SPM Additional Mathematics 2017, Paper 2 (Question 4 – 6)


Question 4:
Diagram 2 shows part of a rectangular wall painted with green, G, blue, B and purple, P subsequently.
The height of the wall is 2 m. The side length of the first coloured rectangle is 5 cm and the side length of each subsequent coloured rectangle increases by 3 cm.


It is given that the total number of the coloured rectangles is 54.
(a) Find
(i) the side length, in cm, of the last coloured rectangle,
(ii) the total length, in cm, of the painted wall.
(b) Which coloured rectangle has an area of 28000 cm2?
  Hence, state the colour of that particular rectangle.

Solution: 
(a)
5, 8, 11, …
a = 5, d = 3

(i)
T54 = 1 + (54 – 1)d
= 5 + 53(3)
= 164 cm

(ii)
S n = n 2 ( a+l ) S 54 = 54 2 ( 5+164 )  =4563 cm


(b)
Area of the first rectangle
= 2 m × 5 cm
= 200 × 5
= 1000 cm

Area of the second rectangle
= 200 × (5 + 3)
= 1600 cm

Area of the third rectangle
= 200 × (5 + 3 + 3)
= 2200 cm

1000, 1600, 2200, …
a = 1000, d = 600
Tn = 28 000
a + (n – 1)d = 28 000
1000 + (n – 1)600 = 28 000
600(n – 1) = 27 000
n – 1 = 45
n = 46

The colour of that particular rectangle is green.



Question 6:
Diagram 4 shows a cylindrical container with the length of 20 cm placed on the floor against the wall. W is the point on the edge of the base of the container. It is given that the distance of point W is 2 cm from the wall and 1 cm from the floor.

Karen wants to keep the container in a box with a dimension of 21 cm × 7 cm × 7 cm.
Determine whether the container can be kept in that box or otherwise. Give a reason for your answer.

Solution: 


By using Pythagoras’ Theorem, r 2 = ( r1 ) 2 + ( r2 ) 2 r 2 =( r 2 2r+1 )+( r 2 4r+4 ) r 2 6r+5=0 ( r5 )( r1 )=0 r=5 only, r>1 Diameter=2( 5 )=10 cm Diameter>7 cm. Container cannot be kept in that box.


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