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Algebra, STD2 EQ-Bank 29

The formula below is used to estimate the number of hours you must wait before your blood alcohol content (BAC) will return to zero after consuming alcohol.

\(\text{Number of hours}\ =\dfrac{\text{BAC}}{0.015}\)

The spreadsheet below has been created by Ben so his 21st birthday attendees can monitor their alcohol consumption if they intend to drive, given their BAC reading. 

  1. By using appropriate grid references, write down a formula that could appear in cell B5.   (2 marks)

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  2. Ben's friend Ryan's reading reflects that he will have to wait 11 hours for his BAC to return to zero. Using the formula, calculate the value Ryan entered into cell B3.   (2 marks)

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a.    \(=\text{B3}/0.015\)

b.    \(0.165\)

Show Worked Solution

a.    \(=\text{B3}/0.015\)

b.    \(11\) \(=\dfrac{\text{BAC}}{0.015}\)
  \(\text{BAC}\) \(=11\times 0.015=0.165\)

  
\(\text{Ryan entered 0.165 into cell B3.}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 3, Band 5, smc-6235-10-\(BAC\) formula, smc-6235-60-Spreadsheets, smc-6509-10-BAC, smc-6509-60-Spreadsheets, syllabus-2027

Algebra, STD2 EQ-Bank 28

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

\(\text{Dosage}=\dfrac{\text{Age of infant (months)}\  \times \ \text{adult dose}}{150}\)

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.
 

Amber, a 12 month old child, is being discharged from hospital with two medications. Medicine A has an adult dosage of 325 milligrams and she is to take 26 milligrams. She must also take 9.6 milligrams of Medicine B but the equivalent adult dosage has been left off the spreadsheet.

  1. By using appropriate grid references, write down a formula that could appear in cell B10.   (2 marks)

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  2. Calculate the equivalent adult dosage for Medicine B (cell B7) using the information in the spreadsheet.    (2 marks)

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a.    \(=\text{B5}^*\text{B6}/150\)

b.    \(120 \ \text{milligrams}\)

Show Worked Solution

a.     \(=\text{B5}^*\text{B6}/150\)
 

b.    \(\text{Let} \ A= \text{Adult dose}\)

\(\text{Using given formula:}\)

\(9.6\) \(=\dfrac{12 \times A}{150}\)
\(A\) \(=\dfrac{9.6 \times 150}{12}=120 \ \text{milligrams}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, Band 5, smc-6235-40-Medication Formulas, smc-6235-60-Spreadsheets, smc-6509-30-Medication Formulas, smc-6509-60-Spreadsheets, syllabus-2027

Algebra, STD2 EQ-Bank 12 MC

Young's formula for determining the medicine dosage for children aged 1 - 12 years is:

\(\text{Dosage} = \dfrac{\text{Age of child (years)}\ \times\ \text{Adult dose}}{\text{Age of child (years) } +\  12}\)

The spreadsheet below is used as a calculator for determining an child's medicine dosage according to Young's formula.

 

Young's formula for calculating an 8 year old child's dosage has been used in cell B9. Using appropriate cell references, the correct formula to input into cell B9 is:

  1. \(=\text{B5}^*\text{B6}/\text{B5}+12\)
  2. \(=(\text{B5}^*\text{B6})/\text{B5}+12\)
  3. \(=\text{B5}^*\text{B6}/(\text{B6}+12)\)
  4. \(=(\text{B5}^*\text{B6})/(\text{B5}+12)\)
Show Answers Only

\(D\)

Show Worked Solution

\(=(\text{B5}^*\text{B6})/(\text{B5}+12)\)

\(\Rightarrow D\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 5, smc-6235-40-Medication Formulas, smc-6235-60-Spreadsheets, smc-6509-30-Medication Formulas, smc-6509-60-Spreadsheets, syllabus-2027

Algebra, STD2 EQ-Bank 26

Fried's formula for determining the medicine dosage for children aged 1 - 2 years is:

\(\text{Dosage}=\dfrac{\text{Age of infant (months)}\  \times \ \text{adult dose}}{150}\)

The spreadsheet below is used as a calculator for determining an infant's medicine dosage according to Fried's formula.
 

