Chapter 4 Expressions using Letter-Numbers (Class 7 - Latest Maths NCERT (Ganita Prakash I) NCERT Solutions)

Given:

Cost of one coconut = $\textsf{₹}35$

Cost of $1\text{ kg}$ jaggery = $\textsf{₹}60$

To Find:

Total amount to be paid for $8$ coconuts and $9\text{ kg}$ jaggery.

Solution:

The total cost is the sum of the cost of the coconuts and the cost of the jaggery.

Now, we add these two amounts to get the total payment:

Total Amount = $\textsf{₹}820$

Final Answer:

Ketaki should pay a total of $\textsf{₹}820$.

Solution:

The perimeter of any closed figure is the total length of its boundary. For a regular polygon with $n$ sides, where each side length is $s$, the perimeter is $n \times s$.

(a) Triangle with all sides equal (Equilateral Triangle):

A triangle has $3$ sides. If each side is $s$, then:

$\text{Perimeter} = 3 \times s$

(b) Regular Pentagon:

A pentagon has $5$ sides. For a regular pentagon with side length $s$:

$\text{Perimeter} = 5 \times s$

(c) Regular Hexagon:

A hexagon has $6$ sides. For a regular hexagon with side length $s$:

$\text{Perimeter} = 6 \times s$

Given:

Length of the first pipe = $20\text{ m}$

Length of the second pipe = $k\text{ m}$

Solution:

To find the combined length, we need to add the lengths of the two pipes that have been joined together.

Combined length = Length of first pipe $+$ Length of second pipe

Combined length = $(20 + k)\text{ m}$

In this algebraic expression, $20$ is a constant and $k$ is a variable representing the unknown length of the added pipe.

Solution:

To find the total amount, we multiply the number of notes by their respective denominations and sum them up.

In the final row, the total amount is expressed as an algebraic expression using the variables $x, y,$ and $z$.

To Find: The algebraic expression representing the total time taken to grind $y$ kg of grain.

Solution:

According to the problem:

Fixed time to start the mill = $10$ seconds.

Grinding time per kg of grain = $8$ seconds.

Time taken to grind $y$ kg of grain = $8 \times y$ seconds.

Total time taken = (Fixed start time) + (Time for $y$ kg grain)

Total time = $10 + 8 \times y$

Final Answer:

Comparing this with the given options, the correct expression is (d) $10 + 8 \times y$.

Solution:

Let the number be represented by the letter-number $n$.

(a) $5$ more than a number:

This means we add $5$ to the number.

Expression: $n + 5$

(b) $4$ less than a number:

This means we subtract $4$ from the number.

Expression: $n - 4$

(c) $2$ less than $13$ times a number:

$13$ times a number is $13n$. Subtracting $2$ from this gives:

Expression: $13n - 2$

(d) $13$ less than $2$ times a number:

$2$ times a number is $2n$. Subtracting $13$ from this gives:

Expression: $2n - 13$

Solution:

(a) Expression: $8 \times x + 3 \times y$

Situation: Suppose a student buys $8$ notebooks costing $\textsf{₹} x$ each and $3$ pens costing $\textsf{₹} y$ each. The expression represents the total amount of money the student spends at the stationary shop.

(b) Expression: $15 \times j - 2 \times k$

Situation: A fruit seller has $15$ boxes of apples, and each box contains $j$ apples. While checking, he finds that $2$ boxes are completely rotten, and each of those boxes had $k$ apples. The expression represents the total number of good apples left with the seller.

Given:

A $2 \times 3$ grid from a calendar. The bottom middle cell is $w$.

