(a) Sasha invests £1000 at a rate of 5% per year compound interest. Sasha says
After one year, my investment will get £50 in interest and will be worth £1050. Therefore, after two years, my investment will get another £50 in interest and will be worth £1100.
Is Sasha correct? Give a reason for your answer. [1]
(b) Sasha buys a car. The value, £\(V\), of the car after \(n\) years is given by the formula
\(V = a \times b^n\).
The graph shows some information about the value of the car.
Not to scale
Find the value of \(a\) and the value of \(b\). [4]
Mark scheme (a)
Answer
Marks
Part marks and guidance
No oe AND correct valid reason or correct supporting values e.g. • The value of the interest changes each year as the amount grows • It is exponential growth • Compound interest means the interest grows each year
1
e.g. Accept e.g. • There will 5% interest on the £50 as well as an extra £50 oe • It will increase by 5% of 1050 • Finds £1102.5[0] or 102.5[0] or 52.50 for 2nd year
If they show a calculation in their reason it must be correct
See appendix 3
Appendix 3: Question 12(a)
Response
Mark
A
No, compound interest does not increase by the same amount each year, just the same %
1
B
No, compound interest increases exponentially
1
C
No, it is 5% of the amount at the end of the first year
1
D
No, The 5% is calculated on the previous year
1
E
No, it is 5% of the 2nd year, not the first again
1
F
Incorrect, interest is taken on the total including added interest
1
G
Incorrect, she gets 5% of the new amount
1
H
No it is compound interest not simple interest
0
I
No, She has used simple interest not compound interest
0
J
No its not simple interest
0
K
No in the second year the interest is more than £50
0
L
No she gets £51 interest in 2nd year (incorrect value for calculation)
0
Mark scheme (b)
Answer
Marks
Part marks and guidance
[\(a =\) ] 8000
[\(b =\) ] 0.8
4
B1 for [\(a =\) ] 8000
AND
B3 for [\(b =\) ] 0.8 oe or M2 for \(\frac{6400}{8000}\) oe or 80% oe or \(\frac{8000 - 6400}{8000}\) oe 0.2 oe
Allow M2 for e.g. \(a = 0.8\)
M2 for e.g. 20%
or M1 for \(6400 = a \times b^{[1]}\) soi or better
e.g. For M1 \(6400 = \textit{their } a \times b^{[1]}\) seen or \(6400 = 8000 \times b^{[1]}\) For M1 accept \(8000 - 6400 = 8000b^{[1]}\) seen
