Chapter 2: Arithmetic Expressions

Step 1 – Identify the daily expense
Mallika spends ₹$$25$$ every school day on lunch.

Step 2 – Identify (or define) the number of days she has lunch at school
From Monday to Friday there are 5 school days. Let us denote this number by the variable $$d$$, so for this situation $$d = 5$$.

Step 3 – Form the algebraic expression for the total cost
If one day costs ₹$$25$$, then $$d$$ days will cost $$25 \times d$$.

Step 4 – Substitute $$d = 5$$ (Monday to Friday)

Therefore, Mallika spends ₹$$125$$ on lunch in one school week.

We have to compare two sums:

$$1023 + 125$$  and  $$1022 + 128$$

Step 1 – Add the numbers separately.

\[ 1023 + 125 = 1148 \]

\[ 1022 + 128 = 1150 \]

Step 2 – Compare the results.

$$1150 \gt 1148$$

Therefore, $$1022 + 128$$ is greater than $$1023 + 125$$.

We have to compare two numerical expressions:

$$113 - 25 \qquad\text{and}\qquad 112 - 24$$

Method 1 – Direct subtraction

• Compute the first difference:

  $$113 - 25 = 88$$

• Compute the second difference:

  $$112 - 24 = 88$$

Both evaluations give the same result, $$88$$. Hence neither expression is greater; they are equal.

Method 2 – Using a property of subtraction

If we add the same number to the minuend and the subtrahend, the difference does not change:

$$a - b = (a + k) - (b + k)$$

Put $$a = 112,\; b = 24,\; k = 1$$:

$$112 - 24 = (112 + 1) - (24 + 1) = 113 - 25$$

This shows algebraically that the two given expressions are identical, so their values must be equal.

Conclusion

$$113 - 25 = 112 - 24 = 88$$
The two expressions are equal; neither is greater.

Step 1 : Define variables

Let $$m$$ be the number of marbles Mallesh brought.
Let $$b$$ be the number of bags Arun brought.
Let $$n$$ be the number of marbles in each bag.

Step 2 : Write expressions

Mallesh’s marbles: $$m = 30$$.

Arun’s marbles: $$b = 5$$ and $$n = 4$$ so the number of marbles Arun brought is $$b \\times n = 5 \\times 4$$.

Step 3 : Simplify Arun’s marbles

Calculate multiplication first (order of operations):
$$5 \\times 4 = 20$$

Step 4 : Total marbles

Total marbles $$T$$ brought to the playground:
$$T = m + (b \\times n) = 30 + 20$$

Adding gives:

Conclusion

Mallesh and Arun together brought $$50$$ marbles to the playground.

First write down the price of each item Irfan buys:

• Pack of biscuits  $$= 15$$ rupees
• Packet of toor dal $$= 56$$ rupees

Find the total amount he has to pay:

$$15 + 56$$

Now compare this total with the money Irfan actually gives to the shopkeeper (₹100).
Change = Amount given − Total cost

Therefore an algebraic expression for the change is

\[100 - (15 + 56)\]

(If we simplify, Irfan will get back $$29$$ rupees, but the required expression itself is $$100 - (15 + 56)$$.)

Given: Madhu first makes the drone go up $$6\text{ m}$$ from the terrace and then brings it down $$4\text{ m}$$.

We treat upward movement as positive and downward movement as negative.

1. Upward displacement  = $$+6\text{ m}$$.

2. Downward displacement  = $$-4\text{ m}$$.

3. Total (net) displacement from the terrace:
   $$\bigl(+6\text{ m}\bigr) + \bigl(-4\text{ m}\bigr)$$

4. Simplify the arithmetic expression:
   $$6 - 4 = 2$$

Thus, the drone finally hovers $$2\text{ m}$$ above the terrace.

First collect the information given in the statement:

• Number of dosas ordered  =  4
• Cost of one dosa  =  $$₹\,23$$
• Tip for the waiter  =  $$₹\,5$$

1. Write the cost of all the dosas together.

The cost of one dosa is $$23\text{ rupees}$$. For 4 dosas the cost will be

$$23 \times 4$$

2. Add the tip.

The tip is a fixed amount of $$5\text{ rupees}$$, so the total cost becomes

$$23 \times 4 + 5$$

3. Present the final arithmetic expression.

The required expression is

One may also write it equivalently as $$4 \times 23 + 5$$.

The teacher asked the 33 children (Ruby is not included) to sit in groups of 5.

To find how many complete groups of 5 can be made from 33, we perform short division:

33 ÷ 5 gives quotient $$6$$ and remainder $$3$$ because

$$5 \times 6 = 30 \lt 33 \text{ and } 33 - 30 = 3.$$

Therefore:

• Number of complete groups  =  $$6$$.
• Children left without a group ("out")  =  $$3$$.

This situation can be written as the arithmetic expression

which Ruby jotted down as “$$6 \times 5 + 3$$” – i.e. “3 more than six fives”.

So Ruby’s expression correctly represents the grouping of the 33 students when the teacher called out 5.

Step 1 – Number of new 2 kg packets prepared from 100 kg rice
Each packet has a mass of 2 kg.
Hence the number of packets that can be made is
$$\dfrac{100}{2}$$.

Step 2 – Total number of packets now
Raghu already had 4 packets.
Therefore, the total number of 2 kg packets he now possesses is
$$\dfrac{100}{2} + 4$$.

Required expression

Identification of terms

• First term : $$\dfrac{100}{2}$$
• Second term : $$4$$

The target is to make a total of $$\text{₹}\,432$$ using the available denominations (₹1, ₹5, ₹10, ₹20, ₹50, ₹100).

To keep the number of pieces small we begin with the largest value and work downwards, keeping track of the unpaid balance at every step.

1. ₹100 notes
   $$\frac{432}{100}=4$$ complete hundreds, remainder $$432-4\times100=32$$.
   Use 4 notes of ₹100; balance left = ₹32.
2. ₹50 notes
   The balance ₹32 is less than ₹50 ⇒ take 0 notes of ₹50.
3. ₹20 notes
   $$\frac{32}{20}=1$$ twenty, remainder $$32-1\times20=12$$.
   Use 1 note of ₹20; balance left = ₹12.
4. ₹10 notes
   $$\frac{12}{10}=1$$ ten, remainder $$12-1\times10=2$$.
   Use 1 note of ₹10; balance left = ₹2.
5. ₹5 coins
   The balance ₹2 is less than ₹5 ⇒ take 0 coins of ₹5.
6. ₹1 coins
   Pay the last ₹2 with ₹1 coins: $$2=2\times1$$.
   Use 2 coins of ₹1.

Summary:

Total paid: $$400+20+10+2=432$$.

Hence Kannan can pay the shopkeeper with

• 4 notes of ₹100,
• 1 note of ₹20,
• 1 note of ₹10,
• 2 coins of ₹1.

Step 1 – Understand the arithmetic expression

The expression is  $$5 \times 2 + 3$$. This reads as:

• "Five times two"  means five equal groups with  2 objects in each group.
• “+ 3”  means we must add three more single objects after forming those groups.

Step 2 – Compute the total number of objects

First do the multiplication, then the addition (as the order of operations tells us):

$$5 \times 2 = 10$$

$$10 + 3 = 13$$

So any drawing that represents this expression must finally contain exactly 13 objects arranged as  five groups of two  plus  three singles.

