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

6.1 Algebra Play & 6.2 Thinking about ‘Think of a Number’ Tricks

NCERT Intext Questions (Page 136)

Question 1.
How would you change this game to make the final answer 3? What about 5?
Solution:
Rules for a final answer of 3.
Think of a number: x
Double it: 2x
Add six: 2x + 6
Divide by two : x + 3
Subtract the original number you thought of: x + 3 – x = 3

Rules for a final answer of 5
Think of a number: x
Double it: 2x
Add ten :2x + 10
Divide by two : x + 5
Subtract the original number you thought of: x + 5 – x = 5
To change the game to make the final answer 3, change “Add four” to “Add six,” and to make the final answer 5, change “Add four” to “Add ten”.

Question 2.
Can you come up with more complicated steps that always lead to the same final value?
Solution:
Yes. Here is an example.
We can understand such tricks through algebra.
Step 1: Think of a number = x
Step 2. 5 times it = 5x
Step 3: Add 25 = 5x + 25
Step 4: Divide by 5 = \(\frac{5 x+25}{5}\) = x + 5
Step 5: Subtract the original number you thought of (x + 5) – x = 5
For Example:
Consider a number 18.
5 times 18 = 90
Add 25 = 90 + 25 = 115
Divide by 5 = 115 ÷ 5 = 23
∴ 23 – 18 = 5

Math Talk (Page 137)

Question 1.
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.

6.3 Number Pyramids

NCERT Intext Questions (Pages 138-139)

Question 1.
Use the same rule to fill these pyramids:

Solution:
Each number is the sum of the two numbers directly below it.
(i)

8 = 6 + 2
(ii)

3 + 4 = 7; 4 + 3 = 7 and 7 + 7 = 14
(iii) 5 + 4 = 9; 4 + 5 = 9 and 5 + 0 = 5;
9 + 9 = 18; 9 + 5 = 14 and 18 + 14 = 32.

Question 2.
Fill in the following pyramids:

Solution:
(i) Let a, b, c, d, e, and f be the missing numbers.

a + 22 = 50
⇒ a = 50 – 22
⇒ a = 28
b + c = 28 …..(i)
c + d = 22 …..(ii)
Adding equations (i) and (ii), we get
b + d + 2c = 50 ……(iii)
Also, 4 + e = b …..(iv)
e + 6 = c …(v)
6 + f = d ….(vi)
Adding equations (iv) and (v), we get
10 + 2e = b + c
10 + 2e = 28 [From (i)]
2e = 18
⇒ e = 9
4 + 9 = b
⇒ b = 13
9 + 6 = c
⇒ c = 15
Putting c = 15 in equation (ii), we get
d = 22 – 15 = 7
⇒ d = 7
Putting d = 7 in equation (vi), we get
6 + f = 7
⇒ f = 1

∴ a = 28, b = 13, c = 15, d = 7, e = 9 and f = 1.

(ii) Let a, b, c, d, e, and f be the missing numbers.

40 + b = a …(i)
c + d = 40 …(ii)
d + 9 = b …(iii)
5 + e = c …(iv)
e + 7 = d …(v)
7 + f = 9
⇒ f = 2
Adding equations (iv) and (v), we get
2e + 12 = c + d
⇒ 2e + 12 = 40 [From (ii)]
⇒ 2e = 28
⇒ e = 14
c = 5 + 14 = 19
⇒ c = 19 [From (iv)]
14 + 7 = d
⇒ d = 21
b = 21 + 9 = 30
⇒ b = 30
a = 40 + 30 = 70
⇒ a = 70

∴ a = 70, b = 30, c = 19, d = 21, e = 14, and f = 2.

