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Ecolint Campus des NationsMathematics
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · Ratio and Proportion

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.4:94:9; HCF = 4
2.3:2:53:2:5
3.25%
4.1:31:3
5.360 g
6.75%
7.0.25 chf
8.12
9.125 chf
10.93.75 chf
11.1.50 chf
12.3.00 chf
13.Yes — y/x is constant (= 4).
14.k=5k = 5
15.54
16.x=8x = 8
17.y=−6y = -6
18.y=5xy = 5x
19.The origin (0,0)(0, 0).
20.y=4xy = 4x (proportional)
21.800 m
22.50 cm
23.4.5 m
24.21 cm
25.3
26.4 km
27.Yes — equal angles → similar (AAA).
28.3 m by 2 m
29.7 cm
30.Corresponding angles are equal; corresponding sides are in the same ratio.
Silver
31.3:4:53:4:5
32.24 chf and 36 chf
33.(a) 13:713:7 (b) 35%
34.7.50 chf
35.18 bars
36.80 km/h ≈ 1.33 km/min
37.(a) 72 L (b) 720 L
38.2.50 chf/kg
39.25%
40.4320 chf
41.54 m in 3 min; 90 m in 5 min
42.5 minutes
43.144 km
44.7 eggs
45.Yes — k=3k = 3.
46.Gradient = 7. The gradient of y=kxy = kx is exactly kk, the constant of proportionality.
47.(a) 4 L/min (b) V=4tV = 4t (c) 28 L
48.12.50 chf
49.b=55b = 55
50.y=2.5xy = 2.5x; y=15y = 15
51.1200 m by 1800 m
52.10 cm
53.100 cm = 1 m
54.15 cm and 20 cm
55.15 cm by 22.5 cm
56.1 : 500
57.1 cm : 0.5 km
58.2 m
59.w=4.5w = 4.5 cm
60.12.5 m
Gold
61.25%
62.797.30 chf
63.15:2815:28
64.Male = 650; Female = 1000
65.93.75 chf
66.Shop B (≈ 0.23 chf/bar vs 0.25 chf/bar)
67.−60-60 m
68.A and C
69.(a) 90 km/h (b) 25 m/s
70.100 chf
71.(a) y=4.5xy = 4.5 x (b) y=45y = 45 (c) x=20x = 20
72.A: yes (k=3k = 3). B: no.
73.(a) 12.50 chf (b) 8 tickets (c) Yes — passes through origin, C/nC/n constant.
74.(a) 45 chf (b) 500 km (c) Straight line through origin, gradient 0.18
75.Alex (8 chf/h vs Brigit 7.50 chf/h).
76.(a) 1 cm : 33.3 cm (b) 72 cm
77.(a) m=2.4ℓm = 2.4 \ell (b) 180 g (c) 100 cm
78.(a) 320 cal (b) 5000 steps (c) Approximately, but actual calories depend on weight, gradient, speed, etc.
79.(a) k=1.5k = 1.5 (b) y=30y = 30 (c) x=20x = 20
80.(a) 14 km/L (b) 392 km (c) 4200 km
81.x=8x = 8 cm
82.(a) 400 cm by 300 cm (4 m by 3 m) (b) 12 m²
83.3 : 5
84.120 cm²
85.(a) 1500 m (b) 1.5 km
86.18 m; similar because both triangles have a vertical line, a horizontal shadow, and the same sun angle.
87.(a) 10 m (b) ≈ 78.5 m²
88.(a) 4 m by 3 m (b) 6 m²
89.64 cm²
90.(a) AB = 6 km, AC = 4.5 km (b) BC = 7.5 km
Platinum
91.1:21:2
92.10% decrease
93.100 chf
94.16
95.(a) 60 (b) 33.3%
96.£13 770
97.254.80 EUR
98.5 litres
99.£800, £1200, £2000
100.£480
101.(a) V=6tV = 6t (b) ≈33.3≈ 33.3 min (c) V=(6−2)t=4tV = (6-2)t = 4t → 5050 min
102.(a) w=10w = 10, ℓ=20\ell = 20 (b) 200 m² (c) Area becomes 800 m² — quadruples (linear scale 2 → area scale 4).
103.A is proportional (y=2xy = 2x); B is not (passes through (0, -4)). Same shade when 2x=3x−42x = 3x - 4, i.e. x=4x = 4, y=8y = 8.
104.Yes — y=k1xy = k_1 x and x=k2zx = k_2 z ⇒ y=k1k2zy = k_1 k_2 z, so y∝zy \propto z with constant k1k2k_1 k_2.
105.k=4,p=12k = 4, p = 12
106.(a) 24 km/h (b) d=24td = 24t (t in h) (c) 42 km (d) Line through origin slope 24
107.(a) 40, 60, 100 cm (b) Yes — scaling all by the same factor preserves ratios.
108.(a) k=4k = 4 (b) y=28y = 28 (c) See working
109.E.g. data (1, 3) and (2, 6) gives ratio 3, but adding (3, 8) breaks proportionality.
110.Perimeter: yes (P=4sP = 4s). Area: no (A=s2A = s^2, a power relationship).
111.(a) 4:14:1 (b) Yes; length ratio 2:12:1, area ratio =4:1= 4:1
112.(a) c=9.4b/ac = 9.4 b/a (b) Area =πr2= \pi r^2, ratio a2:b2a^2:b^2.
113.(a) See working (b) k=3k = 3
114.2≈1.414\sqrt{2} \approx 1.414
115.480 000 m²
116.12
117.0.05 km²
118.(0,0)(0,0), (12,0)(12, 0), (0,8)(0, 8); old area 12, new 48 = 12 × 4 ✓
119.5.625 × 10⁵ m² ≈ 0.5625 km²
120.(a) 216 m³ (b) 216 m²

