Ratio, Proportion, and Rates of Change — Diagnostic Tests

Unit Tests

UT-1: Ratio Problems

Question: (a) Divide $\pounds 360$ in the ratio 2:3:7. (b) The ratio of boys to girls in a class is 5:3. If there are 15 girls, how many boys are there? (c) A recipe for 6 people uses 240 g of flour. How much flour is needed for 10 people? (d) The ratio of the lengths of two rectangles is 3:5. If the smaller rectangle has an area of 27 cm$^2$What is the area of the larger rectangle?

Solution:

(a) Total parts $= 2 + 3 + 7 = 12$. One part $= 360/12 = 30$. Shares: $60, 90, 210$.

(b) $B:G = 5:3$. $\frac{B}{15} = \frac{5}{3}$. $B = 15 \times \frac{5}{3} = 25$ boys.

(c) Direct proportion: $\frac{240}{6} \times 10 = 40 \times 10 = 400$ g.

(d) If lengths are in ratio 3:5, areas are in ratio $3^2:5^2 = 9:25$. Larger area $= 27 \times \frac{25}{9} = 3 \times 25 = 75$ cm$^2$.

UT-2: Compound Measures

Question: (a) A car travels 180 km in 2.5 hours. Calculate its average speed in km/h and m/s. (b) The density of aluminium is 2.7 \text{ g/cm^3. Calculate the mass of an aluminium block measuring 5 \text{ cm \times 3 \text{ cm \times 10 \text{ cm. (c) A pressure of 1500 \text{ N/m^2 is exerted on an area of 0.8 \text{ m^2. Calculate the force. (d) Water flows through a pipe at a rate of 0.5 \text{ m^3/minute. How long does it take to fill a tank of volume 12 \text{ m^3?

Solution:

(a) Speed $= 180 / 2.5 = 72$ km/h. In m/s: $72 \times \frac{1000}{3600} = 72 \times \frac{5}{18} = 20$ m/s.

(b) Volume $= 5 \times 3 \times 10 = 150$ cm$^3$. Mass $= 2.7 \times 150 = 405$ g.

(c) Force = \text{Pressure \times \text{Area = 1500 \times 0.8 = 1200 N.

(d) Time $= 12 / 0.5 = 24$ minutes.

UT-3: Growth and Decay

Question: (a) A population of bacteria doubles every 3 hours. If the initial population is 500, calculate the population after 12 hours. (b) A car depreciates at 15% per year. Calculate its value after 4 years if it was bought for $\pounds 18,000$. (c) The half-life of a radioactive substance is 8 days. If there are 200 g initially, how much remains after 24 days? (d) Explain the difference between simple and compound interest with a numerical example.

Solution:

(a) Number of doublings in 12 hours $= 12/3 = 4$. Population $= 500 \times 2^4 = 500 \times 16 = 8000$.

(b) Value $= 18000 \times (1 - 0.15)^4 = 18000 \times 0.85^4 = 18000 \times 0.5220 = \pounds 9396$.

(c) Number of half-lives in 24 days $= 24/8 = 3$. Remaining $= 200 \times 0.5^3 = 200 \times 0.125 = 25$ g.

(d) Simple interest: Interest is calculated only on the original principal. After 3 years at 10% on $\pounds 1000$: interest $= 1000 \times 0.10 \times 3 = \pounds 300$. Total $= \pounds 1300$.

Compound interest: Interest is calculated on the principal plus accumulated interest. After 3 years at 10% on $\pounds 1000$: $A = 1000 \times 1.10^3 = \pounds 1331$. The extra $\pounds 31$ comes from earning interest on interest.

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Integration Tests

IT-1: Speed, Distance, Time with Graphs (with Algebra)

Question: A car accelerates uniformly from rest to 30 m/s in 10 seconds, then travels at constant speed for 20 seconds, then decelerates uniformly to rest in 8 seconds. (a) Draw a velocity-time graph. (b) Calculate the total distance travelled. (c) Calculate the average speed for the entire journey. (d) A second car travels the same distance at a constant speed of 20 m/s. Which car completes the journey first and by how much?

Solution:

(a) The graph has three sections: (1) a straight line from $(0,0)$ to $(10, 30)$(2) a horizontal line from $(10, 30)$ to $(30, 30)$(3) a straight line from $(30, 30)$ to $(38, 0)$.

