Question

Q.Dinesh borrowed a sum of 20000rs from a finance firm and paid back 26620 in 3 years if the interest is compounded annually find the rate of interest​

Answer 1

Step-by-step explanation:10%

$p = 20000 \\ t = 3years \\ r = we \: have \: to \: find {?} \\ a = 26620 \\ we \: know \: a = p {(1 + \frac{r}{100}) }^{t} \\ 26620 = 20000 {(1 + \frac{r}{100}) }^{3} \\ \frac{26620}{20000} = { ( \frac{100 + r}{100} )}^{3} \\ \frac{1331}{1000} = { ( \frac{100 + r}{100} )}^{3} \\ { \binom{11}{10} }^{3} = { ( \frac{100 + r}{100} )}^{3} \\ \frac{11}{10} = \frac{100 + r}{100} \\ 10(100 + r) = 11 \times 100 \\ 1000 + 10r = 1100 \\ 10r = 1100 - 1000 \\ 10r = 100 \\ r = \frac{100}{10 } \\ r = 10\%$

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Answer 2

Step-by-step explanation:

Step-by-step explanation:10%

\begin{gathered}p = 20000 \\ t = 3years \\ r = we \: have \: to \: find {?} \\ a = 26620 \\ we \: know \: a = p {(1 + \frac{r}{100}) }^{t} \\ 26620 = 20000 {(1 + \frac{r}{100}) }^{3} \\ \frac{26620}{20000} = { ( \frac{100 + r}{100} )}^{3} \\ \frac{1331}{1000} = { ( \frac{100 + r}{100} )}^{3} \\ { \binom{11}{10} }^{3} = { ( \frac{100 + r}{100} )}^{3} \\ \frac{11}{10} = \frac{100 + r}{100} \\ 10(100 + r) = 11 \times 100 \\ 1000 + 10r = 1100 \\ 10r = 1100 - 1000 \\ 10r = 100 \\ r = \frac{100}{10 } \\ r = 10\%\end{gathered}

p=20000

t=3years

r=wehavetofind?

a=26620

weknowa=p(1+

100

r

)

t

26620=20000(1+

100

r

)

3

20000

26620

=(

100

100+r

)

3

1000

1331

=(

100

100+r

)

3

(

10

11

)

3

=(

100

100+r

)

3

10

11

=

100

100+r

10(100+r)=11×100

1000+10r=1100

10r=1100−1000

10r=100

r=

10

100

r=10%

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