Mathematics • Year 9 • Unit 4 • Lesson 20
Multi-Step Synthesis Problems
Build fluency with the lesson's five-step problem-solving routine: identify needed skills → break into steps → apply formulas → check each step → combine results. One fully worked multi-step example, one faded example, then eight graduated synthesis problems pulling from Lessons 16-19.
1. I do — fully worked example
A two-topic synthesis problem combining measurement (Lesson 18) and the scale factor rule.
Problem. A cylindrical water tank has radius 2 m and height 3 m. It is currently 2/3 full. A 1 : 25 scale model is built. Find (a) the volume of water currently in the real tank in m³, then (b) the model's TOTAL volume in cm³.
Step 1 — Identify needed skills.
Need: cylinder volume V = πr²h (Lesson 18), fractional reasoning (2/3), and scale-factor rule (volume scales by k³).
Step 2 — Real tank's full volume.
V_full = π × 2² × 3 = π × 4 × 3 = 12π m³ ≈ 37.70 m³.
Step 3 — Water currently in real tank (2/3 full).
V_water = (2/3) × 12π = 8π m³ ≈ 25.13 m³.
Step 4 — Apply scale factor k = 1/25 to TOTAL real volume (12π).
Volume scales by k³ = (1/25)³ = 1/15 625.
V_model = 12π ÷ 15 625 ≈ 37.70 ÷ 15 625 ≈ 0.002 413 m³.
Step 5 — Convert to cm³ (1 m³ = 1 000 000 cm³).
V_model ≈ 0.002 413 × 1 000 000 ≈ 2 413 cm³.
Step 6 — Reasonableness check.
A 1:25 scale model of a 12 000 L tank holding about 2.4 L makes sense — small enough to sit on a desk. ✓
Answers: (a) 8π ≈ 25.13 m³ of water; (b) model total volume ≈ 2 413 cm³.
2. We do — fill in the missing steps
A trigonometry + scale-factor synthesis. Fill the blanks. 4 marks
Problem. A building casts a 30 m shadow when the sun's angle of elevation is 35°. (a) Find the height of the building. (b) Find the height of a 1 : 100 scale model of the building, in cm.
Step 1 — Identify the trig ratio for (a).
Know: angle (35°), adjacent (30 m). Want: opposite (height). Use ____ (T___).
Step 2 — Substitute and solve.
tan 35° = h ÷ ____
h = ____ × tan 35°
h ≈ ______ m
Step 3 — Apply the linear scale factor k = 1/100 to find the model height in metres.
Heights scale by k¹ (not k² or k³ — it's a LENGTH).
Model height = ______ ÷ 100 = ______ m
Step 4 — Convert to centimetres.
Model height ≈ ______ × 100 = ______ cm
3. You do — independent practice
Foundation = single-topic warm-ups. Standard = two-topic synthesis. Extension = three-topic synthesis.
Foundation — single topic
3.1 A cone has radius 3 cm and height 4 cm. Find its volume in terms of π. (Use V = (1/3)πr²h.) 1 mark
3.2 Find the angle of elevation if a ramp rises 2 m over 10 m horizontal. 1 mark
3.3 A fair coin is tossed and a fair die is rolled. Find P(head AND a 6). 1 mark
3.4 A stratified sample of 100 students from a school with 60% Junior, 30% Middle, 10% Senior should have how many from each group? 1 mark
Standard — combine two topics
3.5 A cone has radius 3 cm and height 4 cm. A similar cone has radius 6 cm. Find the larger cone's volume using the scale factor (do not re-calculate from V = (1/3)πr²h). 2 marks
3.6 Of 200 people surveyed, 120 support a new policy. (a) Find the experimental P(support). (b) Two people are chosen from this sample without replacement. Find P(both support). 2 marks
Extension — push your thinking
3.7 From the top of a 40 m cliff, the angle of depression to a yacht is 22°. (a) Find the horizontal distance to the yacht (1 d.p.). (b) Find the slant (line-of-sight) distance. (c) Pythagoras-check that horizontal² + 40² ≈ slant². 3 marks
3.8 A spinner has 5 equal sections (red, blue, green, yellow, purple). It is spun 200 times. Red appears 35 times. (a) Find theoretical P(red). (b) Find experimental P(red). (c) Expected number of reds in 200 spins? (d) Comment in one sentence on whether the spinner looks fair (use the law of large numbers idea). 3 marks
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What I'll revisit before next class:
Section 2 — We do (building + shadow + model)
Step 1: use tan (TOA).
Step 2: tan 35° = h ÷ 30 → h = 30 × tan 35° ≈ 21.0 m.
Step 3: model = 21.0 ÷ 100 = 0.21 m.
Step 4: 0.21 × 100 = 21 cm.
3.1 — Cone volume
V = (1/3) × π × 3² × 4 = (1/3) × 36π = 12π cm³ ≈ 37.7 cm³.
3.2 — Ramp angle
tan θ = 2/10 = 0.2 → θ = tan⁻¹(0.2) ≈ 11.3°.
3.3 — Coin + die independent events
P(head AND 6) = (1/2) × (1/6) = 1/12 ≈ 0.083.
3.4 — Stratified sample of 100
Junior: 60% × 100 = 60; Middle: 30% × 100 = 30; Senior: 10% × 100 = 10.
3.5 — Similar cone via scale factor
Length scale k = 6 ÷ 3 = 2. Volume scales by k³ = 8.
V_large = 8 × 12π = 96π cm³ ≈ 301.6 cm³.
Quick check via direct formula: (1/3)π(6²)(h_large). The similar cone has h_large = 2 × 4 = 8 cm. V = (1/3)π(36)(8) = 96π. ✓
3.6 — Two supporters without replacement
(a) P(support) = 120 ÷ 200 = 0.6.
(b) P(both support) = (120/200) × (119/199) = (0.6) × (119/199) ≈ 0.6 × 0.598 = ≈ 0.359.
3.7 — Cliff and yacht
(a) tan 22° = 40 ÷ d → d = 40 ÷ tan 22° ≈ 99.0 m.
(b) Slant = 40 ÷ sin 22° ≈ 106.8 m.
(c) Check: 99² + 40² = 9 801 + 1 600 = 11 401; √11 401 ≈ 106.8. ✓
3.8 — Five-section spinner
(a) Theoretical P(red) = 1/5 = 0.2.
(b) Experimental P(red) = 35 ÷ 200 = 0.175.
(c) Expected = 200 × 0.2 = 40 reds.
(d) Difference of 5 reds in 200 spins (0.025 in probability) is well within normal random variation for this sample size, so the spinner looks fair — by the law of large numbers, with more spins we'd expect experimental P to approach 0.2 even more closely.