Section A: (43 marks)

2 x  3 y  3
1. Solve  . (4 marks)
5 x  3 y  45

y5
2. Make y the subject of the formula x  . (3 marks)
y2

17-18 F.3 1st TERM UT1-MATH- 2

3. Simplify the following expression and express your answer with positive indices. (5 marks)
4u 3 2
(a) ( )
t2
x 2 y 0 3
(b) ( )
6 x 1 y

4. (a) Write down the place value of the digit underlined in 100112.
(b) Express FAD16 in expanded form and convert it into a decimal number.
(3 marks)

5. Factorize the following expressions.

(a) 81x2  16
(b) 16h2  24hk  9k 2 (4 marks)

17-18 F.3 1st TERM UT1-MATH- 3

6. Factorize the following expressions.
(a) x2  10 x  16
(b) 10m2  23mn  21n2 (4 marks)

7. On a school’s sports day, each student can only participate in one event. The following table
shows the number of participants from F.3 in each event.

Event 100m 200m 400m 800m 3000m

Number of participants 50 20 16 48 26
(a) Find the total number of F.3 students.
(b) Find the experimental probability that a F.3 student participate in 800m. (3 marks)

8. Two fair dice are thrown.

(a) List out all possible outcomes in the following table.
Dice I
1 2 3 4 5 6
1
2
Dice II

3
4
5
6
(b) Find the probability that
(i) the sum of two numbers obtained from the two dice is larger than 8,
(ii) the sum of two numbers obtained from the two dice is a prime number. (5 marks)

17-18 F.3 1st TERM UT1-MATH- 4

9. The number of cells increases by 40% every minute. If there are 500 000 cells at presents,
(a) find the number of cells after 1 minute,
(b) find the number of cells after 50 minutes. (5 marks)

10. It is given that the speed of light is 3  108 m/s. Take 1 year = 365 days. If a star that is
2.346  1014 km away from the Earth is born, how long after it is born would we see it from the
Earth? Give your answer in scientific notation, correct to 3 significant figures. (4 marks)

11. If n is an integer, simplify 5n1  252n3 . (3 marks)

17-18 F.3 1st TERM UT1-MATH- 5

Section B: (27 marks)

12. Factorize the following expressions.

(a) 32u3 108v3
(b) 15(3k  2)2  22(3k  2)  8 (7 marks)

13. If a mobile phone was depreciated 20% each year in the past two years and its value is $4 992
now.
(a) Find the value of the mobile phone two years ago.
(b) If the value of this mobile phone is $3 120 after two years, find the overall percentage
change of the value of this mobile phone over these four years.
(6 marks)

17-18 F.3 1st TERM UT1-MATH- 6

14. The figure shows a right-angled triangle dartboard with shaded region in it. Now a dart hits a
point at random on the dartboard. Given that the dart does not hit the boundaries, AB = 2 cm
BC = 4 cm and CD = 6 cm.

2 cm

B 4 cm C 6 cm D

(a) (i) Find the probability that the dart hits the shaded region of the dartboard.
(ii) Find the probability that the dart hits the white region of the dartboard.
(4 marks)
(b) Sam will get 50 points if his dart hits the shaded region and 0 points if his dart hits the
white region. Sam has to give 40 points for throwing a dart, is the game favourable to him?
Explain your answer. (3 marks)

17-18 F.3 1st TERM UT1-MATH- 7

15. The interest rate offered by bank A is 4% p.a. on simple interest. The interest rate offered by
bank B is 3% p.a. and the interest is compounded quarterly. Mr. Chan wants to deposit
$400 000 in one of the two banks for 6 years.
(a) Which bank should he choose so that he can receive more interest?
(b) If he deposits the money in bank B , when will be the amount more than $500 000?
(7 marks)

End of Paper

17-18 F.3 1st TERM UT1-MATH- 8

2017-18 F.3 Maths 1st term UT1 marking

2 x  3 y  3
1. 
5 x  3 y  45

2x + 3y = –3 . . . . . . . . . . . . (1)
5x – 3y = 45 . . . . . . . . . . . . (2)
(1) + (2) 1M

