A Level H2 Mathematics Numbers Ratio Proportion Quiz

A-Level Maths H2 Quiz - Numbers Ratio Proportion

Answer Key and Teaching Notes

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Question 1 [2 marks]

Answer: $8\sqrt{3}$

Working: $\sqrt{75} + \sqrt{27} = \sqrt{25 \times 3} + \sqrt{9 \times 3} = 5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$

Teaching Notes: To simplify a surd $\sqrt{ab}$, look for the largest perfect square factor of the number. Here, $75 = 25 \times 3$ and $27 = 9 \times 3$. Then combine like surds (same radicand) just like combining like terms in algebra. Common mistake: Writing $\sqrt{75} + \sqrt{27} = \sqrt{102}$ — surds cannot be added under a single root.

Marking: M1 for correctly simplifying one surd; A1 for $8\sqrt{3}$.

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Question 2 [3 marks]

Answer: $2^{2x-5}$

Working: Rewrite all terms as powers of 2: $8^{x-2} = (2^3)^{x-2} = 2^{3x-6}, \qquad 4^{2x} = (2^2)^{2x} = 2^{4x}$

$\dfrac{2^{3x+1} \cdot 2^{3x-6}}{2^{4x}} = \dfrac{2^{(3x+1)+(3x-6)}}{2^{4x}} = \dfrac{2^{6x-5}}{2^{4x}} = 2^{(6x-5)-4x} = 2^{2x-5}$

Teaching Notes: The key strategy is to express everything as powers of the same base (here, base 2). Then use the laws of indices: $a^m \cdot a^n = a^{m+n}$ and $a^m \div a^n = a^{m-n}$. Common mistake: Treating $2^{3x+1} \cdot 8^{x-2}$ as $16^{4x-1}$ — you cannot multiply the bases when the bases are different.

Marking: M1 for converting one term to base 2; M1 for correct numerator simplification; A1 for $2^{2x-5}$.

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Question 3 [4 marks]

Answer: $x = \sqrt{\dfrac{3}{2}}$ or $x = -\sqrt{\dfrac{3}{2}}$ (i.e. $x = \pm\dfrac{\sqrt{6}}{2}$)

Working: $\dfrac{1}{x+2} + \dfrac{3}{x-1} = 2$

Multiply both sides by $(x+2)(x-1)$: $(x-1) + 3(x+2) = 2(x+2)(x-1)$ $x - 1 + 3x + 6 = 2(x^2 + x - 2)$ $4x + 5 = 2x^2 + 2x - 4$ $0 = 2x^2 - 2x - 9$

Using the quadratic formula: $x = \dfrac{2 \pm \sqrt{(-2)^2 - 4(2)(-9)}}{2(2)} = \dfrac{2 \pm \sqrt{4 + 72}}{4} = \dfrac{2 \pm \sqrt{76}}{4} = \dfrac{2 \pm 2\sqrt{19}}{4} = \dfrac{1 \pm \sqrt{19}}{2}$

Correction: Let me re-check the algebra.

$4x + 5 = 2x^2 + 2x - 4$

$0 = 2x^2 + 2x - 4 - 4x - 5$

$0 = 2x^2 - 2x - 9$

$x = \dfrac{2 \pm \sqrt{4 + 72}}{4} = \dfrac{2 \pm \sqrt{76}}{4} = \dfrac{2 \pm 2\sqrt{19}}{4} = \dfrac{1 \pm \sqrt{19}}{2}$

So $x = \dfrac{1 + \sqrt{19}}{2}$ or $x = \dfrac{1 - \sqrt{19}}{2}$

Teaching Notes: When solving rational equations, multiply through by the lowest common denominator to eliminate fractions. Always check that solutions do not make any denominator zero (here $x \neq -2$ and $x \neq 1$, which is satisfied). Common mistake: Forgetting to check for extraneous solutions or making sign errors when expanding brackets.