Step 3 – Compare the two given pictorial arrangements

1. Picture 1: It shows five small circles drawn in a row, each labelled “2”, meaning five distinct clusters, every cluster containing two dots. Besides those clusters, three individual dots are drawn separately. Counting:
   • Groups:  5 groups × 2 objects = 10 objects
   • Singles:  3 objects
   • Total:  10 + 3 = 13 objects.
2. Picture 2: It shows two clusters labelled “5” (so each of the two clusters contains five dots) and then three single dots. Counting:
   • Groups:  2 groups × 5 objects = 10 objects
   • Singles:  3 objects
   • Total:  10 + 3 = 13 objects.

Step 4 – Decide which picture matches the structure of the expression

Even though both drawings finally have 13 dots, only Picture 1 shows “five groups of two” followed by “+ 3 singles”.
In Picture 2 the grouping is “two groups of five”, which would match the expression $$2 \times 5 + 3$$, not $$5 \times 2 + 3$$.

Conclusion

The arrangement in Picture 1 represents the arithmetic expression  $$5 \times 2 + 3$$.

Let the amount Irfan pays be denoted by the number $$100$$.

The two items bought are

• Biscuit packet  $$15$$ rupees
• Toor dal packet  $$56$$ rupees

We shall find the balance (change) by two different but equivalent ways and check that both yield the same result.

Method 1  – using a single expression with brackets

The total cost of the two packets is added first because the sum is inside brackets:

$$15 + 56 = 71$$

Now subtract this total from the money paid:

$$100 - 71 = 29$$

Hence, by this method the change Irfan should receive is $$29$$ rupees.

Method 2  – successive subtraction (step-by-step)

1. Subtract the price of the biscuit packet from the amount paid:
     $$100 - 15 = 85$$
2. From the remainder, subtract the price of the toor dal packet:
     $$85 - 56 = 29$$

The balance obtained again equals $$29$$ rupees.

Verification that the two procedures are the same algebraically

The numerical example illustrates the general property

Putting $$a = 100$$, $$b = 15$$ and $$c = 56$$ gives

$$100 - (15 + 56) = (100 - 15) - 56$$

Both sides evaluate to $$29$$, confirming that either way of grouping the operations leads to the same final answer.

Thus Irfan must receive ₹ 29 back as change.

Step 1 – Write the given expression

$$500-(250-100)$$

Step 2 – Recall the rule for removing brackets after a minus sign

When a bracket is preceded by a minus sign, the sign of every term inside that bracket is reversed on removing the bracket:

$$a-(b-c)=a-b+c$$

Step 3 – Apply the rule

$$500-(250-100)=500-250+100$$

Step 4 – Verify the result (optional, for confidence)

• Evaluate by keeping the bracket:
  $$250-100=150\ 500-150=350$$
• Evaluate the bracket-free form:
  $$500-250+100=250+100=350$$

Both methods give the same answer, so the transformation is correct.

Therefore, the expression without brackets is

\[500-250+100\]

Given

• Coins in Bag I = $$28$$
• Coins in Bag II = $$35$$

Step 1 – Coins left in Bag II after gifting

Hira gifts $$10$$ coins from the second bag:

$$35 - 10$$ coins remain in Bag II.

Step 2 – Total coins left with Hira

Add the coins still in each bag:

$$28 + (35 - 10)$$

Step 3 – Simplify (optional)

$$28 + 25 = 53$$

Thus, the required arithmetic expression (with its simplified value) is

Let us list the prices given in the question.

• Cost of one vegetable cutlet = $$₹43$$
• Cost of one rasgulla = $$₹24$$

Step 1 : Amount for one person

For a single person (either Lhamo or Norbu), the bill consists of

$$43 + 24$$

So, the amount one person has to pay is $$43 + 24$$.

Step 2 : Amount for two persons

Both Lhamo and Norbu order the same items. Hence we multiply the amount for one person by $$2$$ :

Amount for two persons = $$2\times(43 + 24)$$

Step 3 : Final expression

Thus, an algebraic (arithmetic) expression for the total amount is

\[2(43 + 24)\]

You may expand or simplify it further if required:

$$2(43 + 24) = 2\times43 + 2\times24 = 86 + 48 = 134$$

Therefore, the hotel will charge $$₹134$$ in total, and the required expression is $$2(43 + 24)$$.

Step 1 — Number of boy scouts

The scouts are marching in $$4$$ rows with $$5$$ scouts in each row.

\[\text{Total scouts}=4 \times 5\]

Therefore, $$4 \times 5 = 20$$ scouts.

Step 2 — Number of girl guides

The guides are marching in $$3$$ rows with $$5$$ guides in each row.

\[\text{Total guides}=3 \times 5\]

Therefore, $$3 \times 5 = 15$$ guides.

Step 3 — Total participants

Add the two totals:

$$\text{Total participants}=20+15$$

Alternatively, combine the common factor $$5$$ first:

$$4\times5+3\times5=(4+3)\times5=7\times5$$

\[7 \times 5 = 35\]

Therefore, $$35$$ scouts and guides are marching in the parade.

We already know the product

• $$53 \times 18 = 954$$.

To make use of this result, rewrite the required number $$63$$ with the help of $$53$$:

$$63 = 53 + 10$$.

Now apply the distributive law of multiplication over addition:

$$63 \times 18 = (53 + 10) \times 18$$

According to the distributive property,

$$ (a + b) \times c = a \times c + b \times c $$.

Comparing, let $$a = 53$$, $$b = 10$$ and $$c = 18$$. Therefore

$$ (53 + 10) \times 18 = 53 \times 18 + 10 \times 18 $$

We already have $$53 \times 18 = 954$$, and $$10 \times 18$$ is easy to find:

$$10 \times 18 = 180$$

Add the two partial products:

$$954 + 180 = 1134$$

Hence,

\[63 \times 18 = 1134\]

Goal: Evaluate $$97 \times 25$$ quickly by using properties of numbers instead of ordinary long multiplication.

Method 1 — Using $$25=\dfrac{100}{4}$$

1. Since $$25=\dfrac{100}{4}$$, rewrite the product:
   $$97 \times 25 = 97 \times \dfrac{100}{4}$$.
2. By the associative property of multiplication, group $$97\times100$$ first:
   $$97 \times \dfrac{100}{4} = \dfrac{97 \times 100}{4}$$.
3. Multiplying by $$100$$ is easy: just append two zeros:
   $$97 \times 100 = 9700$$.
4. Now divide by $$4$$ (or think “take a quarter”):
   $$\dfrac{9700}{4} = 2425$$, because $$4 \times 2425 = 9700$$.

Method 2 — Using the distributive property

1. Write $$97$$ as $$100-3$$.
2. Apply the distributive law $$a(b-c)=ab-ac$$:
   $$(100-3) \times 25 = 100 \times 25 - 3 \times 25$$.
3. Calculate each product:
   $$100 \times 25 = 2500$$ and $$3 \times 25 = 75$$.
4. Subtract:
   $$2500 - 75 = 2425$$.

Therefore:

\[97 \times 25 = 2425\]

Let us pick 7 as the favourite number. We now create many different arithmetic expressions and verify, step by step, that each one has the value 7.