Solution:

In any calendar:

1. Dates in the same row differ by $1$. The date to the left is $1$ less, and the date to the right is $1$ more.

2. Dates in the same column differ by $7$. The date in the row above is $7$ less than the date below it.

Filling the Bottom Row:

Middle cell $= w$

Left cell $= w - 1$

Right cell $= w + 1$

Filling the Top Row:

Middle cell (above $w$) $= w - 7$

Left cell (above $w-1$) $= (w - 1) - 7 = w - 8$

Right cell (above $w+1$) $= (w + 1) - 7 = w - 6$

Final Answer:

The completed grid table is:

To Find: To evaluate the expressions $5u$ and $5 + u$ for given values of $u$ and compare them.

Solution:

We substitute the values of $u$ into the center expression to find the result for each outer box.

Evaluations for $5u$:

Evaluations for $5 + u$:

Comparison:

When we compare the values of $5u$ (multiplication) and $5 + u$ (addition), we observe that for larger values of $u$, the expression $5u$ increases much faster than $5 + u$. For example, at $u = 11$, $55$ is significantly greater than $16$.

Evaluations for $10y - 3$:

Evaluations for $10(y - 3)$:

Conclusion:

No, the two expressions are not equal. According to the distributive property, $10(y - 3)$ is equal to $10y - 30$, which is different from $10y - 3$. The evaluation results confirm this as none of the pairs of output values match.

To Find: To write and simplify the algebraic expressions for the three groups of terms shown in the image using different methods of addition.

Group 1: Terms involving $x$ and $y$

Way 1: Adding by rows

First row terms: $5y$, $-6$, $x$

Second row terms: $x$, $2$, $5y$

Expression: $(5y - 6 + x) + (x + 2 + 5y)$

Combining like terms: $(5y + 5y) + (x + x) + (-6 + 2)$

Simplified Result: $10y + 2x - 4$

Way 2: Adding like terms directly (Alternate Solution)

Total $y$ terms: $5y + 5y = 10y$

Total $x$ terms: $x + x = 2x$

Total constants: $-6 + 2 = -4$ Combined result: $10y + 2x - 4$

Group 2: Terms involving $p$ and $q$

Way 1: Adding by clusters

Top section: $(2p + 3q - 2 + 3) = 2p + 3q + 1$

Middle section: $(3q + 2p + 3 - 2) = 2p + 3q + 1$

Bottom sections: $(2p + 3q) + (3q + 2p) = 4p + 6q$

Total Sum: $(2p + 3q + 1) + (2p + 3q + 1) + (4p + 6q)$

Simplified Result: $8p + 12q + 2$

Way 2: Counting total occurrences

There are four $2p$ terms, four $3q$ terms, and constants $-2, 3, 3, -2$.

Sum: $4(2p) + 4(3q) + (-2 + 3 + 3 - 2)$

Result: $8p + 12q + 2$

Group 3: Terms involving $g$ and $k$

Way 1: Row-wise addition

Row 1: $-5g + 5k + 5k - 5g = -10g + 10k$

Row 2: $5k + 5k + 5k + 5k = 20k$

Row 3: $5k + 5k + 5k + 5k = 20k$

Row 4: $-5g + 5k + 5k - 5g = -10g + 10k$

Total: $(-10g + 10k) + 20k + 20k + (-10g + 10k)$

Simplified Result: $60k - 20g$

Way 2: Direct count

Total $5k$ circles $= 12 \implies 12 \times 5k = 60k$

Total $-5g$ circles $= 4 \implies 4 \times (-5g) = -20g$

Simplified Result: $60k - 20g$

Solution:

(a) Simplification:

$\bullet$ $p + p + p + p = \mathbf{4p}$

$\bullet$ $p + p + p + q = \mathbf{3p + q}$

$\bullet$ $p + q + p - q = (p + p) + (q - q) = \mathbf{2p}$

(b) Simplification:

$\bullet$ $p - q + p - q = (p + p) + (-q - q) = \mathbf{2p - 2q}$

$\bullet$ $p + q - p + q = (p - p) + (q + q) = \mathbf{2q}$

(c) Simplification:

$\bullet$ $p + q - (p + q) = p + q - p - q = \mathbf{0}$

$\bullet$ $p - q - p - q = (p - p) + (-q - q) = \mathbf{-2q}$

(d) Simplification:

$\bullet$ $2d - d - d - d = 2d - 3d = \mathbf{-d}$

$\bullet$ $2d - d - d - c = (2d - 2d) - c = \mathbf{-c}$

(e) Simplification:

$\bullet$ $2d - d - (d - c) = d - d + c = \mathbf{c}$

$\bullet$ $2d - (d - d) - c = 2d - (0) - c = \mathbf{2d - c}$

(f) Simplification:

$\bullet$ $2d - d - c - c = (2d - d) + (-c - c) = \mathbf{d - 2c}$

To Find: Identifying mistakes in algebraic simplifications, explaining the errors, and providing the correct simplest forms.

Solution:

In algebra, we must remember that Like Terms (terms with the same letter-numbers) can be added or subtracted, but Unlike Terms cannot. Also, when opening brackets, a negative sign outside changes the signs of all terms inside.

Step-by-step correction for $(4x + 3y) - (3x + 4y)$ as an example:

Analysis of the relation:

When we look at the Correct Simplest Forms, we observe a clear pattern: In a fully simplified expression, the number of terms is equal to the number of distinct letter-numbers (variables) plus a single constant term (if any numbers are left).

For example, in $6p + 3q$, there are $2$ letter-numbers ($p, q$) and $2$ terms. In $3 + 6z$, there is $1$ letter-number and $2$ terms (including the constant $3$).

In the Indian Perspective of Mathematics, this process is fundamental to solving daily life problems involving multiple variables like cost of items or quantities, ensuring we only combine what is truly similar.

Given:

1. Top Row (Blue Machines): Pairs of inputs $(5, 2)$, $(8, 1)$, $(9, 11)$, and $(10, 10)$ result in outputs $5, 7, 18,$ and $18$ respectively.

2. Bottom Row (Purple Machines): Pairs of inputs $(4, 1)$, $(6, 0)$, and $(3, 2)$ result in outputs $5, 1,$ and $7$ respectively.

To Find:

1. The mathematical rule (formula) and the general expression for the Blue Machines.

2. The mathematical rule (formula), the missing output for $(10, 3)$, and the general expression for the Purple Machines.

Solution:

Analysis of the Blue Number Machines (Top Row):

Let the two inputs be $a$ and $b$. We observe the following relationships:

For inputs $5$ and $2$: $5 + 2 = 7$; and $7 - 2 = 5$ (Output)

For inputs $8$ and $1$: $8 + 1 = 9$; and $9 - 2 = 7$ (Output)

For inputs $9$ and $11$: $9 + 11 = 20$; and $20 - 2 = 18$ (Output)

For inputs $10$ and $10$: $10 + 10 = 20$; and $20 - 2 = 18$ (Output)

In each case, the output is found by adding the inputs and subtracting $2$.

The expression for each set of inputs is:

1. $5 + 2 - 2$

2. $8 + 1 - 2$

3. $9 + 11 - 2$

4. $10 + 10 - 2$

5. $a + b - 2$

Analysis of the Purple Number Machines (Bottom Row):

Let the two inputs be $a$ and $b$. We observe the following relationships:

For inputs $4$ and $1$: $4 \times 1 = 4$; and $4 + 1 = 5$ (Output)

For inputs $6$ and $0$: $6 \times 0 = 0$; and $0 + 1 = 1$ (Output)

For inputs $3$ and $2$: $3 \times 2 = 6$; and $6 + 1 = 7$ (Output)

In each case, the output is found by multiplying the inputs and adding $1$.

Using this rule for the missing output (inputs $10$ and $3$):

The expression for each set of inputs is:

1. $(4 \times 1) + 1$

2. $(6 \times 0) + 1$

3. $(3 \times 2) + 1$

4. $(10 \times 3) + 1 = 31$

5. $a \times b + 1$

Note: These puzzles are excellent for developing algebraic intuition. By treating numbers as inputs into a machine (Yantra), students learn that variables like $a$ and $b$ represent any number, and the formula represents the logic of the machine.