  1. By using appropriate grid references, write down a formula that could appear in cell B9.   (2 marks)

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  2. Another infant requiring the same medicine has been recommended a dosage of 2 millilitres. What is the age of the infant?   (2 marks)

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a.    \(=\text{B5}^*\text{B6}/150\)

b.    \(15 \ \text{months}\)

Show Worked Solution

a.     \(=\text{B5}^*\text{B6}/150\)
 

b.    \(\text{Let} \ n= \text{age of infant}\)

\(\text{Using given formula:}\)

\(2\) \(=\dfrac{n \times 20}{150}\)
\(n\) \(=\dfrac{2 \times 150}{20}=15 \ \text{months}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, Band 5, smc-6235-40-Medication Formulas, smc-6235-60-Spreadsheets, smc-6509-30-Medication Formulas, smc-6509-60-Spreadsheets, syllabus-2027

Algebra, STD2 EQ-Bank 31

Clark's formula for determining the medicine dosage for children is:

\(\text{Dosage}=\dfrac{\text{weight in kilograms}\  \times \ \text{adult dosage}}{70}\)

The spreadsheet below is used as a calculator for determining a child's medicine dosage according to Clark's formula.
 

  1. By using appropriate grid references, write down a formula that could appear in cell E5.   (2 marks)

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  2. Another child requiring the same medicine has been recommended a dosage of 62.5 milligrams. How much does the child weigh?   (2 marks)

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a.    \(=\text{B6}^*\text{B5}/70\)

b.    \(17.5 \ \text{kilograms}\)

Show Worked Solution

a.    \(=\text{B6}^*\text{B5}/70\)
 

b.    \(\text{Let} \ w= \text{weight of the child.}\)

\(\text{Using given formula:}\)

\(62.5\) \(=\dfrac{w \times 250}{70}\)
\(w\) \(=\dfrac{62.5 \times 70}{250}=17.5 \ \text{kilograms}\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 4, Band 5, smc-6235-40-Medication Formulas, smc-6235-60-Spreadsheets, smc-6509-30-Medication Formulas, smc-6509-60-Spreadsheets, syllabus-2027

Algebra, STD2 EQ-Bank 19

Sharon drinks three glasses of chardonnay over a 180-minute period, each glass containing 1.6 standard drinks.

Sharon weighs 78 kilograms, and her blood alcohol content (BAC) at the end of this period can be calculated using the following formula:

\(\text{BAC}_{\text {female }}=\dfrac{10 N-7.5 H }{5.5 M}\)

where \(N\) = number of standard drinks consumed
\(H\)  = the number of hours drinking
\(M\)  = the person's mass in kilograms

 
The spreadsheet below can be used to calculate Sharon's \(\text{BAC}\).
 

  1. What value should be input into cell B5.   (1 mark)

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  2. Write down the formula that has been used in cell E4, using appropriate grid references.   (2 marks)

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a.    \(\text{Cell B5 value}=3\)

b.  \(=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})\)

Show Worked Solution

a.    \(\text{180 minutes}\ =\ \text{3 hours}\)

\(\therefore \ \text{Cell B5 value}=3\)
 

b.  \(=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})\)

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t Tagged With: Band 3, Band 5, smc-6235-10-\(BAC\) formula, smc-6235-60-Spreadsheets, smc-6509-10-BAC, smc-6509-60-Spreadsheets, syllabus-2027

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

Show Worked Solution

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

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

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

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

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

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

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

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

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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\(\text{15 minutes}\)

Show Worked Solution

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

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

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

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

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

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

Show Worked Solution

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

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

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

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

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

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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\(\text{4 hours}\)

Show Worked Solution

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

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

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

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

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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\(\text{2.5 hours}\)

Show Worked Solution

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

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

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

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

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

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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\(\text{300 km}\)

Show Worked Solution

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

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

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

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

Show Worked Solution

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

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

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 24

Sarah, a 60 kg female, consumes 3 glasses of wine at a family dinner over 2.5 hours.

