Chapter 3 of 15

Updated 6 August 2026

NCERT Solutions

Arithmetic Expressions — NCERT Solutions

CBSE · Class 7 · Mathematics

NCERT Solutions for Arithmetic Expressions, CBSE Class 7 Mathematics: 20 textbook questions solved step by step.

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20 Questions Solved · 6 Sections

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Figure it Out (Comparing Expressions)

1Fill in the blanks to make the expressions equal on both sides of the = sign:
(a)

13 + 4 = \_ + 6

(b)

22 + \_ = 6 \times 5

(c)

8 \times \_ = 64 \div 2

(d)

34 - \_ = 25

Show solution

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) $34 - \_ = 25$

$\_ = 34 - 25 = 9$

$\boxed{34 - 9 = 25}$

2Arrange the following expressions in ascending (increasing) order of their values.
(a)

67 - 19

(b)

67 - 20

(c)

35 + 25

(d)

5 \times 11

(e)

120 \div 3

Show solution

Step 1: Evaluate each expression.

(a) $67 - 19 = 48$

(b) $67 - 20 = 47$

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

Figure it Out (Terms and Brackets)

1Find the values of the following expressions by writing the terms in each case.
(a)

28 - 7 + 8

(b)

39 - 2 \times 6 + 11

(c)

40 - 10 + 10 + 10

(d)

48 - 10 \times 2 + 16 \div 2

(e)

6 \times 3 - 4 \times 8 \times 5

Show solution

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: $+18,\ -160$

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

2Write a story/situation for each of the following expressions and find their values.
(a)

89 + 21 - 10

(b)

5 \times 12 - 6

(c)

4 \times 9 + 2 \times 6

Show solution

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

3For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(i) for four adults and three children?
(ii) for two groups having three adults each?
(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.Show solution

(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.

Figure it Out (Removing Brackets — I)

1Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
(a)

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

(b)

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

(c)

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

(d)

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

(e)

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

(f)

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

Show solution

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: $8 - 3$ .
$\boxed{27 - (8 - 3) = 27 - 8 + 3}$

2Remove the brackets and write the expression having the same value.
(a)

14 + (12 + 10)

(b)

14 - (12 + 10)

(c)

14 + (12 - 10)

(d)

14 - (12 - 10)

(e)

-14 + (12 - 10)

(f)

14 - (-12 - 10)

Show solution

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

3Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(a)

(6 + 10) - 2

and

6 + (10 - 2)

(b)

16 - (8 - 3)

and

(16 - 8) - 3

(c)

27 - (18 + 4)

and

27 + (-18 - 4)

Show solution

(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.

4In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.Show solution

(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.)

Figure it Out (Distributive Property — Products)

Use this methodUse the method shown in Examples 17–18 to find the following products:
(a)

95 \times 8

(b)

104 \times 15

(c)

49 \times 50

Show solution

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.

Figure it Out (Distributive Property — Fill in the Blanks)

1Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.
(a)

3 \times (6 + 7) = 3 \times 6 + 3 \times 7

(b)

(8 + 3) \times 4 = 8 \times 4 + 3 \times 4

(c)

3 \times (5 + 8) = 3 \times 5\ \square\ 3 \times \_

(d)

(9 + 2) \times 4 = 9 \times 4\ \square\ 2 \times \_

(e)

3 \times (\_ + 4) = 3\ \square\ \_ + \_

(f)

(\_+ 6) \times 4 = 13 \times 4 + \_

(g)

3 \times (\_ + \_) = 3 \times 5 + 3 \times 2

(h)

(\_ + \_) \times \_ = 2 \times 4 + 3 \times 4

(i)

5 \times (9 - 2) = 5 \times 9 - 5 \times \_

(j)

(5 - 2) \times 7 = 5 \times 7 - 2 \times \_

(k)

5 \times (8 - 3) = 5 \times 8\ \square\ 5 \times \_

(l)

(8 - 3) \times 7 = 8 \times 7\ \square\ 3 \times 7

(m)

5 \times (12 - \_) = \_\ \square\ 5 \times \_

(n)

