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6.1 Algebra Play

🎓 Class 8 Mathematics CBSE Theory Ch 6 — Algebra Play ⏱ ~16 min
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This MCQ module is based on: 6.1 Algebra Play

This mathematics assessment will be based on: 6.1 Algebra Play
Targeting Class 8 level in General Mathematics, with Basic difficulty.

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6.1 Algebra Play

Over the last two years, we have used algebra? to model different situations. We have learned how to solve algebraic equations? and find the values of unknown letter-numbers. Let's now have some fun with algebra. We shall investigate tricks and puzzles, and explain why they work using algebra. We will also see how to invent new tricks and puzzles to entertain others.

6.2 Thinking about 'Think of a Number' Tricks

In Grade 7, we learned about 'Think of a Number'? tricks, like this one:

Trick 1 — The Steady Trick
  1. Think of a number.
  2. Double it.
  3. Add four.
  4. Divide by 2.
  5. Subtract the original number you thought of.
The answer is always 2!
I predict you get 2. Am I right? Try it with different starting numbers. Do you always end up with the same value — namely 2? Why?

We saw that we can understand such tricks through algebra. Let \(x\)? stand for the number thought of:

  1. Think of a number: \(x\).
  2. Double it: \(2x\).
  3. Add four: \(2x + 4\).
  4. Divide by 2: \(\dfrac{2x + 4}{2} = x + 2\).
  5. Subtract the original number: \((x + 2) - x = 2\).

Therefore, no matter what the starting number is, the end result will always be 2. The \(x\) simply cancels out.

xstart 2x×2 2x+4+4 x+2÷2 = 2−x
The starting "x" flows through the five operations and exits as 2 — a constant, because the x cancels.
How would you change this trick to make the final answer 3? What about 5? Just replace "add four" with "add six" (for answer 3) or with "add ten" (for answer 5). In general, if step 3 is "add \(c\)" then the final answer is \(c/2\).
Can you come up with more complicated steps that always lead to the same final value? Yes — for example: pick \(x\); triple it; add 9; divide by 3; subtract the original. You always get \(3\). Try inventing your own!

Trick 2 — Shubham guesses Mukta's date of birth

Mukta thinks of a date (e.g. 26 Jan — Republic Day). Shubham guides her through these steps and, at the end, reveals the date he never saw.

Shubham's 7-Step Recipe
  1. Multiply the month by 5. (e.g. Jan = 1 → 5)
  2. Add 6. (5 + 6 = 11)
  3. Multiply by 4. (11 × 4 = 44)
  4. Add 9. (44 + 9 = 53)
  5. Multiply by 5. (53 × 5 = 265)
  6. Add the day. (265 + 26 = 291)
  7. Tell Shubham the final number.

Shubham's answer: 291. Now he subtracts 165 → gets 126 → that's 1 | 26 → month 1, day 26 → 26 January!

Let's prove why this recipe works. Let the month be \(M\) and the day be \(D\).

Algebraic Proof
  • Step 1: \(5M\)
  • Step 2: \(5M + 6\)
  • Step 3: \(4(5M + 6) = 20M + 24\)
  • Step 4: \(20M + 24 + 9 = 20M + 33\)
  • Step 5: \(5(20M + 33) = 100M + 165\)
  • Step 6: \(100M + 165 + D\)

Mukta's answer = \(100M + 165 + D\). If Shubham subtracts 165, he is left with \(100M + D\) — a number whose last 1 or 2 digits are \(D\) and whose earlier digits are \(M\). Since \(D \le 31 \le 99\), the split is unambiguous (when \(M > 0\)).

In the worked example, \(291 - 165 = 126\). Reading the last two digits gives \(D = 26\); the earlier digit gives \(M = 1\) — confirming 26/01.

Try this. If someone reports 1390, you compute \(1390 - 165 = 1225\), so \(M = 12\), \(D = 25\) — 25 December! The trick works for any valid calendar date.
Interactive: Try Shubham's Trick
Enter a month and day — the box performs Shubham's 7 steps and then "reveals" the date.
Press "Run the trick" to see the seven steps.
Activity: Invent Your Own Think-of-a-Number Trick
L6 Create
Materials: Paper, pencil, a friend
Predict: You will design steps that always lead to the answer 7, no matter what number your friend starts with.
  1. Let your friend's starting number be \(x\).
  2. Plan at least 5 steps. Use some combination of + / − / × / ÷ and a final "subtract the original number".
  3. Track the expression after every step using algebra on your own scratch sheet.
  4. Adjust the constants so the final expression equals 7.
  5. Test the trick with 3 different starting numbers — does it always give 7?

