Hybrid PBD Matematik TG3_Edisi Guru

Matematik Tingkatan 3 Bab 3
4. Hitung jumlah simpanan yang melibatkan faedah mudah dan faedah kompaun. SP 3.1.2 TP3
Calculate the total savings involving simple interest and compound interest.
(a) Hitung jumlah simpanan yang (b) Hitung nilai matang bagi
akan diperoleh selepas 2 tahun simpanan RM15 000 untuk 5 Tip Penting
dengan simpanan asal RM8 000 tahun, dengan kadar faedah 3.5% • Bagi simpanan yang
dan kadar faedah mudah tahunan setahun, dikompaun setiap bulan. memberi faedah mudah,
2%. Calculate the matured value for RM15 000 For savings which give simple
Calculate the total savings after 2 years savings for 5 years, with the interest rate interest,
with the initial savings of RM8 000 and the 3.5% per annum, compounded monthly. I = Prt
yearly simple interest rate of 2%.
Nilai matang dengan keadaan/ where
I = faedah/ interest
Jumlah simpanan Matured value P = prinsipal/ principal
Total savings 0.035 12(5) r = kadar faedah/ interest
1
= prinsipal + faedah = 15 000 1 + 12 2 rate
principal + interest = RM17 864.14 t = masa/ time
2
1
2
= 8 000 + 8 000 × 100 × 2 • Bagi simpanan yang
= RM8 320 memberi faedah
kompaun,
For savings which give
(c) Suraya membuat simpanan (d) Puan Zubaidah membuat compound interest,
sebanyak RM5 000 di suatu bank simpanan sebanyak RM20 000 MV = P 1 + r 2 nt
1
dengan kadar faedah mudah dalam suatu akaun simpanan n
2.5% setahun. Hitung jumlah tetap dengan kadar faedah 4% dengan keadaan/ where
simpanannya selepas 10 bulan. setahun, dikompaun setiap MV = nilai matang
Suraya makes a savings of RM5 000 in a tahun. Hitung nilai matang dalam matured value
bank with the simple interest rate 2.5% per akaun itu selepas 12 tahun. P = prinsipal
annum. Calculate her total savings after 10 Puan Zubaidah makes a savings of principal
months. r = kadar faedah
RM20 000 in a fixed deposit account tahunan
with the interest rate 4% per annum,
yearly interest rate
Jumlah simpanan compounded yearly. Calculate the matured n = bilangan kali faedah
Total savings value in the account after 12 years. dikompaun setahun
= prinsipal + faedah number of periods the
principal + interest Nilai matang interest is compounded
per year
2.5
1
= 5 000 + 5 000 × 100 × 10 2 Matured value 0.04 2 1(12) t = tempoh dalam
1
12
= 20 000 1 +
tahun
= RM5 104.17 1 term in years
= RM32 020.64
5. Selesaikan masalah berikut. SP 3.1.2 TP4
Solve the following problems.
(a) Su Lin menyimpan RM6 000 di sebuah bank dengan kadar faedah mudah 3.2% setahun. Hitung jumlah
simpanan Su Lin selepas dia menyimpan selama
Su Lin deposits RM6 000 in a bank with simple interest rate of 3.2% per annum. Calculate the total amount of Su Lin’s savings after
she has saved for
(i) 3 tahun/ 3 years (ii) 4 tahun/ 4 years
(i) Jumlah simpanan/ Total savings (ii) Jumlah simpanan/ Total savings
= prinsipal + faedah / principal + interest = prinsipal + faedah / principal + interest
3.2
3.2
2
1
2
1
= 6 000 + 6 000 × 100 × 3 = 6 000 + 6 000 × 100 × 4
= RM6 576 = RM6 768
Semakin lama tempoh simpanan (di bank), semakin tinggi jumlah faedah yang diperoleh.
Dengan ini, jumlah simpanan juga bertambah .
The longer the savings period (at the bank), the higher the total interest earned. With this, the total savings also
increases .




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03_ Hybrid Mate Tg3.indd 27 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
(b) Encik Lee menyimpan RM2 000 di sebuah bank. Berapakah jumlah simpanan Encik Lee selepas 1 tahun
jika kadar faedah mudah yang diberikan ialah
Mr Lee deposits RM2 000 in a bank. What is the total amount of Mr Lee's savings after 1 year if the simple interest rate given is
(i) 2% setahun/ 2% per annum (ii) 3% setahun/ 3% per annum
(i) Jumlah simpanan/ Total savings (ii) Jumlah simpanan/ Total savings
= prinsipal + faedah / principal + interest = prinsipal + faedah / principal + interest
2
3
2
1
1
2
= 2 000 + 2 000 × 100 × 1 = 2 000 + 2 000 × 100 × 1
= 2 000 + 40 = 2 000 + 60
= RM2 040 = RM2 060


Bagi prinsipal yang sama, apabila kadar faedah bertambah, jumlah simpanan akan bertambah .
For the same principal, when the interest rates increase, the total savings will increase .

(c) Pada awal tahun, Encik Jaafar menyimpan RM15 000 dalam suatu akaun simpanan tetap dengan kadar
faedah 4% setahun. Berapakah jumlah wang dalam akaun simpanan tetapnya pada akhir tahun jika
faedah dikompaun
At the beginning of the year, Encik Jaafar saves RM15 000 in a fixed deposit account with the interest rate of 4% per annum. How
much money is in his fixed deposit account at the end of the year if interest is compounded
(i) setiap setengah tahun/ half yearly (ii) setiap suku tahun/ quarterly

(i) dikompaunkan 6 bulan sekali, n = 2 (ii) dikompaunkan 3 bulan sekali, n = 4
compounded once every 6 months compounded once every 3 months
r
r
1
1
MV = P 1 + n 2 nt MV = P 1 + n 2 nt
1
1
= 15 000 1 + 0.04 2 2(1) = 15 000 1 + 0.04 2 4(1)
2
4
= RM15 606 = RM15 609.06
Apabila kekerapan pengkompaunan bertambah, nilai masa hadapan simpanan akan bertambah .
When the compounding frequency increases, the future value of savings will increase .



6. Bulatkan jawapan yang betul. SP 3.1.2 TP2 i-Think Peta Pelbagai Alir
i-Think
i-Think
Circle the correct answers.
Tempoh yang lebih panjang Tinggi / Rendah
Longer period
Higher / Lower
(a) (a) Tip Penting
Nilai masa hadapan
Nilai masa merujuk kepada nilai
Kadar faedah atau pulangan (b) hadapan (b) Tinggi / Rendah sejumlah wang yang
yang lebih rendah simpanan terkumpul pada masa
Lower rate of interest or return Future value Higher / Lower depan.
of savings Future value refers to the
value of the amount of money
(c) (c) accumulated in the future.
Pengkompaunan yang Tinggi / Rendah
lebih kerap
Compounded more frequent Higher / Lower









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03_ Hybrid Mate Tg3.indd 28 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
FOKUS TOPIK


Nilai pulangan pelaburan merujuk kepada nilai pulangan atas setiap ringgit yang dilaburkan oleh pelabur.
Nilai pulangan pelaburan juga merupakan nisbah keuntungan atau kerugian yang diperoleh daripada suatu
pelaburan.
Return on investment (ROI) refers to the return value of each ringgit invested by the investor. Return on investment is also a ratio of profit or loss
derived from an investment.
Jumlah pulangan
Nilai pulangan pelaburan = × 100% Jenis pulangan
Nilai pelaburan awal Types of returns
Total return
Return on investment (ROI) = × 100%
The value of the initial investment INFO

7. Selesaikan masalah berikut.
Solve the following problems.
(a) Encik Yusof telah menggunakan RM8 000 untuk membeli syer syarikat A. Dia telah mendapat dividen
sebanyak RM200 dalam tempoh dia memegang syer itu. Selepas itu, dia menjual semua syer dan
mendapat RM8 800. Hitung nilai pulangan pelaburan Encik Yusof.
Encik Yusof used RM8 000 to buy shares of company A. He received dividend of RM200 during the time he was holding the shares.
Subsequently, he sold all shares and received RM8 800. Calculate the return on investment for Encik Yusof. SP 3.1.3 TP3
Modal awal/ Initial capital = RM8 000 ROI = RM1 000 × 100%
RM8 000
Jumlah pulangan/ Total return = 12.5% jumlah pulangan × 100%
= RM200 + (RM8 800 – RM8 000) nilai pelaburan awal
total return
= RM200 + RM800 Initial investment value × 100%
= RM1 000
(b) Pada tahun 2015, Puan Zainab membeli unit amanah saham dengan bonusnya sebanyak RM10 000.
Selepas dua tahun, dia menjual semua unit amanah saham itu dan mendapat RM10 400. Dia mendapat
dividen RM350 sebanyak dua kali dalam tempoh dia memegang unit amanah saham itu. Hitung nilai
pulangan pelaburan Puan Zainab.
In year 2015, Puan Zainab bought unit trusts with her bonus of RM10 000. After two years, she sold all the unit trusts and received
RM10 400. She received dividend of RM350 twice during the time she was holding the unit trusts. Calculate the return on investment
for Puan Zainab. SP 3.1.3 TP3
Modal awal/ Initial capital = RM10 000 ROI = RM1 100 × 100%
Jumlah pulangan/ Total return RM10 000
= RM350 × 2 + (RM10 400 – RM10 000) = 11%
= RM700 + RM400
= RM1 100


