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

Class 8 Maths Algebra Play Solutions

Class 8 Ganita Prakash Part 2 Chapter 6 Solutions Algebra Play

NCERT (Page 136)

Question 1.
(i) 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”.

(ii) Can you come up with more complicated steps that always lead to the same final value?
Solution:
Yes, many different sequences of operations can be created.

NCERT (Page 137)

Question 1.
(i) Find the dates if the final answers are the following.
(a) 1269
(b) 394
(c) 296
Solution:
(a) Let the month be Mand the day be D.
We know that 1269 = 100 M + 165 + D
⇒ 1269 – 165 = 100M +D
⇒ 1104 = 100M + D
Here, the hundreds and thousands digits represent the month (M) and the last two digits represent the day (D).
Hence, the date is the 4th of November.
(b) Do the same as part(a). Ans 29th February
(c) Do the same as part(a). Ans 31st January

(ii) You can try this trick with your friends. Ask them to choose the starting date as their birthday.
Solution:
Do yourself

Question 2.
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 any of the multipliers or the numbers added.
Let the month be M and the day be D.
Multiply M by 2 :2M
Add 2 : 2M + 2
Multiply by 50 : 100M + 100
Add the day : 100M + D + 100
In this new version, we should subtract 100 from the final answer instead of 165 to find the date.

NCERT (Page 139)

Question 1.
Fill the following pyramids.

Solution:

Question 2.
Fill the following pyramids.

Solution:
(i) Let us fill the empty boxes with letter-numbers.

From the rules for filling up pyramids, we get the following equation.
a + 22 = 50 ⇒ a = 50 – 22 = 28 ………(i)
b + c = a ⇒ b + c = 28 ….(ii)
c + d = 22 …(iii)
4 + e = b …(iv)
e + 6 =c …….(v)
6 + f =4 …….(vi)

On adding Eqs. (iv) and (v), we get
4 + e + e + 6 = b + c
⇒ 2e + 10 =28
⇒ 2e = 28 – 10 [from Eq. (ii)]
⇒ e = \(\frac{18}{2}\) = 9

From Eq. (iv), 4 + e = b
⇒ 4 + 9 = b [∵ e = 9]
⇒ b = 13

From Eq.(ii), b + c = 28
⇒ 13 + c = 28 [∵ b = 13]
⇒ c = 28 -13 =15

From Eq. (iii), c + 4 = 22
⇒ 15 + 4 = 22 [∵ c = 15]
⇒ 4 = 22 – 15 = 7

From Eq. (vi), 6 + f = 4 ⇒ 6 + f = 7 [∵ d = 7]
⇒ f = 7 – 6 = 1
∴ Completed pyramid is

(ii) Do same as part (i).

(iii) Do same as part (i).

Figure it Out (NCERT 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:
The rule for the pyramid is to add the adjacent numbers in the lower row to get the number in the box directly above them.
(i) The bottom row is 4,13, 8.
The first number in the second row is the sum of the • first two numbers in the bottom row i.e. 4 + 13 = 17. The second number in the second row is the sum of the last two numbers in the bottom row i.e. 13 + 8 = 21
The second row is 17, 21.
The topmost number is the sum of the numbers in the second row i.e. 17 + 21 = 38.
(ii) Do the same as part (i). Ans The topmost number is 32.
(iii) Do the same as part (i). Ans The topmost number 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:
In a number pyramid, each number is the sum of the two numbers directly below it.
Let the bottom row values be A, B, C and D from left to right.
∴ The third row values are (A + B),(B + C) and (C + D).
The second row values are
(A + B) + (B + C) = (A + 2B + C) and (B + C) + (C + D) = (B + 2C + D).
The topmost (first) row value is
A + 2B + C + B + 2C + D = 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 the previous question, the expression for the topmost row value, if the bottom row has 4 numbers is A + 3B + 3C + D
(i) Here, A = 8, 5 = 19, C = 21 and D = 13
Therefore, the topmost number
= 8 + 3 × 19 + 3 × 21 + 13 = 8 + 57 + 63 + 13 = 141
So, the number in the topmost is 141
(ii) Do same as part (i). Ans 124
(iii) Do same as part (i). Ans 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:
The Virahanka-Fibonacci sequence starts with 1, where each subsequent number is the sum of the two preceding ones i.e. 1, 2, 3, 5, 8, 13,….
The first three numbers in the sequence are 1, 2 and 3. The pyramid has three rows, so the bottom row has three numbers, the middle row has two, and the top row has one. Each number in the row above is the sum of the two numbers directly below it.
The numbers in the middle row are calculated by summing adjacent numbers in the bottom row
i. e. 1 + 2 = 3 and 2 + 3 = 5.
The middle row numbers are 3 and 5.
The number at the top is the sum of the two numbers in the middle row i.e. 3 + 5 = 8.
So, the top number is 8. Therefore, the numbers in the grid are 1, 2, 3, 5 and 8. All these numbers are part of the Virahanka-Fibonacci sequence.

