This MCQ module is based on: Figure it Out — Final Exercise Set
TOPIC 19 OF 22
Figure it Out — Final Exercise Set
🎓 Class 8
Mathematics
CBSE
Theory
Ch 6 — Linear Equations in One Variable
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Figure it Out — Final Exercise Set
Targeting Class 8 level in Algebra, with Basic difficulty.
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Figure it Out — Final Exercise Set
Q1. Expand the following expressions.
(i) \(3(s+1)(s-3)\)
(ii) \((10a+8)(10a+d)\)
(iii) \((5-x)(3-6)\)
(iv) \((-5a+b)(a+d)\)
(v) \((-5x+y)(3+d)\)
(i) \(3(s+1)(s-3)\)
(ii) \((10a+8)(10a+d)\)
(iii) \((5-x)(3-6)\)
(iv) \((-5a+b)(a+d)\)
(v) \((-5x+y)(3+d)\)
(i) \(3(s^2-2s-3) = 3s^2 - 6s - 9\).
(ii) \(100a^2 + 10ad + 80a + 8d\).
(iii) \(-3(5-x) = 3x - 15\).
(iv) \(-5a^2 - 5ad + ab + bd\).
(v) \(-15x - 5xd + 3y + yd\).
(ii) \(100a^2 + 10ad + 80a + 8d\).
(iii) \(-3(5-x) = 3x - 15\).
(iv) \(-5a^2 - 5ad + ab + bd\).
(v) \(-15x - 5xd + 3y + yd\).
Q2. Find 3 examples where the product of two numbers (unchanged) may not be increased by 1 when one of them is increased by 1 and the other decreased by 1.
Using \((a+1)(b-1) = ab + b - a - 1\), the change is \(b - a - 1\). This is not always \(+1\). Examples (change):
• \(a=5, b=3\): change \(= 3-5-1 = -3\) (decreases).
• \(a=10, b=2\): change \(= 2-10-1 = -9\).
• \(a=7, b=7\): change \(= -1\).
• \(a=5, b=3\): change \(= 3-5-1 = -3\) (decreases).
• \(a=10, b=2\): change \(= 2-10-1 = -9\).
• \(a=7, b=7\): change \(= -1\).
Q3. Expand \((a+b-3b^2)(4+a)\) and \((10a + 3)(a^2 + 9)\).
\((a+b-3b^2)(4+a) = 4a + a^2 + 4b + ab - 12b^2 - 3ab^2\).
\((10a+3)(a^2+9) = 10a^3 + 90a + 3a^2 + 27\).
\((10a+3)(a^2+9) = 10a^3 + 90a + 3a^2 + 27\).
Q4. (i) \((a-3b^2)(4a-b)\), (ii) \((10a+8)(a-b)\), (iii) \(3y+(12z-3)\), (iv) \((-5+c)(-3-d)\).
(i) \(4a^2 - ab - 12ab^2 + 3b^3\).
(ii) \(10a^2 - 10ab + 8a - 8b\).
(iii) \(3y + 12z - 3\).
(iv) \(15 + 5d - 3c - cd\).
(ii) \(10a^2 - 10ab + 8a - 8b\).
(iii) \(3y + 12z - 3\).
(iv) \(15 + 5d - 3c - cd\).
Q5. Use (i) \((a+b)(a+b)\) to find its square. (ii) \((a-b)(a-b)\). (iii) Do you see a pattern? What could have gone wrong if you simply wrote \((a+b)^2 = a^2 + b^2\)?
(i) \(a^2 + 2ab + b^2\). (ii) \(a^2 - 2ab + b^2\). (iii) The middle cross-term is the key; omitting \(2ab\) is the most common pitfall.
Q6. Use identities learnt so far to find each product.
(i) \(46^2\) using \((a+b)^2\)
(ii) \(397 \times 403\) using \((a+b)(a-b)\)
(iii) \(91^2\) using \((a-b)^2\)
(iv) \(43 \times 45\).
(i) \(46^2\) using \((a+b)^2\)
(ii) \(397 \times 403\) using \((a+b)(a-b)\)
(iii) \(91^2\) using \((a-b)^2\)
(iv) \(43 \times 45\).
