TOPIC 5 OF 31

Using and Removing Brackets in Expressions

Q: How do you remove brackets in algebraic expressions?

To remove brackets: if there is a plus sign before the bracket, simply remove the bracket keeping all signs the same. If there is a minus sign, remove the bracket and change every sign inside. For example, minus(3x minus 5) becomes minus 3x plus 5. This is taught in NCERT Class 7 Chapter 2.

🎓 Class 7 Mathematics CBSE Theory Ch 2 — Arithmetic Expressions ⏱ ~30 min

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More Expressions and Their Terms — Examples

📝 Example 8 — Ruby's Class Game

33 students are playing "Fire in the mountain, run, run, run!" Ruby is watching. When the teacher calls out '5', students form groups of 5. Ruby writes the expression: 6 × 5 + 3

📝 Example 10 — Ways to Pay ₹432

Kannan pays ₹432 using ₹1 coins and ₹5 coins, ₹10, ₹20, ₹50, ₹100 notes. There is more than one way!

Terms in Way 1: 4×100, 1×20, 1×10, 2×1. Terms in Way 2: 8×50, 1×10, 4×5, 2×1. Both evaluate to 432.

📝 Example 11 — Which Arrangement Matches 5 × 2 + 3?

Two arrangements of coloured tiles are shown. Which one is described by the expression 5 × 2 + 3?

Arrangement I
5×2 blue + 3 yellow
= 10 + 3 = 13 tiles
✅ Matches 5×2+3

≠

Arrangement II
2 rows of (5 blue + 3 yellow)
= 2×(5+3) = 2×8 = 16 tiles
Also = 5+3+5+3 = 5×2+3×2

💡 Key Insight from the Tile Example

Arrangement I (13 tiles) matches 5×2+3 — "3 more than 5×2"

Arrangement II (16 tiles) matches 2×(5+3) — "2 groups of (5+3)"
Which is the same as: 5+3+5+3 or 5×2+3×2

Note: 5×2+3 ≠ 2×(5+3) since 13 ≠ 16

Removing Brackets — I

What happens when we remove brackets^? from an expression? The key lies in the sign before the bracket.

Case 1: Minus sign before bracket containing a sum

📝 Example 12 — Irfan's Change

Irfan buys a biscuit packet (₹15) and toor dal (₹56). He pays ₹100. What is his change?

Expression: Change = 100 – (15 + 56)

Method 1 (brackets first): 15 + 56 = 71 → 100 – 71 = 29

Method 2 (remove brackets): 100 – 15 – 56 = 85 – 56 = 29 ✔

Key: –(15 + 56) → –15 – 56 Both terms inside flip sign from + to –

Removing Brackets — II

Case 2: Minus sign before bracket containing a difference

📝 Example 13 — The Sign Flip Surprise

Can 500 – (250 – 100) be written without brackets?

❗ MINUS before bracket

a – (b + c) = a – b – c

a – (b – c) = a – b + c

Every sign inside flips: + becomes – and – becomes +

✅ PLUS before bracket

a + (b + c) = a + b + c

a + (b – c) = a + b – c

Signs inside stay the same: no change needed

📝 Example 14 — Hira's Coin Collection

Hira has 28 coins in one bag and 35 coins in another. She gifts 10 coins from the second bag. How many coins remain?

Expression: 28 + (35 – 10)

Remove bracket (+ before bracket → signs stay): = 28 + 35 – 10

Value: 28 + 35 – 10 = 63 – 10 = 53 coins

Note: The + before the bracket means no sign change for terms inside.

🔧 Bracket Removal Checker

Select the type of bracket removal and enter values to see the sign rules applied automatically.

Pattern:

Tinker the Terms — I

What happens to an expression's value when we change one of its terms by a small amount? We can reason without full calculation.

Change in term	Expression	Value	Reasoning
Base case	53 + (–16)	37	Given
First term +1	54 + (–16)	38	54 is 1 more than 53 → value is 1 more
First term –1	52 + (–16)	36	52 is 1 less → value is 1 less
Second term +1	53 + (–15)	38	–15 is 1 more than –16 → value +1
Second term –1	53 + (–17)	36	–17 is 1 less than –16 → value –1
Both terms changed	–87 + (–16)	–103	–87 = –(53+34), so 37 – 140 = –103
Practice	–88 + (–15)	–103	–88 is 1 less than –87, –15 is 1 more → same
Practice	–86 + (–18)	–104	–86+2=–84 vs base; –18–2=–20 vs base → net –2 from –103... recalculate: –86+(–18) = –104
Practice	–97 + (–26)	–123	–97+(–26) = –123

💡 The Tinker Principle

If one term in an expression increases by k, the value of the expression increases by k.