Note: there are 1.2 standard drinks in one glass of wine.

The blood alcohol content \((BAC)\) for females can be estimated by

\(B A C_{\text {female}}=\dfrac{10 N-7.5 H}{5.5 M},\)

  where \(N\)  = number of standard drinks
  \(H\)  = number of hours drinking
  \(M\)  = mass in kilograms
 
  1. Calculate Sarah's \(BAC\) at the end of the dinner, correct to 3 decimal places.   (2 marks)

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  2. The time it takes a person's BAC to reach zero is given by   

                  \(\text {Time}=\dfrac{B A C}{0.015}.\)

    Calculate the time it takes for Sarah's BAC to return to zero, assuming she stopped drinking after 2.5 hours. Give your answer to the nearest minute.  (2 marks)

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a.   \(BAC=0.052\)

b.   \(\text{3 h 28 m}\)

 

Show Worked Solution

a.     \(N=3 \times 1.2=3.6, H=2.5, M=60\)

  \(\therefore B A C\) \(=\dfrac{10 \times 3.6-7.5 \times 2.5}{5.5 \times 60}\)
    \(=0.052 \ \text{(3 d.p.)}\)

 

b.     \(T\) \(=\dfrac{0.052}{0.015}=3.466\)
    \(=3 \text{ h 28 m}\)
COMMENT: Use calculator degrees/minutes function for part (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 3, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

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
Show Answers Only

\(B\)

Show Worked Solution

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

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

 
\(\Rightarrow B\)

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, STD2 A1 2023 HSC 36

The following formula can be used to calculate an estimate for blood alcohol content (`BAC`) for males.
 

`BAC_text{male}=(10N-7.5H)/(6.8M)`

`N` is the number of standard drinks consumed

`M` is the person's weight in kilograms

`H` is the number of hours of drinking
 

Cameron weighs 75 kg. His `BAC` was zero when he began drinking alcohol. At 9:00 pm, after consuming 3 standard drinks, his `BAC` was 0.02.

Using the formula, estimate at what time Cameron began drinking alcohol, to the nearest minute.   (4 marks)

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`text{6:22 pm}`

Show Worked Solution
`BAC` `=(10N-7.5H)/(6.8M)`  
`0.02` `=(10xx3-7.5xxH)/(6.8 xx 75)`  
`0.02xx510` `=30-7.5H`  
`7.5H` `=30-10.2`  
`H` `=19.8/7.5`  
  `=2.64\ text{hours}`  
  `=2\ text{hours}\ 38\ text{minutes (nearest minute)}`  

 
`text{Time Cameron began drinking}`

`=\ text{9 pm less 2 h 38 m}=\ text{6:22 pm}`

Mean mark 56%.

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-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD1 A1 2022 HSC 19

Fried's formula is used to calculate the dosage of medication for children aged 1-2 years based on the adult dosage. The formula is

`text{Dosage}=(text{age (in months)} xx\ text{adult dosage})/(150)`.

The adult dosage of a particular medication is 200 mg.

Betty's dosage is calculated to be 24 mg.

How old is Betty in months?   (2 marks)

Show Answers Only

`18\ text(months)`

Show Worked Solution

`text{Substituting into the formula:}`

`24` `=( \ text{age} xx 200)/150`  
`text{age}` `=(24 xx 150)/200=360/20`  
  `= 18\ text(months)`  

♦ Mean mark 52%.

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medicine and d=s x t, Applications: D=SxT and Other (Std 1) Tagged With: Band 5, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas

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
Show Answers Only

`C`

Show Worked Solution

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 2020 HSC 13 MC

When Jake stops drinking alcohol at 10:30 pm, he has a blood alcohol content (BAC) of 0.08375.