(iii) Let a, b, c, d, e, and f be the missing numbers.

a + b = 35 …(i)
c + d = 40 …(ii)
d + 7 = b …(iii)
Adding equations (ii) and (iii), we get
c + d + d +7 = a + b
⇒ c + 2d = 35 – 7 = 28 [From (i)]
⇒ c + 2d = 28 …(iv)
Also, 3 + 5 = c
⇒ c = 8
Putting c = 8 in equation (iv), we get
8 + 2d = 28
⇒ 2d = 28 – 8 = 2o
⇒ d = 10
8 + 10 = a
⇒ a = 18
10 + 7 = b
⇒ b = 17
5 + e = 10
⇒ e = 10 – 5 = 5
⇒ e = 5

5 + f = 7
⇒ f = 7 – 5 = 2
⇒ f = 2
∴ a = 18, b = 17, c = 8, d = 10, e = 5, and f = 2.

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,

(a) Given, bottom row:

The number in the topmost row = 4 + 2 × 13 + 8
= 4 + 26 + 8
= 38

(b) Given, bottom row:

The number in the topmost row = 7 + 2 × 11 + 3
= 7 + 22 + 3
= 32

(c) Given, bottom row:

The number in the topmost row = 10 + 2 × 14 + 25
= 10 + 28 + 25
= 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:
Let a, b, c, d, e, f, g, h, i, and j be the elements of the pyramid.

∴ e = a + b, f = b + c, g = c + d
h = e + f = (a + b) + (b + c) = a + 2b + c
i = f + g = (b + c) + (c + d) = b + 2c + d
j = h + i = (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d

Thus, the expression for the top 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:
(a) If a, b, c, and d are the bottom row, then the expression of the topmost row of the pyramid is a + 3b + 3c + d.
Given the bottom row;

Here, a = 8, b = 19, c = 21, and d = 13
∴ The number in the topmost row = a + 3b + 3c + d
= 8 + 3(19) + 3(21) + 13
= 8 + 57 + 63 + 13
= 141
Thus, the number in the topmost row is 141.

(b) Given, bottom row:

Here, a = 7, b = 18, c = 19 and d = 6
∴ The number in the topmost row = a + 3b + 3c + d
= 7 + 3(18) + 3(19) + 6
= 7 + 54 + 57 + 6
= 124
Thus, the number in the topmost row is 124.

(c) Given, bottom row:

Here, a = 9, b = 1,c = 5, and d = 11
∴ The number in the topmost row = a + 3b + 3c + d
= 9 + 3(7)+ 3(5) + 11
= 9 + 21 + 15 + 11
= 56
Thus, 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:
We know that the first three Virahanka-Fibonacci number sequence = 1, 2, 3
Here, the bottom row

Let a, b, and c be the missing numbers.

b = 1 + 2 = 3
c = 2 + 3 = 5
and a = b + c = 3 + 5 = 8
The complete pyramid is:

The numbers that appear in the grid = 1, 2, 3, 3, 5, 8
∴ The number at the top = 8
Yes, 1, 2, 3, 3, 5, 8 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) We know that
The first four Virahanka-Fibonacci numbers = 1, 2, 3, 5
Here, the bottom row

Let a, b, c, d, e, and f be the missing numbers.

d = 1 + 2 = 3
e = 2 + 3 = 5
f = 3 + 5 = 8
b = d + e = 3 + 5 = 8
c = e + f = 5 + 8 = 13
and a = b + c = 8 + 13 = 21

The numbers in the pyramid are 1, 2, 3, 5, 8, 8, 13, 21,….
We can say that the numbers are a Virahanka-Fibonacci sequence.
∴ The number at the top = 21

(ii) From the above solution, we get
The number at the top = 2 × (Total number of digits present at the bottom) – 1
= 2x – 1
= 2 × 29 – 1
= 58 – 1
= 57th
Fibonacci numbers.

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:
Every number in the pyramid is a Virahanka-Fibonacci number. The number at the top of the n-row pyramid with the first nthVirahanka-Fibonacci numbers at the bottom is the (2n – 1)th Virahanka-Fibonacci number.

6.4 Fun with Grids

Math Talk (Page 142)

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

Solution:
(i) Add the 5 numbers in this grid and tell the sum.

1 + 7 + 8 + 9 + 15 = 40
Let ‘a’ represent the topmost number.