If one term decreases by k, the value decreases by k.

This works because each term contributes independently to the final sum.

📝 Figure it Out — Removing Brackets (Pages 37–38)

Q1. Fill in the blanks so that both sides are equal:
(a) 24 + (6 – 4) = 24 + 6 ___
(b) 38 + (___ ___) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 ___ 6 – 4
(d) 24 – 6 – 4 = 24 – 6 ___
(e) 27 – (8 + 3) = 27 ___ 8 ___ 3
(f) 27 – (___ ___) = 27 – 8 + 3

(a) 24 + (6–4): + before bracket → signs keep. LHS = 24+6–4 = 26. So blank = – 4

(b) 38 + (9–4) = 38 + 9 – 4. Bracket: 9 – 4

(c) 24 – (6+4): minus before bracket → flip signs. = 24 – 6 – 4. Blank = – (minus sign)

(d) 24 – 6 – 4 = 24 – 6 – 4

(e) 27 – (8+3) = 27 – 8 – 3

(f) 27 – (___) = 27 – 8 + 3 means –(___)= –8+3, so inside bracket = 8 – 3

Q2. Remove the brackets and write an equivalent expression:
(a) 14 + (12 + 10) (b) 14 – (12 + 10) (c) 14 + (12 – 10)
(d) 14 – (12 – 10) (e) –14 + 12 – 10 (f) 14 – (–12 – 10)

(a) 14 + (12+10): + before → keep signs. = 14 + 12 + 10 = 36

(b) 14 – (12+10): – before → flip all signs. = 14 – 12 – 10 = –8

(c) 14 + (12–10): + before → keep signs. = 14 + 12 – 10 = 16

(d) 14 – (12–10): – before → flip all. = 14 – 12 + 10 = 12

(e) –14 + 12 – 10 (no brackets to remove): = –12

(f) 14 – (–12–10): flip signs of –12 and –10 → = 14 + 12 + 10 = 36

Q3. For each pair, guess whether they are equal, then calculate to check:
(a) (6+10)–2 and 6+(10–2) (b) 16–(8–3) and (16–8)–3 (c) 27–(18+4) and 27+(–18–4)

(a) (6+10)–2 = 16–2 = 14; 6+(10–2) = 6+8 = 14. Equal. ✔
Why? Removing the bracket in (6+10)–2: the result is 6+10–2 (same as the second). Addition grouping does not matter.

(b) 16–(8–3) = 16–5 = 11; (16–8)–3 = 8–3 = 5. NOT equal. ✗
Why? 16–(8–3) = 16–8+3 = 11; but (16–8)–3 = 16–8–3 = 5. Subtraction is not associative!

(c) 27–(18+4) = 27–22 = 5; 27+(–18–4) = 27–18–4 = 5. Equal. ✔
Why? –(18+4) = –18–4 = (–18–4), so 27–(18+4) = 27+(–18–4). ✔

Q4. Identify which expressions have the same value (without evaluating):
(a) 319+537, 319–537, –537+319, 537–319
(b) 87+46–109, 87+46–109, 87+46–109, 87–46+109, 87–(46+109), (87–46)+109

(a) Terms of each:
• 319+537: terms 319, 537 → value 856
• 319–537: terms 319, –537 → value –218
• –537+319: terms –537, 319 → same as 319–537 → value –218 ✔ (same as 319–537)
• 537–319: terms 537, –319 → value 218
Equal pairs: 319–537 = –537+319

(b) Rewrite each as sum of terms:
• 87+46–109: terms 87,46,–109 → value 24 (the first three are identical)
• 87–46+109: terms 87,–46,109 → value 150
• 87–(46+109) = 87–46–109: terms 87,–46,–109 → value –68
• (87–46)+109 = 87–46+109: terms 87,–46,109 → value 150
Equal pairs: 87+46–109 = 87+46–109 = 87+46–109 (the first three); 87–46+109 = (87–46)+109

Q5. Add brackets so the expressions give the stated values:
(a) 34 – 9 + 12 = 13 (b) 56 – 14 – 8 = 34 (c) –22 – 12 + 10 + 22 = –22

(a) 34 – 9 + 12 = 37 normally. To get 13: need 34 – (9+12) = 34–21 = 13 ✔

(b) 56 – 14 – 8 = 34 normally. Already 34! Or: 56 – (14–8) = 56–6 = 50 (not 34). So 56 – 14 – 8 = 34 as given (no brackets needed, or add redundant ones).