Marking: M1 for multiplying through by $(x+2)(x-1)$; M1 for correct expansion and simplification to quadratic; M1 for correct use of quadratic formula; A1 for both correct values of $x$.

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Question 4 [3 marks]

Answer: $13 + 7\sqrt{3}$

Working: Rationalise the denominator by multiplying numerator and denominator by the conjugate $2 + \sqrt{3}$:

$\dfrac{5 + \sqrt{3}}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} = \dfrac{(5+\sqrt{3})(2+\sqrt{3})}{(2)^2 - (\sqrt{3})^2}$

Numerator: $(5+\sqrt{3})(2+\sqrt{3}) = 10 + 5\sqrt{3} + 2\sqrt{3} + 3 = 13 + 7\sqrt{3}$

Denominator: $4 - 3 = 1$

Result: $13 + 7\sqrt{3}$

Teaching Notes: To rationalise a denominator of the form $a - \sqrt{b}$, multiply by the conjugate $a + \sqrt{b}$. This uses the difference of squares: $(a-\sqrt{b})(a+\sqrt{b}) = a^2 - b$. Common mistake: Forgetting to multiply both numerator and denominator by the conjugate, or incorrectly expanding the numerator.

Marking: M1 for multiplying by the conjugate; M1 for correct expansion of numerator and denominator; A1 for $13 + 7\sqrt{3}$.

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Question 5

(a) [1 mark]

Answer: $k = 3600$

Working: $P = 3.6 \times 10^8 = 3.6 \times 10^3 \times 10^5 = 3600 \times 10^5$

So $k = 3600$.

(b) [2 marks]

Answer: $1.5 \times 10^{11}$

Working: $\dfrac{P}{Q} = \dfrac{3.6 \times 10^8}{2.4 \times 10^{-3}} = \dfrac{3.6}{2.4} \times 10^{8-(-3)} = 1.5 \times 10^{11}$

Teaching Notes: Standard form is $a \times 10^n$ where $1 \leq a < 10$. When dividing numbers in standard form, divide the coefficients and subtract the exponents. Common mistake: Adding exponents instead of subtracting when dividing.

Marking: (a) A1 for $k = 3600$. (b) M1 for correct division of coefficients and exponent subtraction; A1 for $1.5 \times 10^{11}$.

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Question 6

(a) [2 marks]

Answer: Bella is 25 years old.

Working: Total parts: $3 + 5 + 7 = 15$

Bella's share: $\dfrac{5}{15} \times 75 = 25$ years

(b) [3 marks]

Answer: $n = 3$

Working: Amir's current age: $\dfrac{3}{15} \times 75 = 15$ years
Charlie's current age: $\dfrac{7}{15} \times 75 = 35$ years

In $n$ years: $\dfrac{15 + n}{35 + n} = \dfrac{4}{9}$

Cross-multiply: $9(15 + n) = 4(35 + n)$ $135 + 9n = 140 + 4n$ $5n = 5$ $n = 1$

Correction: Let me recheck.

$9(15+n) = 4(35+n)$

$135 + 9n = 140 + 4n$

$5n = 5$

$n = 1$

Teaching Notes: For ratio problems, first find the value of one part by dividing the total by the number of parts. For part (b), set up an equation using the future ratio. Common mistake: Using current ages instead of future ages when setting up the ratio equation.

Marking: (a) M1 for finding one part (= 5); A1 for 25 years. (b) M1 for correct ages; M1 for setting up the equation; A1 for $n = 1$.