(i) \((40+6)^2 = 2116\). (ii) \(400^2 - 3^2 = 159991\). (iii) \((100-9)^2 = 8281\). (iv) \((44-1)(44+1) = 1935\).
Q7. A town planner wants to plan a residential colony with small gardens of two square parks of sides 60 m and 5 m. She wants to increase each side of the big park by 5 m. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
The identity used is \((60+5)^2 = 60^2 + 2(60)(5) + 5^2\). Both sides equal 4225 — a valid instance of Identity 1A.
Q8. Is the algebraic expression describing the following steps — add any two numbers. Multiply by half the sum of the two numbers. Prove that this result is half the square of the sum of the two numbers.
Let the numbers be \(a, b\). Sum = \(a+b\). Half of sum = \(\frac{a+b}{2}\). Product = \((a+b)\cdot \frac{a+b}{2} = \frac{(a+b)^2}{2}\). Hence proved.
Q9. Which is larger? Find out without actually computing the product.
(i) \(14\times 26\) or \(16 \times 24\)
(ii) \(25 \times 75\) or \(76 \times 74\).
(i) \(14\times 26\) or \(16 \times 24\)
(ii) \(25 \times 75\) or \(76 \times 74\).
(i) \(14\times 26 = (20-6)(20+6) = 400 - 36 = 364\); \(16\times 24 = (20-4)(20+4) = 400 - 16 = 384\). So \(16\times 24\) is larger.
(ii) \(25\times 75 = 1875\) (direct via \((50-25)(50+25) = 2500-625\)); \(76\times 74 = (75+1)(75-1) = 75^2 - 1 = 5624\). So \(76 \times 74\) is larger.
(ii) \(25\times 75 = 1875\) (direct via \((50-25)(50+25) = 2500-625\)); \(76\times 74 = (75+1)(75-1) = 75^2 - 1 = 5624\). So \(76 \times 74\) is larger.
Q10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area \(x^2\) sq. ft, will have a green cover. All the remaining area is a walking path to be tiled. Write an expression for the area that needs to be tiled.
If the total park has dimensions \(\ell \times w\), tiled area = \(\ell w - 2x^2\).
Q11. For each pattern shown below: (i) draw the next figure in the sequence, (ii) how many basic units are there in Step 10, (iii) write an expression to describe the number of basic units in Step \(y\).
Counts: 1, 5, 13, 25, … The differences 4, 8, 12, … follow the arithmetic rule, so Step \(y\) contains \(2y^2 - 2y + 1\) units. At \(y=10\): \(200 - 20 + 1 = 181\) units.
Activity — Coin Conjoin (Puzzle Time)
Puzzle: Coin Conjoin
L5 EvaluateMaterials: 10 coins arranged in a triangle (pointing up), pencil
Predict: What is the minimum number of coins to move so that the triangle points down?
- Arrange 10 coins in a triangle as in the figure below, pointing up. The task is to turn the triangle upside down by moving one coin at a time.
- A triangle of 3 coins can be inverted in a single move; a triangle of 6 coins in 2 moves.
- Figure out the minimum number of moves for 10 coins. (Answer: 3)
- The 10-coin triangle can be flipped with just 3 moves. Try the same for bigger triangles (15, 21, 28 coins…).
- Big question: Is there a simple rule to calculate the minimum moves for any triangular number of coins?
10-coin triangle (pointing up) — flip in 3 moves
For a triangular arrangement of side \(n\), the minimum number of coins to move is about \(\lfloor n(n-1)/6 \rfloor\) (depending on parity). For \(n=4\) (10 coins) it is 3; for \(n=5\) (15 coins) it is 5.
Chapter Summary
📘 What We Learned
- The distributive property? of multiplication over addition: \(a(b+c) = ab+ac\). This is the single foundational rule behind all the identities? in this chapter.
- The general product-identity: \((a+b)(c+d) = ac + ad + bc + bd\).
- Special cases — Identity 1A, 1B, 1C:
\((a+b)^2 = a^2 + 2ab + b^2\)\((a-b)^2 = a^2 - 2ab + b^2\)\((a+b)(a-b) = a^2 - b^2\)
- Extension to three numbers: \((a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca\).
