NCERT Solutions for Class 8 Maths Ganita Prakash Part 2 Chapter 6: Algebra Play

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• Figure it Out: Page 140
• Figure it Out: Page 144
• Figure it Out: Page 145
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Class 8 Ganita Prakash Part 2 Chapter 6 asks which arrangement of three digits gives the largest product and why reversing a two-digit number always produces a difference divisible by 9. The final puzzles include horses and hens, a dosa cart’s profit and Karim and the Genie.
Get here the Solutions and Answers of Figure it Out at Page 140.

Answer:
For a pyramid with three rows, let the bottom row be a, b and c.
Number at the top = a + 2b + c
(i) For 4, 13 and 8:
Number at the top = 4 + 2 × 13 + 8
= 4 + 26 + 8
= 38

(ii) For 7, 11 and 3:
Number at the top = 7 + 2 × 11 + 3
= 7 + 22 + 3
= 32

(iii) For 10, 14 and 25:
Number at the top = 10 + 2 × 14 + 25
= 10 + 28 + 25
= 63

Answer:
Let the numbers in the bottom row be a, b, c and d.
The row above it contains a + b, b + c and c + d.
The next row contains:
(a + b) + (b + c) = a + 2b + c
(b + c) + (c + d) = b + 2c + d
Number at the top = (a + 2b + c) + (b + 2c + d)
= a + 3b + 3c + d
Thus, the required expression is a + 3b + 3c + d.

Answer:
For a pyramid with four rows, let the bottom row be a, b, c and d.
Number at the top = a + 3b + 3c + d
(i) For 8, 19, 21 and 13:
Number at the top = 8 + 3 × 19 + 3 × 21 + 13
= 8 + 57 + 63 + 13
= 141

(ii) For 7, 18, 19 and 6:
Number at the top = 7 + 3 × 18 + 3 × 19 + 6
= 7 + 54 + 57 + 6
= 124

(iii) For 9, 7, 5 and 11:
Number at the top = 9 + 3 × 7 + 3 × 5 + 11
= 9 + 21 + 15 + 11
= 56

Recall the Virahāṅka-Fibonacci number sequence 1, 2, 3, 5, … where each number is the sum of the two numbers before it.

Answer:
The bottom row contains 1, 2 and 3.
Adding the two numbers directly below each box, we get:
1 + 2 = 3
2 + 3 = 5
Number at the top = 3 + 5 = 8

The numbers appearing in the grid are 1, 2, 3, 5 and 8. The number 3 appears twice.
The number at the top is 8. Yes, all the numbers are Virahāṅka-Fibonacci numbers.

(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid.
(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.
Answer:
(i) The bottom row contains 1, 2, 3 and 5.
Adding neighbouring numbers in each row, we get:
Row above the bottom row: 3, 5, 8
Next row: 8, 13
Top row: 21

All the numbers in the pyramid are Virahāṅka-Fibonacci numbers. The number at the top is 21, which is the 7th number in the sequence.

(ii) The sum of two consecutive Virahāṅka-Fibonacci numbers is the next number in the sequence. So, every row contains consecutive numbers from the same sequence.
The bottom row starts with the 1st number. The row above it starts with the 3rd number and the next row starts with the 5th number.
Each time we move up one row, the first number moves two places ahead in the sequence.
There are 28 rows above the bottom row.
Position of the top number = 1 + 2 × 28
= 57
Thus, all the numbers in the pyramid are Virahāṅka-Fibonacci numbers. The number at the top is the 57th number in the sequence, which is 591286729879.

Answer:
The sum of two consecutive Virahāṅka-Fibonacci numbers is the next number in the sequence. Therefore, all the numbers in the pyramid are Virahāṅka-Fibonacci numbers.

The rows, starting from the bottom, begin with the 1st, 3rd, 5th, … numbers of the sequence.
There are (n − 1) rows above the bottom row.
Position of the top number = 1 + 2(n − 1)
= 1 + 2n − 2
= 2n − 1
Thus, the number at the top is the number in position (2n − 1) of the Virahāṅka-Fibonacci sequence.

Get here the Solutions and Answers of Figure it Out at Page 144.

Answer:
Using each digit once, the possible products are:
13 × 7 = 91
17 × 3 = 51
31 × 7 = 217
37 × 1 = 37
71 × 3 = 213
73 × 1 = 73
The largest product is 217.
So, we fill the boxes as 31 × 7.

