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Mathematics 8 Online
OpenStudy (mathmath333):

If the compound interest for a certain sum for 2 years is 21.what could be the simple interest? a. 20 b. 16 c. 18 d. 20.5

OpenStudy (mathmath333):

@Michele_Laino

OpenStudy (anonymous):

ermm i got a formula by googeling :P my brain crashed so i'll just post it without checking if its right or not :P (Hope its help)

OpenStudy (anonymous):

nvm i have no clue :3 @hartnn

OpenStudy (mathmath333):

\(\large \tt \begin{align} \color{black}{21=p\left(1+\dfrac{n}{100}\right )^2-p\\~\\ \normalsize \text{now i want to find}\\~\\ \dfrac{p\times n\times 2}{100}=? }\end{align}\)

OpenStudy (mathmath333):

@Marki

hartnn (hartnn):

looks like you have to isolate p from 1st equation and plug it in 2nd equation to get an equation in only n

OpenStudy (mathmath333):

yes its very painful ,

OpenStudy (mathmath333):

but their must be a trick

hartnn (hartnn):

to simplify a bit, keep n/100 as say N

hartnn (hartnn):

p = 21/(N^2+2N) not so bad

hartnn (hartnn):

you need 42N/(N^2+2N) = 42/(N+2)

OpenStudy (mathmath333):

how did u get the above ^

hartnn (hartnn):

n/100 = N 21 = p [1+2N+N^2-1]

OpenStudy (mathmath333):

yes i got that but how u got that 42N stuff

hartnn (hartnn):

you need 2pN right? 2N times p 2n times 21/[N^2+2N]

OpenStudy (mathmath333):

oh i see ,lol

hartnn (hartnn):

but we don't know the rate of interest

hartnn (hartnn):

i'll say keep p =100 find N use this p and N in SI formula

hartnn (hartnn):

even better, keep p =21 :P

OpenStudy (mathmath333):

yes but why assuming p=100

hartnn (hartnn):

we have one equation and 2 unknowns

OpenStudy (mathmath333):

oh i see the trial and error

OpenStudy (mathmath333):

for \(p=21\) i got \((N+1)^2=2\) how to proceed further

ganeshie8 (ganeshie8):

still stuck ?

OpenStudy (mathmath333):

:( yes

OpenStudy (mathmath333):

this question beats me , i think its somewhat lacking information or either wrong

hartnn (hartnn):

with p =21 i get SI = 17.38 ...

ganeshie8 (ganeshie8):

the question clearly ask "what could be the simple interest" so there is no single answer, we need to try each option i guess

ganeshie8 (ganeshie8):

\(21=p\left(1+\dfrac{r}{100}\right )^2-p\) suppose the interest rate is 100% then \(21 = p(1+1)^2 -p \implies p = 7\) that gives you simple interest = \(14\) so 100% is too optimistic

ganeshie8 (ganeshie8):

lets try interest rate = 10%, then you get \(21 = p(1+0.1)^2-p \implies p = 100\) that gives you simple interest = \(20\) which is one of your options! so you're done!

OpenStudy (mathmath333):

how will my mind think that i have to put \(r=10\%\)

hartnn (hartnn):

with some other interest rate, we could get 16 or 18 ...

ganeshie8 (ganeshie8):

lets try and see if its really possible..

OpenStudy (mathmath333):

yes it does work for \(p=100\)

hartnn (hartnn):

20%, 19 25% , 18.66, 30% ,18.25

ganeshie8 (ganeshie8):

try p = 12.8 and r = 62.5% you will get simple interest = 16 and compound interest = 21

OpenStudy (mathmath333):

so is this question wrong

ganeshie8 (ganeshie8):

i guess so

ganeshie8 (ganeshie8):

but you will need to do ponder on this some more before concluding the question is wrong

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