(c) Encik Tan membuka sebuah kedai farmasi. Pada awal tahun, dia telah membayar RM4 000 untuk
membeli topeng muka untuk simpanan stoknya. Pada akhir tahun, didapati Encik Tan mendapat
RM4 800 selepas semua topeng muka telah habis dijual. SP 3.1.3 TP4
Mr Tan owns a pharmacy. At the beginning of the year, he paid RM4 000 to buy masks as stock. At the end of the year, Mr Tan gained
RM4 800 after selling all the masks.
(i) Hitung nilai pulangan pelaburan Encik Tan.
Calculate the return on investment for Mr Tan.
(ii) Nyatakan satu faktor yang mempengaruhi pulangan pelaburan Encik Tan dan kesannya.
State a factor that can affect the return on investment for Mr Tan and its effect.
RM4 800 – RM4 000
(i) ROI = × 100%
RM4 000
= 20%
(ii) Kejadian pencemaran udara atau jerebu menyebabkan permintaan topeng muka bertambah. Jadi,
harga topeng muka akan bertambah dan pulangan Encik Tan akan bertambah.
The occurrence of air pollution and haze increases masks’ demand. Consequently, mask’s price will hike and this will cause
Mr Tan’s return to increase.



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03_ Hybrid Mate Tg3.indd 29 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
8. Tandakan (3) bagi pernyataan yang benar dan (7) bagi pernyataan yang palsu. SP 3.1.4 TP2
Mark (3) for true statement(s) and (7) for false statement(s).
(a) Pulangan ialah keuntungan yang diperoleh daripada suatu simpanan atau pelaburan.
Return is the profit gained from a savings or investment. ( 3 )
(b) Risiko ialah kemungkinan seseorang mendapat keuntungan yang banyak.
Risk is the possibility a person gains a lot of profit. ( 7 )
(c) Kecairan bermaksud keupayaan menukar suatu simpanan, pelaburan atau aset menjadi wang
tunai. ( 3 )
Liquidity means the ability of converting a savings, investment or asset into cash.

9. Susun jenis simpanan dan pelaburan berikut dari tahap rendah ke tahap tinggi. SP 3.1.4 TP2
Arrange the following type of savings and investments from low level to high level.
akaun simpanan tetap saham hartanah
fixed deposit account shares real estate
Tip Penting
Rendah Sederhana Tinggi Pelaburan yang berisiko
Low Moderate High rendah biasanya mempunyai
kecairan yang tinggi.
(a) Risiko akaun simpanan hartanah saham Low risk investment usually has
Risk tetap real estate shares high liquidity.
fixed deposit account

(b) Kecairan hartanah saham akaun simpanan Perbandingan
Liquidity real estate shares tetap simpanan dan
pelaburan
fixed deposit account Comparison between
INFO savings and investments

10. Isikan tempat kosong dengan perkataan ‘rendah’ atau ‘tinggi’. SP 3.1.4 TP2
Fill in the blanks with the words ‘lower’ or ‘higher’.
(a) Pulangan akaun simpanan tetap adalah lebih rendah daripada hartanah.

Return of fixed deposit account is lower than real estate.
(b) Risiko amanah saham adalah lebih tinggi daripada akaun simpanan.

Risk of unit trust is higher than savings account.

(c) Pulangan akaun simpanan adalah lebih rendah daripada amanah saham.
Return of savings account is lower than unit trust.

(d) Kecairan saham adalah lebih rendah daripada akaun simpanan.
Liquidity of shares is lower than savings account.


11. Suhaimi melabur dalam Amanah Saham Nasional (ASN) sebanyak 2 000 unit yang bernilai RM0.65 seunit.
Suhaimi invested 2 000 units valued at RM0.65 per unit in Amanah Saham Nasional (ASN). SP 3.1.4 TP3
(a) Apakah jenis pelaburan yang dilakukan oleh Suhaimi?
What type of investment was made by Suhaimi?
Amanah saham/ Unit trusts
(b) Nyatakan tahap potensi risiko, pulangan dan kecairan atas pelaburan yang dilakukan oleh Suhaimi.
State the potential risk level, return level and liquidity level of the investment made by Suhaimi.
Risiko – rendah; Pulangan – sederhana; Kecairan – tinggi
Risk – low; Return – moderate; Liquidity – high







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03_ Hybrid Mate Tg3.indd 30 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
12. Selesaikan masalah yang berikut.
Solve the following problems.
(a) Suhaimi melabur RM24 900 dalam saham P dengan strategi pemurataan pada bulan April dan Mei.
Pada bulan April, harga saham P ialah RM2.10 seunit. Pada bulan Mei, harga saham P ialah RM2.04
seunit. Hitung jumlah saham yang dimiliki oleh Suhaimi.
Suhaimi invested RM24 900 in share P with averaging strategy in April and May. In April, the price of share P is RM2.10 per unit. In
May, the price of share P is RM2.04 per unit. Calculate the number of shares owned by Suhaimi. SP 3.1.5 TP3

Bulan Jumlah pelaburan
Month Investment amount Tip Penting
April RM12 450 Strategi pemurataan melibatkan
Mei / May RM12 450 pelaburan amaun yang tetap untuk
tempoh yang khusus (seperti bulanan,
Jumlah / Total RM24 900 suku tahunan atau tahunan) tanpa
mengambil kira keadaan pasaran
Jumlah saham/ Total shares saham.
Cost averaging strategy involves investing
= 12 450 + 12 450 a fixed amount of money for a specific
2.10 2.04 period (such as monthly, quarterly or yearly)
= 5 929 + 6 103 irrespective of share market conditions.
= 12 032 unit/ 12 032 units


(b)
Saya mempunyai RM24 000. Saya melabur amaun yang sama
dalam saham S mengikut bulan-bulan tertentu.
Encik I have RM24 000. I invested the same amount of money in share S in the
Hashim designated months.

Jadual di bawah menunjukkan pembelian saham oleh Encik Hashim mengikut bulan-bulan tertentu.
The table shows the purchase of shares by Encik Hashim in the designated months.
Bulan Jun Ogos Oktober Disember
Month June August October December
Harga saham seunit (RM)
Share price per unit (RM) 2.30 2.10 2.05 2.20

Hitung kos purata seunit saham yang dimiliki oleh Encik Hashim. SP 3.1.5 TP3
Calculate the average cost per share for the shares owned by Encik Hashim.


Jumlah saham/ Total shares Kos purata/ Average cost
6 000 6 000 6 000 6 000 Jumlah pelaburan 24 000
= + + + setiap bulan = jumlah pelaburan
2.30 2.10 2.05 2.20 Investment amount each 11 120 bilangan unit yang
dimiliki
= 2 609 + 2 857 + 2 927 + 2 727 month = RM2.16 investment amount
= 11 120 unit/ 11 120 units = RM24 000 number of share units
4
owned
= RM6 000


(c) Selvi melabur RM8 000 dalam saham Q dengan kos purata seunit saham sebanyak RM3.20. Berapakah
jumlah unit saham Q yang dimiliki oleh Selvi? SP 3.1.5 TP4
Selvi invested RM8 000 in share Q with the average cost per share unit of RM3.20. How many units of share Q owned by Selvi?
8 000
3.20 =
Jumlah unit yang dimiliki
Total units owned
Jumlah unit yang dimiliki = 8 000
Total units owned 3.20
= 2 500 unit / 2 500 units








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03_ Hybrid Mate Tg3.indd 31 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
(d) Jadual di bawah menunjukkan pembelian unit amanah saham oleh Encik Zainal untuk tiga bulan
berturutan. SP 3.1.5 TP4
The table shows the purchase of unit trust by Encik Zainal for three months consecutively.
Januari Februari Mac
January February March
Jumlah pelaburan RM3 000 RM3 000 RM3 000
Total investment
Harga seunit RM1.25 RM1.24 RM1.22
Price per unit

(i) Hitung kos purata seunit amanah saham yang dimiliki oleh Encik Zainal.
Calculate the average cost per share for the unit trust owned by Encik Zainal.
(ii) Apakah strategi yang digunakan oleh Encik Zainal dalam pembelian unit amanah saham? Jelaskan
kebaikan menggunakan strategi ini.
What is the strategy used by Encik Zainal in purchasing unit trust? Explain the benefit of using this strategy.
(i) Jumlah saham / Total shares Kos purata / Average cost

= 3 000 + 3 000 + 3 000 = 9 000
1.25 1.24 1.22 7 278
= 2 400 + 2 419 + 2 459 = RM1.24
= 7 278 unit / 7 278 units

(ii) Strategi pemurataan kos ringgit digunakan. Strategi ini dapat mengurangkan risiko pelaburan
dengan mengurangkan purata harga pembelian.
Ringgit cost averaging strategy is used. This strategy helps in reducing investment risk by bringing down average purchasing
cost.



