Class 8 Mathematics · Part 2 · Chapter 6

6. Algebra Play

Solutions to all Figure it Out and in-text questions. CBSE Class 8 Mathematics, NCERT Ganita Prakash Part 2.

6.2 Thinking about ‘Think of a Number’ Tricks

1How would you change the trick (double, add four, halve, subtract the number) to make the final answer 3? What about 5?

Add 6 instead of 4: (2x + 6) ÷ 2 − x = 3.

Add 10 instead of 4: (2x + 10) ÷ 2 − x = 5.

In general, add 2k to get the final answer k.

2Can you come up with more complicated steps that always give the same value?

Example: Think of a number (x). Multiply by 3 (3x). Add 12 (3x + 12). Divide by 3 (x + 4). Subtract the original number. The answer is always 4.

3Date trick: Find the dates if the final answers are (i) 1269 (ii) 394 (iii) 296.

Subtract 165 from the answer. The last two digits give the day and the rest gives the month.

(i) 1269 − 165 = 1104 → 4 November

(ii) 394 − 165 = 229 → 29 February

(iii) 296 − 165 = 131 → 31 January

4Can you change the steps in this trick and still find the original date?

Yes. For example: multiply the month by 5, add 3, multiply by 4, add 2, multiply by 5, add the day.

This gives 100M + 70 + D. So subtract 70 from the answer to get the date.

6.3 Number Pyramids

5Fill these pyramids: bottom rows (6, 2), (3, 4, 3), (5, 4, 5, 0).

6Fill the following pyramids.

Use letter-numbers for the missing bottom numbers and form equations.

Pyramid 1: Bottom 4, b, 6, d. Second row from top: 10 + 2b and b + 12 + d = 22. Top: 10 + 2b + 22 = 50, so b = 9 and d = 1.

Pyramid 2: Bottom 5, b, 7, d. 7 + d = 9, so d = 2. 5 + 2b + 7 = 40, so b = 14.

Pyramid 3: Bottom 3, 5, c, d. c + d = 7 and 3 + 3 × 5 + 3c + d = 35, so 3c + d = 17. Hence c = 5, d = 2.

Figure it Out (Page 140)

1Find the number at the top of a 3-row pyramid without building it.

Top = a + 2b + c.

(i) 4 + 2 × 13 + 8 = 38

(ii) 7 + 2 × 11 + 3 = 32

(iii) 10 + 2 × 14 + 25 = 63

2Write an expression for the top of a pyramid with 4 rows in terms of the bottom row.

Bottom a, b, c, d → next row a + b, b + c, c + d → next row a + 2b + c, b + 2c + d.

Top = a + 3b + 3c + d

3Find the number at the top (4 rows) without building the pyramid.

(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

4The first three Virahāṅka-Fibonacci numbers (1, 2, 3) are in the bottom row of a 3-row pyramid. Fill it.

The numbers are 1, 2, 3, 5, 8. The top is 8. Yes, all are Virahāṅka-Fibonacci numbers.

5(i) The first four numbers in a 4-row pyramid. (ii) The first 29 numbers in a 29-row pyramid.

(i) Every number is a Virahāṅka-Fibonacci number. The top is 21, the 7th number of the sequence.

(ii) Again every number in the pyramid is a Virahāṅka-Fibonacci number. Each row has consecutive numbers of the sequence. The top is the 57th number of the sequence.

6If the bottom row of an n-row pyramid has the first n Virahāṅka-Fibonacci numbers, what can we say?

Two neighbouring numbers of the sequence add up to the next number. So each row again has consecutive Virahāṅka-Fibonacci numbers.

The number at the top is the (2n − 1)th Virahāṅka-Fibonacci number.

6.4 Fun with Grids

7Create your own calendar trick with a grid of a different size and shape.

3 × 3 grid: If the centre number is c, the numbers are c − 8, c − 7, c − 6, c − 1, c, c + 1, c + 6, c + 7, c + 8. Their sum is 9c.

So if a friend tells the sum (say 135), divide by 9 to get the centre (15). Then all nine numbers are known.

8Algebra Grids: find the values of the shapes and fill the empty squares.

Grid 1 (square, circle)

2 squares + 1 circle = 27 and 2 circles + 1 square = 21.

Adding: 3 squares + 3 circles = 48, so square + circle = 16.

Square = 27 − 16 = 11, circle = 5.

Third row: circle + square + circle = 5 + 11 + 5 = 21

Grid 2 (circle, diamond)

Circle + 2 diamonds = 18 and diamond + 2 circles = 15.