(b) Distance $=$ area under the graph. Section 1: $\frac{1}{2} \times 10 \times 30 = 150$ m. Section 2: $20 \times 30 = 600$ m. Section 3: $\frac{1}{2} \times 8 \times 30 = 120$ m. Total $= 150 + 600 + 120 = 870$ m.

(c) Total time $= 10 + 20 + 8 = 38$ s. Average speed $= 870 / 38 \approx 22.89$ m/s.

(d) Second car: time $= 870 / 20 = 43.5$ s. The first car is faster by $43.5 - 38 = 5.5$ seconds.

IT-2: Density and Proportion in Context (with Geometry)

Question: A solid metal cone has a base radius of 6 cm, height of 10 cm, and density of 8.5 \text{ g/cm^3. (a) Calculate the volume and mass of the cone. (b) The cone is melted down and recast into a sphere of the same metal. Calculate the radius of the sphere. (c) A second cone has the same mass but half the height. If the density is the same, what is the ratio of the radii of the two cones? (d) If the original cone is instead cut in half horizontally (at half its height), calculate the mass of the smaller top piece.

Solution:

(a) Volume $= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(36)(10) = 120\pi \approx 377.0$ cm$^3$. Mass $= 8.5 \times 120\pi = 1020\pi \approx 3204.4$ g.

(b) Mass is conserved, so sphere volume $= 120\pi$ cm$^3$. $\frac{4}{3}\pi r^3 = 120\pi$. $r^3 = 90$. $r = \sqrt[3]{90} \approx 4.48$ cm.

(c) Same mass and density means same volume. Original volume $= \frac{1}{3}\pi r_1^2 h_1$. New volume $= \frac{1}{3}\pi r_2^2 (h_1/2)$. Since volumes are equal: $r_1^2 h_1 = r_2^2 \times \frac{h_1}{2}$. $r_2^2 = 2r_1^2$. $r_2 = r_1\sqrt{2}$. Ratio $r_1 : r_2 = 1 : \sqrt{2}$.

(d) Cutting at half height ($h = 5$ cm): by similarity, the top piece is a cone with radius $r' = 6/2 = 3$ cm and height $h' = 5$ cm. Volume of top $= \frac{1}{3}\pi(9)(5) = 15\pi$ cm$^3$. Mass $= 8.5 \times 15\pi = 127.5\pi \approx 400.6$ g.

IT-3: Exchange Rates and Best Value (with Number)

Question: Product X costs \45$USD or$38$euros or$\pounds 32$. Exchange rates:$$1 = \pounds 0.79$,$\pounds 1 = 1.15$euros. (a) Convert all prices to GBP and identify the cheapest option. (b) A tourist has$\pounds 200$and buys as many units of X as possible at the cheapest price. How many can they buy and what change do they receive? (c) If the pound strengthens by 5$\pounds 1 = $1.27 \times 1.05$), recalculate the USD price in GBP. How much does the tourist save per unit? (d) Explain the concept of purchasing power parity in this context.

Solution:

(a) USD price in GBP: $45 \times 0.79 = \pounds 35.55$. Euro price in GBP: $38 / 1.15 = \pounds 33.04$. GBP price: $\pounds 32$.

Cheapest option: $\pounds 32$ (paying in GBP).

(b) At $\pounds 32$: number of units $= \lfloor 200 / 32 \rfloor = 6$ units. Change $= 200 - 6 \times 32 = 200 - 192 = \pounds 8$.

(c) New rate: \1 = \pounds 0.79 / 1.05 = \pounds 0.7524$. New USD price in GBP:$45 \times 0.7524 = \pounds 33.86$.

Saving per unit (USD price): $35.55 - 33.86 = \pounds 1.69$. The GBP option ($\pounds 32$) is still cheapest.

(d) Purchasing power parity (PPP) suggests that exchange rates should adjust so that identical goods cost the same in all currencies when converted. Here, the same product costs different amounts in different currencies ($\pounds 35.55$ vs $\pounds 33.04$ vs $\pounds 32$), suggesting either: (1) the product is genuinely cheapest in the UK market, (2) there are transaction costs (shipping, taxes) not reflected in the exchange rate, or (3) the market is not perfectly efficient. PPP predicts that over time, either the pound should depreciate or the foreign prices should adjust to equalise costs.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.