(e) Jasraj membeli unit amanah saham R pada 1 Januari 2018 berjumlah RM200 pada harga RM0.85 seunit.
Kemudian, dia meneruskan pembelian unit amanah saham tersebut dengan jumlah RM200 sebulan
selama 1 tahun. Harga pembelian seunit amanah saham adalah berbeza-beza berdasarkan harga
pasaran setiap bulan. Pada 31 Disember 2018, dia memegang 3 105 unit amanah saham tersebut.
Jasraj purchased unit trust R on 1 January 2018 amounted RM200 at RM0.85 per unit. Then, he continually purchased the particular
unit trust amounted RM200 each month for 1 year. Purchase price per unit of the unit trust varies according to market price each
month. On 31 December 2018, he holds 3 105 units of the unit trust. SP 3.1.5 TP4
(i) Hitung kos purata seunit amanah saham yang dipegang oleh Jasraj.
Calculate the average cost per share for the unit trust held by Jasraj.
(ii) Hitung jumlah unit yang diperoleh Jasraj jika dia membeli unit amanah saham sekali gus dengan
RM2 400 pada bulan Januari.
Calculate the total units to be held by Jasraj if he purchased the unit trust with a lump sum of RM2 400 in January.
(iii) Antara pembelian secara berasingan dan berturutan dan pembelian sekali gus, strategi yang
manakah lebih bermanfaat kepada Jasraj? Jelaskan jawapan anda.
Among separately and continually purchase and lump sum purchase, which strategy is more beneficial for Jasraj? Explain your
answer.

(i) Kos purata seunit saham (ii) Jumlah unit yang dibeli
Average cost per share Total units purchased

= 12 × 200 = 2 400
3 105 0.85
= RM0.77 = 2 824 unit/ 2 824 units

(iii) Pembelian secara berasingan dan berturutan lebih bermanfaat kepada Jasraj. Hal ini demikian
kerana Jasraj mendapat purata harga pembelian yang lebih rendah dan jumlah unit yang lebih
banyak berbanding dengan pembelian sekali gus.
Separately and continually purchase is more beneficial for Jasraj. This is because Jasraj manages to get a lower average
purchase price and hold more units when compared to lump sum purchase.






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03_ Hybrid Mate Tg3.indd 32 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
13. Selesaikan masalah berikut yang melibatkan simpanan dan pelaburan.
Solve the following problems involving savings and investments. SP 3.1.6 TP5
(a) Encik Subra mempunyai RM50 000 dalam akaun simpanan. Dia mengeluarkan separuh daripada
simpanan itu untuk membuka akaun simpanan tetap dengan kadar faedah 3.8% setahun, dikompaun
setiap setengah tahun. Baki simpanan itu dilabur di pasaran saham dengan strategi pemurataan pada
bulan Januari dan Februari. Dia membeli saham A dengan harga RM2.25 seunit pada bulan Januari dan
kemudiannya dengan harga RM2.30 seunit pada bulan Februari. Selepas setahun, dia menjual semua
sahamnya dengan harga RM2.45 seunit. KBAT Mengaplikasi
Encik Subra had RM50 000 in a savings account. He uplifted half of the savings to open a fixed deposit account with interest rate of
3.8% per annum, compounded half yearly. The remaining savings was invested in share market with averaging strategy in January
and February. He bought share A at RM2.25 per unit in January and subsequently at RM2.30 per unit in February. After one year, he
sold all the shares at RM2.45 per unit.
(i) Hitung nilai pulangan pelaburan bagi simpanan tetap Encik Subra selepas setahun.
Calculate the return on investment for Encik Subra's fixed deposit after one year.
(ii) Hitung kos purata seunit saham A yang dimiliki oleh Encik Subra.
Calculate the average cost per share for the share A owned by Encik Subra.
(iii) Bandingkan nilai pulangan pelaburan bagi simpanan tetap dan saham Encik Subra, yang manakah
lebih menguntungkan? Jelaskan jawapan anda.
Compare the return on investment for Encik Subra's fixed deposit and share, which is more profitable? Explain your answer.


1
(i) MV = 25 000 1 + 0.038 2 2(1) Separuh simpanan Encik Subra Tip Penting
telah dikeluarkan.
2
= RM25 959.03 Half of Encik Subra’s savings has uplifted. Selain daripada bentuk
peratusan, nilai pulangan
Jumlah pulangan pelaburan juga boleh
Total return dinyatakan dalam bentuk
= 25 959.03 – 25 000 nisbah.
= RM959.03 Besides the percentage form,
the return on investment can
also be expressed in the form
Nilai pulangan pelaburan (simpanan tetap) of ratio.
Return on investment (fixed deposit)
959.03
= × 100%
25 000
= 3.84%


(ii) Jumlah saham Kos purata seunit saham
Total shares Average cost per share
12 500 12 500 25 000
= + =
2.25 2.30 10 991
= 5 556 + 5 435 = RM2.27
= 10 991 unit
10 991 units

(iii) Jumlah pulangan Nilai pulangan pelaburan (saham)
Total return Return on investment (share)
= keuntungan modal = 1 978.38 × 100%
capital gain 25 000
= (2.45 – 2.27) × 10 991 = 7.91%
= RM1 978.38

Pelaburan saham lebih menguntungkan kerana nilai pulangan pelaburannya lebih tinggi.
Share investment is more profitable because it has higher return on investment.











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03_ Hybrid Mate Tg3.indd 33 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
(b) Wilson membahagikan bonusnya sebanyak RM30 000 secara sama rata kepada dua bahagian untuk
membuat simpanan tetap dan pelaburan amanah saham masing-masing. Bagi simpanan tetap, kadar
faedah yang ditawarkan oleh bank ialah 3.6% setahun, dikompaun setiap bulan. Bagi pelaburan amanah
saham, jumlah bilangan unit yang dipegang olehnya selepas setahun ialah 20 000 unit. KBAT Menganalisis
Wilson divided his RM30 000 bonus equally into two parts for fixed deposit and unit trust investment respectively. For fixed deposit,
the interest rate offered by the bank is 3.6% per annum, compounded monthly. For unit trust investment, the total units he hold after
one year are 20 000 units.
(i) Hitung nilai matang bagi simpanan tetap selepas setahun.
Calculate the matured value of the fixed deposit after one year.
(ii) Dalam tempoh setahun itu, Wilson telah menerima dividen sebanyak RM300. Hitung harga jualan
per unit bagi amanah saham supaya nilai pulangan pelaburan bagi amanah saham adalah dua kali
nilai pulangan pelaburan bagi simpanan tetap.
In the one year period, Wilson received dividend of RM300. Calculate the selling price per unit of the unit trust in order to gain
twice return on investment when compared to fixed deposit.

1
(i) MV = 15 000 1 + 0.036 2 12(1)
12
= RM15 549

15 549 – 15 000
(ii) Nilai pulangan pelaburan bagi simpanan tetap = × 100%
Return on investment of fixed deposit 15 000
= 3.66%
Katakan x = harga jualan per unit bagi amanah saham
Let x = selling price per unit of the unit trust
300 + 20 000x – 15 000
2(3.66) = × 100
15 000
1 098 = 300 + 20 000x − 15 000
20 000x = 15 798
x = RM0.79





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Saya berjaya
• Mengenal pelbagai jenis simpanan dan pelaburan.
• Membuat pengiraan yang melibatkan faedah mudah dan faedah kompaun bagi simpanan, dan seterusnya menerangkan kesan perubahan tempoh, kadar
faedah atau pulangan dan kekerapan pengkompaunan terhadap nilai masa hadapan simpanan.
• Membuat pengiraan yang melibatkan nilai pulangan pelaburan, dan seterusnya menerangkan faktor yang mempengaruhi pulangan pelaburan serta
kesannya.
• Membanding dan membeza potensi risiko, pulangan dan kecairan pelbagai jenis simpanan dan pelaburan.
• Mengira purata kos sesyer bagi pelaburan saham menggunakan strategi pemurataan kos ringgit dan menjelaskan manfaat strategi ini.
• Menyelesaikan masalah yang melibatkan simpanan dan pelaburan




PBD 3.2 Pengurusan Kredit dan Hutang Nota Buku Teks
PBD
PBD
Credit and Debt Management

Visual
ms. 73 – 81
14. Isikan tempat kosong. SP 3.2.1 TP1
Fill in the blanks.
(a) Kredit ialah satu kemudahan penangguhan bayaran yang diberikan oleh pembekal kepada
pengguna.
Credit is a postponement of payment facility provided by the supplier to the consumer.

