CLASS 8 · GANITA PRAKASH PART II · CHAPTER 6
Use algebra to explain number tricks, solve pyramids and calendar puzzles, maximise products, and investigate divisibility.
44 worked answers · short steps first, explanations when you need them
Textbook pages 136–137
Question think-number-two
Start with x.
Double: 2x; add 4: 2x+4.
Divide by 2: x+2.
Subtract x: x+2−x.
Always 2. The x cancels, so the result does not depend on the starting number.
Question change-trick-final-three-five
Double x, add 6, divide by 2, subtract x: result 3.
For 5, double x, add 10, divide by 2, subtract x: result 5.
Add twice the desired final answer before dividing by 2. In general, double x, add 2k, halve, then subtract x to get k.
Question date-trick-formula
5M+6.
4(5M+6)+9=20M+33.
5(20M+33)+D=100M+165+D.
Final number = 100M+165+D. Subtract 165; the hundreds-and-higher part gives the month and the final two digits give the day.
Question date-1269
Subtract 165: 1269−165=1104.
Read the month from the remaining hundreds and the day from its last two digits.
November 4.
Question date-394
Subtract 165: 394−165=229.
Read the month from the remaining hundreds and the day from its last two digits.
February 29.
Question date-296
Subtract 165: 296−165=131.
Read the month from the remaining hundreds and the day from its last two digits.
January 31.
Question date-1390
Subtract 165: 1390−165=1225.
Read the month from the remaining hundreds and the day from its last two digits.
December 25.
Question date-291
Subtract 165: 291−165=126.
Read the month from the remaining hundreds and the day from its last two digits.
January 26.
Question invent-date-trick
Choose operations that encode the month as a multiple of 100 and then add the day.
For example, aim to transform M into 100M, then add D, while tracking any fixed constant C.
Tell the player to report 100M+D+C; the decoder subtracts C.
One valid design is multiply M by 5, add 25, multiply by 20, then add D: the result is 100M+500+D, so subtract 500 to decode.
Textbook pages 137–140
Question pyramid-rule
The second row is a+b and b+c.
The top is (a+b)+(b+c).
Second row: a+b, b+c; top: a+2b+c.
Question pyramid-solve-10
The left middle box is 1 plus the bottom-middle number, so 1+x=4 and x=3.
The right middle box is x plus the bottom-right number: 3+y=6, so y=3.
Check the top: 4+6=10.
Bottom row 1, 3, 3; middle row 4, 6; top 10.
Question pyramid-solve-60
a=12+c and b=c+8.
a+b=60, so (12+c)+(c+8)=60.
20+2c=60, hence c=20.
Then a=32 and b=28.
Bottom row 12,20,8; middle row 32,28; top 60.
Question pyramid-top-three-values
For bottom row a,b,c, top=a+2b+c.
(i) 4+2×13+8=38.
(ii) 7+2×11+3=32.
(iii) 10+2×14+25=63.
The top values are 38, 32 and 63.
Question pyramid-four-row-expression
The next row is a+b, b+c, c+d.
The row above is a+2b+c and b+2c+d.
Add these two expressions for the top.
Top = a+3b+3c+d.
Question pyramid-four-row-numeric
For bottom row a,b,c,d, top=a+3b+3c+d.
(i) 8+3×19+3×21+13=8+57+63+13=141.
(ii) 7+3×18+3×19+6=7+54+57+6=124.
(iii) 9+3×7+3×5+11=9+21+15+11=56.
The three top numbers are 141, 124 and 56.
Question pyramid-fibonacci-three
Second row: 1+2=3 and 2+3=5.
Top: 3+5=8.
The pyramid is bottom 1,2,3; middle 3,5; top 8. Every value is in the Virahāṅka-Fibonacci sequence.
Question pyramid-fibonacci-four
Next row: 3,5,8.
Next row: 8,13.
Top: 8+13=21.
Upper rows are 3,5,8; then 8,13; top 21. All entries are Virahāṅka-Fibonacci numbers.
Question pyramid-fibonacci-29
Each upward step replaces adjacent terms by their sum, which is the next Fibonacci term.
For n bottom terms, the top is the term F_(2n−1) when F_1=1,F_2=2.
For n=29, the top is F_57.
