Year 11 · Algebra
Quick tips — memory joggers
Substituting & solving
Changing the subject
Blood alcohol content
Medication dosages
Level 1 · Fluency
The area of a rectangle is \(A = lw\). Find \(A\) when \(l = 8\) cm and \(w = 5\) cm.
40 cm²
\(A = 8 \times 5 =\) 40 cm².
The perimeter of a rectangle is \(P = 2(l + w)\). Find \(P\) when \(l = 9\) cm and \(w = 4\) cm.
26 cm
\(P = 2(9 + 4) = 2 \times 13 =\) 26 cm.
The area of a triangle is \(A = \tfrac{1}{2}bh\). Find \(A\) when \(b = 10\) cm and \(h = 6\) cm.
30 cm²
\(A = \tfrac{1}{2} \times 10 \times 6 =\) 30 cm².
Level 2 · Application
The area of a circle is \(A = \pi r^{2}\). Calculate \(A\) when \(r = 6\) cm, correct to two decimal places.
113.10 cm²
\(A = \pi \times 6^{2} = 36\pi =\) 113.10 cm².
Evaluate \(y = 2x^{2} + 3x - 4\) when \(x = -5\). Remember to keep the negative in brackets.
31
\(y = 2(-5)^{2} + 3(-5) - 4 = 50 - 15 - 4 =\) 31.
Evaluate \(y = 5 - 2x^{2}\) when \(x = -4\). Remember to keep the negative in brackets.
−27
\(y = 5 - 2(-4)^{2} = 5 - 32 =\) −27.
Level 3 · Further Application
The stopping distance of a car is \(d = 0.2v + 0.006v^{2}\), where \(v\) is the speed in km/h. Find \(d\) when \(v = 60\) km/h.
33.6 m
\(d = 0.2(60) + 0.006(60)^{2} = 12 + 21.6 =\) 33.6 m.
The height of a ball is \(h = 20t - 5t^{2}\), where \(t\) is the time in seconds. Find \(h\) when \(t = 2\) s.
20 m
\(h = 20(2) - 5(2)^{2} = 40 - 20 =\) 20 m.
The kinetic energy of an object is \(E = \tfrac{1}{2}mv^{2}\). Find \(E\) when \(m = 4\) kg and \(v = 10\) m/s.
200 J
\(E = \tfrac{1}{2} \times 4 \times 10^{2} =\) 200 J.
Level 1 · Fluency
Using \(v = u + at\), find \(a\) when \(v = 20\), \(u = 8\) and \(t = 3\).
4
\(20 = 8 + 3a \Rightarrow 3a = 12 \Rightarrow a =\) 4.
Using \(v = u + at\), find \(t\) when \(v = 30\), \(u = 6\) and \(a = 4\).
6
\(30 = 6 + 4t \Rightarrow 4t = 24 \Rightarrow t =\) 6.
Using \(F = ma\), find \(m\) when \(F = 48\) and \(a = 6\).
8
\(48 = 6m \Rightarrow m =\) 8.
Level 2 · Application
The perimeter of a rectangle is \(P = 2(l + w)\). Find \(w\) when \(P = 46\) cm and \(l = 15\) cm.
8 cm
\(46 = 2(15 + w) \Rightarrow 23 = 15 + w \Rightarrow w =\) 8 cm.
The area of a triangle is \(A = \tfrac{1}{2}bh\). Find \(h\) when \(A = 54\) cm² and \(b = 12\) cm.
9 cm
\(54 = \tfrac{1}{2} \times 12 \times h = 6h \Rightarrow h =\) 9 cm.
The simple interest formula is \(I = Prn\). Find \(P\) when \(I =\) $720, \(r = 0.06\) and \(n = 3\).
$4000
\(720 = P \times 0.06 \times 3 = 0.18P \Rightarrow P = 720 \div 0.18 =\) $4000.
Level 3 · Further Application
The volume of a cylinder is \(V = \pi r^{2}h\).
9.95
(a) \(500 = \pi \times 4^{2} \times h \Rightarrow h = 500 \div (16\pi) =\) 9.95.
(b) centimetres (cm).
The volume of a cone is \(V = \tfrac{1}{3}\pi r^{2}h\).
