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

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Class 8 Maths Ganita Prakash Part 2 Chapter 6 Solutions

Ganita Prakash Class 8 Chapter 6 Solutions Algebra Play

Class 8 Maths Ganita Prakash Part 2 Chapter 6 Algebra Play Solutions Question Answer

6.2 Thinking about ‘Think of a Number’ Tricks

NCERT In-Text Questions (Pages 136-137)

Question 1.
How would you change this game to make the final answer 3? What about 5?
Solution:
Steps for 3:

1. Think of a number: x
2. Double it: 2x
3. Add six: 2x + 6 (change from four to six)
4. Divide by two: \(\frac{2 x+6}{2}\) = x + 3
5. Subtract the original number: (x + 3) – x = 3

The final answer will always be 3 if we add six in step 3.
Steps for 5:

1. Think of a number: x
2. Double it: 2x
3. Add ten: 2x + 10 (change from four to ten)
4. Divide by two: \(\frac{2 x+10}{2}\) = x + 5
5. Subtract the original number: (x + 5) – x = 5

The final answer will always be 5 if we add ten in step 3.
By adjusting the number added in step 3, we can change the final result of the trick to any number we desire, while keeping the rest of the steps the same.

Question 2.
Can you come up with more complicated steps that always lead to the same final value?
Solution:
Do it yourself.

Question 3.
Find the dates if the final answers are the following:
(i) 1269
(ii) 394
(iii) 296
Solution:
(i) Subtract 165 from 1269, to get:
1269 – 165 = 1104
Here, 04 represents the date, and 11 represents the month, i.e., November.
So, the correct date is 4th November.

(ii) Subtract 165 from 394, to get:
394 – 165 = 229
Here, 29 represents the date, and 2 represents the month, i.e., February.
So, the correct date is 29th February.

(iii) Subtract 165 from 296, to get:
296 – 165 = 131
Here, 31 represents the date, and 1 represents the month, i.e., January.
So, the correct date is 31st January.

Question 4.
Can you change the steps in this trick and still find the original date? Instead of subtracting 165 from the final answer, you might have to subtract some other number.
Solution:
Yes, we can change the steps in this trick and still find the original date.
Let the month be M and the day be D.
Steps for the modified trick are as follows:

1. Multiply M by 10: 10M
2. Add 6: 10M + 6
3. Multiply by 2: 20M + 12
4. Add 9: 20M + 21
5. Multiply by 5: 100M + 105
6. Add the day: 100M + 105 + D
7. Subtract 105: 100M + D

The last two digits of the resultant number will be the D (Day), and what comes before D is M (Month).

Question 5.
Try to devise your own ‘Think of a Number’ trick.
Solution:
Do it yourself.

6.3 Number Pyramids

NCERT In-Text Questions (Pages 137-139)

Question 1.
In a number pyramid, each number is the sum of the two numbers directly below it (see the figure).

Use the same rule to fill these pyramids:

Solution:
In a number pyramid, each number is the sum of the two numbers directly below it.

Question 2.
Fill the following pyramids:

Solution:
(a) Let us put letter-numbers in the empty boxes.

We obtain the following equations using the filling-up pyramid rules.
a + 22 = 50 …(i)
b + c = a …(ii)
c + d = 22 …(iii)
4 + e = b …(iv)
e + 6 = c …(v)
6 + f = d …(vi)
From equation (i), we get
a + 22 = 50
⇒ a = 50 – 22 = 28
For a = 28, from equations (ii), (iv), and (v), we get
4 + e + e + 6 = 28
⇒ 2e = 18
⇒ e = 9
Now, from equations (iv) and (v), we get
4 + e = b
⇒ b = 4 + 9 = 13
e + 6 = c
⇒ c = 9 + 6 = 15
From equations (iii) and (vi), we get
c + d = 22
⇒ 15 + d = 22
⇒ d = 22 – 15 = 7
and 6 + f = d
⇒ 6 + f = 7
⇒ f = 7 – 6 = 1
So, the complete number pyramid is

(b) Let us put letter-numbers in the empty boxes.