Pack B — Answers

Bronze
1.3:43:4; HCF = 6
2.2:3:52:3:5
3.30%
4.2:32:3
5.200 g
6.35%
7.0.30 chf
8.8
9.92 chf
10.72 chf
11.1.50 chf
12.4.50 chf
13.No — y/x is not constant (3, 4.5, 9).
14.k=7k = 7
15.48
16.x=8x = 8
17.y=−6y = -6
18.y=4xy = 4x
19.The origin (0,0)(0, 0).
20.y=4xy = 4x
21.900 m
22.40 cm
23.4.8 m
24.15 cm
25.4
26.7 km
27.Yes — AAA.
28.5 m by 3 m
29.8 cm
30.Same.
Silver
31.3:2:53:2:5
32.30 chf and 50 chf
33.(a) 3:23:2 (b) 40%
34.6.00 chf
35.20 bars
36.90 km/h = 1.5 km/min
37.(a) 96 L (b) 960 L
38.3 chf/kg
39.25%
40.5830 chf
41.75 m in 3 min; 175 m in 7 min
42.7.5 minutes
43.300 km
44.5 eggs
45.Yes — k=4k = 4.
46.Gradient = 3. Same general rule.
47.(a) 2.5 L/min (b) V=2.5tV = 2.5t (c) 22.5 L
48.17.50 chf
49.b=37.5b = 37.5
50.y=0.4xy = 0.4x; y=6y = 6
51.800 m by 1400 m
52.17.5 cm
53.200 cm = 2 m
54.15 cm and 36 cm
55.20 cm by 30 cm
56.1 : 250 000
57.1 cm : 0.25 km
58.1.8 m
59.w=9.6w = 9.6 cm
60.15 m
Gold
61.32%
62.600.00 chf
63.6:56:5 (or 1.2:11.2:1)
64.Male = 400; Female = 500
65.216 chf
66.Shop B (0.16 chf/bar vs 0.25 chf/bar)
67.−96-96 m
68.A and C
69.(a) 72 km/h (b) 20 m/s
70.96 chf
71.(a) y=7xy = 7 x (b) y=84y = 84 (c) x=13x = 13
72.A: yes. B: no.
73.(a) 8.50 chf (b) 11 (with 6 chf left) (c) Yes.
74.(a) 55 chf (b) ≈ 409 km (c) Gradient 0.22
75.Alex (9 chf/h vs Brigit ≈ 8.33 chf/h).
76.(a) 1 cm : 40 cm (b) 50 cm
77.(a) m=2.5ℓm = 2.5 \ell (b) 187.5 g (c) 96 cm
78.See Pack A.
79.(a) k=4/3k = 4/3 (b) y=80/3≈26.67y = 80/3 ≈ 26.67 (c) x=22.5x = 22.5
80.(a) 12 km/L (b) 336 km (c) 2880 km
81.x=15x = 15 cm
82.(a) 500 cm by 400 cm (5 m by 4 m) (b) 20 m²
83.4 : 7
84.108 cm²
85.(a) 5000 m (b) 5 km
86.32 m; same reasoning.
87.(a) 4 m (b) ≈ 12.57 m²
88.(a) 5 m by 4 m (b) 10 m²
89.72 cm²
90.Same approach.
Platinum
91.2:12:1
92.20% increase
93.100 chf
94.32
95.(a) 100 (b) 33.3%
96.£20 655
97.288.12 EUR
98.4 litres
99.£1200, £1800, £3000
100.£490
101.See Pack A.
102.(a) w=15w = 15, ℓ=30\ell = 30 (b) 450 m² (c) Quadruples to 1800 m².
103.A proportional. Same at 3x=4x−53x = 4x - 5, x=5x = 5, y=15y = 15.
104.Same.
105.k=5,p=10k = 5, p = 10
106.(a) 27 km/h (b) d=27td = 27t (c) 60.75 km (d) slope 27
107.See Pack A.
108.(a) k=3k = 3 (b) y=21y = 21 (c) See working
109.Similar — use a parabola or piecewise example.
110.Same.
111.(a) 25:425:4 (b) Yes; length ratio 5:25:2, area ratio 25:425:4
112.See Pack A.
113.See Pack A.
114.3≈1.732\sqrt{3} \approx 1.732
115.80 000 m²
116.36
117.0.0075 km²
118.(0,0)(0,0), (18,0)(18, 0), (0,12)(0, 12); old 12, new 108 = 12 × 9 ✓
119.5 × 10⁵ m² = 0.5 km²
120.(a) 64 m³ (b) 96 m²

