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CLASS 8 • MATHEMATICS • CHAPTER 13

Algebra Play

Explore the published notes, key concepts and questions for this chapter.

Quick answer
Algebra Play (Class 8 Ganita Prakash Part II, Chapter 6) teaches algebra through number tricks, number pyramids, calendar grids, maximum-product digit arrangements, divisibility and word problems. This guide covers all six main sections and 26 worked question items.
Key points
  • Use variables to explain why a number trick works for any starting value.
  • The date trick gives 100M+165+D, allowing the date to be decoded.
  • Three-row number pyramid top is a+2b+c.
  • Four-row pyramid top is a+3b+3c+d.
  • Calendar 2×2 sums equal 4a+16.
  • For digits p<q<r, (10q+p)r is the largest two-digit-by-one-digit product.
  • A two-digit number and its reversal differ by a multiple of 9 and sum to a multiple of 11.
  • Three-digit repetition equals 1001 times the original number.

Use these notes alongside your current textbook. Check important formulas, worked answers and syllabus coverage with your teacher or official learning material.

Algebra Play — NCERT Class 8 Mathematics, Ganita Prakash Part II

Direct answer: Algebra Play (Part II Chapter 6; overall Class 8 Chapter 13) explores number tricks, number pyramids, calendar and shape grids, maximum products, divisibility and word puzzles. Algebra explains why patterns work for every allowed number, not just a few examples.

Can a friend predict your chosen date from a single number? Can you find the top of a pyramid without calculating each row? These puzzles turn algebra into a practical tool for discovering and proving hidden rules.

6.1 Algebra Play: mathematical modelling

Represent an unknown by x, translate operations into expressions, simplify, and check the answer in the original problem. Equal quantities remain equal when the same operation is applied to both sides.

6.2 Thinking about Think of a Number Tricks

Choose x, double to 2x, add 4 to obtain 2x+4, halve to x+2, subtract x to obtain 2. The starting number cancels. To force answer k, add 2k before halving. In the calendar-date trick, month M and day D lead to 100M+165+D. Subtract 165 and read the month and two-digit day. For final answers 1269, 394 and 296, subtract 165 to obtain 1104,229 and 131: these represent 4 November, 29 February (only a valid date in leap years), and 31 January, respectively. For the stated August 2025 calendar, the February date is not an August date; the trick can use any valid date.

6.3 Number Pyramids

Each upper cell is the sum of the two cells directly below. For bottom a,b,c, top=a+2b+c; for bottom a,b,c,d, top=a+3b+3c+d. Coefficients follow Pascal-triangle rows. If the first n Virahāṅka–Fibonacci terms 1,2,3,5,… form the base, the top is term F(2n−1), with F1=1 and F2=2.

6.4 Fun with Grids

In a calendar 2×2 block with top-left a, entries are a,a+1,a+7,a+8; sum=4a+16. A sum of 36 gives a=5, hence 5,6,12,13. Check that the selected cells exist within the calendar month and week. Shape grids translate repeated icons into simultaneous equations; preserve the printed row and column positions before solving.

6.5 The Largest Product

For three distinct positive digits p<q<r, the maximum two-digit-by-one-digit product is (10q+p)×r. Example: 2,3,5 gives 32×5=160, greater than 52×3=156. Comparing algebraic expansions proves the rule.

6.6 Decoding Divisibility Tricks

A two-digit number minus its reversal is a multiple of 9; their sum is a multiple of 11. Three cyclic arrangements of a three-digit number sum to 111 times its digit sum and are divisible by 37 and 3. Repeating a three-digit number yields 1001 times the original; 1001=7×11×13.

Essential formula table

PatternAlgebraic ruleExample
Number trick(2x+4)/2−x=2Always 2
Date decodingResult−165=100M+D291−165=126→26 January
3-row pyramida+2b+c4,13,8→38
4-row pyramida+3b+3c+d8,19,21,13→141
Calendar 2×2 sum4a+1636→a=5
Reverse difference9|a−b|74−47=27
Reverse sum11(a+b)31+13=44
Three-digit repeat1001N123123=1001×123

NCERT Figure it Out — Number Pyramids

Questions are paraphrased from the official chapter or broken into their clearly identified subparts, with calculations shown.

  1. Question 1. Find the top of a three-row pyramid with bottom entries 4, 13, 8.

    Step-by-step solution: Top=a+2b+c=4+26+8=38.

    Final answer: 38.

  2. Question 2. Find the top of a three-row pyramid with bottom entries 7, 11, 3.

    Step-by-step solution: 7+2×11+3=32.

    Final answer: 32.

  3. Question 3. Find the top of a three-row pyramid with bottom entries 10, 14, 25.

    Step-by-step solution: 10+2×14+25=63.

