(Long Questions) – Question 6

Posted on May 18, 2020 by Myhometuition

Question 6 (10 marks):
Table 2 shows the information related to four ingredients, J, K, L and M, used in the production of a type of energy drinks.

The production cost for this energy drinks is RM47600 in the year 2017.
(a) If the price of ingredient K in the year 2013 is RM4.20, find its price in the year 2017.

(b) Percentage of usage for several ingredients was given in Table 2.
Calculate the corresponding production cost in the year 2013.

(c) The production cost is expected to increase by 50% from the year 2017 to the year 2019.
Calculate the percentage of changes in production cost from the year 2013 to the year 2019.

Solution:
(a)

\begin{array}{l} \frac{P_{2017}}{P_{2013}} \times 100 = 120 \\ \frac{P_{2017}}{4.20} \times 100 = 120 \\ P_{2017} = \frac{4.20}{100} \times 120 \\ P_{2017} = RM 5.04 \end{array}

(b)

\begin{array}{l} Percentage of usage of ingredient L \\ = (100 - 10 - 10 - 50) % \\ = 30 % \\ Composite index for the production cost in \\ the year 2017 based on the year 2013, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(140) (10) + (120) (10) + (160) (30) + (90) (50)}{10 + 10 + 30 + 50} \\ \bar{I} = \frac{11900}{100} \\ \bar{I} = 119 \\ \frac{P_{2017}}{P_{2013}} \times 100 = 119 \\ \frac{47600}{P_{2013}} \times 100 = 119 \\ P_{2013} = 47600 \times \frac{100}{119} \\ P_{2013} = RM40000 \\ The production cost in the year 2013 \\ is RM40000 . \end{array}

(c)

\begin{array}{l} year 2013 \overset{+ 19 %}{\to} year 2017 \overset{+ 50 %}{\to} year 2019 \\ ∴ Expected composite index for the year 2019 \\ based on the year 2013, \\ \bar{I} = 119 \times \frac{150}{100} = 178.5 \\ Percentage of changes \\ = [178.5 - 100] % \\ = 78.5 % \end{array}

(Long Questions) – Question 5

Posted on May 18, 2020 by Myhometuition

Question 5 (10 marks):
Table 2 shows the prices and the price indices of three types of ingredients P, Q and R, used in the production of a type of instant noodle.

(a) The price of ingredient Q increased by 20% from the year 2014 to the year 2016.
(i) State the value of x.
(ii) Find the value of y.

(b) Calculate the composite index for the cost of making the instant noodles for the year 2016 based on the year 2014.

(c) It is given that the composite index for the cost of making the instant noodles increased by 40% from the year 2012 to the year 2016.

(i) Calculate the composite index for the cost of making the instant noodles in the year 2014 based on the year 2012.

(ii) The cost of making a packet of instant noodle is 10 sen in the year 2012.
Find the maximum number of packet of instant noodles that can be produced using an allocation of RM80 in the year 2016.

Solution:
(a)(i)
x = 120

(a)(ii)

\begin{array}{l} \frac{P_{2016}}{P_{2014}} \times 100 = 120 \\ \frac{3.00}{y} \times 100 = 120 \\ y = \frac{3.00}{120} \times 100 \\ y = 2.50 \end{array}

(b)

\begin{array}{l} Composite index for the cost of making the instant noodles \\ in the year 2016 based on the year 2014, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(132.8) (50) + (120) (20) + (190) (1)}{50 + 20 + 1} \\ \bar{I} = \frac{9230}{71} \\ \bar{I} = 130 \end{array}

(c)(i)

\begin{array}{l} Given \frac{I_{2016}}{I_{2012}} \times 100 = 140 \to \frac{I_{2016}}{I_{2012}} = \frac{140}{100} \\ and \frac{I_{2016}}{I_{2014}} \times 100 = 130 \to \frac{I_{2016}}{I_{2014}} = \frac{130}{100} \\ Composite index for the cost of making the instant noodle \\ in the year 2014 based on the year 2012, \\ \bar{I} = \frac{I_{2014}}{I_{2012}} \times 100 \\ \bar{I} = \frac{I_{2014}}{I_{2016}} \times \frac{I_{2016}}{I_{2012}} \times 100 \\ \bar{I} = \frac{100}{130} \times \frac{140}{100} \times 100 \\ \bar{I} = 107.69 \end{array}

