Chapter 6 Algebra Play (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

Given:

We are given the bottom rows of three different pyramids, each with 3 rows. The rule for the number pyramid is that each number is the sum of the two numbers immediately below it.

To Find:

The topmost number for each case without filling the entire pyramid.

Solution:

Let the bottom row be represented by variables $a, b, c$. Then the second row will be $(a + b)$ and $(b + c)$. The topmost row will be the sum of these two:

Now, we apply this general expression to the given cases:

Case 1: Bottom row $[4, 13, 8]$

Case 2: Bottom row $[7, 11, 3]$

Case 3: Bottom row $[10, 14, 25]$

Final Answer:

The topmost numbers for the given pyramids are 38, 32, and 63 respectively.

To Prove/Find:

A general algebraic expression for the top value of a 4-row pyramid starting with bottom row values $a, b, c, d$.

Proof / Solution:

Let the bottom row (Row 1) be: $a, b, c, d$

The next row (Row 2) is formed by summing adjacent pairs:

The next row (Row 3) is formed by summing the values of Row 2:

The topmost row (Row 4) is the sum of the two values in Row 3:

Observation (Indian Perspective):

In Class 8 Algebra, we learn that these coefficients ($1, 3, 3, 1$) correspond to the 4th row of the Meru Prastara (also known as Pascal’s Triangle), which was discussed in ancient Indian mathematics. These patterns are fundamental in expanding binomial expressions like $(x+y)^3$.

Given:

Bottom rows of three different 4-row pyramids. We use the formula derived in the previous question.

Formula:

For a bottom row with values $a, b, c, d$, the topmost value is:

Solution:

Case 1: Bottom row $[8, 19, 21, 13]$

Case 2: Bottom row $[7, 18, 19, 6]$

Case 3: Bottom row $[9, 7, 5, 11]$

Note: The mention of the Virahāṅka sequence highlights the recurring pattern of addition in mathematical structures. Virahāṅka (an Indian mathematician from the 6th-7th century) described these summation patterns long before they were known in the West.

Given:

The first three numbers of the Virahāṅka-Fibonacci sequence are $1, 2,$ and $3$. These are placed in the bottom row of a 3-row pyramid.

To Find:

The numbers in the remaining rows and determine if they all belong to the same sequence.

Solution:

Let us construct the pyramid by adding adjacent numbers to find the level above:

Row 1 (Bottom): $1, 2, 3$

Row 2 (Middle): $(1+2)$ and $(2+3)$ which gives $3$ and $5$

Row 3 (Top): $(3+5)$ which gives $8$

The numbers appearing in the grid (pyramid) are: $1, 2, 3, 3, 5, 8$.

The number at the top is $8$.

Conclusion:

The Virahāṅka-Fibonacci sequence is $1, 2, 3, 5, 8, 13, 21, \dots$

Comparing the numbers in our pyramid ($1, 2, 3, 5, 8$) with the sequence, we find that yes, they are all Virahāṅka-Fibonacci numbers.

(i) Four-row Pyramid with first four numbers:

Solution:

The first four numbers are $1, 2, 3, 5$. Let's build the layers:

Row 1: $1, 2, 3, 5$

Row 2: $3, 5, 8$

Row 3: $8, 13$

Row 4 (Top): $21$

In this case, every number in the pyramid is a Virahāṅka-Fibonacci number. The number at the top is $21$, which is the 7th number in the sequence.

(ii) 29-row Pyramid with first 29 numbers:

Solution:

Based on the pattern observed, we can generalize that if we start with the first $n$ numbers of the sequence, every number generated by the summation rule will also be a member of that sequence.

From our previous observations:

For $n = 3$ rows, the top is $F_5 = 8$ (where $F_k$ is the $k^{th}$ Fibonacci number).

For $n = 4$ rows, the top is $F_7 = 21$.

The relationship for the index of the top number appears to be $(2n - 1)$.

For $n = 29$, the number at the top will be the $57^{th}$ Virahāṅka-Fibonacci number, i.e., $F_{2(29)-1} = F_{57}$.

Conclusion:

All numbers in the $29$-row pyramid will be Virahāṅka-Fibonacci numbers, and the top number will be $F_{57}$.

Given:

A number pyramid with $n$ rows where the bottom row consists of $F_1, F_2, \dots, F_n$ (the first $n$ Virahāṅka-Fibonacci numbers).

Observations and Properties:

1. Numbers within the Pyramid: Every number within the pyramid will be a Virahāṅka-Fibonacci number. This happens because the sequence is defined as $F_{k} = F_{k-1} + F_{k-2}$. The pyramid construction follows a similar summation logic, which maps directly onto the properties of the sequence.

2. Diagonal Pattern: Each diagonal in the pyramid (going upwards) consists of consecutive Fibonacci numbers starting from a higher index.

