17. Compound Growth

All new questions.

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Compound Growth FSQ1

The population of a city increases by $3\%$ per year.

In $2020$, the population is $250000$ people.

 

Compound Growth FS1(a)

Work out what the population will be in $2021$

Answer type: simple

Answer: 257500

Workings:

$3\% = 0.03$ so a $3$ increase can be calculated as:

$(1+0.03)\times250000 \\ 1.03\times 250000 \\=257500$

2 marks

 

Compound Growth FS1(b)

Using your answer from part(a), work out what the population will be in $2022$

Answer type: simple

Answer: 265225

Workings:

$3\% = 0.03$ so a $3\%$ increase can be calculated as:

$(1+0.03)\times250000 \\ 1.03\times 257500 \\=265225$

2 marks

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Compound Growth FSQ2

 

Suzy deposits $£1750$ into a bank account with an interest rate of $1.2\%$ per year.

How much money will be in her account after $1$ year?

Answer: (£) 1771

Workings:

$1.2 \% = 0.012$ so a $1.2 \%$ increase can be calculated as:

$(1+0.012)\times1750 = £1771$

2 marks

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Compound Growth FSQ3

E. Coli is a species of bacteria which can double in size every $20$ minutes.

 

Compound Growth FS3(a)

What percentage increase is required for the population to double?

Select the correct answer.

Answer type: multiple choice

Answer: $100\%$

Wrong answers:

$2\%$ $20\%$ $200\%$

Workings:

A $100\%$ increase in size is the same as doubling the original value.

As a decimal, $100\% = 1.00$ so a $100 \%$ increase is the same as multiplying by $1+1.00=2$

 

Compound Growth FS3(b)

If there are initially $20000$ bacteria, calculate how many there will be after $20$ minutes.

Answer type: simple

Answer: $40000$

Workings:

The population doubles every $20$ minutes, so the new population can be calculated as:

$20000 \times 2 = 40000$

1 mark

 

Compound Growth FS3(c)

How many bacteria will there be $1$ hour after the initial measurement?

Answer type: simple

Answer: $160000$

Workings:

$1$ hour $= 3\times 20$ minutes

The population doubles every $20$ minutes, so:

After $20$ minutes, the population is $20000 \times 2 = 40000$

$20$ minutes later, the population is $40000 \times 2 = 80000$

$20$ minutes later (after $1$ hour in total), the population is $80000 \times 2 = 160000$

3 marks

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Compound Growth FSQ4

Jackie is starting a new job.

 

Compound Growth FS4(a)

Her salary is $£24000$ and increases by $6\%$ per year.

How much will her salary be after $1$ year?

 

Answer type: simple

Answer: (£) 25440

Workings:

$6 \% = 0.06$ so a $6 \%$ increase can be calculated as:

$(1+0.06)\times 24000 = £25440$

2 marks

 

Compound Growth FS4(b)

Using your answer to part(b), calculate Jackie’s salary after $2$ years in her new job.

Answer: (£) 26966.40

Workings:

$6 \% = 0.06$ so a $6 \%$ increase can be calculated as:

$(1+0.06)\times 25440 = £26966.40$

2 marks

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Compound Growth FSQ5

A company is expanding its team at a rate of $10$ per year.

 

Compound Growth FS5(a)

Calculate how many members of staff there will be after $1$ year if there are initially $200$ members of staff.

 

Answer type: simple

Answer: 220

Workings:

$10 \% = 0.1$ so a $10 \%$ increase can be calculated as:

$(1+0.1)\times 200 = 220$

2 marks

 

Compound Growth FS5(b)

Using your answer from part (a), how many staff members will there be after another year?

Answer type: simple

Answer: 242

Workings:

$10 \% = 0.1$ so a $10 \%$ increase can be calculated as:

$(1+0.1)\times 220 = 242$

2 marks

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Compound Growth FSQ6

A house is bought for $£350000$

Its value increases by $4.4$ per year.

 

Compound Growth FS6(a)

Calculate the value of the house after $1$ year

 

Answer type: simple

Answer: (£) 365400

Workings:

$4.4 \% = 0.044$ so a $4.4 \%$ increase can be calculated as:

$(1+0.044)\times 350000 = £365400$

2 marks

 

Compound Growth FS6(b)

Using your answer from part (a), calculate the value of the house after another year.

 

Answer type: simple

Answer: (£) 381477.60

Workings:

$4.4 \% = 0.044$ so a $4.4 \%$ increase can be calculated as:

$(1+0.044)\times 365400 = £381477.60$

2 marks

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