7x = 42
x=6

1A
sub x = 6 into (1) ,
2(6) + 3y = –3 1M

3y= –15
y = –5
 The solution of the simultaneous equations is x = 6and y = – 5. 1A

2.

y5
x
y2
x( y  2)  y  5 1M

xy  2 x  y  5
xy  y  2 x  5 1M
y ( x  1)  2 x  5
2x  5
y 1A
x 1
4u 3 2 16u 6
3. (a) ( )  4 1M + 1A
t2 t
(b)
x 2 y 0 3 x3 3
( )  ( )
6 x 1 y 6y
1M + 1M
6y
 ( 3 )3
x
216 y 3
 1A
x9

4. (a) 22= 4 1A

FAD16  15 162  10 161  13 160 1M

(b)
 401310 1A
5. (a)

81x 2  16
 (9 x) 2  42 1M

 (9 x  4)(9 x  4) 1A

(b) 16h2  24hk  9k 2 = (4h + 3k)2 1M + 1A

6.
(a) x2  10x  16  ( x  8)( x  2) 1M + 1A
(b)

10m2  23mn  21n2

 (10m2  23mn  21n2 )
 (10m  7n)(m  3n) or (7n  10m)(m  3n) 1M + 1A

7.(a) total number = 160 1A

48 3
(b) the required probability  
160 10 1M + 1A

8. (a)
Dice I
1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
Dice II

3 3,1 3,2 3,3 3,4 3,5 3,6 1A + 1A

4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6

10 5
(b)(i) P(sum > 8)   1A
36 18

(b)(ii) P(sum = prime no.)

1 2  4  6  2
 1M
36
5
 1A
12

9. (a) the number of cells after 1 minute = 500000  (1  40%)  700000

(b) the number of cells after 50 minutes
 500000  (1  40%)50
 1.011013
10
The time for star travel to the Earth
2 . 3 46 114km 0 1M

3  1 0m s/
8

2 . 3 46 114km
0
 years 1M +1A
3  1 0km  60 6 0  2 4 3 6 5
5

 2 4 . years
8 (corr. to 3 sig. fig.) 1A

11.

5n 1  252 n 3
1M
 5n 1  52(2 n 3)
 5n 1  54 n 6 1M
 5n 1 4 n 6
 55 n 5 1A

12. (a)

32u 3  108v 3
 4(8u 3  27v 3 ) 1M
 4[(2u )3  (3v)3 ]
 4(2u  3v)(4u 2  6uv  9v 2 ) 1M +1A

(b) Let a = 3k - 2

15(3k  2)2  22(3k  2)  8

 15a 2  22a  8
 (3a  2)(5a  4)
 [3(3k  2)  2][5(3k  2)  4] 1M +1A
 (9k  6  2)(15k  6  4)
 (9k  4)(15k  6) 1M
 3(9k  4)(5k  2)
1A
13.(a)
 4992  (1  20%) 2 1M +1A
value of the mobile phone two years ago
 $7800 1A
(b) overall % change
3 1 2 0 7 8 0 0
  100%
7800
 60%
14. (a)(i)
0.5  2  6
P(shaded region) = 1M
0.5  2 10
 0.6
1A
(a)(ii)
1M
P(white region) = 1  0.6
 0.4 1A

(b) Expected value = 0.6  50  0.4  0

1M
= 30
<40 1A

∴ not favourable.
1

15.(a)
The amount received after 5 years in bank A
 $400 000 1  4%  6  1M

 $496000 1A

The amount received after 5 years in bank B

6 4
 3%  1M
 $400 000 1  
 4 
1A
 $478565.4
 $496000

 In order to receive more interest, Mr. Chan should choose bank A. 1A

(b)
29
 3% 
Amount after 29 quarter  $400 000 1  
 4 
 $496782.8 1M
30
 3% 
Amount after 30 quarter  $400 000 1  
 4 
 $500508.7

 after 30 quarter , i.e. 7.5 years , the amount in bank B > $500 000. 1