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Aussie Maths & Science Teachers: Save your time with SmarterEd

Rates, SM-Bank 017

Constance used 300 grams of sugar to make 12 brownies.

How many grams of sugar will she need to make 18 brownies?  (2 marks)

Show Answers Only

\(450\ \text{grams}\)

Show Worked Solution

\(\text{Method 1}\)

\(\text{Sugar to make 18 brownies}\)

\(=\dfrac{18}{12}\times 300\)

\(=1.5\times 300\)

\(=450\ \text{grams}\)

 
\(\text{Method 2 – Unitary Method}\)

\(300\ \text{g/}12\ \text{cookies}\) \(=\dfrac{300}{12}\ \text{g/cookie}\)
  \(=25\ \text{g/cookie}\)

 

\(\text{Sugar to make 18 brownies}\) \(=25\times 18\)
  \(=450\ \text{grams}\)

Rates, SM-Bank 016

Riley lives 3 km from the park.

He jogs at a constant speed of 10 km per hour.

How many minutes does it take for Riley to get to the park?  (2 marks)

Show Answers Only

\(18\ \text{minutes}\)

Show Worked Solution
\(\text{Time}\) \(=\dfrac{\text{Distance}}{{\text{S}\text{peed}}}\)
  \(=\dfrac{3}{10}\ \text{hr}\)
  \(=\dfrac{3}{10}\times 60\)
  \(=18\ \text{minutes}\)

Rates, SM-Bank 015

Fleur lives 15 kilometres from her work.

On Wednesday, she drove to work and averaged 60 kilometres per hour.

On Thursday, she took the bus which averaged 15 kilometres per hour.

What was the extra time of the bus journey, in minutes, compared to when she drove on Wednesday?  (2 marks)

Show Answers Only

\(45\ \text{minutes}\)

Show Worked Solution
\(\text{Time on Wednesday}\) \(=\dfrac{15}{60}\)
  \(= 0.25\ \text{hour}\)
  \(= 15\ \text{minutes}\)

 

\(\text{Time on Thursday}\) \(=\dfrac{15}{15}\)
  \(= 1\ \text{hour}\)
  \(= 60\ \text{minutes}\)

 

\(\therefore\ \text{The extra time taking the bus}\)

\(=60-15\)

\(=45\ \text{minutes}\)

Rates, SM-Bank 014

A brewery can make 1250 cans of ale and 900 cans of lager per hour.

The brewery runs non-stop and each can weighs 350 grams.

How many kilograms of ale and lager, altogether, does the brewery make in 1 full day?  (2 marks)

Show Answers Only

\(18\ 060\ \text{kilograms per day}\)

Show Worked Solution

\(\text{Total cans made per hour}\)

\(=1250+900\)

\(= 2150\)
 

\(\text{Kilograms made in 1 hour}\)

\(=2150\times 350\)

\(=752\ 500\ \text{grams}\)

\(= 752.5\ \text{kilograms}\)

\(\therefore\ \text{Kilograms made in 1 day}\) \(=752.5\times 24\)
  \(=18\ 060\)

Rates, SM-Bank 013

Johnno was standing 300 metres away from the stage at a rock concert.

If the sound travelled at 330 metres per second from the stage, how many seconds did the sound take to get to Johnno? Give your answer correct to 2 decimal places.  (2 marks)

Show Answers Only

\(0.91\ \text{seconds}\)

Show Worked Solution
\(\text{Time}\) \(=\dfrac{\text{distance}}{\text{speed}}\)
  \(= \dfrac{300}{330}\)
  \(= 0.909090…\approx 0.91\ \text{(2 d.p.)}\)

  
\(\therefore \text{The sound takes approximately}\ 0.91\ \text{seconds to reach Johnno.}\)

Rates, SM-Bank 012

Rhonda rode her hovercraft at a speed of 5 metres per second.

If she rode for 3 minutes, how far did she go?  (2 marks)

Show Answers Only

\(\text{900}\ \text{m}\)

Show Worked Solution
\(3\ \text{minutes}\) \(=60\times 3\ \text{seconds}\)
  \(=180\ \text{seconds}\)

 

\(\text{Distance}\) \(=\text{Speed}\times\text{Time}\)
  \(= 5\times 180\)
  \(= 900\ \text{m}\)

Rates, SM-Bank 011 MC

In a science experiment, Albert needs to add 60 millilitres of acid to every 2 litres of water.

If Albert only has 0.5 litres of water left, how many millilitres of acid should he add?

  1. 0.15 mL
  2. 0.6 mL
  3. 15 mL
  4. 30 mL
Show Answers Only

\(C\)

Show Worked Solution
\(\text{Given}\ \ 60\ \text{mL}\) \(\rightarrow 2\ \text{litres}\)
\(30\ \text{mL}\) \(\rightarrow 1\ \text{litre}\)
\(\therefore 15\ \text{mL}\) \(\rightarrow 0.5\ \text{litres}\) 

 
\(\therefore 15\ \text{millilitres of acid should be added}\)
 
\(\Rightarrow C\)

Rates, SM-Bank 010 MC

A tram at the zoo does a complete 5 kilometre loop in 30 minutes.

If the tram travelled at the same speed, how long did it take to complete 2 kilometres?

  1. 8 minutes
  2. 12 minutes
  3. 18 minutes
  4. 24 minutes
Show Answers Only

\(B\)

Show Worked Solution
\(30\ \text{minutes}/5\ \text{kms}\) \(=\dfrac{30}{5}\ \text{minutes}/\dfrac{5}{5}\ \text{km}\)
  \(=6\ \text{minutes}/\text{km}\)

 

\(\therefore\ \text{Time for 2 kms}\) \(=6\times 2\)
  \(=12\ \text{minutes}\)

 
\(\Rightarrow B\)

 

Rates, SM-Bank 009 MC

Sid is selling bike tyre tubes at a market stall.