The Top Number:

As derived from the pattern in previous problems:

Where $n$ is the number of rows in the pyramid. For example, if $n = 5$, the top number would be $F_{2(5)-1} = F_9 = 55$.

Indian Perspective:

The Virahāṅka numbers were originally described by the Indian mathematician Virahāṅka (around 6th-7th century) to count the number of ways to form a meter in Sanskrit poetry using long and short syllables. This exercise shows how ancient Indian patterns in prosody (poetry) can be modeled using modern algebraic structures like number pyramids.

Given:

The available digits are $1, 3,$ and $7$. The required format is a two-digit number multiplied by a one-digit number: $\square \ \square \times \square$.

To Find:

The arrangement of these digits that results in the largest product.

Solution:

To obtain the largest product in the form $(10A + B) \times C$, the most significant positions (the multiplier $C$ and the tens digit $A$) should be filled with the largest available digits. Let's compare the most promising arrangements:

1. Arrangement 1: Use $7$ as the multiplier and $3$ as the tens digit ($31 \times 7$):

$\begin{array}{cc}& 3 & 1 \\ \times & & 7 \\ \hline 2 & 1 & 7 \\ \hline \end{array}$

2. Arrangement 2: Use $3$ as the multiplier and $7$ as the tens digit ($71 \times 3$):

$\begin{array}{cc}& 7 & 1 \\ \times & & 3 \\ \hline 2 & 1 & 3 \\ \hline \end{array}$

Comparing the two results:

Final Answer:

The digits should be filled as $3 \ 1 \times 7$ to get the largest product, which is $217$.

Observation: From an Indian perspective of mental math, a useful rule is that the multiplier should be the largest digit, and the tens place should be the second-largest digit to maximize the "tens" contribution.

Given:

The available digits are $3, 5,$ and $9$. The format is $\square \ \square \times \square$.

To Find:

The arrangement that yields the maximum product.

Solution:

Applying the logic that the largest digit should generally act as the multiplier to scale the entire multiplicand, we test the following primary arrangements:

1. Case A: Largest digit $9$ as the multiplier and $5$ in the tens place ($53 \times 9$):

$\begin{array}{cc}& 5 & 3 \\ \times & & 9 \\ \hline 4 & 7 & 7 \\ \hline \end{array}$

2. Case B: Second largest digit $5$ as the multiplier and $9$ in the tens place ($93 \times 5$):

$\begin{array}{cc}& 9 & 3 \\ \times & & 5 \\ \hline 4 & 6 & 5 \\ \hline \end{array}$

Comparing these two products:

Final Answer:

The digits should be filled as $5 \ 3 \times 9$ to result in the largest product, $477$.

Alternate Solution (Algebraic Verification):

Let digits be $x, y, z$ such that $x < y < z$. We compare $(10y+x)z$ and $(10z+x)y$.

$(10y+x)z = 10yz + xz$

$(10z+x)y = 10zy + xy$

Since $z > y$, it follows that $xz > xy$. Therefore, placing the largest digit in the multiplier position ($z$) and the second-largest in the tens position ($y$) will always yield the higher value.

Given:

The "trick" involves taking a $2$-digit number, reversing its digits, and finding the difference between the two numbers. Let the digits be $x$ and $y$.

To Find:

The quotient when the difference is divided by $9$ and its relationship with the digits.

Solution:

Let the $2$-digit number be $10x + y$. When we reverse the digits, the new number is $10y + x$.

The difference between these numbers is:

When we divide this difference by $9$:

Conclusion:

The quotient is always $x - y$. The relationship is that the quotient is equal to the difference between the two digits of the original number.

Given:

A $2$-digit number is added to its reverse.

To Prove:

The sum of a $2$-digit number and its reverse is always divisible by $11$.

Proof:

Let the $2$-digit number be $ab$, which can be written as $10a + b$.

When we reverse the digits, the new number is $ba$, which is $10b + a$.

Now, finding the sum of the two numbers:

Conclusion:

Since the sum is expressed as $11 \times (a + b)$, where $(a + b)$ is an integer (the sum of the digits), it is always divisible by $11$. The quotient will be the sum of the digits.

Given:

Three $3$-digit numbers formed by cyclic permutations of digits: $abc$, $bca$, and $cab$.

To Prove:

The sum of these three numbers is divisible by both $37$ and $3$.

Solution:

Let's write the expanded form of the three numbers:

Adding the three expressions:

We can factor $111$ as follows:

Therefore, we can rewrite the sum as:

Conclusion:

Since the sum has factors of both $3$ and $37$, it is always divisible by $37$ and also always divisible by $3$. This mathematical beauty is often used in Vedic Math puzzles in India to quickly find factors of repeated digit sums.

Given:

A $6$-digit number formed by repeating a $3$-digit number, $abcabc$.

To Find:

The result after dividing by $7$, $11$, and $13$, and the reason why this result occurs.

