Chapter 4: Expressions using Letter-Numbers

Step 1 – Introduce letters
Let $$A$$ represent Aftab’s present age (in years) and $$S$$ represent Shabnam’s present age (in years).

Step 2 – Translate the information into an equation
We are told that Shabnam is 3 years older than Aftab, so

$$S = A + 3$$

Step 3 – Check the relation with the given example
When Aftab’s age is 10 years (i.e. $$A = 10$$), the relation predicts
$$S = 10 + 3 = 13$$,
exactly as stated. Hence the equation is correct.

Step 4 – Use the equation for the required case
We are now told that Aftab’s present age is 18 years, so $$A = 18$$. Substitute this value in the equation:

$$S = 18 + 3$$

$$S = 21$$

Step 5 – Conclusion
Shabnam’s present age is therefore 21 years.

The figure shows that one L-shape is made from exactly 2 matchsticks   (one vertical, one horizontal).

Step 1 – Look at a few specific cases

• For 1 L  →  $$1 \times 2 = 2$$ matchsticks
• For 5 Ls →  $$5 \times 2 = 10$$ matchsticks
• For 7 Ls →  $$7 \times 2 = 14$$ matchsticks
• For 45 Ls →  $$45 \times 2 = 90$$ matchsticks

Step 2 – Introduce a letter for the general case

Let $$n$$ stand for the number of Ls Parthiv makes.

Step 3 – Write an expression

Because each L needs 2 sticks, the total number of sticks is

With the letter $$n$$ this becomes the algebraic expression

$$2n$$.

Conclusion

If Parthiv makes $$n$$ Ls, he needs $$2n$$ matchsticks. Thus, the required relation is:

Let us denote Aftab’s present age by a letter, say $$x$$ years.

The statement in the question (given just before this part of the exercise) tells us that Shabnam is 3 years younger than Aftab.

'Younger by 3 years' means we have to subtract 3 from Aftab’s age to obtain Shabnam’s age.

Therefore

\[\text{Shabnam\’s age} = x - 3\]

So, once Aftab’s age is known, simply subtract 3 to get Shabnam’s age.

Yes. Whenever a statement involves an (as yet) unknown number, we introduce a letter to stand for that number and then translate the words into the usual arithmetic symbols. The resulting symbolic sentence is called an algebraic expression.

How it is done

1. Choose a letter for the unknown quantity.
   Example: let the unknown number be $$x$$.

2. Translate each phrase into a mathematical operation.
   • “sum” → $$+$$  • “difference” → $$-$$  • “times” → $$\times$$  • “quotient” → $$\div$$, etc.

3. Write the parts in the given order to obtain the required expression.

Illustration

• Verbal statement: “Twice a number increased by 5.”
     Unknown number → $$x$$.
     “Twice a number” → $$2x$$.
     “increased by 5” → add 5 ⇒ $$2x + 5$$.

Hence we can (and do) write such statements as algebraic expressions.

Known relation between the two ages
If Aftab’s present age is denoted by the variable $$a$$, then Shabnam’s present age is expressed as
$$\frac{a}{2} - 2$$     (half of Aftab’s age, 2 years less).

Substitute the given value of $$a$$
Aftab’s actual age is given to be $$a = 23$$ years, so

$$\text{Shabnam\’s age} = \frac{a}{2} - 2 = \frac{23}{2} - 2$$

Simplify step by step

• First divide: $$\frac{23}{2} = 11.5$$
• Then subtract: $$11.5 - 2 = 9.5$$

Conclusion
Shabnam is, therefore, $$9.5\text{ years}$$ old.

Step 1 – Choose a variable for Shabnam’s age.
Let Shabnam’s present age be denoted by the letter-number (variable) $$x$$ years.

Step 2 – Read the relation given in words.
The statement tells us that Aftab is 3 years younger than Shabnam.

Step 3 – Convert the verbal relation into algebra.
“3 years younger” means we subtract $$3$$ from Shabnam’s age. Hence Aftab’s age is

\[ x - 3 \]

Conclusion.
The required algebraic expression for Aftab’s age, when Shabnam’s age is $$x$$ years, is $$x - 3$$.

Earlier we wrote Aftab’s age in terms of Shabnam’s age.

Let Shabnam’s age be represented by the variable $$s$$ (in years).

If Aftab is 2 years older than Shabnam, then

Now we are told that Shabnam is 20 years old, so

Substituting this value of $$s$$ in the expression for Aftab’s age:

Hence, Aftab is 22 years old.

Let the cost (in rupees) of

• 1 coconut be $$x$$,
• 1 kg jaggery be $$y$$.

Step 1 – Cost of 10 coconuts
For each coconut the cost is $$x$$ rupees, so for 10 coconuts:

$$\text{Cost of 10 coconuts}=10 \times x = 10x$$ rupees.

Step 2 – Cost of 5 kg jaggery
For each kilogram the cost is $$y$$ rupees, so for 5 kg jaggery:

$$\text{Cost of 5 kg jaggery}=5 \times y = 5y$$ rupees.

Step 3 – Total amount to be paid
Add the two individual costs:

\[\text{Total payment}=10x + 5y \;\text{rupees}\]

Thus, she has to pay $$10x + 5y$$ rupees altogether.

Let the price of one coconut be $$c$$ rupees and the price of 1 kg jaggery be $$j$$ rupees.

She buys:

• 8 coconuts
• 9 kg jaggery

Total cost of coconuts   = $$8 \times c = 8c$$ rupees

Total cost of jaggery   = $$9 \times j = 9j$$ rupees

Therefore, the amount she has to pay is

\[T = 8c + 9j\]

So, she should pay $$8c + 9j$$ rupees in all.

Step 1 : Choose suitable letters for every variable

• Number of coconuts to be purchased → $$c$$
• Quantity of jaggery to be purchased (in kilograms) → $$j$$
• Price of one coconut (in rupees) → $$p_c$$
• Price of one kilogram of jaggery (in rupees) → $$p_j$$

Step 2 : Express the cost of each item

• Cost of all coconuts: $$p_c \times c = p_c c$$
• Cost of the jaggery: $$p_j \times j = p_j j$$

Step 3 : Form the required expression for the total amount

Adding the two individual costs, the total amount to be paid is

Hence, the algebraic expression that gives the total amount to be paid for buying $$c$$ coconuts and $$j$$ kilograms of jaggery is $$p_c c + p_j j$$.

Let the cost of one coconut be denoted by $$c$$ rupees and that of one kilogram of jaggery by $$j$$ rupees.

The given expression for the total amount to be paid when buying $$n$$ coconuts and $$m$$ kg jaggery is

\[ \text{Amount} = nc + mj \]

For the present purchase we have

• number of coconuts, $$n = 7$$,
• kilograms of jaggery, $$m = 4$$.

Substituting these numbers in the formula:

$$\text{Amount} = (7)c + (4)j$$

Simplifying:

$$\text{Amount} = 7c + 4j$$

Hence, the total amount to be paid for 7 coconuts and 4 kg jaggery is $$7c + 4j$$ rupees.

Step 1 — Recall the algebraic expression for the perimeter of a square
For any square whose side–length is represented by the letter $$s$$ (in centimetres), its perimeter $$P$$ is obtained by adding the four equal sides:
$$P = s + s + s + s$$.
Grouping like terms, this becomes the compact literal–number expression
$$P = 4 \times s$$.