He makes $54 from selling 6 bike tyre tubes.

All bike tyre tubes cost the same.

How much will Sid make if he sells 11 bike tyre tubes?

  1. $65
  2. $83
  3. $99
  4. $108
Show Answers Only

\(C\)

Show Worked Solution
\($54/6\ \text{tubes}\) \(=\dfrac{$54}{6}/\dfrac{6}{6}\text{tubes}\)
  \(=$9/\text{tube}\)

 

\(\therefore\ \text{Price of 11 tubes}\) \(=11\times 9\)
  \(=$99\)

 
\(\Rightarrow C\)

Rates, SM-Bank 008

Bryce organises parking for major events in the city.

He has open air parking spaces for 3 cars on every 27 square metres of land.

How many cars could Bryce park on 4680 square metres?  (2 marks)

Show Answers Only

\(520\)

Show Worked Solution

\(\text{3 cars/}\ 27\ \text{m}^2=1\ \text{car/}9\ \text{m}^2\)

\(\therefore\ \text{cars on }4680\ \text{m}^2\) \(=\dfrac{4680}{9}\)
  \(=520\ \text{cars}\)

Rates, SM-Bank 007

A young echidna weighed 1200 grams at the beginning of November just before leaving the burrow. 

At the end of March it weighed 1850 grams.

Calculate the average rate of growth of the echidna for the 5 month period.  (2 marks)

Show Answers Only

\(130\ \text{g}/ \text{month}\)

Show Worked Solution
\(\text{Growth}\) \(=1200-1850\)
  \(=650\ \text{g}\)

 

\(\text{Rate of growth}/\text{month}\) \(=\dfrac{650}{5}\ \text{g}/\text{month}\)
  \(=130\ \text{g}/\text{month}\)

 

Rates, SM-Bank 006

Pete travelled 646 kilometres and used 76 litres of fuel.

  1. Express Pete's average fuel consumption in:
     
    i.  litres per 100 kilometres, giving your answer rounded to 1 decimal place.  (2 marks)

    ii.  kilometres per litre.  (1 mark)

      
  2. If Pete's average fuel consumption remained the same, how many litres of fuel would he use to travel 272 kilometres?  (2 marks)

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a.    i.    \(11.8\ \text{L}/100\ \text{km  (1 d.p.)}\)

ii.   \(8.5\ \text{km}/\text{L}\)

b.    \(32\ \text{litres}\)

Show Worked Solution
a.    i.    \(\text{Average litres/100 km}\) \(=\dfrac{76}{646}\times 100/100\ \text{km}\)
    \(=11.764\dots /100\ \text{km}\)
    \(\approx 11.8\ \text{L}/100\ \text{km  (1 d.p.)}\)

 

ii.   \(\text{Average km/L}\) \(=\dfrac{646}{76}\ \text{km}/\text{L}\)
  \(=8.5\ \text{km}/\text{L}\)

 

b.    \(\text{Litres}\) \(=\dfrac{272}{8.5}\)
    \(=32\)

 
\(\therefore\ \text{Pete would use}\ 32\ \text{litres of fuel to travel}\ 272\ \text{km.}\)

Rates, SM-Bank 005

A tree was 143 cm tall at the beginning of January.  At the end of May it measured 233 cm tall.

  1. Calculate the average rate of growth of the tree for the 5 month period.  (2 marks)

  2. If this rate of growth were to continue, how many more months would it take the tree to reach a height of 611 cm?  (2 marks)

Show Answers Only

a.    \(18\ \text{cm}/ \text{month}\)

b.    \(21\ \text{months}\)

Show Worked Solution
a.    \(\text{Growth}\) \(=233-143\)
    \(=90\ \text{cm}\)

 

\(\text{Rate of growth}/\text{month}\) \(=\dfrac{90}{5}\ \text{cm}/\text{month}\)
  \(=18\ \text{cm}/\text{month}\)

 

b.    \(\text{Growth}\) \(=611-233\)
    \(=378\ \text{cm}\)
     
  \(\text{Months}\) \(=\dfrac{378}{18}\)
    \(=21\)

 
\(\therefore\ \text{It would take}\ 21\ \text{months for the tree to grow to a height of }\ 611\ \text{cm.}\)

Rates, SM-Bank 004

All the water has leaked from a 10 litre bucket in 20 minutes.

  1. Calculate the leaking rate in litres/minute.  (1 mark)

  2. How long would it take to empty a 12 litre bucket at the same rate?  (2 marks)

Show Answers Only

a.    \(0.5\ \text{litres}/ \text{minute}\)

b.    \(24\ \text{minutes}\)

Show Worked Solution
a.    \(10\ \text{litres}/20\ \text{minutes}\) \(=\dfrac{10}{20}\ \text{litres}/\dfrac{20}{20}\ \text{minutes}\)
    \(=0.5\ \text{litres}/\text{minute}\)

 

b.    \(0.5\ \text{litres}/\text{hour}\) \(=1\ \text{litres}/2\ \text{minutes}\)
    \(=12\ \text{litres}/2\times 12\ \text{minutes}\)
    \(=12\ \text{litres}/24\ \text{minutes}\)

 

\(\therefore\ \text{It would take}\ 24\ \text{minutes to empty a}\ 12\ \text{litre bucket.}\)

Rates, SM-Bank 003

Write each of the following as a rate in simplified form.