The number of hours required for a person to reach zero BAC after they stop consuming alcohol is given by the formula

`text(Time) = frac{ text(BAC)}{0.015}`.

At what time on the next day should Jake expect his BAC to be 0.05?

  1. 12:45 am
  2. 1:50 am
  3. 2:15 am
  4. 4:05 am
Show Answers Only

`A`

Show Worked Solution

♦♦♦ Mean mark 17%.
COMMENT: The rates aspect of this question proved extremely challenging.

`text(Time from  0.08375 → 0  BAC)`

`= frac(0.08375)(0.015)`

`approx 5.58 \ text(hours)`
 

`text(Time from  0.08375 → 0.05  BAC)`

`approx frac{(0.08375-0.05)}{0.08375} xx 5.58` 

`approx 0.4 xx 5.58`

`approx 2.25\ text(hours)`
 

`therefore \ text(Time)` `approx \ 10:30 \ text(pm) \ + 2 \ text(hr) \ 15 \ text(min)`
  `approx \ 12:45 \ text(am)`

 
`=> \ 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 Tagged With: Band 6, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2019 HSC 28

The formula below is used to calculate an estimate for blood alcohol content `(BAC)` for females.

`BAC_text(female) = (10N-7.5H)/(5.5M)`

The number of hours required for a person to reach zero `BAC` after they stop consuming alcohol is given by the following formula.

`text(Time) = (BAC)/0.015`

The number of standard drinks in a glass of wine and a glass of spirits is shown.
 

Hannah weighs 60 kg. She consumed 3 glasses of wine and 4 glasses of spirits between 6:15 pm and 12:30 am the following day. She then stopped drinking alcohol.

Using the given formulae, calculate the time in the morning when Hannah's `BAC` should reach zero.   (4 marks)

Show Answers Only

`text(6:23 am)`

Show Worked Solution

`text(Standard drinks consumed)\ (N) = 3 xx 1.2 + 4 = 7.6`

`text(Hours drinking)\ (H) = text(6 h 15 min = 6.25 hours)`

`BAC(text(Hannah))` `= (10 xx 7.6-7.5 xx 6.25)/(5.5 xx 60)`
  `= 0.08825…`

COMMENT: Convert a decimal answer into hours and minutes using the calculator degree/minute function.

`text(Time(to zero))` `= (0.08825…)/0.015`
  `= 5.883…\ text(hours)`
  `= 5\ text(hours 53 minutes)`

 
`:.\ text(Hannah should reach zero)\ BAC\ text(at 6:23 am)`

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-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2018 HSC 28e

Sophie is driving at 70 km/h. She notices a branch on the road ahead and decides to apply the brakes. Her reaction time is 1.5 seconds. Her braking distance (`D` metres) is given by  `D = 0.01v^2`, where  `v ` is speed in km/h.

Stopping distance can be calculated using the following formula
 

`text(stopping distance = {reaction time distance} + {braking distance})`
 

What is Sophie’s stopping distance, to the nearest metre?   (3 marks)

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`78\ text{m  (nearest m)}`

Show Worked Solution
`text(70 km/hr)` `= 70\ 000\ text(m/hr)`
  `= (70\ 000)/(60 xx 60)\ text(m/sec)`
  `= 19.44…\ text(m/sec)`

 

`:.\ text(Total stopping distance)`

♦ Mean mark 46%.

`=\ text(reaction time distance + braking distance)`

`= 1.5 xx 19.44… + 0.01 xx 70^2`

`= 78.166…=78\ text{m  (nearest m)}`

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) Tagged With: Band 5, smc-1117-40-Stopping Distance, smc-6235-30-Stopping Distance, smc-6509-40-Stopping Distance, smc-791-40-Stopping Distance

Algebra, STD2 A1 2018 HSC 26b

Clark’s formula, given below, is used to determine the dosage of medicine for children.
 

`text(Dosage) = (text(weight in kg × adult dosage))/70`

 
For a particular medicine, the adult dosage is 325 mg and the correct dosage for a specific child is 90 mg.

