Arithmetic Expressions

Fill in the blanks to make the expressions equal on both sides of the = sign:

(a) 13 + 4 = ............ + 6
(b) 22 + ............ = 6 × 5
(c) 8 × ............ = 64 ÷ 2
(d) 34 – ............ = 25

Answer

We first find the value of the side that is fully known, then choose the missing number so that both sides are equal.

(a) 13 + 4 = ............ + 6

13 + 4 = 17

So, ............ + 6 = 17

The missing number is 17 – 6 = 11

Hence, 13 + 4 = 11 + 6

(b) 22 + ............ = 6 × 5

6 × 5 = 30

So, 22 + ............ = 30

The missing number is 30 – 22 = 8

Hence, 22 + 8 = 6 × 5

(c) 8 × ............ = 64 ÷ 2

64 ÷ 2 = 32.

So, 8 × ............ = 32

The missing number is 32 ÷ 8 = 4.

Hence, 8 × 4 = 64 ÷ 2

(d) 34 – ............ = 25

34 – 25 = 9

Hence, 34 – 9 = 25

Arrange the following expressions in ascending (increasing) order of their values.

(a) 67 – 19
(b) 67 – 20
(c) 35 + 25
(d) 5 × 11
(e) 120 ÷ 3

Answer

We first find the value of each expression:

(a) 67 – 19 = 48

(b) 67 – 20 = 47

(c) 35 + 25 = 60

(d) 5 × 11 = 55

(e) 120 ÷ 3 = 40

Arranging these values in ascending order:

40 < 47 < 48 < 55 < 60

Hence, the ascending order of the expressions is 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 + 25.

Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

(a) 245 + 289 $\square$ 246 + 285
(b) 273 – 145 $\square$ 272 – 144
(c) 364 + 587 $\square$ 363 + 589
(d) 124 + 245 $\square$ 129 + 245
(e) 213 – 77 $\square$ 214 – 76

Answer

In each case we compare the two sides by looking at how much each number has changed, instead of finding the exact values.

(a) 245 + 289 $\square$ 246 + 285

First number increases by 1.
Second number decreases by 4.
Net change = −3.

So the right side is 3 less than the left side.

∴ 245 + 289 > 246 + 285

(b) 273 – 145 $\square$ 272 – 144

First number decreases by 1.
Second number also decreases by 1.
Therefore, the difference remains unchanged.

So both are equal.

∴ 273 – 145 = 272 – 144

(c) 364 + 587 $\square$ 363 + 589

First number decreases by 1.
Second number increases by 2.
Net change = +1.

So the right side is greater.

∴ 364 + 587 < 363 + 589

(d) 124 + 245 $\square$ 129 + 245

Second addend is unchanged.
First addend increases by 5.

So the right side is greater.

∴ 124 + 245 < 129 + 245

(e) 213 – 77 $\square$ 214 – 76

First number increases by 1.
Second number decreases by 1.
Difference increases by 2.

So the right side is greater.

∴ 213 – 77 < 214 – 76

Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.

Answer

Replacing a subtraction by adding the inverse means writing a – b as a + (–b). We check this with a few examples.

Example 1:

83 – 14 = 69
83 + (–14) = 69

Example 2:

20 – 35 = –15
20 + (–35) = –15

Example 3:

–18 – 3 = –21
–18 + (–3) = –21

In every example, the value stays the same after replacing the subtraction by adding the inverse.

Hence, replacing a subtraction by adding the inverse of the number does not change the value of the expression.

Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

Answer

In the Token Model, one positive token (+) and one negative token (–) together form a zero pair. A zero pair has value 0, so adding or removing zero pairs does not change the value.

Consider 83 – 14

This means we start with 83 positive tokens and remove 14 positive tokens. After removing them, 69 positive tokens remain.

Now consider 83 + (–14)

We start with 83 positive tokens and add 14 negative tokens. Each negative token forms a zero pair with one positive token. Thus, 14 positive tokens are cancelled by the 14 negative tokens, leaving 69 positive tokens.

So both expressions have the same value:

83 – 14 = 83 + (–14)

Hence, subtracting a number is the same as adding its inverse because the opposite tokens cancel each other as zero pairs.

In the following table, some expressions are given. Complete the table.

Answer

Each subtraction is rewritten as the addition of the inverse, and a multiplication is kept together as a single term. The completed table is:

Hence, the table is completed as shown above, with each subtraction written as the addition of an inverse and each multiplication kept as a single term.

Does changing the order in which the terms are added give different values?

Answer

No. Changing the order in which the terms are added does not change the value. This is the commutative property of addition.

For example, consider the expression:

13 – 2 + 6 = 13 + (–2) + 6

Its terms are 13, –2 and 6

Adding in the given order:

13 + (–2) + 6 = 11 + 6 = 17

Adding in a different order:

6 + 13 + (–2) = 19 + (–2) = 17

Both orders give the same value, 17.

Hence, changing the order in which the terms are added does not change the value of the expression.

Madhu is flying a drone from a terrace. The drone goes 6 m up and then 4 m down. Write an expression to show how high the final position of the drone is from the terrace.

Answer

Going 6 m up is represented by +6 and going 4 m down is represented by –4.

So, the height of the drone above the terrace is:

6 – 4

Writing it as a sum of terms:

6 + (–4) = 2

Therefore, the drone is 2 m above the terrace.

Hence, the expression is 6 + (–4), and the drone is 2 m above the terrace.

Will swapping the terms also hold when there are terms having negative numbers as well? Take some more expressions and check.

Answer

We check whether swapping the terms gives the same value when negative numbers are involved.

Example 1:

6 + (–4) = 2
(–4) + 6 = 2.

Example 2:

(–7) + 10 = 3
10 + (–7) = 3.

Example 3:

(–5) + (–8) = –13
(–8) + (–5) = –13.

In every case, swapping the terms gives the same value.

Hence, swapping the terms does not change the value even when the terms include negative numbers: Term 1 + Term 2 = Term 2 + Term 1.

