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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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 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+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% |
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