Number — Diagnostic Tests

Unit Tests

UT-1: Fractions, Decimals, and Percentages

Question: (a) Calculate $\frac{3}{8} + \frac{5}{6} - \frac{1}{4}$. (b) Convert $0.3\dot{6}$ (recurring) to a fraction in its simplest form. (c) A shirt originally costs $\pounds 45$. It is discounted by 20%, then the discounted price is increased by 20%. Calculate the final price and explain why it is not $\pounds 45$. (d) Calculate 15% of 360 without using a calculator, showing your working.

Solution:

(a) Common denominator of 8, 6, 4 is 24. \frac{3}{8} = \frac{9}{24}$$\frac{5}{6} = \frac{20}{24}$$\frac{1}{4} = \frac{6}{24}. $\frac{9}{24} + \frac{20}{24} - \frac{6}{24} = \frac{23}{24}$.

(b) Let $x = 0.363636...$. Then $100x = 36.3636...$. Subtracting: $99x = 36$So $x = \frac{36}{99} = \frac{4}{11}$.

(c) After 20% discount: $45 \times 0.80 = \pounds 36$. After 20% increase: $36 \times 1.20 = \pounds 43.20$.

The final price is not $\pounds 45$ because the 20% increase is applied to a smaller base ($\pounds 36$) than the 20% decrease ($\pounds 45$). A 20% decrease followed by a 20% increase equals $0.80 \times 1.20 = 0.96$A net 4% decrease.

(d) $10\%$ of $360 = 36$. $5\%$ of $360 = 18$. $15\% = 36 + 18 = 54$.

UT-2: HCF, LCM, and Prime Factorisation

Question: (a) Find the prime factorisation of 180 and 252. (b) Hence find the HCF and LCM of 180 and 252. (c) Three bells ring at intervals of 12, 18, and 30 minutes respectively. If they all ring together at noon, at what time will they next all ring together? (d) Explain why 1 is not a prime number.

Solution:

(a) $180 = 2^2 \times 3^2 \times 5$. $252 = 2^2 \times 3^2 \times 7$.

(b) \text{HCF = 2^2 \times 3^2 = 4 \times 9 = 36. \text{LCM = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1260.

(c) Time = \text{LCM(12, 18, 30). 12 = 2^2 \times 3$$18 = 2 \times 3^2$$30 = 2 \times 3 \times 5. \text{LCM = 2^2 \times 3^2 \times 5 = 180 minutes $= 3$ hours. They next all ring together at 3:00 pm.

(d) A prime number has exactly two factors: 1 and itself. The number 1 has only one factor (itself). The definition of prime requires exactly two distinct factors, which 1 does not satisfy. Additionally, the Fundamental Theorem of Arithmetic (every integer $\gt 1$ has a unique prime factorisation) would break down if 1 were prime, since $6 = 2 \times 3 = 1 \times 2 \times 3 = 1 \times 1 \times 2 \times 3$ would not have a unique factorisation.

UT-3: Bounds and Standard Form

Question: (a) A rectangle has length 8.4 cm (correct to 1 decimal place) and width 5.2 cm (correct to 1 decimal place). Calculate the upper bound of the area. (b) Write $0.000376$ in standard form. (c) Calculate $\frac{4.2 \times 10^5}{2.8 \times 10^{-3}}$Giving your answer in standard form. (d) The population of a city is $3.45 \times 10^6$ to the nearest $10,000$. Write down the lower and upper bounds.

Solution:

(a) Upper bound of length $= 8.45$ cm. Upper bound of width $= 5.25$ cm. Upper bound of area $= 8.45 \times 5.25 = 44.3625$ cm$^2$.

(b) $0.000376 = 3.76 \times 10^{-4}$.

(c) $\frac{4.2 \times 10^5}{2.8 \times 10^{-3}} = \frac{4.2}{2.8} \times 10^{5-(-3)} = 1.5 \times 10^8$.

(d) Nearest $10,000$: half of $10,000 = 5000$. Lower bound $= 3,445,000 = 3.445 \times 10^6$. Upper bound $= 3,455,000 = 3.455 \times 10^6$.