Adding: 3 circles + 3 diamonds = 33, so circle + diamond = 11.

Diamond = 18 − 11 = 7, circle = 4.

Third row: 7 + 4 + 4 = 15. Column totals: 4 + 7 + 7 = 18, 7 + 4 + 4 = 15, 7 + 4 + 4 = 15. Grand total = 48.

Figure it Out (Page 144) — Largest Product

1Fill the digits 1, 3 and 7 in □□ × □ to make the largest product.

Use the largest digit as the multiplier and the other two in decreasing order.

31 × 7 = 217

2Fill the digits 3, 5 and 9 in □□ × □ to make the largest product.

53 × 9 = 477

6.6 Decoding Divisibility Tricks

9Can you work out what happens if a > b?

Then ab > ba. The difference = (10a + b) − (10b + a) = 9a − 9b = 9(a − b). It is again divisible by 9.

Figure it Out (Page 145)

1What is the quotient when you divide by 9? Is there a relationship with the two digits?

Quotient = the difference of the two digits. For 47: 27 ÷ 9 = 3 = 7 − 4.

2Instead of the difference, find the sum of a 2-digit number and its reverse. Is it always divisible by 11?

Yes. (10a + b) + (10b + a) = 11a + 11b = 11(a + b).

The quotient is the sum of the digits.

3Add abc, bca and cab. Show that the sum is always divisible by 37. Is it divisible by 3?

Sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c = 111(a + b + c).

111 = 3 × 37. So the sum is always divisible by 37 and also by 3.

4Make abcabc from abc. Divide by 7, then 11, then 13. What do you get? Why?

We get back abc. Example: 254254 ÷ 7 ÷ 11 ÷ 13 = 254.

abcabc = abc × 1000 + abc = abc × 1001, and 1001 = 7 × 11 × 13.

5Magical ponds double the flowers. He places an equal number in each of 3 shrines and has none left. How many did he start with?

Let him start with x flowers and place s flowers in each shrine.

After pond 1: 2x, places s → 2x − s. After pond 2: 4x − 2s, places s → 4x − 3s. After pond 3: 8x − 6s, places all → 8x − 6s = s.

So 8x = 7s. The smallest answer: he started with 7 flowers and placed 8 flowers in each shrine.

Check: 7 → 14, give 8 → 6 → 12, give 8 → 4 → 8, give 8 → 0 ✔

655 heads and 150 legs of horses and hens. How many of each?

If all 55 were hens: 110 legs. Extra legs = 150 − 110 = 40. Each horse adds 2 extra legs.

Horses = 40 ÷ 2 = 20, hens = 55 − 20 = 35

7A mother is 5 times her daughter’s age. In 6 years, she will be 3 times her daughter’s age. How old is the daughter?

Let the daughter be d. 5d + 6 = 3(d + 6), so 2d = 12 and d = 6 years (mother 30).

8Naina has twice as many cows as Gauri. If Naina gives 3 cows to Gauri, they have the same number. Find the cows.

Let Gauri have g. Naina has 2g. 2g − 3 = g + 3, so g = 6.

Gauri has 6 cows and Naina has 12 cows.

9Dosa cart: rent ₹5000 per day, cost ₹10 per dosa.

(i) For 100 dosas, cost = 5000 + 1000 = ₹6000. For ₹2000 profit, sales must be ₹8000. Price = 8000 ÷ 100 = ₹80

(ii) At ₹50, profit per dosa = ₹40. Needed: 5000 + 2000 = ₹7000. Dosas = 7000 ÷ 40 = 175

10Evaluate 13, 1 + 35 + 7, 1 + 3 + 57 + 9 + 11. What do you observe? Why?

13, 412 = 13, 927 = 13. They are all 13.

The first n odd numbers add up to n². The first 2n odd numbers add up to 4n². So the next n odd numbers add up to 4n² − n² = 3n².

The fraction is n²3n² = 13 always.

11Karim and the Genie.

(i) Let Karim start with x coins. After round 1: 2x − 8. Round 2: 4x − 24. Round 3: 8x − 56 = 0 (he gave his last 8). So x = 7 coins

(ii) In a round, coins go from x to 2x − c. They increase only if 2x − c > x, that is, c < x. With 7 coins, Karim should agree only if the cost is less than 7 coins (at most 6).

(iii) For 3 rounds, the money becomes 8x − 7c. To take everything, the genie sets 8x − 7c = 0, that is, c = 8x7. (For 7 coins, c = 8.) For just one round, c = 2x.

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