1. Expression : $$5 + 2$$

   Calculation : $$5 + 2 = 7$$

2. Expression : $$10 - 3$$

   Calculation : $$10 - 3 = 7$$

3. Expression : $$14 \div 2$$

   Calculation : $$14 \div 2 = 7$$

4. Expression : $$4 \times 2 - 1$$

   Step 1 : $$4 \times 2 = 8$$
   Step 2 : $$8 - 1 = 7$$

5. Expression : $$2^3 - 1$$

   Step 1 : $$2^3 = 8$$
   Step 2 : $$8 - 1 = 7$$

6. Expression : $$(6 + 8) \div 2$$

   Step 1 : $$6 + 8 = 14$$
   Step 2 : $$14 \div 2 = 7$$

7. Expression : $$3 \times 4 - 5$$

   Step 1 : $$3 \times 4 = 12$$
   Step 2 : $$12 - 5 = 7$$

8. Expression : $$-7 + 14$$

   Calculation : $$-7 + 14 = 7$$

9. Expression : $$\dfrac{63}{9}$$

   Calculation : $$63 \div 9 = 7$$

10. Expression : $$1 + 2 + 4$$

    Calculation : $$1 + 2 + 4 = 7$$

11. Expression : $$3^2 - 2$$

    Step 1 : $$3^2 = 9$$
    Step 2 : $$9 - 2 = 7$$

12. Expression : $$(18 - 4) \div 2$$

    Step 1 : $$18 - 4 = 14$$
    Step 2 : $$14 \div 2 = 7$$

Every calculation has been carried out clearly, showing that each listed expression indeed equals 7. You can, of course, invent many more by combining addition, subtraction, multiplication, division, and parentheses in creative ways.

Recall the basic property of integers:

So a subtraction sign “−” may be replaced by “+” followed by the negative of the next term. We now check, with different examples, that this rewriting leaves the value unchanged.

Example 1 : two numbers

Original: $$13 - 7 = 6$$.

After the change: $$13 + (-7) = 6$$.

Both give $$6$$ ⇒ no change.

Example 2 : several numbers

Expression: $$25 - 12 + 3 - 8$$.

• As written: 25 − 12 = 13; 13 + 3 = 16; 16 − 8 = 8.

Re-written: $$25 + (-12) + 3 + (-8)$$.

• 25 + (–12) = 13; 13 + 3 = 16; 16 + (–8) = 8.

Result still $$8$$.

Example 3 : negative numbers present

Expression: $$-5 - (-8) - 6$$.

• Original: –5 – (–8) = –5 + 8 = 3; 3 – 6 = –3.

Re-written: $$-5 + (-(-8)) + (-6) = -5 + 8 + (-6)$$.

• –5 + 8 = 3; 3 + (–6) = –3.

Same answer $$-3$$.

Example 4 : algebraic expression

Let $$E = x - y + 7 - 3y$$.

Replacing subtraction: $$E = x + (-y) + 7 + (-3y)$$ – clearly the same number for every $$x, y$$ because of (1).

All examples confirm that replacing each subtraction by the addition of the additive inverse does not change the value of the expression.

Background : the Token Model recalled from Class 6

• We represent the integer $$+1$$ by one positive token (for example, a blue square marked “+”).
• We represent the integer $$-1$$ by one negative token (for example, a red square marked “–”).
• Putting one positive and one negative token together makes a zero-pair because $$+1 + (-1) = 0$$. In pictures the two opposite coloured tokens cancel each other.

Any integer is shown by collecting the required number of positive or negative tokens. For instance, $$+4$$ is four positive tokens, $$-3$$ is three negative tokens, and $$0$$ can be any collection of zero-pairs.

1. What does “subtract $$b$$” mean in the Token Model ?

For an integer $$a$$, the expression $$a - b$$ asks us to remove a collection of tokens that represents $$b$$ from the collection that represents $$a$$.

Problem : Sometimes the tokens for $$a$$ do not contain the exact tokens of $$b$$ that we are supposed to remove. The remedy is to add zero-pairs (which do not change the value) until the required tokens are present; then we can remove them.

2. Removing tokens is the same as adding the opposite tokens

Suppose $$b$$ itself is represented by $$b_+$$ positive tokens and $$b_-$$ negative tokens (exactly one of the two numbers can be non-zero since $$b$$ is a single integer). Removing those tokens is the same as adding tokens of the opposite colour:

• Removing one positive token has the same overall effect as adding one negative token, because a zero-pair disappears and leaves a negative token behind.
• Removing one negative token is the same as adding one positive token, for the same reason.

Therefore, removing the whole set of tokens that makes $$b$$ is equivalent to adding the set of opposite tokens, which represent $$-b$$. That is precisely what we mean by “adding the inverse”. Symbolically,

3. Two worked examples

1. Example 1 : $$5 - 3$$

   Start with 5 positive tokens.

   Step 1 : We need to remove 3 positive tokens. They are present, so simply take them away.

   Step 2 : 2 positive tokens remain. Hence $$5 - 3 = 2$$.

   Now do it by “add the inverse”. Instead of removing three positive tokens, add three negative tokens (their opposites):

   Begin with 5 positive tokens; add 3 negative tokens.

   They form 3 zero-pairs, leaving exactly 2 positive tokens. So $$5 + (-3) = 2$$ — the same result.

2. Example 2 : $$2 - (-4)$$

   Start with 2 positive tokens.

   Step 1 : We must remove 4 negative tokens, but none are present. Add 4 zero-pairs (each pair is $$+1$$ and $$-1$$) so the value is still 2. Now the collection has 2 positive and 4 negative tokens.

   Step 2 : Remove the 4 negative tokens. What is left? 2 positive tokens plus the 4 positive partners of the zero-pairs, that is 6 positive tokens. Thus $$2 - (-4) = 6$$.

   Do it as addition : $$2 + 4 = 6$$. Again the results match, confirming that removing $$-4$$ is identical to adding $$4$$.

4. The general algebraic statement

For every integer $$a$$ and $$b$$ there exists an integer $$-b$$ (called the additive inverse of $$b$$) such that $$b + (-b) = 0$$. Using the token model, subtracting $$b$$ means removing the tokens that make up $$b$$, which produces the same final token collection as adding the opposite tokens, namely $$-b$$.

Hence, both in pictures and algebraically,

Key idea: In an algebraic (or arithmetic) expression every quantity that is added is called a term. If the original symbol is “−”, we first rewrite it as the addition of a negative number; multiplication or division symbols that sit inside parentheses stay untouched because they are not separators of terms.

Step-by-step rewriting

1. $$5 + 6 \times 3$$ already has only one plus sign. There is no subtraction, so the expression is itself a sum: $$5 + (6 \times 3)$$. Hence the two terms are $$5$$ and $$6 \times 3$$.

2. For $$4 + 15 - 9$$, change “− 9” to “+ (−9)”:
   $$4 + 15 + (-9)$$. The three terms are $$4,\;15,\;-9$$.

3. For $$23 - 2 \times 4 + 16$$, change “− 2 × 4” to “+ (−2 × 4)”:
   $$23 + (-2 \times 4) + 16$$. So the terms are $$23,\;-2 \times 4,\;16$$.

