To increase a number by r%, multiply the number by (1 ); to decrease a number by r% multiply by (1–

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SkillSHEET

answers

SkillSHEET 1.1

Percentages

To find a certain percentage of a quantity, change the percentage into a fraction or a decimal and multiply by that quantity.

To increase or decrease some amount by a certain percentage is a calculation which is often required. An increase yields a larger number and a decrease yields a smaller number compared with the original value.

To increase a number by r%, multiply the number by (1 + ); to decrease a number by r% multiply by (1 – ).

The number (1 + ) or (1 – ) is often referred to as the multiplying factor.

Try these

1 Find:

a 10% of 60 b 5% of 130 c 40% of $800 d 25% of 700 students

e 16% of 75 f 53% of $20 g 4% of 132 h 7.5% of 260 articles.

2 If 95% of students passed their mid-year examination, how many of the 140 students passed?

3 Enrolments this year increased by 12% from last year. If there were 510 enrolments recorded last year, what is

the total recorded enrolment for this year ?

4 a Wages will be increased by r%. What is the multiplying factor if wages increased by:

i 10% ii 20% iii 25% iv 7% v 12 %?

b If a person is earning $260 per week, calculate the new wage if it increased by:

i 10% ii 20% iii 25% iv 7% v 12 %.

5 a The price of an item is reduced by r%. What is the multiplying factor if the reduction was:

i 10% ii 20% iii 25% iv 7% v 12 %?

b If the printer cost $360, calculate its value if the initial cost depreciated by:

i 10% ii 20% iii 25% iv 7% v 12 %.

Find 15% of $270.

Solution

Express 15% as a fraction and multiply by 270.

15% of $270 = ×

=

= $ 40.50 15 100 --- 270 1 ---4050 100

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1

r 100 ---r 100 ---r 100 --- r 100

---Increase $50 by 15%. Decrease 340 by 5%.

Solution Solution

To increase by 15%, multiply $50 by (1 + ).

50 × (1 + ) = 50 × (1 + 0.15)

= 50 × 1.15

= $57.50

To decrease by 5%, multiply 340 by (1 − ).

340 × (1 − ) = 340 × (1 − 0.95)

= 340 × 0.95

= 323

15 100 ---15 100 ---5 100 ---5 100

---2

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3

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---SkillSHEET

answers

SkillSHEET 1.2

Expressing one number as a percentage of

another

To express one number as a percentage of another, form a fraction using these two numbers and multiply it by 100.

Note that the number that is being expressed as a percentage is placed in the numerator of the fraction. For example, if we need to express 5 as a percentage of 20, we put 5 in the numerator and 20 in the denominator of a

fraction (that is, ).

Try these

For each of the following, express the first number as a percentage of the second number, giving your answer correct to two decimal places.

1 42, 53 2 13, 75 3 34, 150

4 47, 95 5 3, 21 6 12, 35

7 23, 60 8 256, 780 9 7, 65

10 5, 41 11 12, 28 12 37, 61

13 341, 730 14 11, 25 15 15, 24

5 20

---Express 18 as a percentage of 34.

Solution

Form a fraction by placing 18 in the numerator and 34 in the denominator.

To change a fraction into a percentage, multiply by 100.

× 100 = × (Write 100 as a fraction by putting it over 1.)

= (Multiply numerators together and denominators together.)

= 52.94% (Divide the numerator by the denominator, giving the answer correct to two

decimal places.) 18

34

---18 34

--- 18

34

--- 100

1

---1800 34

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Ratio

A ratio is a comparison of two quantities using the same units. A rate is a comparison of two quantities using different units. For example the concentrate of a pesticide, Xenon, must be diluted before use. The instructions suggest that 15 mL of pesticide be mixed with 750 mL of water. This is written:

Pesticide : Water 15 mL : 750 mL.

Clearly if one were to mix 5 mL of pesticide with 250 mL of water the strength of the mixture would be the same. The ratio

5 mL : 250 mL is equivalent to

15 mL : 750 mL.

The instructions for this pesticide say that 500 mL will cover 20 m2_{in area. Because these quantities have}

dif-ferent units their comparison is called a rate:

500 mL : 20 m2

is a rate which is equivalent to

250 mL : 10 m2_.

Alternatively one can write 500 mL per 20 m2_{and then reduce it to 25 mL per 1 m}2_{or 25 mL/m}2_.

If these ratios are all equivalent to 400 : 600, find A car travels 350 km on 24 L of petrol. Simplify this the missing number. rate and calculate how many kilometres the same car

a 200 :x will travel on 40 L of fuel.

b x: 200 c x: 150

a 400 ÷ 2 = 200, so x= 600 ÷ 2 = 300 b 600 ÷ 3 = 200, so x= 400 ÷ 3 = 133·3 c 600 ÷ 4 = 150, so x= 400 ÷ 4 = 100

The rate is 350 kM per 24 litres

= 350/24 km per litre

= 14·58 km/L.

On 40 L the car will travel 40 × 14·6 km = 583 km.

1

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2

Jan and Dean built a bird-bath which they sold at the markets for $140. They agreed that Jan had spent 4 hours building it and that Dean had worked for 3 hours. If the money is split in the same ratio as the time they spent, how much does each of them receive?

Solution

Jan : Dean 4 : 3

Total number of hours = 7

So Jan gets of $140 = $80

and Dean gets of $140 = $60

4 7

---3 7

---3

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1 If these ratios are all equivalent to 30 : 48, find the missing number.

a 120 :x b x: 240 c x: 36

a 200 :x b x: 2 c x: 150

3 If a car travels 250 km on 22 L of petrol, express this rate in kilometres per litre and calculate the amount of

petrol needed by this car for a journey of 600 km.

5 A weed-killer concentrate should be mixed with water in the ratio of 25 mL to 2000 mL (2 L). If there is only

15 mL of the concentrate remaining, how much water should be mixed with it?

6 Patsy and Crystal set up a screen printing business, with Patsy contributing $1200 and Crystal $700. How

should the first month’s profit of $400 be distributed?

7 The Green Pines resort charges $624 per person for a 7-night stay. At the same rate what would you expect to

pay for a 2-night stay?

8 To stay 5 nights at the Cool Breeze Caravan Park costs $47·50. What would you pay for 2 nights at the same

rate?

9 A car travels at 80 km/h. Express this rate in these units.

a metres/hour b metres/minute c metres/second

d hours/100 kilometres

10 Hewey Lewis can run 100 m in 12 s. How fast is this in km/h?

11 The speed of sound is 340 m/s.

a What is this speed in km/h?

b In the 200-metre event the starting gun is 170 m from the timers. If we assume that the smoke from the gun

is seen as soon as the gun is fired, how much of a time lag is there between when the firing is seen and heard by the timers? If a timer were to start his watch on the sound of the gun, would the athletes’ recorded time be faster (lower) or slower (higher)?

c For investigation: The speed of sound given here is its speed in air at sea level. How does it change as you

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quantity

The most common application of percentages is to find a percentage of a quantity. In many cases this will be a percentage of a money amount.

