Mathematics • Year 10 • Unit 2 • Lesson 10
Formulas in the Real World
Rearrange and substitute real-world formulas — circle circumference, gravitational drop, BMI, simple interest and Pythagoras. Explain why rearranging first (then substituting) is often the easier method.
1. Word problems
For each problem: (i) write the formula, (ii) make the requested variable the subject, (iii) substitute and solve, (iv) state units in the answer.
1.1 — Wheel diameter. The circumference of a bike wheel is C = 2πr. A wheel has a circumference of 220 cm.
(a) Make r the subject.
(b) Use it to find the radius (use π ≈ 3.14, give the answer to 1 dp).
(c) State the diameter (= 2r). 3 marks
1.2 — Falling object. The distance fallen from rest is s = (1/2)gt², where g ≈ 9.8 m/s² and t is time in seconds.
(a) Make t the subject.
(b) A stone is dropped from a cliff and lands after falling 45 m. Find t (to 1 dp). 3 marks
1.3 — Body Mass Index. BMI = m / h², where m is mass in kg and h is height in metres.
(a) Make h the subject (note: height is positive, so take the positive root).
(b) A person has BMI = 22 and mass 65 kg. Find h to 2 dp. 4 marks
1.4 — Simple interest. I = PRT/100, where P is the principal ($), R is the interest rate (%) and T is years.
(a) Make R the subject.
(b) A $4000 deposit earns $480 interest over 3 years. Find R. 3 marks
1.5 — Pythagoras. For a right triangle, c² = a² + b².
(a) Make a the subject (taking the positive root since a is a length).
(b) A ladder of length 5 m leans against a wall with its foot 3 m from the wall. How high up the wall does it reach? 3 marks
2. Explain your thinking
Communication, not just numbers. 4 marks
2.1 A classmate is told: "Use I = PRT/100 to find R when P = $2000, T = 5 years and I = $400." They substitute first to get 400 = 2000 × R × 5 / 100 = 100R, then solve to find R = 4%. Their friend, however, first rearranges the formula to R = 100I/(PT) before substituting. Using the words subject and inverse operation, explain (i) why both methods give the same answer, (ii) when it is faster to rearrange first, (iii) when it is faster to substitute first.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Wheel
(a) r = C/(2π). (b) r = 220/(2 × 3.14) = 220/6.28 ≈ 35.0 cm. (c) Diameter = 2r ≈ 70.0 cm.
1.2 — Falling object
(a) Multiply by 2: 2s = gt². Divide by g: 2s/g = t². Take √: t = √(2s/g). (b) t = √(2 × 45 / 9.8) = √(9.18) ≈ 3.0 s.
1.3 — BMI
(a) Multiply both sides by h²: BMI × h² = m. Divide by BMI: h² = m/BMI. Take √: h = √(m/BMI). (b) h = √(65/22) = √2.955 ≈ 1.72 m.
1.4 — Simple interest
(a) Multiply by 100: 100I = PRT. Divide by PT: R = 100I/(PT). (b) R = 100 × 480 / (4000 × 3) = 48000/12000 = 4%.
1.5 — Pythagoras
(a) Subtract b²: a² = c² − b². Take √: a = √(c² − b²). (b) a = √(5² − 3²) = √(25 − 9) = √16 = 4 m.
2.1 — Explain (sample response)
(i) Both methods produce the same answer because every step is an inverse operation applied to both sides of the equation — the order in which you do "substitute" and "rearrange" does not change the final value of R, only the intermediate algebra. (ii) Rearranging first (making R the subject) is faster when you have to compute R for several different (P, T, I) trios — you only do the algebra once, then plug in. (iii) Substituting first is faster when you only need one numerical answer and the numbers cancel cleanly (as in this example, where 2000 × 5 / 100 simplifies straight away). Both methods give R = 4%. Working: R = 100(400)/(2000 × 5) = 40000/10000 = 4 ✓.
Marking: 1 mark for explaining equivalence via inverse operations; 1 mark for "rearrange first" being faster for repeated calculations; 1 mark for "substitute first" being faster for a single clean number; 1 mark for showing R = 4% by either route.