(15 - \_) \times 7 = \_\ \square\ 6 \times 7

(o)

5 \times (\_ - \_) = 5 \times 9 - 5 \times 4

(p)

(\_ - \_) \times \_ = 17 \times 7 - 9 \times 7

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

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2In the boxes below, fill '<', '>' or '=' after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.
(a)

(8 - 3) \times 29\ \square\ (3 - 8) \times 29

(b)

15 + 9 \times 18\ \square\ (15 + 9) \times 18

(c)

23 \times (17 - 9)\ \square\ 23 \times 17 + 23 \times 9

(d)

(34 - 28) \times 42\ \square\ 34 \times 42 - 28 \times 42

(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: $(34-28)\times42 = 34\times42 - 28\times42$ .

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

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3Here is one way to make 14:

2 \times (1 + 6) = 14

. Are there other ways of getting 14? Fill them out below:
(a)

\_ \times (\_ + \_) = 14

(b)

\_ \times (\_ + \_) = 14

(c)

\_ \times (\_ + \_) = 14

(d)

\_ \times (\_ + \_) = 14

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:

$2 \times (1+6)=14$
$2 \times (2+5)=14$
$2 \times (3+4)=14$
$7 \times (1+1)=14$
$14 \times (1+0)=14$
$1 \times (7+7)=14$

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4Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

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.

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Figure it Out (Final Exercise)

1Read the situations given below. Write appropriate expressions for each of them and find their values.
(a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?

(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.

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

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2Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?
(a)

5 \times 2 \times 8

(b)

(7 - 2) \times 8

(c)

8 \times 7

(d)

7 \times 2 \times 8

(e)

7 \times 5 - 2

(f)

(7 + 2) \times 8

(g)

7 \times 8 - 2 \times 8

(h)

(7 - 5) \times 8

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.

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

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3Find different ways of evaluating the following expressions:
(a)

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10

(b)

1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1

(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:
$\text{Five } +1\text{'s and five } -1\text{'s: } 5 - 5 = \boxed{0}$

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4Compare the following pairs of expressions using

<, >

=

or by reasoning.
(a)

49 - 7 + 8\ \square\ 49 - 7 + 8

(b)

83 \times 42 - 18\ \square\ 83 \times 40 - 18

(c)

145 - 17 \times 8\ \square\ 145 - 17 \times 6

(d)

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

(e)

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

(f)

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

(g)

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

(h)

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

(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.
$\boxed{36 \times (28-16) > 35 \times (27-15)}$

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5Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(a)

83 - 37 - 12

(i)

84 - 38 - 12

(ii)

84 - (37 + 12)

(iii)

83 - 38 - 13

(iv)

-37 + 83 - 12

(b)

93 + 37 \times 44 + 76

(i)

37 + 93 \times 44 + 76

(ii)

93 + 37 \times 76 + 44

(iii)

(93 + 37) \times (44 + 76)

(iv)

37 \times 44 + 93 + 76

(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). ✓

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

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6Choose a number and create ten different expressions having that value.

Let the chosen number be 20.

Here are ten different expressions each equal to 20:

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

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

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What are the important topics in Arithmetic Expressions for CBSE Class 7 Mathematics?

Key topics in Arithmetic Expressions include Simple Expressions and Values, Terms in Expressions, Swapping and Grouping Terms, Brackets and Order of Evaluation. Study these first, then practise questions on each for Class 7 exams.

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How should I revise Arithmetic Expressions for Class 7 exams?

Learn the core ideas first, then work through the 43 practice questions on Arithmetic Expressions. Revise definitions regularly and use flashcards for quick recall before the exam.

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Arithmetic Expressions — NCERT Solutions

Studying Arithmetic Expressions? Get the full interactive chapter.

Figure it Out (Comparing Expressions)

Figure it Out (Terms and Brackets)

Figure it Out (Removing Brackets — I)

Figure it Out (Distributive Property — Products)

Figure it Out (Distributive Property — Fill in the Blanks)

Figure it Out (Final Exercise)

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