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A.B. deposits m*$100 into a bank account. The bank credits interest at nominal annual rate j, convertible semiannually, for the first seven years and nominal annual rate 2j, convertible quarterly, for all years thereafter. The accumulated value of the account at the end of 5 years is X. The accumulated value of the account at the end of 10.5 years is 1.98*m*$100...Given that m=1 Find X

Guest Dec 6, 2014

Best Answer 

 #1
avatar+91437 
+5

 

 I have left out M because down the bottom it says M=1

I have not included all of the initial working, but most is here. 

 

After 5 years            

 $$X=100(1+0.5J)^{10}\qquad(1)$$

 

After 7 years 

 

$$FV=100(1+0.5J)^{14}$$

 

After 10.5 years    (3.5 years more)

 

$$FV=[100(1.05J)^{14}](1+0.5J)^{14}=100(1.05J)^{28}$$

 

$$\\1.98*100=100(1+0.5J)^{28}\\\\
1.98=(1+0.5J)^{28}\\\\
(1.98)^{(1/28)}=1+0.5J\\\\
(1.98)^{(1/28)}-1=0.5J\\\\
2[(1.98)^{(1/28)}]-2=J\\\\
J=2[(1.98)^{(1/28)}]-2 \qquad(2)\\\\
$Sub into (1)$\\\\$$

 

$$\\X=100(1+0.5J)^{10}\qquad(1)\\\\
X=100(1+0.5*(2[(1.98)^{(1/28)}]-2)\\\\
X=100(1+(1.98)^{(1/28)}-1)\\\\
X=100(1.98)^{(1/28)}\\\\$$

 

$${\mathtt{100}}{\mathtt{\,\times\,}}\left({{\mathtt{1.98}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{28}}}}\right)}\right) = {\mathtt{102.469\: \!634\: \!086\: \!312\: \!44}}$$    

 

X=$102.47

 

I did the maths while I was writing tht LaTex so there could easily be mistakes.

I have not checked it.

Melody  Dec 7, 2014
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1+0 Answers

 #1
avatar+91437 
+5
Best Answer

 

 I have left out M because down the bottom it says M=1

I have not included all of the initial working, but most is here. 

 

After 5 years            

 $$X=100(1+0.5J)^{10}\qquad(1)$$

 

After 7 years 

 

$$FV=100(1+0.5J)^{14}$$

 

After 10.5 years    (3.5 years more)

 

$$FV=[100(1.05J)^{14}](1+0.5J)^{14}=100(1.05J)^{28}$$

 

$$\\1.98*100=100(1+0.5J)^{28}\\\\
1.98=(1+0.5J)^{28}\\\\
(1.98)^{(1/28)}=1+0.5J\\\\
(1.98)^{(1/28)}-1=0.5J\\\\
2[(1.98)^{(1/28)}]-2=J\\\\
J=2[(1.98)^{(1/28)}]-2 \qquad(2)\\\\
$Sub into (1)$\\\\$$

 

$$\\X=100(1+0.5J)^{10}\qquad(1)\\\\
X=100(1+0.5*(2[(1.98)^{(1/28)}]-2)\\\\
X=100(1+(1.98)^{(1/28)}-1)\\\\
X=100(1.98)^{(1/28)}\\\\$$

 

$${\mathtt{100}}{\mathtt{\,\times\,}}\left({{\mathtt{1.98}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{28}}}}\right)}\right) = {\mathtt{102.469\: \!634\: \!086\: \!312\: \!44}}$$    

 

X=$102.47

 

I did the maths while I was writing tht LaTex so there could easily be mistakes.

I have not checked it.

Melody  Dec 7, 2014

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