 + 3) = 27\ \square\ 8\ \square\ 3$

Removing brackets preceded by $-$: signs flip.
$$27 - 8 - 3$$
$$\boxed{27 - (8+3) = 27 - 8 - 3}$$

(f) $27 - (\underline{\quad}) = 27 - 8 + 3$

For $27 - (\text{something}) = 27 - 8 + 3$, when we remove the bracket the $-8$ stays $-8$ (so $8$ is positive inside) and $+3$ becomes $-3$ inside (sign flips). So inside: $8 - 3$.
$$\boxed{27 - (8 - 3) = 27 - 8 + 3}$$

Concept: $+(\ldots)$ keeps signs; $-(\ldots)$ flips signs.

(a) $14 + (12 + 10)$
$$= 14 + 12 + 10$$

(b) $14 - (12 + 10)$
$$= 14 - 12 - 10$$

(c) $14 + (12 - 10)$
$$= 14 + 12 - 10$$

(d) $14 - (12 - 10)$
$$= 14 - 12 + 10$$

(e) $-14 + (12 - 10)$
$$= -14 + 12 - 10$$

(f) $14 - (-12 - 10)$
$$= 14 + 12 + 10$$

(a) $(6 + 10) - 2$ and $6 + (10 - 2)$

Guess: They look like they might be equal since subtraction is involved in a similar way.

$(6 + 10) - 2 = 16 - 2 = 14$

$6 + (10 - 2) = 6 + 8 = 14$

Both equal 14. They are equal.

Reason: $(6+10)-2 = 6+10-2$ and $6+(10-2)=6+10-2$. Same expression!

(b) $16 - (8 - 3)$ and $(16 - 8) - 3$

Guess: They may not be equal because the bracket placement changes what is subtracted.

$16 - (8 - 3) = 16 - 8 + 3 = 8 + 3 = 11$

$(16 - 8) - 3 = 8 - 3 = 5$

They are NOT equal: $11 \neq 5$.

They would be equal only if the $3$ inside the bracket had no effect, i.e., if $3 = 0$.

(c) $27 - (18 + 4)$ and $27 + (-18 - 4)$

Guess: They should be equal.

$27 - (18 + 4) = 27 - 18 - 4 = 9 - 4 = 5$

$27 + (-18 - 4) = 27 - 18 - 4 = 9 - 4 = 5$

Both equal 5. They are always equal because removing the bracket in the first gives the same expression as the second.

(Note: The full list of expressions for Question 4 was cut off in the OCR. Students should apply the following reasoning principle:

Concept: Two expressions have the same value if, when written as a sum of terms (applying bracket-removal rules), they produce identical sets of terms.

For example, $a - (b + c)$ and $a - b - c$ have the same terms $+a, -b, -c$ and are therefore equal.

Students should expand each expression by removing brackets and compare the resulting terms to identify matching expressions.)

Concept (Distributive Property): $a \times b = (m \pm n) \times b = m \times b \pm n \times b$

(a) $95 \times 8$
$$= (100 - 5) \times 8 = 100 \times 8 - 5 \times 8 = 800 - 40 = \boxed{760}$$

(b) $104 \times 15$
$$= (100 + 4) \times 15 = 100 \times 15 + 4 \times 15 = 1500 + 60 = \boxed{1560}$$

(c) $49 \times 50$
$$= (50 - 1) \times 50 = 50 \times 50 - 1 \times 50 = 2500 - 50 = \boxed{2450}$$

Yes, this method is quicker than the standard multiplication procedure for numbers close to round numbers.