Question

Rohit took a loan of ₹19500 from a bank on 24th January at 8.75% per annum and he paid it back from a bank on 18th june 2020 find the total amount paid by rohit

Answer 1

$\sf\large\underline\blue{Given:}$

$\sf{\implies Principal=Rs.19500}$

$\sf{\implies Rate=8.75\%}$

$\sf{\implies Time=24th\:Jan\:to\:18th\:June\:2020}$

$\sf\large\underline\blue{To\: Find:}$

• Total amount paid by Rohit to the bank:]

$\sf\large\underline\blue{Solution:}$

• To calculate amount at first we have to calculate the time then applying formula of simple interest after that to find the amount with the help of formula of amount:]

$\sf\small\underline\purple{Calculation\:for\: time}$

• As given in the question, a bank on 24th January at 8.75% per annum and he paid it back from a bank on 18th june 2020]

$\tt{\implies January=31-24=>7\:days}$

$\tt{\implies February=(leap\: years)=29\:days}$

$\tt{\implies March=31\:days}$

$\tt{\implies April=30\:days}$

$\tt{\implies May=31\:days}$

$\tt{\implies June=18\:days}$

• Now , we have to sum of all the days above to calculate time:]

$\tt{\implies Time=7+29+31+30+31+18}$

$\tt{\implies Time=146\:days}$

• Now calculate interest here]

$\tt\purple{\implies SI=\dfrac{P\times\:T\times\:R}{100}}$

$\tt{\implies SI=\dfrac{19500\times\:146\times\:8.75}{100\times\:366}}$

$\tt{\implies SI=\dfrac{24911250}{36600}}$

$\tt{\implies SI=Rs.680.63}$

• Now , calculate Amount here:]

$\tt\green{\implies Amount=P+SI}$

$\tt{\implies Amount=19500+680.63}$

$\tt{\implies Amount=Rs.20180.63}$

$\sf\large{Hence,}$

• The amount paid by Rohit to the bank at end of 18th June 2020 is Rs.20180.63:]

Answer 2

principal=19500

time=146/365

rate=8.75

so,

si=19500*8.75*146/100*366

si=24911*250*146/100*366=680.63

therefore,

Amount =19500+680.63=20180.63