11.46
(a) \(300 = \tfrac{1}{3}\pi \times 5^{2} \times h \Rightarrow h = \dfrac{3 \times 300}{25\pi} = \dfrac{900}{25\pi} =\) 11.46.
(b) centimetres (cm).
The surface area of a sphere is \(A = 4\pi r^{2}\).
3.99
(a) \(200 = 4\pi r^{2} \Rightarrow r^{2} = \dfrac{200}{4\pi} = 15.915 \Rightarrow r = \sqrt{15.915} =\) 3.99.
(b) centimetres (cm).
Level 1 · Fluency
Make \(x\) the subject of \(y = 3x + 5\).
\(y - 5 = 3x \Rightarrow x = \dfrac{y - 5}{3}\).
Make \(x\) the subject of \(y = 4x - 7\).
\(x = \dfrac{y + 7}{4}\)
\(y + 7 = 4x \Rightarrow x =\) \(\dfrac{y + 7}{4}\).
Make \(a\) the subject of \(v = u + at\).
\(a = \dfrac{v - u}{t}\)
\(v - u = at \Rightarrow a =\) \(\dfrac{v - u}{t}\).
Level 2 · Application
Make \(r\) the subject of the formula \(C = 2\pi r\).
\(\dfrac{C}{2\pi}\)
\(r = C \div (2\pi) =\) \(\dfrac{C}{2\pi}\).
Make \(h\) the subject of the formula \(A = \tfrac{1}{2}bh\).
\(h = \dfrac{2A}{b}\)
\(2A = bh \Rightarrow h =\) \(\dfrac{2A}{b}\).
Make \(b\) the subject of the trapezium-area formula \(A = \tfrac{1}{2}(a + b)h\).
\(b = \dfrac{2A}{h} - a\)
\(2A = (a + b)h \Rightarrow \dfrac{2A}{h} = a + b \Rightarrow b =\) \(\dfrac{2A}{h} - a\).
Level 3 · Further Application
The formula for simple interest is \(I = Prn\).
5% p.a.
(a) \(r = I \div (Pn) = \dfrac{I}{Pn}\).
(b) \(r = 600 \div (4000 \times 3) = 0.05 =\) 5% p.a.
The area of a circle is \(A = \pi r^{2}\).
3.99 cm
(a) \(r^{2} = \dfrac{A}{\pi} \Rightarrow r = \sqrt{\dfrac{A}{\pi}}\).
(b) \(r = \sqrt{\dfrac{50}{\pi}} = \sqrt{15.915} =\) 3.99 cm.
The volume of a cylinder is \(V = \pi r^{2}h\).
5.09 cm
(a) \(h = \dfrac{V}{\pi r^{2}}\).
(b) \(h = \dfrac{400}{\pi \times 5^{2}} = \dfrac{400}{25\pi} =\) 5.09 cm.
Level 1 · Fluency
Using \(\text{speed} = \text{distance} \div \text{time}\), find the speed of a car travelling 150 km in 2 hours.
75 km/h
\(150 \div 2 =\) 75 km/h.
Using \(\text{distance} = \text{speed} \times \text{time}\), find the distance travelled by a car at 80 km/h for 3 hours.
240 km
\(80 \times 3 =\) 240 km.
Using \(\text{time} = \text{distance} \div \text{speed}\), find the time for a car to travel 200 km at 50 km/h.
4 hours
\(200 \div 50 =\) 4 hours.
Level 2 · Application
A car travels at an average speed of 90 km/h. Using \(\text{time} = \text{distance} \div \text{speed}\), find the time to travel 270 km.
3 hours
\(270 \div 90 =\) 3 hours.
A cyclist travels 45 km in 1 hour 30 minutes. Find the cyclist's average speed in km/h.
30 km/h
1 h 30 min \(= 1.5\) h, so \(\text{speed} = 45 \div 1.5 =\) 30 km/h.
A train travels at 120 km/h for 2 hours 15 minutes. Find the distance travelled.
270 km
2 h 15 min \(= 2.25\) h, so \(\text{distance} = 120 \times 2.25 =\) 270 km.
Level 3 · Further Application
The stopping distance is \(d = 0.2v + 0.006v^{2}\) metres, where \(v\) is in km/h.
54.4 m
(a) \(d = 0.2 \times 80 + 0.006 \times 80^{2} = 16 + 38.4 =\) 54.4 m.
