Writing Expressions for Real-World Situations

How to turn a situation in words (ages, money, lengths, consecutive numbers, speed and rate) into an algebraic expression with its unit, and how to write a formula from a situation and use it.

Before you start

A letter stands for a number. "The length is $x$ cm" means that $x$ is the number of centimetres.

The short forms: $3 \times p$ is written $3p$ , $1 \times n$ is written $n$ , $k \div 6$ is written $\frac{k}{6}$ , and $3 \times (w + 5)$ is written $3(w + 5)$ , where the brackets keep $w + 5$ together.

From earlier chapters: distance = speed × time; 1 hour = 60 minutes; $1 = 100 cents. For example, at 60 km/h for 2 hours a car travels $60 \times 2 = 120$ km.

From Words to an Expression

Many everyday quantities depend on a number we do not know or that can change: a height after a growth spurt, a bill for some number of meals, the age of a person in a few years. An algebraic expression describes such a quantity for every value of the number at once.

The surest way to write the expression is to work one or two cases with numbers first. Watch what you do to the number, then do exactly the same to the letter.

Method: writing an expression for a situation

1. Say what the letter stands for, with its unit ("Mei Ling grows $x$ cm": $x$ is a number of centimetres).

2. Work a case with a number in place of the letter, and notice the operation you use.

3. Do the same operation to the letter, and write the result in its short form.

4. Write the unit after the expression. If the expression has more than one term, put it in brackets first: $(151 + x)$ cm, $(3p + 4)$ dollars.

Example 1: Mei Ling's height

Mei Ling is 151 cm tall. How tall will she be after she grows
(a) 3 cm,
(b) 10 cm,
(c) $x$ cm?

Solution

(a) Growing adds to her height: $151 + 3 = 154$ . She will be 154 cm tall.

(b) In the same way, $151 + 10 = 161$ . She will be 161 cm tall.

(c) Whatever she grows is added to 151, so growing $x$ cm gives $151 + x$ . The expression has two terms, so it goes in brackets before the unit.

(a) 154 cm (b) 161 cm (c) $(151 + x)$ cm. The cases (a) and (b) are this expression with $x = 3$ and $x = 10$ .

The words of a situation tell you the operation. The table gives the common ones, with $n$ as the number. Take care with less than: "5 less than $n$ " starts from $n$ and takes 5 away, so it is $n - 5$ , written in the opposite order from the words.

Words	Expression
5 more than $n$ ; $n$ increased by 5; 5 older, taller or heavier than $n$	$n + 5$
5 less than $n$ ; $n$ decreased by 5; 5 younger, shorter or lighter than $n$	$n - 5$
$n$ less than 5; $n$ subtracted from 5	$5 - n$
5 times $n$ ; $n$ multiplied by 5	$5n$
half of $n$ ; $n$ shared equally between 2	$\frac{n}{2}$
$n$ shared equally among 5; $n$ divided by 5	$\frac{n}{5}$

Example 2: Phrases into expressions

Write an algebraic expression for each of these.
(a) 11 more than $n$
(b) 11 less than $n$
(c) $n$ less than 11
(d) 6 less than three times $n$
(e) 4 times the sum of $n$ and 9
(f) $n$ sweets shared equally among 4 children, then 2 more sweets given to each child

Solution

(a) Add 11 to $n$ : $n + 11$ .

(b) Start from $n$ and take 11 away: $n - 11$ . (With $n = 15$ : 11 less than 15 is 4, and $15 - 11 = 4$ .)

(d) Three times $n$ is $3n$ . Start from $3n$ and take 6 away: $3n - 6$ .

(e) The sum of $n$ and 9 is $n + 9$ . Four times the whole sum needs brackets: $4 \times (n + 9)$ , written $4(n + 9)$ . Without brackets, $4n + 9$ would multiply only $n$ by 4.