Step 2 — Substitute the given side–length
The question says the side–length is $$7\,\text{cm}$$, so put $$s = 7$$ in the expression:
$$P = 4 \times 7\,\text{cm}$$.

Step 3 — Simplify the numerical product
Calculate $$4 \times 7$$:
$$4 \times 7 = 28$$.
Therefore
\[P = 28\,\text{cm}\]

Conclusion
The perimeter of the square is $$28\,\text{cm}$$.

We observe the list of numbers:

$$4,\;8,\;12,\;16,\;20,\;24,\;28,\;\ldots$$

Let us write them together with their position numbers (term numbers):

Step 1: Look for a common difference.

Each number is larger than the previous one by

$$8-4 = 4,\; 12-8 = 4,\; 16-12 = 4,\;\dots$$

So the common difference is $$d = 4$$. This makes the sequence an arithmetic progression (AP).

Step 2: Write the general formula for an AP.

For an AP with first term $$a$$ and common difference $$d$$, the $$n$$-th term $$T_n$$ is

Step 3: Substitute the values from our sequence.

The first term is $$a = 4$$ and the difference is $$d = 4$$, so

$$T_n = 4 + (n-1) \times 4$$

Step 4: Simplify the expression.

$$T_n = 4 + 4(n-1) = 4 + 4n - 4 = 4n$$

Therefore, the algebraic expression for the $$n$$-th term is

That is, the rule "multiply the term number by 4" produces every member of the sequence.

Step 1 – Write the algebraic expression for the earning
One pencil is sold for ₹ $$c$$.
If a total of $$10$$ pencils are sold, the money collected is represented by
$$\text{Amount earned}=10c$$

Step 2 – Substitute the given value of $$c$$
We are told that $$c = 50$$ (i.e. each pencil is ₹ 50). Replacing $$c$$ by 50 gives
$$\text{Amount earned}=10 \times 50$$

Step 3 – Calculate
$$\text{Amount earned}=500$$

The total amount obtained from the sale of the pencils is therefore ₹ 500.

Step 1 – Choose the letters (variables).
Let

• $$p$$ be the price (in rupees) of one eraser,
• $$n$$ be the number of erasers sold.

Step 2 – Translate the statement into an algebraic expression.
Total money earned = (number of erasers sold) × (price of one eraser).
$$\text{Total money earned}= n \times p$$

Step 3 – Simplify the expression.
In algebra, the multiplication sign is omitted when two factors are written together;
$$n \times p = np$$

Result.
Hence, the simplified expression for the total money earned by selling erasers is
$$np$$ rupees.

Step 1 – Identify the terms and their literal (letter) parts.
The expression has two terms:

• First term : $$18c$$ → literal part is $$c$$.
• Second term : $$11d$$ → literal part is $$d$$.

Step 2 – Check if the terms are like terms.
Two terms are like terms only when their literal parts are identical (the same letters with the same powers).
Here the first term contains $$c$$ while the second term contains $$d$$. Because the letters are different, the two terms are unlike terms.

Step 3 – Combine only like terms.
Rules of algebra allow us to add or subtract only like terms. Since $$18c$$ and $$11d$$ are unlike terms, they cannot be combined or simplified by addition or subtraction.

Conclusion
No further simplification is possible. The given expression is already in its simplest form:

\[18c + 11d\]

Given expressions

Expression I: $$E_1 = c(c+1)$$
Expression II: $$E_2 = c^{2}+c$$

Step 1 : Evaluate the two expressions for different numerical values of $$c$$

For every trial value of $$c$$, both expressions give exactly the same numerical answer.

Step 2 : Show algebraically that the two expressions are identical

Start with Expression I:

$$E_1 = c(c+1)$$

Distribute (multiply) $$c$$ over the bracket: $$E_1 = c\times c + c\times 1$$

But $$c\times c = c^{2}$$ and $$c\times1 = c$$, so

Thus $$E_1 = c^{2}+c = E_2$$ for every value of $$c$$.

Conclusion

Whether we verify by substituting particular numbers or prove through algebra, the two expressions always produce the same value. Hence they are equivalent.

The variable symbols $$x$$ (chairs) and $$y$$ (tables) stand for numbers. To get the required money amounts we follow the same two operations again and again:

1. Multiply the amount for one article by the number of such articles to get the total for that article.
2. Add the two totals (one for chairs, one for tables).

Denote the amounts given in the question as follows:

• deposit on one chair  = Rs $$d_1$$
• deposit on one table  = Rs $$d_2$$
• refund on one chair   = Rs $$r_1$$
• refund on one table   = Rs $$r_2$$

Step 1 : Total amount paid at the beginning

• For chairs: $$d_1 \times x = d_1x$$
• For tables: $$d_2 \times y = d_2y$$

Step 2 : Amount one gets back after returning the furniture

• For chairs: $$r_1 \times x = r_1x$$
• For tables: $$r_2 \times y = r_2y$$

Thus, once the numerical values of $$d_1,\,d_2,\,r_1,\,r_2$$ are read from the question, substituting them in the two boxed expressions instantly gives the required answers in terms of $$x$$ and $$y$$.

We are asked whether the expression $$(40x + 75y) - (6x + 10y)$$ can be simplified. Yes, it can, by following the usual algebraic rules for brackets and like terms.

1. Remove the brackets.
   Remember that a minus sign in front of a bracket changes the sign of every term inside that bracket:

   $$ (40x + 75y) - (6x + 10y) = 40x + 75y - 6x - 10y $$

2. Group like terms.
   Collect the terms containing x together and the terms containing y together:

   $$ (40x - 6x) + (75y - 10y) $$

3. Simplify each pair.
   Subtract the coefficients:

   $$ 40x - 6x = 34x, \;\; 75y - 10y = 65y $$

4. Write the final simplified form. \[ 34x + 65y \]

No further simplification is possible because 34 and 65 have no common factor other than 1, and the terms involve different variables (x and y).

Given: We begin with an expression in which one bracket is being subtracted from another:

$$ (40x + 75y) - (6x + 10y) $$.

Step 1  Recall the rule about subtraction.
To subtract any quantity, we may instead add its additive inverse (its “negative”). In symbols, for any algebraic expression $$A$$,

$$ -A = +(-A). $$

Step 2  Identify the subtrahend and form its additive inverse.
Here the subtrahend is the entire bracket $$ (6x + 10y) $$.
The additive inverse of that bracket is obtained by changing the sign of every term inside it:

$$ (6x + 10y) \longrightarrow (-6x - 10y). $$

Step 3  Rewrite the original subtraction as an addition.
Replace the single “minus” sign in front of the bracket by “plus” and use the additive inverse found in Step 2:

$$ (40x + 75y) - (6x + 10y) = (40x + 75y) + (-6x - 10y). $$

Step 4  Verify (optional but reassuring).
Distribute the signs to check that both forms really give the same expanded expression:

• From the subtraction form:
$$ (40x + 75y) - (6x + 10y) = 40x + 75y - 6x - 10y. $$

• From the addition-of-inverse form:
$$ (40x + 75y) + (-6x - 10y) = 40x + 75y + (-6x) + (-10y) \;=\; 40x + 75y - 6x - 10y. $$

Both give the identical simplified result, so the two bracketed forms are equivalent.

