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

Solution:

---

1. To make the final answer $3$:

Let us modify step $3$ (adding $6$ instead of $4$) or adjust the steps using algebra:

1. Think of a number: $x$

2. Double it: $2x$

3. Add six: $2x + 6$

4. Divide by $2$: $x + 3$

5. Subtract the original number: $x + 3 - x = 3$

---

2. To make the final answer $5$:

Let us modify step $3$ by adding $10$:

1. Think of a number: $x$

2. Double it: $2x$

3. Add ten: $2x + 10$

4. Divide by $2$: $x + 5$

5. Subtract the original number: $x + 5 - x = 5$

Solution:

---

1. Designing a more complicated number trick:

Let us create a new set of steps using algebra to ensure the variable $x$ cancels out and leaves a constant final value (say, $7$).

1. Think of a number: $x$

2. Multiply by $3$: $3x$

3. Add $14$: $3x + 14$

4. Add your original number: $3x + 14 + x = 4x + 14$

5. Divide by $2$: $\frac{4x + 14}{2} = 2x + 7$

6. Subtract twice your original number: $2x + 7 - 2x = 7$

No matter what number you start with, following these steps will always result in $7$!

Solution:

---

1. Using algebra to decode the trick:

Let the month be $m$ and the day be $d$.

Step 1: Think of a date: Month $m$, Day $d$

Step 2: Multiply the month by $5$: $5m$

Step 3: Add $6$: $5m + 6$

Step 4: Multiply by $4$: $4(5m + 6) = 20m + 24$

Step 5: Add $9$: $20m + 24 + 9 = 20m + 33$

Step 6: Multiply by $5$: $5(20m + 33) = 100m + 165$

Step 7: Add the day: $100m + 165 + d$

---

2. How Shubham finds the answer:

The final expression Mukta gets is $100m + 165 + d$.

When Mukta tells Shubham her final answer (e.g., $291$), Shubham simply subtracts $165$ from that number:

$291 - 165 = 126$

In the resulting number $126$, the hundreds digit ($1$) represents the month ($m$), and the remaining digits ($26$) represent the day ($d$).

Thus, Shubham easily figures out the date as $26/01$ (January $26^{\text{th}}$).

Solution:

---

1. Decoding the final answer:

From the algebraic derivation of the date trick, the final reported number is given by:

$\text{Final Answer} = 100m + 165 + d$

Where $m$ is the month and $d$ is the day.

2. Calculating the month and day:

First, subtract the fixed constant $165$ from the reported answer ($1390$):

$1390 - 165 = 1225$

Now, separate the result into the month and the day:

The hundreds digit ($12$) represents the month: $m = 12$ (December)

The remaining digits ($25$) represent the day: $d = 25$

Therefore, Mukta started with the date $25/12$ (December $25^{\text{th}}$).

Solution:

---

(i) For the final answer $1269$:

Subtract $165$ from $1269$:

$1269 - 165 = 1104$

Month ($m$) = $11$ (November)

Day ($d$) = $04$ ($4^{\text{th}}$)

Date: $04/11$ (November $4^{\text{th}}$)

---

(ii) For the final answer $394$:

Subtract $165$ from $394$:

$394 - 165 = 229$

Month ($m$) = $2$ (February)

Day ($d$) = $29$ ($29^{\text{th}}$)

Date: $29/02$ (February $29^{\text{th}}$)

---

(iii) For the final answer $296$:

Subtract $165$ from $296$:

$296 - 165 = 131$

Month ($m$) = $1$ (January)

Day ($d$) = $31$ ($31^{\text{st}}$)

Date: $31/01$ (January $31^{\text{st}}$)

Solution:

---

1. Modifying the steps:

Yes, we can change the numbers added in the intermediate steps, as long as the sequence of multiplications still results in multiplying the month by $100$ overall.

Let us try a modified version using algebra where month is $m$ and day is $d$:

Step 1: Think of a month: $m$

Step 2: Multiply by $5$: $5m$

Step 3: Add $4$ (instead of $6$): $5m + 4$

Step 4: Multiply by $4$: $4(5m + 4) = 20m + 16$

Step 5: Add $5$ (instead of $9$): $20m + 16 + 5 = 20m + 21$

Step 6: Multiply by $5$: $5(20m + 21) = 100m + 105$

Step 7: Add the day: $100m + 105 + d$

---

2. Finding the original date:

In this modified trick, the final algebraic expression is $100m + d + 105$.

