Algebra Play — Chapter 6 Solutions
Ganita Prakash · Grade 8 · Part II — Step-by-step, fully worked solutions for every in-text question and exercise.
Think of a Number — Changing the Final Answer
The original trick is: double the number, add 4, divide by 2, subtract the original number — which always leaves 2, because $\frac{2x+4}{2}-x = (x+2)-x = 2$.
The “2” in the final answer comes from half of the number we added (4 ÷ 2 = 2). So to change the final answer, we just change the number added in Step 3:
Rule: To get a final answer of $k$, add $2k$ in Step 3.
More Complicated Steps
Yes — as long as the unknown $x$ cancels out completely at the end, any chain of operations works. Example:
Result is always 2 — no matter how many steps you add, the answer stays fixed as long as every term containing $x$ cancels out.
Date Trick — Find the Original Date
From the chapter, the trick’s algebra gives: Final answer $= 100M + 165 + D$, where $M$ = month, $D$ = day. So:
| Final Answer | Subtract 165 | Month (M) | Day (D) | Date |
|---|---|---|---|---|
| 1269 | 1269 − 165 = 1104 | 11 | 04 | 4th November |
| 394 | 394 − 165 = 229 | 2 | 29 | 29th February |
| 296 | 296 − 165 = 131 | 1 | 31 | 31st January |
(i) 4 November (ii) 29 February (iii) 31 January
Changing the Subtracted Number
The constant 165 came from the fixed operations: Add 6 → ×4 → Add 9 → ×5, which contributed $(6\times4+9)\times5 = (24+9)\times5 = 165$.
If you change any of these fixed numbers (the “Add 6”, “Add 9”, or the multipliers), the constant to subtract at the end will change accordingly. For example, if you replace “Add 9” with “Add 5”:
Then you would subtract 145 instead of 165 to recover $100M+D$.
General rule: whatever constant builds up from the fixed steps (independent of $M$ and $D$) is exactly the number you must subtract at the end.
Devise Your Own Trick
No matter what number you start with, the answer is always 3.
Number Pyramids — Simple Fills
Advanced Pyramid Fills & Algebra Grids
We use letter-numbers for unknown boxes and build equations from the given clues, exactly like the worked example for the pyramid with top 60.
Pyramid A (top = 50): Let bottom row be $4, x, 6, y$.
Bottom: 4, 9, 6, 1 | Next row: 13, 15, 7 | Row 2: 28, 22 | Top: 50
Pyramid B (one entry 9 in the middle, bottom starts 5…7): Let bottom be $5, x, 7, y$.
Bottom: 5, 2, 7, 24 | Next row: 7, 9, 31 | Row 2: 16, 40 | Top: 56
Pyramid C (top = 35, bottom starts 3, 5 … 7): Let bottom be $3,5,p,q$.
Bottom: 3, 5, 2, 11 | Next row: 8, 7, 13 | Row 2: 15, 20 | Top: 35
Method note: in every pyramid puzzle, write unknown cells as letters, build equations from the sum-rule and any visible numbers, then solve the equations exactly as shown for the textbook’s own worked example (top = 60 pyramid).
Algebra Grids (Shapes Represent Numbers)
Grid 1 — let ■ (blue square) = s and ● (red circle) = c:
■ = 11, ● = 5, missing sum = 21
Grid 2 — let ● (blue circle) = o and ◆ (purple diamond) = d:
● = 4, ◆ = 7, Row 3 sum = 15
Apply the same two-equation, two-unknown method to any new grid: turn each row into an equation, then solve by elimination or substitution.
Calendar Magic — How It Works
Let $a$ = top-left number. Since calendar rows differ by 1 (going right) and by 7 (going down), the grid is:
| $a$ | $a+1$ |
| $a+7$ | $a+8$ |
Grid: 5, 6 / 12, 13
Try your own: for a 1×3 horizontal strip $a, a+1, a+2$, the sum is $3a+3$. For a 3×1 vertical strip, the sum is $3a+21$. You can build similar tricks for any shape by adding up the offsets.
The Largest Product — Why the Rule Works
For three digits $p
Largest product = (largest digit) × (other two digits, in decreasing order)
Reversal Trick — What if $a > b$?