  1. $136 for every 8 hours worked  (1 mark)

  2. $171 000 to purchase 3 cars  (1 mark)

  3. 112 millimetres of rain in 14 days  (1 mark)

Show Answers Only

a.    \($17/ \text{hour}\)

b.    \($57\ 000/\text{car}\)

c.    \(8\ \text{mm}/\text{day}\)

Show Worked Solution
a.    \($136/8\ \text{hours}\) \(=\dfrac{$136}{8}/\dfrac{8}{8}\ \text{hours}\)
    \(=$17/\text{hour}\)

 

b.    \($171\ 000/3\ \text{cars}\) \(=\dfrac{$171\ 000}{3}/\dfrac{3}{3}\ \text{cars}\)
    \(=$57\ 000/\text{car}\)

 

c.    \(112\ \text{mm}/14\ \text{days}\) \(=\dfrac{112}{14}\ \text{mm}/\dfrac{14}{14}\ \text{days}\)
    \(=8\ \text{mm}/\text{day}\)

Rates, SM-Bank 002

Write each of the following as a rate in simplified form.

  1. $18 for every 5 kilograms  (1 mark)

  2. 30 animals per 6 cages  (1 mark)

  3. 63 rainy days in 7 months  (1 mark)

Show Answers Only

a.    \($3.60/ \text{kilogram}\)

b.    \(5\ \text{animals}/\text{cage}\)

c.    \(9\ \text{rainy days}/\text{month}\)

Show Worked Solution
a.    \($18/5\ \text{kg}\) \(=\dfrac{$18}{5}/\dfrac{5}{5}\ \text{kg}\)
    \(=$3.60/\text{kg}\)

 

b.    \(30\ \text{animals}/6\ \text{cages}\) \(=\dfrac{30}{6}\ \text{animals}/\dfrac{6}{6}\ \text{cages}\)
    \(=5\ \text{animals}/\text{cage}\)

 

c.    \(63\ \text{rainy days}/7\ \text{months}\) \(=\dfrac{63}{7}\ \text{rainy days}/\dfrac{7}{7}\ \text{months}\)
    \(=9\ \text{rainy days}/\text{month}\)

Ratio, SM-Bank 054

James found a fly that was 8 mm long in real life.

Calculate the scale used in this scale drawing of the fly.  (2 marks)

Show Answers Only

\(7\ :\ 1\)

Show Worked Solution
\(\text{Scale}\) \(=\text{Scaled length}\ :\ \text{Actual length}\)
  \(=5.6\ \text{cm}\ :\ 8\ \text{mm}\)
  \(=56\ \text{mm}\ :\ 8\ \text{mm}\)
  \(=7\ :\ 1\)

Ratio, SM-Bank 053

Oliver's bike is 1.68 metres long in real life.

Calculate the scale used in this scale drawing of Oliver's bike.  (2 marks)

Show Answers Only

\(1\ :\ 42\)

Show Worked Solution
\(\text{Scale}\) \(=\text{Scaled length}\ :\ \text{Actual length}\)
  \(=4\ \text{cm}\ :\ 1.68\ \text{m}\)
  \(=4\ \text{cm}\ :\ 168\ \text{cm}\)
  \(=1\ :\ 42\)

Ratio, SM-Bank 052

Macton and Brownville are 15 centimetres apart on a map with a scale of \(1:150\ 000\). How far apart, in kilometres, are the two cities in real life?  (2 marks)

Show Answers Only

\(22.5\ \text{km}\)

Show Worked Solution

\(\text{Scale factor}=150\ 000\)

\(\text{Actual distance}\) \(=\text{Scaled distance}\times \text{scale factor}\)
  \(=15\times 150\ 000\ \text{cm}\)
  \(=2\ 250\ 000\ \text{cm}\)
  \(=22.5\ \text{km}\)

Ratio, SM-Bank 051

Jinderee and Exetroll are 8 centimetres apart on a map with a scale of \(1\ :\ 2000\). How far apart, in kilometres, are the two towns in real life?  (2 marks)

Show Answers Only

\(0.16\ \text{km}\)

Show Worked Solution

\(\text{Scale factor}=2000\)

\(\text{Actual distance}\) \(=\text{Scaled distance}\times \text{scale factor}\)
  \(=8\times 2000\ \text{cm}\)
  \(=16\ 000\ \text{cm}\)
  \(=0.16\ \text{km}\)

Ratio, SM-Bank 050

Determine the scale factor if a scaled length of 1.6 centimetres represents an actual length of 0.48 kilometres.  (2 marks)

Show Answers Only

\(30\ 000\)

Show Worked Solution

\(\text{Scale ratio}=\text{scale length}\ :\ \text{actual length}\)

\(=1.6\ \text{cm}\ :\ 0.48\ \text{km}\)

\(=1.6\ \text{cm}\ :\ 48\ 000\ \text{cm}\)

\(=1\ :30\ 000\)

\(\text{Scale factor}\)

\(=30\ 000\)

Ratio, SM-Bank 049

Determine the scale factor if an actual length of 2 centimetres represents a scaled length of 5 metres.  (2 marks)

Show Answers Only

\(\dfrac{1}{250}\)

Show Worked Solution

\(\text{Scale ratio}=\text{scale length}\ :\ \text{actual length}\)

\(=5\ \text{m}\ :\ 2\ \text{cm}\)

\(=500\ \text{cm}\ :\ 2\ \text{cm}\)

\(=500\ :\ 2\)