(c) –22 – 12 + 10 + 22 = –2 normally. To get –22: need –22 – (12–10+22) = –22–24 = –46 (no). Try: –22 – (12–10–22) = –22–(–20) = –22+20 = –2 (no). Try: (–22–12+10+22) still not. Need: –22–12+(10+22): –22–12+32 = –2 (no). Try: –22–(12+10+22) = –22–44 = –66 (no). Try: –22–12+10+22 with brackets on (10+22) gives –22–12+32 = –2. Hmm. The answer is –22 – (12–10–22) = –22 – 0 = –22? 12–10–22 = –20, so –22–(–20) = –22+20 = –2. Not right either. Try (–22–12+10+22) = –2 ≠ –22. The correct bracketing: –22–(12–10+22) = –22–24 = –46 ≠ –22. Actually: try –22–12+10+22 as is = –2. To get –22: one answer is to add an expression: it appears the given expression cannot easily reach –22 with just rearranging brackets of the given expression with standard arithmetic. The intended answer per the answer key is bracket placement such that: maybe the expression is meant differently. Taking the given verbatim from NCERT: (–22) – (12+10+22) = –22–44 = –66, or –22–12+10+22 with (–22–12)+(10+22) = –34+32 = –2. The NCERT likely intends: –22–(12–10+22) = –22–24 = –46 or perhaps a typographical variant. Students should explore this.

Q6. Using only reasoning (no calculation), fill in the blanks:
(a) 423 + ___ = 419 + ___
(b) 207 – 68 = 210 – ___

(a) 423+___ = 419+___. The left side starts 4 more than right (423 vs 419). To keep equal, right side must be 4 more than left's blank. So: 423+x = 419+(x+4). One answer: 423 + 419 = 419 + 423 (both sides equal, commutativity). Or more practically: if one blank = k, other = k+4. E.g. 423 + 0 = 419 + 4.

(b) 207–68 = 210–___. Left side: minuend increased by 3 (207→210). To keep difference same, subtrahend must also increase by 3: ___ = 71. Check: 207–68 = 139; 210–71 = 139 ✔

Q7. Using numbers 2, 3, 5 and operators +, –, and brackets, generate as many different values as possible. (e.g., 2–3+5=4; 3–(5–2)=0)

Systematically:

2+3+5 = 10
2+3–5 = 0
2–3+5 = 4
2–3–5 = –6
–2+3+5 = 6
–2+3–5 = –4
–2–3+5 = 0
–2–3–5 = –10
5–(3–2) = 4
3–(5–2) = 0
2–(5–3) = 0
(5+2)–3 = 4
5–(3+2) = 0
3+(5–2) = 6
5–(2+3) = 0
(3–2)+5 = 6
2×3+5 would need ×

Unique values attainable: –10, –6, –4, 0, 4, 6, 10 (and more with different orderings).

Q8. Jasoda subtracts 9 from a number by subtracting 10 and adding 1. Example: 36 – 9 = 26 + 1 = 27.
(a) Does this always work? Why?
(b) Think of similar strategies.

(a) Yes, always works. Reason: subtracting 10 and adding 1 = subtracting 9.
Algebraically: n – 9 = n – 10 + 1 (since –9 = –10+1). This uses the bracket rule: n–9 = n–(10–1) = n–10+1 ✔

(b) Similar strategies:
• Subtract 19: subtract 20, add 1 (e.g. 45–19 = 45–20+1 = 26)
• Subtract 99: subtract 100, add 1 (e.g. 250–99 = 250–100+1 = 151)
• Subtract 8: subtract 10, add 2 (e.g. 37–8 = 37–10+2 = 29)
• Add 9: add 10, subtract 1 (e.g. 23+9 = 23+10–1 = 32)

Q9. For each of (a) 73 – 14 + 1 and (b) 73 – 14 – 1, identify which of the following expressions are equal:
(i) 73 – (14+1) (ii) 73 – (14–1) (iii) 73 + (–14+1) (iv) 73 + (–14–1)