(b) No — the braking term depends on \(v^{2}\), so doubling the speed multiplies it by \(2^{2} = 4\) (from \(0.006 \times 40^{2} = 9.6\) m to \(0.006 \times 80^{2} = 38.4\) m).
The stopping distance is \(d = 0.2v + 0.006v^{2}\) metres, where \(v\) is in km/h.
80 m; 25%
(a) \(d = 0.2 \times 100 + 0.006 \times 100^{2} = 20 + 60 =\) 80 m.
(b) Reaction term \(= 20\) m, so \(\dfrac{20}{80} =\) 25% of the total.
A car travels at 90 km/h. The stopping distance is \(d = 0.2v + 0.006v^{2}\) metres.
66.6 m; 48.6 m
(a) \(d = 0.2 \times 90 + 0.006 \times 90^{2} = 18 + 48.6 =\) 66.6 m.
(b) Braking term \(= 0.006 \times 90^{2} =\) 48.6 m.
Level 1 · Fluency
Solve \(4x - 7 = 21\).
7
\(4x = 28 \Rightarrow x =\) 7.
Solve \(3x + 8 = 26\).
6
\(3x = 18 \Rightarrow x =\) 6.
Solve \(5x + 4 = 2x + 19\).
5
\(5x - 2x = 19 - 4 \Rightarrow 3x = 15 \Rightarrow x =\) 5.
Level 2 · Application
Solve \(x + \dfrac{x - 1}{2} = 9\).
\(\dfrac{19}{3} \approx 6.33\)
Multiply by 2: \(2x + (x - 1) = 18 \Rightarrow 3x - 1 = 18 \Rightarrow 3x = 19 \Rightarrow x =\) \(\dfrac{19}{3} \approx 6.33\).
Solve \(\dfrac{x}{2} + \dfrac{x}{3} = 10\).
12
Multiply by 6: \(3x + 2x = 60 \Rightarrow 5x = 60 \Rightarrow x =\) 12.
Solve \(2(x + 3) = 5x - 9\).
5
\(2x + 6 = 5x - 9 \Rightarrow 15 = 3x \Rightarrow x =\) 5.
Level 3 · Further Application
Solve the equation \(\dfrac{2x + 1}{3} - \dfrac{x - 2}{4} = 3\).
5.2
(a) Multiply by 12: \(4(2x + 1) - 3(x - 2) = 36\).
(b) \(8x + 4 - 3x + 6 = 36 \Rightarrow 5x + 10 = 36 \Rightarrow 5x = 26 \Rightarrow x =\) 5.2.
Solve the equation \(\dfrac{x + 5}{3} + \dfrac{x - 3}{2} = 6\).
7
(a) Multiply by 6: \(2(x + 5) + 3(x - 3) = 36\).
(b) \(2x + 10 + 3x - 9 = 36 \Rightarrow 5x + 1 = 36 \Rightarrow 5x = 35 \Rightarrow x =\) 7.
Solve the equation \(\dfrac{2x + 1}{2} + \dfrac{x - 2}{4} = 5\).
4
(a) Multiply by 4: \(2(2x + 1) + (x - 2) = 20\).
(b) \(4x + 2 + x - 2 = 20 \Rightarrow 5x = 20 \Rightarrow x =\) 4.
Level 1 · Fluency
A number is multiplied by 5 and the result is 45. Write and solve an equation for the number \(x\).
9
\(5x = 45 \Rightarrow x =\) 9.
When 7 is added to a number the result is 22. Write and solve an equation for the number \(x\).
15
\(x + 7 = 22 \Rightarrow x =\) 15.
A number is divided by 4 to give 6. Write and solve an equation for the number \(x\).
24
\(\dfrac{x}{4} = 6 \Rightarrow x =\) 24.
Level 2 · Application
A number is doubled and then 5 is subtracted; the result is 1 less than the original number. Write an equation and solve for the number \(x\).
4
\(2x - 5 = x - 1 \Rightarrow x =\) 4.
Three times a number, increased by 4, equals 25. Write an equation and solve for the number \(x\).
7
\(3x + 4 = 25 \Rightarrow 3x = 21 \Rightarrow x =\) 7.
A number is increased by 6, and the result is three times the original number. Write an equation and solve for the number \(x\).
