Algebra Play

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CLASS VIII Mathematics ~4–5 marks Ch 13 of 14
Algebra Play

Class 8 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • This chapter is about using algebra to explain why number tricks work — not just computing answers. A letter such as $x,\ a,\ M$ stands for "any number", so one equation proves the trick for every starting value.
  • 'Think of a Number' tricks: follow the steps with a variable instead of a number; the variable cancels, leaving a fixed answer.
  • Number pyramids: each box is the sum of the two below it. Writing the bottom row as $a,b,c$ gives the top as $a+2b+c$.
  • Calendar & algebra grids, largest-product and divisibility tricks are all "set up an expression, then solve a simple linear equation".
  • Core skills tested: forming an expression, simplifying (collecting like terms), and solving a one-variable linear equation (do the same operation to both sides).
  • Weightage: ~4–5 marks — usually one "explain/justify the trick" question and one word problem reduced to a linear equation.
Detailed notes

1. What "Algebra Play" means

Over the last two years you used algebra to model situations and to solve equations for an unknown letter-number. This chapter does something more powerful: it uses algebra to explain why tricks and puzzles work. The key idea:

A letter such as $x$ stands for any number. If a trick gives the same final answer no matter what $x$ is, then doing the steps with $x$ and watching it cancel out proves the trick once and for all.

So a single line of algebra replaces checking hundreds of examples. Three tools are used again and again:

  • Form an expression for each step.
  • Simplify it by collecting like terms (e.g. $2x+x=3x$, $2x-x=x$).
  • Solve a linear equation by doing the same operation to both sides.

2. 'Think of a Number' tricks

Here is the classic trick: (1) Think of a number, (2) double it, (3) add four, (4) divide by two, (5) subtract the original number. The claim is that you always get 2. Let us prove it by letting the chosen number be $x$.

Worked example — why the answer is always 2

Follow the steps with the letter $x$:

  • Think of a number: $x$
  • Double it: $2x$
  • Add four: $2x+4$
  • Divide by two: $\dfrac{2x+4}{2}=x+2$
  • Subtract the original: $(x+2)-x=2$

The $x$ cancels, so the result is $2$ for every starting number. To make the final answer $3$ instead, "add four" becomes add six (then $\tfrac{2x+6}{2}-x=3$). To get $5$, add ten.

The "guess the date" trick

A harder version: think of a date; multiply the month by $5$, add $6$, multiply by $4$, add $9$, multiply by $5$, then add the day. Let the month be $M$ and day be $D$.

$5M \to 5M+6 \to 4(5M+6)=20M+24 \to 20M+33 \to 5(20M+33)=100M+165 \to 100M+165+D$
NCERT worked example — answer 291

Final answer $=100M+165+D$. Mukta says $291$. So $291=100M+165+D$, giving $291-165=126=100M+D$. Since the day $D\le 31$ takes the last two digits and $M$ is what comes before, $M=1,\ D=26$ — the date is 26 January (Republic Day). The fixed number to subtract is always $165$.

Quick practice: if the answer is $1390$, then $1390-165=1225$, so $M=12,\ D=25$ — 25 December.

3. Number pyramids

In a number pyramid, each box is the sum of the two boxes directly below it. For example a bottom row $1,9,4$ gives a middle row $10,13$ and a top of $23$.

Bottom $a,\ b$ $\Rightarrow$ top $a+b.$
Bottom $a,\ b,\ c$ $\Rightarrow$ middle $a+b,\ b+c$ $\Rightarrow$ top $(a+b)+(b+c)=a+2b+c.$

So for a three-row pyramid the top equals (left) + 2×(middle) + (right) of the bottom row. The middle number is "counted twice". For four rows the top is $a+3b+3c+d$ (the coefficients are $1,3,3,1$ — like Pascal's triangle).

NCERT worked example — top is 60, fill the gaps

Top $60$, bottom $12,\ c,\ 8$, middle $a,\ b$. The rules give three equations:

$a+b=60$,   $12+c=a$,   $c+8=b$. Adding the last two: $(12+c)+(c+8)=a+b=60$, so $20+2c=60\Rightarrow 2c=40\Rightarrow c=20$. Then $a=12+20=32$ and $b=20+8=28$. The pyramid is bottom $12,20,8$; middle $32,28$; top $60$.

Worked example — find the top without building the whole pyramid

Bottom row $4,\ 13,\ 8$. Use the formula directly: top $=a+2b+c=4+2(13)+8=4+26+8=\boxed{38}$. (Building it up: $17,21$ then $38$ — same answer, less work.)

4. Calendar magic (Fun with Grids)

Pick a $2\times2$ block from a calendar; your friend adds the four numbers and tells you the sum — you find all four. Let the top-left number be $a$. Because of how a calendar is laid out, the others are $a+1$ (next day), $a+7$ (next week), $a+8$.

Sum $=a+(a+1)+(a+7)+(a+8)=4a+16.$
NCERT worked example — sum is 36

Set $4a+16=36$. Subtract $16$ from both sides: $4a=20$. Divide both sides by $4$: $a=5$. The four numbers are $5,\ 6,\ 12,\ 13$. (Check: $5+6+12+13=36$.) This shows the standard solving method — same operation on both sides.

