Determine the simple interest rate under which a sum of money will double in 5 years.
IS my following solution correct below?
$A = p(1+r)^{t}$
$A/p = 2 =(1+r)^{5}$
$ln2 = 5ln(1+r)$
$ln2/5 = 0.13863 = ln(1+r)$
$e^{0.13863} = 1+r$
$r = 14.87\%$
Determine the simple interest rate under which a sum of money will double in 5 years.
IS my following solution correct below?
$A = p(1+r)^{t}$
$A/p = 2 =(1+r)^{5}$
$ln2 = 5ln(1+r)$
$ln2/5 = 0.13863 = ln(1+r)$
$e^{0.13863} = 1+r$
$r = 14.87\%$
On
Simple interest is a technical term that means we don't get interest on accrued interest, just on the principal. Suppose we are using simple interest, with interest rate $r$. If we start with $1$ dollar, after $1$ year we have $1+r$, after $2$ years we have $1+r+r$, and so on. So after $5$ years we have $1+5r$.
We want $1+5r=2$, so $r=0.20$ ($20$ percent).
Remark: You solved the doubling time equals $5$ problem for compound interest, nominal yearly rate $r$, compounding period $1$ year. Your answer to this more complicated problem is correct, but that is not what the question is asking for.
I hope this would suffice your query,
SI formula is SI=PRT/100 and the Amount=P+SI.
so if we deduce with given data.
a/p=2 t=5
a=p+si =a/p =1+si/p=2
2p=1+si
si=prt/100
2p=1+prt/100 200p=100+prt 200p=100+5pr 200p-5pr-100=0 5p(40-r)=100
5p=100 p=20 or r=40%