4. For $$28 + 19 - 8$$, turn “− 8” into “+ (−8)”:
   $$28 + 19 + (-8)$$. The terms are $$28,\;19,\;-8$$.

Completed table

We want to investigate whether re-ordering the addends (the numbers or algebraic terms being added) changes the final result of an addition.

1. Starting with two numbers

Take any two natural numbers, e.g. 4 and 7.

Inline calculation in the usual left–to–right order:
$$4 + 7 = 11$$

Reverse their order and add again:

$$7 + 4 = 11$$

Both sums are equal to 11. Thus, with two numbers, changing the order does not affect the answer.

Algebraically, for any two numbers $$a$$ and $$b$$ we have

This property is called the commutative property of addition.

2. Extending to three numbers

Take three numbers, say 3, 5 and 9.

Original left–to–right grouping

$$(3 + 5) + 9 = 8 + 9 = 17$$

Change the grouping (add 5 and 9 first)

$$3 + (5 + 9) = 3 + 14 = 17$$

Same sum 17 again.

Algebraically, for any three numbers $$a, b, c$$,

This is the associative property of addition and shows that the way we group the numbers (which effectively changes their order of addition) does not change the result.

3. Applying to algebraic terms

Let the addends be algebraic symbols: $$x, y,$$ and $$z$$.

• Re-order two terms: $$x + y = y + x$$
• Re-order three terms: $$(x + y) + z = x + (y + z)$$, which further equals $$y + (x + z)$$, etc.

No matter how we permute the terms or where we place the plus signs first, the final sum is identical.

4. Conclusion

Because addition satisfies both the commutative and associative properties, changing the order or the grouping of the terms being added never changes the value of the sum.

Background — the property we are testing

If an expression has two terms that are joined by addition (or by multiplication), then interchanging the two terms does not change the value: in symbols, for any numbers $$a$$ and $$b$$,

addition    :   $$a + b = b + a$$
multiplication :   $$a \times b = b \times a$$

This is called the commutative property.

The textbook has already verified it for positive numbers. Now we examine whether the same rule works when one or both of the terms are negative integers.

Step 1 : One positive and one negative term

Both sums are $$4$$, so swapping the terms has made no difference.

Step 2 : Both terms negative

Again the two values are equal.

Step 3 : One large positive and one large negative

Yet again the result is the same.

Step 4 : Confirming with multiplication

The product also remains unchanged after interchanging the two factors.

Step 5 : General reason

The rules for adding integers were proved using the number–line idea and by extending the properties of whole numbers. A negative integer is still an integer, so the statement “for any integers $$a$$ and $$b$$ we have $$a+b=b+a$$” automatically covers all possible sign combinations. The same argument holds for multiplication: the product of integers is commutative irrespective of signs.

Conclusion

Swapping the two terms (or factors) does not change the value of the expression even when one or both of the terms are negative. Hence the commutative property is valid for all integers, positive as well as negative.

Step 1 : Recall the token model
In Class 6 we agreed to represent every integer by tiny square cards called tokens.

• one ‘+’-token represents $$+1$$
• one ‘–’-token represents $$-1$$

Step 2 : What does “adding two integers” mean?
If we want to find $$a+b$$ we simply pour the tokens for $$a$$ and the tokens for $$b$$ into the same tray. Nothing else is done at this stage – we have only combined the two piles.

Step 3 : Why the order of pouring is irrelevant
Whether we empty the $$a$$-pile first and then the $$b$$-pile, or the other way round, the final tray contains exactly the same multiset of individual cards, because no card is changed while we are pouring.

Symbolically, the content of the tray after the two pourings is
$$\{\;\text{tokens of }a\;\}\cup\{\;\text{tokens of }b\;\}$$
The union of two sets (or multisets) clearly does not depend on the order in which the sets were thrown in, so the tray is identical in the two cases.

Step 4 : Cancelling opposite pairs
After combining, we look for a ‘+’-token together with a ‘–’-token; each such pair has value $$+1+(-1)=0$$, so the two cards can be removed (they annihilate). Again, the list of possible pairs is determined only by how many ‘+’ and how many ‘–’ cards we finally possess, not by the order in which they originally arrived. Hence the number of tokens left after cancellation – that is, the numerical answer – is the same in both arrangements.

Step 5 : A concrete example

The final answer is $$+1$$ in both sequences; the swap has made no difference.

Step 6 : Summary – the commutative law of addition
Because combining two piles of tokens and removing opposite pairs are processes that do not depend on which pile entered first, we always obtain the same integer. Thus

Remark on multiplication. In Class 7 we often meet products like $$a\times b$$. If each factor is interpreted as “groups of”, then swapping the factors merely exchanges “number of groups” with “size of each group”, which again leaves the total number of tokens unchanged. Hence the token model also justifies $$a\times b=b\times a$$.

Objective
To find out whether changing the grouping (that is, changing the brackets) in an addition expression still gives the same value when some of the terms are negative.

Key idea – Associative property of addition
For any three whole numbers, the associative property says \[(a+b)+c = a+(b+c).\] The property is actually true for all integers (positive, negative or zero). We shall verify it with several numerical examples that contain negative numbers.

Example 1  Expression: $$5 + (-3) + 2$$

• Grouping I: $$(5+(-3))+2 = 2+2 = 4$$
• Grouping II: $$5+((-3)+2) = 5+(-1) = 4$$

Both groupings give the same value, $$4$$.

Example 2  Expression: $$(-7) + 4 + (-2)$$

• Grouping I: $$((-7)+4)+(-2) = (-3)+(-2) = -5$$
• Grouping II: $$(-7)+(4+(-2)) = (-7)+2 = -5$$

The value is $$-5$$ in each case.

Example 3 (four terms)
Expression: $$8 + (-6) + (-3) + 1$$

• First way: $$((8+(-6))+(-3))+1 = (2+(-3))+1 = (-1)+1 = 0$$
• Second way: $$8+((-6)+(-3)+1) = 8+(-8) = 0$$
• Third way: $$(8+(-6))+((-3)+1) = 2+(-2) = 0$$

The result remains $$0$$, no matter how the terms are grouped.

General fact
Because the associative property holds for every integer, we always have \[(a+b)+c = a+(b+c)\] when a, b and c are negative, positive or zero. This extends to any number of terms: you may place the brackets anywhere, and the sum will not change.

Conclusion
Yes, grouping (placing brackets) in different ways gives exactly the same value even when the expression contains negative numbers. Our numerical checks confirm the associative law of addition for integers.

Recall the Token Model

• One black token stands for $$+1$$.
• One red token stands for $$-1$$.
• Whenever a red and a black token appear together they cancel, because $$+1+(-1)=0$$.
• The value of a collection of tokens is therefore $$\text{(number of black tokens)}-\text{(number of red tokens)}.$$

Representing three integers

Let the integers $$a$$, $$b$$ and $$c$$ be represented by the following bags:

• Bag A: $$a$$ is made of $$(\hbox{B}_a)$$ black tokens and $$(\hbox{R}_a)$$ red tokens.
• Bag B: $$b$$ is made of $$(\hbox{B}_b)$$ black tokens and $$(\hbox{R}_b)$$ red tokens.
• Bag C: $$c$$ is made of $$(\hbox{B}_c)$$ black tokens and $$(\hbox{R}_c)$$ red tokens.