My own calendar trick.
Sum = a + (a + 6) + (a + 7) + (a + 8) + (a + 14) = 5a + 35.
Consider a 3 × 3 grid. Add the 9 numbers.

(10 + 11 + 12) + (17 + 18 + 19) + (24 + 25 + 26) = 33 + 54 + 75 = 162
Let ‘a’ represent the top-left number.

My own calendar trick.
Sum = a + (a + 1) + (a + 2) + (a + 7) + (a + 8) + (a + 9) + (a + 14) + (a + 15) + (a + 16)
= 9a + 72
= 9(a + 8)
Consider a 1 × 3 grid.
Add the 3 numbers.

28 + 29 + 30 = 87
Let ‘a’ represent the left number.

My own calendar trick.
Sum = a + (a + 1) + (a + 2)
= 3a + 3
= 3(a + 1)

(ii)

Let ‘a’ represent the topmost number.

My own trick.

Let ‘a’ represent the top-left number.

This trick works.

Let ‘a’ be the topmost number.

This trick works.

Question 2.
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:
There are six ways to place three digits:
We can fill the first box with 1, 3, or 7.
For each of these choices, we have 2 ways of filling the remaining 2 digits.
The six choices are:

In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.

Thus, the first term in both expressions is equal.
The second term shows that it is the largest.

Question 2.
Fill in the digits 3, 5, and 9 in to make the largest product possible.
Solution:
There are six ways to place three digits:
We can fill the first box with 3, 5, or 9.
For each of these choices, we have 2 ways of filling the remaining 2 digits.
The six choices are:

In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.

Thus, the first term in both expressions is equal.
The second term shows that it is the largest.

6.6 Decoding Divisibility Tricks

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:
44, 110, 33 are divisible by 11.
Yes, it is always true.
44 = 4 – 4 = 0, divisible by 11.
110 = (1 + 0) – 1 = 0, divisible by 11.
33 = 3 – 3 = 0, divisible by 11.
Using Algebra
Original number = 10a + b
Reversed number = 10b + a
Sum = 10a + b + 10b + a = 11(a + b)
Hence, 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 and make two other 3-digit numbers from there digits by cycling these digits around, yielding bca and cab.
N₁ = abc = 100a + 10b + c
N₂ = bca = 100b + 10c + a and
N₃ = cab = 100c + 10a + b

On adding above three numbers, we get
N₁ + N₂ +N₃
= 100a + 10b + c + 100b + 10c + a +100c + 10a + b
= 111a + 111b + 111c
= 11(a + b + c)
= 3 × 37 (a + b + c)
Since, the sum can be expressed as 37 multiplied by an integer 3 × (a + b + c), the sum is always divisible by 37.
Similarly, the sum can be expressed as 3 multiplied by an integer 37 × (a + b+c). The sum is always 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:
Given that abc is a 3-digit number.
abc = 100a + 10b + c
Make it a 6-digit number = abcabc
= 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
= 1001(100a + 10b + c)
The smallest number, divisible by 7, 11, and 13 = LCM (7, 11, 13)
= 7 × 11 × 13
= 1001
∴ abcabc = 1001(100a + 10b + c), is divisible by 7, 11, and 13.
Consider 836 a 3-digit number.
Make it 6-digit number = 836836 = 1001 × 836
∴ 836836 is divisible by 7, 11, and 13.
This works because 10001 = 7 × 11 × 13, and repeating a 3-digit number creates a multiple of 1001.

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 an 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 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:
Method 1: Using Algebra
Let x and y be the number of horses and hens, respectively.
According to the questions,
x + y = 55 ……..(i)
And, 4x + 2y = 150
⇒ 2x + y = 75 ……(ii)
Subtracting (i) from (ii), we get
2x + y – x – y = 75 – 55
⇒ x = 20
Putting x = 20 in equation (i), we get
20 + y = 55
⇒ y = 55 – 20 = 35
Thus, the number of horses = 20 and the number of hens = 35.
Method 2: (without letter numbers)
If all 55 animals were hens
Total legs would be 55 × 2 = 110 legs
But actual legs = 150
Difference = 150 – 110 = 40 legs
Each time we replace a hen with a horse.
We remove 2 legs (hen) and add 4 legs (horse).
Net increase = 2 legs
Number of horses needed = 40 ÷ 2 = 20
Number of hens = 55 – 20 = 35