Final Answer:

$\bullet$ The formula for the Blue machine is $a + b - 2$.

$\bullet$ The formula for the Purple machine is $a \times b + 1$ (or $ab + 1$).

$\bullet$ The missing value for the Purple machine with inputs $10$ and $3$ is $31$.

Given:

A cross-shaped pattern on a calendar. The first set of numbers has $15$ at the center.

To Find:

1. The sum of the numbers in the shape and its relation to the center number.

2. Verification with another set of numbers.

3. Algebraic proof for the observation.

Solution:

We calculate the sum of the numbers in the shape for two different cases and organize them in the table below. In a standard calendar, the number above is $7$ less, the number below is $7$ more, the left is $1$ less, and the right is $1$ more than the center number.

Observation: In both cases, the total sum of the five numbers is exactly $5$ times the number at the center.

Proof:

To show that this will always happen, let us represent the numbers using a variable. Let the number at the center be $a$.

The other numbers in the shape can be expressed in terms of $a$ as follows:

Now, we find the Total Sum ($S$) by adding these expressions:

$S = (a - 7) + (a - 1) + a + (a + 1) + (a + 7)$

By rearranging the terms to group the variables and the constants together:

$S = (a + a + a + a + a) + (-7 + 7) + (-1 + 1)$

Conclusion:

Since the sum $S$ is equal to $5a$, where $a$ is the center number, we have mathematically proven that the total sum of this calendar shape will always be 5 times the number in the center, regardless of which part of the calendar is chosen.

To Find: The number of horizontal and diagonal matchsticks in Step 3 and Step 4, and the general algebraic expressions for Step $y$.

Solution:

Let us analyze the number of matchsticks in each step based on the provided image:

Step 3 and Step 4 details:

$\bullet$ In Step 3, there are 3 horizontal matchsticks and 4 diagonal matchsticks.

$\bullet$ In Step 4, there are 4 horizontal matchsticks and 5 diagonal matchsticks.

General expressions for Step '$y$':

By observing the table, we can see a clear pattern for both orientations:

1. Horizontal Matchsticks: The number of horizontal matchsticks is exactly equal to the step number.

2. Diagonal Matchsticks: The number of diagonal matchsticks is always one more than the step number.

Verification of Total sum:

To find the total number of matchsticks, we add the expressions for both orientations:

$\text{Total} = \text{Horizontal} + \text{Diagonal}$

$\text{Total} = y + (y + 1)$

$\text{Total} = 2y + 1$

Yes, the two expressions add up to $2y + 1$. This formula allows us to find the number of matchsticks for any number of triangles in the pattern.

Given:

Cost of one plate of Jowar roti = $\textsf{₹} 30$

Cost of one plate of Pulao = $\textsf{₹} 20$

Number of Jowar roti plates = $x$

Number of Pulao plates = $y$

Solution:

To find the total amount earned, we calculate the cost for each item separately and then add them together.

The total amount earned is the sum of these two costs:

Final Answer:

The expression that describes the total amount is (a) $30x + 20y$.

Given:

Pushpita gives one flag to every customer. We need to find the total number of unique customers.

Solution:

The total number of customers is the sum of all the different groups of people who visited her shop:

1. Customers who bought only champak = $p$

2. Customers who bought only marigold = $q$

3. Customers who bought both = $r$

Since each customer receives one flag, the number of flags given away is equal to the total number of customers.

Final Answer:

The correct expression is (a) $p + q + r$.

(a) Algebraic Expression:

The net distance the snail covers in one full day and night cycle is the difference between the distance climbed up and the distance slipped down.

Since this pattern repeats for $10$ days and $10$ nights, we multiply the daily net progress by 10.