To find the percentage of an amount, write the percentage as a fraction or a decimal and then multiply by the amount. The simplest way to do this is to divide the percentage by 100, which converts the percentage to a decimal before multiplying.

Try these

1 Calculate each of the following.

a 25% of $32 b 50% of $460 c 75% of $84

d 4% of $1400 e 8% of $520 f 3% of $624

g 33 % of $540 h 66 % of $360 i 12 % of $400

j % of $640 k % of $585 l 6.5% of $734

2 Ricky works as a clerk and receives $315.00 per week. He receives a 4% pay rise. How much extra does Ricky

receive each week?

3 Sally is an accountant with an annual salary of $42 000. She receives a Christmas bonus of 3% of her salary.

Calculate the size of her Christmas bonus.

4 Mr and Mrs Hamilton sell their house for $192 000. They must pay a commission to the real estate agent of

2.5%. Calculate the amount of commission received by the real estate agent.

5 Dagmar has shares in a company to the value of $23 500. The dividend paid to the shareholders is 2.5% of the

value of shares held. Calculate the amount Dagmar receives as her dividend.

6 The value of the Australian economy is $2560 billion. The government announces that the economy will grow

by 2.1% over the next year. Calculate the amount by which the government predicts the economy to grow.

Find 48% of $380.

THINK WRITE

Divide 48 by 100 and then multiply by $380. 48% of $380

= 48 ÷ 100 × $380

= $182.40

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xample

1 3

--- 2

3

--- 1

2 ---1

2

--- 3

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---percentage

When asked to increase by a percentage, add the percentage to 100% before finding the total percentage of the quantity.

If decreasing by a percentage the method is similar, except subtract the percentage from 100%.

Try these

a Increase $380 by 5%.

b Increase $850 by 12%.

c Increase $85 by 65%.

d Increase $750 by 2.5%.

a Decrease $150 by 5%.

b Decrease $386 by 40%.

c Decrease $1250 by 90%.

d Decrease $75 by 1%.

3 A drill set sells at the hardware store for $136.00. The price is then increased by 8%. Calculate the new price

of the drill set.

4 A motel room is advertised at $120 per night. A 12% surcharge is added on weekends. Calculate the cost per

night of the motel on weekends.

5 Kristan receives $430 per week in her part-time job as a journalist. She pays 25% of this in tax. Calculate the

amount of money that Kristan has left in her pay packet after tax has been deducted.

6 The cost of a restaurant bill has been reduced by 7.5% due to poor service. If the original bill was $98.50,

cal-culate the amount to be paid.

A sports store buys softball bats for $35 each and the mark-up on each bat is 40%. Calculate the selling price of the softball bats.

THINK WRITE

Add 40% to 100% to find the percentage that we are finding of $35.

140% of $35

Divide 140 by 100 and then multiply by $35. = 140 ÷ 100 × $35

= $49

Give a written answer. The softball bats sell for $49 each.

1

2

3

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Ratio

15 mL : 750 mL.

500 mL : 20 m2

250 mL : 10 m2_.

If these ratios are all equivalent to 400 : 600, find A car travels 350 km on 24 L of petrol. Simplify this the missing number. rate and calculate how many kilometres the same car

b x: 200 c x: 150

a 400 ÷ 2 = 200, so x= 600 ÷ 2 = 300 b 600 ÷ 3 = 200, so x= 400 ÷ 3 = 133·3 c 600 ÷ 4 = 150, so x= 400 ÷ 4 = 100

= 14·58 km/L.

1

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2

Jan and Dean built a bird-bath which they sold at the markets for $140. They agreed that Jan had spent 4 hours building it and that Dean had worked for 3 hours. If the money is split in the same ratio as the time they spent, how much does each of them receive?

Solution

Jan : Dean 4 : 3

4 7

---3 7

---3

(8)

a 120 :x b x: 240 c x: 36

a 200 :x b x: 2 c x: 150

rate?

(9)

quantity

The most common application of percentages is to find a percentage of a quantity. In many cases this will be a percentage of a money amount.

To find the percentage of an amount, write the percentage as a fraction or a decimal and then multiply by the amount. The simplest way to do this is to divide the percentage by 100, which converts the percentage to a decimal before multiplying.

Try these

a 25% of $32 b 50% of $460 c 75% of $84

d 4% of $1400 e 8% of $520 f 3% of $624

g 33 % of $540 h 66 % of $360 i 12 % of $400

j % of $640 k % of $585 l 6.5% of $734

2 Ricky works as a clerk and receives $315.00 per week. He receives a 4% pay rise. How much extra does Ricky

receive each week?

3 Sally is an accountant with an annual salary of $42 000. She receives a Christmas bonus of 3% of her salary.

Calculate the size of her Christmas bonus.

4 Mr and Mrs Hamilton sell their house for $192 000. They must pay a commission to the real estate agent of

2.5%. Calculate the amount of commission received by the real estate agent.

5 Dagmar has shares in a company to the value of $23 500. The dividend paid to the shareholders is 2.5% of the

value of shares held. Calculate the amount Dagmar receives as her dividend.

6 The value of the Australian economy is $2560 billion. The government announces that the economy will grow

by 2.1% over the next year. Calculate the amount by which the government predicts the economy to grow.

Find 48% of $380.

THINK WRITE

Divide 48 by 100 and then multiply by $380. 48% of $380

= 48 ÷ 100 × $380

= $182.40

WORKED

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xample

1 3

--- 2

3

--- 1

2 ---1

2

--- 3

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---percentage

When asked to increase by a percentage, add the percentage to 100% before finding the total percentage of the quantity.

If decreasing by a percentage the method is similar, except subtract the percentage from 100%.

Try these

a Increase $380 by 5%.

b Increase $850 by 12%.

c Increase $85 by 65%.

d Increase $750 by 2.5%.

a Decrease $150 by 5%.

b Decrease $386 by 40%.

c Decrease $1250 by 90%.

d Decrease $75 by 1%.

3 A drill set sells at the hardware store for $136.00. The price is then increased by 8%. Calculate the new price

of the drill set.

4 A motel room is advertised at $120 per night. A 12% surcharge is added on weekends. Calculate the cost per

night of the motel on weekends.

5 Kristan receives $430 per week in her part-time job as a journalist. She pays 25% of this in tax. Calculate the

amount of money that Kristan has left in her pay packet after tax has been deducted.