Conclusion. Yes, we are perfectly free to write the given subtraction as the addition of the additive inverse, hence as $$ (40x + 75y) + (-6x - 10y). $$

Step 1 : Meaning of the three expressions

• $$7p - 3q$$ is the total score when 7 answers are correct (each earns $$p$$ marks) and 3 answers are wrong (each loses $$q$$ marks).
• $$8p - 4q$$ is the score for 8 correct and 4 wrong answers.
• $$6p - 2q$$ is the score for 6 correct and 2 wrong answers.

Step 2 : Scores when one correct answer gives 4 marks and one wrong answer loses 1 mark

Given $$p = 4$$ and $$q = 1$$:

First round:
$$7p - 3q = 7\times4 - 3\times1 = 28 - 3 = 25$$

Second round:
$$8p - 4q = 8\times4 - 4\times1 = 32 - 4 = 28$$

Third round:
$$6p - 2q = 6\times4 - 2\times1 = 24 - 2 = 22$$

Step 3 : If there is no penalty for a wrong answer

No penalty ⇒ $$q = 0$$.

• First round: $$7p - 3q = 7\times4 - 3\times0 = 28$$
• Second round: $$8p - 4q = 8\times4 - 4\times0 = 32$$
• Third round: $$6p - 2q = 6\times4 - 2\times0 = 24$$

Step 4 : Final score without penalty

Total score  $$= 28 + 32 + 24 = 84$$

Step 1 – Understand the requirement
Krishita plays three rounds. If her scores in the three rounds are, respectively,

• $$S_1$$ in Round 1,
• $$S_2$$ in Round 2,
• $$S_3$$ in Round 3,

then the condition given in the question is

Step 2 – Write a general form for each score
Because every score itself must be expressed by letters and numbers, let us take

• $$S_1 = x_1p - y_1q$$,
• $$S_2 = x_2p - y_2q$$,
• $$S_3 = x_3p - y_3q$$,

where $$x_1, x_2, x_3$$ and $$y_1, y_2, y_3$$ are ordinary (numerical) coefficients.

Step 3 – Match the required total
For the three expressions to add up to $$23p - 7q$$ we must have

• $$x_1 + x_2 + x_3 = 23$$ (for the coefficient of $$p$$), and
• $$y_1 + y_2 + y_3 = 7$$ (for the coefficient of $$q$$).

Step 4 – Choose any convenient numbers
One easy choice is

• $$x_1 = 10,\; x_2 = 8,\; x_3 = 5$$  (they add to $$23$$),
• $$y_1 = 3,\; y_2 = 2,\; y_3 = 2$$  (they add to $$7$$).

Step 5 – Write down the three scores

Step 6 – Verify the sum

Adding the three scores:

$$S_1 + S_2 + S_3 = (10p - 3q) + (8p - 2q) + (5p - 2q)$$

$$= (10p + 8p + 5p) - (3q + 2q + 2q)$$

$$= 23p - 7q$$

which is exactly the required total. Therefore the chosen scores are valid.

Remark: Many other answers are possible as long as the three $$p$$-coefficients total $$23$$ and the three $$q$$-coefficients total $$7$$.

Restatement of the question
Two players have their scores written with the help of a letter (say $$x$$). We are asked whether, on the basis of those letter-expressions alone, we can tell who has the larger score.

Step 1 – Understand what a letter (variable) means
A letter like $$x$$ does not stand for one particular number; it can take any number the situation allows. Therefore a score written, for example, as $$x+7$$ is not a single fixed number – it changes whenever the value of $$x$$ changes.

Step 2 – Compare two typical score-expressions

• Put $$x=2$$ (one possible value).
  Player A: $$2+7=9$$;   Player B: $$2\times2+3=7$$ → A scores more.

• Put $$x=6$$ (another possible value).
  Player A: $$6+7=13$$;   Player B: $$2\times6+3=15$$ → B scores more.

The result flips when we change the value of $$x$$. This shows that without knowing the actual value of $$x$$ we cannot reach a definite conclusion.

Step 3 – General reason
An algebraic expression with a variable does not have one permanent value; it can represent many different numbers. Therefore two such expressions usually cannot be compared once for all, unless their difference is clearly always positive or always negative (for example $$x+5$$ is always larger than $$x+2$$ because their difference is the fixed number $$3$$). In the present question no such information is given.

Conclusion
Because the value of the variable is unknown, the two score-expressions can take different numerical values on different occasions. Hence we cannot say who scored more merely from the expressions; extra information about the variable’s value would be necessary.

Given expression:

$$23p - 7q - (21p - 9q)$$

Step 1 – Open the brackets.

The minus sign before the bracket reverses every sign inside it:

$$-(21p - 9q)= -21p + 9q$$

So,

$$23p - 7q - (21p - 9q)=23p - 7q -21p + 9q$$

Step 2 – Collect like terms.

Terms in $$p$$: $$23p - 21p$$
Terms in $$q$$: $$-7q + 9q$$

Step 3 – Simplify each group.

$$23p - 21p = 2p$$
$$-7q + 9q = 2q$$

Step 4 – Write the simplified expression.

[2p + 2q]

Optional factorisation: $$2p + 2q = 2(p + q)$$.

Hence, the simplified form is $$2p + 2q$$.

Step 1  Evaluate  $$5u$$

The four blanks around the central circle marked 5u are therefore 55, 10, 40, 25.

Step 2  Evaluate  $$5 + u$$

The four blanks around the central circle marked 5 + u are therefore 16, 7, 13, 10.

Step 3  Comparison of the two expressions

For each of the given values $$5u > 5 + u$$. In fact, whenever $$u > 1$$ we have

so the product expression always produces the greater number (when $$u = 1$$ the two expressions are equal).

Step 1 – Read the two expressions carefully
We are comparing the expressions $$10y - 3$$ and $$10(y - 3)$$.

Step 2 – Understand what each expression means

• In $$10y - 3$$ we first find the product of 10 and $$y$$, then subtract 3.
• In $$10(y - 3)$$ we first find the difference $$(y-3)$$, then multiply the result by 10.

Step 3 – Expand $$10(y-3)$$ using the distributive law

Distributive law: $$a(b-c)=ab-ac$$. Here $$a=10$$, $$b=y$$, $$c=3$$.

So:

Step 4 – Compare the simplified forms

We now have

• First expression: $$10y-3$$;
• Second expression after expansion: $$10y-30$$.

The two are different because $$-3\neq -30$$. The only way they could match would be if $$-3=-30$$, which is impossible.

Step 5 – Numerical check (optional but convincing)

Take $$y=5$$:

• $$10y-3=10\times5-3=50-3=47$$;
• $$10(y-3)=10\times(5-3)=10\times2=20$$.

Since $$47\neq20$$, the expressions give different results.

Conclusion
Therefore the two expressions are not equal in general.

Step 1 : Recall the meaning of “letter-number”
In Class 7 the word letter-number is used for every product in which a number is multiplied by at least one letter, for example $$3x,\; -5ab,\; \tfrac12p^2$$. A term that contains only numbers, like $$7\text{ or }-3$$, is not called a letter-number.

Step 2 : Write some expressions, reduce them to the simplest form and count

Step 3 : Observe the counts

• Whenever the simplest form does not contain any pure number (examples 1 and 5) the two counts are exactly the same.
• Whenever the simplest form does contain a pure number (examples 2, 3 and 4) the number of terms is one more than the number of letter-numbers because of that extra constant term.

Step 4 : General statement

Let an expression be reduced fully. Call

• $$t$$ = number of terms in the simplified expression,
• $$\,l$$ = number of letter-numbers in it.