F_57 = 591,286,729,879 (using 1,2,3,5,…).
Question pyramid-fibonacci-general
Each new entry is the sum of two adjacent sequence terms.
By the recurrence, that sum is the next sequence term.
Following the successive rows leads to F_(2n−1) at the top.
Every entry is a Virahāṅka-Fibonacci number; the top is F_(2n−1), with F1=1 and F2=2.
Question pyramid-figure-dependent
(i) Bottom 6,2: top=6+2=8.
(ii) Bottom 3,4,3: middle 7,7; top=14.
(iii) Bottom 5,4,5,0: next row 9,9,5; next row 18,14; top=32.
For the top-down pyramid, the given rows are 10 / 4,6 / 1,3,3. Check: 1+3=4, 3+3=6, and 4+6=10.
For each missing-value four-row pyramid, use the shared box sums to find the bottom row, then add upwards.
| Diagram | Bottom row | Rows above (from bottom upwards) | Top |
|---|---|---|---|
| First set (i) | 6, 2 | 8 | 8 |
| First set (ii) | 3, 4, 3 | 7, 7 | 14 |
| First set (iii) | 5, 4, 5, 0 | 9, 9, 5; then 18, 14 | 32 |
| Missing-value (i) | 4, 9, 6, 1 | 13, 15, 7; then 28, 22 | 50 |
| Missing-value (ii) | 5, 14, 7, 2 | 19, 21, 9; then 40, 30 | 70 |
| Missing-value (iii) | 3, 5, 5, 2 | 8, 10, 7; then 18, 17 | 35 |
The first three pyramids have top numbers 8, 14 and 32. The top-down pyramid is consistent with bottom row 1,3,3. The three four-row bottom rows are
Textbook pages 141–142
Question calendar-2x2-rule
Calendar numbers increase by 1 across and 7 down.
The block is a, a+1 / a+7, a+8.
Sum=4a+16.
The four-number sum is 4a+16.
Question calendar-sum-36
4a+16=36.
4a=20, so a=5.
The other values are 6,12,13.
5, 6, 12 and 13.
Question calendar-sum-40
4a+16=40.
4a=24, so a=6.
The other values are 7,13,14.
6, 7, 13 and 14.
Question design-calendar-trick
Use a 2×3 calendar rectangle. If its top-left date is a, its rows are a,a+1,a+2 and a+7,a+8,a+9.
Add the six dates: 6a+27. If the sum is 81, solve 6a+27=81, so a=9.
The dates are 9,10,11,16,17,18; their sum is 81.
Example: a 2×3 block with sum 81 starts on the 9th and contains 9, 10, 11, 16, 17 and 18. In general, subtract 27 from the sum and divide by 6 to find the first date.
Question algebra-grid-solve
Let square=s and circle=c. The rows give 2s+c=27 and s+2c=21.
Double the second equation: 2s+4c=42. Subtract the first equation: 3c=15, so c=5.
Substitute c=5 into 2s+c=27: 2s=22, so s=11.
The third row is circle+square+circle=5+11+5=21.
| Grid row | Equation | Result |
|---|---|---|
| 1 | square+square+circle=27 | 2s+c=27 |
| 2 | circle+circle+square=21 | s+2c=21 |
| 3 | circle+square+circle | 5+11+5=21 |
Blue square=11, red circle=5, and the missing row total is 21.
Question algebra-grid-practice
Let blue circle=m and purple diamond=d. The rows give m+2d=18 and 2m+d=15.
Double the first equation: 2m+4d=36. Subtract the second equation: 3d=21, so d=7.
Substitute d=7 into m+2d=18: m+14=18, so m=4.
The third row is diamond+circle+circle=7+4+4=15. The three row totals add to 18+15+15=48.
| Grid row | Equation | Result |
|---|---|---|
| 1 | circle+diamond+diamond=18 | m+2d=18 |
| 2 | diamond+circle+circle=15 | 2m+d=15 |
| 3 | diamond+circle+circle | 7+4+4=15 |
| All three row totals | 18+15+15 | 48 |
Blue circle=4, purple diamond=7, third row total=15, and the sum of the three row totals is 48. The blank fourth row is an open design task.
Textbook pages 142–144
Question largest-product-235
List the six possibilities: 23×5, 25×3, 32×5, 35×2, 52×3, 53×2.