6 The cost of a restaurant bill has been reduced by 7.5% due to poor service. If the original bill was $98.50,

cal-culate the amount to be paid.

A sports store buys softball bats for $35 each and the mark-up on each bat is 40%. Calculate the selling price of the softball bats.

THINK WRITE

Add 40% to 100% to find the percentage that we are finding of $35.

140% of $35

Divide 140 by 100 and then multiply by $35. = 140 ÷ 100 × $35

= $49

Give a written answer. The softball bats sell for $49 each.

1

2

3

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Ratio

15 mL : 750 mL.

500 mL : 20 m2

250 mL : 10 m2_.

If these ratios are all equivalent to 400 : 600, find A car travels 350 km on 24 L of petrol. Simplify this the missing number. rate and calculate how many kilometres the same car

b x: 200 c x: 150

a 400 ÷ 2 = 200, so x= 600 ÷ 2 = 300 b 600 ÷ 3 = 200, so x= 400 ÷ 3 = 133·3 c 600 ÷ 4 = 150, so x= 400 ÷ 4 = 100

= 14·58 km/L.

1

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2

Jan and Dean built a bird-bath which they sold at the markets for $140. They agreed that Jan had spent 4 hours building it and that Dean had worked for 3 hours. If the money is split in the same ratio as the time they spent, how much does each of them receive?

Solution

Jan : Dean 4 : 3

4 7

---3 7

---3

(12)

a 120 :x b x: 240 c x: 36

a 200 :x b x: 2 c x: 150

rate?

(13)

Rounding decimal numbers to 2 decimal

places

To round a decimal number correct to 2 decimal places, follow these steps: 1. Consider the digit in the third decimal place (that is, the thousandth’s place).

2. If it is less than 5, simply omit this digit and all digits that follow. (That is, omit all digits beginning from the third decimal place.)

3. If it is 5 or greater than 5, add 1 to the preceding digit (that is, the one in the hundredth’s place) and omit all

digits beginning from the third decimal place.

Note that the sign ≈ is read as ‘is approximately equal to’.

Try these

Round each of the following numbers correct to 2 decimal places.

1 0.322 2 0.257 3 1.723 4 2.555 5 4.308

6 12.195 7 8.4678 8 25.033 78 9 18.333 333 10 0.166 666 6

Round each of the following numbers correct to 2 decimal places.

a 0.239 b 4.5842

THINK WRITE

a The digit in the third decimal place is 9, which is greater than 5. So add 1 to

the preceding digit (that is, to 3) and omit 9.

a 0.239 ≈ 0.24

b The digit in the third decimal place is 4. Since it is less than 5, simply omit all

digits beginning from the third decimal place (that is, omit 4 and 2).

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Multiplying decimal numbers by powers of

10

To multiply a decimal number by powers of 10, move the decimal point to the right one space for each zero in the power of 10. For example, to multiply by 10, move the decimal point one place to the right, while to multiply by 1000 move it three places to the right. Note that if there are not enough digits after the decimal point, we can always add extra zeros.

Try these

Calculate each of the following.

1 2.56 × 10 2 7.6 × 10 3 0.98 × 10 4 3.49 × 100

5 2.6 × 100 6 70.1 × 100 7 0.2 × 100 8 5.321 × 1000

9 10.2 × 1000 10 0.758 × 1000 11 2.5 × 10 000 12 3.576 × 10 000

13 0.003 × 1000 14 0.000 6 × 10 000 15 0.000 08 × 100 000 16 0.04 × 10 000

Calculate each of the following.

a 5.67 × 10 b 0.7 × 100

THINK WRITE

a To multiply a decimal by 10, move the decimal

point one place to the right (as there is one zero in 10).

a 5.67 × 10 = 56.7

b To multiply a decimal by 100, we need to

move the decimal point two places to the right. However, there is only one digit after the decimal (7). So add a zero first (to create two decimal places), then move the decimal point two places to the right. Note that we write the answer as 70, rather than 070.

b 0.7 × 100

= 0.70 × 100

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Multiplying decimal numbers by 100

To multiply a decimal number by 100, move the decimal point two places to the right. Note that if there are not enough digits after the decimal point, we can always add extra zeros.

Try these

Calculate each of the following.

1 0.56 × 100 2 0.76 × 100 3 0.98 × 100 4 0.49 × 100

5 2.6 × 100 6 1.1 × 100 7 0.2 × 100 8 5.321 × 100

9 10.2 × 100 10 0.758 × 100 11 0.5 × 100 12 0.0006 × 100

Calculate each of the following.

a 0.67 × 100 b 0.7 × 100

THINK WRITE

a To multiply a decimal number by 100, move

the decimal point two places to the right. Note that we do not write the zero in front of the number and the decimal point at the end of the number (that is, we write 67, rather than 067.).

a 0.67 × 100 = 67

b We need to move the decimal point two places

to the right; however, there is only one digit after the decimal point (7). So add a zero first (to create two decimal places), then move the decimal point.

b 0.7 × 100

= 0.70 × 100

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Dividing whole numbers and decimals by

100

To divide a whole or a decimal number by 100, move the decimal point two places to the left. Note that although a whole number does not have a decimal point, we can always add it at the end of the number. (For example, 35 and 35. are the same numbers.) Also note that if there are not enough digits to move the decimal point the required number of places, we can always add extra zeros.

Try these

a 28 ÷ 100 b 60 ÷ 100 c 34 ÷ 100 d 2 ÷ 100

e 15 ÷ 100 f 7 ÷ 100 g 560 ÷ 100 h 721 ÷ 100

i 3 ÷ 100 j 75 ÷ 100 k 600 ÷ 100 l 250 ÷ 100

a 9.2 ÷ 100 b 52.3 ÷ 100 c 0.5 ÷ 100 d 8.19 ÷ 100

e 4.9 ÷ 100 f 123.4 ÷ 100 g 0.3 ÷ 100 h 71.1 ÷ 100

i 155.6 ÷ 100 j 4.25 ÷ 100 k 75.3 ÷ 100 l 100.5 ÷ 100

Calculate each of the following.

a 34 ÷ 100 b 350 ÷ 100 c 75.6 ÷ 100 d 4.1 ÷ 100

THINK WRITE

a Put a decimal point at the end of the whole

number.

a 34 ÷ 100 = 34. ÷ 100

To divide by 100, move the decimal point 2 places to the left. Add a zero in front of the decimal point.

34 ÷ 100= 0.34

b Put a decimal point at the end of the whole

number.

b 350 ÷ 100 = 350. ÷ 100

Move the decimal point 2 places to the left. You may wish to rewrite your answer, omitting the zero at the end of the resulting decimal (as 3.50 = 3.5).

