Interest word problems

[ch.kridgers]

Ann borrowed some money @ 9.5% per year in simple interest. At the end of 5 years 6 months, she owed a total amount of $6090, including principal and interest. if A=p+prt, find out how much money Ann borrowed. help! how do i set this up?

 

[Deleted member 4993]

Ann borrowed some money @ 9.5% per year in simple interest. At the end of 5 years 6 months, she owed a total amount of $6090, including principal and interest. if A=p+prt, find out how much money Ann borrowed. help! how do i set this up?

What is the problem asking for?

how much money Ann borrowed

Name that unknown (variable) = (it has been already named for you) p

According to the problem what are the values of A, r and t?

Now solve it...

Please show us your work, including exactly where you are stuck - so that we know where to begin to help you.

 

[ch.kridgers]

i understand that "r" must be .095 and "t" must be 5.5 (correct me if i'm wrong) but what is "a"? $6090?

using the A=p+prt formula...

6090=p+.495p ......?

 

[Deleted member 4993]

i understand that "r" must be .095 and "t" must be 5.5 (correct me if i'm wrong) but what is "a"? $6090?

using the A=p+prt formula...

6090=p+.495p ----You are correct......?

6090 = 1*p + 0.495*p

6090 = 1.495 * p

Now continue....

 

[ch.kridgers]

so....p=4174?

Ann borrowed $4174 ?

 

[mmm4444bot]

… 6090 = p + .495p …

I get a different value when I multiply 0.095 times 5.5.

6090 = p + 0.5225p

 

[mmm4444bot]

p = 4174 ?

Ann borrowed $4174 ?

Do you know how to verify whether or not a possible solution to an equation is correct?

Substitute your possible value for p into the original equation, then do the arithmetic, and see if you end up with a true statement.

EG:

Is x = 4000 a solution for 0.5225x + x = 6090 ?

Substitute 4000 for x, and do the arithmetic.

(0.5225)(4000) + 4000 = 6090

2090 + 4000 = 6090

6090 = 6090

Since the last equation is a true statement, x = 4000 is a solution.