This MCQ module is based on: Grouping and Patterns
TOPIC 14 OF 31
Grouping and Patterns
🎓 Class 7
Mathematics
CBSE
Theory
Ch 4 — Expressions Using Letter-Numbers
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Grouping and Patterns
Targeting Class 7 level in Algebra, with Basic difficulty.
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4.5 Grouping Like Terms
When an algebraic expression has many terms, we group like terms together and add their coefficients. Two practical methods:
Method 1 — Group by Like Terms
Simplify \(5y + x + 5y + (-6) + 1\).
Group: \((5y + 5y) + (x) + ((-6) + 1) = 10y + x - 5\).
Method 2 — Group by Rows
Write repeated coefficients vertically, add column by column. Same answer.
Simplify \(2p + 3q - 5p + 5q\) by term type.
Column 1 (p): \(2p - 5p = -3p\). Column 2 (q): \(3q + 5q = 8q\). Final: \(-3p + 8q\).
Simplify \(-5g - 2g + 5k + 10k\).
g terms: \(-5g - 2g = -7g\). k terms: \(5k + 10k = 15k\). Final: \(-7g + 15k\).
Two equivalent ways to simplify an expression
4.6 Matchstick Patterns and Formulas
Look at a pattern with 1 triangle, 2 triangles, 3 triangles made of matchsticks joined side-by-side:
| Step Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| No. of Matchsticks | 3 | 5 | 7 | 9 | 11 | 13 |
Each step adds 2 matchsticks. To find Step \(n\), start with 3 and add \(2(n-1)\). The general rule: \(2n + 1\) matchsticks in Step \(n\).
Verification for Step 4: \(2 \times 4 + 1 = 9\) ✓. Step 33: \(2 \times 33 + 1 = 67\). Step 84: \(169\). Step 108: \(217\).
Another pattern — squares joined in a row
For \(n\) squares joined side-by-side sharing edges, matchsticks needed = \(3n + 1\). Two ways to see this:
- First square uses 4; each new square adds 3. Total = \(4 + 3(n-1) = 3n + 1\).
- Add 3 per square and 1 extra at the end.
Row of joined squares — each new square adds 3 matchsticks
Cross Pattern of Squares (Advanced)
Consider a cross-shaped arrangement that grows step by step. In Step 1 there are 5 squares, Step 2 has 9, Step 3 has 13. The rule: Step \(n\) has \(5 + 4(n-1) = 4n + 1\) squares.
For Step 4: \(4 \times 4 + 1 = 17\). Step 10: 41. Step 50: 201. Each square has 4 vertices, but adjacent squares share vertices. Total vertices formula: \(16n + 4\).
4.7 Using Letter-Numbers for General Rules
The formula? for the number of matchsticks at step \(n\) in a triangle pattern, \(2n + 1\), is a mathematical rule expressed with a letter-number. Formulas summarise patterns.
Formula
A formula is an algebraic expression (often with an equals sign) that gives the value of one quantity in terms of others. Example: Perimeter of square = \(4s\), where \(s\) is the side length.Example — Snail climbing a well
A snail is trying to climb the wall of a deep well. During the day it climbs up \(u\) cm and during the night it slips down \(d\) cm, with \(d < u\). In 10 days and 10 nights:
Net progress per day–night cycle = \(u - d\). Total progress in 10 complete cycles = \(10(u - d) = 10u - 10d\) cm. If the well is deep enough that the snail never reaches the top, the distance from the starting point after 10 days and 10 nights is \(10u - 10d\) cm.
Example — Radha's Cycling
Radha is training for a race. The first week she cycles 5 km every day. Every week she increases daily distance by \(z\) km. How many kilometres does she cycle in 3 weeks?
Week 1: 5 km/day × 7 days = 35 km.
Week 2: \((5+z)\) km/day × 7 = \(7(5+z) = 35 + 7z\) km.
Week 3: \((5+2z)\) km/day × 7 = \(35 + 14z\) km.
Total in 3 weeks = \(35 + (35+7z) + (35+14z) = 105 + 21z\) km.
Activity: Find My Pattern
L3 ApplyMaterials: Matchsticks or toothpicks, graph paper.
Predict: If you make a row of hexagons sharing edges, how many sticks does the 10th hexagon add?
- Make 1 hexagon. Count sticks: 6.
- Add a second hexagon sharing one side: new sticks added = 5.
- Continue to 5 hexagons. Record the counts.
- Find the rule Step \(n\) = ? matchsticks.
First hexagon 6, then +5 for each additional. Formula: \(6 + 5(n-1) = 5n + 1\).
Figure it Out
Q1. In the figure, observe how the expression \(w + 2\) becomes \(6w + 20\) along the path. Fill in missing expressions on the remaining paths, given the operations shown.
Accepted sample: start \(3w\), ×3 gives \(9w\), −5 gives \(9w-5\), +4 gives \(9w-1\), ×4 gives \(36w-4\), +20 gives \(36w+16\).