350 ÷ 100= 3.50

350 ÷ 100= 3.5

c Move the decimal point 2 places to the left and

add a zero in front of it.

c 75.6 ÷ 100 = 0.756

d Move the decimal point two places to the left.

(Add some extra zeros in front of the number as you go.)

d 4.1 ÷ 100 = 0.041 1

2

1

2

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Finding the size of a sector in degrees, given

its size as a fraction of a circle

There are 360° in a full circle. So to find the size of a sector in degrees, given its size as a fraction of a full circle, multiply the fraction by 360°.

To multiply a fraction by 360°:

1. change 360 into a fraction by writing it over 1 2. simplify as much as possible

3. multiply the numerators together and the denominators together 4. if the answer is an improper fraction, convert it to a mixed number.

Try these

Find the size of each of the following sectors in degrees, given that their size as a fraction of a circle is:

1 2 3 4

5 6 7 8

9 10 11 12

Finding the size of a sector in degrees, if the sector is of a circle.

THINK WRITE

To express a fraction of a circle in degrees, multiply by 360.

× 360

Convert 360 into a fraction by writing it over 1. = ×

Cross-cancel 5 and 360 by dividing each by 5 (that is, 360 ÷ 5 = 72; 5 ÷ 5 = 1).

= ×

Multiply the numerators together and the denominators together.

=

Convert the improper fraction into a mixed number (which in this case is actually a whole number) and include the degree sign.

= 216° 3

5

---1 3₅

---2 3₅--- 360

1

---3 3₁--- 72

1

---4 216---₁

5

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---Trigonometry

review 1

A trigonometric function is a function of any angle (θ) inside a right angled triangle and can be defined as the

ratio of two sides of a right angled triangle.

Sine (abbreviated as sin), cosine (abbreviated as cos) and tangent (abbreviated as tan) are used to define the ratios of the sides.

Try these

1 For each of the following right angled triangles, identify the side labelled x in respect to the given angle θ.

Where appropriate, also identify the side labelled y.

a b c

d e f

g h i

2 For each of the triangles in question 1, identify which trigonometric function (sine, cosine or tangent) could be

used to write an equation to find x. sin θ =

or

cos θ =

or

tan θ =

or opposite hypotenuse ---Hypotenuse Opposite side θ Hypotenuse Opposite side θ adjacent hypotenuse ---Hypotenuse

Adjacent sideθ

Hypotenuse Adjacent side θ opposite adjacent ---Opposite side Adjacent side θ Opposite side Adjacent side θ

θ _{10 cm}

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Trigonometry review 2

The sine, cosine or tangent of an angle can be obtained simply from a scientific or graphics calculator. For example, to calculate sin 30°:

1. with a scientific calculator: Enter 30 then press .

2. with a graphics calculator: Press then enter 30 and press .

Caution: Before entering the angle, check that the calculator is in DEGree mode.

Try these

1 Find the value of the following, giving answers correct to 4 decimal places:

a sin 15° b sin 34° c sin 79° d sin 20°46′ e sin 37°25′ f sin 83°17′

g cos 15° h cos 34° i cos 79° j cos 20°46′ k cos 37°25′ l cos 83°17′

m tan 15° n tan 34° o tan 79° p tan 20°46′ q tan 37°25′ r tan 83°17′

2 Find the value of the following, giving answers correct to 3 decimal places:

a sin 90° b cos 90° c tan 180° d sin 220°50′ e cos 139°25′ f tan 283°30′

g sin 115° h cos 234° i tan 370° j sin 180° k cos 180° l tan 90°

SIN

SIN ENTER

Find: a sin 35∞ (to 4 decimal places) b sin 35∞30¢ (to 4 decimal places).

a sin 35°= 0.5736 b 30′ means 30 minutes or 0.5° so sin 35°30′ is the

same as sin 35.5°. sin 35°30′= 0.5807

1

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Find: a cos 45∞ (to 3 decimal places) b cos 45∞20¢ (to 3 decimal places).

a cos 45°= 0.707 b cos 45°20′= 0.703

(Note: cos 45°20′ is the same as cos 45.333°)

2

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xample

Find: a tan 60∞ (to 2 decimal places) b tan 60∞52¢ (to 2 decimal places).

a tan 60°= 1.73 b tan 60°52′= 1.79

(Note: tan 60°52′ is the same as tan 60.867°)

3

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Try these

3 Find the value of θin each of the following. Where appropriate, give your answer in degrees and minutes.

a sin θ°= 0.5 b sin θ°= 0.3145 c sin θ°= 0.7974 d sin θ°= 0.866

e cos θ°= 0.5 f cos θ°= 0.3145 g cos θ°= 0.7974 h cos θ°= 0.866

i tan θ°= 0.5 j tan θ°= 0.3145 k tan θ°= 0.7974 l tan θ°= 0.866

m tan θ°= 0.95 n tan θ°= 3.145 o tan θ°= 1.79 p tan θ°= 2.567

Find the value of angle θ, if sin θ∞ = 0.55, giving your answer in degrees and minutes.

Solution θ° = sin-1_0.55

θ° = 33.3670° (degrees only)

or 33°22' (degrees and minutes)

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Find the value of angle θ, ifcos θ∞ = 0.55, giving your answer in degrees and minutes.

Solution

θ° = cos−1_0.55 θ° = 56.63299° or 56°38′

5

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xample

Find the value of angle θ, iftan θ∞ = 55, giving your answer in degrees and minutes.

Solution θ° = tan−1_0.55 θ° = 28.8108° or 28°49′

6

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Conversion of units — length

Common units are: millimetre (mm), centimetre (cm), metre (m), kilometre (km)

10 mm = 1 cm

100 cm = 1 m

1000 m = 1 km

÷ 10 ÷ 100 ÷ 1000

mm cm m km

× 10 × 100 × 1000

Remember: Converting to a smaller unit → Multiply

Converting to a larger unit → Divide

Try these

Convert each of the following measurements to the unit shown in the brackets.

1 2 m (cm) 2 53 mm (cm)

3 610 km (m) 4 0.0003 km (cm)

5 5600 mm (m) 6 11.3 cm (mm)

7 12 304 m (km) 8 0.0007 m (mm)

9 6300 mm (m) 10 0.8 km (m)

11 0.011 km (cm) 12 0.0042 km (mm)

13 9000 mm (m) 14 765 m (km)

15 0.000 089 km (mm) 16 14 683 mm (km)

Convert 5 centimetres to millimetres. Convert 80 000 centimetres to kilometres.

Converting to a smaller unit so we need to multiply.

To convert cm to mm, multiply by 10.

5 cm × 10 = 50 mm

Converting to a larger unit so we need to divide. To convert cm to km, divide first by 100 (to convert to m) then by 1000 (to convert to km).