How much does the child weigh, to the nearest kg?   (2 marks)

Show Answers Only

`19\ text(kg)`

Show Worked Solution

`90 = (text(weight) xx 325)/70`

`:.\ text(weight)` `= (70 xx 90)/325=19.38…`
  `=19\ text(kg  (nearest kg))`
   

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) Tagged With: Band 3, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

Algebra, STD2 A1 EQ-Bank 17

Fried's formula is used to calculate the medicine dosages for children aged 1-2 years.

`text(Child dosage) = {text{Age(in months)}\ xx\ text(adult dosage)}/150`

Ben is 1.5 years old and receives a daily dosage of 450 mg of a medicine.

According to Fried's formula, what would the appropriate adult daily dosage of the medicine be?   (2 marks)

Show Answers Only

`3750\ text(mg)`

Show Worked Solution

`text(Substituting into the formula:)`

`450= (18 xx text{adult dosage})/150`

`:.\ text(Adult dosage)= (450 xx 150)/18= 3750\ text(mg)`

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) Tagged With: Band 3, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

Algebra, STD2 A1 EQ-Bank 32

Doris is driving in a school zone at a speed of 35 kilometres per hour and needs to stop immediately to avoid an accident.

It takes her 1.25 seconds to react and her breaking distance is 5.3 metres.

Stopping distance can be calculated using the following formula

`text(stopping distance = {reaction time distance} + {braking distance})`

What is Doris' total stopping distance? Give your answer to 1 decimal place.   (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`17.5\ text{metres  (to 1 d.p.)}`

Show Worked Solution
`35\ text(km/hr)` `= 35\ 000\ text(m/hr)`
  `= (35\ 000)/(60 xx 60)\ text(m/sec) =9.722…\ text(m/sec)`

 

`:.\ text(Total stopping distance)`

`=\ text(Reaction time distance) + text(braking distance)`

`= 1.25 xx 9.722… +5.3= 17.452…`

`= 17.5\ text{metres  (to 1 d.p.)}`

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-1117-40-Stopping Distance, smc-6235-30-Stopping Distance, smc-6509-40-Stopping Distance, smc-791-40-Stopping Distance

Algebra, STD2 A1 2017 HSC 27e

Rhys is drinking low alcohol beer at a party over a five-hour period. He reads on the label of the low alcohol beer bottle that it is equivalent to 0.8 of a standard drink.

Rhys weighs 90 kg.

The formula below  can be used to calculate a Rhys's blood alcohol content.
 

`BAC_text(Male) = (10N-7.5H)/(6.8M)`

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

What is the maximum number of complete bottles of the low alcohol beer he can drink to remain under a Blood Alcohol Content (BAC) of 0.05?   (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`8`

Show Worked Solution
`text(BAC)_text(male)` `= (10N-7.5H)/(6.8M)`
`0.05` `= (10N-7.5 xx 5)/(6.8 xx 90)`
`10N` `= (0.05 xx 6.8 xx 90) + 7.5 xx 5= 68.1`
`N` `= 6.81\ \ text(standard drinks)`

 

`:.\ text(Number of low alcohol bottles)`

`= 6.81/0.8=8.51` 

`:.\ text(Complete bottles for Max to stay under 0.05)= 8`

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t, Safety: D=ST & BAC Tagged With: Band 4, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2017 HSC 19 MC

Young’s formula, shown below, is used to calculate the dosage of medication for children aged 1−12 years based on the adult dosage.

`D = (yA)/(y + 12)`

where     `D`   = dosage for children aged 1−12 years
`y`   = age of child (in years)
`A`   = Adult dosage

 
A child’s dosage is calculated to be 20 mg, based on an adult dosage of 40 mg.

How old is the child in years?

