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Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
10. (a) FALSE (b) TRUE (ii) Outstanding balance (b) (i) Total repayment
(c) TRUE (d) TRUE = RM308.82 – RM16 = RM40 000 + (RM40 000 × 0.05 × 4)
= RM292.82 = RM48 000
11. Finance charge RM48 000
Make personal budget. Always late in making repayment. Start saving. Instalment =
Membuat belanjawan peribadi. Selalu membuat pembayaran balik dengan lewat. Memulakan simpanan. 18 18 4 × 12

= RM308.82 × ×  = RM1 000
100 365 (ii) Total repayment
18 20
+ (RM292.82 × ×  = RM40 000 + (RM40 000 × 0.05 × 3)
100 365
Smart management of credits and debts. = RM2.74 + RM2.89 = RM46 000
Pengurusan kredit dan hutang yang bijak. RM46 000
= RM5.63 Instalment =
3 × 12
Late payment charge = RM0 = RM1 277.78
Current amount Amount James needs to top up
Use auto billing to pay bill. Manage credit card wisely. Excessive loan. = RM292.82 + RM5.63+ RM0 = RM1 277.78 – RM1 000
Membayar bil secara automatik. Menguruskan kad kredit dengan bijak. Meminjam dengan banyak. = RM298.45 = RM277.78
15. (a) Total repayment Total
= RM110 000 + (RM110 000 × 0.05 × 9) 1 repayment
12. Late payment charge = RM0 = RM159 500 (c) (i) 4 200 × = 7 × 12
4
Total
A Current amount in April statement Instalment = RM159 500 repayment
= RM1 710 + RM30.60 + 0 9 × 12 1 050 =
= RM1 740.60 = RM1 476.85 84
Weakness (b) Total repayment Total repayment = RM88 200
of credit (b) (i) Finance charge 46 = RM80 000 + (RM80 000 × 0.07 × 5) Let r = annual interest rate
1.5
card usage = RM2 500 × 100 × 30 = RM108 000 RM88 200 = RM64 000 +
Kekurangan (RM64 000 × r × 7)
penggunaan = RM57.50 Instalment = RM108 000 r = 0.054
kad kredit D Late payment charge 5 × 12 r = 5.4%
C = (RM2 500 + RM57.50) × 1% = RM1 800
= RM25.58 (c) Total repayment (ii) Total repayment
= RM50 000 + (RM50 000 × 0.06 × 4) = RM64 000 + (RM64 000 × 0.035
Current amount in May statement = RM62 000 × 5)
= RM2 500 + RM57.50 + RM25.58 RM62 000 = RM75 200
13. (a) Finance charge = RM2 583.08 Instalment = 4 × 12 RM75 200
Caj kewangan Instalment =
(ii) Additional repayment amount = RM1 291.67 5 × 12
(b) Late payment charge = RM2 583.08 – RM2 500 = RM1 253.33
Caj bayaran lewat 16. (a) (i) Total repayment
= RM83.08 = RM81 000 + (RM81 000 × 0.08 × 5) Dahniah cannot shorten the loan
(c) Total amount of outstanding balance period because the instalment for 5
Jumlah baki tertunggak (c) (i) USD1.00 = RM3.92 = RM113 400 years is more than of her salary.
1
14. (a) Outstanding balance Toy price in Ringgit Malaysia Instalment = RM113 400 4
5 × 12
= RM1 800 – RM90 = USD78 × RM3.92 = RM1 890
= RM1 710 USD1.00
= RM305.76 (ii) Let t = new loan period
Finance charge Current amount on July statement = RM81 000 + (RM81 000

1.5
= RM1 800 × 100 × 15  = RM305.76 + (RM305.76 × 1%) × 0.08 × t)
= RM81 000 + 6 480t
30
= RM308.82
1.5

+ RM1 710 × 100 × 20  1 290 = RM81 000 + 6 480t
t × 12
30
= RM13.50 + RM17.10 15 480t = 81 000 + 6 480t
= RM30.60 t = 9 years

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BOOKLET ANS MATH F3.indd 7 03/01/2020 10:20 AM