80 000 cm ÷ 100 = 800 m

800 m ÷ 1000 = 0.8 km

1

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Reading scales (How much is each interval

worth?)

When reading scales it is important to remember that the intervals between the adjacent marks are equal. To find the value of each interval, find the value of the section of the scale whose end points are known (that is, the length between the adjacent major marks) and then divide by the number of intervals along this section.

Try these

For each of the following scales, find how much each interval is worth.

1 2

3 4

5 6

7 8

9 10

For each of the following scales, find how much each interval is worth.

a b c

THINK WRITE

a Find the value of the section of the scale

between the major marks by calculating the difference between the end points.

a 30 − 20 = 10

There are 10 intervals between the major marks. So, to find the value of each interval, divide the value of the section of the scale by 10.

10 ÷ 10 = 1

Write the answer in words. Each interval is worth one unit.

b Repeat steps 1–3 as in part a. b 6 − 5 = 1

1 ÷ 10 = 0.1

Each interval is worth 0.1 of a unit.

c Repeat steps 1–3 as in part a. c 200 − 100 = 100

100 ÷ 5 = 20

Each interval is worth 20 units.

20 30 5 6 100 200

1

2

3

WORKED

E

xample

40 50 70 80

11 12 23 24

10 20 300 400

8 9 20 30

(23)

Pythagoras’ theorem

In any right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. This known as Pythagoras’ theorem.

For the triangle shown, Pythagoras’ theorem can be written as c2=a2+b2.

b

c _a

Find the length of x to 1 decimal place. Find the length of y to 1 decimal place.

The hypotenuse is x (which relates to c in the formula) and the other side lengths are 5 and 6 (which relate to a and b in the formula).

c2=a2+b2 x2= 52+ 62

= 25 + 36

= 61

x= (Take the square root of both sides)

= 7.8

The hypotenuse is 15 (which relates to c in the

formula) and the other side lengths are y and 11

(which relate to a and b in the formula).

c2=a2+b2

152=y2+ 112 225 =y2+ 121

y2+ 121 = 225 (Swap sides to bring the y

term to the left side)

y2= 225 − 121 (Subtract 121 from both sides)

= 104

y = (Take the square root of both

sides)

= 10.2 6

5 x

11 15

y

61

104

1

(24)

For each of the following triangles, find the value of the pronumeral correct to 1 decimal place.

1 2

3 4

5 6

7 8

11 x

7

19

y 28

10 13

a

25

14 x

17

3 k

135 73

m

19.8 8.5

x

0.6 0.7

(25)

Using trigonometric ratios

Trigonometric ratios deal with right-angled triangles. The sides of the triangle are named according to their pos-ition with respect to a specific angle. The hypotenuse is the longest side length of the right-angled triangle.

Trigonometric ratios: SOH CAH TOA means

sin θ= , cos θ= and tan θ= Hypotenuse

Opposite side with respect to the angleθ

θ

Adjacent side with respect to the angleθ

opposite

hypotenuse

--- adjacent

hypotenuse

--- opposite

adjacent

---Find the value of x. Express your answer Find the value

correct to 3 decimal places. of x.

Express your answer correct to 2 decimal places.

Given information is:

Angle is 47°, hypotenuse length is 18 cm.

The side marked x is the opposite side.

Use the sine ratio: sin θ =

sin 47°=

= sin 47°

x= 18 × sin 47°

x= 13.164 cm

Angle = 30°, hence the adjacent side = 45 mm.

The side marked x is the hypotenuse.

Use the cosine ratio: cos θ=

cos 30°=

x cos 30°= 45

x=

x = 51.96 mm

45 mm 30° x 18 cm x 47° opposite hypotenuse ---x 18 ---x 18 ---adjacent hypotenuse ---45 x ---45 cos 30°

---1

(26)

Try these

1 For the following right-angled triangles state which trigonometric ratio would be appropriate in finding the side

length marked x.

a b c

d e

2 Find the side length marked x for each of the triangles in question 1. Express your answers correct to 3 decimal

places.

Find the value of the angle θ, to the nearest degree.

Solution

Opposite side is 24 mm and adjacent side is 37 mm. Use tangent ratio:

tan θ =

tan θ = tanθ= 33°

24 mm

37 mm θ

opposite adjacent

---24 37

---3

WORKED

6 cm x

45°

7 cm x

15°

4 cm x

20°

13 cm

x 73°

10 cm

(27)

a b

c d

e

22 cm

15 cm θ

31 mm

39 mm θ

10 cm 12 cm

θ

8.3 cm 5.9 cm

θ

0.52 m 1.34 m

(28)

Identifying sides of a right-angled triangle

with respect to the given angle.

In a right-angled triangle, the longest side (which is opposite the right angle) is called the hypotenuse. The other

two sides are named depending on their position with respect to the given angle. The side opposite to the given angle is called the opposite side and the side next to the given angle is called the adjacent side.

Observe the two triangles below.

In the triangle at left, side a is opposite and side b is adjacent, while in the triangle at right side a is adjacent and

side b is opposite. This is because sides are named according to their position in relation to a given angle and in

the two triangles above the given angle is not the same.

Try these

For each of the following right-angled triangles identify the side labelled x with respect to the given angle θ.

Where necessary, identify also the side labelled y.

1 2 3

4 5 6

7 8 9

Side a

Side b θ

Opposite side

Hypotenuse

Adjacent side

Side a

Side b θ

Adjacent side

Hypotenuse

Opposite side

For each of the following right-angled triangles, identify the side labelled x with respect to the given angle θ. Where necessary, also identify the side labelled y.

a b

a The side labelled x is directly opposite angle θ. So x is the opposite side.

b The side labelled x is next to given angle θ, and the side labelled y is opposite the right angle.

So x is the adjacent side and y is the hypotenuse.

x

θ

12 cm

x y

θ

WORKED

E

xample

x 10 cm θ x 10 cm θ x 10 cm θ

x 10 cm

(29)

Finding trigonometric values and angles

The sine, cosine and tangent of any angle can be obtained from a scientific or graphics calculator by simply pressing the button for the appropriate function.

Note that the calculator must be in the same mode as the given angle. That is, if the angle is in degrees, the

calculator must also be in the DEG (i.e. degree) mode.

If the value of sin or cos or tan of a certain angle is known, the size of that angle can be found by using an appro-priate inverse function of your calculator.

Use a calculator to find the value of each of the following, giving answers correct to 4 decimal places.

a sin 45° b cos 8° c tan 53°

Solution

a Ensure that your calculator is in the degree mode. To find the sine of a given angle, press the following

sequence of buttons on your scientific calculator: (or press instead of ,

if using a graphics calculator).

