smc-791-10-BAC

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

Calculate Sarah's \(BAC\) at the end of the dinner, correct to 3 decimal places. (2 marks)

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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).

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%.

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?

12:45 am
1:50 am
2:15 am
4:05 am

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`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`

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)

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`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)`

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)

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`8`

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`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(Max complete bottles to stay under 0.05)`

`= 8`

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

`0.081`
`0.065`
`0.0017`
`0.0014`

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`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`

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

`0.030`
`0.037`
`0.046`
`0.057`

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`D`

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`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`

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)

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`5`

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

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

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) = (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.

1.03
0.03
0.04
0.01

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`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`