Exactly two situations can occur:

1. No constant term present  →  $$t = l$$.
2. One constant term present  →  $$t = l + 1$$.

Hence the relation

The difference “+1” appears precisely when the expression also contains a term made of only numbers.

Step 1 : Look for a simple operation that joins the two inputs.

Take the first pair $$(5,2)$$. If we double the first number and then subtract the second number we get

$$2\times5-2=10-2=8,$$ which matches the given output.

Step 2 : Test the same idea on the other examples.

• Pair $$(8,1)$$: $$2\times8-1=16-1=15\;(\checkmark)$$
• Pair $$(9,11)$$: $$2\times9-11=18-11=7\;(\checkmark)$$
• Pair $$(10,10)$$: $$2\times10-10=20-10=10\;(\checkmark)$$

The rule works for every given pair, so it must be the correct formula.

Step 3 : Write the general formula.

Let $$x$$ be the first input and $$y$$ be the second input. The number machine performs

\[\text{Output}=2x-y.\]

Step 4 : Use the formula for the missing output $$(6,4)$$.

$$2\times6-4=12-4=8.$$

So the required output is $$8$$.

Step 1 — Study the examples for Machine 1 (blue)

Look at the sum $$a+b$$ in each row:

• $$5+2=7$$, but the machine gives $$5=7-2$$.
• $$8+1=9$$, machine gives $$7=9-2$$.
• $$9+11=20$$, machine gives $$18=20-2$$.
• $$10+10=20$$, machine gives $$18=20-2$$.

In every case the machine’s result is the sum minus 2. Hence

\[\text{Machine 1 rule: }\boxed{\;\;\text{output}=a+b-2\;}\]

For any pair $$(a,b)$$ the expression is therefore $$a+b-2$$.

Step 2 — Study the examples for Machine 2 (purple)

Check the product $$ab$$ and compare with the output:

• $$4\times1=4$$; output $$5=4+1$$.
• $$6\times0=0$$; output $$1=0+1$$.
• $$3\times2=6$$; output $$7=6+1$$.

Each time, the machine adds 1 to the product. So

\[\text{Machine 2 rule: }\boxed{\;\;\text{output}=ab+1\;}\]

Step 3 — Find the missing outputs

• Machine 2 for inputs $$(10,3):\;10\times3+1=30+1=31$$.
• Machine 1 for $$(a,b):\;a+b-2$$ (already found).
• Machine 2 for $$(a,b):\;ab+1$$.

Final expressions

• Machine 1 (blue): $$a+b-2$$
• Machine 2 (purple): $$ab+1$$ with $$(10,3)\to31$$

Step 1 – Decide the chain of operations

To invent a new “secret rule” (formula), pick any short sequence of simple operations that Class 7 students already know. For example, let us decide to do the following four operations to whatever number the machine is fed with:

1. Add 7
2. Multiply the result by 3
3. Subtract 4
4. Stop – the answer that comes out is the machine’s output

Step 2 – Translate the operations into algebra

Let the input number be denoted by the letter $$n$$ (you may use $$x$$, $$y$$, etc.; any letter is allowed). Carry out the operations on $$n$$ one by one, writing each intermediate step:

Thus the formula produced by this number machine is

\[\boxed{\;3n+17\;}\]

In words: “Multiply the input by 3 and then add 17.”

Step 3 – Check the formula with a test value

Take any convenient test value, say $$n=2$$:

According to the long description: 2 + 7 = 9, 9 × 3 = 27, 27 – 4 = 23.
According to the compact formula: $$3(2)+17=6+17=23$$. Both give the same answer, so the algebra is correct.

Step 4 – Prepare the “number-machine” challenge table

Choose any five distinct inputs (positive, zero, negative – your choice). Feed them into the secret machine and list only the outputs. Do not print the rule beside the table when you hand the sheet to your classmates!

Now write a short instruction like:

• “The same secret rule was used each time. Can you find that rule?”
• “Hint: first look for the pattern in how the outputs are rising.”

Step 5 – How a solver might work it out

Your friends will probably notice that every time the input increases by 1, the output increases by 3. That is a strong clue that the rule involves ‘multiply by 3’. Then they can try to discover that 17 has been added at the end.

Congratulations – you have designed and fully explained your own algebraic formula and number machine!

Step 1   Look at the first few tiles in the row

According to the drawing given in the textbook, the designs follow this order, repeating without a break:

A, B, C, A, B, C, A, B, C  …

If we write the positions underneath, we get

Step 2   Observe the jump (the “gap”) between successive occurrences

• Design A appears at 1, 4, 7, 10, …  The gap between two successive A’s is 3 positions.
• Design B appears at 2, 5, 8, 11, …  Again the gap is 3.
• Design C appears at 3, 6, 9, 12, …  Gap = 3.

So every third tile we come back to the same design.

Step 3   Use a letter (say n) to write the rule

Let $$n$$ stand for any natural number $$1,2,3,\ldots$$

• To reach the 1^st, 4^th, 7^th, … positions (all A’s) we need to move ahead in steps of 3, starting from 1.
  That sequence can be written as $$1, 1+3, 1+2\times3, 1+3\times3,\ldots$$
  Compactly: $$3n-2\;(n=1,2,3,\ldots)$$
• For B we start from 2 and again add multiples of 3: $$2, 2+3, 2+2\times3,\ldots$$
  This is $$3n-1\;(n=1,2,3,\ldots)$$
• For C we start from 3: $$3, 3+3, 3+2\times3,\ldots$$
  This is simply $$3n\;(n=1,2,3,\ldots)$$

Step 4   Summarise

• All positions of Design A: $$3n-2$$  (remainder 1 when divided by 3)
• All positions of Design B: $$3n-1$$  (remainder 2 when divided by 3)
• All positions of Design C: $$3n$$     (exact multiples of 3)

Here $$n$$ can take any natural number value 1, 2, 3, …, giving the infinite set of positions for each design.

Step 1 — Observe the repeating block.
The designs repeat in the fixed order
  A, B, C   |   A, B, C   |   A, B, C  |  …
Each block contains exactly three designs.

Step 2 — List the positions of design C.
From the pattern we read:

The position numbers form the sequence
$$3,\,6,\,9,\,12,\,\ldots$$

Step 3 — Recognise the arithmetic sequence.
The first term is $$a = 3$$ and the common difference is $$d = 3$$.

Step 4 — Write the general term.
For an arithmetic sequence the $$n$$th term is given by
$$a_n = a + (n - 1)d$$.

Putting $$a = 3$$ and $$d = 3$$, we get
$$a_n = 3 + (n - 1)\times 3$$
    $$ = 3 + 3n - 3$$
    $$ = 3n$$.

Step 5 — Interpret the result.
Therefore, the design C will be seen at position number $$3n$$ when it appears for the $$n$$th time.

Step 1 – Write down the positions that each design occupies.

In every row the difference between consecutive positions is $$4$$. Hence, for each design we get an arithmetic progression (A.P.) with common difference $$d = 4$$.

Step 2 – Recall the nth-term formula for an A.P.

If the first term of an A.P. is $$a$$ and the common difference is $$d$$, the nth term is

\[T_n = a + (n - 1)d\]

Step 3 – Find the formula for each design.

