**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 2

*y*= 8

*x*+ 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)**

$\begin{array}{l}2y=8x+3\\ y=4x+\frac{3}{2}\end{array}$

*m*= 4

4 = 4 (6) +

*c**c*= –2

Using formula,

*y*=*mx*+*c*equation of the straight line

*PQ*is*y*= 4*x***– 20.**

(b)

(b)

*y*= 4

*x*– 20, at

*x*-intercept

*, y*= 0

4

*x*– 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 $\pi =\frac{22}{7}$ , calculate

(a) the perimeter, in cm, of the whole diagram,

(b) the area, in cm

^{2}, of the shaded region. [6 marks]

*Answer and solution:***(a)**

*KM*

^{2}=

*JK*

^{2}+

*JM*

^{2}

*KM*

^{2 }= 3

^{2}+ 4

^{2}

*KM*

^{2 }= 25

*KM*

^{ }= 5 cm

Perimeter of the whole diagram

=

$\begin{array}{l}=4+3+\left(\frac{1}{4}\times 2\times \frac{22}{7}\times 5\right)+5+\left(\frac{30}{360}\times 2\times \frac{22}{7}\times 4\right)\\ =7+\left(\frac{55}{7}\right)+5+\left(\frac{44}{21}\right)\\ =21\frac{20}{21}\text{}cm\end{array}$
*NJ*+*JK*+*KL*+*LM*+*MN*

(b)

(b)

Area of the shaded region

= Area of

*KLM*+ Area of*JMN*
$\begin{array}{l}=\left(\frac{1}{4}\times \frac{22}{7}\times {5}^{2}\right)+\left(\frac{30}{360}\times \frac{22}{7}\times {4}^{2}\right)\\ =\frac{274}{14}+\frac{88}{21}\\ =23\frac{16}{21}\text{}c{m}^{2}\end{array}$

**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)

(b)

Rate of change of speed of particle

$=\frac{26-0}{13-0}=2m{s}^{-2}$
*J*for the first 13 s**(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 (2

*t*– 13) ] – 13*t*= 169( 26

*t*– 169 – 13*t*) = 16913

*t*= 338

*t*= 26

very useful..thank you very much 🙂