(f) Each child first gets $n \div 4 = \frac{n}{4}$ sweets, then 2 more: $\frac{n}{4} + 2$ . It has two terms, so it goes in brackets before its unit.

(a) $n + 11$ (b) $n - 11$ (c) $11 - n$ (d) $3n - 6$ (e) $4(n + 9)$ (f) $\left(\frac{n}{4} + 2\right)$ sweets for each child

Ages

Every year each person gets one year older. So "in 4 years" adds 4 to an age, and "3 years ago" takes 3 away. The difference between two people's ages never changes.

Example 3: Ravi and his family

Ravi is $r$ years old. Write an expression for
(a) Ravi's age in 4 years,
(b) Ravi's age 3 years ago,
(c) the age of his sister, who is 3 years older than Ravi,
(d) the age of his father, who is 4 times as old as Ravi,
(e) the age of his father in 5 years.

Solution

(a) If Ravi were 12, in 4 years he would be $12 + 4 = 16$ . So in 4 years he will be $(r + 4)$ years old.

(b) If he were 12, 3 years ago he was $12 - 3 = 9$ . So 3 years ago he was $(r - 3)$ years old.

(d) 4 times Ravi's age is $4 \times r = 4r$ years.

(e) The father is $4r$ years old now. In 5 years he will be 5 years older: $(4r + 5)$ years. (Not $4(r + 5)$ : that would be 4 times Ravi's age in 5 years, a different number. With $r = 12$ , the father is 48 now and 53 in 5 years; $4(12 + 5) = 68$ is wrong.)

(a) $(r + 4)$ years (b) $(r - 3)$ years (c) $(r + 3)$ years (d) $4r$ years (e) $(4r + 5)$ years

Money

A total cost is the number of items times the price of one, added up for each kind of item. The money left, or the change, is the amount you had minus the amount you spent.

Amounts can be added only when they are in the same unit. Change cents into dollars by dividing by 100, and write the result as a decimal: $x$ cents is $\frac{x}{100}$ dollars, which is $0.01x$ dollars, and $3x$ cents is $0.03x$ dollars. Change dollars into cents by multiplying by 100.

Example 4: Burgers and hot dogs

At a stall a burger costs $3 and a hot dog costs $1. Find the total price of
(a) 2 burgers and 5 hot dogs,
(b) $p$ burgers and 4 hot dogs,
(c) $m$ burgers and $n$ hot dogs.

Solution

(a) Burgers: $3 \times 2 = 6$ dollars. Hot dogs: $1 \times 5 = 5$ dollars. Total: $6 + 5 = 11$ dollars.

(b) Burgers: $3 \times p = 3p$ dollars. Hot dogs: $1 \times 4 = 4$ dollars. Total: $(3p + 4)$ dollars.

(c) Burgers: $3 \times m = 3m$ dollars. Hot dogs: $1 \times n = n$ dollars (we write $n$ , not $1n$ ). Total: $(3m + n)$ dollars.

In $3p + 4$ the two terms cannot be joined into one: $3p$ is a number of dollars that depends on $p$ , and 4 is a fixed number of dollars.

(a) $11 (b) $(3p + 4)$ dollars (c) $(3m + n)$ dollars

Example 5: Change, and cents into dollars

Siti buys 3 notebooks at $x$ dollars each and 6 pencils at $p$ cents each.
(a) She pays for the notebooks alone with a $50 note. Write her change.
(b) Write the cost of the 6 pencils in cents, and then in dollars.
(c) Write the total cost of the notebooks and the pencils in dollars.

Solution

(a) The notebooks cost $3 \times x = 3x$ dollars. Change = amount paid − cost = $(50 - 3x)$ dollars.

(b) The pencils cost $6 \times p = 6p$ cents. In dollars, divide by 100: $\frac{6p}{100}$ dollars, which is $0.06p$ dollars.

