12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 144, Q1
Fill the digits 1, 3 and 7 into a two-digit number times a one-digit number to make the largest product possible.
Hint. Apply the rule proved in the previous section, then check it against the alternatives.
Apply the rule. With p < q < r, the largest product is qp × r — the biggest digit as the multiplier, the other two in decreasing order.
Here p = 1, q = 3, r = 7, so the answer should be 31 × 7.
Compute. 31 × 7 = 217
Check every arrangement.
| Product | Value |
|---|
| 31 × 7 | 217 |
| 71 × 3 | 213 |
| 13 × 7 | 91 |
| 73 × 1 | 73 |
| 17 × 3 | 51 |
| 37 × 1 | 37 |
The rule holds ✓
Worth noticing. The winner beats the runner-up by only 4 (217 against 213). The two share the term 10 × 3 × 7 = 210 and differ only in 1 × 7 = 7 against 1 × 3 = 3 — which is exactly the difference of 4 that the algebra predicted.
✦ The largest product is 31 × 7 = 217.
22 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 144, Q2
Fill the digits 3, 5 and 9 into a two-digit number times a one-digit number to make the largest product possible.
Hint. Same rule — largest digit as the multiplier, other two in decreasing order.
Apply the rule with p = 3, q = 5, r = 9: the answer should be qp × r = 53 × 9.
Compute. 53 × 9 = 477
Check every arrangement.
| Product | Value |
|---|
| 53 × 9 | 477 |
| 93 × 5 | 465 |
| 35 × 9 | 315 |
| 95 × 3 | 285 |
| 39 × 5 | 195 |
| 59 × 3 | 177 |
The rule holds again ✓
The margin, checked against the algebra. The top two both contain 10 × 5 × 9 = 450. Their remaining terms are 3 × 9 = 27 and 3 × 5 = 15, a difference of 12 — and indeed 477 − 465 = 12 ✓ The prediction is exact, not approximate.
A common wrong instinct. Many students reach for 95 × 3, putting the two largest digits into the two-digit number. That gives only 285, well behind. The reason is that the multiplier acts on both digits of the multiplicand, so making the multiplier large matters more than making the two-digit number large.
✦ The largest product is 53 × 9 = 477.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp206.pdf), Chapter 6 'Algebra Play', pages 135-147. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the three p.138 pyramids (bottom rows 6,2 / 3,4,3 / 5,4,5,0 giving tops 8, 14 and 32); the p.138 top-down pyramid 10 / 4,6 / 1,3,3; the p.139 letter-number pyramid 60 / 32,28 / 12,20,8; the three p.139 four-row pyramids, whose bottom rows solve to 4,9,6,1 (top 50), 5,14,7,2 (top 70) and 3,5,5,2 (top 35); the p.140 Figure it Out bottom rows 4,13,8 / 7,11,3 / 10,14,25 (tops 38, 32, 63) and 8,19,21,13 / 7,18,19,6 / 9,7,5,11 (tops 141, 124, 56); and the two p.142 algebra grids, which solve to square 11 and circle 5 with a third-row total of 21, and circle 4 and diamond 7 with a third-row total of 15 (the fourth row of that grid is printed blank for the student to design). The Virahanka-Fibonacci result was proved and then checked by computation for n = 2 to 6: every entry of the pyramid is a Virahanka-Fibonacci number, row k from the bottom is the consecutive run V(2k-1) to V(n+k-1), and the top is V(2n-1) - so a 29-row pyramid tops out at V57 = 591286729879. All six arrangements were enumerated for each Largest Product question: 32 x 5 = 160, 31 x 7 = 217 and 53 x 9 = 477, each beating its runner-up by exactly the margin the algebra predicts (4, 4 and 12). The date trick decodes as 100M + 165 + D, giving 4 November, 29 February and 31 January for 1269, 394 and 296. The puzzle answers were each verified by substitution: 7 flowers with 8 per shrine, 20 horses and 35 hens, a daughter of 6 and mother of 30, Gauri 6 cows and Naina 12, a dosa price of Rs 80 or a target of 175 dosas, the fraction sequence equal to 1/3 because n^2/(4n^2 - n^2) = 1/3, and Karim starting with 7 coins with the genie's general ruinous charge being c = 2^k x / (2^k - 1).. Questions are referenced from the NCERT textbook for identification.