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 650 (b) 1000 (c) 50%
Full working
Total parts =20+13=33= 20 + 13 = 33. One part =1650÷33=50= 1650 \div 33 = 50 competitors. (a) Male finishers =13×50=650= 13 \times 50 = 650. (b) Female finishers =20×50=1000= 20 \times 50 = 1000. (c) Female finishers as % of all 2000: 10002000×100=50%\frac{1000}{2000} \times 100 = 50\%.
2Problem 2
Answer
(a) 797.30 chf (b) New price = 743.40 chf; difference = 53.90 chf
Full working
(a) Sale price is 100%−26%=74%100\% - 26\% = 74\% of original. Original =590÷0.74≈797.297→= 590 \div 0.74 \approx 797.297 \rightarrow **797.30 chf**.

(b) Raising 590 by 26%: 590×1.26=743.40590 \times 1.26 = 743.40 chf. This is **less** than the original 797.30 chf. Difference =797.30−743.40=53.90= 797.30 - 743.40 = 53.90 chf. A decrease of 26% followed by an increase of 26% does not return to the start because the increase is taken from a smaller base — combined multiplier 0.74×1.26=0.93240.74 \times 1.26 = 0.9324, about 6.76% below original.
3Problem 3
Answer
Plane: 80 kg (≈ 48%). Train: 15.316 kg (≈ 9%). Car (solo): 165.356 kg (100%). Train ≈ 9.3%, plane ≈ 48.4%, car solo 100%.
Full working
Plane: 80 kg per passenger (round-trip). Train: round-trip =2×547=1094= 2 \times 547 = 1094 km. CO₂ =1094×0.014=15.316= 1094 \times 0.014 = 15.316 kg. Car: round-trip =2×617=1234= 2 \times 617 = 1234 km. CO₂ =1234×0.134=165.356= 1234 \times 0.134 = 165.356 kg. Largest = car (165.36 kg) → 100%. Plane =48.4%= 48.4\%. Train =9.3%= 9.3\%. Train is by far the lowest-carbon option.
4Problem 4
Answer
(a) 1000 mL (b) 1500 mL of blue and 2500 mL of yellow (c) 2.88 L, limited by yellow
Full working
(a) Blue : Yellow =3:5= 3:5. Each part: blue 600 mL means 1 part =200= 200 mL, so yellow =5×200=1000= 5 \times 200 = 1000 mL.

(b) Total parts =8= 8. For 4 L =4000= 4000 mL: 1 part =500= 500 mL. Blue =1500= 1500 mL; yellow =2500= 2500 mL.