• Design I
  Here $$a = 1,\; d = 4$$.
  $$T_n = 1 + (n - 1)\times 4 = 4n - 3$$

• Design II
  Here $$a = 2,\; d = 4$$.
  $$T_n = 2 + (n - 1)\times 4 = 4n - 2$$

• Design III
  Here $$a = 3,\; d = 4$$.
  $$T_n = 3 + (n - 1)\times 4 = 4n - 1$$

• Design IV
  Here $$a = 4,\; d = 4$$.
  $$T_n = 4 + (n - 1)\times 4 = 4n$$

Step 4 – State the required formulas.

Therefore, the position at which each design appears for the $$n$$th time is:

Observe the row of designs reproduced from the textbook:

We notice that the three designs I, II and III keep repeating in the same order. Therefore the cycle length is 3.

To know which design will appear at any position number n, we divide n by 3 and look at the remainder:

Let the quotient be $q$ and the remainder be $r$.

$$n = 3q + r, \qquad 0 \le r \le 2$$

• if $r = 1$ → Design I
• if $r = 2$ → Design II
• if $r = 0$ → Design III  (because a remainder 0 means the position ends exactly at the third design of a cycle)

Now take position 122:

Divide 122 by 3.

$$122 \div 3 = 40 \text{ with remainder } 2$$
That is, $$122 = 3\times40 + 2$$

Here the remainder $r = 2$, so the design at position 122 is the second design of the cycle, namely Design II.

Hence, given any position number, we can always find the corresponding design by finding the remainder on dividing by 3. For position 122 the answer is Design II.

We want to know which letter of the repeating pattern
$$A\;B\;C\;A\;B\;C\;A\;B\;C\;\ldots$$
occurs at any given position number.

Step 1 – Write the position number in the form
Let the required position be $$n$$. Dividing $$n$$ by 3 always gives
\[n = 3q + r, \quad 0 \le r < 3\]
where $$q$$ is the quotient and $$r$$ the remainder.

Because the pattern repeats after every 3 letters, each group of three consecutive positions is

Thus the remainder itself tells the letter:

• $$r = 1 \;\Rightarrow\; \text{letter } A$$
• $$r = 2 \;\Rightarrow\; \text{letter } B$$
• $$r = 0 \;\Rightarrow\; \text{letter } C$$

Step 2 – Check with the given positions

The letters C, B and A obtained by using the remainders are exactly what the pattern gives at those positions, so our rule works.

Conclusion
Yes, to find the letter at any position in the pattern ABC ABC ABC …, simply divide the position number by 3 and use its remainder. Remainder 1 means A, remainder 2 means B, and remainder 0 means C.

Step 1 – Identify the cycle
The border repeats the three–design cycle in exactly the same order:

Step 2 – Translate the pattern into arithmetic
Number any position by $$n$$. On dividing $$n$$ by 3 we get one of the three possible remainders.

• Remainder 1 ⇒ Triangle
• Remainder 2 ⇒ Square
• Remainder 0 ⇒ Circle

Step 3 – Find each remainder

• For position 99: \(99 = 3 \times 33 + 0\)  ⇒ remainder $$0$$, so Circle.
• For position 122: \(122 = 3 \times 40 + 2\)  ⇒ remainder $$2$$, so Square.
• For position 148: \(148 = 3 \times 49 + 1\)  ⇒ remainder $$1$$, so Triangle.

Step 4 – State the results

The required designs are therefore

• 99th place – Circle
• 122nd place – Square
• 148th place – Triangle

Objective : Prove that the two diagonal sums are the same in every $$2 \times 2$$ square of the endless number-grid.

Pick any $$2 \times 2$$ square and call the number in its top-left corner $$n$$.

• Moving one step to the right always adds $$1$$, so the entry to the right of $$n$$ is $$n+1$$.
• Moving one step down adds the same fixed amount to every number in that column; call this vertical increase $$d$$. Hence the numbers in the second row are $$n+d$$ and $$n+d+1$$.

Now add the numbers on each diagonal:

• Main diagonal (\,top-left \( \to \) bottom-right):
  $$n+(n+d+1)=2n+d+1$$
• Other diagonal (\,top-right \( \to \) bottom-left):
  $$(n+1)+(n+d)=2n+d+1$$

Both totals reduce to the same expression $$2n+d+1$$; therefore

No specific value of $$n$$ or $$d$$ was needed, so this equality holds for every position in the grid. Thus, the diagonal sums are always equal in every $$2 \times 2$$ square of the endless grid.

Step 1 : Name the known (top–left) number
Let the number written in the top–left cell be denoted by the letter $$x$$.

Step 2 : Introduce the (equal) row–sum / column–sum
Because every row as well as every column of the square is to have the same total, let us call that common total $$S$$ (you may be told its value in the actual problem).

Step 3 : Work out the three unknown entries one by one

1. Top–right cell
   The first row must add to $$S$$, so $$x + \text{(top–right)} = S\;\;\Rightarrow\;\;\text{top–right} = S - x.$$
2. Bottom–left cell
   The first column must also add to $$S$$, hence $$x + \text{(bottom–left)} = S\;\;\Rightarrow\;\;\text{bottom–left} = S - x.$$
   Notice that the top–right and bottom–left numbers automatically turn out to be equal.
3. Bottom–right cell
   Use the second column now: $$(S - x) + \text{(bottom–right)} = S.$$ Therefore $$\text{bottom–right} = S - (S - x) = x.$$

Step 4 : Write the finished square

Thus, once we know the top-left number $$x$$ and the common sum $$S$$, all the other three numbers are fixed as shown above. If the problem supplies the numerical value of $$S$$, simply substitute it in the expressions $$S - x$$ and $$x$$ to obtain the actual numbers.

Given. We have to check the algebraic rule obtained earlier for the sum of the two diagonals of any consecutive–number $$2\times 2$$ square. We are asked to do this verification by denoting the number in the top-left position by the letter $$a$$.

Step 1 : Write the four entries in algebraic form

Since the four numbers are consecutive and are arranged row-wise, each step to the right increases the number by 1 and each step downwards also increases the number by 1 after finishing the row. Thus:

Step 2 : Find the sum of the first diagonal

The first diagonal joins the top-left and the bottom-right numbers:

$$\text{First diagonal sum}=a+(a+3)=2a+3$$

Step 3 : Find the sum of the other diagonal

The other diagonal joins the top-right and the bottom-left numbers:

$$\text{Second diagonal sum}=(a+1)+(a+2)=2a+3$$

Step 4 : Compare the two sums

Both diagonal sums are identical:

$$2a+3=2a+3$$

Conclusion. The two diagonals of every $$2\times 2$$ consecutive-number square add up to the same value, namely $$2a+3$$. Hence the expression for the diagonal sums is verified.

Step 1 – Read the five numbers
From the calendar shape we have

• just above the centre = 8
• just to the left = 14
• centre = 15
• just to the right = 16
• just below = 22

Step 2 – Add them
$$8+14+15+16+22=75$$

Step 3 – Compare with the middle number
The middle number is 15 and
$$75 = 5\times 15$$
Thus the total is five times the centre number.

Step 4 – Try another set
Choose any other date, say $$20$$, as the centre.
Calendar neighbours are then

• above: $$20-7=13$$
• left: $$20-1=19$$
• centre: $$20$$
• right: $$20+1=21$$
• below: $$20+7=27$$

Sum
$$13+19+20+21+27=100$$
and $$100=5\times20$$, so the same relation holds.

Step 5 – General explanation with letters
Let the centre be $$x$$.

