Question 1

How would you change this game to make the final answer 3? What about 5?

Accepted Answer

(a) We can understand such tricks through algebra.
Step 1: Think of a number = x
Step 2: Triple it = 3x
Step 3: Add 9 = 3x + 9
Step 4: Divide by 3 = $\frac{3 x+9}{3}$ = x + 3
Step 5: Subtract the original number you thought of (x + 3) – x = 3.
For Example:
Consider a number 23.
Triple it = 3 × 23 = 69
Add 9 = 69 + 9 = 78
Divide by 3 = 78 ÷ 3 = 26
∴ 26 – 23 = 3
(b) To make the final answer 5.
Step 1: Think of a number = x
Step 2: Double it = 2x
Step 3: Add 10; 2x + 10
Step 4: Divide by 2 = $\frac{2 x+10}{2}$ = x + 5
Step 5: Subtract the original number = x + 5 – x = 5

Question 2

Can you come up with more complicated steps that always lead to the same final value?

Accepted Answer

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 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.

Accepted Answer

(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, 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?

Accepted Answer

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 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.

Accepted Answer

(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?

Accepted Answer

When the bottom row uses the first n Virahanka-Fibonacci numbers = (2n – 1)th
∴ The number at the top of the pyramid = (2n – 1)th Virahanka-Fibonacci number.
6.4 Fun with Grids
Math Talk (Page 142)

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?

Accepted Answer

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?

Accepted Answer

Let x, y be the number of cows of Gauri and Naina.
According to the question,
2x = y …(i)
Also, x + 3 = y – 3
x – y = -3 – 3 = -6 ……(ii)
From equations (i) and (ii), we get
x – 2x = -6
⇒ -x = -6
⇒ x = 6
Putting x = 6 in equation (i), we get,
y = 2 × 6 = 12
Thus, Gauri and Naina have 6 and 12 cows, respectively.

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 andfuel) 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?

Accepted Answer

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 dosa.
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 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.]

Accepted Answer

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, 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 st

Accepted Answer

(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 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

Algebra Play