5. Algebra grids

In an algebra grid, each shape stands for a number, and the last entry of every row is the sum of the values to its left. You read off equations and solve them.

NCERT worked example — squares and circles

Row 1: $\blacksquare+\blacksquare+\blacksquare=27\Rightarrow 3\blacksquare=27\Rightarrow \blacksquare=9$.

Row 2: $\bullet+\bullet+\blacksquare=19$. Substitute $\blacksquare=9$: $2\bullet+9=19\Rightarrow 2\bullet=10\Rightarrow \bullet=5$.

So square $=9$, circle $=5$. Once known, every empty cell can be filled.

6. The largest product

Use digits once each in a (2-digit) × (1-digit) product, e.g. $\square\square\times\square$ with $2,3,5$. There are six arrangements; algebra finds the best without checking all of them.

Comparing pairs with the same multiplier, the larger multiplicand wins, leaving $53\times2,\ 52\times3,\ 32\times5$. Clearly $52\times3>53\times2$. Expand the last two: $32\times5=(3\times10\times5)+(2\times5)$ and $52\times3=(5\times10\times3)+(2\times3)$ — the first terms are equal, and $2\times5>2\times3$, so $32\times5=160$ is the largest.

Rule (proved by algebra): make the largest digit the single multiplier, and arrange the other two in decreasing order for the 2-digit multiplicand.
Worked example — digits 1, 3, 7

Largest digit $7$ is the multiplier; remaining $3,1$ in decreasing order give $31$. Largest product $=31\times7=\boxed{217}$. (For $3,5,9$: $53\times9=477$.)

7. Decoding divisibility tricks

Take a two-digit number, reverse its digits, find the difference — it is always divisible by $9$. Why? Write the number as $\overline{ab}=10a+b$; reversed it is $10b+a$.

$(10b+a)-(10a+b)=9b-9a=9(b-a).$

The factor $9$ is plain to see, so the difference is always a multiple of $9$ (e.g. $74-47=27=9\times3$). The quotient on dividing by 9 is exactly $b-a$, the difference of the digits.

Worked example — sum instead of difference

Now add the number and its reverse: $(10a+b)+(10b+a)=11a+11b=11(a+b)$ — always divisible by $11$. Check $28+82=110=11\times10$. The factor $11$ comes straight out of the algebra.

Three-digit cycling: for $\overline{abc}+\overline{bca}+\overline{cab}$, each of $a,b,c$ appears once in the hundreds, tens and units place, giving $111(a+b+c)=3\times37\times(a+b+c)$ — always divisible by $37$ (and by $3$).

8. Turning word puzzles into equations

The chapter's Figure-it-Out problems are solved by naming an unknown and forming one linear equation.

Horses and hens (heads & legs)

$55$ heads, $150$ legs. Let horses $=h$, hens $=55-h$. Legs: $4h+2(55-h)=150\Rightarrow 4h+110-2h=150\Rightarrow 2h=40\Rightarrow h=20$ horses, $35$ hens. Without algebra: if all $55$ were hens that is $110$ legs; the extra $150-110=40$ legs need $40\div2=20$ horses.

Mother and daughter (ages)

Mother is $5$ times the daughter; in $6$ years she is $3$ times. Let daughter $=x$, mother $=5x$. Then $5x+6=3(x+6)\Rightarrow 5x+6=3x+18\Rightarrow 2x=12\Rightarrow x=6$. The daughter is now 6 (mother $30$).

Karim and the Genie

Money doubles each round, then he pays $8$ coins. After three rounds he is left with exactly $8$. Let the start be $x$. Round 1: $2x-8$; round 2: $2(2x-8)-8=4x-24$; round 3: $2(4x-24)-8=8x-56$. Set $8x-56=8\Rightarrow 8x=64\Rightarrow x=8$. So Karim started with 8 coins. To gain money the doubling must beat the toll, i.e. the cost per round must be less than the starting amount; setting cost so the final coins are $0$ drains him completely.

9. Common mistakes to avoid

  • Forgetting to let the unknown be a letter — you cannot "prove for all numbers" by testing two or three cases.
  • In a 3-row pyramid, writing the top as $a+b+c$ instead of $a+2b+c$ (the middle is counted twice).
  • Not doing the same operation to both sides when solving (e.g. subtracting $16$ from one side only).
  • Place value slips: a 2-digit number $\overline{ab}$ is $10a+b$, not $a+b$ or $ab$ (product).
  • In word problems, mixing up "$5$ times as old" ($m=5x$) with "$5$ years older" ($m=x+5$).