(c) With 1.2 L blue (3 parts) max paint = 1.23×8=3.2\frac{1.2}{3} \times 8 = 3.2 L. With 1.8 L yellow (5 parts) max = 1.85×8=2.88\frac{1.8}{5} \times 8 = 2.88 L. Limiting colour: **yellow** (smaller max) → green =2.88= 2.88 L. Uses 1.08 L blue (under 1.2 ✓) and 1.80 L yellow (all of it).
5Problem 5
Answer
(a) 260 EUR (b) 300 chf (c) 254.80 EUR
Full working
(a) 250×1.04=260250 \times 1.04 = 260 EUR.

(b) 312÷1.04=300312 \div 1.04 = 300 chf.

(c) Without commission you would receive 260 EUR. Commission removes 2%: 260×0.98=254.80260 \times 0.98 = 254.80 EUR.
6Problem 6
Answer
(a) 4:3:24:3:2 (b) 300 g flour, 225 g sugar, 150 g butter, 5 eggs (c) 25 brownies
Full working
(a) 120:90:60120:90:60. HCF = 30: 4:3:24:3:2.

(b) Scale factor =15÷6=2.5= 15 \div 6 = 2.5. Flour: 300300 g. Sugar: 225225 g. Butter: 150150 g. Eggs: 55.

(c) Flour per brownie =120÷6=20= 120 \div 6 = 20 g. 500÷20=25500 \div 20 = 25 brownies.
7Problem 7
Answer
(a) 110\tfrac{1}{10} (b) 14135\tfrac{14}{135}
Full working
(a) Boys split 1:91:9, so left-handed fraction = 11+9=110\tfrac{1}{1 + 9} = \tfrac{1}{10}.

(b) Whole school = 9 parts (4 boys + 5 girls). Left-handed boys: 49×110=490\tfrac{4}{9} \times \tfrac{1}{10} = \tfrac{4}{90}. Left-handed girls: 59×112=5108\tfrac{5}{9} \times \tfrac{1}{12} = \tfrac{5}{108}. Common denominator 540: 490=24540\tfrac{4}{90} = \tfrac{24}{540} and 5108=25540\tfrac{5}{108} = \tfrac{25}{540}. Sum: 49540\tfrac{49}{540}. Hmm — let's recompute. Boys 49\frac{4}{9}, of whom 110\frac{1}{10} left-handed → 490\frac{4}{90} of school. Girls 59\frac{5}{9}, of whom 112\frac{1}{12} → 5108\frac{5}{108}. To combine: 490+5108\frac{4}{90} + \frac{5}{108}. LCM(90, 108) = 540: 24540+25540=49540\frac{24}{540} + \frac{25}{540} = \frac{49}{540}. (Approx 9.1%9.1\%.)
8Problem 8
Answer
(a) 1077 chf (b) 500 chf (c) 1163.16 chf
Full working
(a) 1000×1.077=10771000 \times 1.077 = 1077 chf.

(b) Pre-tax price = 538.90÷1.077=500538.90 \div 1.077 = 500 chf.

(c) Discounted pre-tax: 1200×0.90=10801200 \times 0.90 = 1080. With VAT: 1080×1.077=1163.161080 \times 1.077 = 1163.16 chf.
9Problem 9
Answer
(a) 1.28, 1.18, 1.12 chf per 100 g (b) 1 kg pack (c) Two paid + one free → 1500 g for 11.80 → 0.787 chf / 100 g (best value)
Full working
(a) 250 g: 3.20÷2.5=1.283.20 \div 2.5 = 1.28 chf/100 g. 500 g: 5.90÷5=1.185.90 \div 5 = 1.18 chf/100 g. 1 kg: 11.20÷10=1.1211.20 \div 10 = 1.12 chf/100 g.

(b) 1 kg pack at 1.12 chf/100 g is cheapest per gram.

(c) Buy two 500 g packs (2 × 5.90 = 11.80) get one free → 1500 g for 11.80 chf → 11.80÷15≈0.78711.80 \div 15 \approx 0.787 chf/100 g. Easily the best deal.
10Problem 10
Answer
(a) 864 (b) ≈ 1007 (c) 2035
Full working
(a) 800×1.08=864800 \times 1.08 = 864.

(b) After 4 years: 800×1.084≈800×1.3605≈1088800 \times 1.08^4 \approx 800 \times 1.3605 \approx 1088. Hmm — let me recompute: 1.082=1.16641.08^2 = 1.1664; 1.084=1.16642≈1.36051.08^4 = 1.1664^2 \approx 1.3605; so 2030 has 800×1.3605≈1088800 \times 1.3605 \approx 1088 trees.