- Geometric interpretations (area models) give a visual proof of every identity.
- The distributive property powers fast-multiplication shortcuts: multiplication by 11, 101, 99, 999, and Sridharacharya's "square using neighbours" formula \(a^2 = (a+b)(a-b)+b^2\).
- We considered different patterns, and explored how to understand them using algebra — one problem often has many correct paths.
Takeaway
Finding different methods to solve the same problem is a creative mathematical activity. In mathematics, there are often multiple ways of looking at a pattern, and different ways of approaching and solving the same problem. Finding such ways often requires a great deal of creativity and imagination!
Competency-Based Questions
Scenario: A craftswoman sells beaded bracelets. A small bracelet needs \(x\) beads; a large one needs \(x+3\) beads. If she makes \(x-2\) small and \(x+2\) large bracelets in one day, she wants to count the total beads algebraically.
Q1. Total beads used today =
L3 Apply(b). Each kind of bracelet contributes (count) × (beads per bracelet).
Q2. Simplify the expression in Q1 and analyse: for which values of \(x\) is the total a perfect square?
L4 Analyse\((x-2)x + (x+2)(x+3) = x^2 - 2x + x^2 + 5x + 6 = 2x^2 + 3x + 6\). For this to be a perfect square, \(x\) must satisfy \(2x^2 + 3x + 6 = k^2\) for some integer \(k\). Trial: \(x=5\) gives 71 (no); \(x=7\) gives 125 (no). Perfect-square values are rare.
Q3. Evaluate: if she can spare only 100 beads in total, what is the maximum value of \(x\)?
L5 EvaluateSolve \(2x^2 + 3x + 6 \leq 100 \Rightarrow 2x^2 + 3x - 94 \leq 0\). At \(x=6\): \(72 + 18 + 6 = 96 \leq 100\) ✓. At \(x=7\): \(98 + 21 + 6 = 125 > 100\). Max x = 6.
Q4. Create: design a new bracelet model and write a quadratic expression for total beads if counts and sizes are related by a pattern of your choice. State the identity you'd use to shortcut the multiplication.
L6 CreateSample: Suppose small = \(x-5\) and large = \(x+5\), each needing \(x\) beads. Total = \(x\cdot(x-5) + x\cdot(x+5) = 2x^2\). Shortcut: \((x-5)+(x+5) = 2x\), so the total is \(x\cdot 2x = 2x^2\) — factoring out \(x\) first uses the distributive property.
Assertion–Reason Questions
Assertion (A): If \(a^2 + b^2 = (a+b)^2\), then \(ab = 0\).
Reason (R): \((a+b)^2 - (a^2+b^2) = 2ab\).
Reason (R): \((a+b)^2 - (a^2+b^2) = 2ab\).
Answer: (a) — If the two are equal, their difference \(2ab = 0\), so \(ab=0\). R explains A.
Assertion (A): Every product \((a+b)(a-b)\) is a difference of two squares.
Reason (R): Identity 1C states \((a+b)(a-b) = a^2 - b^2\).
Reason (R): Identity 1C states \((a+b)(a-b) = a^2 - b^2\).
Answer: (a) — The identity itself is the reason.
Assertion (A): \(101 \times 99 = 10000\).
Reason (R): \(101 \times 99 = (100+1)(100-1) = 100^2 - 1\).
Reason (R): \(101 \times 99 = (100+1)(100-1) = 100^2 - 1\).
Answer: (d) — A is false (\(101\times 99 = 9999\), not 10000). R is true and correctly shows that the value is \(100^2 - 1 = 9999\).
Frequently Asked Questions — Linear Equations in One Variable
What is Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool in NCERT Class 8 Mathematics?
Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 6: Linear Equations in One Variable. 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 Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool step by step?
To solve problems on Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 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 6: Linear Equations in One Variable?
The essential formulas of Chapter 6 (Linear Equations in One Variable) 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 Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool important for the Class 8 board exam?
Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool is part of the NCERT Class 8 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 Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool?
Common mistakes in Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool 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 Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — Exercises & Summary | Class 8 Maths Ch 6 We Distribute, Yet Things Multiply | MyAiSchool 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 6, and solve at least one previous-year board paper to consolidate your understanding.
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