Can you explain why swapping the terms gives the same value using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

Answer

In the Token Model, each term is a collection of tokens: positive tokens (+) for a positive number and negative tokens (–) for a negative number. A positive token and a negative token together form a zero pair, worth 0.

Adding two terms means putting both collections of tokens into a single pile and then cancelling all the zero pairs. The number left over depends only on how many positive and negative tokens there are in total — not on which pile was placed down first.

So whether we combine Term 1 with Term 2, or Term 2 with Term 1, we end up with exactly the same tokens in the pile, and hence the same value after cancelling zero pairs.

That is why:

Term 1 + Term 2 = Term 2 + Term 1

Hence, swapping the terms gives the same value because combining the two piles of tokens in either order leaves the same collection of tokens.

Does adding the terms of an expression in any order give the same value? Take some more expressions and check. Consider expressions with more than 3 terms also.

Answer

Yes, adding the terms of an expression in any order gives the same value. This is true even for expressions with more than three terms and with negative terms.

Consider the expression 15 + (–8) + 5 + 12, which has four terms.

Adding in the given order:

15 + (–8) + 5 + 12

= (15 + (–8)) + 5 + 12

= 7 + 5 + 12

= 24

Adding in a different order (positive terms first, then the negative term):

15 + 5 + 12 + (–8)

= (15 + 5 + 12) + (–8)

= 32 + (–8)

= 24

Both orders give the same value, 24.

Let us check with another four-term expression, (–3) + 9 + (–4) + 6.

(–3) + 9 + (–4) + 6 = 8

9 + 6 + (–3) + (–4) = 15 + (–7) = 8

Again the value is the same, 8.

Hence, the terms of an expression can be added in any order, even for more than three terms, and the value remains the same.

Can you explain why adding the terms in any order gives the same value using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

Answer

Using the Token Model, each term is represented by its own collection of +1 tokens (for a positive number) and –1 tokens (for a negative number).

To find the value of the whole expression, we gather the tokens of every term into one big pile and then remove all the zero pairs (each +1 with a –1). The tokens that remain give the value.

⇒ The order in which we pick up the terms only changes the order in which the tokens enter the pile.

⇒ It does not add or remove any token from the pile.

∴ The final collection of tokens, and hence the number left after removing zero pairs, stays exactly the same.

Hence, adding the terms in any order gives the same value, because the total collection of tokens does not depend on the order in which they are gathered.

Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again? (The numbers she is adding are: 1342, 774, 8611, 9055, 1022.)

Answer

No, Manasa does not have to start all over again.

Since the terms of an expression can be added in any order, she can add the missing number later.

The sum she got without 9055 is 11749.

She only needs to add the missing number 9055 to this sum:

11749 + 9055 = 20804

Check by adding all five numbers directly:

1342 + 774 + 8611 + 9055 + 1022 = 20804

Both give the same total, 20804.

Hence, Manasa need not start again; she can simply add 9055 to 11749 and get 20804.

Amu, Charan, Madhu, and John went to a hotel and ordered four dosas. Each dosa cost ₹23, and they wish to thank the waiter by tipping ₹5. If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms.

Answer

When there are 7 friends, 7 dosas are ordered.

Cost of 1 dosa = ₹23

Cost of 7 dosas = 7 × 23.

The tip of ₹5 is added to this.

The expression for the total amount to be paid is:

7 × 23 + 5

Evaluating the expression:

7 × 23 + 5

= 161 + 5

= 166

The terms of the expression 7 × 23 + 5 are 7 × 23 and 5.

Hence, the 7 friends will have to pay ₹166, and the terms of the expression 7 × 23 + 5 are 7 × 23 and 5.

Children in a class are playing “Fire in the mountain, run, run, run!”. Whenever the teacher calls out a number, students are supposed to arrange themselves in groups of that number. Whoever is not part of the announced group size, is out. Ruby wanted to rest and sat on one side. The other 33 students were playing the game in the class. The teacher called out ‘5’. Once children settled, Ruby wrote 6 × 5 + 3 (understood as 3 more than 6 × 5). Think and discuss why she wrote this.

Answer

There are 33 students playing, and the teacher called out ‘5’, so the students try to arrange themselves in groups of 5.

When 33 students form groups of 5, they can make 6 complete groups of 5 students each:

6 × 5 = 30

This uses up 30 students, and 33 – 30 = 3 students are left over. These 3 students cannot form a complete group of 5, so they are out.

Ruby represented this as 6 groups of 5, together with the 3 left-over students:

6 × 5 + 3

= 30 + 3

= 33

This matches the total number of 33 students.

Hence, Ruby wrote 6 × 5 + 3 because 33 students make 6 groups of 5 with 3 students left over.

For each of the cases below (from the “Fire in the mountain” game, with 33 students playing), write the expression and identify its terms:

If the teacher had called out ‘4’, Ruby would write ............
If the teacher had called out ‘7’, Ruby would write ............

Answer

There are 33 students playing.

If the teacher had called out ‘4’:

Making groups of 4, we get 8 complete groups since 8 × 4 = 32

Number of students left over = 33 – 32 = 1

Expression: 8 × 4 + 1

Terms: 8 × 4 and 1

Hence, for group size 4 the expression is 8 × 4 + 1 (terms 8 × 4 and 1).

If the teacher had called out ‘7’:

Making groups of 7, we get 4 complete groups since 4 × 7 = 28

Number of students left over = 33 – 28 = 5

Expression: 4 × 7 + 5

Terms: 4 × 7 and 5

Hence, for group size 7 the expression is 4 × 7 + 5 (terms 4 × 7 and 5).

Kannan has to pay ₹432 to a shopkeeper. Identify the terms in the two expressions below:

432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1
432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1

Answer

The terms of an expression are the parts separated by the ‘+’ sign.

First expression:

432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1

The terms are 4 × 100, 1 × 20, 1 × 10 and 2 × 1.

Second expression:

432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1

The terms are 8 × 50, 1 × 10, 4 × 5 and 2 × 1.