Check with numbers: if a notebook costs $2 and a pencil 50 cents, the notebooks cost $6, the pencils cost 300 cents = $3, and the total is $9. The expression gives $3 \times 2 + 0.06 \times 50 = 6 + 3 = 9$ .

(a) $(50 - 3x)$ dollars (b) $6p$ cents, or $0.06p$ dollars (c) $(3x + 0.06p)$ dollars

Example 6: Sharing a cost

A group of $n$ friends rent a chalet for $a$ dollars and spend $b$ dollars on food. They share the total cost equally. How much does each friend pay?

Solution

The total cost is $(a + b)$ dollars.

Sharing equally among $n$ people means dividing the whole total by $n$ : $(a + b) \div n$ , written $\frac{a + b}{n}$ . The fraction line acts as a bracket: the whole of $a + b$ is divided.

Check with numbers: with $a = 240$ , $b = 96$ and $n = 8$ , the total is $336 and each pays $336 \div 8 = 42$ dollars; $\frac{240 + 96}{8} = \frac{336}{8} = 42$ .

Each friend pays $\frac{a + b}{n}$ dollars. (Not $\left(a + \frac{b}{n}\right)$ dollars, which would share only the food.)

Lengths and Shapes

The perimeter of a shape is the sum of the lengths of its sides; the area of a rectangle is length × breadth. When a side is an expression, write it in brackets and use it exactly as you would use a number.

Example 7: A rectangle 5 cm longer than it is broad

A rectangular card has breadth $w$ cm. Its length is 5 cm more than its breadth. Write expressions for
(a) its length,
(b) its perimeter,
(c) its area.

Solution

(a) "5 cm more than $w$ cm" is $w + 5$ . The length is $(w + 5)$ cm.

(b) The perimeter is two lengths and two breadths: $2 \times (w + 5) + 2 \times w = 2(w + 5) + 2w$ . The bracket keeps $w + 5$ together: both the $w$ and the 5 are doubled.

We leave the perimeter in this form. Putting the terms together into one shorter expression is done in the next chapter.

(a) $(w + 5)$ cm (b) $[2(w + 5) + 2w]$ cm (c) $w(w + 5)$ cm²

Consecutive Whole Numbers

Consecutive whole numbers follow one another, each 1 more than the one before: 14, 15, 16. Consecutive even numbers (or odd numbers) go up in steps of 2: 14, 16, 18, or 15, 17, 19.

Example 8: Consecutive numbers

(a) The smallest of three consecutive whole numbers is $n$ . Write the other two.
(b) The middle one of three consecutive whole numbers is $m$ . Write all three.
(c) The smallest of three consecutive even numbers is $k$ . Write all three.

Solution

(a) Each number is 1 more than the one before: after $n$ come $n + 1$ and then $n + 2$ . (With $n = 14$ : 14, 15, 16.)

(b) The number before $m$ is $m - 1$ and the number after it is $m + 1$ .

(a) $n + 1$ and $n + 2$ (b) $m - 1$ , $m$ , $m + 1$ (c) $k$ , $k + 2$ , $k + 4$

Time, Speed and Rate

The rules of speed and rate hold with letters as they do with numbers: distance = speed × time, time = distance ÷ speed, speed = distance ÷ time, and an amount = rate × time. Make sure the units agree: a speed in km/h needs a time in hours.

Example 9: A car and a printer

A car travels at a steady 60 km/h.
(a) How far does it travel in $t$ hours?
(b) How long does it take to travel $d$ km?
(c) How far does it travel in $t$ minutes?
A printer prints 12 pages a minute.
(d) How many pages does it print in $m$ minutes?

Solution

(a) Distance = speed × time $= 60 \times t = 60t$ km.

(b) Time = distance ÷ speed $= d \div 60 = \frac{d}{60}$ hours.

(c) The speed is per hour, so change the time into hours first: $t$ minutes $= \frac{t}{60}$ hours. Distance $= 60 \times \frac{t}{60} = t$ km. (Check: in 30 minutes, half an hour, the car goes 30 km.)