Convert to sum of terms:
• 73 – 14 + 1: terms 73, –14, 1 → value 60
• 73–(14+1) = 73–14–1 → terms 73, –14, –1 → value 58 ✗
• 73–(14–1) = 73–14+1 → terms 73, –14, +1 → value 60 ✔ Equal to (a)
• 73+(–14+1) = 73–14+1 → value 60 ✔ Equal to (a)
• 73+(–14–1) = 73–14–1 → value 58 ✗

• 73 – 14 – 1: terms 73, –14, –1 → value 58
• 73–(14+1) = 73–15 = 58 ✔ Equal to (b)
• 73+(–14–1) = 73–15 = 58 ✔ Equal to (b)
Expressions (ii) and (iii) equal (a). Expressions (i) and (iv) equal (b).

🎯 Activity — Sign Detective Challenge

L4 Analyse

Without evaluating, predict whether each pair of expressions are equal, greater, or less — using only your understanding of terms and brackets.

Write each expression as a sum of terms (convert all subtractions).
Compare the sets of terms — if same set, expressions are equal.
If one term differs (e.g. one is k more), the whole expression is k more.

Practice: Without calculating, identify equal expressions in each group:

57 – 23 + 8 vs 57 + 8 – 23
100 – (45+30) vs 100 – 45 – 30
200 – (60–25) vs 200 – 60 – 25
48 + (12–7) vs 48 + 12 + 7

57–23+8 = 57+8–23 (same 3 terms, different order) → Equal ✔
100–(45+30) = 100–45–30 (bracket rule: –(a+b)=–a–b) → Equal ✔
200–(60–25) = 200–60+25 ≠ 200–60–25 → NOT Equal ✗ (first = 165, second = 115)
48+(12–7) = 48+12–7 ≠ 48+12+7 → NOT Equal ✗ (first = 53, second = 67)

🏆

Competency-Based Questions — Removing Brackets & Term Tinkering

Context: Radhika runs a school canteen. She tracks daily income and expenses using arithmetic expressions. She found that understanding how to remove brackets and reason about terms helps her quickly check totals and identify errors without recalculating from scratch. The same skills help students in higher algebra — where removing brackets is a daily task.

Q1. Which of the following is equal to 85 – (20 + 15)?

L2 Understand

(A) 85 – 20 + 15
(B) 85 – 20 – 15
(C) 85 + 20 – 15
(D) 85 + 20 + 15

Answer: (B) 85 – 20 – 15
Rule: –(20+15) → –20–15. So 85–(20+15) = 85–20–15 = 50. Verify: 85–35 = 50 ✔

Q2. Radhika earns ₹500 in the canteen. She spends ₹120 on ingredients and ₹80 on electricity. Write a bracket expression for her remaining amount, then remove the brackets.

L3 Apply

Expression with brackets: 500 – (120 + 80)
Remove brackets: 500 – 120 – 80 = 380 – 80 = ₹300
The –(120+80) = –120–80 because the minus sign distributes to both terms inside.

Q3. True or False (with reason):
(i) 200 – (50 – 30) = 200 – 50 – 30
(ii) 16 – (8–3) ≠ (16–8)–3
(iii) If 53+(–16) = 37, then 54+(–15) = 39

L5 Evaluate

(i) False. 200–(50–30) = 200–50+30 = 180 (not 200–50–30=120). When bracket preceded by –, signs flip: –50 stays –50 but –30 becomes +30.
(ii) True. 16–(8–3) = 16–8+3 = 11; (16–8)–3 = 8–3 = 5. Since 11≠5, they are indeed not equal. Subtraction is not associative.
(iii) False. 54+(–15): 54 is 1 more than 53 (+1), but –15 is 1 MORE than –16 (+1). So value = 37+1+1 = 39. Wait — that IS 39. So (iii) is True. ✔ Each term went up by 1, so total went up by 2: 37+2=39.

Q4. (HOT) A student writes: "500 – (250 – 100) = 500 – 250 – 100 = 150." Find the error. What is the correct value and the correct expanded form?

L4 Analyse

Error: The student wrote –(250–100) as –250–100, but the correct expansion is –250+100 (the sign of –100 flips to +100 when the outer minus distributes).
Correct expansion: 500 – (250–100) = 500 – 250 + 100 = 350
Mnemonic: "Minus before bracket → all signs inside flip." So –(250–100) = –250+100.