(b) Hutang membawa maksud suatu amaun yang telah dipinjam tetapi belum dilunaskan.

Debt means an amount that has been borrowed but has not been settled.





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Matematik Tingkatan 3 Bab 3
15. Tuliskan ‘B’ untuk pernyataan benar dan ‘P’ untuk pernyataan palsu. SP 3.2.1 TP1
Write ‘T’ for true statement(s) and ‘F’ for false statement(s).
Pernyataan B / P
Statement T / F

(a) Ali mendapat pinjaman perumahan daripada bank. Ali telah menampung hutang. B
Ali got housing loan from a bank. Ali was in debt. T

(b) Susan membayar balik pinjaman kereta. Susan telah menampung hutang. P
Susan made repayment for her car loan. Susan borne debt. F

(c) Haris mendapat pinjaman peribadi daripada bank. Haris telah menerima kredit. B
Haris got the personal loan from a bank. Haris received credit. T
(d) Adam membuat pembayaran balik ke atas pinjaman keretanya. Adam telah menyelesaikan B
hutang. T
Adam made repayment for his car loan. Adam cleared his debt.

16. Pilih pengurusan kredit dan hutang yang bijaksana dan tuliskan hurufnya di dalam rajah di bawah.
Choose the smart management of credit and debt and write the letter in the diagram below. SP 3.2.1 TP2
Peta bulatan
A Membuat belanjawan B Meminjam dengan banyak
Do budgeting Excessive loan
Tip Penting
Kredit termasuk kad kredit
C Mengurangkan hutang D Selalu membuat pembayaran dan pinjaman.
Reduce debts balik lewat Credits include credit card and
loan.
Always late in making repayment

E Menggunakan cara auto debit A
untuk membayar bil
Use auto debit payment method to pay
bill
Pengurusan kredit F
F Menjelaskan hutang dalam dan hutang yang
tempoh yang ditetapkan bijaksana
Settle the debt within the period C Smart management of
stipulated credit and debt

G Membayar bayaran minimum E
sahaja untuk penggunaan kad
kredit
Pay only the minimum amount for the
usage of credit card


17. Padankan pernyataan berikut kepada kelebihan atau kelemahan kad kredit. SP 3.2.2 TP1
Match the following statements to the benefits or weaknesses of credit card.

Memudahkan proses pembelian atas talian
Ease online purchase process
Kelebihan
Benefits
Mudah terbelanja melebihi kemampuan
Easy to spend beyond affordability

Boleh dikenakan faedah dan caj-caj lain
Interest and other charges could be imposed
Kelemahan
Memberi tempoh bayar balik tanpa faedah Weaknesses
Provide interest free period for repayment




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Matematik Tingkatan 3 Bab 3
18. Antara berikut, yang manakah tidak boleh dilakukan oleh seorang pemegang kad kredit? Bulatkan jawapan
yang betul. SP 3.2.2 TP1
Which of the following cannot be done by a credit card holder? Circle the correct answer.
A Tandatangani kad kredit sebaik sahaja diterima.
Sign the credit card as soon as it is received.
B Tidak memberikan butir-butir kad kredit kepada orang yang tidak dikenali.
Do not give credit card details to strangers.
C Catat nombor pin di belakang kad kredit.
Record the pin number on the back of the credit card.
D Semak urus niaga dalam penyata kad kredit.
Check the transactions in the credit card statement.

19. Selesaikan masalah berikut.
Solve the following problems.

(a) Min Hui menggunakan kad kredit untuk membuat bayaran pembelian jam tangan sebanyak RM1 500.
Amaun tersebut juga merupakan jumlah terkini pada penyata bulan Jun. Dia tidak membuat sebarang
bayaran balik untuk bulan Jun. Bank A mengenakan caj kewangan 18% setahun ke atas baki belum
jelas untuk 35 hari dan caj bayaran lewat 1% daripada baki belum jelas atau minimum RM10. Hitung
caj kewangan dan caj bayaran lewat yang perlu ditanggung oleh Min Hui. Anggapkan Min Hui tidak
menggunakan kad kredit selepas pembelian jam tangan itu. SP 3.2.3 TP4
Min Hui used credit card to make a payment of RM1 500 for a watch. The amount is also the current amount in June statement. She
didn’t make any repayment for June. Bank A charged a finance charge of 18% per annum on the outstanding balance for 35 days
and late payment charge of 1% of the outstanding balance or minimum RM10. Calculate the finance charge and late payment
charge that Min Hui had to pay. Assume that Min Hui did not use credit card after purchasing the watch.
Baki belum jelas / Outstanding balance = RM1 500
Tip Penting
18
35
Caj kewangan/ Finance charge = RM1 500 × 1 100 2 1 365 2 Tempoh tanpa faedah yang
×
= RM25.89 diberikan biasanya selama
20 hari.
1
Caj bayaran lewat/ Late payment charge = 1 100 2 × (RM1 500 + RM25.89) Interest free period normally is
20 days.
= RM15.26
(b) Eric membeli set perabot yang berharga RM2 000 dengan menggunakan kad kredit pada bulan April.
Dia hanya membuat pembayaran minimum sebanyak RM100 selepas mendapat penyata bulan April.
Bank B mengenakan caj kewangan 1.5% sebulan ke atas transaksi pembeliannya untuk 14 hari dan ke
atas baki belum jelas untuk 18 hari. Caj bayaran lewat yang dikenakan pula ialah minimum RM10 atau
1% daripada baki belum jelas. Hitung jumlah terkini pada penyata bulan Mei. Anggapkan Eric tidak
menggunakan kad kredit sebelum dan selepas pembelian set perabot itu. SP 3.2.3 TP4
Eric bought a set of furniture with a cost of RM2 000 using credit card in April. He only made a minimum payment of RM100 after
he received the April statement. Bank B charged a finance charge of 1.5% per month on his transaction for 14 days and on the
outstanding balance for 18 days. Late payment charge imposed is minimum RM10 or 1% of the outstanding balance. Calculate the
current amount in May statement. Assume that Eric did not use credit card before and after purchasing the furniture.

Baki belum jelas/ Outstanding balance = RM2 000 – RM100
= RM1 900 Tip Penting
Caj kewangan/ Finance charge Jika pemegang kad
kredit membayar hanya
1.5 14 1.5 18
= RM2 000 × × + RM1 900 × × sebahagian daripada
100 30 100 30 hutang kad kredit, dia akan
= RM14 + RM17.10 kehilangan tempoh tanpa
= RM31.10 faedah. Baki yang belum
dijelaskan akan dikenakan
Caj bayaran lewat/ Late payment charge = RM0 faedah.
If a credit card holder pay only
Jumlah terkini pada penyata bulan Mei part of his credit card debt, he
Current amount in May statement will lose the interest free period.
= RM1 900 + RM31.10 + RM0 The outstanding balance will be
charged with interest.
= RM1 931.10




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Matematik Tingkatan 3 Bab 3
(c) Sharmila membeli sebuah telefon bimbit yang berharga RM3 500 dengan menggunakan kad kredit
pada bulan Julai. Dia tidak membuat sebarang bayaran balik selepas menerima penyata bulan Julai.
Bank C mengenakan caj kewangan 18% setahun ke atas baki belum jelas untuk 35 hari dan caj
bayaran lewat ialah minimum RM10 atau 1% daripada baki belum jelas. Anggapkan Sharmila tidak
menggunakan kad kredit sebelum dan selepas pembelian telefon bimbit itu. SP 3.2.4 TP5
Sharmila bought a mobile phone with a cost of RM3 500 using credit card in July. She did not make any repayment after received July
statement. Bank C charged a finance charge of 18% per annum on the outstanding balance for 35 days and the late payment charge
is minimum RM10 or 1% of the outstanding balance. Assume that Sharmila did not use credit card before and after purchasing the
mobile phone.
(i) Hitung jumlah terkini pada penyata bulan Ogos.
Calculate the current amount in August statement.
(ii) Jika Sharmila membeli telefon bimbit itu secara tunai, apakah beza amaun bayaran antara membeli
secara tunai dengan membeli secara kad kredit bagi Sharmila?
If Sharmila bought the mobile phone by cash, what would be the difference in amount paid between the cash payment and
credit card payment by Sharmila?
(iii) Nyatakan satu kebaikan dan satu kelemahan penggunaan kad kredit.
State a benefit and a weakness of using credit card.