Can you think of some more ways of giving ₹432 to someone?

Answer

Yes. There are many ways of giving ₹432.

One way:

432 = 4 × 100 + 6 × 5 + 2 × 1

Check: 400 + 30 + 2 = 432

Terms: 4 × 100, 6 × 5 and 2 × 1

Another way:

432 = 43 × 10 + 2 × 1

Check: 430 + 2 = 432

Terms: 43 × 10 and 2 × 1

Hence, ₹432 can be given in many ways, for example 4 × 100 + 6 × 5 + 2 × 1 or 43 × 10 + 2 × 1.

Here are two pictures. Which of these two arrangements matches with the expression 5 × 2 + 3?

Answer

5 × 2 + 3 = 10 + 3 = 13

The expression 5 × 2 + 3 means 3 more than 5 × 2.

The arrangement on the left shows 5 groups of 2 squares and 3 extra squares.

Therefore, it matches the expression 5 × 2 + 3.

Hence, the arrangement on the left matches the expression 5 × 2 + 3.

Find the values of the following expressions by writing the terms in each case.

(a) 28 – 7 + 8
(b) 39 – 2 × 6 + 11
(c) 40 – 10 + 10 + 10
(d) 48 – 10 × 2 + 16 ÷ 2
(e) 6 × 3 – 4 × 8 × 5

Answer

We first write each expression as a sum of its terms (converting every subtraction into the addition of an inverse), evaluate each term, and then add the terms.

(a) 28 – 7 + 8

28 + (– 7) + 8

= 29

Terms: 28, – 7, 8

Hence, the value is 29.

(b) 39 – 2 × 6 + 11

= 39 + (– 2 × 6) + 11

= 39 + (– 12) + 11

= 38

Terms: 39, – 2 × 6, 11

Hence, the value is 38.

(c) 40 – 10 + 10 + 10

= 40 + (– 10) + 10 + 10

= 50

Terms: 40, – 10, 10, 10

Hence, the value is 50.

(d) 48 – 10 × 2 + 16 ÷ 2

= 48 + (– 10 × 2) + (16 ÷ 2)

= 48 + (– 20) + 8

= 36

Terms: 48, – 10 × 2, 16 ÷ 2

Hence, the value is 36.

(e) 6 × 3 – 4 × 8 × 5

= (6 × 3) + (– 4 × 8 × 5)

= 18 + (– 160)

= – 142

Terms: 6 × 3, – 4 × 8 × 5

Hence, the value is -142.

Write a story/situation for each of the following expressions and find their values.

(a) 89 + 21 – 10
(b) 5 × 12 – 6
(c) 4 × 9 + 2 × 6

Answer

(a) 89 + 21 – 10

Story: A library had 89 books on its shelves. The librarian added 21 new books to the collection, and later 10 books were borrowed by students. How many books remain on the shelves?

89 + 21 – 10

= 89 + 21 + (– 10)

= 110 - 10

= 100

Hence, the value of the expression is 100.

(b) 5 × 12 – 6

Story: A shopkeeper packs eggs in 5 trays with 12 eggs in each tray. On checking, 6 eggs are found to be broken and are removed. How many good eggs are left?

5 × 12 – 6

= (5 × 12) + (– 6)

= 60 – 6

= 54

Hence, the value of the expression is 54.

(c) 4 × 9 + 2 × 6

Story: In a classroom there are 4 rows with 9 chairs in each row, and 2 extra benches with 6 seats each. How many seats are there in all?

4 × 9 + 2 × 6

= (4 × 9) + (2 × 6)

= 36 + 12

= 48

Hence, the value of the expression is 48.

For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.

(a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.

(b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(i) for four adults and three children?
(ii) for two groups having three adults each?

(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture. (Border = 3 cm, Grill = 2 cm, Gap = 5 cm; find the Total Height.)

Answer

(a)

Princess Elsa doubled her 100 coins. So, she has 2 × 100 coins.

Princess Anna has half of her 100 coins left. So, she has $\dfrac{100}{2}$ coins.

Together they have:

$2 \times 100 + \dfrac{100}{2}$

Terms: 2 × 100, $\dfrac{100}{2}$

Value:

$2 \times 100 + \dfrac{100}{2}$

= 200 + 50

= 250

Hence, together they have 250 gold coins.

(b)

The cost of one adult ticket = ₹40

The cost of one child ticket = ₹20.

(i) For four adults and three children:

Expression = 4 × 40 + 3 × 20

Terms: 4 × 40, 3 × 20

4 × 40 + 3 × 20

= 160 + 60

= 220

Hence, the total cost is ₹220.

(ii) For two groups having three adults each:

Expression = 2 × (3 × 40)

Terms: 2 × (3 × 40)

2 × (3 × 40)

= 2 × 120

= 240

Hence, the total cost is ₹240.

(c)

The total height of the window is made up of:

2 borders of 3 cm each,
6 grills of 2 cm each,
7 gaps of 5 cm each.

Expression = 7 × 5 + 6 × 2 + 2 × 3

Terms: 7 × 5, 6 × 2, 2 × 3

7 × 5 + 6 × 2 + 2 × 3

= 35 + 12 + 6

= 53

Hence, the total height of the window is 53 cm.

What happens to the value of an expression if we increase or decrease the value of one of its terms?

Some expressions are given in the following three columns. In each column, one or more terms are changed from the first expression. Go through the example (in the first column) and fill the blanks, doing as little computation as possible.

Answer

We use the fact that changing a term by a certain amount changes the value of the whole expression by the same amount.

Column 1

53 + (– 16) = 37

54 + (– 16) = 38 (54 is one more than 53, so the value is 1 more than 37.)

53 + (– 15) = 38 (– 15 is one more than – 16, so the value is 1 more than 37.)

Column 2

53 + (– 16) = 37

52 + (– 16) = 36 (52 is one less than 53, so the value is 1 less than 37.)

53 + (– 17) = 36 (– 17 is one less than – 16, so the value is 1 less than 37.)

