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Accounting Questions?
Accounting Questions?
1. You have set up an ordinary annuity that will pay you $650.00 a month for the next 25 years. You will earn interest at a rate of 5.5% compounded monthly. What amount did you invest to accomplish this goal?
2. If an annuity was set up for semi-annual payments at the end of each period in the amt of $1350, what would be the value of this annuity after 15 1/2 yrs with interest compounded semiannually at a rate of 4%?
3. A company requires the amount of $850,000 in twenty(20) years to retire a bond issue. Assume they earn 5% interest compounded quarterly. What amount would they have to pay quarterly to be able to retire this debt in 20 years?
4. Jeff and Darleen bought a house for $74,000. After making a $4000 down payment, they took out a $70,000 mortgage at 7% compounded semiannually for 30 years. How large are their semiannual payments to cover the principal and interst? and How much total money will they wind up paying fo th house?
Cans someone help with the formulas/get me start
Answers:
Nobody: 1. The present value of an ordinary annuity is PV= P*(1- (1/1+i)^n)/i with PV = present Value, P= Payment amount i = interest and n=number of payments. So for this specific problem, n=300 P=650 and i=.0045833. The answer is 105848.11 (note: since interest is 5.5% compunded monthly, i= .055/12)
2. You cannot find a current value for this equation, but If this question is asking for the accumulated value, then you would use the formula AV= P*(((1+i)^n - 1)/i)). For this problem, P=1350 n=31 and i=.02. The answer would be 57212.25. (note: since interest is 4% compounded semiannually, i=.04/2)
3. You can use the same formula for AV, but now you are tryiong to find P. So, AV=850000 n=80 i=.0125. The answer should be 6244.55. (note: i = .05/4)
4. Part 1, use the formula from number 1 but we are now trying to find P. So, PV = 70000 i=.035 n=60. The answer is 2806.20. (note: i= .07/2)
Part2. For the total amount they paid, you have 60 payments of $2806.20 + a $4000 down payment, so the total is 4000+60*2806.20 which equals: 172372.00.
Hope it helps
2006-10-22 20:54:37
1. $120,000.
2. Data is insufficient. If an annuity certain, either the premium or the total number of payments would be needed to figure this. If a life annuity, the age of the annuitant must be known.
3. About $7500.
4. About $3000, and $180,000.
2006-10-22 21:09:29
Data:
r = 5.5% pa = 0.0055/12 per month; $M = $850; t = 25 years = 300 months
S = 850(1 + 0.0055/12)^300 + 850(1 + 0.0055/12)^299 + ... + 850(1 + 0.0055/12)^1
= 850((1 + 0.0055/12)^1 + (1 + 0.0055/12)^2 + ... + (1 + 0.0055/12)^300) (on reversinbg the order)
= 850((1 + 0.0055/12)((1 + 0.0055/12)^300 - 1)/((1 + 0.0055/12) - 1) (on summing the GP using S = a(rn - 1)(r - 1))
? 273421.39
ie amount invested = $273,421.39 (to nearest cent)
2, Superannuation again
Data:
r = 4% = 0.04 per half year; $M = $1350; t = 15.5 years = 31 half years
S = 1350(1 + 0.04) + 1350(1 + 0.04)? + 1350(1 + 0.04)? + ...... + 1350(1 + 0.04)^31
= 1350(1.04 + 1.04? + 1.04? + ... + 1.04^31)
= 1350(1.04(1.04^31 - 1)/(1.04 - 1) (Summing GP)
? 118396.98 (nearest cent)
Thus after 15.5 years its value will be $118,396.98 (nearest cent)
3. Superannuation again only this time we know $S and need $A
Data:
$S = $850,000; r = 5%pa = 0.0125 per quarter; t = 20 years = 80 quarters
850000 = S{1.0125 + 1.0125? + 1.0125? + ..... + 1.0125^80)
= S(1.0125)(1.0125^80 -1)(1.0125 - 1) (Summing GP)
Therefore S = 850000 x 0.0125/(1.0125(1.0125^80 -1))
? 6167.45 (nearest cent)
Thus the company would pay $6,167.45 (nearest cent) per quarter over 20 years to retire the debt
4. Let each payment = $S.
Amount borrowed = $70,000.
Interest rate = 7% pa = 3.5% per half year = 0.035.
Time = 30 years = 60 half years (ie 60 installments)
After 1st installment
Amount owing (in $) = A1 = 70000 (1 + 0.035) - S
After 2nd installment
Amount owing (in $) = A2 = A1 (1 + 0.035) - S
= (70000 (1 + 0.035) - S)(1 + 0.035) - S
= 70000 (1 + 0.035)? - S(1 + 0.035) - S
After 3rd installment
Amount owing (in $) = A3 = A2 (1 + 0.035) - S
=(70000 (1 + 0.035)? - S(1 + 0.035) - S)(1 + 0.035) - S
=70000 (1 + 0.035)? - S(1 + 0.035)? - S(1 + 0.035) - S
........
After 60th installment
Amount owing (in $) = A60 = A59 (1 + 0.035) - S
=70000(1 + 0.035)^60 - S(1 + 0.035)^59 - ... - S(1 + 0.035)? - S(1 + 0.035)? - S(1 + 0.035) - S
= 0 (pay off in 60 installments!!)
Thus 70000(1 + 0.035)^60 = S(1 + 0.035)^59 + ...+ S(1 + 0.035)? + S(1 + 0.035)? + S(1 + 0.035) + S
= S(1 + 1.035 + 1.032? + 1.035? + .... + 1.035^59)
= S (1.035^60 - 1)/(1.035 - 1)
So S = 70000(1 .035)^60*0.035/(1.035^60 - 1)
Divide top and bottom by (1 .035)^60
S = 0.035 * 70000/(1 - (1.035)^(-60))
= 2806.20 (nearest cent)
So they pay $2806.20 every half year (nearest cent).
Total amount repaid = 60 x $2806.20
= $168,372
2006-10-22 21:21:36