(i) Baki belum jelas/ Outstanding balance = RM3 500
Caj kewangan/ Finance charge = RM3 500 × 18 × 35
100 365
= RM60.41
Caj bayaran lewat/ Late payment charge = 1 × (RM3 500 + RM60.41)
100
= RM35.60
Jumlah terkini pada penyata bulan Ogos
Current amount in August statement
= RM3 500 + RM60.41 + RM35.60
= RM3 596.01


(ii) Beza/ Difference = RM3 596.01 – RM3 500
= RM96.01



(iii) Kebaikan/ Benefit:
Sharmila tidak perlu membawa tunai yang banyak.
Sharmila does not need to carry a lof of cash.
Kelemahan/ Weakness:
Sharmila terpaksa menanggung faedah yang lebih tinggi apabila dia tidak dapat membuat bayaran
balik.
Sharmila bears higher interest when she cannot make a repayment
(Mana-mana jawapan lain yang sesuai)
(Any other possible answers)



20. Isikan tempat kosong. SP 3.2.4 TP4
Fill in the blanks.
Jika pemegang kad kredit lewat membuat pembayaran atau hanya membuat pembayaran minimum,
If a credit card holder makes late repayment or only makes minimum payment,
(a) faedah akan dikenakan dan jumlah yang perlu dibayarnya akan bertambah jika
dibandingkan dengan harga asal barangan atau perkhidmatan yang diperoleh,
interest will be charged and the amount he/ she needs to pay will be increased if compared with the
original price of the goods or services received,
(b) dia akan mengambil tempoh yang lebih panjang untuk menyelesaikan hutang kad kredit.
he/ she will take longer period to settle the credit card debt.


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03_ Hybrid Mate Tg3.indd 37 20/08/2021 5:30 PM

Matematik
ingkatan 3
T


Bab 3
Matematik Tingkatan 3 Bab 3
21. Selesaikan masalah berikut yang melibatkan faedah sama rata.
Solve the following problems involving flat interest. SP 3.2.5 TP4 Tip Penting
(a) Harga kereta A ialah RM80 000. James menggunakan simpanannya • Faedah sama rata dikira
pada jumlah pinjaman asal
sebanyak RM8 000 sebagai wang pendahuluan semasa membeli sepanjang tempoh pinjaman.
kereta itu. Kemudian, dia mendapat pinjaman jumlah baki itu Flat interest is calculated on the
daripada bank dengan kadar faedah tahunan 4% selama 8 tahun. original loan amount over the
loan period.
Hitung jumlah bayaran balik pinjaman. • Faedah atas baki yang
The price of car A is RM80 000. James used his savings of RM8 000 for down payment dikenakan setiap bulan ke
when buying the car. Then, he got a loan for the balance from a bank with the yearly atas pinjaman bergantung
interest rate 4% for 8 years. Calculate the total amount of loan repayment. kepada jumlah baki pinjaman
Prinsipal, P/ Principal, P = 80 000 − 8 000 pada bulan tersebut.
= RM72 000 Interest on balance charged each
month on the loan depends on the
Jumlah bayaran balik/ Total repayment amount of the loan balance for
that month.
= P + Prt
= 72 000 + (72 000 × 0.04 × 8)
= RM95 040 Tip Penting
Rumus untuk pinjaman dengan
(b) Encik Tan membuat pinjaman peribadi sebanyak RM100 000 daripada faedah sama rata ialah
The formula for the loan with flat
sebuah bank untuk membeli peralatan kilangnya. Diberi bahawa interest is
kadar faedah yang dikenakan ialah 5% setahun. Hitung bayaran A = P + Prt
ansuran bulanan Encik Tan jika tempoh pinjaman ialah 9 tahun. dengan keadaan/ where
Mr Tan made a personal loan of RM100 000 from a bank to buy the equipments for A = jumlah bayaran balik
his factory. It is given that the interest rate is 5% per annum. Calculate the monthly total repayment
instalment of Mr Tan if the loan period is 9 years. P = prinsipal
Jumlah bayaran balik/ Total repayment principal
= 100 000 + (100 000 × 0.05 × 9) r = kadar faedah
interest rate
= RM145 000 t = masa
time
Bayaran ansuran/ Instalment = 145 000 Bayaran ansuran/Instalment
9 × 12 Jumlah bayaran balik
= RM1 342.59 = Jumlah bilangan ansuran
Total repayment

Total number of instalment

22. Selesaikan masalah berikut yang melibatkan faedah atas baki. SP 3.2.5 TP4
Solve the following problems involving interest on balance.

Encik Law membuat pinjaman peribadi sebanyak RM20 000 daripada sebuah bank dengan kadar faedah
5.5% atas baki setahun. Tempoh bayaran balik ialah 5 tahun dan ansuran bulanan ialah RM425. Andaikan
Encik Law membayar ansuran bulanan tersebut setiap akhir bulan. Lengkapkan jadual di bawah dengan
jumlah baki pinjaman pada awal bulan dan faedah yang perlu dibayar oleh Encik Law setiap bulan bagi
3 bulan pertama.
Mr Law obtained a personal loan of RM20 000 from a bank with an interest rate of 5.5% on the balance per annum. The repayment
period is 5 years and the monthly instalment is RM425. Assume that Mr Law pays the monthly instalment at the end of every month.
Complete the table below with the balance of the loan at the beginning of the month and the amount of interest payable by Mr Law
each month for the first 3 months.
Bulan pertama Bulan kedua Bulan ketiga
First month Second month Third month
Jumlah baki pinjaman RM20 000 20 000 + 91.67 – 425 19 666.67 + 90.14 – 425
pada awal bulan = RM19 666.67 = RM19 331.81
Balance of the loan at the
beginning of the month
Faedah 0.055 0.055 0.055
Interest 20 000 × 12 19 666.67 × 12 19 331.81 × 12
= RM91.67 = RM90.14 = RM88.60





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Matematik Tingkatan 3 Bab 3
23. Selesaikan setiap masalah yang berikut. SP 3.2.6 TP5
Solve each of the following problems.
(a) Encik Johnson meminjam RM20 000 dari sebuah bank untuk memulakan perniagaan runcitnya. Bank
tersebut mengenakan kadar faedah sama rata 5% setahun dan tempoh pembayaran balik ialah 7 tahun.
Berapakah jumlah faedah yang akan dibayar oleh Encik Johnson kepada pihak bank?
Mr Johnson borrowed RM20 000 from a bank to start his retail business. The bank charges a flat interest rate of 5% per annum and
the repayment period is 7 years. How much interest will be paid to the bank by Mr Johnson? KBAT Mengaplikasi


Jumlah faedah/ Total interest = 20 000 × 5 × 7
100
= RM7 000


(b) Fatimah ingin membeli perabot untuk rumah baharunya. Dia bercadang untuk membuat pinjaman
peribadi sebanyak RM40 000 daripada bank dengan kadar faedah tahunan 5.5% dan tempoh pinjaman
ialah 6 tahun. KBAT Mengaplikasi
Fatimah wants to buy furniture for her new house. She plans to make a personal loan of RM40 000 from a bank with the yearly
interest rate of 5.5% and the loan period is 6 years.
(i) Hitung bayaran ansuran bulanan bagi pinjaman peribadi Fatimah.
Calculate the monthly instalment of Fatimah’s personal loan.
(ii) Jika Fatimah ingin mengurangkan tempoh bayar balik pinjaman kepada 5 tahun, berapakah wang
yang Fatimah perlu menambah ke atas ansuran bulanannya?
If Fatimah wants to reduce the loan repayment period to 5 years, how much money does Fatimah need to add on her monthly
instalment?
(i) Jumlah bayaran balik/ Total repayment (ii) Jumlah bayaran balik/ Total repayment
= 40 000 + (40 000 × 0.055 × 6) = 40 000 + (40 000 × 0.055 × 5)
= RM53 200 = RM51 000
Bayaran ansuran/ Instalment Bayaran ansuran/ Instalment

= 53 200 51 000
6 × 12 = 5 × 12
= RM738.89 = RM850

Wang yang perlu ditambah
Money to be added
= 850 – 738.89
= RM111.11



(c) Alias membuat pinjaman peribadi sebanyak RM60 000 daripada sebuah institusi kewangan. Diberi
bahawa tempoh pinjamannya ialah 8 tahun dan kadar faedah tahunan ialah 6%. KBAT Menganalisis
Alias made a personal loan of RM60 000 from a financial institution. It is given that the loan period is 8 years and the yearly interest
rate is 6%.
(i) Hitung bayaran ansuran bulanan bagi pinjaman peribadi Alias.
Calculate the monthly instalment of Alias's personal loan.
(ii) Jika Alias boleh menambah RM175 ke atas ansuran bulanannya, apakah tempoh bayaran balik
baharu bagi pinjamannya, dalam tahun?
If Alias could add in RM175 to his monthly instalment, what is the new loan repayment period, in years?
(i) Jumlah bayaran balik/ Total repayment (ii) Katakan t = tempoh bayaran balik baharu dalam tahun
= 60 000 + (60 000 × 0.06 × 8) Let t = new loan repayment period in years
= RM88 800 60 000 + (60 000 × 0.06 × t)
Bayaran ansuran/ Instalment 925 + 175 = t × 12
= 88 800 13 200t = 60 000 + 3 600t
8 × 12 9 600t = 60 000
= RM925 t = 6.25