Sample: Pick \(x\); multiply by 4 → \(4x\); add 28 → \(4x + 28\); divide by 4 → \(x + 7\); subtract \(x\) → 7.

Trick 3 — Mukta guesses Shubham's birthday

Mukta flips the trick and adds an extra check. She asks Shubham to do:

Mukta's Steps for Shubham
  • Add 6: \(5M + 6\)
  • Multiply by 4: \(20M + 24\)
  • Add 9: \(20M + 33\)
  • Multiply by 5: \(100M + 165\)
  • Add the day: \(100M + 165 + D\)

Shubham reports 1390. Mukta notes the 2 right-most digits always come from \(165 + D\) (at most \(165 + 31 = 196\)), so she subtracts 165 and reads month then day: \(1390 - 165 = 1225 \to M = 12, D = 25\) — 25 December.

Find the dates if the final answers are: (i) 1269, (ii) 394, (iii) 296. Subtract 165 from each: (i) 1104 → 11/4 = 11 April. (ii) 229 → 2/29 = 29 February! (iii) 131 → 1/31 = 31 January.
Can you change the steps in this trick and still find the original date? Yes — what matters is that the month-coefficient becomes 100 and the added constant is known, so the final number is \(100M + \text{constant} + D\).

Competency-Based Questions

Scenario: A magician on a stage asks an audience member to "think of a number" and perform the following steps: pick \(x\); multiply by 6; add 18; divide by 3; subtract \(2x\). The magician then announces the final value without knowing \(x\).
Q1. Apply algebra to compute the value the magician will announce.
L3 Apply
6. Steps: \(6x \to 6x+18 \to \frac{6x+18}{3} = 2x+6 \to (2x+6) - 2x = 6\). The answer is always 6, independent of \(x\).
Q2. Analyse: which step is responsible for the "\(x\)" vanishing? Explain the mechanism.
L4 Analyse
The subtraction of \(2x\) cancels the \(2x\) that survived the division. Dividing by 3 turned \(6x\) into \(2x\) and \(18\) into \(6\); the "constant" part (6) had no \(x\) attached, so only the \(x\)-part was removed by the subtraction.
Q3. Evaluate: suppose the magician accidentally says "subtract \(x\)" instead of "subtract \(2x\)" at the last step. Is the trick ruined? Justify numerically.
L5 Evaluate
Yes, the trick is ruined. Final expression = \(2x + 6 - x = x + 6\), which depends on \(x\). For \(x=5\) the answer is 11; for \(x=10\) it is 16 — the magician cannot announce one fixed value.
Q4. Create: design a five-step "think-of-a-number" trick that always gives the answer 100, no matter what number is chosen.
L6 Create
One design: (1) pick \(x\); (2) multiply by 10 → \(10x\); (3) add 1000 → \(10x + 1000\); (4) divide by 10 → \(x + 100\); (5) subtract \(x\) → 100. Many other designs work — just ensure the \(x\)-coefficient at the end is zero and the remaining constant is 100.

Assertion–Reason Questions

Assertion (A): Every "think-of-a-number" trick whose final value does not depend on the chosen number must, in algebraic form, have all \(x\) terms cancel.
Reason (R): The final value depends on \(x\) only if some \(x\) term survives the operations.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — the cancellation of \(x\) is precisely why the answer is constant.
Assertion (A): In Shubham's trick, subtracting 165 always recovers \(100M + D\).
Reason (R): The number \(165\) is the result of the constants 6, 9, 5 that are introduced in the recipe.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — 165 = (6 × 4 + 9) × 5 = 33 × 5, which traces back to the added constants in the recipe.
Assertion (A): To design a trick that always gives the answer 4, we can simply double all constants in a "trick always gives 2" recipe.
Reason (R): Doubling a constant doubles the final answer, because the constant is the only part that survives cancellation of \(x\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (c) — A is true in many recipes (e.g. add 4 → add 8 gives final 4), but R is misleading: if there is a division step after the constant is added, simply doubling the added constant may still work, but it is the effective constant that matters, not every constant in the trick. The cancellation is about \(x\), not about doubling all constants blindly.

Frequently Asked Questions — Chapter 6

What is Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool in NCERT Class 8 Mathematics?

Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 8 Mathematics, Chapter 6: Chapter 6. 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 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool step by step?

To solve problems on Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | 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: Chapter 6?

The essential formulas of Chapter 6 (Chapter 6) 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 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool important for the Class 8 board exam?

Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | 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 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool?

Common mistakes in Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | 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 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | MyAiSchool?

End-of-chapter NCERT exercises for Part 1 — Algebra Play & Think-of-a-Number Tricks | Class 8 Maths Ch 6 | 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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