Thus

Adding by grouping the first two integers first

1. Put together the tokens of Bag A and Bag B.
   We now have $$\hbox{B}_a+\hbox{B}_b$$ black and $$\hbox{R}_a+\hbox{R}_b$$ red tokens.
2. Cancel as many red–black pairs as possible. After cancellation the remaining bag still represents the integer $$(a+b).$$ Call this new bag Bag AB.
3. Now pour in the tokens of Bag C. The combined bag has $$\bigl(\hbox{B}_a+\hbox{B}_b+\hbox{B}_c\bigr)$$ black and $$\bigl(\hbox{R}_a+\hbox{R}_b+\hbox{R}_c\bigr)$$ red tokens.
4. Perform a fresh round of cancellations. The number left is exactly $$(a+b)+c.$$

Adding by grouping the last two integers first

1. First merge Bag B and Bag C. Cancel red–black pairs; the resulting bag represents $$(b+c)$$ and will be called Bag BC.
2. Add the tokens of Bag A. The mixed bag again contains $$\bigl(\hbox{B}_a+\hbox{B}_b+\hbox{B}_c\bigr)$$ black and $$\bigl(\hbox{R}_a+\hbox{R}_b+\hbox{R}_c\bigr)$$ red tokens.
3. After cancellation we obtain the integer $$a+(b+c).$$

Why the two answers are identical

Notice that before the final cancellation both methods produce precisely the same multiset of tokens: every black or red token from the three original bags is present exactly once. Since the set of tokens is identical, the number of red–black pairs we can cancel is also identical, and therefore the value left in the hand is the same.

This shows, using only the Token Model, that the way we group the integers when adding does not change the value. The property is called the associative property of addition of integers.

Illustrative numerical example

Take $$a=-3$$, $$b=5$$, $$c=-2$$.

• Bag A: 3 red.
• Bag B: 5 black.
• Bag C: 2 red.

Method 1: $$(a+b)+c$$

• Combine A and B: 5 black + 3 red → cancel 3 pairs → 2 black (value =2).
• Add C: 2 black + 2 red → cancel 2 pairs → 0 tokens → value =0.

Method 2: $$a+(b+c)$$

• Combine B and C: 5 black + 2 red → cancel 2 pairs → 3 black (value =3).
• Add A: 3 black + 3 red → cancel 3 pairs → 0 tokens → value =0.

Both give the same result, confirming the explanation.

Aim: Verify whether changing the order of the terms in a sum alters its value.

Fundamental facts

• Commutative property of addition:   $$a+b=b+a$$
• Associative property of addition:   $$(a+b)+c=a+(b+c)$$

Because of these two properties, the result of addition should stay the same, no matter how the terms are arranged. We will check this with several examples.

Example 1 (three numerical terms)

Expression: $$2+5+7$$

Original order:   $$2+5+7=7+7=14$$

Different order:   $$5+7+2=12+2=14$$

Both give $$14$$.

Example 2 (three algebraic terms)

Expression: $$3x+4y+5z$$

Re-order: $$5z+3x+4y$$ or $$4y+5z+3x$$

Since only the order has changed, all forms still represent the same sum $$3x+4y+5z$$.

Example 3 (four terms: numbers and variables)

Expression: $$8a+3+2a+7$$

1. Original order: $$8a+3+2a+7=10a+10$$
2. Changed order: $$3+7+8a+2a = 10 + 10a = 10a+10$$

The value is unchanged.

Example 4 (five numerical terms including negatives)

Expression: $$12+(-5)+4+9+(-2)$$

1. Original order: $$12+(-5)+4+9+(-2)=18$$
2. Re-ordered: $$9+4+(-2)+12+(-5)=18$$

Again, the sum is the same: $$18$$.

Reason

Each rearrangement is obtained by repeatedly using the commutative and associative properties, which do not change the total. Therefore, the value of an addition expression depends only on its terms, not on their order.

Conclusion

Yes, adding the terms of an expression in any order always yields the same value. This holds for expressions with three, four, five or any larger number of terms.

1. Recap of the Token Model (Class 6)

• One positive token represents $$+1$$.
• One negative token represents $$-1$$.
• A positive token together with a negative token forms a neutral pair worth $$0$$, so the pair can be removed without changing the total.

2. Writing an arithmetic expression with tokens

Suppose the expression is

$$a_1 + a_2 + a_3 + \\dots + a_n,$$

where each $$a_k$$ is an integer (positive, negative or zero).

For every addend $$a_k$$ we place on the table:

• $$|a_k|$$ positive tokens whenever $$a_k \\gt 0$$,
• $$|a_k|$$ negative tokens whenever $$a_k \\lt 0$$.

After all the terms have been placed, the table holds

• $$P$$ positive tokens in total, and
• $$N$$ negative tokens in total.

3. Why the order of addition is irrelevant

1. Re-arranging the terms only changes when a particular bunch of tokens is laid down; it does not change which tokens finally end up on the table. Hence the final counts $$P$$ and $$N$$ remain exactly the same.
2. Next we remove neutral pairs, each one using the fact \[(+1) + (-1) = 0.\] The order in which we pick pairs makes no difference to the number of pairs that can be removed: we always cancel $$\\min(P,N)$$ such pairs.
3. After cancellation, either only positive or only negative tokens are left, never both. The number of leftover tokens is \[|P - N|,\] each worth $$+1$$ when $$P \\gt N$$ or $$-1$$ when $$N \\gt P$$.
4. So the value of the whole expression is \[(P - N) \\times (+1) = P - N.\] This depends only on how many positive and negative tokens were present, not on the order in which they appeared or were cancelled.

4. Concrete example

Consider $$3 + (-2) + 5 + (-4).$$ Two possible orders are

• Original: $$3 + (-2) + 5 + (-4)$$
• Re-ordered: $$5 + (-4) + 3 + (-2)$$

Token table for either order:

• $$3$$ gives 3 positive tokens.
• $$-2$$ gives 2 negative tokens.
• $$5$$ gives 5 positive tokens.
• $$-4$$ gives 4 negative tokens.

Total positive $$P = 3 + 5 = 8;$$ total negative $$N = 2 + 4 = 6.$$ Remove $$\\min(8,6) = 6$$ neutral pairs. $$2$$ positive tokens remain, so the value of the sum is $$+2$$ in every order.

5. Conclusion

Each integer addend contributes its own fixed set of $$+1$$ and $$-1$$ tokens, and neutral pairs cancel independently of the order in which the tokens arrived. So the leftover count of tokens—and hence the numerical value of the whole expression—remains the same no matter how we rearrange the terms. This is exactly the commutative and associative property of addition, explained visually with tokens.

First write down the five numbers that really have to be added:

• First number  $$1342$$
• Second number $$774$$
• Third number $$8611$$
• Fourth number $$9055$$ ← this one was forgotten
• Fifth number $$1022$$

Because Manasa forgot the fourth number, the total she wrote, $$11749$$, is only the sum of the other four numbers.

Check this quickly to be sure:

$$1342+774=2116$$   and   $$2116+8611=10727$$   and   $$10727+1022=11749$$ ✔️

So the correct (complete) sum must be

\[11749 + 9055 = 20804\]

She can obtain the right answer just by adding the missing $$9055$$ to her earlier result. There is no need to start the five-minute addition all over again.

