In-text — Number Pyramids — Algebra Play

1. 13 marksGanita Prakash Cl-8 Part 2, Section 6.3, pages 137-138

   In a number pyramid each number is the sum of the two numbers directly below it. Fill these pyramids: (a) bottom row 6, 2; (b) bottom row 3, 4, 3; (c) bottom row 5, 4, 5, 0.

   Hint. Work upwards, one row at a time, adding neighbouring pairs.

   The rule. Each box holds the sum of the two boxes immediately below it, so a bottom row of n numbers gives a row of n − 1 above it, and so on to a single number at the top.

   (a) Bottom row 6, 2. Top = 6 + 2 = 8

         8
       6   2
   

   (b) Bottom row 3, 4, 3. Second row: 3 + 4 = 7 and 4 + 3 = 7 Top: 7 + 7 = 14

         14
       7    7
     3   4   3
   

   (c) Bottom row 5, 4, 5, 0. Third row from bottom: 5 + 4 = 9, 4 + 5 = 9, 5 + 0 = 5 Second row: 9 + 9 = 18, 9 + 5 = 14 Top: 18 + 14 = 32

           32
        18    14
      9    9    5
    5   4    5    0
   

   A check worth making. In (b) the bottom row is symmetric (3, 4, 3), and so is every row above it — symmetry in the bottom row is always inherited upwards, because the pyramid is built by the same rule from both ends.

   ✦ (a) top = 8; (b) rows 7, 7 then top 14; (c) rows 9, 9, 5 then 18, 14 then top 32.

2. 23 marksGanita Prakash Cl-8 Part 2, Section 6.3, page 138

   A pyramid has 10 at the top, 4 in the left box of the middle row, and 1 in the left box of the bottom row; the other three boxes are empty. Fill it in, and explain the method.

   Hint. You cannot add upwards here — but the rule can be run backwards as a subtraction.

   Why the usual method stalls. The rule says a box is the sum of the two below it. Here the bottom row is almost empty, so there is nothing to add. But the same rule read backwards is a subtraction: if a box is the sum of two boxes below, then either lower box equals the box above minus its neighbour.

   Step 1 — the missing box in the middle row. The top box is 10 and one middle box is 4, so the other middle box is

   10 − 4 = 6

   Step 2 — the middle box of the bottom row. The left middle box is 4 and the left bottom box is 1, so

   4 − 1 = 3

   Step 3 — the right bottom box. The right middle box is 6 and the middle bottom box is now 3, so

   6 − 3 = 3

   The completed pyramid.

         10
       4     6
     1    3    3
   

   Check by rebuilding upwards. 1 + 3 = 4 ✓, 3 + 3 = 6 ✓, 4 + 6 = 10 ✓.

   The general lesson. A pyramid can be filled in either direction. Go upwards by adding when the row below is complete, and downwards by subtracting when a box and one of its two children are known. Most mixed pyramids are solved by alternating the two.

   ✦ Middle row 4, 6 and bottom row 1, 3, 3, obtained by running the rule backwards: 10 − 4 = 6, then 4 − 1 = 3, then 6 − 3 = 3.

3. 34 marksGanita Prakash Cl-8 Part 2, Section 6.3, pages 138-139

   A three-row pyramid has 60 at the top, both middle boxes empty, and bottom row 12, ?, 8. Where do you even start? Fill it in using letter-numbers.

   Hint. Neither subtraction nor addition can start here — so name the unknowns and let the rule give you equations.

   Why neither direction works on its own. Going upwards needs the whole bottom row, and the middle bottom box is missing. Going downwards needs the top box and one of the boxes below it, and both middle boxes are missing. Nothing can be filled by a single step — which is exactly the moment algebra earns its place.

   Step 1 — name the unknowns.

           60
         a     b
      12    c     8
   

   Step 2 — write what the rule says.

   • a + b = 60
   • 12 + c = a
   • c + 8 = b

   Step 3 — substitute. Replace a and b in the first equation:

   (12 + c) + (c + 8) = 60

   Step 4 — solve. 20 + 2c = 60 2c = 60 − 20 = 40 c = 20

   Step 5 — back-substitute. a = 12 + c = 12 + 20 = 32 b = c + 8 = 20 + 8 = 28

   The completed pyramid.

