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.

Concept (Distributive Property): $a \times (b + c) = a \times b + a \times c$ and $a \times (b - c) = a \times b - a \times c$

(a) Already complete: $3 \times (6+7) = 3\times6 + 3\times7$ ✓

(b) Already complete: $(8+3)\times4 = 8\times4 + 3\times4$ ✓

(c) $3 \times (5+8) = 3\times5\ \boxed{+}\ 3\times\boxed{8}$

(d) $(9+2)\times4 = 9\times4\ \boxed{+}\ 2\times\boxed{4}$

(e) $3\times(\boxed{3}+4) = 3\ \boxed{\times}\ \boxed{3} + 3\times4$

(Since $3\times3 + 3\times4 = 9+12=21$ and $3\times(3+4)=3\times7=21$ ✓)

(f) $(\boxed{13}+6)\times4 = 13\times4 + \boxed{6\times4}$

(Since $(13+6)\times4 = 13\times4 + 6\times4$)

(g) $3\times(\boxed{5}+\boxed{2}) = 3\times5 + 3\times2$

(h) $(\boxed{2}+\boxed{3})\times\boxed{4} = 2\times4 + 3\times4$

(i) $5\times(9-2) = 5\times9 - 5\times\boxed{2}$

(j) $(5-2)\times7 = 5\times7 - 2\times\boxed{7}$

(k) $5\times(8-3) = 5\times8\ \boxed{-}\ 5\times\boxed{3}$

(l) $(8-3)\times7 = 8\times7\ \boxed{-}\ 3\times7$

(m) $5\times(12-\boxed{4}) = \boxed{5\times12}\ \boxed{-}\ 5\times\boxed{4}$

(One valid answer: $5\times(12-4)=5\times12-5\times4$)

(n) $(15-\boxed{9})\times7 = \boxed{15\times7}\ \boxed{-}\ 6\times7$

(Since $(15-9)\times7 = 15\times7 - 9\times7$; but RHS shows $-6\times7$, so the blank $= 9$: $(15-9)\times7 = 15\times7 - 9\times7$. Alternatively if RHS is $15\times7 - 6\times7$ then blank $=6$: $(15-6)\times7=15\times7-6\times7$ ✓)

Answer: $\boxed{6}$; expression: $(15-6)\times7 = 15\times7 - 6\times7$

(o) $5\times(\boxed{9}-\boxed{4}) = 5\times9 - 5\times4$

(p) $(\boxed{17}-\boxed{9})\times\boxed{7} = 17\times7 - 9\times7$

(a) $(8-3)\times29$ vs $(3-8)\times29$

8-3 = 5 > 0 and 3-8 = -5 < 0.

Multiplying by the same positive number 29: 5\times29 > -5\times29.

\boxed{(8-3)\times29 > (3-8)\times29}

(b) $15 + 9\times18$ vs $(15+9)\times18$

LHS $= 15 + 9\times18$ (only $9\times18$ is a term multiplied by 18).

RHS $= (15+9)\times18 = 15\times18 + 9\times18$ (both 15 and 9 are multiplied by 18).

RHS has an extra $15\times18$ compared to LHS which only has $+15$. Since 15\times18 > 15, RHS > LHS.

\boxed{15 + 9\times18 < (15+9)\times18}

(c) $23\times(17-9)$ vs $23\times17 + 23\times9$

LHS $= 23\times17 - 23\times9$ (by distributive property).

RHS $= 23\times17 + 23\times9$.

RHS has $+23\times9$ while LHS has $-23\times9$, so RHS > LHS.

\boxed{23\times(17-9) < 23\times17 + 23\times9}

(d) $(34-28)\times42$ vs $34\times42 - 28\times42$

By distributive property: $(34-28)\times42 = 34\times42 - 28\times42$.

$$\boxed{(34-28)\times42 = 34\times42 - 28\times42}$$

Concept: Find factor pairs of 14 and split one factor as a sum.

