Question

) Kamala borrowed Rs.26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?​

Answer 1

S O L U T I O N :

$\sf Given$

• Principal, (P) = Rs.26400
• Rate, (R) = 15% p.a.
• Time, (T) = 2 years & 4 months

$\sf To\:find$

✯The amount will she paid.✯

$\bf Explanation$

Using formula of the compounded annually :

$\sf A=P\bigg(1+\frac{R}{100} \bigg)^{n}$

$\begin{gathered}\longrightarrow\tt{A=26400\bigg(1+\cancel{\dfrac{15}{100}} \bigg)^{2} }\\\\\\\longrightarrow\tt{A=26400\bigg(1+\dfrac{3}{20} \bigg)^{2} }\\\\\\\longrightarrow\tt{A=26400\bigg(\dfrac{20+3}{20} \bigg)^{2} }\\\\\\\longrightarrow\tt{A=26400\bigg(\dfrac{23}{20} \bigg)^{2} }\\\\\\\longrightarrow\tt{A=\cancel{26400}\times \dfrac{23}{\cancel{20}} \times \dfrac{23}{\cancel{20}} }\\\\\\\longrightarrow\tt{A=Rs.(66\times 23\times 23)}\\\\\\\longrightarrow\bf{A=Rs.34914}\end{gathered}$

Using formula of the Simple Interest :

$\sf S.I.=\frac{P\times R\times T}{100}$

• ⇝Principal = Rs.34914

• ⇝Rate = 15% p.a.

• ⇝Time = 4 months = 4/12 years

Therefore;

$\begin{gathered}\longrightarrow\tt{S.I.=\dfrac{34914\times 15\times \dfrac{4}{12} }{100}}\\\\\\\longrightarrow\tt{S.I.=\dfrac{34914\times \cancel{ \dfrac{60}{12}} }{100}}\\\\\\\longrightarrow\tt{S.I.=\dfrac{34914\times 5}{100}}\\\\\\\longrightarrow\tt{S.I.=\cancel{\dfrac{174570}{100} }}\\\\\\\longrightarrow\bf{S.I.=Rs.1745.7}\end{gathered}$

Thus;

$\sf {The\:Total\:amount\:she\:will\:paid=[Rs.34914+Rs.1745.7}{Rs.36659.7}$

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$\begin{gathered}\small\boxed{ \begin{array}{cc}\large\sf\dag \: {\underline{☆More \: Formulae☆}} \\ \\ \bigstar \: \sf{Gain = S.P – C.P} \\ \\ \bigstar \:\sf{Loss = C.P – S.P} \\ \\ \bigstar \: \sf{S.P = \dfrac{100+Gain\%}{100} \times C.P} \\ \\ \bigstar \: \sf{ C.P =\dfrac{100}{100+Gain\%} \times S.P} \\ \\\bigstar \: \sf{ S.P = \dfrac{100-loss\%}{100} \times C.P} \\ \\ \bigstar \: \sf{ C.P =\dfrac{100}{100-loss\%} \times S.P}\end{array}}\end{gathered}$