(d) Pages = rate × time $= 12 \times m = 12m$ .

(a) $60t$ km (b) $\frac{d}{60}$ hours (c) $t$ km (d) $12m$ pages

Writing a Formula

When a quantity in a situation has its own letter, the expression for it becomes a formula: the letter, an equals sign, and the expression. Once the formula is written, the quantity can be found for any value of the other letters by substitution.

Definition: formula

A formula connects two or more quantities; the letter on its own on one side of the equals sign is its subject. In $q = 10p$ , the subject is $q$ .

Example 10: Two incomes

Mr Kumar earns $p$ dollars a year from a part-time job. His brother, Mr Rajan, earns 10 times as much a year.
(a) Write Mr Rajan's yearly income in terms of $p$ .
(b) Mr Rajan earns $q$ dollars a year. Write a formula connecting $q$ and $p$ .
(c) Use the formula to find Mr Rajan's yearly income when Mr Kumar earns $7200 a year.

Solution

(a) 10 times $p$ dollars is $10 \times p = 10p$ dollars.

(b) Mr Rajan's income, $q$ dollars, is equal to $10p$ dollars, so $q = 10p$ . Here $p$ is Mr Kumar's yearly income in dollars and $q$ is Mr Rajan's, in dollars.

(a) $10p$ dollars (b) $q = 10p$ (c) Mr Rajan earns $72 000 a year.

Example 11: Groups at a holiday camp

Every June, 100 children go to a holiday camp. They are split into $z$ groups of equal size, where $z$ is a factor of 100.
(a) How many children are there in each group?
(b) There are $y$ children in each group. Write a formula for $y$ in terms of $z$ .
(c) Use the formula to find the number of children in each group when there are 4 groups.

Solution

(a) Sharing 100 children equally among $z$ groups: $100 \div z = \frac{100}{z}$ children.

(b) $y$ is the number in each group, so $y = \frac{100}{z}$ . The subject is $y$ .

(a) $\frac{100}{z}$ children (b) $y = \frac{100}{z}$ (c) There are 25 children in each group.

Example 12: Hiring a hall

A community centre charges $200 to book its hall, plus $15 for every hour it is used.
(a) Write a formula for the cost, $C$ dollars, of using the hall for $h$ hours.
(b) Find the cost of using the hall for 4 hours.

Solution

(a) With numbers first: for 3 hours the cost is $200 + 15 \times 3 = 245$ dollars. For $h$ hours it is $200 + 15 \times h = (200 + 15h)$ dollars. So $C = 200 + 15h$ , where $C$ is the cost in dollars and $h$ the number of hours.

(b) Substitute $h = 4$ : $C = 200 + 15 \times 4 = 200 + 60 = 260$ .

(a) $C = 200 + 15h$ (b) Using the hall for 4 hours costs $260.

Common Mistakes

Writing "less than" in the order of the words. "5 less than $n$ " is $n - 5$ , not $5 - n$ . Try a number: 5 less than 12 is 7, which is $12 - 5$ .
Leaving out brackets. Three shirts at $(p - 7)$ dollars each cost $3(p - 7)$ dollars, not $3p - 7$ ; $(a + b)$ dollars shared among $n$ people is $\frac{a + b}{n}$ , not $a + \frac{b}{n}$ .
Adding amounts in different units. 5 rulers at $x$ cents and a $2 notebook do not cost $5x + 2$ of anything: write $(5x + 200)$ cents, or $(0.05x + 2)$ dollars.
Writing the unit after only part of the expression. " $3p + 4$ dollars" reads as $3p$ plus 4 dollars; the total is $(3p + 4)$ dollars.
Mixing up "older" and "younger", "ago" and "in". 3 years ago Ravi was $(r - 3)$ years old; in 3 years he will be $(r + 3)$ years old.
Writing consecutive numbers as $n$ , $2n$ , $3n$ . Consecutive whole numbers go up by 1: $n$ , $n + 1$ , $n + 2$ .
Using minutes with a speed in km/h. At 60 km/h for $t$ minutes the distance is not $60t$ km: change $t$ minutes into $\frac{t}{60}$ hours first.

