Question

A man borrowed a sum of Rs. 10000 for 1 year at 37.5% per annum compounded every four months. Find the amount that he will have to repay at the end of 1 year. Please answer it is urgent i will surely mark as brainliest ​​

Answer 1

GIVEN :-

• Principal , P = Rs. 10000.
• Time , n = 1 years.
• Rate = 37 . 5 %.

TO FIND :-

• The amount that he will how to repay at the end of 1st year.

SOLUTION :-

Firstlywe will convert the given rate and time according to question.

$\\ \\ : \implies \displaystyle \sf \: Time = 12 \: month.$

$: \implies \displaystyle \sf \:Time = \dfrac{12}{4} \: month$

$: \implies \underline{ \boxed{\displaystyle \sf \:Time = 3 \: month}}$

Now,

$\\ \\ : \implies \displaystyle \sf \:Rate= 37.5 \: \%$

$: \implies \displaystyle \sf \:Rate= \dfrac{37.5}{3} \: \%$

$: \implies \underline{ \boxed{ \displaystyle \sf \:Rate=12.5\%}}$

_____________________

$\dashrightarrow\displaystyle \sf Amount = p\bigg( 1 + \dfrac{R}{100} \bigg) ^{n}$

$\dashrightarrow\displaystyle \sf Amount = 10000\bigg( 1 + \dfrac{12.5}{100} \bigg) ^{3}$

$\dashrightarrow\displaystyle \sf Amount = 10000\bigg( 1 + \dfrac{125}{1000} \bigg) ^{3}$

$\dashrightarrow\displaystyle \sf Amount = 10000\bigg( \dfrac{1000 + 125}{1000} \bigg) ^{3}$

$\dashrightarrow\displaystyle \sf Amount = 10000\bigg( \dfrac{1125}{1000} \bigg) ^{3}$

$\dashrightarrow\displaystyle \sf Amount =10000 \times \dfrac{1423828125}{1000000000}$

$\dashrightarrow\displaystyle \sf Amount =\frac{1423828125}{100000}$

$\dashrightarrow \underline{ \boxed{\displaystyle \sf Amount = \: 14238.28125}}$

$\therefore\underline{\displaystyle \sf \:The \: required \: amount \: is \: 14238.28125 }$

Answer 2

Given:

• Principal = rs.10000
• Time = 1 year
• Rate = 37.5%

Find:

• Amount he had to paid after an year

Solution:

Here, convert the Time and Rate according the Question...

we, know that

Time = 1 year = 12months

So,

$\rm \implies \dfrac{ \cancel{12}}{ \cancel{4}} = 3months$

Hence, Time = 3 months

Now,

Rate = 37.5%

$\rm \implies \dfrac{ \cancel{37.5}}{ \cancel{3}} = 12.5\%$

Hence, Rate = 12.5%

Now, we know that

$\underline{ \boxed{ \rm \color{green}Amount = P \times {\bigg(1 + \dfrac{R}{100} \bigg)}^{n} }}$

where,

• P = rs. 10000
• R = 12.5%
• n = 3months

So,

$\rm \to Amount = P \times {\bigg(1 + \dfrac{R}{100} \bigg)}^{n}$

$\rm \to Amount = 10000 \times {\bigg(1 + \dfrac{12.5}{100} \bigg)}^{3}$

now, take the L.C.M

$\rm \to Amount = 10000 \times {\bigg( \dfrac{100 + 12.5}{100} \bigg)}^{3}$

$\rm \to Amount = 10000 \times {\bigg( \dfrac{112.5}{100} \bigg)}^{3}$

$\rm \to Amount = 10000 \times 1.423828125$

$\rm \to Amount = rs.14238.28125$

Hence, The sum of money He had to pay after an year will be rs.14238.28125