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Question 7

(a) [3 marks]

Answer: $0.63 \text{ km}^2$

Working: Actual dimensions: $3.6 \times 25\,000 = 90\,000 \text{ cm} = 0.9 \text{ km}$ $2.8 \times 25\,000 = 70\,000 \text{ cm} = 0.7 \text{ km}$

Actual area: $0.9 \times 0.7 = 0.63 \text{ km}^2$

(b) [2 marks]

Answer: $12 \text{ cm}^2$

Working: $0.75 \text{ km}^2 = 0.75 \times (100\,000)^2 \text{ cm}^2 = 0.75 \times 10^{10} \text{ cm}^2 = 7.5 \times 10^9 \text{ cm}^2$

Map area: $\dfrac{7.5 \times 10^9}{(25\,000)^2} = \dfrac{7.5 \times 10^9}{6.25 \times 10^8} = 12 \text{ cm}^2$

Teaching Notes: For area scale problems, the area ratio is the square of the linear scale ratio. A scale of $1:25\,000$ means areas are in the ratio $1 : (25\,000)^2$. Common mistake: Using the linear scale factor for area conversions instead of squaring it.

Marking: (a) M1 for converting one dimension to km; M1 for correct area calculation; A1 for $0.63 \text{ km}^2$. (b) M1 for correct method; A1 for $12 \text{ cm}^2$.

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Question 8

(a) [2 marks]

Answer: $x = 3y^2$

Working: Since $x \propto y^2$, we have $x = ky^2$ for some constant $k$.

When $y = 4$, $x = 48$: $48 = k(4)^2 = 16k \implies k = 3$

So $x = 3y^2$.

(b) [1 mark]

Answer: $x = 147$

Working: $x = 3(7)^2 = 3 \times 49 = 147$

(c) [2 marks]

Answer: $y = 7.00$

Working: $147 = 3y^2 \implies y^2 = 49 \implies y = 7.00$

Teaching Notes: Direct proportionality $x \propto y^2$ means $x = ky^2$. Use the given pair of values to find the constant $k$, then use the formula for subsequent parts. Common mistake: Confusing $x \propto y^2$ with $x \propto y$ and writing $x = ky$.

Marking: (a) M1 for setting up $x = ky^2$ and substituting; A1 for $x = 3y^2$. (b) A1 for 147. (c) M1 for correct substitution; A1 for $y = 7.00$.

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Question 9

(a) [2 marks]

Answer: $T = \dfrac{540}{n}$

Working: Since $T \propto \dfrac{1}{n}$, we have $T = \dfrac{k}{n}$.

When $n = 12$, $T = 45$: $45 = \dfrac{k}{12} \implies k = 540$

So $T = \dfrac{540}{n}$.

(b) [1 mark]

Answer: $T = 27$ hours

Working: $T = \dfrac{540}{20} = 27 \text{ hours}$

(c) [1 mark]

Answer: $n = 20$ workers

Working: $27 = \dfrac{540}{n} \implies n = \dfrac{540}{27} = 20$

Teaching Notes: Inverse proportionality means the product $T \times n$ is constant. This is a common real-world relationship — more workers means less time. Common mistake: Setting up $T = kn$ (direct proportion) instead of $T = k/n$.

Marking: (a) M1 for setting up $T = k/n$ and finding $k$; A1 for $T = 540/n$. (b) A1 for 27 hours. (c) A1 for 20 workers.

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Question 10

(a) [3 marks]

Answer: $z = \dfrac{4x^2}{\sqrt{y}}$

Working: Since $z \propto \dfrac{x^2}{\sqrt{y}}$, we have $z = k \cdot \dfrac{x^2}{\sqrt{y}}$.

When $x = 3$, $y = 16$, $z = 9$: $9 = k \cdot \dfrac{9}{\sqrt{16}} = k \cdot \dfrac{9}{4} \implies k = 4$

So $z = \dfrac{4x^2}{\sqrt{y}}$.

(b) [2 marks]

Answer: $z = 20$

Working: $z = \dfrac{4(5)^2}{\sqrt{25}} = \dfrac{4 \times 25}{5} = \dfrac{100}{5} = 20$

Teaching Notes: Combined variation problems involve both direct and inverse proportionality. The general form is $z = k \cdot \dfrac{x^2}{\sqrt{y}}$. Substitute the given values to find $k$, then use the formula. Common mistake: Writing $\sqrt{y}$ in the numerator instead of the denominator.

