Jake borrowed $6,500 for 18 months at 5.05%. How much interest will he pay? Sarah borrowed $1,000...

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Finance

  1. Jake borrowed $6,500 for 18 months at 5.05%. How much interestwill he pay?
  1. Sarah borrowed $1,000 on March 12, 2007 and repaid it onSeptember 4, 2010. For how many days was the loan outstanding?

Page 2 of5                                                                                                                           A

  1. A loan of $1,100 can be repaid in 8 months by paying theprincipal sum borrowed plus $44 in interest. What was the rate ofinterest charged?

4. How many days would it take for$1,500 to earn $20.96 at 8.5%?

5. What amount of money would earn$54.69 in 3 months at 8.75%?

  1. Helen borrowed $500 at 4% from her Mom on April 4, and promisedto repay the money in 90 days.
  1. What is the due date?
  1. How much will she owe her Mom on that date?

Page 3 of5                                                                                                                           A

  1. Dave borrowed money from Steve at 8% for 9 months. He now owesSteve $424. How much did be borrow?
  1. Debt payments of $600 each are due 3 months and 6 months fromnow respectively. If interest is at 10%, what single payment isrequired to settle the debt today?

FORMULA SHEET

i = j / m

I = Prt

t = I / Pr

P = I / rt

S = P(1 + i)n

f = (1 + i)m- 1

n = ln (S / P)

ln (1 + i)

Sn = R[(1 + p)n- 1]

p

R=          Sn

[(1 + p)n- 1] / p

n = ln [1 + pSn/R]

ln (1 + p)

Sn(due) = R[(1 + p)n- 1](1+ p)

p

n = ln [1 +[pSn(due) / R(1 + p)] ln(1 + p)

  1. = -ln[1 - (p[1 + p]dAn(def))/R] ln(1 + p)

An(def) = R [1 - (1+ p)-n] p(1 + p)d

A = R / p

m = j / i

S = P(1 + rt)

r = I / Pt

P = S / (1 + rt) = S(1 + i)-n

c = # of compoundings/# of payments

p = (1 + i)c- 1

i = [S / P] 1/n- 1

An = R[1 - (1 + p)-n]

p

R=          An

[1 - (1 + p)-n] / p

  1. = -ln [1 - pAn/R] ln (1 + p)

An(due) = R[1 - (1 + p)-n](1 + p)

p

n = -ln[1 - [pAn(due) / R(1 +p)]

ln(1 + p)

  1. = -ln{R[1-(1 + p)-n] / pAn(def)}ln(1 + p)

Sn(def) = Sn

A(due) = (R / p)(1 + p)

Answer & Explanation Solved by verified expert
4.0 Ratings (554 Votes)
Jake borrowed 6500 for 18 months at 505 How much interest will he pay Interest Prt 6500 x 505 x 18 12 492375 Sarah borrowed 1000 on March 12 2007 and repaid it on September 4 2010 For how many days was the loan outstanding Outstanding days Sep 4 2010    See Answer
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Transcribed Image Text

Jake borrowed $6,500 for 18 months at 5.05%. How much interestwill he pay?Sarah borrowed $1,000 on March 12, 2007 and repaid it onSeptember 4, 2010. For how many days was the loan outstanding?Page 2 of5                                                                                                                           AA loan of $1,100 can be repaid in 8 months by paying theprincipal sum borrowed plus $44 in interest. What was the rate ofinterest charged?4. How many days would it take for$1,500 to earn $20.96 at 8.5%?5. What amount of money would earn$54.69 in 3 months at 8.75%?Helen borrowed $500 at 4% from her Mom on April 4, and promisedto repay the money in 90 days.What is the due date?How much will she owe her Mom on that date?Page 3 of5                                                                                                                           ADave borrowed money from Steve at 8% for 9 months. He now owesSteve $424. How much did be borrow?Debt payments of $600 each are due 3 months and 6 months fromnow respectively. If interest is at 10%, what single payment isrequired to settle the debt today?FORMULA SHEETi = j / mI = Prtt = I / PrP = I / rtS = P(1 + i)nf = (1 + i)m- 1n = ln (S / P)ln (1 + i)Sn = R[(1 + p)n- 1]pR=          Sn[(1 + p)n- 1] / pn = ln [1 + pSn/R]ln (1 + p)Sn(due) = R[(1 + p)n- 1](1+ p)pn = ln [1 +[pSn(due) / R(1 + p)] ln(1 + p)= -ln[1 - (p[1 + p]dAn(def))/R] ln(1 + p)An(def) = R [1 - (1+ p)-n] p(1 + p)dA = R / pm = j / iS = P(1 + rt)r = I / PtP = S / (1 + rt) = S(1 + i)-nc = # of compoundings/# of paymentsp = (1 + i)c- 1i = [S / P] 1/n- 1An = R[1 - (1 + p)-n]pR=          An[1 - (1 + p)-n] / p= -ln [1 - pAn/R] ln (1 + p)An(due) = R[1 - (1 + p)-n](1 + p)pn = -ln[1 - [pAn(due) / R(1 +p)]ln(1 + p)= -ln{R[1-(1 + p)-n] / pAn(def)}ln(1 + p)Sn(def) = SnA(due) = (R / p)(1 + p)

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