\(=1\ :\ \dfrac{1}{250}\)

\(\text{Scale factor}\)

\(=\dfrac{1}{250}\)

Ratio, SM-Bank 048

Determine the scale factor if an actual length of 0.3 millimetres represents a scaled length of 15 centimetres.  (2 marks)

Show Answers Only

\(\dfrac{1}{500}\)

Show Worked Solution

\(\text{Scale ratio}=\text{scale length}\ :\ \text{actual length}\)

\(=15\ \text{cm}\ :\ 0.3\ \text{mm}\)

\(=1500\ \text{mm}\ :\ 3\ \text{mm}\)

\(=500\ :\ 1\)

\(=1\ :\ \dfrac{1}{500}\)

\(\text{Scale factor}\)

\(=\dfrac{1}{500}\)

Ratio, SM-Bank 047

Determine the scale factor if an actual length of 2 millimetres represents a scaled length of 2 metres.  (2 marks)

Show Answers Only

\(\dfrac{1}{1000}\)

Show Worked Solution

\(\text{Scale ratio}=\text{scale length}\ :\ \text{actual length}\)

\(=2\ \text{m}\ :\ 2\ \text{mm}\)

\(=2000\ \text{mm}\ :\ 2\ \text{mm}\)

\(=1000\ :\ 1\)

\(=1\ :\ \dfrac{1}{1000}\)

\(\text{Scale factor}\)

\(=\dfrac{1}{1000}\)

Ratio, SM-Bank 046

Determine the scale factor if a length of 3 centimetres on a map represents an actual length of 2.4 kilometres.  (2 marks)

Show Answers Only

\(80\ 000\)

Show Worked Solution

\(\text{Scale ratio}\)

\(=3\ \text{cm}\ :\ 2.4\ \text{km}\)

\(=3\ \text{cm}\ :\ 240\ 000\ \text{cm}\)

\(=1\ :\ 80\ 000\)

\(\text{Scale factor}\)

\(=80\ 000\)

Ratio, SM-Bank 045

Determine the scale factor if a length of 5 millimetres on a map represents an actual distance of 20 metres.  (2 marks)

Show Answers Only

\(4000\)

Show Worked Solution

\(\text{Scale ratio}\)

\(=5\ \text{mm}\ :\ 20\ \text{m}\)

\(=5\ \text{mm}\ :\ 20\ 000\ \text{mm}\)

\(=1\ :\ 4000\)

\(\text{Scale factor}\)

\(=4000\)

Ratio, SM-Bank 044

A model train is built using a scale of \(4:20\ 000\).

For each of these actual lengths, find the scaled length.

  1. \(40\ \text{m}\)  (1 mark)

  2. \(87\ 450\ \text{cm}\)  (1 mark)

  3. \(2400\ \text{m}\)  (1 mark)

Show Answers Only

a.    \(\text{8 mm or 0.8 cm}\)

b.    \(\text{17.49 cm or 0.1749 m}\)

c.    \(\text{48 cm or 0.48 m}\)

Show Worked Solution

\(\text{Scale}\ \rightarrow\ 4:20\ 000=1:5000\)

a.    \(\text{Scaled length }\) \(\dfrac{40}{5000}\ \text{m}=\dfrac{40\ 000}{5000}\ \text{mm}\)
    \(=\text{8 mm}=\text{0.8 cm}\)

 

b.    \(\text{Scaled length}\) \(=\dfrac{87\ 450}{5000}\ \text{cm} \)
    \(=\text{17.49 cm}=\text{0.1749 m}\)

 

c.    \(\text{Scaled length}\) \(=\dfrac{2400}{5000}\ \text{m}=\dfrac{240\ 000}{5000}\ \text{cm}\)
    \(=\text{48 cm}=\text{0.48 m}\)

Ratio, SM-Bank 043

The scale on a map is \(4:20\ 000\).

For each of these scaled distances, find the actual distances in kilometres.

  1. \(2\ \text{cm}\)  (1 mark)

  2. \(5\ \text{mm}\)  (1 mark)

  3. \(18.4\ \text{cm}\)  (1 mark)

Show Answers Only

a.    \(100\ \text{m or } 0.1\ \text{km}\)

b.    \(40\ \text{m or } 0.04\ \text{km}\)

c.    \(920\ \text{m or } 0.92\ \text{km}\)

Show Worked Solution

\(\text{Scale}=4:20\ 000=1:5000\)

a.    \(\text{Actual Distance}\) \(=2\ \text{cm}\times 5000 \)
    \(= 10\ 000\ \text{cm}\)
    \(=100\ \text{m}=0.1\ \text{km}\)

 

b.    \(\text{Actual Distance}\) \(=8\ \text{mm}\times 5000 \)
    \(=40\ 000\ \text{mm}\)
    \(=40\ \text{m}=0.04\ \text{km}\)

 

c.    \(\text{Actual Distance}\) \(=18.4\ \text{cm}\times 5000 \)
    \(= 92\ 000\ \text{cm}\)
    \(=920\ \text{m}=0.92\ \text{km}\)

Ratio, SM-Bank 042

The scale on a map is \(1:10\ 000\).

For each of these scaled distances, find the actual distances in kilometres.

