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Algebra, STD2 EQ-Bank 35

A train departs from Town X at 1:00 pm to travel to Town Y. Its average speed for the journey is 80 km/h, and it arrives at 4:00 pm. A second train departs from Town X at 1:20 pm and arrives at Town Y at 3:30 pm.

What is the average speed of the second train? Give your answer to the nearest kilometre per hour.   (2 marks)

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\(\text{111 km/h}\)

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\(\text{Distance between towns X and Y using first train}\)

\(\text{Time taken by first train = 3 hours}\)

\(\text{Distance}=\text{speed}\times\text{time}=80\times 2=240\ \text{km}\)
 

\(\text{Time taken by second train = 2 hours 10 minutes.}\)

\(\text{2 hours 10 minutes = }\dfrac{130}{60}=\dfrac{13}{6}\ \text{hours.}\)

\(\text{Find speed of second train using}\ \ s=\dfrac{d}{t}:\)

\(s=\dfrac{240}{\frac{13}{6}}=240\times \dfrac{6}{13}=110.769…\)

\(\therefore\ \text{The average speed of the second train is 111 km/h (nearest km/h).}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 5, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 30

James lives 20 kilometres from his workplace.

On Monday, he drove to work and averaged 80 kilometres per hour.

On Tuesday, he took the train which averaged 40 kilometres per hour.

What was the extra time of the train journey, in minutes, compared to when he drove on Monday?   (2 marks)

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\(\text{15 minutes}\)

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\(\text{Time driving on Monday:}\)

\(t=\dfrac{\text{distance}}{\text{speed}}=\dfrac{20}{80}=0.25\ \text{hours}=15\ \text{minutes}\)
 

\(\text{Time taken on train on Tuesday:}\)

\(t=\dfrac{20}{40}=0.5\ \text{hours}=30\ \text{minutes}\)

\(\text{Extra time}=30-15=15\ \text{minutes}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 27

A train left Strathfield at 7:18 am and arrived at Central Station at 8:52 am. The distance travelled by the train from Strathfield to Central was 94 km.

What was the average speed of this train, correct to the nearest km/h?   (2 marks)

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\(\text{60 km/h}\)

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\(\text{Journey time}=8:52-7:18=1\ \text{hour}\ 34\ \text{minutes}=94\ \text{minutes}\)

\(\text{Convert to hours:}\)

\(\text{94 minutes} =\dfrac{94}{60}=\dfrac{47}{30}\ \text{hours}\)

\(\text{Using}\ \ d=s \times t\ \ \Rightarrow \ \ s=\dfrac{d}{t}:\)

\(s_{\text{avg}}=\dfrac{94}{\frac{47}{30}}=94\times \dfrac{30}{47}=60\ \text{km/h}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 25

A car takes 5 hours to complete a journey when travelling at 72 km/h.

How long would the same journey take if the car were travelling at 90 km/h?   (2 marks)

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\(\text{4 hours}\)

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\(\text{Distance of the trip:}\)

\(d=s \times t = 72 \times 5= 360\ \text{km}\)

\(\text{Time at new speed:}\)

\(\text{Time at 90 km/h}\ =\dfrac{360}{90}= 4\ \text{hours}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 21

Maria drove 420 km in 6 hours. Her average speed for the first 280 km was 80 km per hour.

How long did she take to travel the last 140 km?   (2 marks)

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\(\text{2.5 hours}\)

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\(\text{First 280 km:}\)

\(t=\dfrac{\text{distance}}{\text{speed}}=\dfrac{280}{80}= 3.5\ \text{hours}\)

\(\text{Total time}=6\ \text{hours}\)

\(\text{Time for first 280 km}=3.5\ \text{hours}\)

\(\text{Time taken for last 140 km}=6-3.5=\text{2.5 hour}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 3, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 22

A car is travelling at 80 km/h.

How far will it travel in 3 hours and 45 minutes?   (1 mark)

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\(\text{300 km}\)

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\(\text{3 hours 45 minutes}=3.75\ \text{hours}\)

\(\text{Using}\ \ d=s \times t:\)

\(d=80 \times 3.75=300\ \text{km}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 EQ-Bank 16

The distance between Newcastle and Sydney is 165 km. A person travels from Newcastle to Sydney at an average speed of 110 km/h.

How long does it take the person to complete the journey in hours and minutes?   (2 marks)

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\(\text{1 hour 30 minutes}\)

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\(\text{Using}\ \ d=s \times t\ \ \Rightarrow\ \ t=\dfrac{d}{s}:\)

\(t=\dfrac{165}{110}=1.5\ \text{hours}\ =\ \text{1 hour 30 minutes}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 3, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\)

Algebra, STD2 A1 2024 HSC 11 MC

A train left Richmond at 6:42 am and arrived at Central Station at 8:04 am. The distance travelled by the train from Richmond to Central was 61 km.

