SPM Mathematics (Model Test Paper)



SPM Mathematics (Model Test Paper)

Section A
 (52 marks)
Answer all questions in this section.

7.      Diagram 4 shows a straight line ST and a straight line PQ drawn on a Cartesian plane.  ST is  parallel to PQ. Given that equation of the straight line ST is 2y = 8x + 3.
      

Diagram 4
Find,
(a)  the equation of the straight line PQ,
(b)  the x-intercept of the straight line PQ.
                   [ 5 marks ]
Answer and solution:
(a)
2y=8x+3 y=4x+ 3 2  
m = 4

4 = 4 (6) + c 
c = 2

Using formula, y = mx + c   
equation of the straight line PQ is y = 4x 20. 

(b)
y = 4x 20, at x-intercept, y = 0
4x – 20 = 0,   y = 0
x = 5   or   x-intercept = 5


8.      Diagram 5, shows a quadrant KLM with centre  M and sector JMN with centre J.


Diagram 5        

Using π= 22 7 , calculate
(a)    the perimeter, in cm, of the whole diagram,
(b)   the area, in cm2, of the shaded region.
                                                                                                             [6 marks]
Answer and solution:
(a)
KM2 = JK2 + JM2
KM2 = 32 + 42
KM2 = 25
KM = 5 cm

Perimeter of the whole diagram
= NJ + JK + KL + LM + MN
=4+3+( 1 4 ×2× 22 7 ×5 )+5+( 30 360 ×2× 22 7 ×4 ) =7+( 55 7 )+5+( 44 21 ) =21 20 21  cm

(b)
Area of the shaded region
= Area of KLM + Area of JMN
=( 1 4 × 22 7 × 5 2 )+( 30 360 × 22 7 × 4 2 ) = 274 14 + 88 21 =23 16 21  c m 2


9.      Diagram 6 shows the speed-time graph for the movement of two particles, J and K, for a period of t s. The graph ABCD represents the movement of J and the graph AE represents the movement of K. Both particles start at the same point and move along the same route.

Diagram 6

(a)    State the uniform speed, in ms-1, of particle J.
(b)   Calculate the rate of change of speed, in ms-2, of particle J for the first 13 s.
(c)    At t s, the difference between the distance travelled by J and K is 169 m. Calculate the value of t.
[ 6 marks ]
Answer and solution:
(a)
Uniform speed of particle J = 26 ms-1

(b)
Rate of change of speed of particle J for the first 13 s
= 260 130 =2m s 2

(c)
Given at t s, the difference between the distance travelled by J and K is 169
(distance travelled by particle J) – (distance travelled by particle K) = 169
[ ½ (t – 13 + t) × 26 ] – [ ½ (26) (t )] = 169
[ 13 (2t – 13) ] – 13t  = 169
( 26t – 169 – 13t ) = 169
13t  = 338
t  = 26