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Matematik Tingkatan 3 Bab 3
(d) Abdullah hendak membuat pinjaman sebanyak RM90 000 daripada sebuah bank. Dia bercadang untuk
1
menggunakan daripada gaji bulanannya untuk membayar ansuran bulanan selama 8 tahun. Diberi
4
bahawa gaji bulanannya ialah RM5 400. KBAT Menilai
1
Abdullah wants to apply a loan of RM90 000 from a bank. He plans to use of his monthly income to pay for the monthly instalment
4
for 8 years. It is given that his monthly income is RM5 400.
(i) Hitung kadar faedah tahunan maksimum yang dia boleh menanggung.
Calculate the maximum yearly interest rate that he can bear.
(ii) Jika Abdullah berjaya mendapat kadar faedah tahunan yang 1% lebih rendah daripada (i), bolehkah
dia mengurangkan tempoh bayar balik pinjaman kepada 7 tahun? Jelaskan jawapan anda.
If Abdullah manages to get the yearly interest rate 1% lower than in (i), can he reduce the loan repayment period to 7 years?
Explain your answer.

1 Jumlah bayaran balik/ Total repayment
(i) × 5 400 =
4 8 × 12
1 350 = Jumlah bayaran balik/ Total repayment
96
Jumlah bayaran balik/ Total repayment = RM129 600
Katakan r = kadar faedah tahunan
Let r = yearly interest rate
129 600 = 90 000 + (90 000 × r × 8)
720 000r = 39 600
r = 0.055
= 5.5%


(ii) Jumlah bayaran balik/ Total repayment
= 90 000 + (90 000 × 0.045 × 7)
= RM118 350

Bayaran ansuran/ Instalment = 118 350 = RM1 408.93
7 × 12
1
Tidak boleh. Hal ini demikian kerana bayaran ansuran bulanan melebihi daripada gaji bulanannya.
4
1
Cannot. This is because monthly instalment exceeds of his monthly income.
4





































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Matematik Tingkatan 3 Bab 3
(e) Baiyah ingin membuat pinjaman kereta sebanyak RM80 000 daripada suatu bank dengan kadar faedah
tahunan 2.8% dan tempoh pinjaman ialah 8 tahun.
Baiyah wants to make a car loan of RM80 000 from a bank with the yearly interest rate of 2.8% and the loan period is 8 years.
(i) Hitung bayaran ansuran bulanan bagi pinjaman kereta itu.
Calculate the monthly instalment of the car loan.
(ii) Baiyah bercadang mengurangkan bayaran ansuran bulanan kepada RM940 dan memendekkan
tempoh bayaran balik kepada 7 tahun. Berapakah pinjaman yang dia boleh buat daripada bank?
Baiyah plans to reduce the monthly instalment to RM940 and shorten the repayment period to 7 years. How much he can loan
from the bank?

(i) Jumlah bayaran balik
Total repayment
= 80 000 + (80 000 × 0.028 × 8)
= RM97 920

Bayaran ansuran/ Instalment
= 97 920
8 × 12
= RM1 020


(ii) 940 = Jumlah bayaran balik
7 × 12
Total repayment
940 =
7 × 12
Jumlah bayaran balik
Total repayment
= 940 × 7 × 12
= RM78 960

Katakan P = pinjaman yang boleh dibuat,
Let P = the loan amount that can be made,
78 960 = P + (P × 0.028 × 7)
78 960 = 1.196P
P = RM66 020.07



24. Lakukan aktiviti yang berikut. SP 3.2.6 TP5
Carry out the following activity.

Jalan Galeri

(a) Lakukan kerja dalam kumpulan.
Work in groups.
(b) Setiap kumpulan perlu mengumpulkan maklumat tentang kadar faedah dan tempoh bagi akaun
simpanan tetap dan jenis pinjaman yang berlainan daripada beberapa bank.
Each group has to collect information about the interest rate and time period for fixed deposit account and different types of loan
from a few banks.
(c) Organisasikan maklumat yang telah dikumpul.
Organise the information collected.
(d) Persembahkan hasil kerja dalam pelbagai bentuk persembahan.
Present the work in various forms.


Tip Penting
Bank perdagangan di Malaysia: Bank Bank Islam di Malaysia Bank
• AmBank (M) Berhad perdagangan • Bank Islam Malaysia Berhad Islam
• CIMB Bank Berhad Commercial • Bank Muamalat Malaysia Berhad Islamic
• Hong Leong Bank Berhad banks • CIMB Islamic Bank Berhad banks
• Public Bank Berhad INFO • Hong Leong Islamic Bank Berhad INFO



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Matematik Tingkatan 3 Bab 3
25. Lakukan projek STEM di bawah. TP6
Carry out the STEM project below.
Project-Based Learning
Pr ojek
Projek

Objektif aktiviti: Meningkatkan pengetahuan dan pemahaman mengenai konsep simpanan dan
Activity objective: pinjaman.
Improve knowledge and understanding of the savings and loan concepts.
Pernyataan masalah: Azman mempunyai simpanan sebanyak RM20 000. Dia ingin membeli sebuah
Problem statement: kereta baharu yang berharga RM90 000. Gaji bulanannya ialah RM3 800. Dia hendak
menggunakan 25% daripada gaji bulanannya untuk membayar ansuran bulanan
kereta baharunya. Dia mempertimbangkan dua cara untuk membeli kereta itu:
(a) Guna kesemua simpanannya sebagai wang pendahuluan bagi mengurangkan
amaun pinjaman.
(b) Masukkan kesemua simpanannya ke dalam suatu akaun simpanan tetap dan
buat pinjaman penuh.
Andaikan anda sebagai Azman, bagaimanakah anda membuat pilihan?
Azman has a savings amounted RM20 000. He wants to buy a new car at a price of RM90 000. His monthly
salary is RM3 800. He wishes to use 25% of his monthly salary for the monthly instalment of his new car. He
considers two ways to buy the car:
(a) Use all the savings as a down payment to reduce the amount of the loan.
(b) Put all the savings into a fixed deposit account and make a full loan.
Assuming you are Azman, how do you make a choice?
Pencarian fakta: Mencari maklumat daripada laman sesawang tentang kadar faedah simpanan
Fact finding: tetap dan pinjaman kereta daripada bank yang berlainan.
Search information related to the interest rate of fixed deposit and car loan from different banks from the
internet.
Konsep yang Simpanan dan pinjaman
diaplikasikan: Savings and loan
Concepts applied:
Pelan tindakan / Action plan:
(a) Kelas dibahagikan kepada beberapa kumpulan.
Class is divided into several groups.
(b) Berdasarkan maklumat yang dicari, tentukan
Based on the information found, determine
(i) kadar faedah tahunan yang terbaik bagi akaun simpanan tetap yang ditawarkan oleh bank.
the best yearly interest rate of fixed deposit account offered by the banks.
(ii) kadar faedah dan tempoh pinjaman kereta yang sesuai untuk Azman.
the suitable interest rate and car loan period for Azman.
(c) Bandingkan situasi menggunakan kesemua simpanan dan tidak dalam pembelian kereta baharu oleh
Azman.
Compare the situations of using all the savings and not in buying new car by Azman.
(d) Sediakan kertas cadangan kewangan untuk Azman.
Prepare a financial proposal for Azman.
Penyelesaian: Menggunakan pengetahuan yang berkaitan dengan simpanan dan pinjaman
Solution: dalam kertas cadangan kewangan untuk Azman.
Use the knowledge related to savings and loan in the financial proposal for Azman.
Pembentangan: Membuat persembahan kertas cadangan kewangan dengan menggunakan
Presentation: Microsoft Powerpoint dan Microsoft Excel.
Present the financial proposal using Microsoft Powerpoint and Microsoft Excel.

Kriteria Kejayaan: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Saya berjaya
• Menjelaskan maksud kredit dan hutang, dan seterusnya menghuraikan pengurusan yang bijaksana tentang kredit dan hutang.
• Mengkaji dan menghuraikan kelebihan dan kekurangan kad kredit dan penggunaannya secara bijaksana.
• Mengkaji dan menghuraikan kesan pembayaran minimum dan pembayaran lewat bagi penggunaan kad kredit.
• Menyelesaikan masalah yang melibatkan penggunaan kad kredit.
• Mengira jumlah bayaran balik pinjaman dan bayaran ansuran, dengan pelbagai kadar faedah dan tempoh pinjaman yang berbeza.
• Menyelesaikan masalah yang melibatkan pinjaman.


