14 marksGanita Prakash Cl-8 Part 2, Section 6.4, pages 141-142 — 'Calendar Magic'
A friend picks a 2 × 2 block from a calendar page, adds its four numbers and tells you only the sum. Show how to recover all four numbers, and find them when the sum is 40 and when the sum is 36. Then create your own calendar trick using a differently shaped block.
Hint. Name the top-left number and write the other three in terms of it, using the fact that a calendar row is 7 days.
Set up with one letter. In a calendar, moving one square right adds 1 day and moving one square down adds 7 days. So a 2 × 2 block with top-left number a is
The sum. a + (a + 1) + (a + 7) + (a + 8) = 4a + 16
So the sum always leaves a remainder of 0 when 16 is subtracted and the result divided by 4 — and knowing the sum is enough to recover a, and hence all four numbers.
Sum = 40.
4a + 16 = 40 → 4a = 24 → a = 6
The block is 6, 7, 13, 14 ✓ (the example shown in the chapter).
Sum = 36.
4a + 16 = 36 → 4a = 20 → a = 5
The block is 5, 6, 12, 13.
A quicker mental version. The sum is 4a + 16 = 4(a + 4), so divide the sum by 4 and subtract 4 to get the top-left number. For 40: 10 − 4 = 6 ✓. For 36: 9 − 4 = 5 ✓.
Your own trick — some blocks worth using.
| Block | Numbers | Sum | How to invert |
|---|
| Vertical 1 × 3 strip | a, a + 7, a + 14 | 3a + 21 | sum ÷ 3, subtract 7 |
| Horizontal 1 × 3 strip | a, a + 1, a + 2 | 3a + 3 | sum ÷ 3, subtract 1 |
| 3 × 3 block | nine numbers around centre m | 9m | sum ÷ 9 gives the centre directly |
| Plus/cross shape | m and its four neighbours | 5m | sum ÷ 5 gives the centre |
The 3 × 3 block and the cross are the most striking, because the sum is exactly 9 times (or 5 times) the middle number — the neighbours cancel in pairs, since each is as far above the centre as its partner is below.
A caution for real calendars. The block must fit entirely inside the month, or the +1 and +7 relations break at the edges. A 2 × 2 block starting on a Saturday, for instance, does not exist.
✦ A 2 × 2 calendar block sums to 4a + 16 where a is the top-left number, so a = (sum ÷ 4) − 4. Sum 40 gives 6, 7, 13, 14 and sum 36 gives 5, 6, 12, 13. Other blocks work the same way — a 3 × 3 block sums to 9 × the centre, and a plus-shape to 5 × the centre.
23 marksGanita Prakash Cl-8 Part 2, Section 6.4, page 142 — 'Algebra Grids'
In an algebra grid, shapes stand for numbers and the last column of each row is the sum of the values to its left. A grid has row 1 = square, square, square = 27 and row 2 = circle, circle, square = 19. Find the value of each shape.
Hint. One row contains only one kind of shape — start there.
Start with the row that has only one shape.
Row 1: ■ + ■ + ■ = 27
That is 3 × ■ = 27, so ■ = 9.
Use that value in the second row.
Row 2: ● + ● + ■ = 19
Substituting ■ = 9: ● + ● + 9 = 19
2 × ● = 19 − 9 = 10
● = 5
Check. Row 1: 9 + 9 + 9 = 27 ✓. Row 2: 5 + 5 + 9 = 19 ✓.
The general strategy. An algebra grid is a set of simultaneous equations dressed up as pictures. Look for the row with the fewest different shapes — ideally one — since it solves immediately. Substitute the value into the next row, and continue. If no row has a single shape, you must eliminate as with ordinary simultaneous equations, which is what the next question requires.
✦ ■ = 9 and ● = 5.
35 marksGanita Prakash Cl-8 Part 2, Section 6.4, page 142
Find the values of the shapes and fill in the empty squares. Grid A has rows: square, square, circle = 27; circle, circle, square = 21; circle, square, circle = ?. Grid B has rows: circle, diamond, diamond = 18; diamond, circle, circle = 15; diamond, circle, circle = ?; and an empty fourth row.
Hint. Neither grid has a single-shape row, so eliminate one shape between the two given equations.
Grid A. Let ■ = s and ● = c.
- Row 1: s + s + c = 2s + c = 27
- Row 2: c + c + s = s + 2c = 21
Adding the two equations: 3s + 3c = 48, so s + c = 16.
Substituting into row 1: 2s + c = s + (s + c) = s + 16 = 27, so s = 11, and then c = 5.
Check: 2(11) + 5 = 27 ✓ and 11 + 2(5) = 21 ✓
Row 3: ● + ■ + ● = 5 + 11 + 5 = 21
| Row | Shapes | Sum |
|---|
| 1 | ■ ■ ● | 27 |
| 2 | ● ● ■ | 21 |
| 3 | ● ■ ● | 21 |
Rows 2 and 3 both hold two circles and one square, so it is no accident that they have the same total — only the multiset of shapes matters, not the order.
Grid B. Let ● (circle) = b and ◆ (diamond) = d.
- Row 1: b + d + d = b + 2d = 18
- Row 2: d + b + b = 2b + d = 15
From row 1: b = 18 − 2d.
Substituting into row 2: 2(18 − 2d) + d = 15 → 36 − 4d + d = 15 → −3d = −21 → d = 7, and then b = 18 − 14 = 4.
Check: 4 + 7 + 7 = 18 ✓ and 7 + 4 + 4 = 15 ✓
Row 3: ◆ + ● + ● = 7 + 4 + 4 = 15 — the same shapes as row 2, hence the same sum.
Row 4 is left completely blank in the book: it is an invitation to invent your own row. For example ◆ ◆ ◆ would total 21, and ● ● ● would total 12.
| Row | Shapes | Sum |
|---|
| 1 | ● ◆ ◆ | 18 |
| 2 | ◆ ● ● | 15 |
| 3 | ◆ ● ● | 15 |
| 4 | your own | your own |
Why the answers are unique. Each grid gives two independent equations in two unknowns, so there is exactly one solution. Had the two rows contained the same shapes in a different order, they would have given the same equation twice and the puzzle would have had no unique answer — worth checking before you start.
✦ Grid A: square = 11, circle = 5, third row total = 21. Grid B: circle = 4, diamond = 7, third row total = 15; the fourth row is blank for you to design.
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.