• above: $$x-7$$
• left: $$x-1$$
• right: $$x+1$$
• below: $$x+7$$

\[S=(x-7)+(x-1)+x+(x+1)+(x+7)\] \[S=5x\]

Observation
No matter which date you choose, the five numbers in this “plus” shape always add up to five times the number in the centre.

In the textbook a 3 × 3 square made with nine consecutive natural numbers is shown.

If we denote the number written in the centre by $$a$$ then, because the numbers rise by 1 each time as we move left to right and top to bottom, the whole square can be written only in terms of $$a$$ as

\[ \begin{array}{ccc} a-4 & a-3 & a-2\\[2pt] a-1 & a & a+1\\[2pt] a+2 & a+3 & a+4 \end{array} \]

(Check: if $$a = 5$$ we indeed get 1 – 9.)

1. Will the property shown in the book always hold?

The property was: “Add the five numbers lying at the four corners and at the centre; the sum is 25 in the example shown.”
Taking the general array above, the required five numbers are

• top-left corner  $$a-4$$
• top-right corner  $$a-2$$
• bottom-left corner  $$a+2$$
• bottom-right corner  $$a+4$$
• centre  $$a$$

Their sum is

$$ S = (a-4) + (a-2) + (a+2) + (a+4) + a = 5a. $$

Because $$S = 5a$$, the total is always exactly five times the middle number, no matter which consecutive numbers were taken. Hence the phenomenon will always happen.

2. Why does this happen?

Each number we add occurs in a pair equidistant from the centre:

$$ (a-4)+(a+4)=2a, \;\; (a-2)+(a+2)=2a. $$

Adding the centre $$a$$ itself gives $$2a+2a+a=5a$$. Since every term is a multiple of $$a$$, the result is automatically a multiple of the centre number.

3. Other shapes having the same kind of property

The same idea—taking numbers that occur in equal steps on the two sides of $$a$$—works for many other figures in the same 3 × 3 square. A few examples are listed below; in each case the sum turns out to be a fixed multiple of one of the numbers in the figure.

Whenever the chosen numbers can be paired so that each pair is equally above and below the centre value $$a$$, their sum will turn out to be a whole-number multiple of $$a$$. Thus many symmetric shapes in the square possess this attractive property.

What the pattern looks like
In the textbook picture, Step 1 is one complete square made with 4 matchsticks. In Step 2 a second square is placed right next to the first one; because the two squares share one side, the new square needs only 3 more matchsticks. In every new step we simply keep attaching one more square to the row; each additional square therefore contributes exactly 3 fresh sticks.

Collecting the data

The ‘+3’ difference remains the same every time, so $$M$$ grows linearly with $$n$$.

Writing an algebraic rule
The first square already contributes 4 sticks. After that, each of the remaining $$(n-1)$$ squares adds 3 sticks. Hence

$$M = 4 + 3\,(n-1)$$

Simplifying,

$$M = 4 + 3n - 3 = 3n + 1$$

This single formula tells us the number of matchsticks in any step $$n$$ without having to draw the figure.

Calculations for the required steps

• For Step 33: $$M = 3\times33 + 1 = 99 + 1 = 100$$
• For Step 84: $$M = 3\times84 + 1 = 252 + 1 = 253$$
• For Step 108: $$M = 3\times108 + 1 = 324 + 1 = 325$$

So, instead of drawing 33, 84 or 108 squares, the pattern rule $$M = 3n + 1$$ lets us obtain the answers in just a few seconds.

Setting the scene  We already know that in this match-stick pattern, if the step number is called $$x$$ then the required number of sticks is given by the rule

$$\text{Number of sticks}=2x+1$$

The textbook now writes the expression with another letter, namely $$y$$, and asks whether this new expression also works and whether it is really any different from the old one.

1. Check the rule with concrete values

These are exactly the numbers we had earlier (3, 5, 7 …), so $$2y+1$$ gives the correct count at every step.

2. Are the two expressions different?

In algebra a letter is just a placeholder for a number. Whether we write

$$2x+1,\; 2y+1,\; 2n+1 \text{ or } 2k+1$$

does not change the rule at all, provided the letter stands for the same quantity (here, the step number). Only the name of the variable has been changed; the numerical part “$$2\square+1$$” is identical.

Conclusion

Yes, the expression $$2y+1$$ gives the correct number of matchsticks for every step, and it is exactly the same rule as $$2x+1$$ — the difference is only the choice of the variable letter.

From the figures given in the textbook we already know the counts for the first two steps:

The growth from one step to the next is regular: exactly two additional sticks of each kind are needed.

• Horizontal: $$+2$$ each step
• Diagonal  : $$+2$$ each step

If we denote by $$H_n$$ and $$D_n$$ the numbers of horizontal and diagonal sticks in Step $$n$$, then for every $$n\ge 2$$

$$H_n = H_{n-1}+2,\qquad D_n = D_{n-1}+2.$$

Starting with $$H_1 = 3,\; D_1 = 2$$ we obtain successively

Therefore the required numbers are:

• Step 3: $$7$$ horizontal, $$6$$ diagonal
• Step 4: $$9$$ horizontal, $$8$$ diagonal

(In general, $$H_n = 2n + 1$$ and $$D_n = 2n$$.)

Step 1&colon; Observe and record the counts

Place the staircase so that its right edge is vertical&colon; we call this Orientation A.
Turn the same staircase through a right angle so that its left edge is vertical&colon; we call this Orientation B.

The numbers in each column increase by 1 every time the step number goes up by 1.

Step 2&colon; Write an algebraic rule for each orientation

• Orientation A starts with 2 sticks and gains 1 stick per new step.
    So at Step y&comma;

$$\text{sticks in A}=2+(y-1)=y+1$$

• Orientation B starts with 1 stick and also gains 1 stick per new step. Hence

$$\text{sticks in B}=1+(y-1)=y$$

Step 3&colon; Do the two expressions add up to $$2y+1$$?

Add the two expressions just obtained&colon;

$$\bigl(y+1\bigr)+y=2y+1$$

The sum is exactly $$2y+1$$, so yes&comma; the two separate formulae together make the given expression.

We know one pipe is 20 m long.

Let the length of the extra pipe be denoted by the letter-number $$k$$ metres.

When the two pipes are joined end-to-end, the total (combined) length is obtained by adding the two individual lengths:

\[\text{Total length} = 20 + k\]

Hence, the required algebraic expression for the combined length (in metres) is $$20 + k$$.

For every kind of note we multiply the number of notes by the value of one note and then add the three amounts.
Let us do this row by row.

The fully completed table is therefore:

Step 1 – Understand the two parts of the job.

• Starting the machine: always takes $$10$$ seconds, no matter how much grain we grind.
• Grinding the grain: takes $$8$$ seconds for each kilogram.

Step 2 – Write an algebraic expression for the grinding time.

If $$y$$ kg of grain are to be ground, the grinding part needs

$$8 \times y$$ seconds.

Step 3 – Combine the two times to get the total time.

Total time = starting time + grinding time

$$10 + 8 \times y$$

Step 4 – Match with the given options.

• (a) $$10 + 8 + y$$ — adds instead of multiplying; incorrect.
• (b) $$(10 + 8) \times y$$ — multiplies the 10-second start time by $$y$$; incorrect.
• (c) $$10 \times 8 \times y$$ — multiplies everything; incorrect.
• (d) $$10 + 8 \times y$$ — exactly what we obtained; correct.
• (e) $$10 \times y + 8$$ — multiplies the wrong number by $$y$$; incorrect.