Marking: (a) M1 for correct proportionality setup; M1 for substituting values; A1 for $z = \dfrac{4x^2}{\sqrt{y}}$. (b) M1 for correct substitution; A1 for $z = 20$.

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Question 11 [4 marks]

Answer: Revenue in 2022 was $2,400,000.

Working: Let the 2022 revenue be $R$.

2023 revenue: $R \times 1.15$

2024 revenue: $R \times 1.15 \times 0.90 = 2\,484\,000$

$R \times 1.035 = 2\,484\,000$

$R = \dfrac{2\,484\,000}{1.035} = 2\,400\,000$

Teaching Notes: For successive percentage changes, multiply the original value by the respective multipliers. A 15% increase means multiplying by $1.15$; a 10% decrease means multiplying by $0.90$. Work backwards from the final value. Common mistake: Adding/subtracting percentages (e.g., thinking net change is $+15\% - 10\% = +5\%$ and using $1.05$ instead of $1.15 \times 0.90 = 1.035$).

Marking: M1 for setting up the equation with multipliers; M1 for $1.15 \times 0.90 = 1.035$; M1 for correct division; A1 for $2,400,000.

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Question 12 [4 marks]

Answer: $r = 5$

Working: Using the compound interest formula: $A = P\left(1 + \dfrac{r}{100}\right)^n$

$11\,576.25 = 10\,000\left(1 + \dfrac{r}{100}\right)^3$

$\left(1 + \dfrac{r}{100}\right)^3 = 1.157625$

$1 + \dfrac{r}{100} = \sqrt[3]{1.157625} = 1.05$

$\dfrac{r}{100} = 0.05 \implies r = 5$

Teaching Notes: The compound interest formula is $A = P(1 + r/100)^n$. To find $r$, divide both sides by $P$, then take the $n$th root. Note that $1.05^3 = 1.157625$ (a useful value to recognise). Common mistake: Using simple interest formula instead of compound interest.

Marking: M1 for correct substitution into formula; M1 for dividing by 10,000; M1 for taking cube root; A1 for $r = 5$.

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Question 13

(a) [1 mark]

Answer: $1,962

Working: $\text{Price with GST} = 1800 \times 1.09 = \$1\,962$

(b) [3 marks]

Answer: $1,724.64

Working: Discounted price before GST: 1800 \times 0.88 = \1,584$

Price including GST: 1584 \times 1.09 = \1,726.56$

Correction: $1584 \times 1.09 = 1584 + 142.56 = 1726.56$

Answer: $1,726.56

Teaching Notes: GST is applied to the selling price. When a discount is given, it is applied to the marked price first, then GST is calculated on the discounted price. Common mistake: Applying GST first and then the discount, or adding GST and discount percentages together.

Marking: (a) A1 for $1,962. (b) M1 for discount calculation; M1 for GST on discounted price; A1 for $1,726.56.

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Question 14

(a) [1 mark]

Answer: $7 : 6 : 8$

Working: Scores: $84 : 72 : 96$

Divide by HCF (12): $7 : 6 : 8$

(b) [3 marks]

Answer: Deepa: $1,512, Eshan: $1,296, Farah: $1,728

Working: Total parts: $7 + 6 + 8 = 21$

Deepa: \dfrac{7}{21} \times 5400 = \1,800$

Eshan: \dfrac{6}{21} \times 5400 = \1,542.86$

Farah: \dfrac{8}{21} \times 5400 = \2,057.14$

Correction: Let me recalculate.

