Compound growth and decay

(a) £3183.62

1. Each year the value is multiplied by $1.02$.
   $$3000 \times 1.02^3 = 3183.624$$
2. To the nearest penny: £3183.62.

• M1 Writing $3000 \times 1.02^3$, or working year by year ($3060$, $3121.20$, ...).
• A1 The correct answer, £3183.62.

(a) $6750$

1. Increase by half each day.
   $$2000 \to 3000 \to 4500 \to 6750$$

• M1 Finding the number after 1 or 2 days: 3000 or 4500.
• A1 The correct answer, $6750$.

(a) £$1936$

1. $$2500\times 0.88^{2}$$
2. The yearly multiplier is 1 $-$ 0.12 = 0.88.
3. 2500 $\times$ $0.88^{2}$ = £1936.

• P1 Establishing $2500\times 0.88^{2}$ or an equivalent valid method.
• A1 Correct answer: £$1936$

(a) £10\,794.52

1. The multiplier for a $12\%$ decrease is $0.88$.
   $$18\,000 \times 0.88^4 = 10\,794.515\ldots$$
2. To the nearest penny: £10 794.52.

• M1 Writing $18\,000 \times 0.88^4$ (or four repeated decreases).
• A1 The correct answer, £10 794.52.

(b) $6$ years

1. Half the value is £9000, so find the first $n$ with $18\,000 \times 0.88^n < 9000$, that is $0.88^n < 0.5$.
2. Try values of $n$.
   $$0.88^5 = 0.5277\ldots, \qquad 0.88^6 = 0.4644\ldots$$
3. So the value first falls below half after 6 years.
   $$18\,000 \times 0.88^6 = 8359.27$$

• P1 Setting the target as £9000, or comparing $0.88^n$ with $0.5$.
• P1 Testing values of $n$ either side of the answer: $n = 5$ gives about £9499 and $n = 6$ gives about £8359.
• A1 The correct answer, $6$ years.

(a) $4$

1. $$1500\times 1.04^{3}$$
2. $$1500\times 1.04^{4}$$
3. After n years the balance is 1500 $\times$ 1.04ⁿ.
4. After 3 years it is £1687.296; after 4 years it is £1754.78784.
5. The balance increases every year, so the first whole year above £1750 is year 4.

• P1 Establishing $1500\times 1.04^{3}$ or an equivalent valid method.
• P1 Establishing $1500\times 1.04^{4}$ or an equivalent valid method.
• A1 Correct answer: $4$

(a) £$2000$

1. $$1458/0.9^{3}$$
2. Three years of decay gives final value = original value $\times$ $0.9^{3}.$
3. Original value = £1458 $\div$ 0.729 = £2000.

• P1 Establishing $1458/0.9^{3}$ or an equivalent valid method.
• A1 Correct answer: £$2000$

(a) $10$%

1. $$302.5/250$$
2. $$\sqrt{1.21}$$
3. Let the multiplier be m > 1. Then $250m^{2}$ = 302.50.
4. $m^{2}$ = 1.21, so m = 1.1.
5. Each increase is (1.1 $-$ 1) $\times$ 100 = 10%.

• P1 Establishing $302.5/250$ or an equivalent valid method.
• P1 Establishing $\sqrt{1.21}$ or an equivalent valid method.
• A1 Correct answer: $10$%

(a) $20$%

1. $$0.99\times 32/33$$
2. $$1-(r/100)^{2}=0.96$$
3. Account A has multiplier 1.1 $\times$ 0.9 = 0.99.
4. Account B has multiplier 0.99 $\times$ 32/33 = 0.96.
5. Let p = r/100. Then (1 + p)(1 $-$ p) = 1 $-$ $p^{2}$ = 0.96.
6. $p^{2}$ = 0.04 and p > 0, so p = 0.2 and r = 20.

• P1 Establishing $0.99\times 32/33$ or an equivalent valid method.
• P1 Establishing $1-(r/100)^{2}=0.96$ or an equivalent valid method.
• A1 Correct answer: $20$%