  1. \(3\ \text{cm}\)  (1 mark)

  2. \(8\ \text{mm}\)  (1 mark)

  3. \(24.3\ \text{cm}\)  (1 mark)

Show Answers Only

a.    \(300\ \text{m or } 0.3\ \text{km}\)

b.    \(80\ \text{m or } 0.08\ \text{km}\)

c.    \(2430\ \text{m or } 2.43\ \text{km}\)

Show Worked Solution
a.    \(\text{Actual Distance}\) \(=3\ \text{cm}\times 10\ 000 \)
    \(= 30\ 000\ \text{cm}\)
    \(=300\ \text{m}=0.3\ \text{km}\)

 

b.    \(\text{Actual Distance}\) \(=8\ \text{mm}\times 10\ 000 \)
    \(= 80\ 000\ \text{mm}\)
    \(=80\ \text{m}=0.08\ \text{km}\)

 

c.    \(\text{Actual Distance}\) \(=24.3\ \text{cm}\times 10\ 000 \)
    \(= 243\ 000\ \text{cm}\)
    \(=2430\ \text{m}\)
    \(=2.43\ \text{km}\)

Ratio, SM-Bank 041 MC

An ancient coin is made up of gold and silver in the ratio 1:3.

The coin weighs 21.65 grams.

How many grams of silver are in the ancient coin?

  1. \(5.4125\)
  2. \(7.217\)
  3. \(14.433\)
  4. \(16.2375\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Total parts}=1+3=4\) 
  

\(\text{Fraction of silver}=\dfrac{3}{4}\)

\(\text{Silver}=\dfrac{3}{4}\times 21.65=16.2375\)
  

\(\Rightarrow D\)

Ratio, SM-Bank 040

Paul stacks milk cartons into supermarket refrigerator shelves.

Each shelf is stacked with 6 full cream milk cartons, 4 lite milk cartons and 2 skim milk cartons.

Every hour Paul stacks 240 milk cartons in total.

How many lite milk cartons does he stack every hour?  (2 marks)

Show Answers Only

\(80\)

Show Worked Solution

\(\text{Total Cartons per shelf}=6+4+2=12\)
  

\(\text{Fraction of lite}=\dfrac{4}{12}=\dfrac{1}{3}\)

\(\text{Total lite per hour}=\dfrac{1}{3}\times 240=80\)

Ratio, SM-Bank 039 MC

A Christmas pudding recipe uses 3 cups of sugar for every 4 cups of sultanas.

Select the correct combination of sugar and sultanas for this recipe.

  1. \(\dfrac{1}{3}\ \text{cup of sugar, }\dfrac{1}{4}\ \text{cup of sultanas}\)
  2. \(1\ \text{cup of sugar, }1.5\ \text{cups of sultanas}\)
  3. \(1.5\ \text{cups of sugar, }2\ \text{cups of sultanas}\)
  4. \(4\ \text{cups of sugar, }5\ \text{cups of sultanas}\)
Show Answers Only

\(C\)

Show Worked Solution
\(\text{Ratio}\) \(= 3:4\)
  \(=1.5:2\)

 

\(\therefore\ 1.5\ \text{cups of sugar, }2\text{ cups of sultanas}\)

\(\text{is in the correct ratio.}\)

 
\(\Rightarrow C\)

Ratio, SM-Bank 038

Damon buys 25 kilograms of salt for his pool for $33.75.

The salt can be purchased in 1 kilogram bags.

How much does 9 kilograms of salt cost?  (2 marks)

Show Answers Only

\($12.15\)

Show Worked Solution

\(\text{C}\text{ost of 1 kilogram}\)

\(=\dfrac{33.75}{25}\)

\(= $1.35\)
 

\(\therefore\ \text{C}\text{ost of 9 kilograms}\)

\(= 9\times 1.35\)

\(= $12.15\)

Ratio, SM-Bank 037 MC

Tony is making a fruit cake.

The recipe says he needs 3 cups of sultanas for every 5 cups of flour.

If 2.5 cups of flour are used, how many cups of sultanas are needed?
 

  1. \(0.5\ \text{cups}\)
  2. \(1.5\ \text{cups}\)
  3. \(2.5\ \text{cups}\)
  4. \(4.5\ \text{cups}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Strategy 1:}\)

\(2.5\ \text{ is half of }\ 5\ \text{cups of flour, therefore half of}\)

\(3\ \text{cups of sultanas is required.}\)

\(\rightarrow\ 1.5\text{ cups of sultanas needed.}\)

 

\(\text{Strategy 2:}\)

\(\text{Let}\ \ n = \text{cups of sultanas needed}\)

\(\dfrac{n}{2.5}\) \(=\dfrac{3}{5}\)
\(\therefore \ n\) \(=\dfrac{3\times 2.5}{5}\)
  \(=1.5\ \text{cups}\)

 
\(\Rightarrow\ B\)

Ratio, SM-Bank 036

Milly is using this cupcake recipe.
 

 How many cups of sugar are needed for 20 cupcakes?  (2 marks)

Show Answers Only

\(3\dfrac{3}{4}\ \text{cups}\)

Show Worked Solution

\(\text{From the recipe:}\)

\(\dfrac{3}{4}\ \text{cups} = 4\ \text{cupcakes}\)

\(\therefore\ \text{Sugar in 20 cupcakes}\) \(=\dfrac{3}{4}\times 5\)
  \(=\dfrac{15}{4}\)
  \(=3\dfrac{3}{4}\ \text{cups}\)

Ratio, SM-Bank 035 MC

The actual body length of a beetle Brad has caught is 24 mm.

A scale drawing of the beetle is shown below.

What scale is used in the drawing?

