Compound interest, growth and decay

Multiplier for 4% growth: $1.04$

$7500 \times 1.04^2 = 7500 \times 1.0816 = £8112$

Multiplier: $1 + 0.018 = 1.018$

After 3 years: $4500 \times 1.018^3 = 4747.40...$

Interest: $4747.40 - 4500 = £247.40$

Multiplier for a 6% increase: $1.06$

$25\,000 \times 1.06^3 = 29\,775.4$

So about $29\,775$ workers.

Multiplier for 3.4% interest: $1.034$

$4500 \times 1.034^2 = 4811.202$, so £4811.20

Tamsin: $150\,000 \times 1.04^2 = £162\,240$

Rachel: $160\,000 \times 1.015^2 = £164\,836$

$164\,836 > 162\,240$, so Rachel's house is worth more.

After year 1: $7000 \times 1.03 = 7210$

After year 2: $7210 \times 1.015 = £7318.15$

The multiplier for a 4% decrease is $0.96$.

$679 \times 0.96^3 = 600.7357...$

Value after 3 years: £600.74

Multiplier for 1.5% interest: 1.015

$200\,000 \times 1.015^4 = 212\,272.71$

Interest: $212\,272.71 - 200\,000 = £12\,272.71$

(a) Ali: $2000 \times 1.025^3 = £2153.78$, interest $£153.78$

Ben: $1600 \times 1.035^3 = £1773.95$, interest $£173.95$

Ben gets more.

(b) Ben: $1600 \times 1.035^2 \times 1.04 = £1782.52$, interest $£182.52$. Still Ben, so no change.

Multiplier for 4% growth is 1.04

$7500 \times 1.04^2 = 7500 \times 1.0816 = £8112$

Before the 2023 withdrawal: $2937.14 + 1000 = 3937.14$

Start of 2023: $3937.14 \div 1.035 = 3804$

Before the 2022 withdrawal: $3804 + 750 = 4554$

Start of 2022: $4554 \div 1.035 = £4400$

(a) $4500 \times 1.024 = 4608$

$4608 \times 1.018^2 = 4775.38...$, so £4775.38

(b) Multiplier over 2 years $= 4107 \div 7500 = 0.5476$

Yearly multiplier $= \sqrt{0.5476} = 0.74$

$1 - 0.74 = 0.26$, so $x = 26$

Multipliers: $1 - 0.23 = 0.77$ and $1 - 0.19 = 0.81$

Original $\times 0.77 \times 0.81 = 10\,914.75$

Original $= 10\,914.75 \div 0.6237 = £17\,500$

(a) Try powers of 1.13:

$1.13^5 = 1.84...$ (not yet doubled)

$1.13^6 = 2.08...$ (more than 2)

So it takes 6 years.

(b) If $k$ keeps growing, the population grows faster, so it would double in fewer years.

Multiplier after 3 years: $0.875 \times 0.9 \times 0.9 = 0.70875$

Original value $= 17\,010 \div 0.70875 = £24\,000$

Louise: $x \times 1.025^3 = 344\,605$, so $x = 344\,605 \div 1.025^3 = 320\,000$.

Sadiq: $320\,000 \times 1.02^2 \times 1.035 = 344\,580.48$.

Sadiq's investment is worth £344 580.48.

After 2 years: $3550 \times 1.026^2 = £3736.9998$

Multiplier for year 3: $3819.21 \div 3736.9998 = 1.02199...$

So $R = 2.199... \approx 2.2$

Multiplier for year 1: $12\,336 \div 12\,000 = 1.028$, so $x = 2.8$

Year 2 rate: $2.8 \div 2 = 1.4\%$

$12\,336 \times 1.014 = £12\,508.70$

$P_1 = 400$

$P_2 = 1.01 \times 400 = 404$

$P_3 = 1.01 \times 404 = 408.04$, so about $408$ animals.

(a) Each year multiplies the amount by $1.03$, so after 2 years it is multiplied by $1.03^2 = 1.0609$.

Original $= 2652.25 \div 1.0609 = £2500$

(b) Saver Account: $1.04^5 = 1.2167$, so the interest is about $21.67\%$.

Investment Account: $21\%$.

$21.67\% > 21\%$, so the Saver Account is best.

(a) $P_2 = kP_1$: $6 = 8k$, so $k = 0.75$

$P_3 = 0.75 \times 6 = 4.5$ million

(b) New $k = 0.75 + 0.3 = 1.05$. Multiplying by a number greater than 1 each year makes the population grow, so Louise is correct.

$1040 = K \times 1200 + 20$, so $K = \frac{1020}{1200} = 0.85$.

At 09 17: $0.85 \times 1040 + 20 = 904$.

At 09 18: $0.85 \times 904 + 20 = 788.4$ m.

(a) $10^6 \times 2^{-8} = \frac{1\,000\,000}{256} = 3906.25$, so about 3906 errors.

(b) In the model the number halves every hour, and halving a positive number never gives 0, so the model never reaches zero. (You could also argue that in reality the count must eventually drop below 1, and there can't be a fraction of an error.)

2017 to 2019 is two years, so $P_{2019} = k^2 P_{2017}$.

$3610 = 4000k^2$

$k^2 = 0.9025$, so $k = \sqrt{0.9025} = 0.95$