3
\(x + 6 = 3x \Rightarrow 6 = 2x \Rightarrow x =\) 3.
Level 3 · Further Application
Two consecutive even numbers add to 74.
36 and 38
(a) \(x + (x + 2) = 74\).
(b) \(2x + 2 = 74 \Rightarrow 2x = 72 \Rightarrow x = 36\), so the numbers are 36 and 38.
The length of a rectangle is 4 cm more than its width. The perimeter is 52 cm.
width 11 cm, length 15 cm
(a) \(2(x + (x + 4)) = 52\).
(b) \(4x + 8 = 52 \Rightarrow 4x = 44 \Rightarrow x = 11\), so the width is 11 cm and the length is 15 cm.
Three consecutive whole numbers add to 96.
31, 32 and 33
(a) \(x + (x + 1) + (x + 2) = 96\).
(b) \(3x + 3 = 96 \Rightarrow 3x = 93 \Rightarrow x = 31\), so the numbers are 31, 32 and 33.
Level 1 · Fluency
The BAC of a driver is 0.045. Is this above or below the general legal limit of 0.05?
Below
Below the 0.05 limit (but above zero — a learner/P-plate limit of 0.00 would be exceeded).
The BAC of a driver is 0.068. Is this above or below the general legal limit of 0.05?
Above
Above the 0.05 limit (\(0.068 > 0.05\)), so this driver is over the legal limit.
A full-licence driver has a legal limit of 0.05, while a learner driver's limit is 0.00. A driver records a BAC of 0.02. For which of these two drivers is this over the limit?
The learner driver
0.02 is above the learner limit of 0.00 but below the full-licence limit of 0.05, so it is over the limit only for the learner driver.
Level 2 · Application
The BAC for a male is \(BAC = \dfrac{10N - 7.5H}{6.8M}\), where \(N =\) standard drinks, \(H =\) hours drinking and \(M =\) mass in kg. Find the BAC of a 72 kg man who has 5 standard drinks over 2 hours, correct to three decimal places.
0.071
\(BAC = \dfrac{10 \times 5 - 7.5 \times 2}{6.8 \times 72} = 35 \div 489.6 =\) 0.071.
Using the male formula \(BAC = \dfrac{10N - 7.5H}{6.8M}\), find the BAC of a 90 kg man who has 4 standard drinks over 1 hour, correct to three decimal places.
0.053
\(BAC = \dfrac{10 \times 4 - 7.5 \times 1}{6.8 \times 90} = 32.5 \div 612 =\) 0.053.
The BAC for a female is \(BAC = \dfrac{10N - 7.5H}{5.5M}\). Find the BAC of a 55 kg woman who has 3 standard drinks over 2 hours, correct to three decimal places.
0.050
\(BAC = \dfrac{10 \times 3 - 7.5 \times 2}{5.5 \times 55} = 15 \div 302.5 = 0.0496 \approx\) 0.050.
Level 3 · Further Application
Using the male formula \(BAC = \dfrac{10N - 7.5H}{6.8M}\):
0.069
(a) \(BAC = \dfrac{60 - 22.5}{6.8 \times 80} = 37.5 \div 544 =\) 0.069.
(b) 0.069 is above 0.05, so he is over the limit.
Using the male formula \(BAC = \dfrac{10N - 7.5H}{6.8M}\):
0.043; under the limit
(a) \(BAC = \dfrac{50 - 30}{6.8 \times 68} = 20 \div 462.4 =\) 0.043.
(b) 0.043 is below 0.05, so he is under the limit.
A 75 kg woman has 6 standard drinks over 2 hours. Use the female formula \(BAC = \dfrac{10N - 7.5H}{5.5M}\).
0.109; over the limit
(a) \(BAC = \dfrac{60 - 15}{5.5 \times 75} = 45 \div 412.5 =\) 0.109.
(b) 0.109 is well above 0.05, so she is over the limit and must not drive.
Level 1 · Fluency
In the female BAC formula \(BAC = \dfrac{10N - 7.5H}{5.5M}\), what does \(M\) represent?
The person’s body mass in kilograms.
In the male BAC formula \(BAC = \dfrac{10N - 7.5H}{6.8M}\), what does \(N\) represent?
The number of standard drinks.
\(N\) is the number of standard drinks consumed, while \(H\) is the hours of drinking and \(M\) the mass in kg.
