Question

A sum of money put at 11% per annum amounts to ₹4491 in 2 years 3 months. What will it amount to in 3 years at the same rate

Answer 1

Answer:

3600

Step-by-step explanation:

Rate of interest = 11% and Time = 2 yrs 3 months = 2 yrs 3/12 = 2.25 yrs. X = - 449100/- 124.75 = 3600. So the Principal amount is equal to Rs. 3600.

Answer 2

$\sf\large\underline{Let:-}$

$\sf{\implies The\:sum\:of\: money=X}$

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

$\sf{\implies Amount=Rs.4491}$

$\sf{\implies Time=2\: years\:3\: months}$

$\sf{\implies Interest=11\%}$

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

$\sf{\implies Amount\:_{(at\:the\:end\:of\:3\: years)}=?}$

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

To calculate amount at the end of 3 years , at first we have to find out simple interest then calculate principal or sum of money. At last we have to find amount for 3 years at same interest which is given in the question. Before calculate simple interest at first we have to calculate time as per the annual interest per year ,so 2years 3months which is equal to 2 whole 3/12 or 27/12=> 9/4 years:

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

$\tt{\implies SI=\dfrac{x\times\:9\times\:11}{100\times\:4}}$

$\tt{\implies SI=\dfrac{99x}{400}}$

Now calculate calculate P here:]

$\sf{\implies Amount=P+SI}$

$\tt{\implies 4491=x+\dfrac{99x}{400}}$

$\tt{\implies 4491=\dfrac{400x+99x}{400}}$

$\tt{\implies 4491=\dfrac{499x}{400}}$

$\tt{\implies 4491\times\:400=499x}$

$\tt{\implies 9\times\:400=x}$

$\tt{\implies x=Rs.3600}$

Now calculate amount for 3 years:]

$\tt{\implies SI=\dfrac{3600\times\:3\times\:11}{100}}$

$\tt{\implies SI=36\times\:3\times\:11}$

$\tt{\implies SI=Rs.1188}$

$\sf{\implies Amount\:_{(for\:3\: years)}=P+SI}$

$\sf{\implies Amount=3600+1188}$

$\sf{\implies Amount=Rs.4788}$

$\sf\large{Hence,}$

$\sf{\implies Amount\:_{(at\:the\:end\:of\:3\: years)}=Rs.4788}$