There is no practical restriction that forces Manasa to put on the hat before the shoes or the shoes before the hat. The two actions are independent — neither one blocks the other.

Therefore, in everyday life she can proceed in either order:

• hat → shoes, or
• shoes → hat.

This little story is meant to prepare us for a later point in the chapter: some processes (like these two) can be carried out in any sequence, while others (for example, wearing socks and shoes) have a fixed order. In arithmetic we will discover that certain operations must also follow a fixed order, called the order of operations.

One dosa costs ₹23.

The number of friends is 7, so each friend orders one dosa.

Cost of the dosas: $$23 \times 7 = 161$$.

Tip (fixed): ₹5.

Therefore the required algebraic expression for the total bill is

\[ 23 \times 7 + 5 \]

Identifying the terms (parts separated by the ‘+’ sign):

• First term = $$23 \times 7$$
• Second term = $$5$$

On calculating, $$23 \times 7 = 161$$ and $$161 + 5 = 166$$.

Hence the friends will have to pay ₹166.

Let us recall the situation described in Example 8.

Ruby had to find the total number of objects that could be thought of as

• five equal groups, each group containing six objects, and
• three extra objects that do not belong to any group.

Written only with addition, the count looks like

$$6 + 6 + 6 + 6 + 6 + 3$$

Class 7 students already know that repeated addition is the same as multiplication. The first five sixes are therefore replaced by one product:

$$6 \times 5$$ (because there are five 6’s)

After taking care of the grouped objects, the three single objects are still left to be added. Hence the compact arithmetic expression becomes

$$6 \times 5 + 3$$

Ruby also keeps the BODMAS/PEMDAS rule in mind: multiplication is done before addition, so the calculation proceeds as

$$6 \times 5 = 30,\; 30 + 3 = 33$$

Thus the expression $$6 \times 5 + 3$$ correctly represents and evaluates the required total, which is why Ruby wrote it in Example 8.

Step 1 – Choose a letter for the unknown.
Let the total number of students in Ruby’s class be denoted by the variable $$c$$.
(Any letter may be chosen; we have picked $$c$$ to remind us of “class size”.)

Step 2 – Translate the teacher’s call into an algebraic expression.
Whenever the teacher calls out a whole-number, Ruby writes the sum of that number and the class size. In symbols:

$$\text{Ruby\'s expression} = c + n$$

Here $$n$$ stands for the number that the teacher has just called out.

Case (i) – The teacher calls out 4.
Replace $$n$$ by $$4$$ in the general expression $$c + n$$:

$$c + 4$$

Identifying the terms: the two terms are $$c$$ and $$4$$.

Case (ii) – The teacher calls out 7.
Replace $$n$$ by $$7$$:

$$c + 7$$

Identifying the terms: the two terms are $$c$$ and $$7$$.

Step 3 – Write similar expressions for your class size.
Suppose there are, say, 42 students in your class (change the number to match your real class strength). Then $$c = 42$$ and

• for the call 4: $$42 + 4 = 46$$
• for the call 7: $$42 + 7 = 49$$

If your class contains a different number of students, simply replace 42 by that number and evaluate in the same way.

Step 1 : Recall the meaning of a term.

In an arithmetic or algebraic expression every part that is added (or subtracted) is called a term.
Thus in the expression $$a+b-c$$ we have three terms: $$a$$, $$b$$ and $$-c$$.

Step 2 : Write each of the two given expressions clearly.

(i) $$432 = 4 \times 100 + 1 \times 20 + 1 \times 10 + 2 \times 1$$

(ii) $$432 = 8 \times 50 + 1 \times 10 + 4 \times 5 + 2 \times 1$$

Step 3 : Locate the ‘+’ signs in each expression.

• In the first expression the ‘+’ signs separate
  $$4 \times 100,\; 1 \times 20,\; 1 \times 10,\; 2 \times 1$$
• In the second expression the ‘+’ signs separate
  $$8 \times 50,\; 1 \times 10,\; 4 \times 5,\; 2 \times 1$$

Step 4 : State the terms.

(i) Terms of $$4 \times 100 + 1 \times 20 + 1 \times 10 + 2 \times 1$$ are
$$4 \times 100,\; 1 \times 20,\; 1 \times 10,\; 2 \times 1.$$

(ii) Terms of $$8 \times 50 + 1 \times 10 + 4 \times 5 + 2 \times 1$$ are
$$8 \times 50,\; 1 \times 10,\; 4 \times 5,\; 2 \times 1.$$

Hence we have identified all the terms in both expressions.

One straightforward way has already been shown in the textbook:

• $$4\times 100+3\times 10+2\times 1=432$$   ⇒ four ₹100 notes, three ₹10 notes and two ₹1 coins.

Below is a systematic search for other combinations, using the most common Indian denominations (₹100, ₹50, ₹20, ₹10, ₹5, ₹2, ₹1). A Class 7 student only needs to be able to add and subtract multiples of these denominations.

1. Start with ₹100 notes.
   If we use $$x$$ hundred–rupee notes, then the amount still to be paid is $$432-100x$$. Obviously $$0\le x\le 4$$ because five ₹100 notes would exceed ₹432.
   • (a) Take $$x=3$$: remaining amount $$=432-300=132$$.
     • Try two ₹50 notes: $$132-2\times 50 = 32$$(rupees left).
     • 32 can be paid as three ₹10 notes and two ₹1 coins.
     • Hence
     $$3\times 100+2\times 50+3\times 10+2\times 1 = 432.$$
   • (b) Take $$x=2$$: remaining amount $$=432-200=232$$.
     • Four ₹50 notes give ₹200, leaving ₹32 again.
     • As before, ₹32 = three ₹10 + two ₹1.
     $$2\times 100+4\times 50+3\times 10+2\times 1 = 432.$$
   • (c) Take $$x=0$$: remaining amount is the full ₹432.
     • Eight ₹50 notes give ₹400, leaving ₹32.
     • Use three ₹10 notes (₹30) and one ₹2 coin.
     $$8\times 50+3\times 10+1\times 2 = 432.$$
2. Try a combination of ₹20 notes and ₹2 coins.
   If we insist on ending with an even number, the last digit ‘2’ in 432 suggests ₹2 coins. Suppose we use six ₹2 coins (₹12). The balance is $$432-12 = 420$$. Since 420 is a multiple of 20, we can take $$\tfrac{420}{20}=21$$ twenty-rupee notes.
   $$21\times 20 + 6\times 2 = 432.$$
3. All ₹10 notes except the last ₹2.
   Use forty-three ₹10 notes (₹430) and one ₹2 coin.
   $$43\times 10 + 1\times 2 = 432.$$

Thus we have already found five fresh ways, besides the textbook’s original one.

Clearly, many more combinations are possible; the important skill is to keep subtracting convenient multiples of the available denominations until the remainder becomes 0, 1, or 2 and can be paid with ₹1 or ₹2 coins.

Step 1 – Choose variables
Let $$y$$ denote one yellow square and $$b$$ denote one blue square.