If $a > b$, then the original number $\overline{ab} = 10a+b$ is bigger than the reversed number $\overline{ba}=10b+a$. So instead we take the difference the other way round:
The difference is still divisible by 9 — we just subtract the smaller number from the larger one. (If $a=b$, the difference is simply 0, which is also divisible by 9.)
Figure It Out (Page 140) — Number Pyramids
38, 32, 63
$\text{Top} = a+3b+3c+d$
141, 124, 56
Grid contains: 1, 2, 3, 3, 5, 8 — all of these are Virahāṅka–Fibonacci numbers. Top = 8 (also a Fibonacci number).
(i) Bottom: 1, 2, 3, 5. Row above: 3, 5, 8. Next: 8, 13. Top: 21.
(ii) The same pattern continues: every number that appears anywhere inside a pyramid built from consecutive Fibonacci numbers is itself a Fibonacci number, and the topmost number is also a term far along the sequence.
Every single number that appears anywhere in the pyramid is itself a term of the Virahāṅka–Fibonacci sequence. This follows because each pyramid entry is built by repeatedly adding two Fibonacci numbers, and the sequence is defined exactly by that addition rule.
The topmost number is also a Virahāṅka-Fibonacci number — specifically, it turns out to be the $(2n-1)^{\text{th}}$ term of the sequence (e.g. for $n=3$, top is the 5th term = 8; for $n=4$, top is the 7th term = 21).
Figure It Out (Page 144) — Largest Product
Rule: largest digit as the single multiplier; remaining two digits in decreasing order as the multiplicand.
Largest product = $31 \times 7 = 217$
Largest product = $53 \times 9 = 477$
Figure It Out (Page 145–147) — Divisibility & Word Problems
The quotient always equals the difference between the two digits of the original number.
This is always a multiple of 11, since it equals $11 \times (a+b)$ exactly — confirmed by the examples: 44 = 11×4, 110 = 11×10, 33 = 11×3.
Since $111 = 3 \times 37$, the sum is always divisible by 37 — and always divisible by 3, because $111(a+b+c) = 3\times[37(a+b+c)]$.
Dividing $abcabc$ successively by 7, 11, and 13 always brings you back to the original number $abc$ — because you’re simply undoing the multiplication by $1001$.
Let initial flowers $= x$ and flowers placed in each shrine $= k$ (same each time).
He started with 7 flowers and placed 8 flowers in each shrine.
Check: $7\to14$(pond), give 8, left 6 → $6\to12$(pond), give 8, left 4 → $4\to8$(pond), give all 8. ✓
Using algebra: let $h$ = horses, $n$ = hens.
Without algebra (using the hint):
Horses = 20, Hens = 35
Daughter is currently 6 years old (mother is 30).
Gauri has 6 cows, Naina has 12 cows.
(i)
(ii) Let $n$ = number of dosas sold at ₹50 each.
(i) ₹80 per dosa (ii) 175 dosas/day
Why: the sum of the first $k$ odd numbers is $k^2$. The next $k$ odd numbers (from the $(k+1)^{\text{th}}$ to the $2k^{\text{th}}$) sum to $(2k)^2-k^2 = 3k^2$.
All the fractions are equal to $\dfrac{1}{3}$, always.
Let Karim start with $x$ coins; each round his coins double, then he pays 8 to the genie.
(i) Karim started with 7 coins.
Check: 7 → double 14, pay 8 → 6 → double 12, pay 8 → 4 → double 8 (exactly what’s owed). ✓
(ii) After one round, coins become $2y – c$ where $y$ is the current amount and $c$ is the fixed charge. For his coins to increase:
As long as the charge per round is less than the number of coins he currently holds, his total keeps growing.
(iii) If the genie wants to take all of Karim’s coins after exactly $n$ rounds, the amount after $n$ rounds is $2^n x – c(2^n-1)$. Setting this to 0:
The genie should fix the charge per round at $c = \dfrac{2^n x}{2^n-1}$ for a chosen number of rounds $n$ — this guarantees Karim’s coins hit exactly zero after round $n$. (In the story above, $n=3,\, x=7$ gives $c = \dfrac{8\times7}{7}=8$, matching exactly what happened!)