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 present age of the daughter = x years
And the present age of her mother = y years
According to the question,
5(x) = y
⇒ 5x = y …..(i)
In 6 years,
3(x + 6) = y + 6
⇒ 3x + 18 = y + 6
⇒ 3x + 18 – 6 = y
⇒ 3x + 12 = y …..(ii)
From equations (i) and (ii), we get
3x + 12 = 5x
⇒ 5x – 3x = 12
⇒ 2x = 12
⇒ x = 6
The present age of the daughter = 6 years

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 x be the number of cows Gauri has, and y be the number of cows Naina has.
According to Gauri, we have
y = 2x …(i)
According to Naina, we have
y – 3 = x + 3
⇒ 2x-3 = x + 3 [from Eq. (i)]
⇒ 2x – x = 3 + 3
⇒ x = 6
From Eq.(i), we get
y = 2 × 6 = 12
Hence, 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:
Given, rent for the dosa cart = ₹ 5000/day.
The total cost of making one dosa = ₹ 10
(i) Given,
Number of dosas = 100
∴ The cost of making 100 dosas = 100 × ₹ 10 = ₹ 1000
Total cost price = Rent for the dosa cart + the cost of making 100 dosas
= ₹ 5000 + ₹ 1000
= ₹ 6000
Profit = ₹ 2000
∴ Total selling price = ₹ 6000 + ₹ 2000 = ₹ 8000
The selling price of one dosa = \(\frac {8000}{100}\) = ₹ 80

(ii) Let n be the number of dosas.
Then total cost price = n × ₹ 10 + ₹ 5000
Total selling price = n × ₹ 50
Profit = ₹ 2000
S.P = C.P + profit
⇒ 50n = 10n + 5000 + 2000
⇒ 50n – 10n = 5000 + 2000
⇒ 40n = 7000
⇒ n = \(\frac {7000}{40}\)
⇒ n = 175
So, I should sell 175 dosas.

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:

Thus, the given sequences are equivalent fractions.
We know that the sum of the first n odd numbers is n2.
Numerators:
1 = 12 = 1
1 + 3 = 22 = 4
1 + 3 + 5 = 32 = 9
Denominators:
3 = 3 × 12
5 + 7 = 12 = 3 × 22
7 + 9 + 11 = 27 = 3 × 32
Thus, each fraction is \(\frac{n^2}{3 n^2}=\frac{1}{3}\)

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, O 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?
Solution:
(i) Let Karim have n coins initially
No. of coins after 1st round = 2n – 8
No. of coins after 2nd round = [2(2n – 8)] – 8
= 4n – 16 – 8
= 4n – 24
No. of coins after 3rd round = 2(4n – 24) – 8 = 0
⇒ 2(4n – 24) – 8 = 0
⇒ 8n – 48 – 8 = 0
⇒ 8n = 56
⇒ n = 7
So, Karim initially had 7 coins.

(ii) Let c = cost per round (coins to give the genie)
Starting with 7 coins:
After round 1: 2(7) – c = 14 – c
After round 2: 2(14 – c) – c = 28 – 3c
After round 3: 2(28 – 3c) – c = 56 – 7c
For Karim to increase his coins:
56 – 7c > 7
⇒ 56 – 7 > 7c
⇒ 49 > 7c
⇒ c < 7
The cost per round should be less than 7 coins.
For example, if c = 6:
After 3 rounds: 56 – 7(6) = 56 – 42 = 14 coins (doubled his money!)

(iii) Let Karim start with n coins, and let c = cost per round.
After 3 rounds, Karim has: 8n – 7c coins
For the genie to get all coins:
8n – 7c = 0
⇒ 7c = 8n
⇒ c = 8n/7
The genie should charge (8n)/7 coins per round, where n is Karim’s starting amount.
For this to be a whole number, n must be a multiple of 7.

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