Q2. A local train takes \(t\) minutes between consecutive stations and stops 2 minutes at each of 3 intermediate stations. Total travel time from Yahapur to Vahapur is 45 min. Find \(t\).
Running legs: 4 × \(t\). Stops: 3 × 2 = 6 min. So \(4t + 6 = 45\Rightarrow 4t = 39 \Rightarrow t = 9.75\) min.
Competency-Based Questions
Scenario: A tiling pattern uses \(n\) identical square tiles placed in a straight row, with grout along every edge. Each tile has side \(s\) cm.
Q1. Write an expression for the total length of grout lines (horizontal plus vertical) along a straight row of \(n\) tiles.
L3 ApplyTop + bottom = \(2ns\). Vertical grout lines = \((n+1)\) lines, each \(s\) cm long = \((n+1)s\). Total = \(2ns + (n+1)s = (3n+1)s\).
Q2. For \(n=5, s=20\), find the grout length. Analyse how the answer changes when \(n\) doubles.
L4 Analyse\((3 \times 5 + 1) \times 20 = 320\) cm. When \(n = 10\): \(31 \times 20 = 620\) cm. The result roughly doubles but grows by \(6s\) less than double because of the "+1" in the formula.
Q3. Evaluate a student's claim: "Total grout length grows linearly with \(n\)". Is this correct?
L5 EvaluateYes. The formula \((3n+1)s\) is linear in \(n\) (highest power of \(n\) is 1). Grout length increases by \(3s\) for each extra tile — a constant rate.
Q4. Design a 2D \(m \times n\) grid of square tiles and write a formula for the total grout length. Verify for \(m=n=3\).
L6 CreateHorizontal lines: \((m+1)\) of length \(ns\). Vertical lines: \((n+1)\) of length \(ms\). Total = \((m+1)ns + (n+1)ms = s(mn+n+mn+m) = s(2mn+m+n)\). For \(m=n=3\): \(s(18+3+3) = 24s\). A 3×3 grid of side 20 cm needs \(24 \times 20 = 480\) cm of grout. ✓
Assertion–Reason Questions
A: The number of matchsticks to build \(n\) joined triangles in a row is \(2n+1\).
R: Each new triangle after the first uses exactly 2 more matchsticks.
R: Each new triangle after the first uses exactly 2 more matchsticks.
(a)
A: The expression \(3n+1\) and \(3(n+1)-2\) give the same value for every \(n\).
R: Distributing in the second form gives \(3n+3-2 = 3n+1\).
R: Distributing in the second form gives \(3n+3-2 = 3n+1\).
(a)
Frequently Asked Questions
How do you group terms in an algebraic expression?
Use brackets to group terms that share a factor or that should be added/subtracted together. For example, (3x + 2) + (5x - 1) groups in pairs. Grouping helps apply the distributive property or simplify. NCERT Class 7 Chapter 4 practises grouping.
What is the distributive property?
The distributive property states a(b + c) = ab + ac. It lets you multiply a factor across a sum. For example, 3(x + 4) = 3x + 12. NCERT Class 7 Ganita Prakash Chapter 4 uses this property to expand expressions.
How can number patterns be written algebraically?
Spot a rule, then use a letter-number for the position. The pattern 3, 6, 9, 12... has nth term 3n. Odd numbers 1, 3, 5... have nth term 2n - 1. NCERT Class 7 Chapter 4 teaches this translation.
What is the general form of an even number?
Any even number can be written as 2n where n is a whole number. For n = 0, 1, 2, 3 the values are 0, 2, 4, 6. NCERT Class 7 Ganita Prakash Chapter 4 uses this general form to prove properties.
Why do we use algebra for patterns?
Algebra lets a single formula describe infinitely many terms of a pattern. Instead of listing 1, 4, 9, 16, we write n^2. This compactness and generality is algebra's power, as emphasised in NCERT Class 7 Chapter 4.
How do you expand 2(x + 3)?
Apply the distributive property: 2(x + 3) = 2*x + 2*3 = 2x + 6. Each term inside the bracket is multiplied by the factor outside. NCERT Class 7 Ganita Prakash Chapter 4 works through many such expansions.
Frequently Asked Questions — Expressions Using Letter-Numbers
What is Grouping and Patterns in NCERT Class 7 Mathematics?
Grouping and Patterns is a key concept covered in NCERT Class 7 Mathematics, Chapter 4: Expressions Using Letter-Numbers. 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 Grouping and Patterns step by step?
To solve problems on Grouping and Patterns, 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 4: Expressions Using Letter-Numbers?
The essential formulas of Chapter 4 (Expressions Using Letter-Numbers) 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 Grouping and Patterns important for the Class 7 board exam?
Grouping and Patterns 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 Grouping and Patterns?
Common mistakes in Grouping and Patterns 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 Grouping and Patterns?
End-of-chapter NCERT exercises for Grouping and Patterns 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 4, and solve at least one previous-year board paper to consolidate your understanding.
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