To find the original date, instead of subtracting $165$, you would now subtract $105$ from the final answer.

The remaining number will easily reveal the month (in the hundreds or thousands place) and the day (in the tens and ones place).

Solution:

---

1. My 'Think of a Number' trick:

Here is a trick where the final answer will always end up being $10$.

Step 1: Think of a number.

Step 2: Multiply the number by $2$.

Step 3: Add $20$ to the result.

Step 4: Divide the new number by $2$.

Step 5: Subtract the original number you thought of.

Your final answer is $10$!

---

2. How it works algebraically:

Let the starting number be $x$.

Step 1: $x$

Step 2: $2x$

Step 3: $2x + 20$

Step 4: $\frac{2x + 20}{2} = x + 10$

Step 5: $x + 10 - x = 10$

By using algebra, we can see that subtracting the original number ($x$) causes the variable to cancel out completely, leaving a constant value of exactly $10$ every single time.

Solution:

---

1. First Pyramid (2 tiers):

Bottom row numbers: $6$ and $2$.

Top box = $6 + 2 = 8$.

---

2. Second Pyramid (3 tiers):

Bottom row numbers: $3, 4, 3$.

Middle row left box = $3 + 4 = 7$.

Middle row right box = $4 + 3 = 7$.

Top box = $7 + 7 = 14$.

---

3. Third Pyramid (4 tiers):

Bottom row numbers: $5, 4, 5, 0$.

Second row from bottom (left to right):

Box $1 = 5 + 4 = 9$

Box $2 = 4 + 5 = 9$

Box $3 = 5 + 0 = 5$

Third row from bottom (left to right):

Box $1 = 9 + 9 = 18$

Box $2 = 9 + 5 = 14$

Top box = $18 + 14 = 32$.

Solution:

---

1. First Pyramid:

Let us assign a variable to each of the empty boxes in the first pyramid.

Let the bottom row be $4, a, 6, b$.

Let the second row from the bottom be $c, d, e$.

Let the third row from the bottom be $f, 22$.

The top row is $50$.

In a number pyramid, each box is the sum of the two boxes directly below it.

From the top row, we have:

$f + 22 = 50$

$f = 50 - 22 = 28$

From the second and third rows, since $f = 28$, we have the following relationships:

From the bottom and second rows, we have:

$c = 4 + a$

$d = a + 6$

$e = 6 + b$

Substitute the expressions for $c$ and $d$ into equation (i):

$(4 + a) + (a + 6) = 28$

$2a + 10 = 28$

$2a = 18$

$a = 9$

Now, substitute the expressions for $d$ and $e$ into equation (ii):

$(a + 6) + (6 + b) = 22$

$a + b + 12 = 22$

Since $a = 9$:

$9 + b + 12 = 22$

$b + 21 = 22$

$b = 1$

Now we can find the remaining values:

$c = 4 + 9 = 13$

$d = 9 + 6 = 15$

$e = 6 + 1 = 7$

---

2. Second Pyramid:

Let us assign variables to the empty boxes in the second pyramid.

Let the bottom row be $5, a, 7, b$.

Let the second row from the bottom be $c, d, 9$.

Let the third row from the bottom be $40, e$.

Let the top box be $f$.

Using the sum rule from the bottom right:

$7 + b = 9$

$b = 2$

Now, express $c$ and $d$ in terms of $a$:

$c = 5 + a$

$d = a + 7$

We know that the sum of $c$ and $d$ gives the box above them ($40$):

Substitute the expressions for $c$ and $d$ into equation (iii):

$(5 + a) + (a + 7) = 40$

$2a + 12 = 40$

$2a = 28$

$a = 14$

Now we can find $c$ and $d$:

$c = 5 + 14 = 19$

$d = 14 + 7 = 21$

Next, find $e$ using the second row:

$e = d + 9$

$e = 21 + 9 = 30$

Finally, find the top box $f$:

$f = 40 + e$

$f = 40 + 30 = 70$

---

3. Third Pyramid:

Let us assign variables to the empty boxes in the third pyramid.