Column 3

– 87 + (– 16) = – 103

– 88 + (– 15) = – 103 (– 88 is one less than – 87 and – 15 is one more than – 16, so the value stays the same.)

– 86 + (– 18) = – 104 (– 86 is one more than – 87 and – 18 is two less than – 16, so the value is 1 less than – 103.)

– 97 + (– 26) = – 123 (– 97 is ten less than – 87 and – 26 is ten less than – 16, so the value is 20 less than – 103.)

Hence, when a term of an expression increases (or decreases) by a certain amount, the value of the expression increases (or decreases) by exactly the same amount.

Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.

(a) 24 + (6 – 4) = 24 + 6 $\square$ ............
(b) 38 + (............ $\square$ ............) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 $\square$ 6 – 4
(d) 24 – 6 – 4 = 24 – 6 $\square$ ............
(e) 27 – (8 + 3) = 27 $\square$ 8 $\square$ 3
(f) 27 – (............ $\square$ ............) = 27 – 8 + 3

Answer

A bracket that is not preceded by a minus sign can be removed without changing the signs of the terms inside. A bracket that is preceded by a minus sign changes the sign of each term inside when it is removed.

(a) 24 + (6 – 4) = 24 + 6 $\boxed{–}$ 4

(b) 38 + ( 9 $\boxed{–}$ 4 ) = 38 + 9 – 4

(c) 24 – (6 + 4) = 24 $\boxed{–}$ 6 – 4

(d) 24 – 6 – 4 = 24 – 6 $\boxed{–}$ 4

(e) 27 – (8 + 3) = 27 $\boxed{–}$ 8 $\boxed{–}$ 3

(f) 27 – ( 8 $\boxed{–}$ 3 ) = 27 – 8 + 3

Remove the brackets and write the expression having the same value.

(a) 14 + (12 + 10)
(b) 14 – (12 + 10)
(c) 14 + (12 – 10)
(d) 14 – (12 – 10)
(e) –14 + (12 – 10)
(f) 14 – (–12 – 10)

Answer

If the bracket is preceded by a ‘+’ sign, the terms keep their signs. If the bracket is preceded by a ‘–’ sign, every term inside changes its sign.

(a) 14 + (12 + 10)

= 14 + 12 + 10

= 36

(b) 14 – (12 + 10)

= 14 – 12 – 10

= – 8

(c) 14 + (12 – 10)

= 14 + 12 – 10

= 16

(d) 14 – (12 – 10)

= 14 – 12 + 10

= 12

(e) – 14 + (12 – 10)

= – 14 + 12 – 10

= – 12

(f) 14 – (– 12 – 10)

= 14 + 12 + 10

= 36

The expressions 14 + (12 + 10) and 14 – (– 12 – 10) have the same value.

Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?

(a) (6 + 10) – 2 and 6 + (10 – 2)
(b) 16 – (8 – 3) and (16 – 8) – 3
(c) 27 – (18 + 4) and 27 + (–18 – 4)

Answer

(a)

(6 + 10) – 2 = 6 + 10 – 2 = 14

6 + (10 – 2) = 6 + 10 – 2 = 14

∴ (6 + 10) – 2 = 6 + (10 – 2)

(b)

16 – (8 – 3) = 16 – 8 + 3 = 11

(16 – 8) – 3 = 16 – 8 – 3 = 5

∴ 16 – (8 – 3) ≠ (16 – 8) – 3

(c)

27 – (18 + 4) = 27 – 18 – 4 = 5

27 + (– 18 – 4) = 27 – 18 – 4 = 5

∴ 27 – (18 + 4) = 27 + (– 18 – 4)

In these pairs, (a) and (c) are equal because removing the brackets produces the same terms with the same signs. In (b), the sign of 3 is different, so the expressions are not equal.

Hence, the pairs in (a) and (c) are equal, while the pair in (b) is not.

In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.

(a) 319 + 537, 319 – 537, –537 + 319, 537 – 319
(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109

Answer

Two expressions have the same value if they are made up of exactly the same terms (with signs), since the order of the terms does not change the sum.

(a) Writing each as a sum of terms:

319 + 537 → terms 319, 537
319 – 537 → terms 319, – 537
– 537 + 319 → terms – 537, 319
537 – 319 → terms 537, – 319

The expressions 319 – 537 and – 537 + 319 have the same terms (319 and – 537), so they have the same value.

(b) Writing each as a sum of terms:

87 + 46 – 109 → terms 87, 46, – 109
87 + 46 – 109 → terms 87, 46, – 109
87 + 46 – 109 → terms 87, 46, – 109
87 – 46 + 109 → terms 87, – 46, 109
87 – (46 + 109) = 87 – 46 – 109 → terms 87, – 46, – 109
(87 – 46) + 109 = 87 – 46 + 109 → terms 87, – 46, 109

The expressions 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109 have the same value, and 87 – 46 + 109 and (87 – 46) + 109 have the same value.

Add brackets at appropriate places in the expressions such that they lead to the values indicated.

(a) 34 – 9 + 12 = 13
(b) 56 – 14 – 8 = 34
(c) –22 – 12 + 10 + 22 = –22

Answer

(a) 34 – 9 + 12 = 13

= 34 – (9 + 12)

= 34 – 21

= 13

(b) 56 – 14 – 8 = 34

(56 – 14) – 8

= 42 – 8

= 34

(c) –22 – 12 + 10 + 22 = –22

– 22 – (12 + 10) + 22

= – 22 – 22 + 22

= – 22

Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal
(a) 423 + ............ = 419 + ............
(b) 207 – 68 = 210 – ............

Answer

(a)

Since 423 is 4 more than 419, the term added to 419 should be 4 more than the term added to 423.

One possible answer is:

423 + 419 = 419 + 423 (Commutative property)

(b)

Since 210 is 3 more than 207, we must subtract 3 more than 68 to keep the value unchanged.

68 + 3 = 71

∴ 207 – 68 = 210 – 71

Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘–’, and brackets, as necessary, generate expressions to give as many different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.