Step 2 – Count the number of each colour in one column
In a single column there are
  • 5 yellow squares ⇒ $$5y$$
  • 3 blue squares ⇒ $$3b$$
Hence one column is represented by the expression
$$5y + 3b$$

Step 3 – Account for the two identical columns
The right-hand arrangement has two such columns, so multiply the above expression by 2:
$$2 \times (5y + 3b)$$

Step 4 – (Optional) expand the bracket
$$2(5y + 3b) = 2 \times 5y + 2 \times 3b = 10y + 6b$$

Therefore, using $$y$$ and $$b$$ for yellow and blue squares, the required algebraic expression is

\[2(5y + 3b)\]
(or, after simplification, $$10y + 6b$$).

Recall. To keep track of the size of an integer we use the number line: if we replace one number by the next greater number (move one step to the right) the value of the sum also increases by 1; if we replace it by the next smaller number (move one step to the left) the value of the sum decreases by 1.

Column 1

1. Given: $$53 + (-16) = 37$$.
2. Here 54 is one more than 53, so the sum must be one more than 37:
   $$54 + (-16) = 38$$.
3. Compare $$-15$$ with $$-16$$.
     $$-15$$ is one more (less negative) than $$-16$$, hence the total will again be one more:
   $$53 + (-15) = 38$$.

Column 2

1. Given once more: $$53 + (-16) = 37$$.
2. Now 52 is one less than 53, so the value must be one less than 37:
   $$52 + (-16) = 36$$.
3. Next we keep 53 but replace $$-16$$ by $$-17$$. Since $$-17$$ is one less (more negative) than $$-16$$, the result drops by 1:
   $$53 + (-17) = 36$$.

Column 3

Start with $$-87 + (-16)$$ as the reference and change one term at a time; note that moving one step right on the number line adds 1, moving left subtracts 1.

1. Direct addition:
   $$-87 + (-16) = -87 - 16 = -103$$.
2. From the result above replace both numbers: $$-87 \to -88$$ (one less, so −1) and $$-16 \to -15$$ (one more, so +1). The two changes cancel each other out, therefore the total stays −103:
   $$-88 + (-15) = -103$$.
3. Again start from $$-103$$. Now $$-87 \to -86$$ (one more, +1) and $$-16 \to -18$$ (two less, −2). Net change: $$+1-2=-1$$, so
   $$-86 + (-18) = -103 - 1 = -104$$.
4. Here both numbers are quite different; we may simply add:
   $$-97 + (-26) = -97 - 26 = -(97+26) = -123$$.

Filled–in table

Step 1 – Recall the original expression
Let $$x$$ be the number of vegetable cutlets and $$y$$ be the number of rasgullas ordered in Example 15.
Price of one vegetable cutlet = ₹43, so the cost of $$x$$ cutlets = $$43x$$.
Price of one rasgulla = ₹24, so the cost of $$y$$ rasgullas = $$24y$$.
Hence, the total amount before Sangmu joins is $$43x + 24y$$.

Step 2 – Add Sangmu’s identical order
Sangmu orders the same items: another $$x$$ cutlets and another $$y$$ rasgullas.
Cost of Sangmu’s order = $$43x + 24y$$.

Step 3 – Form the new total
New total amount $$ (43x + 24y) + (43x + 24y) $$ Add the like terms:
$$43x + 43x = 86x, \; 24y + 24y = 48y$$
So the expression becomes $$86x + 48y$$.

Step 4 – Optional factor form
The same result can be written compactly as $$2(43x + 24y)$$.

Result

Step 1  Evaluate $$5 \times 4 + 3$$ (no parentheses)

According to the order of operations, multiplication is done before addition.

First multiply: $$5 \times 4 = 20$$

Then add: $$20 + 3 = 23$$

So the left-hand side equals $$23$$.

Step 2  Evaluate $$5 \times (4 + 3)$$ (with parentheses)

With parentheses, we must first simplify what is inside the brackets.

Inside the brackets: $$4 + 3 = 7$$

Now multiply: $$5 \times 7 = 35$$

The right-hand side equals $$35$$.

Step 3  Compare the two results

That is why $$5 \times 4 + 3 \neq 5 \times (4 + 3)$$. Parentheses change the order in which the operations are carried out.

Step 4  Check whether the three bracketed expressions are equal

• Commutative law of addition: $$4 + 3 = 3 + 4$$, so $$5 \times (4 + 3) = 5 \times (3 + 4)$$.
• Let the common sum be $$7$$. Using the commutative law of multiplication, $$5 \times 7 = 7 \times 5$$.
• But $$7 = 3 + 4$$, hence $$7 \times 5 = (3 + 4) \times 5$$.

Therefore all three bracketed expressions are equal.

Step 1 — Recall why the earlier questions were “quick”

In the examples just before this question we saved time because at least one factor

• was very close to a power of ten, such as $$99, 101, 999,$$ so that we could write it as $$100 \\pm 1$$ or $$1000 \\pm 1$$ and then use the distributive law;
• was an exact fraction of a power of ten, such as $$25 = \\dfrac{100}{4}, \\; 50 = \\dfrac{100}{2}, \\; 125 = \\dfrac{1000}{8},$$ so multiplying by 100 (or 1000) and then dividing was easy;
• could be split into small factors, such as $$15 = 3 \\times 5$$ or $$16 = 2 \\times 2 \\times 2 \\times 2,$$ so the calculation could be broken into a chain of little steps.

Step 2 — Other pairs that fit the same patterns

Any product in which at least one factor is

• close to $$10, 100, 1000$$ (for example $$98, 99, 101, 1001$$),
• equal to $$25, 50, 75, 125,$$ or
• easy to split into small factors ($$15, 16, 24, 32$$ and so on)

will usually be quicker to evaluate mentally than by long multiplication.

Step 3 — Some concrete illustrations

1. A factor just below a power of ten: $$99 \\times 68 = (100 - 1) \\times 68 = 6800 - 68 = 6732.$$
2. A factor just above a power of ten: $$101 \\times 47 = (100 + 1) \\times 47 = 4700 + 47 = 4747.$$
3. Both factors equidistant from the same power of ten: $$97 \\times 103 = (100 - 3)(100 + 3) = 100^2 - 3^2 = 10000 - 9 = 9991.$$
4. Using $$25 = \\dfrac{100}{4}$$: $$47 \\times 25 = \\dfrac{47 \\times 100}{4} = \\dfrac{4700}{4} = 1175.$$
5. Using $$125 = \\dfrac{1000}{8}$$: $$32 \\times 125 = \\dfrac{32 \\times 1000}{8} = \\dfrac{32}{8} \\times 1000 = 4 \\times 1000 = 4000.$$
6. Breaking one factor into small parts: $$15 \\times 68 = 3 \\times 5 \\times 68 = 3 \\times 340 = 1020.$$

Step 4 — A short list of other “quick” products

• $$98 \\times 57$$ — write $$98 = 100 - 2$$.
• $$999 \\times 36$$ — write $$999 = 1000 - 1$$.
• $$1001 \\times 48$$ — write $$1001 = 1000 + 1$$.
• $$25 \\times 68$$ — quarter of $$100 \\times 68$$.
• $$50 \\times 87$$ — half of $$100 \\times 87$$.
• $$125 \\times 24$$ — one-eighth of $$1000 \\times 24$$.
• $$32 \\times 75$$ — use $$75 = \\dfrac{3 \\times 100}{4}$$, so $$32 \\times 75 = \\dfrac{32}{4} \\times 3 \\times 100 = 8 \\times 3 \\times 100 = 2400.$$
• $$15 \\times 96$$ — use $$15 = 3 \\times 5$$, so $$15 \\times 96 = 3 \\times (5 \\times 96) = 3 \\times 480 = 1440.$$

Any product whose factor fits one of these patterns can usually be done in one line of mental arithmetic instead of full long multiplication.