Let the bottom row be $3, 5, a, b$.

Let the second row from the bottom be $c, d, 7$.

Let the third row from the bottom be $e, f$.

The top row is $35$.

First, find $c$ from the bottom row:

$c = 3 + 5 = 8$

Express $d$ and the relationship for $7$ in terms of $a$ and $b$:

$d = 5 + a$

Express $e$ and $f$ using the second row:

$e = c + d = 8 + (5 + a) = 13 + a$

$f = d + 7 = (5 + a) + 7 = 12 + a$

We know the top box is $35$, so:

Substitute the expressions for $e$ and $f$ into equation (v):

$(13 + a) + (12 + a) = 35$

$2a + 25 = 35$

$2a = 10$

$a = 5$

Now substitute the value of $a$ into equation (iv):

$5 + b = 7$

$b = 2$

Finally, find the remaining boxes:

$d = 5 + 5 = 10$

$e = 13 + 5 = 18$

$f = 12 + 5 = 17$

Solution:

---

1. General Algebraic Formulation:

Let us presume the numbers at the bottom row are $a$, $b$, and $c$.

In a number pyramid, each box is the sum of the two boxes directly below it.

Therefore, the middle row boxes will be $(a + b)$ and $(b + c)$.

The topmost box will be the sum of the middle row boxes: $(a + b) + (b + c) = a + 2b + c$.

So, the number in the topmost row is always given by the formula: $a + 2b + c$, where $b$ is the middle number.

---

2. Case 1: Bottom row is $4, 13, 8$

Here, $a = 4$, $b = 13$, and $c = 8$.

Using the formula for the topmost number:

$\text{Topmost number} = a + 2b + c$

$\text{Topmost number} = 4 + 2(13) + 8$

$\text{Topmost number} = 4 + 26 + 8 = 38$

The number in the topmost row is $38$.

---

3. Case 2: Bottom row is $7, 11, 3$

Here, $a = 7$, $b = 11$, and $c = 3$.

Using the formula for the topmost number:

$\text{Topmost number} = a + 2b + c$

$\text{Topmost number} = 7 + 2(11) + 3$

$\text{Topmost number} = 7 + 22 + 3 = 32$

The number in the topmost row is $32$.

---

4. Case 3: Bottom row is $10, 14, 25$

Here, $a = 10$, $b = 14$, and $c = 25$.

Using the formula for the topmost number:

$\text{Topmost number} = a + 2b + c$

$\text{Topmost number} = 10 + 2(14) + 25$

$\text{Topmost number} = 10 + 28 + 25 = 63$

The number in the topmost row is $63$.

Solution:

---

1. Setting up the base variables:

Let the $4$ values in the bottom row be $a, b, c,$ and $d$.

2. Building the pyramid row by row:

In a number pyramid, each box is the sum of the two boxes directly below it.

Row 1 (Bottom):

$a \quad b \quad c \quad d$

Row 2 (3 boxes):

Box 1 = $a + b$

Box 2 = $b + c$

Box 3 = $c + d$

Row 3 (2 boxes):

Box 1 = $(a + b) + (b + c) = a + 2b + c$

Box 2 = $(b + c) + (c + d) = b + 2c + d$

Row 4 (Topmost row, 1 box):

Top box = $(a + 2b + c) + (b + 2c + d)$

Top box = $a + 2b + b + c + 2c + d$

Top box = $a + 3b + 3c + d$

---

3. Conclusion:

The expression for the topmost row of a pyramid with $4$ rows, given the bottom row values $a, b, c, d$, is $a + 3b + 3c + d$.

Solution:

---

1. General Algebraic Formula:

From the previous question, we know that for a $4$-row number pyramid with bottom row values $a, b, c,$ and $d$, the topmost number can be directly calculated using the expression:

$\text{Topmost number} = a + 3b + 3c + d$

---

2. Case 1: Bottom row is $8, 19, 21, 13$

Here, $a = 8$, $b = 19$, $c = 21$, and $d = 13$.

Using the formula:

$\text{Topmost number} = a + 3b + 3c + d$

$\text{Topmost number} = 8 + 3(19) + 3(21) + 13$

$\text{Topmost number} = 8 + 57 + 63 + 13 = 141$

The number in the topmost row is $141$.