Practice

Try each question before you open its solution. The questions get harder as you go.

Question 1

Write an algebraic expression for each of these.
(a) 12 more than $m$
(b) 12 less than $m$
(c) $m$ less than 12
(d) one third of $m$
(e) 4 less than 5 times $m$
(f) twice the sum of $m$ and 9

ShowHide solution

(a) Add 12 to $m$ : $m + 12$ .

(b) Start from $m$ and take 12 away: $m - 12$ .

(d) One third of $m$ is $m \div 3 = \frac{m}{3}$ .

(e) 5 times $m$ is $5m$ ; take 4 away: $5m - 4$ .

(f) The sum is $m + 9$ ; twice the whole sum is $2(m + 9)$ .

(a) $m + 12$ (b) $m - 12$ (c) $12 - m$ (d) $\frac{m}{3}$ (e) $5m - 4$ (f) $2(m + 9)$

Question 2

Siti is $s$ years old. Write an expression for
(a) her age in 6 years,
(b) her age 2 years ago,
(c) the age of her brother, who is 5 years younger than Siti,
(d) the age of her mother, which is 4 more than 3 times Siti's age,
(e) Siti's age in months.

ShowHide solution

(a) In 6 years she is 6 years older: $s + 6$ .

(b) 2 years ago she was 2 years younger: $s - 2$ .

(d) 3 times Siti's age is $3s$ ; 4 more than that is $3s + 4$ .

(e) Each year has 12 months: $12 \times s = 12s$ .

(a) $(s + 6)$ years (b) $(s - 2)$ years (c) $(s - 5)$ years (d) $(3s + 4)$ years (e) $12s$ months

Question 3

Hui Min is 20 cm shorter than Ravi, and Ravi is 10 cm shorter than Siti. Ravi is $x$ cm tall. Write expressions for the heights of Hui Min and Siti.

ShowHide solution

Hui Min is shorter than Ravi by 20 cm: $x - 20$ .

Ravi is 10 cm shorter than Siti, so Siti is 10 cm taller than Ravi: $x + 10$ . (Check: if Ravi is 150 cm, Siti is 160 cm and Ravi is indeed 10 cm shorter.)

Hui Min is $(x - 20)$ cm tall and Siti is $(x + 10)$ cm tall.

Question 4

At a bakery, a bun costs $2 and a box of tarts costs $9. Ali has $50. He buys $b$ buns and $t$ boxes of tarts.
(a) Write the total cost.
(b) Write the amount of money Ali has left.

ShowHide solution

(a) Buns: $2 \times b = 2b$ dollars. Tarts: $9 \times t = 9t$ dollars. Total: $(2b + 9t)$ dollars.

(b) From $50 he spends $2b$ dollars on buns and $9t$ dollars on tarts, so he has $(50 - 2b - 9t)$ dollars left.

Check: with 4 buns and 2 boxes he spends $8 + 18 = 26$ dollars and has $24 left; $50 - 2 \times 4 - 9 \times 2 = 50 - 8 - 18 = 24$ .

(a) $(2b + 9t)$ dollars (b) $(50 - 2b - 9t)$ dollars

Question 5

Ravi buys 4 pens at $x$ cents each and a file at $3. Write the total cost
(a) in cents,
(b) in dollars.

ShowHide solution

(a) The pens cost $4 \times x = 4x$ cents. The file costs $3 = 300 cents. Total: $(4x + 300)$ cents.

(b) The pens cost $4x$ cents, which is $\frac{4x}{100}$ dollars, or $0.04x$ dollars. Total: $(0.04x + 3)$ dollars.

