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P5100, A=5227.50,  t=3 months

 

A=(1+rt) t=0.25 5227.50 = 5100(1+0.25)

this is as far as I can get

 Jun 15, 2014

Best Answer 

 #1
avatar+33661 
+10

I guess you are trying to find the values of r in the equation A = P*(1+r*t) where all the other terms are known.

A = 5227.5

P = 5100

t = 3 months = 0.25 years

5227.5 = 5110*(1 + r*0.25)   where r will be in units of "per year".

Divide both sides by 5110

5227.5/5110 = 1 + r*0.25

Subtract 1 from both sides.

5227.5/5110 - 1 = r*0.25

Divide both sides by 0.25

(5227.5/5110 - 1)/0.25 = r

$${\mathtt{r}} = {\frac{\left({\frac{{\mathtt{5\,227.5}}}{{\mathtt{5\,110}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.25}}}} \Rightarrow {\mathtt{r}} = {\mathtt{0.091\: \!976\: \!516\: \!634\: \!050\: \!9}}$$

So r ≈ 0.092 per year (or 0.092*12 ≈1.1 per month) 

 Jun 15, 2014
 #1
avatar+33661 
+10
Best Answer

I guess you are trying to find the values of r in the equation A = P*(1+r*t) where all the other terms are known.

A = 5227.5

P = 5100

t = 3 months = 0.25 years

5227.5 = 5110*(1 + r*0.25)   where r will be in units of "per year".

Divide both sides by 5110

5227.5/5110 = 1 + r*0.25

Subtract 1 from both sides.

5227.5/5110 - 1 = r*0.25

Divide both sides by 0.25

(5227.5/5110 - 1)/0.25 = r

$${\mathtt{r}} = {\frac{\left({\frac{{\mathtt{5\,227.5}}}{{\mathtt{5\,110}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{0.25}}}} \Rightarrow {\mathtt{r}} = {\mathtt{0.091\: \!976\: \!516\: \!634\: \!050\: \!9}}$$

So r ≈ 0.092 per year (or 0.092*12 ≈1.1 per month) 

Alan Jun 15, 2014
 #2
avatar+24 
0

Thank you so much for your assistance, I was completely lost.  I am actually beginning to understand what I am doing now as i see the steps broken down.

 Jun 19, 2014
 #3
avatar+24 
0

FV=4148; n=16, i=0.07

can someone tell me how to solve for PMT?  I get the first part correct but I keep getting the result wrong when I do the division so I cant make it to the second part

 Jun 19, 2014

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