Deepa: $\dfrac{7}{21} \times 5400 = \dfrac{1}{3} \times 5400 = 1800$

Eshan: $\dfrac{6}{21} \times 5400 = \dfrac{2}{7} \times 5400 = \dfrac{10800}{7} = 1542.86$

Farah: $\dfrac{8}{21} \times 5400 = \dfrac{8 \times 5400}{21} = \dfrac{43200}{21} = 2057.14$

Check: $1800 + 1542.86 + 2057.14 = 5400$ ✓

Answer: Deepa: $1,800, Eshan: $1,542.86, Farah: $2,057.14

Teaching Notes: To divide a quantity in a given ratio, find the total number of parts, then calculate each share as (individual parts / total parts) × total amount. Common mistake: Dividing by the number of people instead of the total number of parts.

Marking: (a) A1 for $7:6:8$. (b) M1 for finding total parts (= 21); M1 for correct method for one person; A1 for all three correct amounts.

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Question 15

(a) [4 marks]

Answer: $P = 50\,000 \, e^{0.0262t}$ (or $P = 50\,000 \, e^{kt}$ where $k = \dfrac{\ln 1.3}{10} \approx 0.0262$)

Working: At $t = 0$ (year 2010): $P_0 = 50\,000$

At $t = 10$ (year 2020): $65\,000 = 50\,000 \, e^{10k}$

$e^{10k} = \dfrac{65\,000}{50\,000} = 1.3$

$10k = \ln 1.3 \implies k = \dfrac{\ln 1.3}{10} \approx 0.02623$

So $P = 50\,000 \, e^{0.0262t}$ (to 3 s.f.)

(b) [2 marks]

Answer: $P \approx 85\,000$ (to nearest hundred)

Working: At $t = 20$ (year 2030): $P = 50\,000 \, e^{0.02623 \times 20} = 50\,000 \, e^{0.5246} = 50\,000 \times 1.6894 \approx 84\,470$

To nearest hundred: $P \approx 84\,500$

Teaching Notes: Exponential growth follows $P = P_0 e^{kt}$ where $k > 0$. Use logarithms to find $k$. The value $\ln 1.3 \approx 0.2624$ is useful to compute. Common mistake: Using $P = P_0(1+r)^t$ instead of the continuous model $P = P_0 e^{kt}$ when the question specifies the exponential form.

Marking: (a) M1 for setting up $65000 = 50000 e^{10k}$; M1 for solving for $k$ using $\ln$; M1 for correct $k$ value; A1 for complete formula. (b) M1 for substituting $t = 20$; A1 for 84,500.

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Question 16

(a) [1 mark]

Answer: 750 g

Working: $\text{Flour for 30 cupcakes} = \dfrac{30}{12} \times 300 = 2.5 \times 300 = 750 \text{ g}$

(b) [3 marks]

Answer: 28 cupcakes

Working: Maximum from flour: $\dfrac{800}{300} \times 12 = 32$ cupcakes
Maximum from sugar: $\dfrac{550}{200} \times 12 = 33$ cupcakes
Maximum from butter: $\dfrac{400}{150} \times 12 = 32$ cupcakes

The limiting ingredient is flour (or butter), giving a maximum of 32 cupcakes.

Correction: Let me recheck butter: $\dfrac{400}{150} \times 12 = \dfrac{400 \times 12}{150} = \dfrac{4800}{150} = 32$

So flour allows 32, sugar allows 33, butter allows 32. Maximum = 32 cupcakes.

Answer: 32 cupcakes

(c) [3 marks]

Answer: $33.28

Working: Cost per cupcake: \dfrac{6.50}{12} = \0.5417$

Selling price per cupcake: 0.5417 \times 1.60 = \0.8667$

Total revenue from 32 cupcakes: 32 \times 0.8667 = \27.73$

Correction: Let me recalculate more carefully.

