E-BOOK BUZZMATH PRESENTED BY : GROUP 1
MEET THE GROUP ZULAIKHA SYAZWANI NAJWA IRDINA SITI NURAQILAH 2024936359 2024573551 SITI FATIMAH 2024572991 KHUDZAIRI KHALID 2024936291 Advisor
Group Member Introduction Annuity Notes TABLE OF CONTENT Simple Interest notes Question Answer
INTRODUCTION BuzzMath is an innovative teaching tool that created by part 1 UiTM student.Creative to improve the level of understanding of students in the field of social sciences mainly business mathematics subjects because of their low cost and easy to do anywhere.
SIMPLE NOTES ANNUITY ANNUITY FORMULA S=R [( 1+i) - 1 ] / i I = S-R ( n ) n S= future value R = periodic payment i = k/m ( interest rate per interest period ) / ( no of payments ) n = mt( term of investment for n interest period I= interest amount
SIMPLE NOTES SIMPLE INTEREST SIMPLE INTEREST FORMULA SIMPLE AMOUNT FORMULA I = Prt S = P ( 1+rt ) P= Principal r = rate of simple interest t = time or term in years
PART A
QUESTION 1 Mikayla deposited rm7,000 in an account at 8% compounded every three months. After t years, the amount in her account becomes RM22,967.22. Find the value of t. PART A
ANSWER 1 INFORMATION p = RM7,000 k= 8/100 = 0.08 m= 4 t= ? s= RM22,967.22 2,967.22=7000(1+0.08/4) S=P(1+k/m) 4(t) 22,967.22/7000=( 1+0.08/4) 4(t) m(t) 3.281031429=( 1.02) 4(t) log 3.281031429= 4t log 1.02 log 3.281031429/log1.02= 4t 60=4t 60/4 =t 15 years = t #
QUESTION 2 Osman invest RM50,000.00 in Asia Bank with k% compounded every two moths for 20 years. If the balance in his account after 20 years is RM538258.15, find k%
ANSWER INFORMATION P= RM50,000 k= ? m= 6 t= 20 s= rm538285.15 538258.15 = 50000( 1+ k/6 ) 538258.15/50000 = ( 1+k/6) 10.765163 = ( 1+k/6) 6(20) S=P( 1+k/m)m(t) 1.02=1+k/6 1.02-1=k/6 0.02x6=k 0.12x100=k 12%=k 6(20) ¹²⁰√10.765163=1+k/6 # 6(20)
QUESTION 3 Sheila deposited a certain sum of money into a savings account 4.4% compounded every six months. After 6 1/2 years, the bank changed the rate to 5.7% compounded quarterly. If the accumulated amount in her savings account at the end of 12 years is RM10, 970.25, find the initial deposit and the total interest obtained by Sheila.
ANSWER k=4.4/100 =0.044 m=2 p=? 6 1/2 y k=5.7% m=4 k=5.7/100 m=4 12 y S=10970.25 1. S=P(1+k/m) m(t) =P(1+0.044/2) 13 P=1.326971653 2. S=P(1+k/m) m(t) 10970.25=1.326971653P(1+0.057/4) 4(5.5) 10970.25/1.326971653=1.326971653P 8035.745328/1.326971653=P RM 6055.70=P# Interest I=S-P I=10970.75-6055.70 I=RM 4915.05 #
PART B
QUESTION 1 Kamal invested RM450 every month into an investmen scheme for 10 years with an interest rate of 4,7% compounded monthly. Find the future value at the end of the investment period.
ANSWER INFORMATION k= 0.047 t=10 years R= RM450 m = 12 S = R [ 1+k/m ] - 1 _____________ k/m s= 450 [ (1_+_0__._0_4__7_ ) 12 12(10) -1 ___________ 0.047 ________________ 12 S= RM68766.82#
QUESTION 2 The cash price of a townhouse is RM 350 000.Rajit paid a down payment at 10% and took a loan at 3.25% compounded monthly to finance the balance.The loan is to be repaid over 30 years by monthly installments .Find the amount of monthly payment.
ANSWER m=12 i=3.25>0.0325>0.0325/12 t=30y 350 000x0.1 =35 000 A=R(1-(1+i) i 315000=R[(1-(1+0.0325/12) 0.0325/12 315000 _______ -mt _______________________ ] -360 _____________ 229.7760754 =R RM1370.90=R
QUESTION 3 Fairuz buys an apartment that cost RM409,630.64 and pays a downpayment at RM30,000.The balance of the apartment is settled through a loan from a bank which charges interest of 2.5% compounded monthly. The repayment of the loan made through monthly payment for 30 years. i) Find the monthly payment ii) If Fairuz fails to pay the first three monthly installments, how much should he pay on the fourth payment to settle all the outstanding arrears
ANSWER 1.balance= cash price - deposit = 409630.64 - 30,000 = RM 379,630.64 A= R [ 1-(1+i)/ i ) 379630.644 = R ( 1- 1+1/480) n _______________ 1/480 379630.64 = R ( 253.0870938) _3_7__9_6_3_0__._6_4 253.0870938 = R RM 1449.10 = R# i) INFORMATION A= 379630.64 m= 12 k=0.025 i= 0.025/12 = 1/480 t=30 n=12(30)=360
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