           60
        32    28
     12    20    8
   

   Check. 12 + 20 = 32 ✓, 20 + 8 = 28 ✓, 32 + 28 = 60 ✓.

   Why one letter was enough. Notice that c appears in both of the middle boxes, so adding the two middle boxes counts c twice — which is why the equation came out as 20 + 2c rather than 20 + c. That doubling is the same fact that will reappear in the next section as a + 2b + c for the top of a three-row pyramid.

   ✦ Bottom middle c = 20, middle row 32 and 28. The route is to name the unknowns, write the three sum equations the rule gives, and substitute to get 20 + 2c = 60.

4. 45 marksGanita Prakash Cl-8 Part 2, Section 6.3, page 139

   Fill these four-row pyramids: (a) top 50, right box of the second row 22, bottom row 4, ?, 6, ?; (b) second row 40 and ?, right box of the third row 9, bottom row 5, ?, 7, ?; (c) top 35, right box of the third row 7, bottom row 3, 5, ?, ?.

   Hint. Put letters in the empty bottom boxes, build upwards symbolically, then match the given numbers.

   Method. In every case put letters in the unknown bottom boxes, build the pyramid upwards in terms of those letters, and set the resulting expressions equal to the numbers printed higher up.

   ---

   (a) Top 50, second-row right box 22, bottom 4, p, 6, q.

   Third row: 4 + p, p + 6, 6 + q Second row: (4 + p) + (p + 6) = 2p + 10, and (p + 6) + (6 + q) = p + q + 12 Top: (2p + 10) + (p + q + 12) = 3p + q + 22

   Two equations:

   • p + q + 12 = 22 → p + q = 10
   • 3p + q + 22 = 50 → 3p + q = 28

   Subtracting: 2p = 18, so p = 9 and q = 1.

              50
           28    22
        13   15    7
      4    9    6    1
   

   Check: 4 + 9 = 13, 9 + 6 = 15, 6 + 1 = 7; 13 + 15 = 28, 15 + 7 = 22; 28 + 22 = 50 ✓

   ---

   (b) Second row 40 and ?, third-row right box 9, bottom 5, p, 7, q.

   Third row: 5 + p, p + 7, 7 + q. The right box gives 7 + q = 9, so q = 2. Second row left: (5 + p) + (p + 7) = 2p + 12 = 40, so 2p = 28 and p = 14. Second row right: (p + 7) + (7 + q) = 21 + 9 = 30. Top: 40 + 30 = 70.

              70
           40    30
        19   21    9
      5   14    7    2
   

   Check: 5 + 14 = 19, 14 + 7 = 21, 7 + 2 = 9; 19 + 21 = 40, 21 + 9 = 30; 40 + 30 = 70 ✓

   ---

   (c) Top 35, third-row right box 7, bottom 3, 5, p, q.

   Third row: 3 + 5 = 8, 5 + p, p + q. The right box gives p + q = 7. Second row: 8 + (5 + p) = 13 + p, and (5 + p) + (p + q) = 5 + p + 7 = p + 12. Top: (13 + p) + (p + 12) = 2p + 25 = 35, so 2p = 10 and p = 5, hence q = 2.

              35
           18    17
         8   10    7
      3    5    5    2
   

   Check: 3 + 5 = 8, 5 + 5 = 10, 5 + 2 = 7; 8 + 10 = 18, 10 + 7 = 17; 18 + 17 = 35 ✓

   ---

   What to notice. In each case two pieces of information were enough to pin down two unknowns, because the pyramid rule turns every printed number into a linear equation. That is the whole point of the section: a puzzle that looks like guesswork is really a pair of simultaneous equations.

   ✦ (a) bottom 4, 9, 6, 1 with rows 13, 15, 7 and 28, 22; (b) bottom 5, 14, 7, 2 with rows 19, 21, 9 and 40, 30, top 70; (c) bottom 3, 5, 5, 2 with rows 8, 10, 7 and 18, 17.

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.