O Level Math Paper 1 Ans Key
Answer all the questions
The total number of marks for this paper is 80.
1 (a) 6.571
1 (b) 2.93 x 100% = 293 %
2. Kilo = 103
q = 7.056  10-3 or 0.007056
3.
4. (a) Total mass of four boys = 4  66.3 = 265.2 kg
Mass of fourth boy = 265.2 – 68 – 68 – 69 = 60.2 kg
69 kg, 68 kg, 68 kg, 60.2 kg
(b) total mass of five boys = 5 x 68.7 = 343.5 kg
Mass of fifth boy = 343.5 – 265.2 = 78.3 kg
1
5. After 6 years, amount in
6
 1.5 
Plan A : amount to be repaid = 250001 
 = $27 336
 100 
25000  4  6
Plan B : interest =
= 6000
100
Amount to be repaid = 25000 + 6000 = $31 000
Plan B because she will receive more for her investment compared to Plan A.
6. (a) 315  32  5  7
(bi) 315 
32  5  7
90  2  32  5
HCF 
32  5
 45
(bii) 90  45  2 flags
7. (a) Cross sectional area of prism =
42 o
   82
360o
42o
   82  h
0
360
42o
100 
   82  h
360o
100  360
h
42    8 2
Volume of prism =
= 4.26308
h = 4.26 cm (2 decimal places)
(b) Arc length =
42o
 2  8 = 5.86431 cm
360o
Curved surface area = 5.86431  4.26308 = 25.00002267 cm2
Total surface area
42o
=2
   82 + (2  8  4.26308) + 25.00002267 cm2
o
360
= 140.1238 cm2
= 140 cm2 (3 sf)
2
8.
(a) -57 - 36 = -93oC
(b) New melting point =  93  x
New boiling point =  57  y
Temperature difference =  57  y    93  x 
= -57 + y + 93 + x
= 36 + x + y
3
h 
216 27

9. (a)  R  
512 64
 hs 
hR 3

hs 4
3:4
(b)
(c)
3h
4
Radius of X = 3r
81 2
3
2  3h 
r h
Volume of X = 3r    =
5
 4  20
height of X =
3 2 81 2
r h: r h
5
20
3 81
:
5 20
12 : 81
4: 27
10. (a)
5 cm2 : 125 km2
1 cm2 : 25 km2
1 cm : 5 km = 5000 m = 500000 cm
Scale 1: 500 000
(b)
1 cm : 5 km
? cm : 7.86 km
Length of track on map = 1.572 cm
(c)
1 cm2 : 25 km2
1.2 cm2 : 30 km2
3
11. Price per square foot in New York =
370000
= USD 528.57
700
USD 528.57 = 3435.705 CNY
100 m2 = 1075.26882 square feet
Price per square foot in Shanghai =
850000
= 490 CNY
1075.26882
3435.705 > 490,
The apartment in Shanghai is cheaper.
12.
(a) Acceleration =
50  0
= 2.5 m/s2
20  0
(b) Distance from t = 0 to t = 20,
1
 20  50 = 500m
2
Distance from t = 20 to t =40,
50  20 = 1000 m
Total distance = 1500 m > 1200 m, therefore time, T is within 20 to 40 s.
Remaining distance,
1200 – 500 = 700m
Let the time be T
(T – 20)  50 = 700
50T – 1000 = 700
50T = 1700
1700
 34s
T=
50
13.
14.
(a)
5y – 2x + 6y – 2 = 11y – 2x – 2
(b)
3p2 – pq – 4q2
(a)
10 min =
= (3p – 4q)(p + q)
1
hr
6
1
Distance A to B = 15  = 2.5 km
6
5 units = 2.5km
4
11 units =
2.5
 11 = 5.5km
5
(b)
Time for half charge =
45
= 22.5 min
2
60 min  $1.50
1.50
 22.5 = $0.5625 = $0.56 (2 dp)
22.5 min 
60
15.
(a)
When x = 0, y = 3.
3 = kan
k=3
y = 3ax
96 = 3a-5
32 = a-5
25 =a5 -5
1
5
  a
2
a=
1
2
(b)(i)
No line of symmetry
5
(b)(ii)
Line of symmetry: x = 0
16.
17
(a)
75
 40 = 30 students
100
(b)
Agree. The median of Class A is higher than the median of Class B.
Agree. The 25 percent of Class A scored 70 marks or less whereas 50% of
Class B scored 70 marks of less.
(a)
1.74  104 kg
(b)
density =
(c)
1000 cm3 weighs 1450  1000 = 1450 000 grams
mass
volume
1.74  107
=
1.2  10 4
= 1450 g / cm3 = 1.45  103 g /cm3
Remaining mass = 1.74  10 7 - 1450 000 = 15 950 kg
18
(a)
5 – x < 4x – 1
-5x < -6
6
x>
5
x > 1.2
,
4x – 1 ≤ 3x + 1
x≤2
1.2  x  2
1.2
2
6
(b) 2
19
(a)
(b)
20
21
2a 1b 1

4b
2a
43
42a
3h = 3 – 2a
3  2a
h
3
4 3h 
(a)
– (x – 4)2 + 3 = 0
x4 3
x intercepts: 2.27 or 5.73 ( 3 sf)
y intercepts : - 13
(b)
turning point (4, 3)
(c)
x=4
(a) AB  AO  OB
  9    4    13 
     

= 
  12   1    11
(b)
 132   112 = 17.0 units
7
22
(a) exterior angle = 180 – 156 = 24o
(b) number of sides = 360 / 24 = 15
23
(a)
(b)
1
2
cos RTU   cos RTS
1

2
Legend for question paper:
(Level of difficulty)
[s] – Simple
[m] – Moderate
[c] – Challenging
8