Q5. Jasoda's trick: subtract 19 by subtracting 20 and adding 1. Extend this idea: what would she do to subtract 98 from a number? Write the expression and verify with n = 150.

L6 Create

Strategy for –98: –98 = –100+2 (subtract 100, add 2)
Expression: n – 98 = n – (100–2) = n – 100 + 2
Verify with n=150: 150–98 = 52; Using trick: 150–100+2 = 50+2 = 52 ✔
General rule: To subtract (100k–r), subtract 100k and add r. Works because –(100k–r) = –100k+r.

🔬 Assertion–Reason Questions

(A) Both true; Reason is correct explanation (B) Both true; Reason is NOT correct explanation (C) Assertion true; Reason false (D) Assertion false; Reason true

Assertion (A): 14 – (12 – 10) = 14 – 12 + 10 = 12.
Reason (R): When a bracket preceded by a minus sign is removed, every sign inside the bracket changes.

Answer: (A)

Both correct. –(12–10): +12 becomes –12, and –10 becomes +10. So 14–12+10 = 12 ✔. The Reason is the exact rule that explains the Assertion.

Assertion (A): The expression 5 × 2 + 3 and 2 × (5 + 3) have the same value.
Reason (R): Multiplication distributes over addition, so 2×(5+3) = 2×5 + 2×3 = 10+6 = 16.

Answer: (D)

The Assertion is false: 5×2+3 = 13, but 2×(5+3) = 16. They are NOT equal.
The Reason is true: the distributive property correctly gives 2×(5+3) = 16. But the Reason proves the Assertion false, not true.

Assertion (A): 16 – (8 – 3) and (16 – 8) – 3 give different values.
Reason (R): Subtraction is not associative — changing the grouping of subtracted terms changes the result.

Answer: (A)

Both correct. 16–(8–3) = 16–8+3 = 11; (16–8)–3 = 8–3 = 5. Different values ✔. The Reason correctly explains why: unlike addition, subtraction is not associative, so regrouping changes the answer.

Frequently Asked Questions

How do you remove brackets in algebraic expressions?

When a plus sign precedes the bracket, remove the bracket without changing signs inside. When a minus sign precedes the bracket, change every sign inside: plus becomes minus and minus becomes plus. For example, minus (2x minus 3) becomes minus 2x plus 3. NCERT Class 7 Chapter 2 explains this rule.

What is the order of operations in algebra?

The order of operations follows BODMAS: Brackets first, then Orders (powers), Division and Multiplication (left to right), Addition and Subtraction (left to right). This ensures every expression has a unique value. NCERT Class 7 Ganita Prakash Chapter 2 applies BODMAS to algebraic expressions.

Why are brackets important in mathematical expressions?

Brackets indicate which operations should be performed first, overriding the standard order of operations. Without brackets, an expression like 2 times (x plus 3) would be calculated differently as 2 times x plus 3. NCERT Class 7 Maths Chapter 2 emphasises correct bracket usage.

How do you simplify expressions with multiple brackets?

Work from the innermost bracket outward. Simplify the expression inside the innermost bracket first, then move to the next level. Apply sign rules when removing each bracket layer. Combine like terms at the end. NCERT Class 7 Ganita Prakash Chapter 2 provides step-by-step examples.

What mistakes do students make when removing brackets?

Common mistakes include forgetting to change all signs when a minus precedes the bracket, changing signs incorrectly, not distributing multiplication to every term inside the bracket, and removing brackets in the wrong order. NCERT Class 7 Chapter 2 addresses these errors explicitly.

Frequently Asked Questions — Arithmetic Expressions

What is Using and Removing Brackets in Expressions in NCERT Class 7 Mathematics?

Using and Removing Brackets in Expressions is a key concept covered in NCERT Class 7 Mathematics, Chapter 2: Arithmetic Expressions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Using and Removing Brackets in Expressions step by step?

To solve problems on Using and Removing Brackets in Expressions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 2: Arithmetic Expressions?

The essential formulas of Chapter 2 (Arithmetic Expressions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Using and Removing Brackets in Expressions important for the Class 7 board exam?

Using and Removing Brackets in Expressions is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Using and Removing Brackets in Expressions?

Common mistakes in Using and Removing Brackets in Expressions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Using and Removing Brackets in Expressions?

End-of-chapter NCERT exercises for Using and Removing Brackets in Expressions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.

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