Cost per cupcake: $\dfrac{6.50}{12} = \dfrac{6.5}{12}$

Selling price per cupcake: $\dfrac{6.5}{12} \times 1.6 = \dfrac{6.5 \times 1.6}{12} = \dfrac{10.4}{12} = \dfrac{10.4}{12}$

Total revenue: $32 \times \dfrac{10.4}{12} = \dfrac{332.8}{12} = 27.73$

Answer: $27.73

Teaching Notes: For recipe scaling, find the scaling factor (desired quantity ÷ original quantity). For part (b), check all ingredients and the smallest result is the limiting factor. For part (c), a 60% markup means multiplying the cost by $1.60$. Common mistake: Not checking all ingredients for the limiting factor in part (b).

Marking: (a) A1 for 750 g. (b) M1 for checking at least two ingredients; M1 for finding all three; A1 for 32 cupcakes. (c) M1 for cost per cupcake; M1 for selling price; A1 for $27.73.

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Question 17

(a) [1 mark]

Answer: US$1,850

Working: $2500 \times 0.74 = \$1\,850$

(b) [3 marks]

Answer: ¥297,000

Working: S$1,800 to euros: $1800 \times 0.68 = €1\,224$

Euros to yen: $1224 \times 165 = ¥201\,960$

Correction: $1224 \times 165 = 1224 \times 165 = 201\,960$

Answer: ¥201,960

(c) [2 marks]

Answer: S$1,149

Working: $\text{Cost in SGD} = \dfrac{850}{0.74} = 1148.65 \approx \text{S}\$1\,149$

Teaching Notes: When converting currencies, multiply by the exchange rate to convert from SGD to foreign currency, and divide to convert from foreign currency to SGD. For part (b), perform the conversion in two steps. Common mistake: Multiplying instead of dividing (or vice versa) when converting currencies.

Marking: (a) A1 for US$1,850. (b) M1 for SGD to euros; M1 for euros to yen; A1 for ¥201,960. (c) M1 for dividing by 0.74; A1 for S$1,149.

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Question 18

(a) [3 marks]

Answer: $41,614

Working: $V = 95\,000 \times (0.82)^4 = 95\,000 \times 0.45212176 = 42\,951.57$

Correction: $0.82^4 = 0.82 \times 0.82 \times 0.82 \times 0.82$

$0.82^2 = 0.6724$

$0.82^4 = 0.6724^2 = 0.45212176$

$95\,000 \times 0.45212176 = 42\,951.57$

Answer: $42,952 (to nearest dollar)

(b) [3 marks]

Answer: 7 years

Working: We need $95\,000 \times (0.82)^n < 30\,000$

$(0.82)^n < \dfrac{30\,000}{95\,000} = 0.3158$

Taking logarithms: $n \ln(0.82) < \ln(0.3158)$

$n > \dfrac{\ln(0.3158)}{\ln(0.82)} = \dfrac{-1.1527}{-0.19845} = 5.81$

So $n = 6$ full years.

Correction: Let me verify.

$n > 5.81$, so the smallest integer $n$ is 6.

Check: $95\,000 \times 0.82^6 = 95\,000 \times 0.304006 = 28\,880.57 < 30\,000$ ✓

Check $n = 5$: $95\,000 \times 0.82^5 = 95\,000 \times 0.37074 = 35\,220.30 > 30\,000$ ✓

Answer: 6 years

Teaching Notes: Depreciation is modelled by $V = V_0(1-r)^n$ where $r$ is the depreciation rate. For part (b), use logarithms to solve the inequality. Since $\ln(0.82)$ is negative, the inequality direction reverses when dividing. Common mistake: Forgetting to reverse the inequality sign when dividing by a negative logarithm.

Marking: (a) M1 for using $V = 95000 \times 0.82^4$; M1 for correct calculation; A1 for $42,952. (b) M1 for setting up inequality; M1 for using logarithms; A1 for 6 years.

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Question 19

(a) [2 marks]

Answer: $p = 3q^3$

Working: Since $p \propto q^3$, we have $p = kq^3$.

When $q = 2$, $p = 24$: $24 = k(8) \implies k = 3$

So $p = 3q^3$.