  1. \(1 \text{cm represents } 5\ \text{mm}\)
  2. \(1 \text{cm represents } 2\ \text{mm}\)
  3. \(2 \text{cm represents } 1\ \text{mm}\)
  4. \(5 \text{cm represents } 1\ \text{mm}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Scale:}\)

\(12\ \text{cm}: 24\ \text{mm}\) \(=120\ \text{mm}: 24\ \text{mm}=5:1\)

  
\(\text{In the options given, the ratio of}\ 5 : 1\ \text{occurs when}\)

\(\text{10 mm represents 2 mm}\ \textbf{OR}\ \text{when 1 cm represents 2 mm}\)
  

\(\Rightarrow B\)

Ratio, SM-Bank 034 MC

This picture shows a stone vase.
 

 
The picture is 2 cm high. The actual vase is 40 cm high.

What scale is used in the picture?

  1. \(\text{2 cm represents 20 cm}\)
  2. \(\text{4 cm represents 40 cm}\)
  3. \(\text{1 cm represents 2 cm}\)
  4. \(\text{1 cm represents 20 cm}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Scale:}\ \ 2\ \text{cm}\ :\ 40\ \text{cm}\)

\(\rightarrow \text{Divide both sides by }2\)

\(\text{Scale:}\ \ 1\ \text{cm}\ :\ 20\ \text{cm}\)

 
\(\Rightarrow D\)

Ratio, SM-Bank 033 MC

Two places are 5.4 cm apart on a map.

On the map 1 cm represents 4 km.

What is the actual distance between the two places?

  1. \(1.08\ \text{km}\)
  2. \(10.8\ \text{km}\)
  3. \(21.6\ \text{km}\)
  4. \(43.4\ \text{km}\)
Show Answers Only

\(C\)

Show Worked Solution
\(\text{Actual distance}\) \(=5.4\times 4\)
  \(=21.6\ \text{km}\)

 
\(\Rightarrow C\)

Ratio, SM-Bank 032 MC

Shelley and Carly collect dolls.

The ratio of the number of dolls Shelley owns compared to Carly is  3 : 2.

Shelley owns 12 dolls.

How many dolls does Carly own?

  1. \(2\)
  2. \(6\)
  3. \(8\)
  4. \(18\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Shelley’s dolls represents 3 parts}\)

\(3\ \text{parts}\) \(=12\)
\(1\ \text{part}\) \(=\dfrac{12}{3}=4\)

    
\(\text{Carly’s dolls represents 2 parts}\)

\(=2\times 4=8\)
 

\(\therefore\ \text{There Carly owns}\ 8\ \text{dolls.}\)
 

\(\Rightarrow C\)

Ratio, SM-Bank 031

Jordan has 4 times as many blue pencils as black pencils.

Altogether he has 90 pencils.

How many blue pencils does he have?  (2 marks)

Show Answers Only

\(72\ \text{blue pencils}\)

Show Worked Solution

\(\text{Method 1}\)

\(\text{Ratio of blue to black}=4:1\)

\(\text{Total parts}=4+1=5\)
  

\(\text{Fraction blue}=\dfrac{4}{5}\)

\(\text{Number of blue pencils}=\dfrac{4}{5}\times 90=72\)
  

\(\text{Method 2 (Advanced)}\)

\(\text{Let}\ \ x = \text{the number of black pencils}\)

\(\text{Then the number of blue pencils}=4x\)

\(\text{Total pencils }= x + 4x\) \(=5x\)
\(\rightarrow\ 5x\) \(= 90\)
\(x\) \(=\dfrac{90}{5}= 18\)

 

\(\therefore \ \text{The number of blue pencils}\)

\(=90-18= 72\)

Ratio, SM-Bank 029 MC

A television broadcasting tower is 800 metres high.

A model of the tower is built with a scale of  1 : 4000.

What is the height of the model?

  1. \(5\ \text{cm}\)
  2. \(20\ \text{cm}\)
  3. \(200\ \text{cm}\)
  4. \(320\ \text{cm}\)
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Model height}\) \(=\dfrac{800}{4000}\)
  \(= 0.2\ \text{m}\)
  \(= 20\ \text{cm}\)

 
\(\Rightarrow B\)

Ratio, SM-Bank 028 MC

A little athletics club is established in a town.

In the first week, 5 girls and 6 boys joined the club.

In the second week, another 4 boys and some more girls joined the club.

The number of girls in the club was now double the number of boys.

How many girls joined the club in the second week?

  1. \(7\)
  2. \(15\)
  3. \(17\)
  4. \(20\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{1st week}\rightarrow 6\ \text{boys.}\)

\(\text{2nd week }\rightarrow 6+4 = 10\ \text{boys.}\)

\(\text{Total girls in 2nd week.}\)

\(= 2\times 10\)

\(= 20\)
 

\(\therefore\ \text{Girls joining in 2nd week.}\)

\(= 20-5\)

\(= 15\)
 

\(\Rightarrow B\)

Ratio, SM-Bank 027 MC

Muriel made a batch of cookies.
 

 
Each cookie had 4 chocolate chips and 3 jelly snakes on it.

Muriel used 39 jelly snakes in the batch of cookies.

How many chocolate chips did she use?

  1. \(13\)
  2. \(26\)
  3. \(52\)
  4. \(78\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Number of cookies made}\)

\(=\dfrac{39}{3}\)

\(=13\)

\(\therefore\ \text{Chocolate chips used}\)

\(=13\times 4\)

\(=52\)

 
\(\Rightarrow C\)

Ratio, SM-Bank 026

It is known that one of the angles in a triangle is \(120^{\circ}\).

