ADDITION MATHEMATIC

PROJECT WORK2014

NAME : CHEN SHEE XIONG

I/C NUMBER : 970203-01-6457

FORM : 5SN2/2014

TEACHER : MISS TAN KEAT

HONG

CONTENTNO

TOPIC PAGE

1 Content 12 Introduction 23 Part A

a)Way of using money 3-5b)Collecting Data : Amount of saving from

all 50 classmates receiving the aid. 6

C) i)Statistical Graph 7-9ii)Calculation of mean and standard

deviation of total savings. 10

d)Calculation of new mean and new variance of the new total savings. 11

4 Part B*Total saving after 10 years in Bank X. 12*Total saving after 10 years in Bank Y. 13-14*Comparison between Bank X and Bank Y. 14Part Ca)Plot a graph of y against x. 15b)Plotting graph with best fit line. 16c)Calculation :

17-18i)Value of k and p.ii)Mother’s savings when Ali’s savings is Rm55.00

5 Further Explanation 19-206 Reflection 21

1

INTRODUCTION On October 25, 2013, the Prime Minister cum finance minister, Datuk Seri Najib Tun Razak has approved an allocation of RM540 million for Bantuan Khas Awal Persekolahan 1 Malaysia while presenting the Budget Year 2014. All student from Year 1 to Form 5 for 2014 school session will receive RM100 each.

The main objection of Bantuan Khas Awal Persekolahan 1 Malaysia is to give the allowance to the students for the daily expenses. This is because the students need an amount of money to prepare for school opening. These money can help to reduce the burden or the family that faced financial problems.

However, demand for data from survey, students used the RM100 for several purposes. Bantuan Khas Awal Persekolahan 1 Malaysia has became an invaluable source of information for school authorities and government.

2

PART Aa)Way of Using Money

Item Amount Percentage(%)

Saving 30.00 30

Education 50.00 50

Other Expenses

20.00 20

Total 100.00 100

3

i)Presentation of The Uses of Rm100 by A Student In Different Statistical Graph.

Saving30%

Education50%

Other Exprnses20%

Saving

Education

Other Exprnses

4

ii)Percentage of Usage of Money

Savings Education Other Expenses0

10

20

30

40

50

60

Item

Percentage(%)

5

b)Assume that the minimum saving is RM1.00

Total Savings(Rm

)

≤10

≤20

≤30

≤40

≤50

≤60

≤70

≤80

≤90

≤100

Number of Pupils 10 7 10 4 7 3 6 1 1 1

Total Savings(RM) Number of Puplis1-10 1011-20 721-30 1031-40 441-50 751-60 361-70 671-80 181-90 191-100 1

6

c)i)Histogram

5.5 15.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.50

1

2

3

4

5

6

7

8

9

10

Total Savings(RM)

Number of Puplis

7

Bar Chart

Graph of number of pupils against total savings

1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-1000

1

2

3

4

5

6

7

8

9

10

Total Savings(RM)

Number of Pupils

8

Pie Chart

1-1020%

11-2014%

21-3020%

31-408%

41-5014%

51-606%

61-7012%

71-802%

81-902%

91-1002%

Total Savings(%)

1-10

11-20

21-30

31-40

41-50

51-60

61-70

71-80

81-90

91-100

9

c)

ii)Mean x̅ = (5.5×10 )+ (15.5×7 ) + (25.5×7 )+ (35.5×4 )+ (45.5×7 )+ (55.5×3 ) + (65.5×6 ) (75.5×1 ) (85.5×1 ) (95.5×1 )

50

= 1695 50

=33.9

Standard Deviation, σ = √ 103332.550

− (33−9 )

= RM30.29

∴ Mean is the average of the amount of the sample value. The mean of total of amount of savings from 50 classmates receiving the aid (RM) is RM33.90.

∴ Standard deviation is a measure of how spread out number are. In this case, the standard deviation of the amount of savings from 50 classmates receiving the aid is RM30.29.