Check: if a pen costs 50 cents, the pens cost 200 cents = $2 and the total is $5; $0.04 \times 50 + 3 = 2 + 3 = 5$ .

(a) $(4x + 300)$ cents (b) $(0.04x + 3)$ dollars

Question 6

(a) The smallest of four consecutive whole numbers is $n$ . Write the other three.
(b) The largest of three consecutive odd numbers is $p$ . Write the other two.
(c) Mei Ling says that $n$ , $2n$ and $3n$ are three consecutive whole numbers. Is she right? Explain.

ShowHide solution

(a) Each is 1 more than the one before: $n + 1$ , $n + 2$ , $n + 3$ .

(b) Odd numbers go in steps of 2. Going down from $p$ : $p - 2$ , then $p - 4$ . (With $p = 15$ : 11, 13, 15.)

(c) Try $n = 5$ : $n$ , $2n$ , $3n$ are 5, 10, 15, which go up in steps of 5, not 1. The step from each number to the next is $n$ , so they are consecutive only when $n = 1$ (1, 2, 3). For any other whole number they are not, so in general she is wrong; three consecutive whole numbers are $n$ , $n + 1$ , $n + 2$ .

(a) $n + 1$ , $n + 2$ , $n + 3$ (b) $p - 4$ and $p - 2$ (c) No: it works only for $n = 1$ ; for example 5, 10, 15 are not consecutive.

Question 7

A rectangular garden is $l$ m long. Its breadth is 3 m less than its length. Write expressions for
(a) its breadth,
(b) its perimeter,
(c) its area.

ShowHide solution

(a) 3 m less than $l$ m is $l - 3$ .

(b) Two lengths and two breadths: $2 \times l + 2 \times (l - 3) = 2l + 2(l - 3)$ .

(a) $(l - 3)$ m (b) $[2l + 2(l - 3)]$ m (c) $l(l - 3)$ m²

Question 8

A piece of wire is 60 cm long. Siti cuts off part of it and bends that part into a square of side $s$ cm.
(a) How much wire does she use for the square?
(b) How much wire is left?
(c) She bends all of the wire that is left into an equilateral triangle. Write the length of one side of the triangle.

ShowHide solution

(a) A square has 4 equal sides: $4 \times s = 4s$ cm.

(b) The wire left is 60 cm minus the $4s$ cm used: $(60 - 4s)$ cm.

(c) An equilateral triangle has 3 equal sides, so the wire left is divided by 3: $(60 - 4s) \div 3 = \frac{60 - 4s}{3}$ cm. The whole of $60 - 4s$ is divided.

Check: with $s = 6$ , the square uses 24 cm, 36 cm is left, and each side of the triangle is 12 cm; $\frac{60 - 24}{3} = \frac{36}{3} = 12$ .

(a) $4s$ cm (b) $(60 - 4s)$ cm (c) $\frac{60 - 4s}{3}$ cm

Question 9

(a) Ravi walks at 5 km/h for $t$ hours and then rides a bus for 20 km. Write the total distance he travels.
(b) Mei Ling cycles at 15 km/h for $m$ minutes. Write the distance she cycles.
(c) A train travels $d$ km in 2 hours. Write its average speed.

ShowHide solution

(a) Walking: $5 \times t = 5t$ km. Total: $(5t + 20)$ km.

(b) $m$ minutes $= \frac{m}{60}$ hours. Distance $= 15 \times \frac{m}{60} = \frac{15m}{60} = \frac{m}{4}$ km (dividing 15 and 60 by 15). Check: in 60 minutes she cycles 15 km, and $\frac{60}{4} = 15$ .

(a) $(5t + 20)$ km (b) $\frac{m}{4}$ km (c) $\frac{d}{2}$ km/h

Question 10

Mr Tan's choir books a hall for $x$ sessions. Each session is booked for $y$ hours, and the hall costs $40 an hour. Every session ends 1 hour early, and the choir pays only for the hours it uses.
(a) How many hours does the choir use in each session?
(b) Write the total cost.
(c) How much less is this than the cost of all the hours booked?