Answer:
Using each digit once, the possible products are:
35 × 9 = 315
39 × 5 = 195
53 × 9 = 477
59 × 3 = 177
93 × 5 = 465
95 × 3 = 285
The largest product is 477.
So, we fill the boxes as 53 × 9.

Get here the Solutions and Answers of Figure it Out at Page 145.

Answer:
The quotient is the difference between the two digits of the number.
For example, 74 − 47 = 27 and 27 ÷ 9 = 3.
Also, 7 − 4 = 3.
Let the larger digit be a and the smaller digit be b.
The larger number is 10a + b and the smaller number is 10b + a.
Their difference = (10a + b) − (10b + a)
= 9a − 9b
= 9(a − b).
On dividing by 9, we get a − b, which is the difference between the digits.

• We start with 31. After reversing we get 13. Adding 31 and 13, we get 44.
• We start with 28. After reversing we get 82. Adding 28 and 82, we get 110.
• We start with 12. After reversing we get 21. Adding 12 and 21, we get 33.
Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?
Answer:
Yes, the sum is always divisible by 11.
Let the tens digit be a and the ones digit be b.
The number is 10a + b. On reversing the digits, we get 10b + a.
Their sum = (10a + b) + (10b + a)
= 11a + 11b
= 11(a + b).
So, the sum is a multiple of 11. On dividing it by 11, we get a + b, the sum of the two digits.

[Hint: Look at some multiples of 37.]
Answer:
The three numbers are:
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
Their sum = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)
= 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3 × (a + b + c).
Therefore, the sum is always divisible by both 37 and 3.
For example, 123 + 231 + 312 = 666.
666 ÷ 37 = 18 and 666 ÷ 3 = 222.

[Hint: Multiply 7, 11 and 13.]
Answer:
We get the original 3-digit number.
Taking 123:
123123 ÷ 7 = 17589
17589 ÷ 11 = 1599
1599 ÷ 13 = 123

Taking 248:
248248 ÷ 7 = 35464
35464 ÷ 11 = 3224
3224 ÷ 13 = 248

Let the original 3-digit number be n.
Repeating its digits gives 1000n + n = 1001n.
Also, 7 × 11 × 13 = 1001.
Dividing successively by 7, 11 and 13 is the same as dividing by 1001.
So, 1001n ÷ 1001 = n.
Hence, the result is always the original number.

Answer:
Let the number of flowers at the start be x and the number placed in each shrine be y.
After dipping the flowers in the first pond and placing y flowers in shrine 1, the flowers left are 2x − y.

After dipping these in the second pond and placing y flowers in shrine 2, the flowers left are:
2(2x − y) − y = 4x − 3y.

After dipping these in the third pond, the number of flowers becomes:
2(4x − 3y) = 8x − 6y.

All these flowers are placed in shrine 3. Therefore,
8x − 6y = y
8x = 7y.
The smallest positive whole-number values are x = 7 and y = 8.

Thus, he could have started with 7 flowers and placed 8 flowers in each shrine.
These are the least possible numbers. He could also have started with 14 flowers and placed 16 in each shrine or started with 21 and placed 24 in each shrine. In general, starting with 7k flowers and placing 8k flowers in each shrine works for any positive whole number k.

Can you solve this without letter-numbers?
[Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]

Answer:
Each animal has one head, so there are 55 animals.
If all 55 animals were hens, the number of legs would be:
55 × 2 = 110.
Extra legs = 150 − 110 = 40.
A horse has 4 legs, which is 2 more than a hen.
So, each horse accounts for 2 of the extra legs.

Number of horses = 40 ÷ 2 = 20.
Number of hens = 55 − 20 = 35.
Therefore, there are 20 horses and 35 hens.

Answer:
Let the daughter’s present age be x years.
Then the mother’s present age is 5x years.
After 6 years, their ages will be x + 6 years and 5x + 6 years.

Therefore,
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x − 3x = 18 − 6
2x = 12
x = 6.
The daughter is 6 years old now.

Answer:
Let Gauri have x cows. Then Naina has 2x cows.
If Naina gives 3 cows to Gauri, Gauri will have x + 3 cows and Naina will have 2x − 3 cows.
Their numbers of cows will be equal. So,
x + 3 = 2x − 3
⇒ 3 + 3 = 2x − x
⇒ x = 6.
Therefore, Gauri has 6 cows and Naina has 2 × 6 = 12 cows.