$14 = 1 \times 14 = 2 \times 7 = 7 \times 2 = 14 \times 1$

(a) $\boxed{7 \times (1 + 1) = 14}$ (since $7\times2=14$)

(b) $\boxed{7 \times (3 + (-1+1)...}$ — simpler: $\boxed{2 \times (3 + 4) = 14}$ (since $2\times7=14$)

(c) $\boxed{7 \times (2 + 0)}$ — or $\boxed{14 \times (1 + 0)}$ — better: $\boxed{1 \times (10 + 4) = 14}$

(d) $\boxed{2 \times (5 + 2) = 14}$

Note: Many valid answers exist. Some examples:
- $2 \times (1+6)=14$
- $2 \times (2+5)=14$
- $2 \times (3+4)=14$
- $7 \times (1+1)=14$
- $14 \times (1+0)=14$
- $1 \times (7+7)=14$

(Note: The figures are not visible in the OCR. The general approach is described below.)

Method: When a set of numbers is arranged in a pattern (e.g., rows and columns of dots or a grid), use the distributive property to find the sum in two ways.

Example approach (assuming a grid of numbers):

Way 1: Add row by row, then sum the rows.
$$\text{Total} = (\text{Row 1 sum}) + (\text{Row 2 sum}) + \ldots$$

Way 2: Use the distributive property — if each row has the same numbers, multiply the number of rows by the sum of one row.
$$\text{Total} = \text{(number of rows)} \times (\text{sum of one row})$$

Students should apply these methods to the specific numbers shown in their textbook figures.

(a)

Given: Rahim supplies 9 kg/day, Shyam supplies 11 kg/day, market runs 7 days.

Expression: $(9 + 11) \times 7$

Value:
$$= 20 \times 7 = \boxed{140 \text{ kg}}$$

Alternatively: $9 \times 7 + 11 \times 7 = 63 + 77 = 140$ kg.

(b)

Given: Monthly income = ₹20,000; Monthly expenses = ₹5,000 + ₹5,000 + ₹2,000 = ₹12,000.

Monthly savings $= 20000 - (5000 + 5000 + 2000) = 20000 - 12000 = ₹8000$.

Expression for yearly savings: $(20000 - 5000 - 5000 - 2000) \times 12$

Value:
$$= 8000 \times 12 = \boxed{₹96,000}$$

(c)

Given: Climbs 3 cm in day, slips 2 cm at night. Net gain per full day-night cycle = $3 - 2 = 1$ cm. Post height = 10 cm.

Analysis: After 7 complete day-night cycles, the snail is at $7 \times 1 = 7$ cm. On Day 8, it climbs 3 cm: $7 + 3 = 10$ cm — it reaches the top!

Expression: The snail reaches the top on Day 8.

$$\text{Net progress after } n \text{ nights} = n \times (3-2) = n \text{ cm}$$

After 7 nights: 7 cm. On the 8th day it climbs 3 cm to reach 10 cm.

$$\boxed{\text{The snail gets the treat in 8 days.}}$$

Given: Melvin reads on all days except Tuesday and Saturday, i.e., he reads on $7 - 2 = 5$ days per week. He reads 1 story (2 pages) per day.

Number of stories in 8 weeks $= 5 \times 8 = 40$ stories.

Identifying correct expressions:

- (a) $5 \times 2 \times 8$: This gives $80$ — incorrect (the 2 here would mean 2 stories/day, but he reads 1 story/day).
- (b) $(7-2) \times 8 = 5 \times 8 = 40$ ✓ — reads on $(7-2)$ days per week for 8 weeks.
- (c) $8 \times 7 = 56$ — incorrect.
- (d) $7 \times 2 \times 8 = 112$ — incorrect.
- (e) $7 \times 5 - 2 = 33$ — incorrect.
- (f) $(7+2)\times8 = 72$ — incorrect.
- (g) $7 \times 8 - 2 \times 8 = 56 - 16 = 40$ ✓ — total days minus days not reading.
- (h) $(7-5)\times8 = 16$ — incorrect.

$$\boxed{\text{Correct expressions: (b) and (g); Number of stories} = 40}$$

(a) $1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10$

Way 1: Group consecutive pairs:
$$= (1-2) + (3-4) + (5-6) + (7-8) + (9-10)$$
$$= (-1) + (-1) + (-1) + (-1) + (-1) = -5$$

Way 2: Separate positive and negative terms:
$$\text{Positive terms: } 1+3+5+7+9 = 25$$
$$\text{Negative terms: } 2+4+6+8+10 = 30$$
$$= 25 - 30 = \boxed{-5}$$

(b) $1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1$

Way 1: Group consecutive pairs:
$$= (1-1)+(1-1)+(1-1)+(1-1)+(1-1) = 0+0+0+0+0 = \boxed{0}$$