ShowHide solution

(a) Each session uses 1 hour less than booked: $y - 1$ hours.

(b) One session costs $40 \times (y - 1) = 40(y - 1)$ dollars. For $x$ sessions: $x \times 40(y - 1)$ , written with the number first, $40x(y - 1)$ dollars.

Check: 3 sessions booked for 4 hours each use 3 hours each, 9 hours in all, costing $360; $40 \times 3 \times (4 - 1) = 360$ . The 12 hours booked would cost $480, and $480 - 360 = 120 = 40 \times 3$ .

(a) $(y - 1)$ hours (b) $40x(y - 1)$ dollars (c) $40x$ dollars

Question 11

A phone plan costs $18 a month, plus 5 cents for every minute of overseas calls.
(a) Write a formula for the monthly bill, $B$ dollars, when $m$ minutes of overseas calls are made.
(b) Use your formula to find the bill for a month with 120 minutes of overseas calls.

ShowHide solution

(a) $m$ minutes at 5 cents cost $5m$ cents, which is $\frac{5m}{100} = 0.05m$ dollars. Adding the fixed $18: $B = 18 + 0.05m$ , where $B$ is the bill in dollars and $m$ the number of minutes.

(b) Substitute $m = 120$ : $B = 18 + 0.05 \times 120 = 18 + 6 = 24$ .

(a) $B = 18 + 0.05m$ (b) The bill is $24.

Question 12

In a class there are $c$ boys and $d$ girls. In a test, the average mark of the boys is $x$ and the average mark of the girls is $y$ .
(a) Write the total of the boys' marks.
(b) Write the total of the marks of the whole class.
(c) Write the average mark of the whole class.
(d) There are 18 boys with an average of 60 and 22 girls with an average of 70. Use your answer to (c) to find the average mark of the class.

ShowHide solution

(a) Average = total ÷ number, so total = average × number: $x \times c$ , written $cx$ with the letters in alphabetical order. (Check: 10 boys with an average of 60 have 600 marks in all.)

(b) The girls' total is $dy$ in the same way. The class total is $cx + dy$ .

(c) There are $c + d$ students. Average = total ÷ number of students $= \frac{cx + dy}{c + d}$ . Both the top and the bottom are whole sums.

(d) Substitute $c = 18$ , $x = 60$ , $d = 22$ , $y = 70$ : $\frac{18 \times 60 + 22 \times 70}{18 + 22} = \frac{1080 + 1540}{40} = \frac{2620}{40} = 65.5$ .

(a) $cx$ (b) $cx + dy$ (c) $\frac{cx + dy}{c + d}$ (d) The average mark of the class is 65.5.

Key Points

Say what the letter stands for, with its unit, before you write an expression.
Work a case with numbers first, then do the same to the letter.
"5 more than $n$ " is $n + 5$ ; "5 less than $n$ " is $n - 5$ ; " $n$ less than 5" is $5 - n$ .
"In 4 years" adds 4 to an age; "3 years ago" takes 3 away.
A total cost is number × price for each kind of item, added; change is the amount paid minus the cost.
Put amounts in the same unit before adding them: $x$ cents is $\frac{x}{100}$ dollars, which is $0.01x$ dollars.
A whole sum that is multiplied or shared goes in brackets or over a fraction line: $4(n + 9)$ , $\frac{a + b}{n}$ .
An expression with more than one term goes in brackets before its unit: $(151 + x)$ cm, $(3p + 4)$ dollars.
Consecutive whole numbers are $n$ , $n + 1$ , $n + 2$ ; consecutive even or odd numbers go up by 2.
Distance = speed × time, with the time in the unit of the speed.
A formula connects quantities: its subject, an equals sign, and an expression, such as $q = 10p$ . Its value is found by substituting numbers for the other letters.