Direct proportion and scale

$k = \dfrac{20}{4} = 5$ . Equation: $y = 5x$ .
$y = 5 \times 7 = 35$ .
Yes. $\dfrac{y}{x} = 3$ for every pair, so $y = 3x$ .
No. $\dfrac{5}{2} = 2.5$ , $\dfrac{9}{4} = 2.25$ , $\dfrac{13}{6} \approx 2.17$ , $\dfrac{17}{8} = 2.125$ . The ratio is not constant (this is $y = 2x + 1$ ).
Actual distance $= 6.4 \times 25\,000 = 160\,000$ cm $= 1\,600$ m.
$3$ km $= 300\,000$ cm. Drawing length $= \dfrac{300\,000}{50\,000} = 6$ cm.
Distance $= 80 \times 2.5 = 200$ km.
Time $= \dfrac{120}{5} = 24$ minutes.
$250 \times 0.70 = 175$ USD.
Mass $= 7.87 \times 15 = 118.05$ g.

$y = 2.5x$ . Yes, $y$ is directly proportional to $x$ because the line passes through the origin and $k = 2.5$ .
Flour per serve $= \dfrac{450}{6} = 75$ g. For 10 serves: $75 \times 10 = 750$ g. Assumption: the amount of flour is directly proportional to the number of serves (the recipe scales linearly).
Drawing dimensions: $3.5$ cm $\times$ $2.8$ cm. Actual: $3.5 \times 200 = 700$ cm $= 7$ m and $2.8 \times 200 = 560$ cm $= 5.6$ m. Area $= 7 \times 5.6 = 39.2$ m $^2$ .
Cyclist: $\dfrac{36}{1.5} = 24$ km/h. Runner: $\dfrac{15}{1.25} = 12$ km/h. The cyclist is faster.
Volume $= 25 \times 5 \times 2 = 250$ cm $^3$ . Mass $= 19.3 \times 250 = 4\,825$ g $= 4.825$ kg.
$k = \dfrac{9}{6} = 1.5$ cm/kg. Extension for 10 kg: $1.5 \times 10 = 15$ cm. Load for 15 cm: $\dfrac{15}{1.5} = 10$ kg.
Net flow rate $= 40 - 15 = 25$ litres per minute. Time $= \dfrac{5000}{25} = 200$ minutes $= 3$ hours $20$ minutes.
Scale factor for length $= 10\,000$ . Area scale factor $= 10\,000^2 = 10^8$ . Actual area $= 12 \times 10^8$ cm $^2 = 12 \times 10^4$ m $^2 = 120\,000$ m $^2$ (or $0.12$ km $^2$ ).

Calculate $\dfrac{y}{x}$ for every pair. If the ratio is the same each time, the relationship is directly proportional. Example of a near-miss: $x$ : 1, 2, 3, 4; $y$ : 3, 6, 9, 13. The first three ratios are 3, but the last is 3.25, so it is not proportional.
No. The total cost $C = 5 + 2w$ has a non-zero $y$ -intercept ($5 base fee), so $C$ is not directly proportional to $w$ . The graph is a straight line crossing the $C$ -axis at 5, not through the origin. Doubling the weight does not double the cost.
Road actual length $= 8 \times 20\,000 = 160\,000$ cm. On the second map: $\dfrac{160\,000}{50\,000} = 3.2$ cm.
$k = \dfrac{600}{2} = 300$ W/m $^2$ . Maximum output $= 300 \times 14 = 4\,200$ W. Real-world factors: panels may not all face the sun at the optimal angle; shading, temperature, and panel efficiency losses reduce actual output.
The student confuses a constant rate of change (gradient) with direct proportionality. Direct proportion requires $y = kx$ (the line passes through the origin). Since $y = 3x + 1$ has $y = 1$ when $x = 0$ , it does not pass through the origin and is not a proportional relationship.

Linear scale factor $= 18$ . Volume scale factor $= 18^3 = 5\,832$ . Estimated real car mass $= 1.2 \times 5\,832 = 6\,998.4$ kg $\approx 7\,000$ kg. (In practice, real cars are not solid like models, so the actual mass would be much less — around 1,200—1,800 kg. The calculation shows what would happen if density were identical throughout.)
$y$ is not directly proportional to $x$ (doubling $x$ quadruples $y$ ). However, $y$ is directly proportional to $x^2$ with constant $k$ . Distance in 3 s: $d = 4.9 \times 9 = 44.1$ m. Time to fall 100 m: $100 = 4.9t^2$ , so $t^2 = \dfrac{100}{4.9} \approx 20.41$ and $t \approx 4.52$ s.
Let rate $= k \times \text{depth}$ . When depth $= 2$ : $0.5 = k \times 2$ , so $k = 0.25$ litres per minute per metre. Equation: rate $= 0.25d$ . When $d = 3.5$ : rate $= 0.25 \times 3.5 = 0.875$ litres per minute.
Scale factor $= \dfrac{35}{15} = \dfrac{7}{3} \approx 2.333$ . Height $= 10 \times \dfrac{7}{3} = \dfrac{70}{3} \approx 23.3$ cm. Enlarged area $= 35 \times \dfrac{70}{3} = \dfrac{2450}{3} \approx 816.7$ cm $^2$ . Cost $= 816.7 \times 0.08 \approx 65.33$ dollars.

Tier 1