(c)(ii)

\begin{array}{l} \frac{P_{2016}}{P_{2012}} \times 100 = 140 \\ \frac{P_{2016}}{10} \times 100 = 140 \\ P_{2016} = \frac{140 \times 10}{100} \\ P_{2016} = 14 sen \\ Number of pack of instant noodles \\ = \frac{RM80}{RM 0.14} \\ = 571.4 \\ Maximum number of pack of instant noodles = 571. \end{array}

(Long Questions) – Question 4

Posted on April 22, 2020 by user

Question 4:

Table below shows the prices and the price indices for the four ingredients P, Q, R and S, used in making biscuits of a particular kind.

Diagram below is a pie chart which represents the relative amount of the ingredients P, Q, R and S, used in making biscuits.

(a) Find the value of x, y and z.

(b)(i) Calculate the composite index for cost of making these biscuits in the year 2008 based on the year 2006.

(ii) Hence, calculate the corresponding cost of making these biscuits in the year 2008 if the cost in the year 2006 was RM 585.

(c) The cost of making these biscuits is expected to increase by 40 % from the year 2008 to the year 2010. Find the expected composite index for the year 2010 based on the year 2006.

Solution:
(a)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ I = \frac{P_{2008}}{P_{2006}} \times 100 \\ 125 = \frac{x}{0.80} \times 100 \\ x = R M 1.00 \\ y = \frac{2.80}{2.00} \times 100 \\ y = 140 \\ 150 = \frac{0.60}{z} \times 100 \\ z = RM0 .40 \end{array}

(b)(i)

Sector P = 360^o– 120^o – 100^o – 60^o = 80^o

\begin{array}{l} Composite index for cost of making biscuits \\ in the year 2008 based on the year 2006, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(125) (80) + (140) (120) + (150) (100) + (80) (60)}{80 + 120 + 100 + 60} \\ \bar{I} = \frac{46600}{360} \\ \bar{I} = 129.44 \end{array}

(b)(ii)

\begin{array}{l} \bar{I} = \frac{P_{2008}}{P_{2006}} \times 100 \\ 129.44 = \frac{P_{2008}}{585} \times 100 \\ ∴ Cost of making biscuit in the year 2008 based on the year 2006 = R M 757.22. \end{array}

(c)

\begin{array}{l} year 2006 \overset{+ 29.44 %}{\to} year 2008 \overset{+ 40 %}{\to} year 2010 \\ ∴ Expected composite index for the year 2010 based on the year 2006, \\ \bar{I} = 129.44 \times \frac{140}{100} = 181.22 \end{array}

(Long Questions) – Question 3

Posted on April 22, 2020 by user

Question 3:

Diagram below is a bar chart indicating the weekly cost of the items A, B, C, D and E for the year 2000. Table below shows the prices and the price indices for the items.

(a) Find the value of

(i) x

(ii) y

(iii) z

(b) Calculate the composite index for items in the year 2006 based on the year 2000.

(c) The total monthly cost of the items in the year 2000 is RM456. Calculate the corresponding total monthly cost for the year 2006.

(d) The cost of the items increases by 15 % from the year 2006 to the year 2009.
Find the composite index for the year 2009 based on the year 2000.

Solution:
(a)(i)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ I = \frac{P_{2006}}{P_{2000}} \times 100 \\ 140 = \frac{1.75}{x} \times 100 \\ x = RM 1.25 \end{array}

(a)(ii)

\begin{array}{l} I = \frac{P_{2006}}{P_{2000}} \times 100 \\ y = \frac{4.80}{4.00} \times 100 \\ y = 120 \end{array}

(a)(iii)

\begin{array}{l} I = \frac{P_{2006}}{P_{2000}} \times 100 \\ 125 = \frac{z}{6.00} \times 100 \\ z = RM7 .50 \end{array}

(b)

\begin{array}{l} Composite index for items in the year 2006 based on the year 2000, \bar{I} \\ \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(140) (15) + (125) (30) + (120) (24) + (125) (33) + (110) (12)}{15 + 30 + 24 + 33 + 12} \\ \bar{I} = \frac{14175}{114} \\ \bar{I} = 124.34 \end{array}

(c)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ Lets total monthly cost for the year 2006 is P_{2006} \\ I = \frac{P_{2006}}{P_{2000}} \times 100 \\ 124.34 = \frac{P_{2006}}{RM 456} \times 100 \\ P_{2006} = RM 567 \end{array}

(d)

\begin{array}{l} Lets composite index for the year 2009 based on year 2000 is {\bar{I}}_{2009} \\ {\bar{I}}_{2009} = 124.34 \times \frac{115}{100} \\ {\bar{I}}_{2009} = 143 \end{array}

(Long Questions) – Question 2

Posted on April 22, 2020 by user

Question 2:

Table shows the prices, the price indices and weightage for four types of stationery A, B, C and D.