We obtain the following equations using the filling-up pyramid rules.
40 + b = a …(i)
c + d = 40 …(ii)
d + 9 = b …(iii)
5 + e = c …(iv)
e + 7 = d …(v)
7 + f = 9 …(vi)
From*(vi), f = 9 – 7 = 2
By solving equations (iv) and (v), we get
d – c = 2 …(vii)
By solving equations (ii) and (vii), we get c = 19 and d = 21
Substituting the value of d = 21 in equation (v), we get e = 14
Now, substituting the value of d = 21 in equation (iii), we get b = 30
Now, substituting the value of b = 30 in equation (i), we get a = 70
So, the complete number pyramid is

(c) Let us put letter-numbers in the empty boxes.

We obtain the following equations using the filling-up pyramid rules.
a + b = 35 …(i)
c + d = a …(ii)
d + 7 = b …(iii)
3 + 5 = c …(iv)
5 + e = d …(v)
e + f = 7 …(vi)
From equation (iv), we get c = 8
Substituting the value of c = 8 in equation (ii), we get
8 + d = a …(vii)
By solving equations (iii) and (vii), we get
(d + 7) – (8 + d) = b – a
⇒ b – a = -1 …(viii)
By solving equations (i) and (viii), we get a = 18 and b = 17
Substituting the value of a = 18 in equation (vii), we get d = 10
Now, substituting d = 10 in equation (v), we get e = 5.
Now, from equation (vi), we get f = 7 – 5 = 2.
So, the complete number pyramid is

Figure it Out (Page 140)

Question 1.
Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

Solution:
We know that if a b c form the bottom row of a pyramid, then the number in the top-most row will be a + 2b + c.
For 4 13 8,
a = 4, b = 13, c = 8
⇒ a + 2b + c = 4 + 26 + 8
Therefore, the number in the topmost row is 38.
For 7 11 3,
a = 7, b = 11, c = 3
⇒ a + 2b + c = 7 + 22 + 3
Therefore, the number in the topmost row is 32.
For 10 14 25,
a = 10, b = 14, c = 25
⇒ a + 2b + c = 10 + 28 + 25
Therefore, the number in the topmost row is 63.

Question 2.
Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.
Solution:
For a pyramid with 4 rows, the numbers in each row are generated by summing adjacent numbers in the row directly beneath it.
Let the bottom row be denoted by a, b, c, and d.
So, the expression for the topmost row is a + 3b + 3c + d.

Question 3.
Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

Recall the Virahanka-Fibonacci number sequence 1, 2, 3, 5, … where each number is the sum of the two numbers before it.
Solution:
From question 2, the topmost row of a pyramid with the bottom row a b c d is given by a + 3b + 3c + d.
For 8 19 21 13,
a = 8, b = 19, c = 21, d = 13
⇒ a + 3b + 3c + d = 8 + 57 + 63 + 13
Therefore, the number in the topmost row is 141.
For 7 18 19 6,
a = 7, b = 18, c = 19, d = 6
⇒ a + 3b + 3c + d = 7 + 54 + 57 + 6
Therefore, the number in the topmost row is 124.
For 9 7 5 11,
a = 9, b = 7, c = 5, d = 11
⇒ a + 3b + 3c + d = 9 + 21 + 15 + 11
Therefore, the number in the topmost row is 56.

Question 4.
If the first three Virahanka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahanka-Fibonacci numbers?
Solution:
In the Virahanka-Fibonacci number sequence, each number is the sum of the two numbers before it.
The first three Virahanka-Fibonacci numbers are: 1, 2, 3.
Now, the numbers in the number pyramid are:
Bottom row: 1, 2, 3
Second row: 1 + 2 = 3, 2 + 3 = 5
Top row: 3 + 5 = 8
Thus, the numbers in the pyramid are:
Bottom row: 1, 2, 3
Second row: 3, 5
Top row: 8
Yes, all these numbers are Virahanka-Fibonacci numbers.

Question 5.
What can you say about the numbers in the pyramid and the number at the top in the following cases?
(i) The first four Virahanka-Fibonacci numbers are written in the bottom row of a four-row pyramid.
(ii) The first 29 Virahanka-Fibonacci numbers are written in the bottom row of a 29-row pyramid.
Solution:
(i) The first four Virahanka-Fibonacci numbers are: 1, 2, 3, 5.
The pyramid will look like this:
Bottom row: 1, 2, 3, 5
Second last row: 3, 5, 8
Top row: 8, 13
The final number in the topmost row is 21.
Also, all the numbers in the pyramid are Virahanka-Fibonacci numbers.

