Question

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Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively.
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Thank you for the answers!~​

Answer 1

Answer:

Given:-

• Principal = ₹ 17000 (P)
• Time period= 3 yrs (n)
• Rate of interest for successive years = 10%, 10% and 14% respectively (r)

To find:-

• Amount and compound interest

First we find amount

$\tt \large A = p ({1 + \frac{r}{100} })^{n}$

Substituting the values in above formula

$\tt \implies A = 17000(1 + \frac{10}{100} )(1 + \frac{10}{100} )(1 + \frac{14}{100} )$

$\tt \implies A \: = 17000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{57}{50} \\ \tt{ \red{ \boxed {A = 23449.8 \: \: \: \: rupees}}}$

Then we have to find compound interest

As we know that,

C.I. = Amount - Principal

C.I. = 23449.8 - 17000

C.I. = ₹6449.80

So,

• Amount= ₹23449.80
• Compound interest= ₹6449.80

Answer 2

Given : A Sum of Rs.17000 in 3 years at the rate for Successive years is 10 % , 10 % and 14 % .

To Find : Find the Amount and Compound Interest

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SolutioN : As the Rate is Successive .So, we'll apply the formula Compound Interest at Successive rate . Let's Solve :

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ A = P \bigg( 1 + \dfrac{R}{100} \bigg)^n }}}}}$

• ${\underline{\boxed{\pmb{\sf{ C.I = Amount - Principal }}}}}$

Where :

• A = Amount
• P = Principal
• R = Rate
• n = Time
• C.I = Compound Interest

$\maltese$ Calculating the Amount :

$\begin{gathered} \dashrightarrow \; \; \sf { A = P \bigg( 1 + \dfrac{R}{100} \bigg)^n } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = P \bigg( 1 + \dfrac{R}{100} \bigg)^1 \times \bigg( 1 \times \dfrac{R}{100} \bigg)^1 \times \bigg( 1 + \dfrac{R}{100} \bigg)^1 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \bigg( 1 + \dfrac{10}{100} \bigg)^1 \times \bigg( 1 + \dfrac{10}{100} \bigg)^1 \times \bigg( 1 + \dfrac{14}{100} \bigg)^1} \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \bigg( 1 + \cancel\dfrac{10}{100} \bigg)^1 \times \bigg( 1 + \cancel\dfrac{10}{100} \bigg)^1 \times \bigg( 1 + \cancel\dfrac{14}{100} \bigg)^1 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \bigg( 1 + 0.10 \bigg)^1 \times \bigg( 1 + 0.10 \bigg)^1 \times \bigg( 1 + 0.14 \bigg)^1 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \bigg( 1.10 \bigg)^1 \times \bigg( 1.10 \bigg)^1 \times \bigg( 1.14 \bigg)^1 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \bigg( 1.10 \bigg)^1 \times \bigg( 1.10 \bigg)^1 \times \bigg( 1.14 \bigg)^1 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \times 1.10 \times 1.10 \times 1.14 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \times 1.21 \times 1.14 } \\ \\ \end{gathered}$

$\begin{gathered} \dashrightarrow \; \; \sf { A = 17000 \times 1.3794 } \\ \\ \end{gathered}$

$\begin{gathered} {\dashrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Amount = ₹ \; 23449.8 }}}}}} \; {\pink{\bigstar}} \end{gathered}$

$\maltese$ Calculating the Compound Interest :

$\begin{gathered} \longmapsto \; \; \sf { Compound \; Interest = Amount - Principal } \\ \\ \end{gathered}$

$\begin{gathered} \longmapsto \; \; \sf { Compound \; Interest = 23449.8 - 17000 } \\ \\ \end{gathered}$

$\begin{gathered} {\longmapsto \; \; {\underline{\boxed{\pmb{\sf{ Compound \; Interst = ₹ \; 6449.8 }}}}}} \; {\green{\bigstar}} \end{gathered}$

$\therefore \; \;$ Amount is ₹ 23449.8 and Compound Interest is ₹ 6449.8 .

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