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v1 Algebra, STD2 A1 SM-Bank 1 MC

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

\(BAC_{\text{male}}=\dfrac{10N - 7.5H}{6.8M}\)  , 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 5 standard drinks in 2.5 hours and weighs 72 kgs, correct to 2 decimal places. 

  1.    1.06
  2.    0.06
  3.    0.04
  4.    0.01
Show Answers Only

\(B\)

Show Worked Solution
\(BAC_{\text{male}}\) \(=\dfrac{10\times 5-7.5\times 2.5}{6.8\times 72}\)
  \(=\dfrac{31.25}{489.6}\)
  \(=0.0638\dots\)

\(\Rightarrow B\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2014 HSC 29b

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

\(BAC_{\text{Male}} = \dfrac{10N-7.5H}{6.8M}\)

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 Jacko, who has a mass of 75 kg, can consume over 5 hours in order to maintain a blood alcohol content (\(BAC\)) of less than 0.05?   (3 marks)

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

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\(6\)

Show Worked Solution

\(BAC_\text{male}=\dfrac{10N-7.5H}{6.8M}\)

\(\text{Find}\ \ N\ \text{for }BAC<0.05,\ \text{given}\ \ H=5\ \text{and}\ \ M = 75\)
 

\(\dfrac{10N-7.5\times 5}{6.8\times 75}\) \(< 0.05\)
\(10N-37.5\) \(< 0.05\times 6.8\times 75\)
\(10N\) \(< 25.5+37.5\)
\(10N\) \(<63\)
\(\therefore\ N\) \(< 6.3\)

 

\(\therefore\ \text{Max number of standard drinks is 6.}\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2015 HSC 23 MC

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

Number of standard drinks
White Wine Red Wine
small glass large glass small glass large glass
0.9 1.4 1.0 1.5
 

A woman weighing 58 kg drinks two small glasses of white wine and three small glasses of red wine between 7 pm and 11 pm.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content (\(BAC\)) estimate at 11 pm, correct to three decimal places?
 

\(BAC_{\text{Female}}=\dfrac{10N-7.5H}{5.5M}\)
 

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. 0.013
  2. 0.023
  3. 0.046
  4. 0.056
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\(D\)

Show Worked Solution
\(N\) \(=2\times 0.9 + 3\times 1\)
  \(=4.8\ \text{standard drinks}\)
\(H\) \(=4\ \text{hours}\)
\(M\) \(=58\ \text{kg}\)

  

\(BAC_f\) \(=\dfrac{10\times 4.8-7.5\times 4}{5.5\times 58}\)
  \(=0.05642\dots\)

  
\(\Rightarrow D\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2016 HSC 10 MC

Anika drinks two small bottles of wine over a four-hour period. Each of these bottles contains 2.4 standard drinks. Anika weighs 55 kg.

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

\(BAC_{\text{Female}}=\dfrac{10N - 7.5H}{5.5M}\)
 

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. 0.013
  2. 0.060
  3. 0.0013
  4. 0.0060
Show Answers Only

\(B\)

Show Worked Solution
\(BAC_f\) \(=\dfrac{10N – 7.5H}{5.5M}\)
  \(=\dfrac{10(2\times 2.4) – 7.5\times 4}{5.5\times 55}\)
  \(= 0.0595\dots\approx 0.060\)

 
\(\Rightarrow B\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2017 HSC 27e

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

Bryce weighs 85 kg.

