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Answers
Answer:
okay answer mine
Explanation:
Solution:-
\begin{gathered} \sf \bold{We \: have \: A = P(1 + \frac{R}{100} {)}} \mathbb{^{n} } \\ \sf \small \red{\:Principal(P) = 12600 } \\ \sf \small \red{Rate(R) = 10\%} \\ \sf \small \red{Number \: of \: years} \mathbb \red{(n) } \sf \small \red{ = 2}\end{gathered}
WehaveA=P(1+
100
R
)
n
Principal(P)=12600
Rate(R)=10%
Numberofyears(n)=2
\sf \orange{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 12600(1 + \frac{10}{100} {)}^{2} = 12600( \frac{11}{10 } {)}^{2} }=12600(1+
100
10
)
2
=12600(
10
11
)
2
\sf \orange{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 12600 \times \frac{11}{10} \times \frac{11}{10} = 15246}=12600×
10
11
×
10
11
=15246
\sf{CI =A - P = 15246 - 12600 =2646}CI=A−P=15246−12600=2646
More:-
\sf \small{(i)Amount \: when \: interest \: is \: compunded annually}(i)Amountwheninterestiscompundedannually
\sf \small\orange{ = p( 1 + \frac{r}{100} {)}^{n} }=p(1+
100
r
)
n
\sf \small{(i)Amount \: when \: interest \: is \: compounded \: half \: yearly}(i)Amountwheninterestiscompoundedhalfyearly
\sf \small \orange{ = p(1 + \frac{r}{200} {)}^{2n} }=p(1+
200
r
)
2n
Answer:
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