All new questions.
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 \\=2575002 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 \\=2652252 marks
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 = £17712 marks
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 = 400001 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
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 = £254402 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 = £26966.402 marks
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 = 2202 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 = 2422 marks
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 = £3654002 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 = £381477.602 marks