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

\(BAC_{\text{Male}}=\dfrac{10N-7.5H}{6.8M}\)

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 Bryce can drink to remain under a Blood Alcohol Content (\(BAC\)) of 0.05?  (4 marks)

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\(7\)

Show Worked Solution
\(BAC_\text{male}\) \(=\dfrac{10N-7.5H}{6.8M}\)
\(0.05\) \(=\dfrac{10N-7.5\times 4}{6.8\times 85}\)
\(10N\) \(=0.05\times 6.8\times 85+7.5\times 4\)
\(10N\) \(=58.9\)
\(N\) \(=5.89\ \text{standard drinks}\)

  
\(\therefore\ \text{Number of low alcohol bottles}\)

\(=\dfrac{5.89}{0.8}\)

\(=7.3625\)
 

\(\therefore\ \text{Max complete bottles to stay under 0.05}\)

\(=7\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2019 HSC 28

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

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

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

\(\text{Time}=\dfrac{BAC}{0.015}\)

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

Georgie weighs 58 kg. She consumed 2 glasses of wine and 4 glasses of spirits between 7:45 pm and 12:15 am the following day. She then stopped drinking alcohol.

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

Show Answers Only

\(\text{6:34 am}\)

Show Worked Solution

\(\text{Standard drinks consumed}\ (N)=2\times 1.2+4=6.4\)

\(\text{Hours drinking}\ (H) = \text{4 h 30 min = 4.5 hours}\)

\(BAC_{\text{Georgie}}\) \(=\dfrac{10\times 6.4-7.5\times 4.5}{5.5\times 58}\)
  \(=0.09482\dots\)

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

\(\text{Time (to zero)}\) \(=\dfrac{0.09482\dots}{0.015}\)
  \(=6.3218\dots\ \text{hours}\)
  \(\approx 6\ \text{hours 19 minutes}\)

 
\(\therefore\ \text{Georgie should reach zero}\ BAC\)

\(=12:15+6:19=6:34\ \text{am}\)

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2023 HSC 36

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

\(BAC_{\text{male}}=\dfrac{10N-7.5H}{6.8M}\)

\(N\) is the number of standard drinks consumed

\(M\) is the person's weight in kilograms

\(H\) is the number of hours of drinking
 

Min weighs 70 kg. His \(BAC\) was zero when he began drinking alcohol. At 10:30 pm, after consuming 4 standard drinks, his \(BAC\) was 0.032.

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

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\(7:12\text{ pm}\)

Show Worked Solution
\(BAC\) \(=\dfrac{10N-7.5H}{6.8M}\)
\(0.032\) \(=\dfrac{10\times 4-7.5\times H}{6.8\times 70}\)
\(0.032\times 476\) \(=40-7.5H\)
\(7.5H\) \(=40-15.232\)
\(H\) \(=\dfrac{24.768}{7.5}\)
  \(=3.3024\ \text{hours}\)
  \(\approx 3\ \text{hours}\ 18\ \text{minutes (nearest minute)}\)

 
\(\text{Time Min began drinking}\)

\(=10:30\text{ pm – 3 h 18 m}\)

\(=7:12\text{ pm}\)

Mean mark 56%.

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 4, smc-5234-10-BAC

v1 Algebra, STD2 A1 2020 HSC 13 MC

When Stuart stops drinking alcohol at 11:30 pm, he has a blood alcohol content (BAC) of 0.08625.

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

\(\text{Time}=\dfrac{BAC}{0.015}\).

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

  1.  1:33 am
  2.  1:55 am
  3.  2:15 am
  4.  5:15 am
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Time from  0.08625 → 0}\ BAC\)

\(=\dfrac{0.08625}{0.015}\)

\(=5.75\ \text{hours}\)
 

\(\text{Time from  0.08625 → 0.05}\ BAC\)

\(=\dfrac{(0.08625 – 0.05)}{0.08625}\times 5.75\) 

\(=\dfrac{29}{69}\times 5.75\)

\(=2.41\dot{6}=2\ \text{h}\ 25\ \text{min}\)
 

\(\therefore\ \text{Time}\) \(=11:30\ \text{pm} \ + 2 \ \text{h} \ 25 \ \text{min}\)
  \(=1:55\ \text{am}\)

 
\(\Rightarrow B\)


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

Filed Under: Applications: BAC, Medication and D=SxT (Std 2-X) Tagged With: Band 6, smc-5234-10-BAC

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