Record the number shown on the display and round off to 4 decimal places.

sin 45° = 0.707 106 781

sin 45°≈ 0.7071

Note that the sign ≈ is read as ‘approximately equal to’.)

b To find the cosine of the given angle, press the following sequence of buttons on your calculator:

. Copy the number from the display and round off to 4 decimal places.

cos 8° = 0.990 268 068

cos 8°≈ 0.9903

c To find tangent of the given angle, press the following sequence of buttons on your calculator:

. Record the number shown on the display and round off to 4 decimal places.

tan 53° = 1.327 044 822

tan 53°≈ 1.3270

SIN 4 5 = ENTER =

COS 8 =

TAN 5 3 =

1

WORKED

E

xample

For each of the following, find the value of θ, giving answers correct to 1 decimal place. a sin θ° = 0.5505 b cos θ° = 0.5505 c tan θ° = 0.5505

Continued over page Solution

a To find the value of an angle when given the value of its sine, press [SIN–1], enter 0.5505 and press

(or press in place of , if you are using a graphics calculator).

Record the number shown on the display and round it off to 1 decimal place. (Your answer is in degrees, provided that the calculator was in the degree mode prior to your making the calculations.)

sin θ° = 0.5505 sin°θ= sin−1 0.5505 sin° θ= 33.401 321 82 sin° θ≈ 33.4°

2nd

= ENTER =

2

(30)

Try these

1 Find the value of each of the following, giving answers correct to 4 decimal places.

a sin 15° b sin 34° c sin 79° d sin 20° e sin 37° f sin 83°

g cos 15° h cos 34° i cos 79° j cos 20° k cos 37° l cos 83°

mtan 15° n tan 34° o tan 79° p tan 20° q tan 37° r tan 83°

2 For each of the following, find the value of θ, giving answers correct to 1 decimal place.

a sin θ° = 0.5 b sin θ° = 0.3145 c sin θ° = 0.7974 d sin θ° = 0.866

e cos θ° = 0.5 f cos θ° = 0.3145 g cos θ° = 0.7974 h cos θ° = 0.866

i tan θ° = 0.5 j tan θ° = 0.3145 k tan θ° = 0.7974 l tan θ° = 0.866

mtan θ° = 0.95 n tan θ° = 3.145 o tan θ° = 1.79 p tan θ° = 2.567

Solution

b To find the value of an angle when given the value of its cosine, press , enter 0.5505 and

press .

Record the number shown on the display and round it off to 1 decimal place. cos θ° = 0.5505

cos°θ= cos−1 0.5505 cos° θ= 56.598 678 18 cos° θ≈ 56.6°

c To find the value of an angle when given the value of its tangent, press , enter 0.5505 and

press .

Record the number shown on the display and round it off to 1 decimal place. tan θ° = 0.5505

tan°θ= tan−1 0.5505 tan° θ= 28.832 783 64 tan° θ≈ 28.8°

2nd [COS-1]

=

2nd [TAN-1]

(31)

Area and perimeter of composite figures

Composite figures are figures made up of a number of distinct shapes.

Area of a composite figure = sum of the areas of each individual shape

Perimeter of a composite figure = length around the outside edge of the composite figure

Calculate a the area and b the perimeter of Calculate a the area and b the perimeter of the the following shape. following shape. State your answers correct to 2

decimal places.

The composite shape is made up of a rectangle and a triangle.

a Area = area of rectangle + area of triangle Area= length × width + × base × height For the rectangle:

Length = 16 cm, width = 12 cm

For the triangle:

Base = 12 cm (same as width of rectangle),

Height = (24 − 16) = 8 cm So area = 16 × 12 + × 12 × 8 So area= 192 + 48

So area= 240 cm2

b Perimeter = sum of the lengths of 5 sides

Perimeter = 16 + 10 + 10 + 16 + 12 Perimeter = 64 cm

The composite shape is made up of a semicircle and a trapezium.

a Area = area of semicircle + area of trapezium

Area= πr2+ (a+b)h

For the semi circle:

r= × 30 = 15 mm For the trapezium:

a= 30 mm, b= 48 mm and h= 15 mm.

So area = × π × 152+ × (30 + 48) × 15 So area = 353.43 + 585

So area = 938.43 mm2

b Perimeter = (circumference of circle) +

Perimeter = 3 side lengths of trapezium To find missing side length of trapezium:

Let c= hypotenuse of right-angled triangle with

side lengths of 15 and 18

c2= 152+ 182 (Using Pythagoras’ theorem)

= 225 + 324

= 549

c= c= 23.43

So perimeter = × 2πr+ 15 + 48 + 23.43 So perimeter =π× 15 + 86.43

So perimeter = 47.12 + 86.43 So perimeter = 133.55 mm 12 cm 16 cm 24 cm 10 cm 15 m 48 m 30 m 1 2 ---1 2 ---1 2 --- 1 2 ---1 2 ---1 2 --- 1 2 ---1 2 ---549 1 2

---1

(32)

Calculate a the area and b the perimeter of each of the following shapes. State your answers correct to 2 decimal places where appropriate.

1 2

3 4

5 6

10 m

18 m

30 m

13 m

20 cm 10 cm

6 cm

18 cm

49 cm

23 cm

35 cm

12 m 9 m

7 m

3 mm 3 mm

5 mm

15 mm

(33)

Volume

The volume of an object is the amount of space that the object occupies. Some examples of volume units are: mm3, cm3 and m3.

Prisms: V= cross-sectional area × height of prism

Tapered objects: V= × area of base × height of object

Sphere: V= πr3

Try these

Find the volume of each of the following shapes. State your answers correct to 2 decimal places (where appropriate).

1 2 3

4 5 6

1 3 ---4 3

---Find the volume of the following shape. Find the volume of the following shape. Express your answer correct to 2 decimal places.

The shape is a rectangular prism.

V= cross-sectional area × height V= area of rectangle × height V= 12 × 5 × 7

V= 420 cm3

The shape is a cone (tapered object).

V= × area of base × height of object

V= × area of circle × height of object

V= ×πr2×H

V = ×π× 32× 10

V= 94.25 m3

7 cm

5 cm 12 cm

6 m

10 m

1 3 ---1 3 ---1 3 ---1 3

---1

WORKED

E

xample

WORKED

E

xample

2

19 m

8 m 26 m

4 cm 9 cm

8 mm

6 mm

16 mm

2 cm

8 cm

13 cm 9 cm

5 cm

27.2 m

(34)

10 11 12

16 cm 14 cm

17 m 38 m

5.7 cm

12.4 cm

6 mm

15 mm 7 mm

4 mm

42 cm

(35)

Conversion of units — volume

Common units for volume are: cubic millimetre (mm3), cubic centimetre (cm3) and cubic metre (m3).