What is the only difference between the male BAC formula and the female BAC formula?
The mass constant: 6.8 for males, 5.5 for females.
Both use \(\dfrac{10N - 7.5H}{\text{constant} \times M}\); only the denominator constant differs — 6.8 for males and 5.5 for females.
Level 2 · Application
Calculate the BAC of a 60 kg female who has 4 standard drinks over 3 hours, correct to three decimal places.
0.053
\(BAC = \dfrac{10 \times 4 - 7.5 \times 3}{5.5 \times 60} = 17.5 \div 330 =\) 0.053.
Calculate the BAC of an 85 kg male who has 6 standard drinks over 2 hours, using \(BAC = \dfrac{10N - 7.5H}{6.8M}\), correct to three decimal places.
0.078
\(BAC = \dfrac{10 \times 6 - 7.5 \times 2}{6.8 \times 85} = 45 \div 578 =\) 0.078.
Calculate the BAC of a 50 kg female who has 2 standard drinks over 1 hour, using \(BAC = \dfrac{10N - 7.5H}{5.5M}\), correct to three decimal places.
0.045
\(BAC = \dfrac{10 \times 2 - 7.5 \times 1}{5.5 \times 50} = 12.5 \div 275 =\) 0.045.
Level 3 · Further Application
A 70 kg man and a 70 kg woman each drink 4 standard drinks over 2 hours.
(a) Man: \(\dfrac{40 - 15}{6.8 \times 70} = 0.053\). Woman: \(\dfrac{40 - 15}{5.5 \times 70} = 0.065\).
(b) The female formula divides by a smaller factor (\(5.5M\) vs \(6.8M\)), reflecting differences in body composition, so the same alcohol gives a higher BAC.
An 80 kg man and a 60 kg woman each have 5 standard drinks over 3 hours.
Man 0.051, woman 0.083; the woman
(a) Man: \(\dfrac{50 - 22.5}{6.8 \times 80} = 27.5 \div 544 = 0.051\). Woman: \(\dfrac{50 - 22.5}{5.5 \times 60} = 27.5 \div 330 = 0.083\).
(b) The woman has the higher BAC.
A 90 kg man drinks 7 standard drinks over 4 hours.
0.065; 4.3 hours
(a) \(BAC = \dfrac{70 - 30}{6.8 \times 90} = 40 \div 612 =\) 0.065.
(b) \(\text{time} = 0.065 \div 0.015 =\) 4.3 hours.
Level 1 · Fluency
BAC falls at about 0.015 per hour. Using \(\text{time} = BAC \div 0.015\), how long to clear a BAC of 0.030?
2 hours
\(0.030 \div 0.015 =\) 2 hours.
Using \(\text{time} = BAC \div 0.015\), how long to clear a BAC of 0.060?
4 hours
\(0.060 \div 0.015 =\) 4 hours.
Using \(\text{time} = BAC \div 0.015\), how long to clear a BAC of 0.045?
3 hours
\(0.045 \div 0.015 =\) 3 hours.
Level 2 · Application
A person stops drinking with a BAC of 0.072. Using \(\text{time} = BAC \div 0.015\), calculate the time to reach zero BAC, correct to two decimal places.
4.80 hours
\(\text{time} = 0.072 \div 0.015 =\) 4.80 hours (\(\approx\) 4 h 48 min).
A person stops drinking with a BAC of 0.096. Using \(\text{time} = BAC \div 0.015\), calculate the time to reach zero BAC, correct to two decimal places.
6.40 hours
\(\text{time} = 0.096 \div 0.015 =\) 6.40 hours (\(\approx\) 6 h 24 min).
A person stops drinking with a BAC of 0.054. Using \(\text{time} = BAC \div 0.015\), calculate the time to reach zero BAC, correct to two decimal places.
3.60 hours
\(\text{time} = 0.054 \div 0.015 =\) 3.60 hours (\(\approx\) 3 h 36 min).
Level 3 · Further Application
A driver stops drinking at 11:00 pm with a BAC of 0.084.
4:36 am
(a) \(0.084 \div 0.015 = 5.6\) hours.
(b) 5.6 h = 5 h 36 min after 11:00 pm, i.e. about 4:36 am.
A driver stops drinking at 10:00 pm with a BAC of 0.075.