(a) Find the value of

(i) x, (ii) y, (iii) z.

(b) The composite index for the price of the stationery in the year 2009 based on the year 2008 is 118.5. Calculate the value of m.

(c) The total expenditure for the stationery in the year 2008 is RM350. Calculate the corresponding total expenditure in the year 2009.

(d) The price index for B in the year 2010 based on the year 2008 is 132. Calculate the price index for B in the year 2010 based on the year 2009.

Solution:
(a)(i)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ x = \frac{P_{2009}}{P_{2008}} \times 100 \\ x = \frac{2}{2.5} \times 100 \\ x = 80 \end{array}

(a)(ii)

\begin{array}{l} 135 = \frac{P_{2009}}{2} \times 100 \\ 135 = \frac{y}{2} \times 100 \\ y = R M 2.70 \end{array}

(a)(iii)

\begin{array}{l} 116 = \frac{5.80}{P_{2008}} \times 100 \\ 116 = \frac{5.80}{z} \times 100 \\ z = RM5 \end{array}

(b)

\begin{array}{l} Composite index for items in the year 2009 based on the year 2008, \bar{I} = 118.5 \\ \bar{I} = \frac{\sum I W}{\sum W} \\ 118.5 = \frac{(80) (2) + (120) (3) + (135) (m) + (116) (4)}{2 + 3 + m + 4} \\ 118.5 = \frac{984 + 135 m}{9 + m} \\ 1066.5 + 118.5 m = 984 + 135 m \\ 16.5 m = 82.5 \\ m = 5 \end{array}

(c)

\begin{array}{l} \bar{I} = \frac{P_{2009}}{P_{2008}} \times 100 \\ 118.5 = \frac{P_{2009}}{350} \times 100 \\ ∴ Total expenditure in the year 2009, P_{2009} = R M 414.75. \end{array}

(d)

\begin{array}{l} For stationery B, \\ It is given \frac{P_{2010}}{P_{2008}} \times 100 = 132 \to \frac{P_{2010}}{P_{2008}} = \frac{132}{100} \\ and \frac{P_{2009}}{P_{2008}} \times 100 = 120 \to \frac{P_{2009}}{P_{2008}} = \frac{120}{100} \\ Price index for stationery B in the year 2010 based on the year 2009, \\ I_{10 / 09} = \frac{P_{2010}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{P_{2010}}{P_{2008}} \times \frac{P_{2008}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{132}{100} \times \frac{100}{120} \times 100 \\ I_{10 / 09} = 110 \end{array}

(Long Questions) – Question 1

Posted on April 22, 2020 by user

Question 1:
Table below shows the price indices and percentage of usage of four items, P , Q , R and S , which are the main ingredients in the production of a type of toy.

(a) Calculate
(i) the price of S in the year 2004 if its price in the year 2007 is RM43.20 ,
(ii) the price index of P in the year 2007 based on the year 2002 if its price index in the year 2004 based on the year 2002 is 110.

(b) The composite index of the cost of toy production for the year 2007 based on the year 2004 is 125.
Calculate
(i) the value of x ,
(ii) the price of the toy in the year 2004 if the corresponding price in the year 2007 is RM75.