Expression: $10 \times (u - d)$

(b) Analysis for $d > u$:

If $d > u$, it means the distance the snail slips down at night is greater than the distance it climbs up during the day.

In this case, the value of $(u - d)$ will be negative. This implies that the snail is not making any upward progress; instead, it is actually ending up lower than its starting position after each cycle, moving deeper into the well.

Given:

Distance in Week 1 = $5 \text{ km/day}$

Increase per week = $z \text{ km/day}$

Solution:

First, let's find the daily distance for each of the three weeks:

$\bullet$ Daily distance in Week 1 = $5 \text{ km}$

$\bullet$ Daily distance in Week 2 = $(5 + z) \text{ km}$

$\bullet$ Daily distance in Week 3 = $(5 + z + z) = (5 + 2z) \text{ km}$

Now, to find the total distance cycled in $3$ weeks (assuming $7$ days per week):

Simplifying the expression inside the bracket:

$\text{Total Distance} = 7 \times [15 + 3z]$

$\text{Total Distance} = 105 + 21z$

Final Answer:

Radha would have cycled a total of $(105 + 21z)$ km after 3 weeks.

Given:

1. A central expression oval containing $w + 2$.

2. A flowchart with four paths, where each path has two operations in boxes and two resulting expressions in ovals.

To Find:

The missing algebraic expressions in the empty ovals for the Top-Left, Bottom-Left, and Bottom-Right paths.

Solution:

We solve for each path by applying the operations given in the boxes to the preceding expression.

1. Top-Left Path:

The starting expression is $w + 2$. The first operation is $-5$.

The next operation is $\times 3$.

2. Bottom-Left Path:

The starting expression is $w + 2$. The first operation is $-8$.

The next operation is $-4$.

3. Top-Right Path (Example):

Starting with $w + 2$, the first operation is $+3$, resulting in $w + 5$. The second operation is $\times 4$, resulting in $4 \times (w + 5) = 4w + 20$.

4. Bottom-Right Path:

The starting expression is $w + 2$. The first operation is $-4$.

The next operation is $\times 3$.

Given:

Number of intermediate stations = $3$

Time for each travel segment = $t$ minutes

Time for each stop = $2$ minutes

Visualisation:

The journey looks like this:

Yahapur $\xrightarrow{t}$ Station 1 $\xrightarrow{\text{stop } 2}$ Station 2 $\xrightarrow{\text{stop } 2}$ Station 3 $\xrightarrow{\text{stop } 2}$ Vahapur

Wait, if there are 3 stations *along the way*, there are 4 segments of travel:

1. Yahapur to Station 1 ($t$)

2. Station 1 to Station 2 ($t$)

3. Station 2 to Station 3 ($t$)

4. Station 3 to Vahapur ($t$)

Total travel segments = $4$. Total stops at intermediate stations = $3$.

(a) Solution for $t = 4$:

Total travel time $= 4 \times t = 4 \times 4 = 16$ minutes

Total stoppage time $= 3 \times 2 = 6$ minutes

Total time $= 16 + 6 = 22$ minutes

(b) Algebraic Expression:

The total time taken is the sum of travel time for all segments and the total stoppage time.

$\text{Total Time} = (4 \times t) + (3 \times 2)$

Expression: $4t + 6$

This is similar to calculating the schedule of a Mumbai Local or a Metro train, where the total journey time is always the sum of running time and fixed dwell times at platforms.