• Rent for the dosa cart is ₹5000 per day.
• The cost of making one dosa (including all the ingredients and fuel) is ₹10.
(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?
(ii) If my customers are willing to pay only ₹50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹2000?
Answer:
(i) Cost of making 100 dosas = 100 × ₹10 = ₹1000.
Total daily expenses = ₹5000 + ₹1000 = ₹6000.
Required amount from sales = ₹6000 + ₹2000 = ₹8000.
Selling price of one dosa = ₹8000 ÷ 100 = ₹80.
So, I should sell each dosa for ₹80.

(ii) Amount left from each dosa after its making cost = ₹50 − ₹10 = ₹40.
This amount must cover the rent and the required profit.
Rent + required profit = ₹5000 + ₹2000 = ₹7000.
Number of dosas = ₹7000 ÷ ₹40 = 175.
So, I should aim to sell 175 dosas a day.

1/3, (1 + 3)/(5 + 7), (1 + 3 + 5)/(7 + 9 + 11)
What do you observe? Can you explain why this happens?
[Hint: Recall what you know about the sum of the first n odd numbers.]
Answer:
First fraction = 1/3.

Second fraction = (1 + 3)/(5 + 7)
= 4/12
= 1/3.

Third fraction = (1 + 3 + 5)/(7 + 9 + 11)
= 9/27
= 1/3.
All the fractions are equal to 1/3.
In each fraction, the numerator is the sum of the first n odd numbers, and the denominator is the sum of the next n odd numbers.

Sum of the first n odd numbers = n².
Sum of the first 2n odd numbers = (2n)² = 4n².
So, the sum of the next n odd numbers = 4n² − n² = 3n².
Therefore, each fraction = n²/(3n²) = 1/3.

Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, “I have come to serve you, Oh master”. He woke up and to his surprise, it was a genie!

“Do you want to make money?”, asked the genie. Karim nodded dumbly in bewilderment. The genie continued, “Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double”.

Karim immediately started towards the tree, only to be stopped by the genie. “One moment!”, said the genie. “Since I am bringing you great riches, you should share some of your gains with me. You must give me 8 coins each time you go around the tree.”

Thinking that was a trifling amount, Karim readily agreed.

He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie!

As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.

(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?

Answer:
(i) Let Karim initially have x coins.
Coins left after the first round = 2x − 8.
Coins left after the second round = 2(2x − 8) − 8 = 4x − 24.
Coins left after the third round = 2(4x − 24) − 8 = 8x − 56.
After paying the genie for the third round, Karim had no coins left.
8x − 56 = 0
⇒ 8x = 56
⇒ x = 56 ÷ 8 = 7
Therefore, Karim initially had 7 coins.

(ii) Let the cost per round be c coins.
After the first round, Karim would have 2 × 7 − c = 14 − c coins.
For his coins to increase,
14 − c > 7
⇒ c < 7
Therefore, Karim should agree only if the cost is less than 7 coins per round. If the cost is a positive whole number of coins, it should be at most 6 coins per round.
For example, at a cost of 6 coins per round, his coins would increase from 7 to 8, then to 10 and then to 14.

(iii) Let Karim initially have x coins.
After one round, his coins double to 2x. The genie can set the cost at 2x coins per round. Then,
Coins left after paying = 2x − 2x = 0.
Thus, charging twice the initial number of coins lets the genie take all the coins after the first round.
To use the same three-round trick, the genie can charge 8k coins per round when Karim starts with 7k coins, where k is a positive whole number:
First round: 2 × 7k − 8k = 6k coins left.
Second round: 2 × 6k − 8k = 4k coins left.
Third round: 2 × 4k − 8k = 0 coins left.
For example, if Karim starts with 14 coins, a cost of 16 coins per round leaves him with 12 coins, then 8 coins and finally no coins.

Explore number tricks and algebraic puzzles with 10 Algebra Play practice papers for Class 8 Maths, Ganita Prakash Part 2 Chapter 6. Revise how algebra explains patterns in number pyramids, calendar grids and divisibility puzzles. Each test PDF includes worked solutions, helping you follow the reasoning behind a result and practise writing clear mathematical explanations.

The surest way to understand this chapter is to invent a trick yourself. Start with x, add steps such as doubling, adding a constant and halving, then subtract x so only a fixed number survives; the book’s 2x + 4 example works exactly this way.
For a date-guessing trick, make sure the month ends up multiplied by 100, so the day fits in the last two digits. If a trick fails, retrace it line by line, the way these Algebra Play Class 8 solutions do. Chapter 7, Area, completes Part 2 and for revision across both parts of the book, head back to our NCERT Solutions for Class 8 Maths.