What was the average speed of this train, correct to the nearest km/h?

  1. 38
  2. 45
  3. 50
  4. 74
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\(B\)

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\(\text {Time of travel:  8:04 less 6:42 = 82 minutes.}\)

\(\text{Speed (avg)}\) \(=\dfrac{\text{distance}}{\text{time}}\)
  \(=\dfrac{61}{82}\ \ \text{km/min}\)
  \(=\dfrac{61}{82} \times 60\ \ \text{km/hr}\)
  \(=44.6\ \text{km/hr}\)

 
\(\Rightarrow B\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t Tagged With: Band 4, smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\), smc-791-20-\(d=s\times t\)

Algebra, STD1 A1 2020 HSC 7 MC

The distance between Bricktown and Koala Creek is 75 km. A person travels from Bricktown to Koala Creek at an average speed of 50 km/h.

How long does it take the person to complete the journey?

  1.  40 minutes
  2.  1 hour 25 minutes
  3.  1 hour 30 minutes
  4.  1 hour 50 minutes
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`C`

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Mean mark 52%.
`text(Time)` `= frac(text(Distance))(text(Speed))`
  `= frac(75)(50)`
  `=1.5 \ text(hours)`
  `= 1 \ text(hour) \ 30 \ text(minutes)`

 
`=> \ C`

Filed Under: Applications: BAC, Medicine and d=s x t, Applications: D=SxT and Other (Std 1) Tagged With: Band 4, smc-1117-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\), std2-std1-common

Algebra, STD2 A1 2017 HSC 2 MC

A car is travelling at 95 km/h.

How far will it travel in 2 hours and 30 minutes?

  1. 38 km
  2. 41.3 km
  3. 218.5 km
  4. 237.5 km
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`D`

Show Worked Solution

`text(Distance)= 95 xx 2.5= 237.5\ text(km)`

`=>D`

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t, Applications: D=SxT and Other (Std 1), Other Linear Modelling, Safety: D=ST & BAC Tagged With: Band 3, smc-1117-20-\(d=s\times t\), smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\), smc-791-20-\(d=s\times t\)

Algebra, STD2 A1 2011 HSC 21 MC

A train departs from Town A at 3.00 pm to travel to Town B. Its average speed for the journey is 90 km/h, and it arrives at 5.00 pm. A second train departs from Town A at 3.10 pm and arrives at Town B at 4.30 pm.

What is the average speed of the second train?

  1. 135 km/h
  2. 150 km/h
  3. 216 km/h
  4. 240 km/h
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`A`

Show Worked Solution

 `text(1st train)`

♦ Mean mark 49%

`text(Travels 2hrs at 90km/h)`

`text(Distance)` `=text(Speed)xxtext(Time)`
  `=90xx2=180\ text(km)`

 
`text(2nd train)`

`text(Travels 180 km in 1 hr 20 min)\ (4/3\ text(hrs))`

`text(Speed)` `=text(Distance)/text(Time)`
  `=180-:4/3=180xx3/4`
  `=135\ text(km/h)`

`=> A`

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t, Applications: D=SxT and Other (Std 1), Safety: D=ST & BAC Tagged With: Band 5, smc-1117-20-\(d=s\times t\), smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\), smc-791-20-\(d=s\times t\)

Algebra, STD2 A1 2012 HSC 15 MC

A car takes 6 hours to complete a journey when travelling at 60 km/h.

How long would the same journey take if the car were travelling at 100 km/h?

  1. 36 minutes
  2. 1 hour and 40 minutes
  3. 3 hours and 6 minutes
  4. 3 hours and 36 minutes
Show Answers Only

`D`

Show Worked Solution

`T = D/S`

`text(S)text(ince)\ \ \ T = 6\ \ \ text(when)\ \ \ S = 60`

`6` `= D/60`
`D` `= 360\ text(km)`

 

`text(Find)\ \ T\ \ text (when)\ \ \ S = 100\ \ text(and)\ \ \ D = 360`

`T= 360/100= 3.6\ text(hours)= 3\ text(hrs)\ \ 36\ text(minutes)`

`=>  D`

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t, Applications: D=SxT and Other (Std 1), Safety: D=ST & BAC Tagged With: Band 4, smc-1117-20-\(d=s\times t\), smc-6235-20-\(d=s\times t\), smc-6509-20-\(d=s \times t\), smc-791-20-\(d=s\times t\)

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