(c) Solve 800×1.08n>1500⇒1.08n>1.875800 \times 1.08^n > 1500 \Rightarrow 1.08^n > 1.875. Take logs (or iterate): 1.088≈1.8511.08^8 \approx 1.851; 1.089≈2.0001.08^9 \approx 2.000. So after 9 years = **2035**.
11Problem 11
Answer
(a) Alex 24 km/h; Brigit ≈ 24.5 km/h (b) ≈ 48 : 49 (c) ≈ 145.5 km apart
Full working
(a) Alex: 60÷2.5=2460 \div 2.5 = 24 km/h. Brigit: 1 h 50 min=1161\,h\,50\,min = \tfrac{11}{6} h. Speed =45÷116=45×611=27011≈24.5= 45 \div \tfrac{11}{6} = 45 \times \tfrac{6}{11} = \tfrac{270}{11} \approx 24.5 km/h.

(b) Ratio =24:27011=264:270=44:45= 24 : \tfrac{270}{11} = 264 : 270 = 44 : 45.

(c) Combined separation rate ≈48.5\approx 48.5 km/h. In 3 h: ≈145.5\approx 145.5 km. (Computed: 24×3+24.5×3=72+73.5=145.524 \times 3 + 24.5 \times 3 = 72 + 73.5 = 145.5 km.)
12Problem 12
Answer
(a) Plan X ≈ 1157.63 chf; Plan Y ≈ 1156.66 chf (b) Plan X by ≈ 0.97 chf — almost equal!
Full working
(a) Plan X multiplier: 1.053=1.1576251.05^3 = 1.157625. Value: 1000×1.157625=1157.631000 \times 1.157625 = 1157.63 chf. Plan Y multiplier: 1.08×1.04×1.03=1.1566561.08 \times 1.04 \times 1.03 = 1.156656. Value: 1156.661156.66 chf.

(b) Difference: 1157.63−1156.66≈0.971157.63 - 1156.66 \approx 0.97 chf in favour of Plan X. They are surprisingly close because the **product** of rates matters more than their order or precise distribution; both plans give an average annual rate near 5%.
13Problem 13
Answer
(a) 18 m/min (b) d=18td = 18t (c) 216 m (d) 30 minutes
Full working
(a) Rate =36÷2=18= 36 \div 2 = 18 m/min.

(b) d=ktd = kt with k=18k = 18, so d=18td = 18t.

(c) d=18×12=216d = 18 \times 12 = 216 m.

(d) t=540÷18=30t = 540 \div 18 = 30 minutes.
14Problem 14
Answer
(a) 138.24 km (b) 429 km (c) 172.8% (d) No — geometric (multiplicative), not proportional.
Full working
Weekly distances: 80,96,115.2,138.2480, 96, 115.2, 138.24.

(a) Week 4 =80×1.23=138.24= 80 \times 1.2^3 = 138.24 km.

(b) Total ≈429\approx 429 km.

(c) 138.24/80=1.728=172.8%138.24 / 80 = 1.728 = 172.8\%.

(d) Not proportional — week-vs-distance is a geometric (exponential) sequence, not a constant-ratio (linear) one. Plotting gives a curve, not a straight line through the origin.
15Problem 15
Answer
(a) Ratios all equal 12 km/L (b) d=12fd = 12f (c) 336 km (d) 25 L
Full working
(a) 60/5=120/10=180/15=240/20=1260/5 = 120/10 = 180/15 = 240/20 = 12. Constant ratio → proportional.

(b) d=12fd = 12f (km/L).

(c) d(28)=336d(28) = 336 km.

(d) f=300/12=25f = 300/12 = 25 L.
16Problem 16
Answer
(a) 60 g flour, 40 g sugar, 0.75 eggs per person (b) 720 g flour, 480 g sugar, 9 eggs (c) Eggs
Full working
(a) Per person: 480/8=60480/8 = 60 g flour; 320/8=40320/8 = 40 g sugar; 6/8=0.756/8 = 0.75 eggs.

(b) For 12 people: flour =720= 720 g; sugar =480= 480 g; eggs =9= 9.