(b) [3 marks]

Answer: $r = \dfrac{120}{q^3}$

Working: Since $r \propto \dfrac{1}{p}$, we have $r = \dfrac{m}{p}$.

When $p = 24$, $r = 5$: $5 = \dfrac{m}{24} \implies m = 120$

So $r = \dfrac{120}{p} = \dfrac{120}{3q^3} = \dfrac{40}{q^3}$

Correction: $r = \dfrac{120}{p}$ and $p = 3q^3$, so $r = \dfrac{120}{3q^3} = \dfrac{40}{q^3}$

Answer: $r = \dfrac{40}{q^3}$

(c) [2 marks]

Answer: $q \approx 2.99$

Working: $1.5 = \dfrac{40}{q^3} \implies q^3 = \dfrac{40}{1.5} = \dfrac{80}{3} = 26.667$

$q = \sqrt[3]{26.667} \approx 2.986 \approx 2.99$

Teaching Notes: This question combines two variation relationships. First find the $p$-$q$ relationship, then the $r$-$p$ relationship, and combine them to get $r$ in terms of $q$. common mistake: Not substituting $p = 3q^3$ into $r = 120/p$ to get the direct relationship between $r$ and $q$.

Marking: (a) M1 for $p = kq^3$ and finding $k$; A1 for $p = 3q^3$. (b) M1 for $r = m/p$; M1 for finding $m = 120$; A1 for $r = 40/q^3$. (c) M1 for correct substitution; A1 for $q = 2.99$.

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Question 20

(a) [2 marks]

Working: After 1 dilution: $V_1 = V_0 - 0.20V_0 = 0.80V_0$

After 2 dilutions: $V_2 = 0.80V_1 = 0.80(0.80V_0) = V_0(0.8)^2$

By induction, after $n$ dilutions: $V_n = V_0(0.8)^n$

(b) [3 marks]

Answer: $n = 6$ dilutions

Working: We need $V_0(0.8)^n < 0.30V_0$

$(0.8)^n < 0.30$

$n \ln(0.8) < \ln(0.3)$

$n > \dfrac{\ln(0.3)}{\ln(0.8)} = \dfrac{-1.2040}{-0.2231} = 5.40$

So $n = 6$ dilutions.

Check: $0.8^5 = 0.32768 > 0.30$ and $0.8^6 = 0.262144 < 0.30$ ✓

(c) [3 marks]

Answer: $n \approx 4$

Working: Initial concentration: 60%, so initial pure chemical = $0.60V_0$

After $n$ dilutions, pure chemical = $0.60V_0(0.8)^n$

Total volume remains $V_0$, so concentration after $n$ dilutions: $\dfrac{0.60V_0(0.8)^n}{V_0} = 0.60(0.8)^n = 0.15$

$(0.8)^n = \dfrac{0.15}{0.60} = 0.25$

$n \ln(0.8) = \ln(0.25)$

$n = \dfrac{\ln(0.25)}{\ln(0.8)} = \dfrac{-1.3863}{-0.2231} = 6.21$

So $n \approx 6$ (to nearest integer).

Correction: $n = 6.21$, so to nearest integer $n = 6$.

Check: $0.60 \times 0.8^6 = 0.60 \times 0.2621 = 0.1573 \approx 0.15$ ✓

Answer: $n = 6$

Teaching Notes: Each dilution removes 20% of the current mixture, leaving 80%. This is a geometric sequence. For parts (b) and (c), use logarithms to solve exponential inequalities/equations. common mistake: Thinking the concentration decreases by a fixed amount each time rather than by a fixed percentage.

Marking: (a) M1 for showing the pattern; A1 for $V_n = V_0(0.8)^n$. (b) M1 for setting up inequality; M1 for using logarithms; A1 for $n = 6$. (c) M1 for setting up $0.60(0.8)^n = 0.15$; M1 for using logarithms; A1 for $n = 6$.

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