Answer

Some possible expressions are:

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

Hence, using the numbers 2, 3 and 5 with ‘+’, ‘–’ and brackets, we can obtain the values –6, –4, 0, 4, 6 and 10.

Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1.

(a) Do you think she always gets the correct answer? Why?
(b) Can you think of other similar strategies? Give some examples.

Answer

(a)

Yes, she always gets the correct answer.

Since 9 = 10 − 1,

subtracting 9 is the same as subtracting 10 and then adding 1.

For example,

36 - 9 = 36 - (10 - 1) = 36 - 10 + 1 = 26 + 1 = 27

Removing the bracket preceded by the ‘–’ sign changes the sign of –1 to +1, which is exactly what Jasoda does.

Hence, Jasoda always gets the correct answer.

(b)

Yes. Similar strategies can be used for other numbers.

For example:

To subtract 8, subtract 10 and add 2 (since 8 = 10 – 2):

55 - 8 = 55 - (10 - 2) = 55 - 10 + 2 = 45 + 2 = 47

To subtract 99, subtract 100 and add 1 (since 99 = 100 – 1):

412 - 99 = 412 - (100 - 1) = 412 - 100 + 1 = 312 + 1 = 313

Hence, similar strategies such as “subtract 10 and add 2” to take away 8, or “subtract 100 and add 1” to take away 99, work in the same way.

Consider the two expressions:
(a) 73 – 14 + 1,
(b) 73 – 14 – 1.
For each of these expressions, identify the expressions from the following collection that are equal to it.

(a) 73 – (14 + 1)
(b) 73 – (14 – 1)
(c) 73 + (–14 + 1)
(d) 73 + (–14 – 1)

Answer

We remove the brackets in each expression of the collection and write it as a sum of terms. Recall that a bracket preceded by a ‘–’ sign changes the signs of the terms inside it.

73 - (14 + 1) = 73 - 14 - 1
73 - (14 - 1) = 73 - 14 + 1
73 + (-14 + 1) = 73 - 14 + 1
73 + (-14 - 1) = 73 - 14 - 1

Now we compare the terms:

The expression 73 – 14 + 1 has terms 73, –14 and 1. This matches:

• (b) 73 – (14 – 1) = 73 – 14 + 1
• (c) 73 + (–14 + 1) = 73 – 14 + 1

The expression 73 – 14 – 1 has terms 73, –14 and –1. This matches:

• (a) 73 – (14 + 1) = 73 – 14 – 1
• (d) 73 + (–14 – 1) = 73 – 14 – 1

Hence, expressions (b) and (c) are equal to 73 – 14 + 1, and expressions (a) and (d) are equal to 73 – 14 – 1.

Lhamo and Norbu went to a hotel. Each of them ordered a vegetable cutlet and a rasgulla. A vegetable cutlet costs ₹43 and a rasgulla costs ₹24. What about the total amount they have to pay? Can it be described by the expression: 2 × 43 + 24?

(i) If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid?

Answer

Each of them ordered one vegetable cutlet (₹43) and one rasgulla (₹24), so each person’s share is 43 + 24.

Since there are two people, the total amount is twice (43 + 24), which is written using brackets:

2 × (43 + 24) = 2 × 67 = 134

The expression 2 × 43 + 24 is not correct.

Writing it as a sum of terms:

2 × 43 + 24 = 86 + 24 = 110

This means “24 more than 2 × 43”, which pays for two vegetable cutlets but only one rasgulla — not what we want.

The correct expression can also be written as the cost of two vegetable cutlets and two rasgullas:

2 × (43 + 24) = 2 × 43 + 2 × 24 = 86 + 48 = 134

Hence, the total amount cannot be described by 2 × 43 + 24; the correct expression is 2 × (43 + 24) = 2 × 43 + 2 × 24 = ₹134.

(i)

Each friend orders one vegetable cutlet and one rasgulla.

Cost for one friend = 43 + 24

For three friends, the total amount is:

3 × (43 + 24)

= 3 × 67

= 201

Since there are three people, the total amount is three times (43 + 24):

Hence, the expression for the total amount is 3 × (43 + 24), which equals ₹201.

(i) 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
(ii) Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?

Answer

(i)

Writing 5 × 4 + 3 as a sum of terms, its terms are 5 × 4 and 3:

5 × 4 + 3 = 20 + 3 = 23

This means “3 more than 5 × 4”.

But,

5 × (4 + 3) = 5 × 7 = 35

Since 23 ≠ 35,

Hence, 5 × 4 + 3 ≠ 5 × (4 + 3)

(ii)

4 + 3 = 3 + 4 (swapping terms does not change a sum)

∴ 5 × (4 + 3) = 5 × (3 + 4) = 5 × 7 = 35

And (3 + 4) × 5 = 7 × 5 = 35

Hence,

5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5

∴ All three expressions are equal.

Hence, all three expressions are equal, and each has the value 35.

Example 18 showed an effective way of evaluating 97 × 25 by writing it as (100 – 3) × 25 = 100 × 25 – 3 × 25. Use this method to find the following products:

(a) 95 × 8
(b) 104 × 15
(c) 49 × 50

Is this quicker than the multiplication procedure you use generally?

Answer

We rewrite one factor as a sum or difference involving a round number, and then use the distributive property.

First, completing the example:

97 × 25 = (100 – 3) × 25

= 100 × 25 – 3 × 25

= 2500 – 75

= 2425

(a) 95 × 8

95 × 8 = (100 – 5) × 8

= 100 × 8 – 5 × 8

= 800 – 40

Hence, 95 × 8 = 760

(b) 104 × 15

104 × 15 = (100 + 4) × 15

= 100 × 15 + 4 × 15

= 1500 + 60

= 1560

Hence, 104 × 15 = 1560

(c) 49 × 50

49 × 50 = (50 – 1) × 50

= 50 × 50 – 1 × 50

= 2500 – 50

= 2450

Hence, 49 × 50 = 2450

Yes, this method is often quicker because it uses multiplication with round numbers such as 100 and 50, followed by a simple addition or subtraction.