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Integration Tests

IT-1: Compound Interest and Percentages (with Ratio and Proportion)

Question: A bank offers compound interest at 3.5% per year. (a) Calculate the value of $\pounds 5000$ after 4 years. (b) Another bank offers simple interest. What simple interest rate would give the same final amount after 4 years? (c) The investor splits their money in the ratio 3:2 between two accounts. Account A pays 3.5% compound interest and Account B pays 4% compound interest. Calculate the total amount after 3 years if the total investment is $\pounds 5000$.

Solution:

(a) $A = P(1 + r)^n = 5000 \times 1.035^4 = 5000 \times 1.14752 = \pounds 5737.60$ (to 2 d.p.).

(b) Simple interest: $A = P(1 + rt)$. $5737.60 = 5000(1 + 4r)$. $1.14752 = 1 + 4r$. $r = 0.14752 / 4 = 0.03688 = 3.69\%$ per year.

(c) Ratio 3:2 means Account A gets $\frac{3}{5} \times 5000 = \pounds 3000$ and Account B gets $\pounds 2000$.

Account A: $3000 \times 1.035^3 = 3000 \times 1.10872 = \pounds 3326.16$. Account B: $2000 \times 1.04^3 = 2000 \times 1.12486 = \pounds 2249.73$.

Total $= 3326.16 + 2249.73 = \pounds 5575.89$.

IT-2: Direct and Inverse Proportion (with Algebra)

Question: The time $t$ taken to fill a swimming pool is inversely proportional to the number of pipes $n$ used. With 4 pipes, it takes 6 hours. (a) Find the formula connecting $t$ and $n$. (b) How long would it take with 10 pipes? (c) The cost of running the pipes is directly proportional to the number of pipes and the time they run, at $\pounds 15$ per pipe per hour. Calculate the minimum cost to fill the pool if at least 3 pipes must be used. (d) If only 2 pipes can be used simultaneously due to water pressure, but 4 pipes are available in total (used in shifts), design a schedule that minimises cost and calculate the total cost.

Solution:

(a) $t = \frac{k}{n}$. When n = 4$$t = 6: $6 = k/4$So $k = 24$. Formula: $t = \frac{24}{n}$.

(b) $t = 24/10 = 2.4$ hours $= 2$ hours 24 minutes.

(c) Cost $C = 15 \times n \times t = 15 \times n \times \frac{24}{n} = \pounds 360$.

The cost is always $\pounds 360$ regardless of the number of pipes (since $n \times t = 24$ is constant). The minimum cost with at least 3 pipes is $\pounds 360$.

(d) With 2 pipes at a time and 4 pipes total, we can run pipes in two shifts:

• Shift 1: Pipes 1 and 2 run for $24/2 = 12$ hours.
• Shift 2: Pipes 3 and 4 run for $24/2 = 12$ hours.

The pool needs 24 pipe-hours total. With 2 pipes for 12 hours $= 24$ pipe-hours, the pool is filled.

But the pipes can be swapped. Using 2 pipes at a time for 12 hours: Cost $= 15 \times 2 \times 12 = \pounds 360$.

Since $n \times t = 24$ regardless, the cost remains $\pounds 360$. There is no way to reduce it below this with the given rate structure.

IT-3: Number Problems in Context (with Statistics)

Question: In a survey of 240 people, the results were: 108 prefer tea, 96 prefer coffee, 60 prefer neither, and the rest prefer both. (a) Draw a Venn diagram and find how many prefer both. (b) A person is chosen at random. Calculate the probability they prefer exactly one drink. (c) Of those who prefer at least one drink, what fraction prefer tea? Give your answer as a fraction in its simplest form. (d) If the survey is extended to 600 people with the same proportions, estimate how many would prefer both tea and coffee.

Solution:

(a) Total who prefer at least one drink $= 240 - 60 = 180$. Let $x$ = prefer both. $(108 - x) + x + (96 - x) = 180$. $204 - x = 180$. $x = 24$.

Only tea: $108 - 24 = 84$. Only coffee: $96 - 24 = 72$. Both: 24. Neither: 60.

(b) P(exactly one) $= \frac{84 + 72}{240} = \frac{156}{240} = \frac{13}{20}$.

(c) Of those who prefer at least one drink (180 people), fraction preferring tea $= \frac{108}{180} = \frac{3}{5}$.

(d) Proportion preferring both $= 24/240 = 0.1$. Estimated from 600 people: $600 \times 0.1 = 60$ people.

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.