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03_ Hybrid Mate Tg3.indd 42 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
Pentaksiran Sumatif
PT3


PRAKTIS PdPR Jawapan
Bab 3 PRAKTIS PdPR Bab 3


Bahagian A 5. Jadual di bawah menunjukkan butiran pinjaman
peribadi yang dibuat oleh Hussein.
1. Hasni membuka suatu akaun di sebuah bank. The table below shows the details of the personal loan made
by Hussein.
Dia membuat perbankan internet dengan akaun
tersebut. Antara yang berikut, yang manakah Jumlah Kadar faedah Tempoh
bukan jenis akaun Hasni yang mungkin? pinjaman sama rata bayaran balik
Hasni opens an account at a bank. He does internet banking Loan Flat interest Repayment
using that account. Which of the following is not a possible amount rate period
type of Hasni’s account? 5% setahun 8 tahun
A Akaun semasa RM12 000 5% per annum 8 years
Current account
B Akaun simpanan Berapakah jumlah bayaran yang perlu dibayar
Savings account oleh Hussein dalam tempoh bayaran balik itu?
C Akaun simpanan tetap What is the total amount payable by Hussein within the
Fixed deposit account repayment period?
D Akaun wadiah A RM600
Wadiah account B RM4 800
2. Pelaburan berikut membawa pulangan dalam C RM12 600
bentuk sewa dan keuntungan modal kecuali D RM16 800
The following investments bring returns in the form of rent and
capital gains except 6. Hasmawi melabur RM20 000 dalam amanah

A tanah saham. Dia menjualkan semua amanah sahamnya
land
B rumah selepas 5 tahun. Nilai pulangan pelaburan bagi
house Hasmawi ialah 45%. Hitung jumlah pulangan
C lot komersial yang diperoleh Hasmawi.
commercial lot Hasmawi invested RM20 000 in a unit trust. He sold all the unit
D amanah saham trust after 5 years. The return on investment for Hasmawi is
unit trusts 45%. Calculate the total return received by Hasmawi.
A RM8 500
3. Encik Lee menyimpan RM16 000 dalam suatu
akaun yang menawarkan kadar faedah mudah B RM9 000
2% setahun. Hitung faedah yang diperoleh Encik C RM9 250
Lee selepas satu tahun. D RM9 500
Mr Lee deposits RM16 000 in an account that offers a simple
interest rate of 2% per annum. Calculate the interest earned by
Mr Lee after one year. Bahagian B
A RM320
B RM350
C RM380 7. Tandakan (3) bagi pernyataan yang betul dan
D RM400 (7) bagi pernyataan yang salah berkaitan dengan
kelebihan kad kredit.
4. Antara berikut, yang manakah bebas risiko? Mark (3) for true statement and (7) for false statement
Which of the following is risk free? regarding advantages of credit card.
A Hartanah [4 markah/ 4 marks]
Real estate Jawapan/ Answer:
B Saham syarikat (a) Tidak perlu membawa wang tunai
Company shares yang banyak. ( 3 )
C Simpanan tetap Do not have to carry a lot of cash.
Fixed deposits
D Unit amanah (b) Boleh berbelanja tanpa had.
Unit trust Can spend without limits. ( 7 )




43 © Penerbitan Pelangi Sdn. Bhd.






03_ Hybrid Mate Tg3.indd 43 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
Cost averaging strategy is a technique of investing in
(c) Dapat menikmati tempoh tanpa
faedah. ( 3 ) shares with fixed amount for a certain period regardless
Can enjoy an interest free period. of the stock market conditions.
(d) Proses pembayaran yang mudah Faedah sama rata
dan cekap. ( 3 ) (c) dikira pada jumlah
Easy and efficient payment process. pinjaman asal sepanjang tempoh pinjaman
dan dikekalkan hingga akhir tempoh
8. Isikan ruang kosong pada jadual berikut dengan pinjaman.
jawapan yang tepat.
Fill in the blanks in the following table with the correct answers. Flat interest is calculated on the original
[4 markah/ 4 marks] loan amount over the loan period and is maintained until
Jawapan/ Answer: the end of loan period.

Jenis Tahap Tahap Tahap
pelaburan kecairan (d) Faedah atas baki yang dikenakan
Type of risiko pulangan Liquidity setiap bulan bergantung kepada jumlah
investment Risk level Return level level
baki pinjaman pada bulan tersebut.
Bebas
Simpanan Rendah Tinggi Interest on balance charged each month
Savings risiko Low High
Risk free depends on the amount of the loan balance for that
month.
Hartanah Rendah Tinggi Rendah
Real estate Low High Low
Bahagian C


Saham Tinggi Tinggi Sederhana 10. (a) Jadual di bawah menunjukkan butiran
Shares High High Moderate
tentang pelaburan Encik Mohamad dalam
tempoh satu tahun.
The table below shows the details of Encik Mohamad’s
9. Isikan tempat kosong dengan perkataan yang investment in the period of one year.
diberikan.
Fill in the blanks with the given words. Jenis pelaburan Amanah saham P
Type of investment Unit trust P
Faedah sama rata Faedah atas baki RM400 sebulan
Flat interest Interest on balance Amaun pelaburan untuk satu tahun
Amount of investment RM400 per month for
Nilai pulangan one year
pelaburan Strategi pemurataan
Cost averaging strategy
Return on investment Kos purata seunit
Average cost per unit RM0.75
[4 markah/ 4 marks]
Jawapan/ Answer: (i) Hitung jumlah bilangan unit amanah
saham P yang dibeli oleh Encik
(a) Nilai pulangan pelaburan merupakan Mohamad.
nisbah keuntungan atau kerugian yang Calculate the total number of units of the unit trust
P bought by Encik Mohamad.
diperoleh daripada suatu pelaburan.
[2 markah / 2 marks]
Return on investment is a ratio of profit or loss (ii) Encik Mohamad mendapat dividen 9.5
derived from an investment. sen seunit selepas satu tahun.
Hitung jumlah dividen yang diterima
oleh Encik Mohamad.
(b) Strategi pemurataan merupakan satu
Encik Mohamad received a dividend of 9.5 sen per
teknik melabur dalam saham dengan unit after one year.
amaun tetap untuk tempoh tertentu tanpa Calculate the total dividend received by Encik
Mohamad.
mengambil kira keadaan pasaran saham. [2 markah / 2 marks]





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03_ Hybrid Mate Tg3.indd 44 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
Jawapan/ Answer: Jawapan/ Answer:
(i) Jumlah bilangan unit amanah saham P Pinjaman yang dibuat / Borrowing amount
yang dibeli oleh Encik Mohamad = 95% × RM98 000
Total number of units of unit trust bought by Encik = RM93 100
Mohamad

= 12 × RM400 Jumlah pinjaman yang perlu dibayar
Total loan needed to pay
RM0.75 = RM93 100 + RM93 100 × 2.85% × 7
= 6 400 unit/units = RM111 673.45

(ii) Jumlah dividen yang diterima oleh
Encik Mohamad
Total dividend received by Encik Mohamad 11. (a) Aminah mempunyai RM5 000 dalam akaun
= 6 400 × RM0.095 simpanannya. Berapa lama diperlukan
= RM608 supaya simpanannya itu berganda dengan
kadar faedah mudah 2% setahun?
Aminah has RM5 000 in her savings account. How long is
needed to make her savings double if the simple interest
rate is 2% per annum?
(b) Rajeswari menyimpan RM40 000 di suatu [2 markah/ 2 marks]
akaun simpanan tetap dengan kadar faedah Jawapan/ Answer:
3% setahun dan pengkompaunan setiap 4 Katakan t = masa/ Let t = time,
bulan. Hitung nilai matang Rajeswari selepas 2(5 000) = 5 000 + (5 000)(0.02)(t)
3 tahun. 10 000 = 5 000 + 100t
Rajeswari deposits RM40 000 in a fixed deposit account 100t = 5 000
with an interest rate of 3% per annum and compounded
every 4 months. t = 50 tahun/ 50 years
Calculate the maturity value of Rajeswari after 3 years.
[2 markah/ 2 marks]
Jawapan/ Answer:
Nilai matang / Maturity value (b) Institusi kewangan A menawarkan satu
pelan pinjaman peribadi dengan faedah
1
= RM40 000 1 + 0.03 2 (3)(3) sama rata kepada pelanggannya. Diberi
3
= RM43 747.41 bayaran ansuran untuk pinjaman peribadi
sebanyak RM150 000 untuk tempoh 10
tahun ialah RM1 576.40 sebulan, cari kadar
faedah tahunan bagi pinjaman peribadi
yang ditawarkan.
(c) Jadual di bawah menunjukkan harga dan Financial Institution A offers a personal loan plan with
flat interest to the customers. Given that the instalment
wang pendahuluan sebuah kereta. for a personal loan of RM150 000 and a loan period of
The table below shows the price and the down payment 10 years is RM1 576.40 per month, find the yearly interest
of a car. rate of the personal loan offered.
Harga Wang pendahuluan [4 markah/ 4 marks]
Price Down payment Jawapan/ Answer:
RM98 000 5%
Jumlah bayaran balik
Annie membeli kereta tersebut. Dia Total repayment
membayar wang pendahuluan dengan kad = 1 576.40 × 12 × 10
debit dan bakinya dibayar secara ansuran = RM189 168
selama 7 tahun dengan kadar faedah sama
rata 2.85% setahun. Katakan r = kadar faedah tahunan,
Let r = yearly interest rate,
Hitung jumlah pinjaman yang perlu dibayar
balik oleh Annie. 189 168 = 150 000 + 150 000(r)(10)
Annie bought the car. She paid the down payment by r = 189 168 – 150 000
debit card and the balance is payable in instalments over 150 000(10)
7 years with a flat interest rate of 2.85% per annum. = 0.026
Calculate the total loan needed to pay by Annie. = 2.6%
[4 markah/ 4 marks]




45 © Penerbitan Pelangi Sdn. Bhd.