3:00 am
(a) \(0.075 \div 0.015 = 5\) hours.
(b) 5 hours after 10:00 pm is 3:00 am.
A person has a BAC of 0.090 at midnight. The legal limit for driving is 0.05.
2:40 am
(a) Drop needed \(= 0.090 - 0.05 = 0.040\), so \(\text{time} = 0.040 \div 0.015 = 2.67\) hours (2 h 40 min).
(b) 2 h 40 min after midnight is 2:40 am.
Level 1 · Fluency
True or false: the BAC formula gives an exact reading identical to a breathalyser.
False
False — it is only an estimate.
The value the BAC formula produces is best described as an estimate or an exact measurement?
An estimate
It is an estimate — it cannot capture every individual factor that affects real BAC.
True or false: two people of the same weight who drink the same amount over the same time will always have exactly the same BAC.
False
False — real BAC also depends on individual factors (metabolism, food, health) that the formula ignores.
Level 2 · Application
State two factors, not included in the BAC formula, that can affect a person’s actual BAC.
Any two of: food eaten, rate of drinking, individual metabolism, fitness/health, medication, tiredness, or the strength/size of the drinks.
The BAC formula does not include the type or strength of the drinks consumed. State one other personal factor it ignores that can change a person’s real BAC.
Any one of: food eaten, metabolism, general health/fitness, medication or tiredness.
The formula uses only drinks, hours and mass, so it cannot allow for factors such as how much food has been eaten (or metabolism, health, medication), which change how fast alcohol is absorbed and removed.
Explain why the BAC formula might underestimate the true BAC of a person who has not eaten.
Food in the stomach slows the absorption of alcohol. On an empty stomach alcohol is absorbed faster, so the real BAC can be higher than the formula's estimate, which does not account for food.
Level 3 · Further Application
A driver calculates their BAC as 0.049 using the formula and decides to drive.
(a) The formula is only an estimate and ignores individual factors (metabolism, food, drink strength), so the true BAC could be higher than 0.049 — and it is very close to the limit.
(b) Not to drive after drinking (the only certain way to be under the limit).
A person uses the formula to plan their drinking so their BAC stays “just under” 0.05 before driving.
(a) The formula is only an estimate and ignores individual factors such as metabolism, food, health and drink strength, so the actual BAC could be higher; and being so close to 0.05 leaves no margin for that error.
(b) Do not drink any alcohol before driving.
Two friends of the same weight drink the same number of standard drinks over the same time, yet a breathalyser gives them different BAC readings. Explain how this is possible even though the formula gives them the same value. (3 marks)
The formula depends only on drinks, time and mass, so it gives them the same value. But real BAC also depends on factors the formula ignores — metabolism, liver function, food eaten, hydration, general health and fitness — which differ between individuals, so their actual readings can differ.
Level 1 · Fluency
A child’s dose is one quarter of the 400 mg adult dose. What is the child’s dose?
100 mg
\(\dfrac{1}{4} \times 400 =\) 100 mg.
A child’s dose is one third of the 600 mg adult dose. What is the child’s dose?
200 mg
\(\dfrac{1}{3} \times 600 =\) 200 mg.
A child’s dose is two fifths of the 500 mg adult dose. What is the child’s dose?
200 mg
\(\dfrac{2}{5} \times 500 =\) 200 mg.
Level 2 · Application
Young’s formula is \(\text{Dose} = \dfrac{\text{age}}{\text{age} + 12} \times \text{adult dose}\). Calculate the dose for an 8-year-old if the adult dose is 500 mg.
200 mg
\(\text{Dose} = \dfrac{8}{8 + 12} \times 500 = \dfrac{8}{20} \times 500 =\) 200 mg.
Clark’s formula is \(\text{Dose} = \dfrac{\text{weight in kg}}{70} \times \text{adult dose}\). Calculate the dose for a 42 kg child if the adult dose is 500 mg.
300 mg
\(\text{Dose} = \dfrac{42}{70} \times 500 = 0.6 \times 500 =\) 300 mg.
Fried’s formula is \(\text{Dose} = \dfrac{\text{age in months} \times \text{adult dose}}{150}\). Calculate the dose for a 15-month-old child if the adult dose is 600 mg.
60 mg
\(\text{Dose} = \dfrac{15 \times 600}{150} = \dfrac{9000}{150} =\) 60 mg.