    Final answer: 63.

  4. Question 4. Write the top expression for a four-row number pyramid with bottom a,b,c,d.

    Step-by-step solution: Row above: a+b,b+c,c+d. Next: a+2b+c,b+2c+d. Top: a+3b+3c+d.

    Final answer: a+3b+3c+d.

  5. Question 5. Find the top of a four-row pyramid with bottom 8,19,21,13.

    Step-by-step solution: 8+3×19+3×21+13=141.

    Final answer: 141.

  6. Question 6. Find the top of a four-row pyramid with bottom 7,18,19,6.

    Step-by-step solution: 7+54+57+6=124.

    Final answer: 124.

  7. Question 7. Find the top of a four-row pyramid with bottom 9,7,5,11.

    Step-by-step solution: 9+21+15+11=56.

    Final answer: 56.

  8. Question 8. Build a three-row pyramid from the first three Virahāṅka–Fibonacci numbers 1,2,3.

    Step-by-step solution: Second row 1+2=3 and 2+3=5. Top 3+5=8. Each number 1,2,3,3,5,8 belongs to the sequence.

    Final answer: Top 8.

  9. Question 9. Build a four-row pyramid from 1,2,3,5 and identify the top.

    Step-by-step solution: Second row 3,5,8; third row 8,13; top 21. All are terms of the Virahāṅka–Fibonacci sequence.

    Final answer: 21.

  10. Question 10. What happens if the first n Virahāṅka–Fibonacci numbers are placed in the bottom of an n-row pyramid?

    Step-by-step solution: Let F1=1,F2=2 and F(k+2)=F(k+1)+F(k). Each upper cell is the sum of adjacent consecutive sequence entries, so remains a sequence term. Each row starts at indices 1,3,5,... and the top is F(2n−1).

    Final answer: All entries are sequence terms; top is F(2n−1)..

NCERT Figure it Out — Largest Product

Questions are paraphrased from the official chapter or broken into their clearly identified subparts, with calculations shown.

  1. Question 1. Arrange digits 1,3,7 in a two-digit number times a one-digit number to maximize product.

    Step-by-step solution: For distinct digits p<q<r, compare (10q+p)r with (10r+p)q: difference=p(r−q)>0. Thus use 31×7.

    Final answer: 31×7=217.

  2. Question 2. Arrange digits 3,5,9 in a two-digit number times a one-digit number to maximize product.

    Step-by-step solution: By the same comparison, place 9 as multiplier and 5,3 as the two-digit number in descending order.

    Final answer: 53×9=477.

NCERT Figure it Out — Divisibility and Word Problems (Part A)

Questions are paraphrased from the official chapter or broken into their clearly identified subparts, with calculations shown.

  1. Question 1. Reverse a two-digit number with distinct digits and divide the positive difference by 9. What is the quotient?

    Step-by-step solution: For tens digit a and units b, the difference is |(10a+b)−(10b+a)|=9|a−b|. Divide by 9.

    Final answer: Absolute difference of the digits, |a−b|.

  2. Question 2. Show that a two-digit number plus its digit reversal is divisible by 11.

    Step-by-step solution: (10a+b)+(10b+a)=11a+11b=11(a+b).

    Final answer: Always divisible by 11.

  3. Question 3. Cycle the digits of a three-digit number abc to form bca and cab. Is their sum divisible by 37 and by 3?

    Step-by-step solution: Sum=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)=111(a+b+c)=37×3(a+b+c).

    Final answer: Divisible by both 37 and 3.

  4. Question 4. Repeat a three-digit number abc to form abcabc, then divide successively by 7, 11 and 13.

    Step-by-step solution: If N=100a+10b+c, then abcabc=1001N. Since 1001=7×11×13, division by all three gives N.

    Final answer: Original three-digit number abc.

NCERT Figure it Out — Word Problems (Part B)

Questions are paraphrased from the official chapter or broken into their clearly identified subparts, with calculations shown.

  1. Question 1. A person passes three ponds that double flowers, placing the same number at each shrine and all remaining flowers at the third. Find a possible initial number and equal offering.

    Step-by-step solution: Let starting flowers x and each offering y. After first: 2x−y; second: 4x−3y; third: 8x−6y=y. Thus 8x=7y. Smallest positive integer solution is x=7,y=8; positive multiples also work.

    Final answer: Smallest: 7 flowers initially, 8 at each shrine; multiples (7k,8k) also work.

  2. Question 2. A farm has 55 heads and 150 legs from horses and hens. Find each count.

    Step-by-step solution: Let horses h and hens k. h+k=55; 4h+2k=150. Subtract 2(h+k)=110 from leg count: 2h=40, so h=20,k=35.

