Question

ramsha borrowed ₹6200 from a bank for 4 years at 15% p. a. the amount she will have to pay at the end of 4 years is​

Answer 1

$Simple \: Intrest \: = \: \frac{P×R×T}{100}$

$\frac{6200 \times 15 \times 4}{100}$

$\huge\frac{ 62 \cancel0 \cancel0 \times 15 \times 4}{1 \cancel0 \cancel0}$

$₹3720$

$Amount\:=\: Principal\:+\: Intrest$

$Amount\:=\:6200+3720$

$\therefore\: Amount\:=\:₹9920$

$Hence \: Ramsha \: will \: pay \: ₹9920 \: at \: the \: end \: of \: 4th \: year.$

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$\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{(100+Gain\%) or(100-Loss\%)}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

Answer 2

Given,

Ramsha borrowed $6200$ rupees from a bank for $4$ years at $15$% p.a.

We have to find the amount she will have to pay at the end of $4$ years

Here, we are using simple interest formula,

Simple interest$= \frac{PTR}{100}$

Here, $P=6200$, $T=4$ years, $R=15$%

Simple interest$= \frac{6200\times15\times4}{100}$

$= 62\times15\times4$

$=3720$

The amount she pay $= 6200+3720$

$=9920$

Therefore, the amount she will have to pay at the end of $4$ years is $9920$.