Step 1 : What does any such expression look like?

We must use the three numbers 2, 3 and 5 exactly once, only the two operators + and − and brackets. After all brackets are removed every expression becomes a sum of the three numbers, each carrying either a plus or a minus sign. For instance,

• $$2 - 3 + 5 = +2 -3 +5$$
• $$3 - (5 - 2) = +3 -5 +2$$
• $$5 - (2 + 3) = +5 -2 -3$$

The number written first is always positive because no sign precedes it. The other two numbers may be written with either a plus or a minus sign.

Step 2 : Try every possible sign choice

For any chosen “first number” there are $$2 \times 2 = 4$$ choices of signs for the remaining two numbers. Repeating the process with each of 2, 3, 5 in first place gives the table below.

The six different results that appear are $$10,\;6,\;4,\;0,\;-4,\;-6$$. No other number is possible because the first term is always positive, so all three numbers can never be negative at the same time and their total can never be −10, −8, −2, 2, 8, etc.

Step 3 : One neat expression for every value

• $$2 + 3 + 5 = 10$$
• $$3 + 5 - 2 = 6$$
• $$2 - 3 + 5 = 4$$
• $$3 - (5 - 2) = 0$$
• $$(3 - 5) - 2 = -4$$
• $$2 - 3 - 5 = -6$$

Therefore, the maximum number of different values that can be produced is 6, namely $$10,\;6,\;4,\;0,\;-4,\;-6$$.

Step 1: Evaluate the two given expressions

Because addition and subtraction have the same priority, we work from left to right.

For $$73 - 14 + 1$$:
  $$73 - 14 = 59$$
  $$59 + 1 = 60$$
So it simplifies to $$60$$.

For $$73 - 14 - 1$$:
  $$73 - 14 = 59$$
  $$59 - 1 = 58$$
So it simplifies to $$58$$.

Step 2: Evaluate each expression in the list (a)–(d)

1. $$(a)\; 73 - (14 + 1)$$
     First simplify inside the bracket: $$14 + 1 = 15$$
     Then $$73 - 15 = 58$$.
2. $$(b)\; 73 - (14 - 1)$$
     Inside the bracket: $$14 - 1 = 13$$
     Then $$73 - 13 = 60$$.
3. $$(c)\; 73 + (-14 + 1)$$
     Inside the bracket: $$-14 + 1 = -13$$
     Then $$73 + (-13) = 60$$.
4. $$(d)\; 73 + (-14 - 1)$$
     Inside the bracket: $$-14 - 1 = -15$$
     Then $$73 + (-15) = 58$$.

Step 3: Match equal values

• The value $$60$$ (from $$73 - 14 + 1$$) also occurs in expressions (b) and (c).
• The value $$58$$ (from $$73 - 14 - 1$$) also occurs in expressions (a) and (d).

Picture (I) : 3 × 3 grid

The nine cells contain five 4’s and four 8’s arranged as

We find the total in two different ways.

Method 1 – Adding row by row

• Row 1 : $$4+8+4 = 16$$
• Row 2 : $$8+4+8 = 20$$
• Row 3 : $$4+8+4 = 16$$

Sum of all three rows:

Method 2 – Grouping equal numbers

• Number of 4’s = $$5$$  ⇒  contribution $$=5\times4$$
• Number of 8’s = $$4$$  ⇒  contribution $$=4\times8$$

Thus, the required sum for Picture (I) is $$52$$.

Picture (II) : 4 × 4 grid

Method 1 – Adding row by row

• Every row has the same numbers: $$5+6+6+5 = 22$$
• There are $$4$$ such rows.

Method 2 – Grouping equal numbers

• Count of 5’s = $$8$$  ⇒  contribution $$=8\times5$$
• Count of 6’s = $$8$$  ⇒  contribution $$=8\times6$$

Thus, the required sum for Picture (II) is $$88$$.

Step 1 — Find the number of reading days in one week

There are $$7$$ days in a week, but Melvin does not read on Tuesdays and Saturdays (that is, on $$2$$ days).

Therefore, the number of days on which he does read in one week is

$$7 - 2 = 5$$ days.

Step 2 — Stories finished in one week

On every reading day he completes exactly one two-page story, so in one week he finishes

$$5 imes 1 = 5$$ stories.

Step 3 — Stories finished in eight weeks

Eight weeks have $$8$$ identical groups of five reading days, hence

$$5 imes 8 = 40$$ stories in all.

Step 4 — Choose the expression(s) that match this calculation

Conclusion

Melvin completes 40 stories in eight weeks, and the scenario is described by any expression equivalent to $$5 \times 8$$; among the given choices those are option (b) $$(7 - 2) \times 8$$ and option (g) $$7 \times 8 - 2 \times 8$$.

Step 1 – Choosing the number
We first pick any whole number that we like. Let us choose the number $$24$$.

Step 2 – Recalling the rules for evaluating an arithmetic expression
A Class 7 student must remember the order of operations (BODMAS):
Brackets → Orders (powers, roots) → Division → Multiplication → Addition → Subtraction.

Step 3 – Constructing and verifying ten different expressions

1. Expression 1: $$12 + 12$$    Evaluate: $$12 + 12 = 24$$    Therefore the value is $$24$$.

2. Expression 2: $$30 - 6$$    Evaluate: $$30 - 6 = 24$$.

3. Expression 3: $$8 \times 3$$    Evaluate: $$8 \times 3 = 24$$.

4. Expression 4: $$48 \div 2$$    Evaluate: $$48 \div 2 = 24$$.

5. Expression 5: $$4 \times (10 - 4)$$    Step 1 (brackets first): $$10 - 4 = 6$$    Step 2 (multiplication): $$4 \times 6 = 24$$.

6. Expression 6: $$(7 + 5) \times 2$$    Step 1 (brackets): $$7 + 5 = 12$$    Step 2: $$12 \times 2 = 24$$.

7. Expression 7: $$(18 \div 3) \times 4$$    Step 1 (division): $$18 \div 3 = 6$$    Step 2: $$6 \times 4 = 24$$.

8. Expression 8: $$90 - (6 \times 10 + 6)$$    Step 1 (inside the bracket, multiplication before addition):
         $$6 \times 10 = 60$$    Step 2 (still inside the bracket): $$60 + 6 = 66$$    Step 3 (subtraction outside the bracket): $$90 - 66 = 24$$.

9. Expression 9: $$144 \div 6$$    Evaluate: $$144 \div 6 = 24$$.

10. Expression 10: $$(3 + 1) \times (8 - 2)$$    Step 1 (left bracket): $$3 + 1 = 4$$    Step 2 (right bracket): $$8 - 2 = 6$$    Step 3 (multiplication): $$4 \times 6 = 24$$.

Step 4 – Final check
Every one of the ten expressions evaluates to our chosen number $$24$$, so the requirement of the question is fully satisfied.