10 mm = 1 cm → 103 mm3= 13 cm3 so 1000 mm3= 1 cm3 100 cm = 1 m → 1003 cm3= 13 m3 so 1 000 000 cm3= 1 m3

÷ 103 ÷ 1003

mm3 cm3 m3

× 103 ×1003

Remember: converting to a smaller unit → multiply

converting to a larger unit → divide

Try these

Convert each of the following measurements to the unit shown in brackets.

1 268 mm3 (cm3) 2 0.035 m3 (cm3)

3 0.000 610 cm3 (mm3) 4 37.569 cm3 (m3)

5 5.03 cm3 (mm3) 6 68 005 624 cm3 (m3)

7 3.1 m3 (mm3) 8 15 000 000 000 mm3 (m3)

9 86 mm3 (m3) 10 0.0704 m3 (mm3)

Convert 2300 mm3 to cm3. Convert 3.057 m3 to cm3.

Solution Solution

Converting to a larger unit so we need to divide.

To convert mm3 to cm3, divide by 1000.

2300 mm3÷ 1000 = 2.3 cm3

Converting to a smaller unit so we need to multiply.

To convert m3 to cm3, multiply by 1 000 000.

3.057 m3× 1 000 000 = 3 057 000 cm3

1

(36)

Labelling right-angled triangles

In trigonometry, the sides of right-angled triangles are named according to their position relative to a specific

angle (not the right angle). The symbol θ (theta) is one of many Greek letters of the alphabet used to represent this

specific angle.

If the specific angle is θ, then

the side of the triangle opposite θis called the opposite side

the longest side of the triangle, opposite the right angle, is called the hypotenuse

the third side, next to the angle θ, is called the adjacent side

Try these

1 Label the sides of each of the following triangles using the words Opposite, Adjacent, and Hypotenuse.

a b

c d

e f

g h

Hypotenuse

θ

45° x

6 cm

22 cm

15 cm θ

18 cm

x

47°

10 cm 12 cm

θ

73°

x

13 cm 27°

10 cm

x

8.3 cm 5.9 cm

θ

24 mm

(37)

2 For each triangle in question 1, list the two sides which provide information. That is, which two of Opposite, Adjacent and Hypotenuse have either a known value or a pronumeral.

x 7 cm

15°

x

(38)

Using trigonometric ratios

Trigonometric ratios deal with right-angled triangles. The sides of the triangle are named according to their pos-ition with respect to a specific angle. The hypotenuse is the longest side length of the right-angled triangle.

Trigonometric ratios: SOH CAH TOA means

sin θ= , cos θ= and tan θ= Hypotenuse

θ

opposite

hypotenuse

--- adjacent

hypotenuse

--- opposite

adjacent

---Find the value of x. Express your answer Find the value

correct to 3 decimal places. of x.

Express your answer correct to 2 decimal places.

Angle is 47°, hypotenuse length is 18 cm.

The side marked x is the opposite side.

Use the sine ratio: sin θ =

sin 47°=

= sin 47°

x= 18 × sin 47°

x= 13.164 cm

Angle = 30°, hence the adjacent side = 45 mm.

The side marked x is the hypotenuse.

Use the cosine ratio: cos θ=

cos 30°=

x cos 30°= 45

x=

x = 51.96 mm

30° x 45 mm 18 cm x 47° opposite hypotenuse ---x 18 ---x 18 ---adjacent hypotenuse ---45 x ---45 cos 30°

---1

(39)

Try these

1 For the following right-angled triangles state which trigonometric ratio would be appropriate in finding the side

length marked x.

a b c

d e

2 Find the side length marked x for each of the triangles in question 1. Express your answers correct to 3 decimal

places.

Find the value of the angle θ, to the nearest degree.

Solution

Opposite side is 24 mm and adjacent side is 37 mm. Use tangent ratio:

tan θ =

tan θ = tanθ= 33°

24 mm

37 mm θ

opposite adjacent

---24 37

---3

WORKED

45° 6 cm x

15° 7 cm x

20°

4 cm x

73° 13 m

x

27° 10 cm

(40)

c d

e

22 cm

15 cm θ

31 mm

39 mm θ

10 cm 12 cm

θ

8.3 cm 5.9 cm

θ

0.52 m 1.34 m

(41)

Finding the size of a sector in degrees, given

its size as a fraction of a circle

There are 360° in a full circle. So to find the size of a sector in degrees, given its size as a fraction of a full circle, multiply the fraction by 360°.

To multiply a fraction by 360°:

1. change 360 into a fraction by writing it over 1 2. simplify as much as possible

3. multiply the numerators together and the denominators together 4. if the answer is an improper fraction, convert it to a mixed number.

Try these

Find the size of each of the following sectors in degrees, given that their size as a fraction of a circle is:

1 2 3 4

5 6 7 8

9 10 11 12

Finding the size of a sector in degrees, if the sector is of a circle.

THINK WRITE

To express a fraction of a circle in degrees, multiply by 360.

× 360

Convert 360 into a fraction by writing it over 1. = ×

Cross-cancel 5 and 360 by dividing each by 5 (that is, 360 ÷ 5 = 72; 5 ÷ 5 = 1).

= ×

Multiply the numerators together and the denominators together.

=

Convert the improper fraction into a mixed number (which in this case is actually a whole number) and include the degree sign.

= 216° 3

5

---1 3₅

---2 3₅--- 360

1

---3 3₁--- 72

1

---4 216---₁

5

WORKED

E

xample

(42)

---Finding a mean

To find the mean of two or more numbers, add the numbers together and then divide the sum by the number of numbers.

Try these

Find the mean of each of the following sets of numbers.

1 3, 5 2 2, 9 3 2, 4, 8 4 5, 6, 9 5 4, 7, 8, 12

6 7, x 7 x, 19 8 x, 5, 12 9 2, 4, x 10 17, x, 25

Find the mean of each of the following sets of numbers.

a 2, 3, 7 b 5, x

THINK WRITE

a To find the mean of three numbers, add

them together and then divide by 3.

a Average =

Evaluate the numerator of the fraction first and then divide.

Average = Average = 4

b To find the mean of two numbers, add them

together and divide the total by 2. (Since the

value of x is not given, the expression can not

be evaluated and should be left as is.)

b Average =

1 2+3+7

3

---2 12

3

---5+x

(43)

Finding the median

The median of a set of scores is the middle score when the data are arranged in order of size.

The position of the median is found using the formula:

Median position = score

where n is the number of scores.

For example, if there are 13 scores, the median position is the score; that is, the 7th score. If there are

16 scores, the median position is the score; that is, the 8 th score. The 8 th score is halfway between

and is equal to the average of the 8th and 9th scores.