(c) Max people from each: flour 1000/60≈16.71000/60 \approx 16.7; sugar 800/40=20800/40 = 20; eggs 10/0.75≈13.310/0.75 \approx 13.3. **Eggs are the limiting ingredient** — max 13 people.
17Problem 17
Answer
(a) Yes (b) No (has yy-intercept 4) (c) No (curve, not linear) (d) Yes
Full working
Direct proportion has the form y=kxy = kx (line through origin, no offset, power 1).

(a) E=12hE = 12h ✓ — proportional.

(b) F=4+1.5dF = 4 + 1.5d — line with yy-intercept 4, not through origin → **not** proportional.

(c) A=s2A = s^2 — curve (parabola), not linear → not proportional.

(d) C=0.30nC = 0.30n ✓ — proportional.
18Problem 18
Answer
(a) No (b) No (c) V/S=s/6V/S = s/6 — grows linearly with ss.
Full working
(a) V=s3V = s^3 is a cube relationship; doubling ss gives V×8V \times 8, not V×2V \times 2. Not proportional.

(b) S=6s2S = 6s^2 — quadratic. Doubling ss gives S×4S \times 4. Not proportional.

(c) V/S=s3/(6s2)=s/6V/S = s^3 / (6s^2) = s/6. As s→∞s \to \infty, this grows without bound. **Bigger cubes have proportionally more volume per surface area** — this is the "square-cube law" of biology and engineering.
19Problem 19
Answer
(a) C=0.92UC = 0.92U (b) 230 CHF (c) 500 USD (d) Yes
Full working
(a) C=0.92UC = 0.92U where CC is CHF and UU is USD.

(b) C=0.92×250=230C = 0.92 \times 250 = 230 CHF.

(c) U=460/0.92=500U = 460 / 0.92 = 500 USD.

(d) Yes — passes through origin (0 USD = 0 CHF) and is linear with constant rate k=0.92k = 0.92.
20Problem 20
Answer
(a) 1, 2, 3, 4 (b) No (c) t=60/vt = 60/v
Full working
(a) t=d/v=60/vt = d/v = 60/v: t=1,2,3,4t = 1, 2, 3, 4 h for v=60,30,20,15v = 60, 30, 20, 15.

(b) Not directly proportional — as vv doubles, tt halves. Direct proportion would mean both double together.

(c) **Inverse proportion**: t=k/vt = k/v with k=60k = 60. The product vt=k=60vt = k = 60 is constant.
21Problem 21
Answer
(a) 600 people/year (b) P=12000+600tP = 12000 + 600t (c) 18 000 (d) No — exponential, not proportional.
Full working
(a) Growth =15000−12000=3000= 15000 - 12000 = 3000 over 5 years → 600 per year.

(b) P=12000+600tP = 12000 + 600t. P0=12000P_0 = 12000, k=600k = 600.

(c) P(10)=12000+6000=18000P(10) = 12000 + 6000 = 18000.

(d) Percentage growth (e.g. 5% per year) makes the population follow P=P0(1.05)tP = P_0(1.05)^t — exponential, not proportional to tt.
22Problem 22
Answer
(a) 2.20, 2.10, 1.90 chf/kg (b) Not a straight line through the origin — bulk discount (c) 5 kg sack (d) Bulk discounts / lower per-unit packaging cost.
Full working
(a) 1 kg: 2.20 chf/kg. 2.5 kg: 5.25/2.5=2.105.25/2.5 = 2.10. 5 kg: 9.50/5=1.909.50/5 = 1.90 chf/kg.

(b) Plotting (1, 2.20), (2.5, 5.25), (5, 9.50) gives points that are roughly on a straight line, but not perfectly through origin — the per-kg rate decreases as size grows, so the points actually bend slightly.

(c) 5 kg sack at 1.90 chf/kg is the best value.

(d) Larger sacks have lower per-unit packaging cost and incentivise bulk buying — common retail strategy. Strict direct proportion would imply equal per-kg cost.
23Problem 23
Answer
(a) C=0.126dC = 0.126d (b) ≈ 40.32 chf (c) ≈ 44.80 chf (d) ≈ 8.40 chf
Full working
(a) Petrol per km: 7/100=0.077/100 = 0.07 L/km. Cost per km: 0.07×1.80=0.1260.07 \times 1.80 = 0.126 chf/km. So C=0.126dC = 0.126d.

(b) C(320)=0.126×320=40.32C(320) = 0.126 \times 320 = 40.32 chf.

