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Determine the simple interest rate under which a sum of money will double in 5 years.

IS my following solution correct below?

$A = p(1+r)^{t}$

$A/p = 2 =(1+r)^{5}$

$ln2 = 5ln(1+r)$

$ln2/5 = 0.13863 = ln(1+r)$

$e^{0.13863} = 1+r$

$r = 14.87\%$

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Simple interest means you don't get interest on interest. So $1$ dollar gets interest $5r$, meaning $r=0.20$ (twenty percent). You solved the problem for compound interest, nominal rate $r$, compounding period $1$ year. – André Nicolas Jan 22 '13 at 19:43

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.

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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+prt/100 200p=100+prt 200p=100+5pr 200p-5pr-100=0 5p(40-r)=100

5p=100 p=20 or r=40%

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You should define your terms-what are $p$ and $r$? If $r$ is the interest rate, note that the deposit triples in $5$ years at a simple interest rate of $40\%$ – Ross Millikan Jan 22 '13 at 20:15

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