---

3. Case 2: Bottom row is $7, 18, 19, 6$

Here, $a = 7$, $b = 18$, $c = 19$, and $d = 6$.

Using the formula:

$\text{Topmost number} = a + 3b + 3c + d$

$\text{Topmost number} = 7 + 3(18) + 3(19) + 6$

$\text{Topmost number} = 7 + 54 + 57 + 6 = 124$

The number in the topmost row is $124$.

---

4. Case 3: Bottom row is $9, 7, 5, 11$

Here, $a = 9$, $b = 7$, $c = 5$, and $d = 11$.

Using the formula:

$\text{Topmost number} = a + 3b + 3c + d$

$\text{Topmost number} = 9 + 3(7) + 3(5) + 11$

$\text{Topmost number} = 9 + 21 + 15 + 11 = 56$

The number in the topmost row is $56$.

Solution:

---

1. Setting up the base:

The first three numbers of the Virahāṅka-Fibonacci sequence (as given in the prompt context) are $1, 2,$ and $3$.

Place these numbers in the bottom row of a $3$-row pyramid: $1 \quad 2 \quad 3$.

2. Filling in the rest of the pyramid:

In the second row (middle row):

Left box = $1 + 2 = 3$

Right box = $2 + 3 = 5$

In the top row:

Top box = $3 + 5 = 8$

---

3. Analyzing the numbers in the grid:

The numbers that appear in the grid are: $1, 2, 3$ (bottom row), $3, 5$ (middle row), and $8$ (top box).

The number at the top is $8$.

Yes, all the numbers that appear in this specific pyramid ($1, 2, 3, 5, 8$) belong exactly to the sequence of Virahāṅka-Fibonacci numbers.

Solution:

---

(i) Four row pyramid with the first four Virahāṅka-Fibonacci numbers:

The first four numbers of the Virahāṅka-Fibonacci sequence are $1, 2, 3,$ and $5$.

Bottom row: $1, 2, 3, 5$

Second row: $1+2=3$, $2+3=5$, $3+5=8$

Third row: $3+5=8$, $5+8=13$

Top row: $8+13=21$

Observation: All the numbers that appear anywhere in the pyramid ($1, 2, 3, 5, 8, 13, 21$) are Virahāṅka-Fibonacci numbers. The number at the top is $21$, which is the $7^{\text{th}}$ Virahāṅka-Fibonacci number.

---

(ii) $29$ row pyramid with the first $29$ Virahāṅka-Fibonacci numbers:

Based on the pattern observed in the $3$-row and $4$-row pyramids, placing consecutive Virahāṅka-Fibonacci numbers at the base always generates subsequent Virahāṅka-Fibonacci numbers in the rows above because the fundamental rule of the pyramid (adding two adjacent numbers) exactly mirrors the fundamental rule of the Fibonacci sequence.

Observation: We can conclude that every single number appearing in any cell of this $29$-row pyramid will be a Virahāṅka-Fibonacci number.

To find which Fibonacci number is at the top:

For a $3$-row pyramid (base ends at $3^{\text{rd}}$ Fibonacci number), top is the $5^{\text{th}}$ Fibonacci number ($3 \times 2 - 1 = 5$).

For a $4$-row pyramid (base ends at $4^{\text{th}}$ Fibonacci number), top is the $7^{\text{th}}$ Fibonacci number ($4 \times 2 - 1 = 7$).

Following this linear pattern, for an $n$-row pyramid, the top number is the $(2n - 1)^{\text{th}}$ Virahāṅka-Fibonacci number.

For a $29$-row pyramid, the number at the top will be the $(2 \times 29 - 1) = 57^{\text{th}}$ Virahāṅka-Fibonacci number.

Solution:

---

Let us analyze the structure of an $n$-row pyramid whose bottom row consists of the first $n$ Virahāṅka-Fibonacci numbers.

---

1. Numbers in the pyramid:

Every number that appears in any cell of the $n$-row pyramid is a Virahāṅka-Fibonacci number. This occurs because the construction rule of the pyramid—where each entry is the sum of the two numbers directly beneath it—precisely mirrors the recurrence relation of the Virahāṅka-Fibonacci sequence:

---

2. Number at the top of the pyramid:

By generalizing the patterns established in smaller pyramids (such as $3$-row and $4$-row pyramids), the position index of the Fibonacci number at the apex of the pyramid increases by $2$ for each additional row added to the base.