(a) $3a + 9b - 6 + 8a - 4b - 7a + 16$

Grouping like terms:

$= (3a + 8a - 7a) + (9b - 4b) + (-6 + 16)$

$= \mathbf{4a + 5b + 10}$

(b) $3(3a - 3b) - 8a - 4b - 16$

Distributing the 3:

$= 9a - 9b - 8a - 4b - 16$

$= (9a - 8a) + (-9b - 4b) - 16$

$= \mathbf{a - 13b - 16}$

(c) $2(2x - 3) + 8x + 12$

Distributing the 2:

$= 4x - 6 + 8x + 12$

$= (4x + 8x) + (12 - 6)$

$= \mathbf{12x + 6}$

(d) $8x - (2x - 3) + 12$

Removing the bracket (signs change):

$= 8x - 2x + 3 + 12$

$= \mathbf{6x + 15}$

(e) $8h - (5 + 7h) + 9$

Removing the bracket:

$= 8h - 5 - 7h + 9$

$= (8h - 7h) + (9 - 5)$

$= \mathbf{h + 4}$

(f) $23 + 4(6m - 3n) - 8n - 3m - 18$

Distributing the 4 and grouping terms:

$= 23 + 24m - 12n - 8n - 3m - 18$

$= (24m - 3m) + (-12n - 8n) + (23 - 18)$

$= \mathbf{21m - 20n + 5}$

Solution:

To add algebraic expressions, we group the like terms (terms with the same literal coefficients) together and then perform the addition or subtraction on their numerical coefficients.

(a) $(4d - 7c + 9) + (8c - 11 + 9d)$

Grouping like terms: $(4d + 9d) + (-7c + 8c) + (9 - 11)$

Result: $13d + c - 2$

(b) $(-6f + 19 - 8s) + (-23 + 13f + 12s)$

Grouping like terms: $(-6f + 13f) + (-8s + 12s) + (19 - 23)$

Result: $7f + 4s - 4$

(c) $(8d - 14c + 9) + [16c - (11 + 9d)]$

First, simplify the second expression: $16c - 11 - 9d$

Now add: $(8d - 9d) + (-14c + 16c) + (9 - 11)$

Result: $-d + 2c - 2$

(d) $(6f - 20 + 8s) + (23 - 13f - 12s)$

Grouping like terms: $(6f - 13f) + (8s - 12s) + (-20 + 23)$

Result: $-7f - 4s + 3$

(e) $(13m - 12n) + (12n - 13m)$

Grouping like terms: $(13m - 13m) + (-12n + 12n)$

Result: $0$

(f) $(-26m + 24n) + (26m - 24n)$

Grouping like terms: $(-26m + 26m) + (24n - 24n)$

Result: $0$

Solution:

To subtract one expression from another, we change the sign of each term in the expression being subtracted and then add the two expressions.

(a) $(6a + 9b - 18) - (9a - 6b + 14)$

$= 6a + 9b - 18 - 9a + 6b - 14$

$= (6a - 9a) + (9b + 6b) + (-18 - 14)$

Result: $-3a + 15b - 32$

(b) $(7y - 10 + 3x) - (-15x + 13 - 9y)$

$= 7y - 10 + 3x + 15x - 13 + 9y$

$= (3x + 15x) + (7y + 9y) + (-10 - 13)$

Result: $18x + 16y - 23$

(c) $(11 - 10g + 3h) - (17g + 9 - 7h)$

$= 11 - 10g + 3h - 17g - 9 + 7h$

$= (-10g - 17g) + (3h + 7h) + (11 - 9)$

Result: $-27g + 10h + 2$

(d) $[6a - (9b + 18)] - (9a - 6b + 14)$

$= (6a - 9b - 18) - 9a + 6b - 14$

$= (6a - 9a) + (-9b + 6b) + (-18 - 14)$

Result: $-3a - 3b - 32$

(e) $(-3y + 8 - 3x) - (10x + 2 + 10y)$

$= -3y + 8 - 3x - 10x - 2 - 10y$

$= (-3x - 10x) + (-3y - 10y) + (8 - 2)$

Result: $-13x - 13y + 6$

(f) $(7h - 8g + 20) - (8g + 4h - 10)$

$= 7h - 8g + 20 - 8g - 4h + 10$

$= (-8g - 8g) + (7h - 4h) + (20 + 10)$

Result: $-16g + 3h + 30$

In our daily lives, we often use algebraic thinking while shopping in local markets or managing household supplies.