Solution:

Let's represent the $6$-digit number algebraically:

Now, let's look at the product of the three divisors mentioned:

Calculation of $7 \times 11 \times 13$:

Final Observation:

When you divide $abcabc$ by $7$, $11$, and $13$ sequentially, you are effectively dividing the number by $1001$.

Result: You get the original $3$-digit number ($abc$) back. This works because $1001$ is the unique product of these three prime numbers, and multiplying any $3$-digit number by $1001$ creates a repeated pattern ($abcabc$).

Given:

1. There are three shrines with magical ponds that double the flowers.

2. An equal number of flowers are placed in each shrine.

3. After placing flowers in the third shrine, zero flowers are left.

To Find:

The initial number of flowers and the number of flowers placed in each shrine.

Solution:

Let the initial number of flowers be $y$ and the number of flowers placed in each shrine be $x$.

Step 1: At the first shrine

The person dips $y$ flowers, they become $2y$. He places $x$ flowers.

Step 2: At the second shrine

He dips $(2y - x)$ flowers, they double to $2(2y - x)$. He places $x$ flowers.

Step 3: At the third shrine

He dips $(4y - 3x)$ flowers, they double to $2(4y - 3x)$. He places $x$ flowers and is left with $0$.

Final Answer:

The simplest ratio is $7:8$. Therefore, the person started with 7 flowers and placed 8 flowers in each shrine.

Verification:

Start with 7 $\to$ Double (14) $\to$ Place 8 (Left 6) $\to$ Double (12) $\to$ Place 8 (Left 4) $\to$ Double (8) $\to$ Place 8 (Left 0). The condition is satisfied.

Given:

Total heads = $55$

Total legs = $150$

Solution (Using Algebra):

Let the number of horses be $h$ and the number of hens be $e$.

Multiplying equation (i) by $2$:

Subtracting (iii) from (ii):

$\begin{array}{cc} & 4h & + & 2e & = & 150 \\ - & 2h & + & 2e & = & 110 \\ \hline & 2h & & & = & 40 \\ \hline \end{array}$

Substituting $h = 20$ in equation (i): $20 + e = 55 \implies e = 35$.

Alternate Solution (Without Algebra / Indian Mental Math Perspective):

1. Suppose all $55$ animals are hens. Then each has $2$ legs.

2. But the actual number of legs is $150$. Let's find the extra legs:

3. Each horse has $4$ legs, which is $2$ more than a hen. These $40$ extra legs come from horses having $2$ extra legs each.

Final Answer: There are 20 horses and 35 hens on the farm.

Given:

In $6$ years, the mother will be $3$ times the daughter's age.

To Find:

The current age of the daughter.

Solution:

Let the current age of the daughter be $d$ years.

After $6$ years:

According to the problem:

Final Answer:

The daughter is currently 6 years old. (The mother's age is $30$ years).

Given:

1. Naina has twice as many cows as Gauri.

2. If Naina gives $3$ cows to Gauri, they will both have an equal number of cows.

To Find:

The number of cows Gauri and Naina each possess.

Solution:

Let the number of cows Gauri has be $x$.

Then, according to the first statement, the number of cows Naina has is $2x$.

Now, if Naina gives $3$ cows to Gauri:

Number of cows Gauri will have = $x + 3$

Number of cows Naina will have = $2x - 3$

According to the second condition, these quantities are equal:

Transposing $x$ to the Left Hand Side (LHS) and $-3$ to the Right Hand Side (RHS):

Now, finding Naina's cows using equation (i):

Final Answer:

Gauri has $6$ cows and Naina has $12$ cows.

Verification: If Naina (12) gives 3 to Gauri (6), Naina has 9 and Gauri has 9. They are equal.

Given:

Fixed Cost (Rent) = $\textsf{₹}5000$ per day.

Variable Cost per dosa = $\textsf{₹}10$.

(i) To Find: Selling price for $100$ dosas to earn $\textsf{₹}2000$ profit.

Let $S$ be the selling price per dosa.

The selling price should be $\textsf{₹}80$ per dosa.

(ii) To Find: Number of dosas ($n$) to sell at $\textsf{₹}50$ each for $\textsf{₹}2000$ profit.

I should aim to sell $175$ dosas per day.

Indian Perspective:

Running a small business like a dosa cart is a common micro-entrepreneurial activity in India. Understanding the break-even point and pricing is vital for survival and profit in local markets.

To Find: Evaluate the terms and explain the observation.

Evaluation:

Term 1:

Term 2:

Term 3:

Observation:

We observe that every term in the sequence evaluates to $\frac{1}{3}$.

Algebraic Explanation:

We know that the sum of the first $n$ odd numbers is $n^2$.

The denominator is the sum of the next $n$ odd numbers. This is equivalent to (Sum of first $2n$ odd numbers) $-$ (Sum of first $n$ odd numbers).

Taking the ratio:

This explains why the result is always $\frac{1}{3}$ regardless of how many terms are added in the pattern.