03_ Hybrid Mate Tg3.indd 45 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
(c) Baki tertunggak pada penyata kad kredit 12. (a) Khairul menyimpan RM24 000 dalam akaun
Yusnita bertarikh 17 Januari 2020 ialah simpanan di sebuah bank, mengikut prinsip
RM4 800. Tempoh tanpa faedah ialah 20 wadiah. Selepas satu tahun, dia menerima

hari daripada tarikh penyata. Yusnita tidak hibah sebanyak RM624 daripada bank.
membuat sebarang bayaran untuk bulan Hitung peratus hibah itu.
Januari. Pengeluar kad kredit mengenakan Khairul deposits RM24 000 in a savings account in a
caj kewangan 15% setahun ke atas baki bank, according to the principle of wadiah. After one
year, he receives a hibah of RM624 from the bank.
tersebut dalam kadar harian. Pengeluar kad Calculate the percentage of the hibah.
kredit juga mengenakan caj bayaran lewat [2 markah/ 2 marks]
minimum RM10 atau 1% daripada jumlah
baki pada tarikh penyata bulan seterusnya. Jawapan/ Answer:
Jika Yusnita tidak menggunakan kad kredit Peratus hibah
selepas transaksi terakhir pada penyata Percentage of hibah
Januari, hitung baki tertunggak kad kredit RM624
Yusnita pada penyata bertarikh 17 Februari = RM24 000 × 100%
2020. KBAT Mengaplikasi = 2.6%
The outstanding balance of Yusnita’s credit card
statement dated 17 January 2020 is RM4 800. The interest
free period is 20 days from the statement date. Yusnita
does not make any payment for January. The credit card
issuer charges a financial charge of an annual interest
rate of 15% on the balance based on daily rate. The credit
card issuer also charges a minimum late payment charge
of RM10 or 1% of total outstanding balance as on the
next statement date. (b) Baiyah ingin membuat pinjaman kereta
If Yusnita does not use her credit card after the last
transaction on the January statement, calculate her daripada sebuah bank dengan kadar faedah
credit card outstanding balance on the statement dated sama rata 2.8% setahun dan tempoh bayaran
17 February 2020. balik 7 tahun. Baiyah mampu membayar
[4 markah/ 4 marks] ansuran bulanan sehingga RM1 000 sebulan.
Berapakah amaun maksimum yang boleh
Tip KBATKBAT dipinjam oleh Baiyah daripada bank itu?
KBAT
Baiyah wants to make a car loan from a bank with a flat
Hitung caj kewangan dan caj bayaran lewat. interest rate of 2.8% per annum and repayment period of
Seterusnya, hitung baki tertunggak. 7 years. Baiyah is able to pay a monthly instalment of up
Calculate the finance charge and late payment charge. to RM1 000 a month.
Hence, calculate the outstanding balance. What is the maximum amount Baiyah can loan from the
bank?
Jawapan/ Answer: [4 markah/ 4 marks]
Tempoh dikenakan caj kewangan
Period subject to financial charges Jawapan/ Answer:
17 Jan → 17 Feb Jumlah bayar balik / Total repayment
= 11 hari / days = RM1 000 × 12 × 7

Caj kewangan / Financial charge = RM84 000
= RM4 800 × 15% × 11 Katakan amaun maksimum yang dipinjam
= RM21.70 365 ialah RMP.
Let the maximum loan amount be RMP.
Caj bayaran lewat P + P × 2.8% × 7 = 84 000
Late payment charge P + 0.196P = 84 000
= 1% × (RM4 800 + RM21.70) 1.196P = 84 000
= 1% × (RM4 821.70) P = RM70 234.11
= RM48.22

Baki tertunggak pada penyata Februari
Outstanding balance on February statement
= RM4 800 + RM21.70 + RM48.22
= RM4 869.92








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03_ Hybrid Mate Tg3.indd 46 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3
(c) Jadual di bawah menunjukkan butiran Jawapan/ Answer:
pinjaman rumah Encik Wong. Faedah bulan yang pertama
The table below shows the details of Mr Wong housing First month interest
loan. 1
= RM480 000 × 5.4% ×
Amaun pinjaman = RM2 160 12
Loan amount RM480 000
Jumlah pinjaman pada akhir bulan pertama
Kadar faedah Loan at the end of first month
atas baki 5.4% setahun = RM480 000 + RM2 160
5.4% per annum
Interest rate on balance = RM482 160
Tempoh bayaran 30 tahun Baki selepas bayaran ansuran bulan pertama
balik 30 years Balance after the first month of the instalment payment
Repayment period
= RM482 160 – RM2 695.35
Ansuran bulanan RM2 695.35 = RM479 464.65
Monthly instalment
Faedah bulan yang kedua
Hitung jumlah faedah yang dibayar oleh Second month interest
Encik Wong bagi dua bulan yang pertama. = RM479 464.65 × 5.4% × 1
Calculate the total amount of interest payable by Mr = RM2 157.59 12
Wong for the first two months.
[4 markah/ 4 marks] Jumlah faedah bagi dua bulan yang pertama
Total interest for the first two months
= RM2 160 + RM2 157.59
= RM4 317.59























































47 © Penerbitan Pelangi Sdn. Bhd.






03_ Hybrid Mate Tg3.indd 47 20/08/2021 5:30 PM

Matematik Tingkatan 3 Bab 3








1. Rosline ingin membeli sebuah buku secara atas talian. Dia melayari
internet dan menjumpai dua promosi berikut:
Rosline wants to buy a book online. She surfs the Internet and finds two promotions:
Syarikat P di Malaysia Syarikat Q di Amerika Syarikat
menawarkan harga promosi menawarkan harga promosi
RM89. Caj kiriman ke alamat USD15.99. Untuk tempahan
rumah Rosline ialah RM8. daripada luar Amerika Syarikat,
Company P in Malaysia offers a caj kiriman USD11 dikenakan.
promotional price of RM89. The Company Q in United States offers a
shipping charge to Rosline’s house promotional price of USD15.99. For orders
address is RM8. outside United States, USD11 shipping
charges apply.

Rosline bercadang membuat pembayaran dengan kad kredit. Pihak
bank akan mengenakan caj tambahan 1% ke atas setiap transaksi
daripada luar negara. Jika kadar semasa pertukaran mata wang untuk
ringgit Malaysia adalah RM1 bersamaan dengan USD0.24, tawaran
manakah yang Rosline harus pilih? Berikan justifikasi anda.
Rosline intends to make payment by credit card. The bank will charge an additional 1% on
each transaction from abroad. If the current exchange rate for Malaysian ringgit is RM1
equivalent to USD0.24, which offer should Rosline choose? Give your justification.

Syarikat P/ Company P: • Bandingkan harga sebenar
Harga promosi = RM89 yang akan dibayar jika membeli
daripada dua syarikat tersebut.
Promotional price Compare the actual price to be paid if
buying from the two companies.
Harga sebenar yang akan dibayar = RM89 + RM8
Actual price to be paid = RM97



Syarikat Q/ Company Q:
Harga promosi = USD15.99 × (1 ÷ 0.24) = RM66.63
Promotional price

Caj kiriman = USD11 × (1 ÷ 0.24) = RM45.83
Shipping charge

Caj tambahan oleh bank = RM66.63 × 1 = RM0.67
Additional charges by bank 100

Harga sebenar yang akan dibayar = RM66.63 + RM45.83 + RM0.67
Actual price to be paid = RM113.13

Rosline harus pilih tawaran dari syarikat P. Harga sebenar yang akan
dibayar adalah lebih murah. Rosline dapat berjimat RM16.13.
Rosline should choose offer from company P. The actual price to be paid is cheaper. Rosline
can save RM16.13.
Kuiz 3








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