Their products are 115,75,160,70,156,106.
32×5=160 is largest.
Question largest-product-137
For each multiplier, put the larger remaining digit in the tens place.
Compare 31×7=217, 71×3=213 and 73×1=73.
31×7=217.
Question largest-product-359
Best candidate with multiplier 9: 53×9=477.
With multiplier 5: 93×5=465.
With multiplier 3: 95×3=285.
53×9=477 is the largest.
Question largest-product-general-proof
For each fixed multiplier, put the larger remaining digit in the tens place.
Compare rp×q with rq×p: q>p makes rp×q larger.
Compare qp×r with rp×q: both have tens contribution 10qr; qp×r has larger units contribution pr>pq.
The maximum is the two-digit number formed by the two smaller digits in descending order, multiplied by the largest digit: qp×r.
Textbook pages 144–146
Question reverse-difference-divisible-9
Original=10a+b; reversed=10b+a.
Difference in magnitude=|10b+a−(10a+b)|=9|b−a|.
Divide by 9.
The quotient is |b−a|, the difference between the digits; the remainder is 0.
Question reverse-sum-divisible-11
10a+b + 10b+a = 11a+11b.
Factor: 11(a+b).
Yes. The sum is 11(a+b), so it is divisible by 11.
Question cyclic-three-digits-divisible-37
Sum=(100a+10b+c)+(100b+10c+a)+(100c+10a+b).
Collect equal digits: 111a+111b+111c=111(a+b+c).
111=3×37.
The sum is 111(a+b+c), so it is always divisible by both 37 and 3.
Question abcabc-divide-7-11-13
abcabc=1000×abc+abc=1001×abc.
1001=7×11×13.
Dividing successively by 7, 11 and 13 leaves abc.
The result is the original three-digit number abc.
Question magic-pond-flowers
Let x be the starting count and n the equal number left at each shrine.
After first shrine: 2x−n remain. After second: 4x−3n remain.
After third doubling: 8x−6n; this equals n, so 8x=7n.
Smallest positive integer solution is x=7,n=8.
The smallest whole-number case starts with 7 flowers and leaves 8 at each shrine. More generally x=7k and n=8k also works.
Question horses-and-hens
If all 55 were hens, there would be 110 legs.
Each horse has 2 more legs than a hen; extra legs=150−110=40.
Horses=40÷2=20; hens=55−20=35.
20 horses and 35 hens.
Question mother-daughter-ages
Let daughter age be d; mother age=5d.
In 6 years: 5d+6=3(d+6).
5d+6=3d+18, so 2d=12.
The daughter is 6 years old.
Question cowherd-cows
Let Gauri have g cows and Naina 2g.
After transfer, Gauri has g+3 and Naina 2g−3.
Set equal: g+3=2g−3, so g=6.
Gauri has 6 cows and Naina has 12.
Question dosa-profit-price
Total cost=5,000+100×10=₹6,000.
Required revenue=6,000+2,000=₹8,000.
Price per dosa=8,000÷100.
₹80 per dosa.
Question dosa-profit-quantity
Profit per dosa before rent=50−10=₹40.
Required contribution=5,000+2,000=₹7,000.
Quantity=7,000÷40.
175 dosas.
Question odd-sum-fractions
First: 1/3.
Second: 4/12=1/3.
Third: 9/27=1/3.
In general, first n odd numbers sum to n²; the next n odd numbers sum to 3n².
Every fraction equals 1/3. The numerator is n² and the denominator is 3n².
Question genie-coins-initial
Let the initial number be x.
After round 1: 2x−8. After round 2: 4x−24.
After round 3: 8x−56=8.
8x=64.
Karim started with 8 coins.
Question genie-cost-more
After three rounds the balance is 8×8−(4c+2c+c)=64−7c.
Require 64−7c>8.
Thus 7c<56.
The fee must be less than 8 coins per round. At a fee of 8, he ends with 8 coins; below 8, he gains.
Question genie-cost-all
Final balance=8x−7c.
Set 8x−7c=0.
Solve c=8x/7.
Charge 8x/7 coins each round. For x=8, that is 64/7 coins; if fees must be whole coins, an exactly zero final balance is impossible with one fixed integer fee.