Which other products might be quicker to find like the ones above?

Answer

This method is quicker whenever one of the factors is close to a multiple of 10, 50, 100, 1000, and so on.

We rewrite that factor as “round number + a little” or “round number – a little”, and then use the distributive property.

Some examples:

98 × 7 = (100 - 2) × 7

= 100 × 7 - 2 × 7

= 700 - 14

= 686

103 × 9 = (100 + 3) × 9

= 100 × 9 + 3 × 9

= 900 + 27

= 927

49 × 5 = (50 - 1) × 5

= 50 × 5 - 1 × 5

= 250 - 5

= 245

Hence, products in which one factor is close to a multiple of 10, 50, 100 or 1000 — such as 98 × 7, 103 × 9 and 49 × 5 — are quicker to find using this method.

Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.

(a) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 $\square$ 3 × ............
(d) (9 + 2) × 4 = 9 × 4 $\square$ 2 × ............
(e) 3 × (............ + 4) = 3 × ............ + ............
(f) (............ + 6) × 4 = 13 × 4 + ............
(g) 3 × (............ + ............) = 3 × 5 + 3 × 2
(h) (............ + ............) × ............ = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × ............
(j) (5 – 2) × 7 = 5 × 7 – 2 × ............
(k) 5 × (8 – 3) = 5 × 8 $\square$ 5 × ............
(l) (8 – 3) × 7 = 8 × 7 $\square$ 3 × 7
(m) 5 × (12 – ............) = ............ $\square$ 5 × ............
(n) (15 – ............) × 7 = ............ $\square$ 6 × 7
(o) 5 × (............ – ............) = 5 × 9 – 5 × 4
(p) (............ – ............) × ............ = 17 × 7 – 9 × 7

Answer

We use the distributive property: multiplying a number by a sum (or difference) inside brackets is the same as multiplying the number by each term and then adding (or subtracting).

(a) 3 × (6 + 7) = 3 × 6 + 3 × 7

(b) (8 + 3) × 4 = 8 × 4 + 3 × 4

(c) 3 × (5 + 8) = 3 × 5 $\boxed{+}$ 3 × 8

(d) (9 + 2) × 4 = 9 × 4 $\boxed{+}$ 2 × 4

(e) 3 × (10 + 4) = 3 × 10 + 3 × 4

(f) (13 + 6) × 4 = 13 × 4 + 6 × 4

(g) 3 × (5 + 2) = 3 × 5 + 3 × 2

(h) (2 + 3) × 4 = 2 × 4 + 3 × 4

(i) 5 × (9 – 2) = 5 × 9 – 5 × 2

(j) (5 – 2) × 7 = 5 × 7 – 2 × 7

(k) 5 × (8 – 3) = 5 × 8 $\boxed{–}$ 5 × 3

(l) (8 – 3) × 7 = 8 × 7 $\boxed{–}$ 3 × 7

(m) 5 × (12 – 3) = 5 × 12 $\boxed{–}$ 5 × 3

(n) (15 – 6) × 7 = 15 × 7 $\boxed{–}$ 6 × 7

(o) 5 × (9 – 4) = 5 × 9 – 5 × 4

(p) (17 – 9) × 7 = 17 × 7 – 9 × 7

In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.

(a) (8 – 3) × 29 $\square$ (3 – 8) × 29
(b) 15 + 9 × 18 $\square$ (15 + 9) × 18
(c) 23 × (17 – 9) $\square$ 23 × 17 + 23 × 9
(d) (34 – 28) × 42 $\square$ 34 × 42 – 28 × 42

Answer

We reason about the terms and brackets rather than evaluating fully.

(a) (8 – 3) × 29 > (3 – 8) × 29

On the LHS, 8 – 3 is a positive number, so (8 – 3) × 29 is positive. On the RHS, 3 – 8 is a negative number, so (3 – 8) × 29 is negative. A positive value is greater than a negative value.

(b) 15 + 9 × 18 < (15 + 9) × 18

The LHS has terms 15 and 9 × 18, so it is “fifteen more than nine 18s”. The RHS is (15 + 9) × 18 = 15 × 18 + 9 × 18, which is “fifteen 18s plus nine 18s”. Both have the part 9 × 18, but the RHS has 15 × 18 in place of just 15. Since 15 × 18 is much larger than 15, the RHS is greater.

(c) 23 × (17 – 9) < 23 × 17 + 23 × 9

By the distributive property, the LHS is 23 × (17 – 9) = 23 × 17 – 23 × 9. The RHS is 23 × 17 + 23 × 9. Both start from 23 × 17, but the LHS subtracts 23 × 9 while the RHS adds 23 × 9. So the LHS is smaller.

(d) (34 – 28) × 42 = 34 × 42 – 28 × 42

By the distributive property, (34 – 28) × 42 = 34 × 42 – 28 × 42. The two sides are exactly the same, so they are equal.

Hence, the boxes are filled as: (a) >, (b) <, (c) <, (d) =.

Here is one way to make 14: 2 × (1 + 6) = 14. Are there other ways of getting 14? Fill them out below:

(a) ............ × (............ + ............) = 14
(b) ............ × (............ + ............) = 14
(c) ............ × (............ + ............) = 14
(d) ............ × (............ + ............) = 14

Answer

We need expressions of the form ☐ × (☐ + ☐) whose value is 14.

Since 14 = 2 × 7 = 7 × 2, we choose a first factor and split the other factor as a sum inside the bracket.

(a) 2 × (5 + 2) = 2 × 7 = 14

(b) 2 × (3 + 4) = 2 × 7 = 14

(c) 7 × (1 + 1) = 7 × 2 = 14

(d) 2 × (6 + 1) = 2 × 7 = 14

Hence, four ways of making 14 are 2 × (5 + 2), 2 × (3 + 4), 7 × (1 + 1) and 2 × (6 + 1).

Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

Answer

The grid holds five 4s and four 8s.