Level 3 · Further Application
For an adult dose of 600 mg, compare the dose given by Young’s formula (age 6) and Clark’s formula (weight 30 kg), where Clark’s \(\text{Dose} = \dfrac{\text{weight in kg}}{70} \times \text{adult dose}\).
(a) Young’s: \(\dfrac{6}{18} \times 600 = 200\) mg. Clark’s: \(\dfrac{30}{70} \times 600 = 257.1\) mg.
(b) Clark’s gives the larger dose (257.1 mg vs 200 mg).
An adult dose is 480 mg. A 10-year-old child weighs 40 kg.
(a) Young’s: \(\dfrac{10}{22} \times 480 = 218.2\) mg. Clark’s: \(\dfrac{40}{70} \times 480 = 274.3\) mg.
(b) Clark’s gives the larger dose (274.3 mg vs 218.2 mg).
A medicine has an adult dose of 750 mg.
Young’s 150 mg; Clark’s 160.7 mg
(a) \(\text{Dose} = \dfrac{3}{3 + 12} \times 750 = \dfrac{3}{15} \times 750 =\) 150 mg.
(b) \(\text{Dose} = \dfrac{15}{70} \times 750 =\) 160.7 mg (to 1 dp).
Level 1 · Fluency
Fried’s formula is \(\text{Dose} = \dfrac{\text{age in months} \times \text{adult dose}}{150}\). What age unit is used?
months
The child’s age in months.
In Fried’s formula, what age value (in months) is substituted for a child aged 2 years?
24 months
2 years \(= 2 \times 12 =\) 24 months.
In Fried’s formula \(\text{Dose} = \dfrac{\text{age in months} \times \text{adult dose}}{150}\), what number appears in the denominator?
150
The denominator is 150.
Level 2 · Application
Using Fried’s formula, calculate the dose for an 18-month-old child if the adult dose is 500 mg.
60 mg
\(\text{Dose} = \dfrac{18 \times 500}{150} = \dfrac{9000}{150} =\) 60 mg.
Using Fried’s formula, calculate the dose for a 12-month-old child if the adult dose is 450 mg.
36 mg
\(\text{Dose} = \dfrac{12 \times 450}{150} = \dfrac{5400}{150} =\) 36 mg.
Using Fried’s formula, calculate the dose for a 15-month-old child if the adult dose is 800 mg.
80 mg
\(\text{Dose} = \dfrac{15 \times 800}{150} = \dfrac{12000}{150} =\) 80 mg.
Level 3 · Further Application
A medicine has an adult dose of 750 mg.
100 mg
(a) \(\text{Dose} = \dfrac{20 \times 750}{150} =\) 100 mg.
(b) It is calibrated for ages up to about 2 years (24 months); beyond this it would over-estimate the dose, so Young’s formula is used instead.
A medicine has an adult dose of 900 mg.
60 mg; 120 mg
(a) \(\text{Dose} = \dfrac{10 \times 900}{150} = \dfrac{9000}{150} =\) 60 mg.
(b) \(\text{Dose} = \dfrac{20 \times 900}{150} = \dfrac{18000}{150} =\) 120 mg.
A medicine has an adult dose of 600 mg. A dose of 96 mg is given to a child using Fried’s formula.
24 months
(a) \(96 = \dfrac{\text{age} \times 600}{150} = 4 \times \text{age} \Rightarrow \text{age} =\) 24 months.
(b) 24 months is 2 years, at the upper end of the 1–2 year range the formula is designed for.
Level 1 · Fluency
Young’s formula is \(\text{Dose} = \dfrac{\text{age}}{\text{age} + 12} \times \text{adult dose}\). For a 12-year-old, what fraction of the adult dose is given?
\(\dfrac{1}{2}\)
\(\dfrac{12}{12 + 12} = \dfrac{12}{24} =\) \(\dfrac{1}{2}\) of the adult dose.
Using Young’s formula, for a 6-year-old, what fraction of the adult dose is given?
\(\dfrac{1}{3}\)
\(\dfrac{6}{6 + 12} = \dfrac{6}{18} =\) \(\dfrac{1}{3}\) of the adult dose.
Using Young’s formula, for a 4-year-old, what fraction of the adult dose is given?
