Date | November 2019 | Marks available | 3 | Reference code | 19N.1.SL.TZ0.T_8 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Find | Question number | T_8 | Adapted from | N/A |
Question
Siân invests 50000 Australian dollars (AUD) into a savings account which pays a nominal annual interest rate of 5.6 % compounded monthly.
Calculate the value of Siân’s investment after four years. Give your answer correct to two decimal places.
After the four-year period, Siân withdraws 40000 AUD from her savings account and uses this money to buy a car. It is known that the car will depreciate at a rate of 18 % per year.
The value of the car will be 2500 AUD after t years.
Find the value of t.
Markscheme
FV=50000(1+5.6100×12)12×4 (M1)(A1)
Note: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.
OR
N=4
I%=5.6
PV=(∓)50000
P/Y=1
C/Y=12 (A1)(M1)
Note: Award (A1) for C/Y=12 seen, (M1) for all other correct entries.
OR
N=48
I%=5.6
PV=(∓)50000
P/Y=12
C/Y=12 (A1)(M1)
Note: Award (A1) for C/Y=12 seen, (M1) for all other correct entries.
(FV=)62520.97 (AUD) (A1) (C3)
Note: The final (A1) is not awarded if the answer is not correct to 2 decimal places.
[3 marks]
2500=40000(1−18100)t OR 2500=40000(0.82)t−1 (M1)(A1)
Note: Award (M1) for substitution into compound interest formula or term of a geometric sequence formula and equating to 2500, (A1) for correct substitution.
OR
I%=−18
PV=(±)40000
FV=(∓)2500
P/Y=1
C/Y=1 (A1)(M1)
Note: Award (A1) for FV=(∓)2500 seen, (M1) for all other correct entries. PV and FV must have opposite signs.
OR
32800, 26896, 22054.72, 18084.87, 14829.59, … (M1)
t13=3031.38 and t14=2485.73 (A1)
Note: Award (M1) for a list of at least 5 correct terms beginning with 32800, (A1) for identifying t13 and t14.
(t=) 14.0 (13.9711…) (A1) (C3)
[3 marks]