(c) New per-km cost: 0.07×2.00=0.140.07 \times 2.00 = 0.14 chf/km. For 320 km: 44.8044.80 chf.

(d) Old (1.80): 0.126×600=75.600.126 \times 600 = 75.60 chf. New (2.00): 0.14×600=84.000.14 \times 600 = 84.00 chf. Difference: 8.408.40 chf extra.
24Problem 24
Answer
(a) No — piecewise linear with a kink at 20 h (b) Two segments (c) 26 hours
Full working
(a) Not proportional — for h≤20h \leq 20: pay =18h= 18h (proportional in this range). For h>20h > 20: pay =360+25(h−20)= 360 + 25(h - 20), which has a different gradient and a non-zero constant.

(b) Two straight-line segments: from (0, 0) to (20, 360) with slope 18, then from (20, 360) to (30, 610) with slope 25. There is a "kink" at h=20h = 20.

(c) 510 chf is above the 20-hour threshold (360 chf). Extra: 510−360=150510 - 360 = 150 at 25 chf/h → 150/25=6150/25 = 6 extra hours. Total: 20+6=2620 + 6 = 26 hours.
25Problem 25
Answer
(a) R1 1200 × 1800 m, R2 400 × 600 m (b) R1 2.16 km², R2 0.24 km² (c) 9 : 1 (d) Linear sf 3 (e) ≈ 2016 — earlier than 2035, so the prediction is plausible.
Full working
(a) R1: 1 cm represents 200 m. 6×200=12006 \times 200 = 1200 m; 9×200=18009 \times 200 = 1800 m. R2: 1 cm represents 100 m. 400400 m × 600600 m.

(b) R1 area =2.16= 2.16 km²; R2 =0.24= 0.24 km².

(c) Ratio 9:19 : 1.

(d) Area sf 9 → linear sf 9=3\sqrt{9} = 3.

(e) Area lost 1960–2010: 1.921.92 km² in 50 years ⇒ 0.0384 km²/yr. Remaining 0.24 km² lasts ≈6.25\approx 6.25 years from 2010 → **2016**. 2035 conservative.
26Problem 26
Answer
(a) 4 m × 3 m (b) 12 m² (c) Real radius 0.2 m → area ≈ 0.126 m² (d) See working
Full working
(a) 8×50=4008 \times 50 = 400 cm = 4 m; 6×50=3006 \times 50 = 300 cm = 3 m.

(b) 4×3=124 \times 3 = 12 m².

(c) Real radius =0.4×50=20= 0.4 \times 50 = 20 cm = 0.2 m. Area =π×0.04≈0.1257= \pi \times 0.04 \approx 0.1257 m².

(d) Linear scale 1 : 50 (cm:cm). Area scale =502=2500= 50^2 = 2500. So plan : real = 1 : 2500.

Verify: plan area of room = 8×6=488 \times 6 = 48 cm² = 0.0048 m². Real 12 m². Ratio =0.0048:12=1:2500= 0.0048 : 12 = 1 : 2500 ✓.
27Problem 27
Answer
(a) Same sun-angle and both vertical → AAA (b) 10.5 m (c) 3.75 m
Full working
(a) The sun is far away → its rays are parallel. Each object is vertical (90° at the ground), and the sun-angle is the same → both triangles share two angles → similar (AAA).

(b) tree height1.5=71\frac{\text{tree height}}{1.5} = \frac{7}{1} ⇒ tree = 10.5 m.

(c) Stick: 1.5/2=0.751.5 / 2 = 0.75 (vertical/shadow). Flag: 0.75×5=3.750.75 \times 5 = 3.75 m.
28Problem 28
Answer
(a) 2 (b) 20 cm × 30 cm (c) 16 chf
Full working
(a) Small area =150= 150 cm². Area sf =600/150=4= 600/150 = 4. Linear sf =2= 2.

(b) Dimensions × 2: 20 cm × 30 cm.

(c) Cost ∝ area, so cost sf = 4. Larger cost = 4×4=164 \times 4 = 16 chf.
29Problem 29
Answer
(a) 20 cm × 7.5 cm × 6.25 cm (b) Area ratio 1:5761 : 576; model SA =625= 625 cm² (c) Volume ratio 1:138241 : 13824; model volume ≈868\approx 868 cm³
Full working
(a) Divide by 24: 4.8/24=0.24.8/24 = 0.2 m = 20 cm. 1.8/24=0.0751.8/24 = 0.075 m = 7.5 cm. 1.5/24=0.06251.5/24 = 0.0625 m = 6.25 cm.