    Final answer: 20 horses and 35 hens.

  3. Question 3. A mother is five times her daughter's age; in six years she will be three times as old. Find daughter's present age.

    Step-by-step solution: Let daughter be x; mother 5x. In six years 5x+6=3(x+6), so 5x+6=3x+18; 2x=12.

    Final answer: 6 years.

  4. Question 4. Gauri has half as many cows as Naina; after Naina gives Gauri three cows they have equal counts. Find both.

    Step-by-step solution: Let Gauri have x and Naina 2x. After transfer: x+3=2x−3; x=6.

    Final answer: Gauri 6 cows; Naina 12 cows.

  5. Question 5. A dosa cart has daily rent ₹5000, variable cost ₹10 per dosa, sells 100 dosas and wants ₹2000 profit. Find price.

    Step-by-step solution: Revenue=100p; total cost=5000+100×10=6000. Profit=100p−6000=2000; 100p=8000.

    Final answer: ₹80 per dosa.

  6. Question 6. With dosa price ₹50, daily rent ₹5000 and variable cost ₹10 per dosa, how many dosas yield ₹2000 profit?

    Step-by-step solution: For n dosas, profit=50n−(5000+10n)=40n−5000. Set equal to 2000: 40n=7000.

    Final answer: 175 dosas.

  7. Question 7. Evaluate 1/3, (1+3)/(5+7), and (1+3+5)/(7+9+11); explain the pattern.

    Step-by-step solution: First=1/3. Second=4/12=1/3. Third=9/27=1/3. The sum of first n odd numbers is n²; the next n odd numbers sum to 3n², so ratio is 1/3.

    Final answer: Every fraction equals 1/3.

  8. Question 8. Karim's coins double each round and he pays 8 coins after each of three rounds; immediately before the third payment he has 8 coins. Find starting coins.

    Step-by-step solution: Work backwards: before third doubling he had 4 coins; before second payment 12, so before second doubling 6; before first payment 14, so before first doubling 7.

    Final answer: 7 coins.

  9. Question 9. For a three-round doubling trick with a fixed fee f per round and initial x coins, when does the final balance exceed the initial?

    Step-by-step solution: After three rounds balance=8x−7f. To exceed x require 8x−7f>x, hence 7x>7f.

    Final answer: Fee f<x, assuming the process is feasible.

  10. Question 10. What fixed per-round fee allows the genie to take all x initial coins after three rounds?

    Step-by-step solution: Final balance=8x−7f. Set it to zero: f=8x/7. A whole-number coin fee requires x divisible by 7, unless fractional coins are permitted.

    Final answer: f=8x/7 coins per round (with feasibility and integrality conditions).

Activities, errors and revision

Try the date trick with three valid dates and verify by reversing the arithmetic. Build a number pyramid on squared paper. Use a 2×2 calendar block to check its sum. For a shape grid, write one equation per complete row before solving. Typical errors include mixing up digit place values, overlooking that the top of a pyramid weights central base cells more heavily, ignoring leap years, and treating an equation with infinitely many scaled solutions as having a unique answer.

Textbook reference and scope

Source: official NCERT Ganita Prakash Part II Chapter 6 (pages 135–147). All six chapter sections are covered. The chapter has three numbered Figure it Out sets with 6, 2 and 11 main questions. The 26 worked items here include separately numbered subparts and extended applications. Some picture-based number-pyramid and shape-grid prompts require the printed diagrams for precise cell values and are not represented as verified complete solutions.

Frequently asked questions

What is Algebra Play in Class 8?

Ganita Prakash Part II Chapter 6 on number tricks, pyramids, grids, digit products, divisibility and word problems.

Why does the think-of-a-number trick always give 2?

Because (2x+4)/2−x simplifies to 2.

How do you decode a date in the textbook trick?

Subtract 165 from the result to obtain 100M+D.

What is the top of a three-row number pyramid?

For bottom a,b,c, the top is a+2b+c.

What is the top of a four-row pyramid?

a+3b+3c+d.

How do calendar 2×2 grids work?

Their four values are a,a+1,a+7,a+8, summing to 4a+16.

What is the largest product using digits 2,3,5?

32×5=160.

Why is the reversal difference divisible by 9?

Because the difference is 9 times the difference of the digits.

Why is the reversal sum divisible by 11?

Because the sum is 11 times the sum of the digits.

Why is abcabc divisible by 7, 11 and 13?

Because abcabc=1001×abc and 1001=7×11×13.

How many main Figure it Out questions does the textbook contain?

Three sets with 6, 2 and 11 main questions, some with subparts.

Are all diagram-based grid puzzles solved here?

Not completely; diagram-specific cell arrangements still require visual verification.

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