Question 1
1(a) Stephanie is given a 7\% pay rise. She is currently earning £24,000 a year.
Select the calculation which works out her new salary.
ANSWER: Multiple choice (Type 1)
A: 24,000 \times 0.07
B: 24,000 \times 1.07
C: 24,000 \div 1.07
D: 24,000 \div 1.07
Answer: B
Workings:
£24000\times 1.07 = £25680Marks = 1
1(b) Assuming Stephanie gets the same pay rise every year, select the calculation which calculates Stephanie’s salary after 5 years of working there.
ANSWER: Multiple choice (Type 1)
A: 1.07 \times 24,000^5
B: 24,000 \times 5^{1.07}
C: 24,000 \times 1.07^5
D: 24,000^{1.07} \times 5
Answer: C
Workings:
24000\times 1.07^5 = £33661.24</p> <p><strong>Marks</strong> = 1</p> <p> </p> <p>1(c) James starts on a salary of [latex]£23,500. He gets a raise of 6\% every year.
What would James' salary be after 9 years?
ANSWER: Multiple choice (Type 1)
A: £36,190.00
B: £39,702.76
C: £224,190.00
D: £24,910.00
Answer: B
Workings:
£23,500 \times 1.06^9 = £39,702.76
Marks = 1
Question 2
2(a) A car loses 4\% of its value every year.
In 2019, the car is worth £12,000.
How much will the car be worth in 2024?
ANSWER: Simple text answer
Answer: £9784.47
Workings:
£12,000 \times 0.96^5 = £9784.47
Marks = 2
2(b) How much was the car worth in 2018 assuming its rate of loss was consistent?
ANSWER: Simple text answer
Answer: £12500
Workings:
£12000 \div 0.96 = £12,500
Marks = 2
Question 3
Rebecca invests £1000 at a compound interest rate of R\% per annum.
The value, V, of this investment after n years is given by the formula:
V = 1000 \times 1.065^n
3(a) What is the value of R?
ANSWER: Simple text answer
Answer: 6.5\%
Workings:
(1.065-1)\times 100 = 6.5 \%Marks = 1
3(b) What will the value of her investment be in 7 years time?
ANSWER: Simple text answer
Answer: £1553.99
Workings:
£1000 \times 1.065^7 = £1553.99
Marks = 2
3(c) Rebecca wants to double her money.
How many years will she have to wait?
ANSWER: Simple text answer
Answer: 12
Workings:
£1000 \times 1.065^{12} = £2129.10(Calculate by trial and error)
Marks = 2
Question 4
The number of electric cars, N in the United Kingdom after n years is given by the formula:
N = a \times r^n
Where a is the initial number of cars in a given year.
In 2016, there were 10,000 electric cars in the United Kingdom.
In 2019, there were 150,000 electric cars.
4(a) Calculate the value of r.
Give your answer to 2 decimal places.
ANSWER: Simple text answer
Answer: 2.47
Workings:
N = a\times r^n so 150,000 = 10,000 \times r^3
r^3 = \dfrac{150,000}{10,000} = 15
r = \sqrt[3]{15} = 2.47
Marks =3
4(b) Given that there were 150,000 electric cars in 2019, using the answer from part(a), how many electric cars would you expect there to be in 2021?
ANSWER: Simple text answer
Answer: 915135
Workings:
N = a \times r^n = 150,000 \times 2.47^2 = 915,1135
Marks = 2
Question 5
The number of bacteria in a petri dish grew exponentially.
There was 500 in the original bacterial population.
After 5 hours, the number increased to 121,500.
Calculate how many bacteria there was after 8 hours.
ANSWER: Simple text answer
Answer: 3280500
Workings:
121500 \div 500 = 243
\sqrt[5]{243} = 3
500 \times 3^8 = 3280500
Marks = 3
Question 6
A sunflower grows 12\% taller each week.
Currently the sunflower is 2 m tall.
Chris estimates the height after 5 weeks using the following calculation:
12\% of 2 m is 24 cm
5 \times 24 cm = 120 cm
So the plant is 200 cm + 120 cm = 320 cm
6(a) Is Chris’ estimate an overestimate or underestimate?
Give your answer as a single word, either "overestimate" or "underestimate".
ANSWER: Simple text answer
Answer: underestimate
Marks = 1
(b) What is the actual height of the Sunflower after 5 weeks?
Give your answer in metres.
ANSWER: Simple text answer
Answer: 3.52 m
Workings:
2 m \times 1.12^5 = 3.52 m
Marks = 2
Question 7
£4000 is invested in a fund which earns 11\% compound return per year.
After 5 years, you remove half the balance, leaving the remainder in the fund.
How much would the fund be worth after 10 years.
ANSWER: Simple text answer
Answer: £5678.84
Workings:
£4000 \times (1 + \dfrac{11}{100})^5 = £6740.23262
\dfrac{£6740.23262}{2} = £3370.11631
£3370.11631 \times (1 + \dfrac{11}{100})^5 = £5678.84
Marks = 4