(a) Situation for $8x + 3y$

Imagine Rohan goes to a stationery shop in Delhi. He buys $8$ notebooks, and each notebook costs $\textsf{₹} x$. He also buys $3$ pens, and each pen costs $\textsf{₹} y$. The total amount Rohan has to pay the shopkeeper is represented by the expression $8x + 3y$.

(b) Situation for $15x - 2x$

Suppose a fruit seller in Mumbai has $15$ identical crates of mangoes, where each crate contains $x$ mangoes. While inspecting the crates, he discovers that $2$ crates have completely spoiled mangoes and cannot be sold. The total number of good mangoes remaining with the seller is represented by $15x - 2x$ (which simplifies to $13x$).

Given:

$\bullet$ Straight rope (0 folds), 1 cut $\rightarrow$ 2 pieces.

$\bullet$ Folded once (1 fold), 1 cut $\rightarrow$ 3 pieces.

To Find:

1. Number of pieces if the rope is folded $10$ times and cut.

2. The general algebraic expression for $r$ folds.

Solution:

Let us observe the relationship between the number of folds and the number of pieces resulting from a single cut across all layers:

The pattern shows that the number of pieces is always $2$ more than the number of folds.

1. For 10 folds:

Number of pieces = $12$.

2. General Expression:

If the number of folds is represented by the letter-number '$r$', the expression for the number of pieces is:

Indian Perspective:

This is a common observation when children in India fly kites (Patang). When the Manjha (string) is folded multiple times to make a bundle and then cut at one end, the resulting number of small strings follows this exact mathematical pattern.

Given:

A pattern where squares are joined together. The first square uses $4$ matchsticks, and each subsequent square shares a common wall with the previous one.

To Find:

1. Matchsticks required for $10$ squares.

2. Matchsticks required for $w$ squares.

Solution:

Let us count the matchsticks for the first few steps:

$\bullet$ 1 square: $4$ matchsticks.

$\bullet$ 2 squares: $4 + 3 = 7$ matchsticks.

$\bullet$ 3 squares: $7 + 3 = 10$ matchsticks.

We see that the first square requires $4$ sticks, and every additional square requires $3$ extra sticks because they share one matchstick as a common boundary.

General Formula:

For '$w$' squares, the total matchsticks required is:

$\text{Total} = 4 + (w - 1) \times 3$

$\text{Total} = 4 + 3w - 3$

Calculation for 10 squares:

Substituting $w = 10$ into the expression:

$\text{Total sticks} = 3(10) + 1$

$\text{Total sticks} = 30 + 1 = 31$.

Final Answer:

$\bullet$ To make 10 squares, 31 matchsticks are required.

$\bullet$ To make $w$ squares, the algebraic expression is $3w + 1$.

Given:

The sequence follows a repeating pattern of 4 steps:

1. Red, 2. Yellow, 3. Green, 4. Yellow, (repeats: 5. Red, 6. Yellow...)

To Find:

The colour at positions $90$, $190$, and $343$, and general expressions for each colour.

Solution:

The cycle length is $4$. We divide the position number by $4$ and look at the remainder to determine the colour.

$\bullet$ At position 90: $90 \div 4 = 22$ with a remainder of 2. The colour is Yellow.

$\bullet$ At position 190: $190 \div 4 = 47$ with a remainder of 2. The colour is Yellow.

$\bullet$ At position 343: $343 \div 4 = 85$ with a remainder of 3. The colour is Green.

General Expressions (where $k$ is a whole number $0, 1, 2, \dots$):

$\bullet$ Red: $4k + 1$

$\bullet$ Green: $4k + 3$

$\bullet$ Yellow: $4k + 2$ and $4k$ (for $k > 0$)

Given:

By observing the growing pattern (often called a Vinyasa in Indian mathematics) of the squares:

$\bullet$ In Step 1: There is $1$ central square and $4$ arms with $1$ square each. Total squares $= 1 + 4(1) = 5$.