  1. 6
  2. 8
  3. 10
  4. 12
Show Answers Only

`D`

Show Worked Solution

`D= (yA)/(y + 12)`

`20= (40y)/(y + 12)`

`20(y + 12)` `= 40y`
`20y + 240` `= 40y`
`20y` `= 240`
`y` `= 12`

`=> 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), Linear and Other Equations, Medication Tagged With: Band 4, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

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
Show Answers Only

`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 2016 HSC 10 MC

Caroline drinks two small bottles of wine over a three-hour period. Each of these bottles contains 2.3 standard drinks. Caroline weighs 53 kg.

Using the formula below, what is Caroline's approximate blood alcohol content (BAC) at the end of this period?
 

`BAC_text(Female) = (10N-7.5H)/(5.5M)`
 

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

  1. `0.081`
  2. `0.065`
  3. `0.0017`
  4. `0.0014` 
Show Answers Only

`A`

Show Worked Solution
`text(BAC)_f` `= (10N-7.5H)/(5.5M)`
  `= (10(2 xx 2.3)-7.5(3))/(5.5 xx 53)= 0.0806…`

 
`=> 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, Safety: D=ST & BAC Tagged With: Band 4, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2015 HSC 30d

Claire is driving on a motorway at a speed of 110 kilometres per hour and has to brake suddenly. She has a reaction time of 2 seconds and a braking distance of 59.2 metres.

Stopping distance can be calculated using the following formula
 

`text(stopping distance = {reaction time distance} + {braking distance})`

 
What is Claire's stopping distance?   (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`120.3\ text{metres  (to 1 d.p.)}`

Show Worked Solution
♦ Mean mark 34%.
`110\ text(km/hr)` `= 110\ 000\ text(m/hr)`
  `= (110\ 000)/(60 xx 60)\ text(m/sec)`
  `= 30.555…\ text(m/sec)`

 

`text(Reaction time distance)` `=2 xx 30.555…`
  `= 61.11…\ text(metres)`

 

`:.\ text(Stopping distance)`

`=\ text(Reaction time distance + braking distance)`

`= 61.11… + 59.2= 120.311…`

`= 120.3\ text{metres  (to 1 d.p.)}`

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-1117-40-Stopping Distance, smc-6235-30-Stopping Distance, smc-6509-40-Stopping Distance, smc-791-40-Stopping Distance

Algebra, STD2 A1 2015 HSC 26b

Clark’s formula is used to determine the dosage of medicine for children.
 

`text(Dosage) = text(weight in kg × adult dosage)/70`
 

The adult daily dosage of a medicine contains 3150 mg of a particular drug.

A child who weighs 35 kg is to be given tablets each containing 525 mg of this drug.

How many tablets should this child be given daily?   (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`3`

Show Worked Solution

`text(Dosage)= (35 xx 3150)/70= 1575\ text(mg)`

`text(# Tablets per day)= text(Dosage)/text(mg per tablet)= 1575/525= 3`

`:.\ text(The child should be given 3 tablets per day.)`

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), Medication Tagged With: Band 3, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

Algebra, STD2 A1 2015 HSC 23 MC

The number of ‘standard drinks’ in various glasses of wine is shown.
 

A woman weighing 62 kg drinks three small glasses of white wine and two large glasses of red wine between 8 pm and 1 am.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content (BAC) estimate at 1 am, correct to three decimal places?
 

`BAC_text(Female) = (10N-7.5H)/(5.5M)`
 

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

  1. `0.030`
  2. `0.037`
  3. `0.046`
  4. `0.057`
Show Answers Only

`D`

Show Worked Solution
`N` `= 3 xx 0.9 + 2 xx 1.5`
  `= 5.7\ text(standard drinks)`
`H` `=\ text(5 hours)`
`M` `=\ text(62 kg)`
   
`:.BAC_f` `= (10 xx 5.7-7.5 xx 5)/ (5.5 xx 62)`
  `= 0.05718…`

`⇒ 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, Safety: D=ST & BAC Tagged With: Band 4, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2005 HSC 24b

The formula  `D = (2A)/15`  is used to calculate the dosage of Hackalot cough medicine to be given to a child.