Thus, for an $n$-row pyramid, the number located at the top will always be the $(2n - 1)^{\text{th}}$ Virahāṅka-Fibonacci number.

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 $217$.

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, $477$.

To Find:

The quotient when the difference between a two-digit number and its reversed number is divided by $9$, and the relationship between the two digits and the quotient.

---

Solution:

---

From the given problem, the difference between the reversed number $ba$ and the original number $ab$ (where $b > a$) is given by:

When this difference is divided by $9$, we get:

Relationship:

The quotient obtained after dividing the difference by $9$ is precisely equal to the difference between the two digits ($b$ and $a$) 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.

Solution:

---

Let us take any $3$-digit number, say $abc$. By repeating its digits, we form the $6$-digit number $abcabc$.

We can express the $6$-digit number $abcabc$ in expanded form based on its place values as follows:

$abcabc = a \times 100000 + b \times 10000 + c \times 1000 + a \times 100 + b \times 10 + c \times 1$

$abcabc = 100000a + 10000b + 1000c + 100a + 10b + c$

Grouping the terms containing $a$, $b$, and $c$ together:

$abcabc = (100000a + 100a) + (10000b + 10b) + (1000c + c)$

$abcabc = 100100a + 10010b + 1001c$

Factoring out $1001$ from the expression:

$abcabc = 1001 \times (100a + 10b + c)$

Since $(100a + 10b + c)$ is simply the original $3$-digit number $abc$, we can write:

$abcabc = 1001 \times abc$

---

Applying the Hint:

Let us find the product of the given divisors $7$, $11$, and $13$:

$7 \times 11 \times 13 = 77 \times 13 = 1001$

---

Conclusion:

When the $6$-digit number $abcabc$ is divided successively by $7$, $11$, and $13$, it is equivalent to dividing the number by their product, which is $1001$:

$\frac{abcabc}{7 \times 11 \times 13} = \frac{1001 \times abc}{1001} = abc$

Therefore, when you divide the $6$-digit number $abcabc$ by $7$, then by $11$, and finally by $13$, you get back the original $3$-digit number $abc$.

To Find:

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

---

Solution:

---

Let the initial number of flowers the person started with be $x$.

Let the number of flowers placed in each of the three shrines be $E$ (since an equal number of flowers, say $E$, is placed in each shrine).

Let us work backwards from the final shrine or use a step-by-step algebraic approach.

Step 1: After the third shrine

Let the number of flowers remaining after placing flowers in the second shrine be $R_2$.

These remaining flowers are dipped into the third pond, so they double:

$2 \times R_2$

He places all of these remaining flowers in the third shrine. Since he places an equal number of flowers $E$ in each shrine:

$2 R_2 = E \implies R_2 = \frac{E}{2}$

Step 2: After the second shrine

Let the number of flowers remaining after placing flowers in the first shrine be $R_1$.

These flowers are dipped into the second pond, so they double: $2 R_1$.

From this doubled amount, he places $E$ flowers in shrine $2$, leaving $R_2$ flowers:

$2 R_1 - E = R_2$

Substituting $R_2 = \frac{E}{2}$ into the equation:

$2 R_1 - E = \frac{E}{2}$

$2 R_1 = E + \frac{E}{2} = \frac{3E}{2}$

$R_1 = \frac{3E}{4}$

Step 3: After the first shrine

The initial number of flowers $x$ is dipped into the first pond, so they double: $2x$.

From this doubled amount, he places $E$ flowers in shrine $1$, leaving $R_1$ flowers:

$2x - E = R_1$

Substituting $R_1 = \frac{3E}{4}$ into the equation:

$2x - E = \frac{3E}{4}$

$2x = E + \frac{3E}{4} = \frac{7E}{4}$

$x = \frac{7E}{8}$

Since the number of flowers $x$ and $E$ must be whole numbers, $7E$ must be divisible by $8$. The smallest positive integer value for $E$ that makes $\frac{7E}{8}$ a whole number is $8$.

