Page 25 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4
Chapter 10 Index Numbers Additional Mathematics

(d) Solution:
2015 2017 2019 (a) For salt,
1.00
123.15 116 x = × 100
0.80
116 = 125
I2019 = × 123.15
2015 100 For sugar,
= 142.854 101.5 = y × 100
≈ 142.85 2.00
The corresponding cost of all the items in y = 101.5 × 2.00
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the year 2019 based on the year 2015 is 100
RM142.85. = 2.03
Try question 7 in Formative Zone 10.2 (b) ∑I w = 128.5(60) + 125(25) + 101.5(15)
i i
+ 114(20)
= 14 637.5
∑w = 60 + 25 + 15 + 20
i
= 120
The index number of the T based on T :
2 0 – ∑Iw
I T 2 T T T I = i i
I T 2 = T 1 × I T 1 0 1 2 ∑w i
100
T 0 T 0 14 637.5
I T 1 I T 2 =
T 0 T 1
120
= 121.9792
Example 15 ≈ 121.98
The table below shows the prices and the price (c) Given the composite index for the cost of
indices of four items used in the production of a making the crackers increased by 30% from
packet of cracker. the year 2013 to the year 2019.
130
Price (RM) Price index in
per kg the year 2019
Item
based on the 2013 2016 2019
2016 2019
year 2016
Flour 2.00 2.57 Increase 28.5% 121.98
(i) Price index in the year 2016 based on
Salt 0.80 1.00 x the year 2013:
Sugar 2.00 y Increase 1.5% I2019
I2019 = 2016 × I2016
Oil 5.04 3.5 Increase 14% 2013 100 2013
121.98 CHAP
Given the usage of item flour, salt, sugar and oil 130 = 100 × I2016 10
2013
are 60 g, 25 g, 15 g and 20 g respectively. I2016 = 106.5748
(a) State the values of x and y. 2013
(b) Calculate the composite index for the cost ≈ 106.57
of making the crackers in the year 2019 (ii) The cost of making a crackers in the
based on 2016. year 2016:
(c) It is given that the composite index for the P
2019
cost of making the crackers increased by 54 × 100 = 106.57
30% from the year 2013 to the year 2019. P = 57.5478
(i) Calculate the composite index for the 2016
cost of making a packet of cracker in the ≈ 58 cent per packet
year 2016 based on the year 2013. Number of packet of cracker in 2019
(ii) The cost of making a crackers is 54 = RM120 ÷ RM0.58
cent per packet in the year 2013. Find = 206.90
the maximum number of the packet ≈ 207 packets
of cracker that can produced using an Try question 8 in Formative Zone 10.2
allocation of RM120 in the year 2016.

1 8.2.12 205
0
.
2
.