Way 1 — group the equal numbers together, i.e. count how many of each there are:

5 × 4 + 4 × 8

= 20 + 32

= 52

Way 2 — add the grid row by row. The top and bottom rows are each 4 + 8 + 4, and the middle row is 8 + 4 + 8:

2 × (4 + 8 + 4) + (8 + 4 + 8)

= 2 × 16 + 20

= 32 + 20

= 52

Hence, the sum of the numbers in this picture is 52.

The grid holds eight 5s and eight 6s.

Way 1 — group the equal numbers together:

8 × 5 + 8 × 6

= 40 + 48

= 88

Way 2 — pair each 5 with a 6. There are 8 such pairs, each summing to 5 + 6:

8 × (5 + 6)

= 8 × 11

= 88

Hence, the sum of the numbers in this grid is 88.

Read the situations given below. Write appropriate expressions for each of them and find their values.

(a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.

(b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?

(c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?

Answer

(a)

The market runs 7 days a week.

Amount of mangoes supplied each day by Rahim = 9 kg

Amount of mangoes supplied each day by Shyam = 11 kg

Amount of mangoes supplied each day by Rahim and Shyam together = 9 + 11 kg

In a week, they supply:

Expression: 7 × (9 + 11)

7 × (9 + 11)

= 7 × 20

= 140

Hence, 140 kg of mangoes are supplied in a week.

(b)

Amount earned by Binu per month = ₹20,000

Total expenditure = Rent + Food + Other expenses

= 5,000 + 5,000 + 2,000

= ₹12,000

Expression: 12 × 20,000 – 12 × (5,000 + 5,000 + 2,000)

12 × 20,000 – 12 × (5,000 + 5,000 + 2,000)

= 2,40,000 – 1,44,000

= 96,000

The same value comes from 12 × (20,000 – 12,000) = 12 × 8,000 = 96,000.

Hence, Binu saves ₹96,000 in a year.

(c)

Each day the snail climbs 3 cm and slips 2 cm at night.

Net gain in one day-night cycle:

3 − 2 = 1 cm

After 7 days and 7 nights, the snail reaches:

7 × 1 = 7 cm

On the 8th day, it climbs 3 cm more and reaches:

7 + 3 = 10 cm

So, the snail reaches the top on the 8th day.

Expression: 7 × (3 – 2) + 3

7 × (3 – 2) + 3

= 7 + 3

= 10

So the snail reaches the treat on the 8th day.

Hence, the snail reaches the treat on the 8th day.

Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?

(a) 5 × 2 × 8
(b) (7 – 2) × 8
(c) 8 × 7
(d) 7 × 2 × 8
(e) 7 × 5 – 2
(f) (7 + 2) × 8
(g) 7 × 8 – 2 × 8
(h) (7 – 5) × 8

Answer

Melvin reads one story per day on the days he reads.

He reads on 5 days each week = (7 − 2).

So in 8 weeks:

Number of stories = (7 – 2) × 8

= 5 × 8

= 40

Now we check the value of each option:

(a) 5 × 2 × 8 = 80

(b) (7 – 2) × 8 = 40 ✓

(c) 8 × 7 = 56

(d) 7 × 2 × 8 = 112

(e) 7 × 5 – 2 = 33

(f) (7 + 2) × 8 = 72

(g) 7 × 8 – 2 × 8 = 56 – 16 = 40 ✓

(h) (7 – 5) × 8 = 16

Both (b) and (g) give 40. In fact (g) is just the distributive form of (b), since 7 × 8 – 2 × 8 = (7 – 2) × 8.

Hence, Melvin completes 40 stories in 8 weeks, and the expressions describing this scenario are (b) and (g).

Find different ways of evaluating the following expressions:

(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1

Answer

(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10

Way 1 — collect the positive terms and the negative terms separately:

(1 + 3 + 5 + 7 + 9) + (–2 – 4 – 6 – 8 – 10)

= 25 + (–30)

= –5

Way 2 — pair the terms two at a time:

(1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)

= (–1) + (–1) + (–1) + (–1) + (–1)

= –5

(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1

Way 1 — pair the terms two at a time:

(1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)

= 0 + 0 + 0 + 0 + 0

= 0

Way 2 — collect the positive terms and the negative terms separately:

(1 + 1 + 1 + 1 + 1) + (–1 – 1 – 1 – 1 – 1)

= 5 + (–5)

= 0

Hence, whatever grouping is used, expression (a) evaluates to –5 and expression (b) evaluates to 0.

Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning.

(a) 49 – 7 + 8 $\square$ 49 – 7 + 8
(b) 83 × 42 – 18 $\square$ 83 × 40 – 18
(c) 145 – 17 × 8 $\square$ 145 – 17 × 6
(d) 23 × 48 – 35 $\square$ 23 × (48 – 35)
(e) (16 – 11) × 12 $\square$ –11 × 12 + 16 × 12
(f) (76 – 53) × 88 $\square$ 88 × (53 – 76)
(g) 25 × (42 + 16) $\square$ 25 × (43 + 15)
(h) 36 × (28 – 16) $\square$ 35 × (27 – 15)

Answer

(a) 49 – 7 + 8 $\boxed{=}$ 49 – 7 + 8

(b) 83 × 42 – 18 $\boxed{\gt}$ 83 × 40 – 18.

(c) 145 – 17 × 8 $\boxed{\lt}$ 145 – 17 × 6.

(d) 23 × 48 – 35 $\boxed{\gt}$ 23 × (48 – 35).

(e) (16 – 11) × 12 $\boxed{=}$ –11 × 12 + 16 × 12.

(f) (76 – 53) × 88 $\boxed{\gt}$ 88 × (53 – 76).

(g) 25 × (42 + 16) $\boxed{=}$ 25 × (43 + 15).

(h) 36 × (28 – 16) $\boxed{\gt}$ 35 × (27 – 15).

Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.

