Set Model PSPM KMK
Section A [25 marks]
This section consists of 3 questions. Answer all the questions given.
1.
A geometric series has the first term 1 and common ratio r where r < 1. Given that the
sum of the first two terms is 13 times the second term. Find the possible value of r.
Find the fourth term and sum of the first fourth terms.
[8 marks]
2.
Three years ago, Akmal deposited in a Kaya Bank that offered simple interest rate at
6% per annum. The balance of his account today is RM11,500.
3.
(a)
Find the principal value. Hence, calculate the interest obtained.
[4 marks]
(b)
Find the accumulated amount of saving after 18 months from today.
[3 marks]
(c)
If today, he withdraws all his money to from account to invest in Kedah Bank
at the interest rate of 6% compounded monthly. How long does it take for
Akmal’s investment to become RM30,000.
[3 marks]
Ahmad needs RM 20 000 to pay down payment for a new car.
(a)
How much should he borrow from a City Bank which charges simple discount
rate of 3.25% for 285 days?
[3 marks]
(b)
Hence, find the amount of the bank discount.
(c)
Determine the equivalent simple interest rate to the simple discount rate?
[2 marks]
[2 marks]
Section B [75 marks]
This section consists of 7 questions. Answer all the questions given.
1.
(a)
Solve 32 x 1  9  3x 3  3x .
(b)
Simplify


2
2  1 . Hence, determine the exact value of
[6 marks]

1

2 1
2



2
2 1 .
[7 marks]
2.
(a)
Solve log4 x  log x 16  3, x  0 .
(b)
Given that


[6 marks]
1
1
ln 2 y 2  9   lnx  4  0 , express x in terms of y.
2
2
[5 marks]
Set Model PSPM KMK
3.
4.
5.
6.
7.
Solve the following inequalities:
(a)
42 x  3  x  47  5x
[3 marks]
(b)
5  x (2 x  1)  0
[4 marks]
(c)
Hence, determine the solution set that satisfying both inequalities of part (a)
and part (b).
[2 marks]
(a)
Determine the solution set for the inequality x  1  5  6 
(b)
Solve 2 x  3  x  4 .
1
x.
2
[5 marks]
[3 marks]
In an arithmetic series, the eighth terms is 10.5 and the sum of the first four terms is 75.
(a)
Find the first term and the common difference.
[5 marks]
(b)
Hence, find the smallest value of n such that S n  0 .
[6 marks]
(c)
Find the sum of the fourth term till the ninth term.
[3 marks]
Miss Joyah deposited RM 1500 at the end of each month for 10 years into a savings
account paying 2.1% interest compounded monthly.
(a)
However, she deposited an additional RM 1000 at the end of the seventh year.
How much money was in the account at the end of the 10th year?
[7 marks]
(b)
Unfortunately, Miss Joyah withdrew RM 1000 at the end of seventh year instead
of depositing it. How much money was in the account at the end of 10th year?
[3 marks]
(c)
Obtain the effective interest rate.
[2 marks]
Ah Juan purchases a land for RM 1 million by taking a loan from a bank with monthly
payments for 25 years at 4.2% interest compounded monthly.
(a)
What is his monthly payment?
(b)
Suppose that at the end of 5 years the loan is changed to a 10 year term for the
remaining balance. What is the new monthly payment?
[7 marks]
END OF QUESTION PAPER
[3 marks]
Set Model PSPM KMK
Suggested answer:
Section A
1
1
1885
, T4 
, S4 
12
1728
1728
1.
r
2.
(a)
(b)
(c)
RM 9745.76, RM 1754.24
RM 12 377.12
16.02 years
3.
(a)
(b)
(c)
RM 20 528.17
RM 528.17
3.34%
Section B
1.
(a)
(b)
x = -1 , x = 2
3 2 2 , 6
2.
(a)
x = 4 , x = 16
e
x  4
2
2y  9
(b)
3.
(c)
5 
 ,
2 
1

  ,   5,  
2

x : x  5
(a)
x : x  4  x  0
(b)
7 

 x : x  ,1
3 

(a)
a = 21, d = 
(b)
30
153
2
(a)
(b)
4.
5.
(c)
3
2
6.
(a)
(b)
(c)
RM 201 166.39
RM 199 036.46
2.12%
7.
(a)
(b)
RM 5 389.42
RM 2 976.13