10. Quick revision checklist

  • Let the unknown be a variable; simplify; watch terms cancel — that proves a trick.
  • Pyramid: bottom $a,b,c\Rightarrow$ top $a+2b+c$.
  • Calendar $2\times2$ sum $=4a+16$; solve for $a$.
  • Largest product: biggest digit = multiplier; rest in decreasing order.
  • Reverse-subtract $=9(b-a)$ (÷9); reverse-add $=11(a+b)$ (÷11).
  • Word problem $\to$ name unknown $\to$ one linear equation $\to$ same operation both sides.
Practice MCQs
1. Think of $x$, double it, add four, divide by two, subtract the original. The result is:
  1. $x$
  2. $2$
  3. $4$
  4. $2x$
Answer: (B) $\dfrac{2x+4}{2}-x=(x+2)-x=2$ — the variable cancels.
2. In a number pyramid each box equals the:
  1. product of the two below it
  2. difference of the two below it
  3. sum of the two below it
  4. average of the two below it
Answer: (C) each box is the sum of the two boxes directly below it.
3. If a three-row pyramid has bottom row $a,\ b,\ c$, the top number is:
  1. $a+b+c$
  2. $a+2b+c$
  3. $2a+b+2c$
  4. $abc$
Answer: (B) middle row is $a+b,\ b+c$; top $=(a+b)+(b+c)=a+2b+c$.
4. The top of a pyramid with bottom row $4,\ 13,\ 8$ is:
  1. $25$
  2. $30$
  3. $38$
  4. $50$
Answer: (C) $a+2b+c=4+26+8=38$.
5. The sum of a $2\times2$ calendar block with top-left $a$ is:
  1. $4a+8$
  2. $4a+16$
  3. $2a+16$
  4. $4a+7$
Answer: (B) $a+(a+1)+(a+7)+(a+8)=4a+16$.
6. A $2\times2$ calendar block sums to $36$. Its top-left number is:
  1. $4$
  2. $5$
  3. $6$
  4. $9$
Answer: (B) $4a+16=36\Rightarrow 4a=20\Rightarrow a=5$.
7. A two-digit number with digits $a$ (tens) and $b$ (units) equals:
  1. $a+b$
  2. $ab$
  3. $10a+b$
  4. $10b+a$
Answer: (C) place value gives $10a+b$.
8. The difference between a two-digit number and its reverse is always divisible by:
  1. $2$
  2. $5$
  3. $9$
  4. $11$
Answer: (C) $(10b+a)-(10a+b)=9(b-a)$.
9. The sum of a two-digit number and its reverse is always divisible by:
  1. $9$
  2. $11$
  3. $7$
  4. $3$
Answer: (B) $(10a+b)+(10b+a)=11(a+b)$.
10. Using the digits $1,3,7$ once each in $\square\square\times\square$, the largest product is:
  1. $31\times7=217$
  2. $71\times3=213$
  3. $73\times1=73$
  4. $37\times1=37$
Answer: (A) biggest digit ($7$) as multiplier, rest in decreasing order ($31$): $217$.
11. In an algebra grid, $\blacksquare+\blacksquare+\blacksquare=27$. The value of $\blacksquare$ is:
  1. $3$
  2. $9$
  3. $27$
  4. $24$
Answer: (B) $3\blacksquare=27\Rightarrow\blacksquare=9$.
12. The expression $20M+24$ comes from "$5M+6$, then multiply by $4$". This uses:
  1. only addition
  2. the distributive property $4(5M+6)$
  3. collecting like terms
  4. division
Answer: (B) $4(5M+6)=20M+24$ — distributive multiplication over a sum.
Assertion–Reason
A: A 'Think of a Number' trick can give the same answer for every starting number.   R: Replacing the number by a variable lets the variable cancel during simplification.
Answer: Both A and R are true, and R is the correct explanation of A — the cancelling of $x$ is exactly why the answer is fixed.
A: The sum of a two-digit number and its reverse is divisible by $9$.   R: That sum equals $11(a+b)$.
Answer: A is false, R is true — the sum is $11(a+b)$, divisible by $11$ (it is the difference that is divisible by $9$).
Exam-style questions
Q1. Explain, using algebra, why "think of a number, double it, add four, divide by two, subtract the original" always gives $2$. (3 marks)
Solution: Let the number be $x$. Steps: $x\to 2x\to 2x+4\to \dfrac{2x+4}{2}=x+2\to (x+2)-x=2$. Since $x$ cancels, the result is $2$ for every starting number.
Q2. A farm has horses and hens with $55$ heads and $150$ legs in total. Find the number of each. (3 marks)
Solution: Let horses $=h$, hens $=55-h$. Legs: $4h+2(55-h)=150\Rightarrow 2h+110=150\Rightarrow h=20$. So $20$ horses and $35$ hens.
Q3. A mother is $5$ times her daughter's age. In $6$ years she will be $3$ times as old. How old is the daughter now? (3 marks)
Solution: Daughter $=x$, mother $=5x$. Then $5x+6=3(x+6)\Rightarrow 5x+6=3x+18\Rightarrow 2x=12\Rightarrow x=6$. The daughter is $6$ years old.
Q4. Prove that the difference between a two-digit number and the number formed by reversing its digits is always a multiple of $9$. (2 marks)
Solution: Number $=10a+b$, reverse $=10b+a$. Difference $=(10b+a)-(10a+b)=9b-9a=9(b-a)$, which is a multiple of $9$.
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A Square and A Cube · Power Play · A Story of Numbers · Quadrilaterals · Number Play · We Distribute, Yet Things Multiply · Proportional Reasoning – 1 · Fractions in Disguise
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