\(\dfrac{1}{4}\)
\(\dfrac{4}{4 + 12} = \dfrac{4}{16} =\) \(\dfrac{1}{4}\) of the adult dose.
Level 2 · Application
Using Young’s formula, calculate the dose for a 4-year-old if the adult dose is 480 mg.
120 mg
\(\text{Dose} = \dfrac{4}{4 + 12} \times 480 = \dfrac{4}{16} \times 480 =\) 120 mg.
Using Young’s formula, calculate the dose for a 6-year-old if the adult dose is 540 mg.
180 mg
\(\text{Dose} = \dfrac{6}{6 + 12} \times 540 = \dfrac{6}{18} \times 540 =\) 180 mg.
Using Young’s formula, calculate the dose for a 3-year-old if the adult dose is 600 mg.
120 mg
\(\text{Dose} = \dfrac{3}{3 + 12} \times 600 = \dfrac{3}{15} \times 600 =\) 120 mg.
Level 3 · Further Application
The adult dose of a medicine is 400 mg.
171.4 mg
(a) \(\text{Dose} = \dfrac{9}{21} \times 400 =\) 171.4 mg (to 1 dp).
(b) Young’s formula applies only to ages 1–12; a 15-year-old is generally given the full adult dose (400 mg).
A medicine has an adult dose of 350 mg.
140 mg; 50 mg
(a) \(\text{Dose} = \dfrac{8}{20} \times 350 =\) 140 mg.
(b) \(\text{Dose} = \dfrac{2}{14} \times 350 =\) 50 mg.
Using Young’s formula, a 10-year-old is given a dose of 250 mg.
Adult 550 mg; 161.8 mg
(a) \(250 = \dfrac{10}{22} \times A \Rightarrow A = \dfrac{250 \times 22}{10} =\) 550 mg.
(b) \(\text{Dose} = \dfrac{5}{17} \times 550 =\) 161.8 mg (to 1 dp).
Level 1 · Fluency
Clark’s formula uses the child’s weight. In the version \(\text{Dose} = \dfrac{\text{weight in kg}}{70} \times \text{adult dose}\), what does 70 represent?
The assumed average adult weight in kilograms.
Clark’s formula is \(\text{Dose} = \dfrac{\text{weight in kg}}{70} \times \text{adult dose}\). What measurement of the child is needed to use it?
The child’s weight in kilograms.
Clark’s formula is weight-based, so you need the child’s weight (in kg).
In Clark’s formula, a child weighs exactly 70 kg. What fraction of the adult dose would they receive?
The full adult dose
\(\dfrac{70}{70} = 1\), so they receive the full adult dose.
Level 2 · Application
Using Clark’s formula, calculate the dose for a 35 kg child if the adult dose is 700 mg.
350 mg
\(\text{Dose} = \dfrac{35}{70} \times 700 =\) 350 mg.
Using Clark’s formula, calculate the dose for a 14 kg child if the adult dose is 500 mg.
100 mg
\(\text{Dose} = \dfrac{14}{70} \times 500 = 0.2 \times 500 =\) 100 mg.
Using Clark’s formula, calculate the dose for a 49 kg child if the adult dose is 600 mg.
420 mg
\(\text{Dose} = \dfrac{49}{70} \times 600 = 0.7 \times 600 =\) 420 mg.
Level 3 · Further Application
A medicine has an adult dose of 600 mg.
240 mg
(a) \(\text{Dose} = \dfrac{28}{70} \times 600 =\) 240 mg.
(b) Weight is a better guide to body size than age — two children of the same age can differ greatly in weight, so a weight-based dose is more accurate for the individual.
A medicine has an adult dose of 800 mg.
240 mg; 640 mg
(a) \(\text{Dose} = \dfrac{21}{70} \times 800 = 0.3 \times 800 =\) 240 mg.
(b) \(\text{Dose} = \dfrac{56}{70} \times 800 = 0.8 \times 800 =\) 640 mg.
Using Clark’s formula, a child is given a dose of 180 mg from an adult dose of 600 mg.
21 kg; larger
(a) \(180 = \dfrac{\text{weight}}{70} \times 600 \Rightarrow \text{weight} = \dfrac{180 \times 70}{600} =\) 21 kg.
(b) A heavier child receives a larger dose, since the dose increases in proportion to weight.