(a) 83 – 37 – 12
(i) 84 – 38 – 12
(ii) 84 – (37 + 12)
(iii) 83 – 38 – 13
(iv) –37 + 83 – 12

(b) 93 + 37 × 44 + 76
(i) 37 + 93 × 44 + 76
(ii) 93 + 37 × 76 + 44
(iii) (93 + 37) × (44 + 76)
(iv) 37 × 44 + 93 + 76

Answer

(a) The given expression 83 – 37 – 12 has terms 83, –37 and –12.

(i) 84 – 38 – 12:

Compare term by term:

• 84 is 1 more than 83.
• –38 is 1 less than –37.
• –12 is unchanged.

The increase of 1 and decrease of 1 cancel each other.

So the value remains the same.

Equal.

(ii) 84 – (37 + 12)

Removing brackets:

84 − 37 − 12

Compared with 83 − 37 − 12

the first term is 1 more.

Not equal.

(iii) 83 – 38 – 13

Terms are: 83, −38, −13

Compared with the original, the expression is 2 less.

Not equal.

(iv) –37 + 83 – 12

Same terms in a different order.

Equal.

Therefore, in (a) the equal expressions are: (i) and (iv)

(b) The given expression 93 + 37 × 44 + 76 has terms 93, 37 × 44 and 76.

(i) 37 + 93 × 44 + 76

The product term is now 93 × 44, which is not 37 × 44.

Not equal.

(ii) 93 + 37 × 76 + 44

The product term is now 37 × 76, which is not 37 × 44.

Not equal.

(iii) (93 + 37) × (44 + 76)

This is a single product of two sums, an entirely different expression.

Not equal.

(iv) 37 × 44 + 93 + 76

Same terms, different order.

Equal.

So the only expression equal to 93 + 37 × 44 + 76 is (iv).

Choose a number and create ten different expressions having that value.

Answer

Let us choose the number 24.

Ten different expressions, each having the value 24, are:

20 + 4 = 24

30 – 6 = 24

4 × 6 = 24

48 ÷ 2 = 24

8 × 3 = 24

2 × (10 + 2) = 24

25 – 1 = 24

100 – 76 = 24

12 + 12 = 24

2 × 3 × 4 = 24

Using three 3’s along with the four operations (addition, subtraction, multiplication, and division) and brackets as needed we can create several expressions. For example, (3 + 3)/3 = 2, 3 + 3 – 3 = 3, 3 × 3 + 3 = 12, and so on.

(i) Using four 4’s, create expressions to get all values from 1 to 20.

(ii) Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as possible between –10 and +10.

(iii) Using the numbers 0 to 9 exactly once in any order, make an expression with a value 100.

Answer

(i) Using four 4’s with the operations +, –, × and ÷ (and brackets), we can make:

1 = (4 + 4) ÷ (4 + 4)

2 = 4 ÷ 4 + 4 ÷ 4

3 = (4 + 4 + 4) ÷ 4

4 = 4 + 4 × (4 – 4)

5 = (4 × 4 + 4) ÷ 4

6 = 4 + (4 + 4) ÷ 4

7 = 4 + 4 – 4 ÷ 4

8 = 4 + 4 + 4 – 4

9 = 4 + 4 + 4 ÷ 4

12 = 4 × (4 – 4 ÷ 4)

15 = 4 × 4 – 4 ÷ 4

16 = 4 × 4 + 4 – 4

17 = 4 × 4 + 4 ÷ 4

20 = 4 × (4 + 4 ÷ 4)

The values 10, 11, 13, 14, 18 and 19 cannot be obtained if only these four operations and brackets are allowed. To obtain every value from 1 to 20, an additional convention—such as concatenation (44) and decimal notation (0.4)—must be permitted. With those additional conventions:

10 = (44 – 4) ÷ 4

11 = 4 ÷ 4 + 4 ÷ 0.4

13 = 4 + (4 – 0.4) ÷ 0.4

14 = 4 × (4 – 0.4) – 0.4

18 = 4 + 4 + 4 ÷ 0.4

19 = (4 + 4 – 0.4) ÷ 0.4

Hence, the stated operations alone produce 14 of the values from 1 to 20; the remaining six require an additional convention such as concatenation or decimal notation.

(ii)

Using 1, 2, 3, 4 and 5 exactly once each (with +, –, ×, ÷ and brackets), in fact every whole number from –10 to +10 can be reached. One expression for each value is:

–10 = 1 × 2 – 3 – 4 – 5

–9 = 1 + 2 – 3 – 4 – 5

–8 = 1 – 2 – 3 × 4 + 5

–7 = 1 + 3 – 4 – 5 – 2

–6 = 1 × 2 × 3 × (4 – 5)

–5 = 1 – 2 – 3 + 4 – 5

–4 = 1 × 2 + 3 – 4 – 5

–3 = 1 – 2 – 3 – 4 + 5

–2 = 1 + 2 × 3 – 4 – 5

–1 = 1 + 2 – 3 + 4 – 5

0 = 5 – 4 – 3 + 2 × 1

1 = 5 – 4 – 3 + 2 + 1

2 = 5 – 4 + 3 – 2 × 1

3 = 5 – 4 + 3 – 2 + 1

4 = 1 × 2 + 3 + 4 – 5

5 = 1 + 2 + 3 + 4 – 5

6 = 2 × 3 – 4 + 5 – 1

7 = 5 + 4 – 3 + 2 – 1

8 = 5 + 4 – 3 + 2 × 1

9 = 1 + 2 – 3 + 4 + 5

10 = 1 + 2 + 3 × 4 – 5

Hence, using 1, 2, 3, 4 and 5 exactly once, every whole number from –10 to +10 can be made, as shown above.

(iii)

One expression that uses each of the digits 0 to 9 exactly once and has the value 100 is:

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 × 9

Evaluating it (the term 8 × 9 = 72 is found first, then added to the rest):

0 + (1 + 2 + 3 + 4 + 5 + 6 + 7) + 8 × 9 = 0 + 28 + 72 = 100

Hence, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 × 9 = 100, using each digit from 0 to 9 exactly once.