10

d)

New mean = 33.9 + 50

= 83.9

New variance , σ 2 = 400802.7550

– (83.9)2

= 976.845

11

PART BBank X

a)Method 1:

Year RM1 103.002 106.093 109.274 112.555 115.936 119.417 122.998 126.689 130.4810 134.39

b)Method 2:

a = 103 r = 1.03

T10 = 103(1.03)9

=134.39

12

Bank Y1 year = 12 months10 years = 12 x 10 = 120 monthsMethod 1:

Months Saving(RM) Months Saving(RM)

Months Saving(RM)

1 100.00 43 111.06 85 123.342 100.25 44 111.33 86 123.643 100.50 45 111.61 87 123.954 100.75 46 111.89 88 124.265 101.00 47 112.17 89 124.576 101.26 48 112.45 90 124.887 101.51 49 112.73 91 125.208 101.76 50 113.01 92 125.519 102.02 51 113.30 93 125.8210 102.27 52 113.58 94 126.1411 102.53 53 113.86 95 126.4512 102.78 54 114.15 96 126.7713 103.04 55 114.43 97 127.0914 103.30 56 114.72 98 127.4015 103.56 57 115.01 99 127.7216 103.82 58 115.29 100 128.0417 104.08 59 115.58 101 128.3618 104.34 60 115.87 102 128.6819 104.60 61 116.16 103 129.0120 104.86 62 116.45 104 129.3321 105.12 63 116.74 105 129.6522 105.38 64 117.04 106 129.9823 105.65 65 117.33 107 130.3024 105.91 66 117.62 108 130.6325 106.18 67 117.92 109 130.9526 106.44 68 118.21 110 131.2827 106.71 69 118.51 111 131.6128 106.97 70 118.80 112 131.9429 107.24 71 119.10 113 132.2730 107.51 72 119.40 114 132.6031 107.78 73 119.69 115 132.9332 108.05 74 119.99 116 133.2633 108.32 75 120.29 117 133.5934 108.59 76 120.59 118 133.9335 108.86 77 120.90 119 134.2636 109.13 78 121.20 120 134.6037 109.41 79 121.5038 109.68 80 121.8139 109.95 81 122.1140 110.23 82 122.4241 110.50 83 122.7242 110.78 84 123.03

13

Bank Y

Method 2: T1 = 100.250

T2 = 100.501

T3 = 100.752

r = 1.0025 a = 100 n = 120

T120 = 100(1.0025)119

= 134.60

∴ The total saving after 10 years in Bank Y is RM134.60.

Compare with interest in Bank X (3% in year) and Bank Y (0.25% in month) , the amount of savings after 10 years in Bank Y (RM134.60) is higher than Bank X (RM130.48).

14

PART Ca)

Graph of y against x

x 25 35 51 65 77 90 100y 50 51 52 54 61 67 73

20 30 40 50 60 70 80 90 100 1100

10

20

30

40

50

60

70

80

x

y

15

b)

x2 625 1205 2601 4225 5929 8100 10000xy 1250 1785 2652 3510 4697 6030 7300

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 11000 120000

1000

2000

3000

4000

5000

6000

7000

8000

x2

xy

16

From graph a, it shows the curve line. It unable to show the information directly.

From graph b, it able to show the information directly as it is a straight line.∴ Compare with graph a and graph b, graph b is more suitable to show the information.

c)i) py=1x + kx

17

y= 1px

+ kxp

y= kp x+1px

Y=mX+c

From graph , y – intercept = 700

1p=700

p= 1700

p=0.0143

ii) from graph , gradient, m

¿ 7300−70010000−0

¿ 660010000

¿0.66

kp=m

k0.0143

=0.66

k=0.0944

ii)Ali’s savings = RM55.00

Muthu’s savings = RM43.00

FURTHER18

EXPLORATIONFrom the information that I have been searched, the students are recommended to activate an account, which is junior saving account. I have compared with RHB Islamic Bank and CIMB Bank.

Table below shows the comparisms between two banks:

RHB Islamic Bank CIMB BankAttraction and competitive

profit sharing ratio.Has higher interest rate than

normal saving account.

Monthly dividend paymentEnjoy up to 100% bonus per

annual (save regularly via Periodic Payment Instruction)

(PPI)Take Takaful personal

protection coverage up to RM5000 (Based on the value

of saving account)

Enjoy discounts are privileges at participating merchant

outlets.

Required RM1 to activate the account.

Required RM300 to activate the account.

19

From the comparison, I able to distinguish how to choose the bank for activating saving account. If student only receiving RM100 as an aid, it is not enough to open account at CIMB Bank. Moreover, RHB Islamic Bank had offered Takeful protection. As a conclusion, RHB Islamic Bank is recommended for student to have a saving account on it.

REFLECTION20

While doing this project, I have spent countless hours doing this project. I realized that this subject is a compulsory to me. Without it, I can’t fulfill my big dreams and wishes.

I used to hate Additional Mathematics. It always an absolute obstacle to me throughout day and night. I sacrificed my precious time to have fun from Monday, Tuesday, Wednesday, Thursday, Friday and even the weekend that I always looking forward to. From now on, I will do my best on every second that I will learn Additional Mathematics full of effort and I have learnt that not to give up easily when facing the problem.

21