Page 23 - TOP ONE MATHS F3
P. 23

Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd





 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

 PT3 Standard Practice  3    3.  Answer: D  Section C         HOTS Challenge
   4.  Answer: D    1.  (a)  Finance charge
 Section A
   5.  P = RM230, r = 5.5% = 0.055     = RM900 ×   1.5   ×   48
   1.  Answer: C          100    30                    (a)  Total repayment
      Interest = RM177.10                                  = RM90 000 + (RM90 000 × 0.034 × 8)
   2.  Initial capital        Interest = Prt     = RM21.60     = RM114 480
      = (RM2.05 × 6 000) + (RM1.00 × 2 700)        177.10 = 230(0.055)t     Late payment charge  RM114 480
      = RM15 000                                           Instalment =
       177.10 = 12.65t     = (RM900 + RM21.60) × 1%                         8 × 12
      Dividend          t = 14 years     = RM9.22                     = RM1 192.50
      = RM135 × 3       Answer: B        Late payment charge = RM10.00                    0.034  1(8)
                                                                                      
      = RM405                                          (b)  Matured value = RM10 000 1 +    1   
   6.  Finance charge     Current amount in November statement
      Capital gains  1.5  50     = RM900 + RM21.60 + RM10                 = RM13 066.65
       = RM16 500 – RM15 000       = RM1 200 ×   100   ×   30     = RM931.60     Car loan interest for RM10 000
      = RM1 500       = RM30                               = RM10 000 + (10 000 × 0.034 × 8)
                                  5
      Total returns        Late payment charge  (b)  P = RM20 000, r =   100  per annum,         = RM12 720
      = RM1 500 + RM405         = (RM1 200 + RM30) × 1%     t = 3 years     Shamsiah should not use the money in fixed
      = RM1 905         = RM12.30     (i)  Half yearly, n = 2  deposit account because the interest gained
 1 905
                              
      ROI =    × 100       Total outstanding in August  0.05  2(3)  in fixed deposit account is higher than the
 15 000           = RM20 000 1 +    2                     interest in car loan.
         = 12.7%       = RM1 200 + RM30 + RM12.30
      = RM1 242.30  = RM23 193.87
      Answer: A
      Answer: C     (ii)  Quarterly, n = 4
                              
                  = RM20 000 1 +   0.05  4(3)
                                    4
 Section B        = RM23 215.09
   1.   Simple interest   Simple interest   (c)  Down payment
 Prinsipal  Total savings
 Prinsipal  rate per annum  Duration  rate  Jumlah simpanan      = RM3 699 × 10%
 Tempoh
 (RM)  Kadar faedah mudah   Faedah mudah          (RM)     = RM369.90
 setahun  diperoleh
              Amount financed
 (a)  365 000  6.8%  1 year  I = 24 820  389 820     = RM3 699 – RM369.90
 (b)  2 500  14.3%  4months  I = 119.17  2 619.17     = RM3 329.10
 4 Bulan      Finance charges
              = 8% × RM3 329.10
              = RM266.33
   2.   Savings  How easily a savings or an asset can be converted into cash.
 Simpanan  Keupayaan simpanan atau aset diubah menjadi wang tunai.
 An action of gaining profit in the future from income and capital
 Investment  gain.
 Pelaburan  Tindakan mendapat  pulangan pada  masa  hadapan melalui pendapatan dan
 keuntungan modal.

 Liquidity  Money that is kept in the bank or other than bank like piggy
 Kecairan  bank.
 Wang yang disimpan di bank mahupun bukan bank seperti dalam tabung.

 Interest  Charge on the money borrowed or invested.
 Faedah  Caj ke atas wang yang dipinjam atau dilaburkan.





 13  © Penerbitan Pelangi Sdn. Bhd.  © Penerbitan Pelangi Sdn. Bhd.  14

 A8



 BOOKLET ANS MATH F3.indd   8                                                             03/01/2020   10:20 AM
   18   19   20   21   22   23   24   25   26   27   28