Try these

Find the median of each of the following sets of data.

1 7, 5, 8, 9, 2, 4, 5, 1, 8, 2, 5 2 21, 19, 28, 25, 17, 24, 22, 25, 19

3 32, 45, 58, 21, 57, 84, 26 4 52, 59, 48, 53, 49, 56, 58, 42, 60

5 12, 5, 7, 9, 5, 14, 7, 2, 9, 5, 4, 1 6 27, 40, 33, 37, 46, 32, 19, 21

7 9, 7, 8, 12, 6, 13, 14, 11, 5, 10 8 20, 16, 26, 21, 15, 20

n+1

( )th 2

---13+1

( )th

2

---16+1

( )th

2 --- 1 2 --- 1 2

---Find the median of the following set of data: 8, 3, 7, 4, 9, 1, 5, 8, 6, 13, 12, 2, 6.

Find the median of the following set of data: 28, 33, 27, 29, 29, 29, 32, 28, 34, 31, 32, 33.

There are 13 scores.

∴n= 13

Median score

=

= the 7th score.

Arrange the data in order of size and locate the seventh score.

1, 2, 3, 4, 5, 6, , 7, 8, 8, 9, 12, 13. So, median = 6.

There are 12 scores.

∴n= 12

Median score

=

= the 6 th score.

Rewrite the data in order of size. The median is located halfway between the 6th and 7th scores.

27, 28, 28, 29, 29, , 32, 32, 33, 33, 34.

To find the median, calculate the average of the 6th and 7th scores.

Median =

= = 30

n+1

( )th 2

---13+1

( )th

2

---6

n+1

( )th 2

---12+1

( )th

2 ---1 2 ---29, 31

29+31

2 ---60 2

---1

(44)

Measuring the rise and the run

To measure the rise and the run for a straight line, follow these steps:

1. Select two points on the line. If the line goes through the origin, it is best to select the origin and any other point. If the line cuts both axes, select the x-intercept and the y-intercept.

2. Construct the gradient triangle, so that the two points are the vertices.

3. Measure the horizontal distance (that is, the distance along the horizontal side of a triangle) between the two

points. This distance represents the run. Note that the run is always positive.

4. Measure the vertical distance (that is, the distance along the vertical side of a triangle) between the two points.

This distance represents the rise. Note that if the line slopes upward from left to right, the rise is positive, while if the line slopes downward, the rise is negative.

State the rise and the run for each of the following straight lines.

a b

THINK WRITE/DRAW

a Since the line goes through the origin,

select the origin and some other point. Draw the gradient triangle so that the selected points are the vertices.

a

Measure the distance along the horizontal side of the triangle (that is, how much it is from 0 to 2). Hence state the value of the run.

Run = 2

Measure the distance along the vertical side of the triangle (that is, how much it is from 0 to 3) to find the value of the rise. Since the line slopes upward from left to right, the rise is positive.

Rise = 3 y x 4 3 2 1 –1 –2 –3 –4

–1 1 2 3 4

–2 –3 –4 y x 4 2 1 3 –2 –3 –1 –4 2 3

1 4 5 6

–2 –3 –1 1 y 4 3 2 1 –1 –2 –3 –4

–1 1 2 3 4

–2 –3 –4 2 3 2 3

(45)

Try these

State the rise and the run for each of the following straight lines.

1 2

3 4

5 6

THINK WRITE/DRAW

b Since the line cuts both axes, select the

x- and y-intercepts. Draw the gradient triangle so that the selected points are the vertices.

b

Measure the distance along the horizontal side of the triangle (it is from 0 to 6) and hence state the value of the run.

Run = 6

Measure the distance along the vertical side of the triangle (that is, from 0 to 2) to find the value of the rise. Since the line slopes downward from left to right, the rise is negative.

Rise =−2

1 y x 4 2 –2 6 1 3 –2 –3 –1 –4 2 3

1 4 5 6

–2 –3 –1 2 3 y x 4 2 1 3 –2 –3

–1 1 2 3 4 5 6 –2 –3 –1 y x 4 2 1 3 –2 –3

–1 1 2 3 4 5 6 –2 –3 –1 y x 4 5 2 1 3 –2 –3

–1 1 2 3 4 5 6 –2 –3 –1 y x 4 2 1 3 –2 –3

–1 1 2 3 4 5 6 –2

(46)

9 10 x

4

2 1 3

–2 –3

–1 1 2 3 4 5 6 –2

–3 –1 x

4

2 1 3

–2 –3

–1 1 2 3 4 5 6 –2

–3 –1

y

x 4

2 1 3

–2 –3

–1 1 2 3 –2

–3 –1

y

x 2

1 3

–2

–4 –3 –1 1 2 –2

(47)

Substitution into a rule

To substitute a given value of the pronumeral into an algebraic sentence or rule means to replace the pronumeral with that value.

When all pronumerals have been replaced with numbers, the expression can be evaluated. Order of operations must be observed at all times when evaluating.

Try these

1 Substitute 5 for x in each of the following rules and then find the value of y.

a y=x+ 9 b y=x− 3 c y= 12 +x d y= 25 − x e y= 4x f y= 7x g y= 3x− 4 h y= 2x+ 6 i y= 25 − 3x j y= 11 + 6x 2 Substitute 3 for x in each of the rules in question 1 and hence find the value of y.

Substitute 5 for x in each of the following rules and then find the value of y.

a y=x+ 7 b y= 2x− 3

THINK WRITE

a Replace x with the given value (i.e. 5). The

rest of the expression remains unchanged.

a y= 5 + 7

Add 5 and 7 to find the value of y. y= 12

b Substitute 5 for x, remembering that in

algebra 2x means 2 ×x.

b y= 2 × 5 − 3

To find the value of y, perform multiplication first, followed by subtraction.

y= 10 − 3

y= 7 1

2

1

2

(48)

Average rate of change

If a rate is variable, it is sometimes useful to know the average rate of change over a specific interval from x₁ to x₂.

Average rate of change = = =

Try these

1 a For the function f(t) = 5t− 4t2− 6, find f(5) and f(7).

b For the function h(t) = 8t−t2 − 3, find h(1) and h(4).

c For the function y= 7x− 4x2− 1, find the values of y when x = 2 and x = 4.

2 a Consider the function f(t) = 5t − 4t2− 6 which defines the position of a particle at time t (seconds) from a fixed point. Find the average rate of change (velocity) between t= 5 s to t= 7 s.

b A discus is tossed so that its height, h metres above the ground, is given by the rule h(t) = 8t −t2 − 3, where

t represents time in seconds. Find the average rate of change between t= 1 s to t= 4 s.

c If y= 7x− 4x2− 1, find the average rate of change of y as