Question

1.)
2. Kamala borrowed 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?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the
2nd year amount for 4/12 years).​

Answer 1

S O L U T I O N :

$\bf{\large{\underline{\bf{Given\::}}}}}$

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

$\bf{\large{\underline{\bf{To\:find\::}}}}}$

The amount will she paid.

$\bf{\large{\underline{\bf{Explanation\::}}}}}$

Using formula of the compounded annually :

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

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

Now;

Using formula of the Simple Interest :

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

• Principal = Rs.34914
• Rate = 15% p.a.
• Time = 4 months = 4/12 years

Therefore;

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

Thus;

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