    • D is the dosage of Hackalot cough medicine in millilitres (mL).
    • A is the age of the child in months.
  1. If George is nine months old, what dosage of Hackalot cough medicine should he be given?   (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

The correct dosage of Hackalot cough medicine for Sam is 4 mL.

  1. What is the difference in the ages of Sam and George, in months?   (3 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    `text(1.2 mL)`

b.    `30`

Show Worked Solution

a.    `D= (2A)/15= (2 × 9)/15=1.2\ text(mL)`

:.\ text(George should be given a dosage of 1.2 mL)`

 

b.   `text(Find)\ A\ text(when)\ D = text(4 mL)`

`4` `= (2 × A)/15`
 `2A` `= 60`
`A`  `= 30` 

 

`:.\ text(Sam is 30 months old and is 21 months)`

`text(older than George.)`

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), Formula Rearrange, Medication Tagged With: Band 4, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

Algebra, STD2 A1 2014 HSC 29b

Blood alcohol content of males can be calculated using the following formula

`BAC_text(Male) = (10N-7.5H)/(6.8M)`

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms 

What is the maximum number of standard drinks that a male weighing 84 kg can consume over 4 hours in order to maintain a blood alcohol content (BAC) of less than 0.05?   (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`5`

Show Worked Solution

`text(BAC)_text(male) = (10N\-7.5H)/(6.8M)`

`text(Find)\ \ N\ \ text(for BAC)<0.05,\ \ text(given)\ \ H = 4\ \ text(and)\ \ M = 84`
 

` (10N-7.5xx 4)/(6.8xx 84)` `< 0.05`
`10N-30` `< 0.05xx 571.2`
`10N` `< 28.56 + 30< 58.56`
`N` `< 5.856`

 

`:.\ text(Max number of drinks is 5.)`

Filed Under: Applications: BAC, D=SxT and Medication, Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and d=s x t, Safety: D=ST & BAC Tagged With: Band 4, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

Algebra, STD2 A1 2014 HSC 4 MC

Young’s formula below is used to calculate the required dosages of medicine for children aged 1–12 years.

 `text(Dosage) = (text(age of child)\ text{(in years)} xx text(adult dosage))/(text(age of child)\ text{(in years)} + 12)`

How much of the medicine should be given to an 18-month-old child in a 24-hour period if each adult dosage is 45 mL? The medicine is to be taken every 6 hours by both adults and children.

  1.       5 mL
  2.    20 mL
  3.    27 mL
  4.    30 mL
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 42%
`text(Dosage)` `= (1.5 xx 45)/(1.5 + 12)`
  `= 5\ text(mL)`

 
`text(S)text(ince 1 dosage every 6 hrs)`

`text(In 24 hours,)`

`text(Medicine given) = 4 xx 5 = 20\ text(mL)`

`=>  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, Applications: D=SxT and Other (Std 1), Medication Tagged With: Band 5, smc-1117-30-Medication Formulae, smc-6235-40-Medication Formulas, smc-6509-30-Medication Formulas, smc-791-30-Medication Formulae

Algebra, STD2 A1 EQ-Bank 4 MC

The blood alcohol content (`BAC`) of a male's blood is given by the formula;

`BAC_text(male) = (10N-7.5H)/(6.8M)`, where

`N` is the number of standard drinks consumed,

`H` is the number of hours drinking and 

`M` is the person's mass in kgs. 

Calculate the `BAC` of a male who consumed 4 standard drinks in 3.5 hours and weighs 68 kgs, correct to 2 decimal places. 

  1. 1.03
  2. 0.03
  3. 0.04
  4. 0.01
Show Answers Only

`B`

Show Worked Solution
`BAC_text(male)` `= ( (10 xx 4)\-(7.5 xx 3.5) )/( (6.8 xx 68) )`
  `= 13.75/462.4=0.0297…`

`=>  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, Safety: D=ST & BAC Tagged With: Band 4, smc-6235-10-\(BAC\) formula, smc-6509-10-BAC, smc-791-10-BAC

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
Show Answers Only

`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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