Solution:

(a)(i)

\begin{array}{l} 135 = \frac{43.2}{P_{2004}} \times 100 \\ P_{2004} = \frac{43.2 \times 100}{135} = 32 \\ ∴ The price of S in the year 2004 is RM32 \end{array}

(a)(ii)

\begin{array}{l} Given \frac{P_{2004}}{P_{2002}} \times 100 = 110 \to \frac{P_{2004}}{P_{2002}} = \frac{110}{100} \\ and \frac{P_{2007}}{P_{2004}} \times 100 = 125 \to \frac{P_{2007}}{P_{2004}} = \frac{125}{100} \\ Price index of P in the year 2007 based on the year 2002, \\ I = \frac{P_{2007}}{P_{2002}} \times 100 \\ I = \frac{P_{2007}}{P_{2004}} \times \frac{P_{2004}}{P_{2002}} \times 100 \\ I = \frac{125}{100} \times \frac{110}{100} \times 100 \\ I = 137.50 \end{array}

(b)(i)

\begin{array}{l} Given composite index \bar{I} = 125 \\ \bar{I} = \frac{\sum I W}{\sum W} \\ 125 = \frac{(125) (30) + (x) (20) + (115) (10) + (135) (40)}{30 + 20 + 10 + 40} \\ 125 = \frac{10300 + 20 x}{100} \\ 12500 = 10300 + 20 x \\ x = 110 \end{array}

(b)(ii)

\begin{array}{l} Lets price of the toy in the year 2004 is P_{2004} \\ P_{2004} \times \frac{125}{100} = 75 \\ P_{2004} = R M 60 \end{array}

11.2 Composite Index

Posted on April 22, 2020 by user

11.2 Composite Index

Example 1

Noodle	Price index	Weightage
Fried noodle	112	4
Tomyam noodle	104	3
Seafood noodle	109	2
Wantan noodle	111	1

The above table shows the price indices of a few types of noodles sold at a shop for the year 2000 based on the year 1996 and their respective weightages. Calculate the composite index for the year 2000 based on the year 1996.

Solution:
Composite index for the year 2000 based on the year 1996

\begin{array}{l} \bar{I} = \frac{\sum I W}{\sum W} \\ \bar{I} = \frac{(112) (4) + (104) (3) + (109) (2) + (111) (1)}{4 + 3 + 2 + 1} \\ \bar{I} = \frac{448 + 312 + 218 + 111}{10} \\ \bar{I} = \frac{1089}{10} = 108.9 \end{array}

Example 2

Item	Price index	Weightage
P	110	4
Q	x	2
R	120	1
S	115	3

The above table shows the price indices of a few types of items, P, Q, R and S, for the year 2002 based on the year 1997 and their respective weightages. If the composite index for the year 2002 based on the year 1997 is 116.5, find the value of x.

Solution:

\begin{array}{l} \bar{I} = \frac{\sum I W}{\sum W} \\ 116.5 = \frac{(110) (4) + (x) (2) + (120) (1) + (115) (3)}{4 + 2 + 1 + 3} \\ 116.5 = \frac{440 + 2 x + 120 + 345}{10} \\ 116.5 = \frac{905 + 2 x}{10} \\ 1165 = 905 + 2 x \\ 2 x = 260 \\ x = 130 \end{array}

11.1 Index Number

Posted on April 22, 2020 by user

11.1 Index Number

Example 1
The price of the house Sulaiman bought in the year 1998 was RM 110000. In the year 2000, the price of the house went up to RM 130000. Calculate the price index of the house for the year 2000 based on the year 1998.

Solution:

\begin{array}{l} P_{0} = R M 110000 \\ P_{1} = R M 130000 \\ I = \frac{P_{1}}{P_{0}} \times 100 \\ I = \frac{130000}{110000} \times 100 \\ I = 118.18 \end{array}

Example 2
The price of a bottle of cooking oil in the year 1999 was RM15.00. Given that the price index for the year 2000 based on the year 1999 is 105, calculate the price of the cooking oil in the year 2000.

Solution:

\begin{array}{l} P_{0} = R M 15 \\ P_{1} = x {(Price}_{2000}) \\ I = \frac{P_{1}}{P_{0}} \times 100 \\ 105 = \frac{x}{15} \times 100 \\ 1575 = 100 x \\ x = R M 15.75 \end{array}

Example 3
Based on the year 1995, the index number of an oven for the years 1998 and 2001 are 110 and 130 respectively. Find the index number for the year 2001 based on the year 1998.

Solution:

\begin{array}{l} Given, \frac{P_{1998}}{P_{1995}} \times 100 = 110 and \frac{P_{2001}}{P_{1995}} \times 100 = 130 \\ ∴ \frac{P_{2001}}{P_{1998}} \times 100 = \frac{P_{2001}}{P_{1995}} \times \frac{P_{1995}}{P_{1998}} \times 100 \\ = \frac{130}{100} \times \frac{100}{110} \times 100 \\ =118 .18 \end{array}