Therefore, the expression that represents the required time is option (d).

We are looking at any two rows and three consecutive columns of a calendar. Every column in a calendar stands for one particular day-of-the-week, so

• moving one step to the right increases the date by $$1$$,
• moving one step down (to the next week) increases the date by $$7$$.

Let us copy the unknown calendar strip and fill in the dates step by step. The six positions are named by their relative places in the grid.

The question already tells us the number in the bottom middle cell and in the bottom left cell:

We now fill the remaining four cells.

1. Bottom right
   Going one step to the right from $$w$$ adds $$1$$:
   $$\text{bottom right}=w+1$$
2. Top middle
   Going one step up from $$w$$ subtracts $$7$$:
   $$\text{top middle}=w-7$$
3. Top left
   Top left is one step left of top middle, so subtract another $$1$$:
   $$\text{top left}=(w-7)-1=w-8$$
4. Top right
   Top right is one step right of top middle, so add $$1$$:
   $$\text{top right}=(w-7)+1=w-6$$

Putting everything back in the table we get

Thus the required algebraic expressions for the six positions, when the bottom-middle date is $$w$$, are:

• top left : $$w-8$$
• top middle : $$w-7$$
• top right : $$w-6$$
• bottom left : $$w-1$$  (already given)
• bottom middle : $$w$$  (given)
• bottom right : $$w+1$$

The given expression is $$23 - 10 \times 2$$.

Step 1 – Apply the order of operations (BODMAS):

First, carry out the multiplication.

$$10 \times 2 = 20$$

Step 2 – Substitute the result into the expression and subtract:

$$23 - 20 = 3$$

Hence, the value of the expression is $$3$$.

The numerical expression is

$$83 + 28 - 13 + 32$$

Step 1: Combine the first two numbers.

$$83 + 28 = 111$$

Now the expression becomes

$$111 - 13 + 32$$

Step 2: Perform the subtraction.

$$111 - 13 = 98$$

The expression is now

$$98 + 32$$

Step 3: Add the remaining numbers.

$$98 + 32 = 130$$

Therefore, the value of the given expression is

\[130\]

We have to simplify the numerical expression

$$34 - 14 + 20$$

Step 1 – Start from the leftmost operation (because addition and subtraction have the same priority and we proceed left to right).

Subtract: $$34 - 14 = 20$$

After carrying out this subtraction, the expression becomes

$$20 + 20$$

Step 2 – Add the two numbers:

$$20 + 20 = 40$$

Thus the value of the given expression is

\[40\]

We must simplify the numerical expression step by step, following the correct order of operations (brackets first, then addition/subtraction from left to right).

1. Write the given expression.
   $$42 + 15 - (8 - 7)$$

2. Simplify the brackets.
   First evaluate what is inside the inner brackets: $$8 - 7 = 1$$ Insert this result back into the main expression: $$42 + 15 - 1$$

3. Add the first two numbers.
   $$42 + 15 = 57$$ So the expression becomes $$57 - 1$$

4. Subtract to obtain the final value.
   $$57 - 1 = 56$$

Hence, the value of the expression is

\[56\]

Step 1 – Identify the order of operations.
According to the BODMAS/PEMDAS rule, we must simplify the brackets first.

Step 2 – Simplify the expression inside the brackets.
The bracket contains the sum $$18 + 13$$.
Compute it: $$18 + 13 = 31$$.

The whole expression now becomes $$68 - 31$$.

Step 3 – Subtract.
Find the difference: $$68 - 31 = 37$$.

Final value.

\[68 - (18 + 13) = 37\]

The given numerical expression is
$$7 \times 4 + 9 \times 6$$

Step 1 – Multiply  $$7$$ by $$4$$:

$$7 \times 4 = 28$$

Step 2 – Multiply  $$9$$ by $$6$$:

$$9 \times 6 = 54$$

Step 3 – Add the two products:

$$28 + 54 = 82$$

Thus, the value of the given expression is

\[82\]

The expression to be simplified is: $$20 + 8 \times (16 - 6)$$

Step 1 – Simplify the bracket

The first operation (by the BODMAS rule) is inside the brackets:

$$16 - 6 = 10$$

Replace the bracketed part by its value:

$$20 + 8 \times 10$$

Step 2 – Perform the multiplication

Next, carry out the multiplication before addition:

$$8 \times 10 = 80$$

The expression now becomes:

$$20 + 80$$

Step 3 – Perform the addition

Add the two numbers:

$$20 + 80 = 100$$

The simplified value of the given expression is therefore

\[100\]

Let $$x$$ be the number of plates of Jowar roti and $$y$$ be the number of plates of Pulao supplied in one day.

Step 1 – Cost of each item

• Cost of one plate of Jowar roti  =  ₹30.
   Cost of $$x$$ such plates  =  $$30 \times x = 30x$$.
• Cost of one plate of Pulao  =  ₹20.
   Cost of $$y$$ such plates  =  $$20 \times y = 20y$$.

Step 2 – Total earning for the day

Step 3 – Check the given options

Therefore, only option (a) represents the correct expression for the total amount earned that day.

Step 1 – Understand the information
Pushpita has three kinds of customers on Independence day:

• Only champak buyers  →  $$p$$ customers
• Only marigold buyers  →  $$q$$ customers
• Buyers of both champak and marigold  →  $$r$$ customers

Every customer—whoever he or she is—gets exactly one little national flag.

Step 2 – List the distinct customers
A customer can belong to one, and only one, of the above three groups, because the headings already separate them:

• The first group (only champak) contains $$p$$ different people.
• The second group (only marigold) contains $$q$$ different people.
• The third group (both flowers) contains $$r$$ different people.

Notice that a person counted in the third line is not counted in the first or second line, so there is no double-counting.

Step 3 – Add them to get the total customers
Total customers = (only champak) + (only marigold) + (both)
\[\text{Total customers}=p+q+r\]

Step 4 – Translate customers to flags
Each customer receives exactly one flag, so the number of flags is the same as the number of customers:

Step 5 – Match with the given options
The expression $$p+q+r$$ appears as option (a).

Given: In the first week Radha rides 5 km each day and, every new week, she increases the daily distance by $$z$$ km.

Step 1 – Daily and weekly distances

• Week 1: daily distance = $$5$$ km
• Week 2: daily distance = $$5+z$$ km (increased by $$z$$)
• Week 3: daily distance = $$5+2z$$ km (increased again by another $$z$$)

Step 2 – Convert each week’s daily distance into the total for that week

Each week has $$7$$ days, so multiply each daily figure by $$7$$:

• Week 1: $$7 \times 5 = 35$$ km
• Week 2: $$7 \times (5+z) = 35 + 7z$$ km
• Week 3: $$7 \times (5+2z) = 35 + 14z$$ km

Step 3 – Add the three weekly totals

$$\begin{aligned} \text{Total distance} &= 35 + (35 + 7z) + (35 + 14z) \\ &= (35+35+35) + (7z + 14z) \\ &= 105 + 21z \,\text{km}. \end{aligned}$$

How to read the diagram

• Each oval carries an expression.
• Each boxed arrow shows the operation to be performed when we move in the direction of the arrow.