If $E = 8$ (flowers placed in each shrine):

Initial flowers $x = \frac{7 \times 8}{8} = 7$

---

Verification:

Initial flowers: $7$

1. Dip in first pond: $7 \times 2 = 14$ flowers.

Place $8$ flowers in shrine $1$: Remaining = $14 - 8 = 6$ flowers.

2. Dip in second pond: $6 \times 2 = 12$ flowers.

Place $8$ flowers in shrine $2$: Remaining = $12 - 8 = 4$ flowers.

3. Dip in third pond: $4 \times 2 = 8$ flowers.

Place all $8$ flowers in shrine $3$: Remaining = $8 - 8 = 0$ flowers.

---

Conclusion:

The person started with $7$ flowers and placed $8$ flowers in each shrine.

To Find:

The number of horses and hens on the farm.

---

Solution (Using Arithmetic / Without Letter-Numbers):

---

Total number of heads (animals) = $55$

Total number of legs = $150$

We know that a hen has $2$ legs and a horse has $4$ legs.

Step 1: Calculate legs assuming all animals are hens

If all $55$ animals were hens, the total number of legs would be:

$55 \times 2 = 110 \text{ legs}$

Step 2: Find the difference in legs

The actual number of legs given in the problem is $150$. The difference between the actual legs and our assumed legs is:

$150 - 110 = 40 \text{ extra legs}$

Step 3: Calculate the number of horses

Each time we replace a hen with a horse, the number of legs increases by $2$ ($4 - 2 = 2$). To account for the $40$ extra legs, the number of horses must be:

$\frac{40}{2} = 20 \text{ horses}$

Step 4: Calculate the number of hens

Since the total number of animals (heads) is $55$, the number of hens is:

$55 - 20 = 35 \text{ hens}$

---

Alternate Solution (Using Algebraic Method):

---

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

According to the given condition for total heads:

According to the given condition for total legs ($2$ legs for each hen and $4$ legs for each horse):

Multiplying equation (i) by $2$:

Subtracting equation (iii) from equation (ii):

$(2h + 4H) - (2h + 2H) = 150 - 110$

$2H = 40$

$H = 20$

Substituting $H = 20$ into equation (i):

$h + 20 = 55$

$h = 55 - 20 = 35$

---

Conclusion:

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.

To Find:

Evaluation of the given sequence of fractions, observation of any pattern, and explanation of why it happens.

---

Solution:

---

Let us evaluate each fraction in the given sequence one by one:

1. First fraction:

$\frac{1}{3} = \frac{1}{3}$

2. Second fraction:

$\frac{1+3}{5+7} = \frac{4}{12} = \frac{\cancel{4}^1}{\cancel{12}_3} = \frac{1}{3}$

3. Third fraction:

$\frac{1+3+5}{7+9+11} = \frac{9}{27} = \frac{\cancel{9}^1}{\cancel{27}_3} = \frac{1}{3}$

---

Observation:

Every single fraction in the given sequence simplifies to $\frac{1}{3}$.

---

Explanation (Why this happens):

Let us use the hint regarding the sum of the first $n$ odd numbers. We know that the sum of the first $n$ odd natural numbers is always equal to $n^2$, i.e.,

Now, let us examine the $n^{\text{th}}$ term in this sequence of fractions:

• The numerator consists of the sum of the first $n$ odd numbers, which is $n^2$.

• The denominator consists of the sum of $n$ consecutive odd numbers starting from $(2n + 1)$.

• This sum represents the difference between the sum of the first $2n$ odd numbers and the first $n$ odd numbers.

• Using the property of the sum of odd numbers, this can be written as $(2n)^2 - n^2$.

• Simplifying this expression gives $4n^2 - n^2 = 3n^2$.

Thus, any general term in the sequence can be written as:

$\frac{\text{Sum of first } n \text{ odd numbers}}{\text{Sum of } n \text{ odd numbers starting from } 2n+1} = \frac{n^2}{3n^2}$

Canceling $n^2$ from both the numerator and the denominator:

$\frac{n^2}{3n^2} = \frac{\cancel{n^2}^1}{\cancel{3n^2}_3} = \frac{1}{3}$

This algebraic consistency ensures that regardless of how far the sequence extends, every fraction will always evaluate to $\frac{1}{3}$.