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.
Solution:
The Virahanka-Fibonacci sequence starts with 1,
where each subsequent number is the sum of the two preceding ones i.e. 1, 2, 3, 5, 8, 13,….
The numbers in the bottom row are 1,2,3 and 5.
The second row numbers are 1 + 2, 2 + 3 and 3 + 5 i.e. 3, 5 and 8.
The third row numbers are 3 + 5 and 5 + 8 i.e. 8 and 13. So, the top number is 8 + 13 i.e. 21.
Here, we observe that, every number in the pyramid is a Virahanka-Fibonacci number and the topmost number is 21, which is the 7th Virahanka-Fibonacci number.
Hence, the number at the top of the n-row pyramid with the first nth Virahanka-Fibonacci numbers at the bottom is the (2n – 1)th Virahanka- Fibonacci number.

(ii) The first 29 Virahanka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.
Solution:
The top number depends on all 29 bottom numbers, combined through repeated addition.
So, the number at the top = (2n – 1)th = 2 × 29 – 1 = 58 – 1 = 57th number from Virahanka-Fibonacci sequence.

Question 6.
If the bottom row of an n-row pyramid contains the first nth 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.

NCERT (Page 142)

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

Solution:
Do yourself

Question 2.
In the following grids, find the values of the shapes and fill in the empty squares:

Solution:
(i) Let value of black square -x and value of grey circle = y
Row 1: 2x + y = 27 …(i)
Row 2: 2y + x = 21 …(ii)
From Eq. (i),y = 27 – 2x …(iii)
Now, Eq. (ii), 2(27 – 2x) + x = 21 [from Eq. (iii)]
⇒ 54 – 4x + x = 21
⇒ -3x = 21 – 54
⇒ x = \(\frac{33}{3}\) = n

On putting the value of x in Eq. (iii),
we get y = 27 – 2 × 11 = 27 – 22 = 5
So, black square = 11 and grey circle = 5
Row 3: The value of O + □ + O = y + x + y
= 5 + 11 + 5
= 21

(ii) Do same as part (i).
Ans. Row 3: 15 Row 4: 18, 15, 15, 48

Figure it Out (NCERT Page 144)

Question 1.
Fill the digits 1, 3 and 7 in □□ × □ to make the largest product possible.
Solution:
We know that the largest digit is the multiplier. The other two digits were arranged in decreasing order to form the multiplicand.
Therefore, 31 × 7 = 217

Question 2.
Fill the digits 3, 5 and 9 in to make the largest product possible.
Solution:
We know that the largest digit is the multiplier. The other two digits were arranged in decreasing order to form the multiplicand.
Therefore, 53 × 9 = 477

NCERT (Page 144-145)

Question 1.
(i) It is now Mukta’s turn to show a mathematics trick to Shubham.

If we choose other 2-digit numbers and follow the steps, will there always be no remainder?
Solution:
Yes, there will always be no remainder.

(ii) Let’s try to understand how Mukta’s trick works. Suppose, the two-digit number is ab. When it is reversed, the new number is ba.
If b> a then ba> ab. So, the difference is (10b+a) – (10a + b) = 10b – b – 10a + a = 9b-9a = 9(b-a)
The difference is divisible by 9.
Can you work out what happens if a> b?
Solution:
The 2-digit number is ab, which can be written as 10a + b.
The reversed number is ba or 10b + a.
If a>b then ab > ba.
So, the difference is
= (10a + b) -(10b + a)
= 10 a + b – 10 b – a
= 10a – a + b – 10b
= 9a – 9b
= 9 (a – b)
The difference is 9(a – b), which is a multiple of 9 and therefore divisible by 9.

Figure it Out (NCERT Page 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:
The quotient is the difference between the digits of the original number.
If the tens digit is larger, the quotient is a-b.
If the unit digit is larger, the quotient is b-a.

Question 2.
In the trick given above, instead of finding the difference of 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:
Yes, it is always true that the sum is divisible by 11.
Let the original 2-digit number be represented by its digits a and b.
The value of the original number can be expressed as 10a + b.
The number obtained by reversing the digits has the value 10b +a.
The sum of these two numbers is
= (10a + b) + (10b + a)
= 10a + a + b + 10b
= 11a + 11b
= 11(a + b)
Since, the sum can always be expressed as 11 multiplied by the sum of its digits (a + b), thus 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:
We know that a 3-digit number abc can be written as 100a + 10b + c.
So, the 6-digit number abcabc can be written as
abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
= 1001(100a + 10b + c)
⇒ abcabc = 100abc
Since, 1001 = 7 × 11 × 13
When divided sequentially by 7,13 and 13, the result is always the original 3-digit number.