(ii) Because each number in a row is the sum of two consecutive numbers from the row below, every row of the pyramid also consists of Virahanka-Fibonacci numbers. Thus, all the numbers in the pyramid follow the sequence. Thus, all the numbers in the pyramid are Virahanka- Fibonacci numbers, provided they are written in order in the bottom row.
The last number of each row, starting from the bottom, is the sum of the last two numbers of the row below. For a pyramid with 29 rows, these numbers follow the sequence consecutively, starting from the 29th number in the bottom row, then the 30th in the next row, the 31st in the row above, and so on, up to the 29th row at the top of the pyramid. So, the number in the top row will be the 57th Virahanka-Fibonacci number.

Question 6.
If the bottom row of an n-row pyramid contains the first n Virahanka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?
Solution:
Because each number in a row is the sum of two consecutive numbers from the row below, every row of the pyramid also consists of Virahanka-Fibonacci numbers, so all numbers in the pyramid follow the sequence. Thus, all the numbers in the pyramid are Virahanka- Fibonacci numbers, provided they are written in order in the bottom row.
If the bottom row of an n-row pyramid contains the first n Virahanka-Fibonacci numbers 1, 2, 3, 5, 8,…, then the numbers in the pyramid follow a clear shifting pattern. The first (bottom) row consists of the 1st to the nth Virahanka-Fibonacci numbers.
Since each number above is the sum of the two numbers just below it, the second row automatically becomes the 2nd to the (n + 1)th Virahanka-Fibonacci numbers. Similarly, the third row consists of the 3rd to the (n + 2)th Virahanka-Fibonacci numbers.
In general, the kth row from the bottom contains the kth to the (n + k – 1)th Virahanka-Fibonacci numbers.
Continuing this process, the top row is reached at k = n, and it contains only one number, which is the (2n – 1)th Virahanka-Fibonacci number.

6.4 Fun with Grids

Calendar Magic

NCERT In-Text Questions (Page 142)

Question 1.
Create your own calendar trick. For instance, choose a grid of a different size and shape.
Solution:

Algebra Grids

NCERT In-Text Questions (Page 142)

Question 1.
In the following grids, find the values of the shapes and fill in the empty squares:

Solution:

6.5 The Largest Product

Figure it Out (Page 144)

Question 1.
Fill in the digits 1, 3, and 7  to make the largest product possible.
Solution:
We need to place the digits 1, 3, and 7 in the two blanks to form the largest possible product.
The possible combinations are:
13 × 7 = 91
31 × 7 = 217
17 × 3 = 51
71 × 3 = 213
37 × 1 = 37
73 × 1 = 73
The largest product is 31 × 7 = 217.
Alternative Method: If p, q, and r are the three digits such that p < q < r, then the largest product is qp × r.
Here, 1 < 3 < 7, so the largest product is 31 × 7 = 217

Question 2.
Fill in the digits 3, 5, and 9  to make the largest product possible.
Solution:
We need to place the digits 3, 5, and 9 in the two blanks to form the largest possible product.
The possible combinations are:
35 × 9 = 315
53 × 9 = 477
39 × 5 = 195
93 × 5 = 465
59 × 3 = 177
95 × 3 = 285
The largest product is 53 × 9 = 477.
Alternative Method: Here, 3 < 5 < 9, so the largest product is 53 × 9 = 477.

6.6 Decoding Divisibility Tricks

NCERT In-Text Questions (Page 145)

Question 1.
Can you work out what happens if a > b?
Solution:
Suppose the two-digit number is ab.
When it is reversed, the new number is ba.
If a > b, then ab > ba.
So, the difference is (10a + b) – (10b + a) = 10a + b – 10b – a
= 9a – 9b
= 9(a – b)
Thus, the difference is divisible by 9.

Figure it Out (Pages 145-147)

Question 1.
In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?
Solution:
In the trick given above, when we divide the difference by 9, the quotient is the difference of the two digits.
i.e., \(\frac{9(a-b)}{9}\) = (a – b) and \(\frac{9(b-a)}{9}\) = (b – a)
The quotient is always equal to the positive difference between the tens digit and the units digit of the original number.