1. Upper path (moving from the centre towards the left)

1. Start from the centre oval: $$w + 2$$.
2. First arrow is labelled “−5”, so  $$w + 2 - 5 = w - 3$$. The oval just left of the centre therefore contains $$w - 3$$ (this matches the figure).
3. Next arrow is “× 3”, so multiply the last result by 3: $$3\bigl(w - 3\bigr) = 3w - 9$$.

Hence the blank oval on the extreme left of the upper path is $$3w - 9$$.

2. Lower path (moving from the centre towards the right)

1. Start again from the centre: $$w + 2$$.
2. Arrow shows “−4”, so  $$w + 2 - 4 = w - 2$$. This is the first blank oval on the right side.
3. Next arrow is “× 3”, so  $$3\bigl(w - 2\bigr) = 3w - 6$$, which indeed equals the given final oval.

Thus the blank oval just right of the centre is $$w - 2$$.

3. Lower path (moving from the centre towards the left)

1. Start from the centre: $$w + 2$$.
2. Arrow shows “−8”, so  $$w + 2 - 8 = w - 6$$. This fills the middle blank oval on the left.
3. Next arrow (further left) is “−4”, so  $$w - 6 - 4 = w - 10$$.

Hence the extreme-left blank oval on the lower path is $$w - 10$$.

4. Summary of all missing expressions

• Upper path, leftmost oval: $$3w - 9$$
• Lower path, leftmost oval: $$w - 10$$
• Lower path, middle oval (after “−8”): $$w - 6$$
• Lower path, rightmost blank oval (after “−4”): $$w - 2$$

Understanding the situation
The rope is cut exactly once, but before cutting it we may fold it a certain number of times. Whenever we fold, additional layers are placed one over the other, so the scissor (or knife) passes through all the layers together.

Step 1  Relate number of folds to number of layers

• 0 folds → 1 layer
• 1 fold → 2 layers
• 2 folds → 3 layers
• …
• r folds → r + 1 layers

Step 2  How many cut points are produced?
The single cut meets every layer, therefore after unfolding we see one cut point on each layer. Thus the number of cut points along the original straight rope is also r + 1.

Step 3  Convert cut points into pieces
A straight object cut at n different points separates into n + 1 pieces. Here n = r + 1, so

\[ \text{pieces}= (r+1) + 1 = r + 2 \]

Step 4  Check with the given information
r = 0: pieces = 0+2 = 2 (matches the figure)
r = 1: pieces = 1+2 = 3 (matches the figure)

Step 5  Use the formula
For r = 10 folds: $$\text{pieces}=10+2=12$$.

Conclusion
When the rope is folded r times and cut once, we obtain $$r+2$$ pieces. In particular, ten folds give 12 pieces.

Step 1 : Observe the first few figures

We notice that each time we add one more square, we do not need another 4 matchsticks because the new square shares one complete side with the previous square.

Step 2 : Work out the change from one figure to the next

• Going from 1 square to 2 squares, matchsticks increase from 4 to 7  ⇒  increase of $$3$$.
• Going from 2 squares to 3 squares, matchsticks increase from 7 to 10  ⇒  increase of $$3$$ again.
• Every additional square therefore needs exactly $$3$$ new matchsticks.

Step 3 : Build an algebraic rule

Start with 1 square (needs 4 sticks). For each extra square beyond the first, add 3 sticks.

If $$n$$ is the number of squares, we can write

Number of matchsticks  =  $$4 + (n-1)\times 3$$.

Simplify this expression:

$$4 + (n-1)\times 3 = 4 + 3n - 3 = 3n + 1$$.

So the general rule is

\[ \text{Matchsticks}(n) = 3n + 1 \]

Step 4 : Use the rule for the required cases

• For 10 squares  ($$n = 10$$):
  $$3\times 10 + 1 = 30 + 1 = 31$$ matchsticks.
• For $$w$$ squares: simply replace $$n$$ by $$w$$ in the rule:
  $$3w + 1$$ matchsticks.

First, copy a few terms of the list that is given in the question.

$$\text{Position} : 1 \; 2 \; 3 \; 4 \; 5 \; 6 \; 7 \; 8 \; 9 \; 10 \; 11 \; 12 \;\ldots$$
$$\text{Colour} : R \; Y \; G \; Y \; R \; Y \; G \; Y \; R \; Y \; G \; Y \;\ldots$$

We see that after every four places the same order Red, Yellow, Green, Yellow repeats. Hence the sequence has a period (length of one full cycle) equal to 4.

1. A quick way to locate any position

For any position number $$n$$ we divide $$n$$ by 4 and read its remainder (also called the modulus). There are only four possible remainders:

2. Finding the three required colours

1. Position $$90$$

   Divide by 4:

   $$90 = 4 \times 22 + 2$$ (since $$4 \times 22 = 88$$ and $$90-88 = 2$$).

   Remainder $$=2 \;\Rightarrow$$ second place ⇒ Yellow.

2. Position $$190$$

   $$190 = 4 \times 47 + 2$$ (because $$4 \times 47 = 188$$).

   Remainder $$=2 \;\Rightarrow$$ second place ⇒ Yellow.

3. Position $$343$$

   $$343 = 4 \times 85 + 3$$ (because $$4 \times 85 = 340$$).

   Remainder $$=3 \;\Rightarrow$$ third place ⇒ Green.

3. General expressions for every colour

• Red appears at positions $$1,5,9,13,\ldots$$. Each of these numbers can be written as \[ n = 4k + 1 \] where $$k$$ is any whole number $$0,1,2,3,\ldots$$
• Green appears at positions $$3,7,11,15,\ldots$$. Each of these numbers is of the form \[ n = 4k + 3 \]
• Yellow appears at positions $$2,4,6,8,10,12,\ldots$$, i.e. whenever the remainder is $$2$$ or $$0$$. Therefore the positions are \[ n = 4k \quad \text{or} \quad n = 4k + 2 \]

Writing them together:

4. Summary

At position $$90$$ ⇒ Yellow;
At position $$190$$ ⇒ Yellow;
At position $$343$$ ⇒ Green.

Step–by–step analysis of the picture pattern

1  Look for a numerical pattern
The first three figures supplied in the textbook look like this (draw them as small square–grids):

• Step 1  →  1 small square
• Step 2  →  4 small squares (arranged 2 × 2)
• Step 3  →  9 small squares (arranged 3 × 3)

The numbers 1, 4, 9 are the perfect squares of the counting numbers 1, 2, 3.

2  Express the count for any step n
If a step contains $$n$$ little squares along one side, then the arrangement is an $$n \times n$$ grid, so the total number of little squares is obtained by multiplying:

3  Use the rule to answer the specific questions

• Step 4:  $$N(4)=4^{2}=16$$ squares
• Step 10: $$N(10)=10^{2}=100$$ squares
• Step 50: $$N(50)=50^{2}=2500$$ squares

Hence the general formula for the number of squares in Step $$n$$ is

Counting vertices instead of squares

When there are $$n$$ little squares along a side, we need one extra corner to close the row, so each horizontal and vertical line shows $$(n+1)$$ points. Putting these in a rectangular array gives

Therefore the relationship changes from $$n^{2}$$ to $$(n+1)^{2}$$ when we shift our attention from squares to their vertices.

Illustrations:

• Step 4 has $$V(4)=(4+1)^{2}=25$$ vertices
• Step 10 has $$V(10)=121$$ vertices
• Step 50 has $$V(50)=2601$$ vertices

Summary

• Squares in Step $$n$$ : $$n^{2}$$
• Vertices in Step $$n$$ : $$(n+1)^{2}$$