07. Compound Growth and Decay - Cardiff Tutor Company

Question 1

Jane invests $\pounds250,000$ in a company that offers a compound interest of $3.3\%$ a year.

What will be the value of this investment after $6$ years?

Select the correct answer from the list below:

A: $\pounds1,591,366.64$

B: $\pounds303,767.94$

C: $\pounds299,500$

D: $\pounds4,950,000$

CORRECT ANSWER: B: $\pounds303,767.94$

WORKED SOLUTION:

To find out how much Jane’s savings are worth after $6$ years, you take the values given in the question and use the compound interest formula as shown

250000 \times \bigg(1 + \dfrac{3.3}{100} \bigg)^6=\pounds303,767.94

Question 2

Graham buys a motorbike for $\pounds7,500$ .

The value of the bike depreciates $17\%$ every year.

How much will the motorbike be worth at the end of $5$ years?

Select the correct answer from the list below:

A: $\pounds2954.28$

B: $\pounds1125$

C: $\pounds16443.36$

D: $\pounds6375$

CORRECT ANSWER: A: $\pounds2954.28$

WORKED SOLUTION:

To find out how much Graham’s bike is worth after $5$ years, you take the values given in the question and use the compound interest formula as shown

7500 \times \bigg(1 - \dfrac{17}{100} \bigg)^5=\pounds2954.28

Question 3

A culture of germs in a petri dish starts with a population of $1,000,000$ .

The population grows at a rate of $5\%$ every $2$ hours.

How big will the population be at the end of the first $24$ hours?

Select the correct answer from the list below:

A: $1,200,000$

B: $1,795,856$

C: $2,200,000$

D: $3,225,099$

CORRECT ANSWER: B: $1,795,856$

WORKED SOLUTION:

Now, we need to be careful when finding out what power we should use. We need to see how many lots of growth we’re doing by dividing the total time by a single growth period.

24\div2=12

So, we can see that there are $12$ growth periods, and this is the power we should raise to.

Then, to find out how big the population will be at the end of the first $24$ hours, you take the values given in the question and what we have calculated and use the compound interest formula as shown

1,000,000\times \bigg(1 + \dfrac{5}{100} \bigg)^{12}=1,795,856

Question 4

A chocolate bar costs $40$ p. The price of the chocolate bar increases at a rate of $15 \%$ per year.

How many years will it take for the chocolate bar to have a future cost of $60$ p?

Select the correct answer from the list below:

A: $1$ year

B: $2$ years

C: $3$ years

D: $4$ years

CORRECT ANSWER: C: $3$ years

WORKED SOLUTION:

We substitute our known values into the compound growth formula:

40 \times \bigg( 1 + \dfrac{15}{100} \bigg)^n = 60

We now substitute various values of $n$ into this equation, until the left hand side is equal to $60$ p.

$n = 1$ gives $46$ p

$n = 2$ gives $52.9$ p

$n = 3$ gives $60.84$ p

$n = 4$ gives $69.96$ p

$n = 3$ gives the closest answer to $60$ p, so it takes approximately $3$ years for the chocolate bar to increase from $40$ p to $60$ p.

Question 5

Aleena buys a wedding dress, which costs $£1500$ . The price of the wedding dress decreases at a rate of $20 \%$ per year.

How many years will it take for the wedding dress to have a future cost of $£1000$ .

Select the correct answer from the list below:

A: $1$ year

B: $2$ years

C: $3$ years

D: $4$ years

CORRECT ANSWER: B: $2$ years

WORKED SOLUTION:

We substitute our known values into the compound decay formula:

1500 \times \bigg( 1 - \dfrac{20}{100} \bigg)^n = 1000

We now substitute various values of $n$ into this equation, until the left hand side is equal to $£1000$ .

$n = 1$ gives $£1200$

$n = 2$ gives $£960$

$n = 3$ gives $£768$

$n = 4$ gives $£614.40$

$n = 2$ gives the closest answer to $£1000$ , so it takes approximately $2$ years for the wedding dress to decrease from $£1500$ to $£1000$ .