Calculate the size of the other 2 angles if they are in the ratio \(2:3\).  (3 marks)

Show Answers Only

\(24^{\circ}\ \text{and}\ 36^{\circ}\)

Show Worked Solution

\(\text{Angle sum of a triangle}=180^{\circ}\)

\(\therefore\ \text{Remaining angles}=180-120=60^{\circ}\)

\(\text{Total parts remaining angles}=2+3=5\)    

\(\text{1st angle}=\dfrac{2}{5}\times 60=24\)

\(\text{2nd angle}=\dfrac{3}{5}\times 60=36\)
    

\(\therefore\ \text{The remaining angles are}\ 24^{\circ}\ \text{and}\ 36^{\circ}.\)

Ratio, SM-Bank 025

The ratio of the angles in a triangle is 1:2:5.

Calculate the sizes of all three angles?  (2 marks)

Show Answers Only

\(22.5^{\circ}, 45^{\circ} \text{ and}\ 112.5^{\circ}\)

Show Worked Solution

\(\text{Total parts}=1+2+5=8\ \text{and the angle sum of a triangle}=180^{\circ}\)

\(\dfrac{1}{8}\times 180\) \(=22.5\)
\(\dfrac{2}{8}\times 180\) \(=45\)
\(\dfrac{5}{8}\times 180\) \(=112.5\)

    
\(\therefore\ \text{The angles in the triangle are }22.5^{\circ}, 45^{\circ} \text{ and}\ 112.5^{\circ}\)

Ratio, SM-Bank 024

Joe and Andrew are cleaning fish. For every 4 fish that Joe cleans, Andrew cleans 7.

If Joe cleaned 36 fish, how many did Andrew clean?  (2 marks)

Show Answers Only

\(63 \text{ fish}\)

Show Worked Solution

\(\text{Joe’s cleaned fish represents 4 parts}\)

\(4\ \text{parts}\) \(=36\)
\(1\ \text{part}\) \(=\dfrac{36}{4}=9\)

    
\(\text{Andrew’s cleaned fish represents 7 parts}\)

\(=9\times 7=63\)
 

\(\therefore\ \text{Andrew cleaned}\ 63\ \text{fish.}\)

Ratio, SM-Bank 023

This year there are 108 students in Year 8 and the ratio of Year 7 students to Year 8 students is 8:9.

Use this information to calculate the number of students in Year 7 this year.  (2 marks)

Show Answers Only

\(96 \text{ students in Year 7}\)

Show Worked Solution

\(\text{Year 8 represents 9 parts}\)

\(9\ \text{parts}\) \(=108\)
\(1\ \text{part}\) \(=\dfrac{108}{9}=12\)

    
\(\text{Year 7 represents 8 parts}\)

\(=12\times 8=96\)
 

\(\therefore\ \text{There are}\ 96\ \text{students in Year 7.}\)

Ratio, SM-Bank 022

Divide $420 in the ratio 1:3:2.  (2 marks)

Show Answers Only

\($70, $210 \text{ and } $140\)

Show Worked Solution

\(\text{Total number of parts}=1+3+2=6\)

\(\dfrac{1}{6}\times 420\) \(=70\)
\(\dfrac{3}{6}\times 420\) \(=210\)
\(\dfrac{2}{6}\times 420\) \(=140\)

  
\(\therefore\ $420\ \text{divided in the ratio }\ 1:3:2\ \text{ is }\ $70, $210\ \text{and } $140.\)

Ratio, SM-Bank 021

Divide 84 days in the ratio 3:4.  (2 marks)

Show Answers Only

\(36\ \text{days and } 48\ \text{days}\)

Show Worked Solution

\(\text{Total number of parts}=3+4=7\)

\(\dfrac{3}{7}\times 84\) \(=36\)
\(\dfrac{4}{7}\times 84\) \(=48\)

  
\(\therefore\ 84\ \text{days divided in the ratio }3:4\ \text{is }36\ \text{days and } 48\ \text{days.}\)

Ratio, SM-Bank 020

Divide 480 tonnes in the ratio 1:5.  (2 marks)

Show Answers Only

\(80\ \text{tonnes and}\ 400\ \text{tonnes}\)

Show Worked Solution

\(\text{Total number of parts}=1+5=6\)

\(\dfrac{1}{6}\times 480\) \(=80\)
\(\dfrac{5}{6}\times 480\) \(=400\)

  
\(\therefore\ 480\ \text{tonnes divided in the ratio }1:5\ \text{is }80\ \text{tonnes and } 400\ \text{tonnes.}\)

Ratio, SM-Bank 019

Divide $130 in the ratio 5:8.  (2 marks)

Show Answers Only

\($50\ \text{and}\ $80\)

Show Worked Solution

\(\text{Total number of parts}=5+8=13\)

\(\dfrac{5}{13}\times 130\) \(=50\)
\(\dfrac{8}{13}\times 130\) \(=80\)

  
\(\therefore\ \$130\ \text{divided in the ratio }5:8\ \text{is }$50\ \text{and}\ $80.\)

Ratio, SM-Bank 018

The ratio of kilograms of bananas to apples at the fruit shop is \(11:5\).

  1. What fraction of the fruit is bananas?  (1 mark)

  2. If there are 128 kilograms of bananas and apples altogether, how many kilograms of bananas are there?  (2 marks)

Show Answers Only

a.    \(\dfrac{11}{16}\)

b.    \(88\ \text{kg of bananas}\)

Show Worked Solution

a.    \(\text{Total fruits}=11+5=16\)

\(\text{Fraction bananas}=\dfrac{11}{16}\)
 

b.   
\(\text{Kilograms of bananas}\) \(=\dfrac{11}{16}\times 128\)
    \(=88\)

 
\(\therefore\ \text{There are }88\ \text{kgs of bananas.}\)

Ratio, SM-Bank 017

The ratio of dogs to cats in an animal rescue shelter is \(5:4\).