Way 2: Count positive and negative terms:
$$\text{Five } +1\text{'s and five } -1\text{'s: } 5 - 5 = \boxed{0}$$

(a) $49 - 7 + 8\ \square\ 49 - 7 + 8$

Both sides are identical expressions.
$$\boxed{=}$$

(b) $83 \times 42 - 18\ \square\ 83 \times 40 - 18$

83 \times 42 > 83 \times 40 (same number multiplied by a larger value). The $-18$ is the same on both sides.
\boxed{83 \times 42 - 18 > 83 \times 40 - 18}

(c) $145 - 17 \times 8\ \square\ 145 - 17 \times 6$

17 \times 8 > 17 \times 6, so we subtract a larger number on the LHS.
\boxed{145 - 17 \times 8 < 145 - 17 \times 6}

(d) $23 \times 48 - 35\ \square\ 23 \times (48 - 35)$

LHS $= 23 \times 48 - 35$

RHS $= 23 \times 48 - 23 \times 35$

Since 23 \times 35 > 35, RHS subtracts more, so LHS > RHS.
\boxed{23 \times 48 - 35 > 23 \times (48 - 35)}

(e) $(16 - 11) \times 12\ \square\ -11 \times 12 + 16 \times 12$

LHS $= 16 \times 12 - 11 \times 12$ (by distributive property)

RHS $= 16 \times 12 - 11 \times 12$ (rearranging)

Both are equal.
$$\boxed{=}$$

(f) $(76 - 53) \times 88\ \square\ 88 \times (53 - 76)$

LHS = (76-53)\times88 = 23\times88 > 0

RHS = 88\times(53-76) = 88\times(-23) = -23\times88 < 0

\boxed{(76-53)\times88 > 88\times(53-76)}

(g) $25 \times (42 + 16)\ \square\ 25 \times (43 + 15)$

LHS: $42 + 16 = 58$

RHS: $43 + 15 = 58$

Both have the same factor 58.
$$\boxed{=}$$

(h) $36 \times (28 - 16)\ \square\ 35 \times (27 - 15)$

LHS $= 36 \times 12$

RHS $= 35 \times 12$

36 > 35 and both multiplied by 12.
\boxed{36 \times (28-16) > 35 \times (27-15)}

(a) Given expression: $83 - 37 - 12$

Terms: $+83, -37, -12$

Check each option:

(i) $84 - 38 - 12$: Terms $+84, -38, -12$. $84-38 = 46$ vs $83-37=46$. Same result! ✓

(Alternatively: $84-38-12 = 83-37-12$ since both first terms differ by $+1$ and both second terms differ by $+1$, net effect zero.)

(ii) $84 - (37+12) = 84 - 37 - 12$: Terms $+84, -37, -12$. This gives $84-49=35$ vs $83-49=34$. ✗

(iii) $83 - 38 - 13$: Terms $+83, -38, -13$. Sum $= 83-51=32$ vs $83-49=34$. ✗

(iv) $-37 + 83 - 12$: Terms $+83, -37, -12$ — same as given. ✓

$$\boxed{\text{Equal expressions: (i) and (iv)}}$$

(b) Given expression: $93 + 37\times44 + 76$

Terms: $+93,\ +37\times44,\ +76$

(i) $37 + 93\times44 + 76$: The term $93\times44 \neq 37\times44$. ✗

(ii) $93 + 37\times76 + 44$: The term $37\times76 \neq 37\times44$. ✗

(iii) $(93+37)\times(44+76)$: This is $130\times120$, completely different structure. ✗

(iv) $37\times44 + 93 + 76$: Terms $+37\times44, +93, +76$ — same as given (addition is commutative). ✓

$$\boxed{\text{Equal expression: (iv)}}$$

Let the chosen number be 20.

Here are ten different expressions each equal to 20:

1. $10 + 10 = 20$
2. $25 - 5 = 20$
3. $4 \times 5 = 20$
4. $100 \div 5 = 20$
5. $3 \times 6 + 2 = 20$
6. $50 - 15 - 15 = 20$
7. $2 \times (7 + 3) = 20$
8. $40 \div 2 = 20$
9. $(6 + 4) \times 2 = 20$
10. $100 - 4 \times 20 = 20$

(Students may choose any number and create their own valid expressions.)