$\bullet$ In Step 2: There is $1$ central square and $4$ arms with $2$ squares each. Total squares $= 1 + 4(2) = 9$.

$\bullet$ In Step 3: There is $1$ central square and $4$ arms with $3$ squares each. Total squares $= 1 + 4(3) = 13$.

To Find:

1. The number of squares in Step 4, Step 10, and Step 50.

2. A general formula for the number of squares in Step $n$.

3. A formula for the total number of vertices of all the squares.

Solution:

Let the step number be represented by $n$ and the number of squares by $N$. From the given pattern, we see that the number of squares increases by $4$ in each subsequent step.

The general formula for the number of squares is:

Using this formula, we calculate the required values:

For Step 4 ($n = 4$):

For Step 10 ($n = 10$):

For Step 50 ($n = 50$):

Calculation for Vertices:

We need to find the total number of vertices of all the squares. Each individual square has exactly $4$ vertices.

To find the total number of vertices ($V$), we multiply the total number of squares ($N$) by $4$.

Substitute the value of $N$ from equation (i):

$V = 4(4n + 1)$

Final Answer:

$\bullet$ The number of squares in Step 4 is 17, in Step 10 is 41, and in Step 50 is 201.

$\bullet$ The general formula for the number of squares is $N = 4n + 1$.

$\bullet$ The formula for the total number of vertices of all the squares is $16n + 4$.

Given:

A number grid with $4$ columns where numbers are arranged consecutively row by row.

(a) Column Expressions:

Let $r$ represent the row number ($r = 1, 2, 3, \dots$). We observe that each row contains $4$ numbers. The expressions for each column are:

(b) Finding Row and Column for Specific Numbers:

To find the position of any number $N$, we divide it by the number of columns ($4$). The quotient and remainder determine the row and column.

(i) For the number $124$:

$\begin{array}{r} 31\phantom{)} \\ 4{\overline{\smash{\big)}\,124\phantom{)}}} \\ \underline{-~12\phantom{b)}} \\ 04\phantom{)} \\ \underline{-~04} \\ 0\phantom{)} \end{array}$

Since the remainder is $0$, the number is exactly at the end of the quotient row.

Position: Row 31, Column 4.

(ii) For the number $147$:

$\begin{array}{r} 36\phantom{)} \\ 4{\overline{\smash{\big)}\,147\phantom{)}}} \\ \underline{-~12\phantom{b)}} \\ 27\phantom{)} \\ \underline{-~24} \\ 3\phantom{)} \end{array}$

Quotient $= 36$ and Remainder $= 3$. This means $36$ rows are full, and the number is the $3$rd element of the next row.

Position: Row 37, Column 3.

(iii) For the number $201$:

$\begin{array}{r} 50\phantom{)} \\ 4{\overline{\smash{\big)}\,201\phantom{)}}} \\ \underline{-~20\phantom{b)}} \\ 01\phantom{)} \\ \underline{-~00} \\ 1\phantom{)} \end{array}$

Quotient $= 50$ and Remainder $= 1$. This means the number is the $1$st element of the $51$st row.

Position: Row 51, Column 1.

(c) General Formula:

The number $N$ located in row $r$ and column $c$ can be generated by the expression:

(d) Observations of Patterns:

Pattern in Multiples of 3:

The multiples of $3$ ($3, 6, 9, 12, 15, \dots$) appear in a shifting diagonal pattern across the columns:

We see the column index decreases by $1$ in each row, cycling back to $4$ after reaching $1$.

Other Observed Patterns:

1. Multiples of 4: All multiples of $4$ ($4, 8, 12, 16, \dots$) always appear in Column 4.

2. Even and Odd Columns: Columns $1$ and $3$ consist entirely of Odd Numbers, while Columns $2$ and $4$ consist entirely of Even Numbers.

3. Common Difference: In any column, the difference between numbers in consecutive rows is always $4$.