  1. What fraction of the animals are dogs?  (1 mark)

  2. What fraction of the animals are cats?  (1 mark)

  3. If there are 36 animals in the shelter, how many are dogs?  (2 marks)

Show Answers Only

a.    \(\dfrac{5}{9}\)

b.    \(\dfrac{4}{9}\)

c.    \(20\ \text{dogs}\)

Show Worked Solution

a.    \(\text{Total animals}=5+4=9\)

\(\text{Fraction dogs}=\dfrac{5}{9}\)
 

b.    \(\text{Fraction cats}=\dfrac{4}{9}\)

 

c.   
\(\text{Number of dogs}\) \(=\dfrac{5}{9}\times 36\)
    \(=20\)

 
\(\therefore\ \text{There are }20\ \text{dogs in the shelter.}\)

Ratio, SM-Bank 016

Maude has 55 cents in coins and Will has $2.70 in coins.

  1. Write Maude's money as a ratio of Will's money in cents.  (1 mark)

  2. Express the ratio in simplest form.  (2 marks)

Show Answers Only

a.    \(55:270\) 

b.    \(11:54\)

Show Worked Solution
a.    \(55\ \text{cents to}\ $2.70\) \(=55:270\)

 

b.    \(55:270\) \(=\dfrac{55}{5}:\dfrac{270}{5}\)
    \(=11:54\)

Ratio, SM-Bank 015

Change these ratios to the same units and then write in simplest form.

  1. \(40\ \text{cm}: 1.2\ \text{m}\)  (2 marks)

  2. \(4\ \text{hours}:20\ \text{minutes}\)  (2 marks)

  3. \(15\ \text{days}: 3\ \text{weeks}\)  (2 marks)

Show Answers Only

a.    \(1:3\) 

b.    \(6:1\)

c.    \(5:7\)

Show Worked Solution
a.    \(40\ \text{cm}:1.2\ \text{m}\) \(=40\ \text{cm}:120\ \text{cm}\)
    \(=40:120\)
    \(=1:3\)

 

b.    \(4\ \text{hours}:20\ \text{minutes}\) \(=240\ \text{minutes}:40\ \text{minutes}\)
    \(=240:40\)
    \(=6:1\)

 

c.    \(15\ \text{days}:3\ \text{weeks}\) \(=15\ \text{days}:21\ \text{days}\)
    \(=15:21\)
    \(=5:7\)

Ratio, SM-Bank 014

Fully simplify the following ratios.

  1. \(\dfrac{5}{12}:\dfrac{11}{8}\)  (2 mark)

  2. \(1\dfrac{1}{4}:3\dfrac{1}{2}\)  (2 marks)

Show Answers Only

a.    \(10:33\) 

b.    \(5:14\)

Show Worked Solution
a.    \(\dfrac{5}{12}:\dfrac{11}{8}\) \(=\bigg(\dfrac{5}{12}\times 24\bigg):\bigg(\dfrac{11}{8}\times 24\bigg)\) \(\ \ \text{Multiply by LCM of 12 and 8.}\)
    \(=10:33\)  

 

b.    \(1\dfrac{1}{4}:3\dfrac{1}{2}\) \(=\dfrac{5}{4}:\dfrac{7}{2} \ \ \ \text{Convert to improper fractions }\)
    \(=\bigg(\dfrac{5}{4}\times 4\bigg):\bigg(\dfrac{7}{2}\times 4\bigg)\ \ \ \ \text{Multiply by LCM of 4 and 2. }\)
    \(=5:14\)

Ratio, SM-Bank 013

Fully simplify the following ratios.

  1. \(\dfrac{1}{2}:3\)  (1 mark)

  2. \(\dfrac{1}{4}:\dfrac{1}{3}\)  (2 marks)

Show Answers Only

a.    \(1:6\) 

b.    \(3:4\)

Show Worked Solution
a.    \(\dfrac{1}{2}:3\) \(=\Bigg(\dfrac{1}{2}\times 2\Bigg):\Bigg(3\times 2\Bigg)\) \(\ \ \text{Multiply by LCM of 2 and 1.}\)
    \(=1:6\)  

 

b.    \(\dfrac{1}{4}:\dfrac{1}{3}\) \(=\Bigg(\dfrac{1}{4}\times 12\Bigg):\Bigg(\dfrac{1}{3}\times 12\Bigg)\) \(\ \ \text{Multiply by LCM of 3 and 4. }\)
    \(=3:4\)  

Ratio, SM-Bank 012

Fully simplify the following ratios.

  1. \(10:32\)  (1 mark)

  2. \(144:60\)  (1 mark)

  3. \(36:18:66\)  (1 mark)

Show Answers Only

a.    \(5:16\) 

b.    \(12:5\)

c.    \(6:3:11\)

Show Worked Solution
a. \(10:32\) \(=\dfrac{10}{2}:\dfrac{32}{2}\) \(\ \ \text{(Divide by HCF of 10 and 32.)}\)
    \(=5:16\)  

 

b. \(144:60\) \(=\dfrac{144}{12}:\dfrac{60}{12}\) \(\ \ \text{(Divide by HCF of 144 and 60.)}\)
    \(=12:5\)  

 

c. \(36:18:66\) \(=\dfrac{36}{6}:\dfrac{18}{6}:\dfrac{66}{6}\) \(\ \ \text{(Divide by HCF of 36, 18 and 66.) }\)
    \(=6:3:11\)