Question 2.
In the trick given above, instead of finding the difference between the two 2-digit numbers, find their sum. What will happen?
For example:

• 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?
Solution:
Let the two-digit number be ab.
When it is reversed, the new number is ba.
So, the sum of the numbers is (10a + b) + (10b + a)
= (10a + a) + (10b + b)
= 11a + 11b
= 11(a + b)
Thus, the sum is always divisible by 11.

Question 3.
Consider any 3-digit number, say abc(100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3?
[Hint: Look at some multiples of 37.]
Solution:
Let the 3-digit number be abc, which can be expressed as 100a + 10b + c.
The other two numbers formed by cycling the digits are: bca = 100b + 10c + a, and cab = 100c + 10a + b.
The sum of these three numbers is (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)
= (100a + a + 10a) + (100b + 10b + b) + (100c + 10c + c)
= 111a + 111b + 111c
= 111(a + b + c)
Since 111 can be factored as 111 = 3 × 37, the sum is always divisible by 37.
Thus, the sum of the three numbers formed by cycling the digits of any 3-digit number is always divisible by 37.
Since 3 is also a factor of 111, the sum is also divisible by 3.

Question 4.
Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is, abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works.
[Hint: Multiply 7, 11, and 13.]
Solution:
Let the 3-digit number be abc, which can be expressed as 100a + 10b + c.
Repeating the digits to form a 6-digit number, we get:
abcabc = 1000 × (100a + 10b + c) + (100a + 10b + c) = 1001 × (100a + 10b + c)
Now, divide this number by 7, 11, and 13 = \(\frac{1001 \times(100 a+10 b+c)}{7 \times 11 \times 13}\)
Since, 7 × 11 × 13 = 1001, so \(\frac{1001 \times(100 a+10 b+c)}{7 \times 11 \times 13}\) = 100a + 10b + c
Thus, dividing the 6-digit number abcabc by 7, 11, and 13 leaves us with the original 3-digit number abc.

Question 5.
There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?
Solution:
Let the number of flowers the person starts with be x.
After the first pond: The number of flowers doubles, so the total becomes 2x.
The person places y flowers in shrine 1, leaving (2x – y) flowers.
After the second pond: The remaining flowers are doubled, so the total becomes (4x – 2y).
The person places y flowers in shrine 2 since each equal number of flowers in each shrine, leaving (4x – 3y) flowers.
After the third pond: The remaining flowers are doubled again, so the total becomes (8x – 6y).
The person places y flowers in shrine 3, leaving (8x – 7y) flowers.
Since the number of flowers left is zero,
8x – 7y = 0
Solving for y: y = \(\frac {8x}{7}\)
For y to be an integer, x must be a multiple of 7; that is, let x = 7k (where k is a positive integer), then y = 8k.
Final Result: The person starts with 7, 14, 21,… flowers.
The number of flowers placed in each shrine is 8, 16, 24, … respectively.
Thus, the person starts with 7k flowers and places 8k flowers in each shrine, where k is any positive integer.

Question 6.
A farm has some horses and hens. The total number of heads of these animals is 55, and the total number of legs is 150. How many horses and how many hens are on the farm? 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?]

Solution:
Suppose all 55 animals were hens.
Then, the total number of legs would be: 55 × 2 = 110 legs.
The actual total number of legs is 150, so the difference is: 150 – 110 = 40 extra legs.
Each horse has 2 extra legs compared to a hen.
Therefore, the number of horses is \(\frac {40}{2}\) = 20 horses.
The number of hens is: 55 – 20 = 35 hens.
So, there are 20 horses and 35 hens on the farm.

Question 7.
A mother is 5 times her daughter’s age. In 6 years, the mother will be 3 times her daughter’s age. How old is the daughter now?
Solution:
Let the daughter’s present age be x years.
The mother’s present age is 5 times the daughter’s age, i.e., 5x years.
According to the problem, in 6 years, the mother will be 3 times the daughter’s age.
Therefore, we set up the equation:
5x + 6 = 3(x + 6)
Now, solve for x:
5x + 6 = 3x + 18
⇒ 2x = 12
⇒ x = 6
The daughter is 6 years old.