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 x be the initial number of flowers and y be the equal number of flowers placed in each shrine.
After the first dip, the number of flowers becomes 2x.
After placing y flowers in shrine 1, the remaining flowers are 2x – y.
The remaining flowers are dipped in the second pond, so the number doubles i.e. 2(2x – y) = 4x – 2y.

After placing y flowers in shrine 2, the remaining flowers = 4x – 2y – y = 4x – 3y
The remaining flowers are dipped in the third pond.
So, the number is doubled
i. e. 2(4x – 3y) = 8x – 6y.
They all are placed in shrine 3, so this quantity must equal toy.
8x – 6y = y
⇒ 8x = 7y
⇒ \(\frac{x}{y}=\frac{7}{8}\)
∴ The smallest positive integer solution is x = 7 and y = 8. Hence, the person started with 7 flowers and placed 8 flowers in each shrine.

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:
Given, the total number of heads of animals = 55
If all 55 animals were hens, each having 2 legs.
∴ The total number of legs = 55 × 2 = 110
If the total number of legs =150
∴ Remaining legs = 150 – 110 = 40
Each horse has 4 legs, which is 2 more than a hen.
This difference of 40 legs is due to the horse.
The number of horse = \(\frac{40}{2}\) = 20
So, the number of hens = 55 – 20 = 35
Hence, there are 20 horses and 35 hens on the farm.

Question 7.
A mother is 5 times her daughter’s age. In 6 yr’s time, the mother will be 3 times her daughter’s age. How old is the daughter now?
Solution:
Let d be the daughter’s current age and m be the mother’s current age.
If a mother is 5 times her daughter’s age.
∴ m = 5d . …(i)
and after 6 yr’s time, the mother will be 3 times her daughter’s age.
∴ m + 6 = 3(d + 6)
⇒ m + 6 = 3d + 18
⇒ 5d – 3d = 18 – 6 [From Eq. (i)]
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
From Eq. (i), we get
m = 5 × 6 = 30
Hence, the daughter is currently 6 yr 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 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 per day and the cost of one dosa= ₹ 10
(i) The total expenses for making 100 dosas
= Fixed rent + Cost per dosa × Number of dosa = 5000 + 10 × 100
= 5000 + 1000
= 6000

To make a profit of ₹ 2000, total revenue needed
= Total expenses + Desired profit
= 6000 + 2000
= 8000
So, selling price per dosa = \(=\frac{\text { Total revenue needed }}{\text { Number of dosas sold }}\)
Hence, the selling price of the dosa should be ₹ 80 to make a profit of ₹ 2000.

(ii) Let M = Number of dosas to sell
∴ Revenue = 50M
and total cost = 5000 + 10 × n = 5000 + 10M
We know that
Revenue – Cost = Profit
⇒ 50M – (5000 + 10 n) = 2000
⇒ 50n – 5000 – 10n = 2000
⇒ 40M = 7000
⇒ M = 175
Hence, need to sell 175 dosas in a day to reach the profit goal at the given price.

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 nodd numbers.]
Solution:
Here, the first three fraction in the sequence are \(\frac{1}{3}, \frac{1+3}{5+7}\)
and \(\frac{1+3+5}{7+9+11}\) i.e. \(\frac{1}{3}, \frac{4}{12}\) and \(\frac{9}{27}\)
So, the value of each fraction in the sequence is consistently.
Since, the sum of the first n odd numbers = n²
Here, the numerators are sums of the first n odd numbers. The denominator is the sum of the next n odd numbers after the first n. The first 2M odd numbers sum to (2M)². The sum of the first n odd numbers is n².
Therefore, the sum of the next n odd numbers (the denominator) = (2M)² – n²
= 4M² – M²
= 3M²
Thus, the Mth term of the sequence is \(\frac{n^2}{3 n^2}=\frac{1}{3} .\)

Question 11.
Karim and 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?
(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 x be the initial number of coins Karim had.
After the first round, the number of coins doubles and he gives away to Genie 8 coins.
∴ The number of coins remaining = 2x – 8 Similarly, for second round,
The number of coins = 2(2x – 8) – 8
= 4x – 16 – 8
= 4x – 24
After third round,
The number of coins = 2(4x – 24) – 8
= 8x – 48 – 8
= 8x – 56
Since, he was left with exactly 8 coins.
∴ 8x – 56 = 8
⇒ 8x = 64
⇒ x = \(\frac{64}{8}\)
x = 8
Hence, Karim initially had 8 coins.

(ii) Let the cost per round be c.
So, the number of coins after one round = 2x – c
For the number of coins to increase, the final amount must be greater than the initial amount.
2x – c > x ⇒ x > c
Karim should agree to the deal if the cost per round is less than 8 coins.

(iii) For the Genie to get all of Karim’s coins, the number of coins remaining after a round must be zero.
∴ 2x – c = 0
⇒ c = 2x
Hence, the Genie should set the cost per second to be exactly twice the number of coins Karim has at the beginning of that round.