(b) Area sf = 1:242=1:5761 : 24^2 = 1 : 576. Model SA = 36 m2/576=0.062536 \text{ m}^2 / 576 = 0.0625 m² = 625 cm².

(c) Volume sf =1:243=1:13824= 1 : 24^3 = 1 : 13824. Model volume = 12/13824≈8.68×10−412 / 13824 \approx 8.68 \times 10^{-4} m³ ≈868\approx 868 cm³.
30Problem 30
Answer
(a) (2, 2), (10, 2), (2, 8) (m) (b) 24 m² (c) 0.25 cm
Full working
(a) Multiply each coordinate by 2: (2,2),(10,2),(2,8)(2, 2), (10, 2), (2, 8) in metres.

(b) Real base =10−2=8= 10 - 2 = 8 m. Real height =8−2=6= 8 - 2 = 6 m. Area =12×8×6=24= \tfrac{1}{2} \times 8 \times 6 = 24 m².

(c) Drawing scale: 1 cm = 2 m, so 0.5 m = 0.25 cm wide.
31Problem 31
Answer
(a) 2 (b) 60 cm (c) 6 cm
Full working
(a) Area sf = 4 ⇒ linear sf = 2.

(b) Perimeter is linear → multiply by 2: 30×2=6030 \times 2 = 60 cm.

(c) Trim length is linear → multiply by 2: 3×2=63 \times 2 = 6 cm.
32Problem 32
Answer
(a) 6 000 m = 6 km (b) 24 cm (c) 64 cm²
Full working
(a) 12×50000=60000012 \times 50000 = 600000 cm = 6000 m = 6 km.

(b) On a 1 : 25 000 map (twice as detailed), the same real distance: 600000/25000=24600000 / 25000 = 24 cm.

(c) Area sf =500002=2.5×109= 50000^2 = 2.5 \times 10^9. Real area =16= 16 km² =1.6×1011= 1.6 \times 10^{11} cm². Map area =1.6×1011/2.5×109=64= 1.6 \times 10^{11} / 2.5 \times 10^9 = 64 cm².
33Problem 33
Answer
(a) 7.5 cm (b) 20 cm × 15 cm (c) ≈ 4.6 g
Full working
(a) 90/12=7.590 / 12 = 7.5 cm.

(b) 240/12=20240 / 12 = 20 cm; 180/12=15180 / 12 = 15 cm.

(c) Volume sf =1/123=1/1728= 1/12^3 = 1/1728. Mass =8000= 8000 g /1728≈4.63/ 1728 \approx 4.63 g.
34Problem 34
Answer
(a) k3Vk^3 V (b) k=2k = 2 (c) 800 cm²
Full working
(a) Volume scales as the cube of the linear scale factor: Vlarger=k3VV_{\text{larger}} = k^3 V.

(b) k3=2000/250=8k^3 = 2000/250 = 8, so k=2k = 2.

(c) Surface area sf =k2=4= k^2 = 4. 200×4=800200 \times 4 = 800 cm².
35Problem 35
Answer
(a) ≈ 54.9° (b) ≈ 1.19 m (c) See working
Full working
(a) tan⁡(elevation)=541/380≈1.4237\tan(\text{elevation}) = 541/380 \approx 1.4237. Angle ≈54.9°\approx 54.9°.

(b) Pedestrian shadow = 1.7×380/541≈1.191.7 \times 380/541 \approx 1.19 m. (Same ratio.)

(c) Both triangles share the sun-angle, both have a vertical (90°) side and a horizontal shadow. AAA → similar. Corresponding sides have the same ratio (height : shadow ratio).
36Problem 36
Answer
(a) 1.5 (b) 7.5 cm (c) ≈ 5.06 L (= 1.5 × 3.375)
Full working
(a) Linear sf =30/20=1.5= 30/20 = 1.5.

(b) Depth =5×1.5=7.5= 5 \times 1.5 = 7.5 cm.

(c) Volume sf =1.53=3.375= 1.5^3 = 3.375. New volume =1.5×3.375=5.0625= 1.5 \times 3.375 = 5.0625 L. (Note: this assumes the larger tin is geometrically similar — the rule k³ for volume).