Question 8.
Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?
Solution:
Let us assume that Gauri has x cows.
Since Naina has twice as many cows as Gauri, Naina has 2x the cows.
Now, if Naina gives 3 cows to Gauri, then Gauri will have (x + 3) cows, and Naina will have (2x – 3) cows.
After this, they will both have the same number of cows,
so x + 3 = 2x – 3
Solving the equation, we get
2x – x = 3 + 3
⇒ x = 6
⇒ 2x = 12
So, Gauri has 6 cows, and Naina has 12 cows.

Question 9.
I run a small dosa cart, and my expenses are as follows:
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?
Solution:
(i) Rent for the dosa cart: ₹ 5000 per day.
Cost of making 100 dosas: 100 × 10 = ₹ 1000
Total expenses = ₹ 5000 (rent) + ₹ 1000 (cost of making 100 dosas) = ₹ 6000
To make a profit of ₹ 2000, the selling price of 100 dosas should be:
Total expenses + Desired profit = ₹ 6000 + ₹ 2000 = ₹ 8000
∴ Selling price per dosa = \(\frac{\text { Selling price }}{\text { Number of dosas }}\)
= \(\frac {8000}{100}\)
= ₹ 80
So, the selling price per dosa should be ₹ 80 to make a profit of ₹ 2000.

(ii) Now, if the customers are willing to pay ₹ 50 for each dosa,
we need to calculate the number of dosas that need to be sold to make a profit of ₹ 2000.
The total amount required to make a profit of ₹ 2000 = ₹ 8000.
Since each dosa is sold for ₹ 50, the amount from selling n dosas is
Amount = 50 × n
⇒ 8000 = 50 × n [∵ Total amount is ₹ 8000]
⇒ n = \(\frac {8000}{50}\) = 160
So, 160 dosas should be sold in a day to make a profit of ₹ 2000 at ₹ 50 per dosa.

Question 10.
Evaluate the following sequence of fractions:
\(\frac{1}{3}, \frac{(1+3)}{(5+7)}, \frac{(1+3+5)}{(7+9+11)}\)
What do you observe? Can you explain why this happens?
[Hint: Recall what you know about the sum of the first n odd numbers.]
Solution:
Step 1: Simplify term 1: \(\frac {1}{3}\)
Step 2: Simplify term 2: \(\frac{1+3}{5+7}=\frac{4}{12}=\frac{1}{3}\)
Step 3: Simplify term 3: \(\frac{1+3+5}{7+9+11}=\frac{9}{27}=\frac{1}{3}\)
Each fraction in the sequence simplifies to the constant value \(\frac {1}{3}\).
To understand why this happens, we use the property that the sum of the first n odd natural numbers is n².
Let n represent the number of terms in the numerator.
Numerator (N): The sum of the first n odd numbers is given by: N = n²
Denominator (D): The denominator consists of the sum of the next n odd numbers.
This is equivalent to the sum of the first 2n odd numbers minus the sum of the first n odd numbers:
D = (2n)² – n²
= 4n² – n²
= 3n²
The ratio of the two sums is: \(\frac{\mathrm{N}}{\mathrm{D}}=\frac{n^2}{3 n^2}=\frac{1}{3}\)
This proves that for any number of terms n, the ratio will always remain 1 : 3.

Question 11.
Karim and the Genie
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?
Solution:
Let the number of coins Karim initially had be x.
According to the question, the number of coins doubles after each round, and Karim gives 8 coins to the genie.
After 1st round, the number of coins Karim had: 2x – 8
After 2nd round the number of coins Karim had: 2(2x – 8) – 8 = 4x – 16 – 8 = 4x – 24
After 3rd round the number of coins Karim had: 2(4x – 24) – 8 = 8x – 48 – 8 = 8x – 56
The problem states that after the third round, he had exactly 8 coins left to pay the genie (leaving him with 0 coins).
Therefore, 8x – 56 = 0
⇒ 8x = 56
⇒ x = 7
Thus, Karim initially had 7 coins.

(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
Solution:
For Karim to make a profit, the amount gained by doubling must be greater than the amount paid to the genie.
Let x be the initial number of coins, and C be the cost per round.
After one round, the number of coins is 2x – C.
For the number of coins to increase:
2x – C > x
⇒ x > C
So, Karim should agree to the deal only if the cost per round (C) is less than the number of coins he currently has (C < x).

(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?
Solution:
To ensure that Karim is left with zero coins after a certain number of rounds (n), the genie must set the cost per round (C) based on the initial amount (x) such that C > x.