Question

the simple and compound interest on a certain sum of money for 2 years at the same rate are respectively rupees 1800 and rupees 1840. 50 find the sum and the rate of interest​

Answer 1

Answer:

The sum is Rs 88235.29 and rate of interest is 1.02%

Step-by-step explanation:

Given that simple interest on certain amount for 2 years at some rate is Rs 1800

Let the principal amount be $P$

Let the rate of interest be $r\%$

Here, SI = Rs 1800

T = 2 years

Simple Interest is calculated as

$SI = \frac{P \times R \times T}{100}\\\\1800 = \frac{P \times R \times 2}{100}\\\\PR = \frac{180000}{2}\\\\ PR = 90000\\\\P=\frac{90000}{R}------ (\,1)\,$

Also, given that compound interest on same amount for years at same rate is Rs 1840.50

We know,

$CI = Amount - Principal\\\\CI = P(\,1 + \frac{R}{100})\,^n - P\\\\1840.5 = P(\,1 + \frac{R}{100})\,^2 - P ---- (\,2)\,$

Putting the value of R from (1) in (2), we get

$1840.5 = \frac{90000}{R}(\,1 + \frac{R}{100})\,^2 - \frac{90000}{R}\\\\1840.5 = \frac{90000}{R}[\,(\,1 + \frac{R}{100})\,^2 - 1]\,\\\\1840.5 = \frac{90000}{R}[\,1 + \frac{R^2}{10000} + \frac{2R}{100} - 1]\\\\1840.5= \frac{90000}{R}[\,\frac{R^2}{10000} + \frac{2R}{100}]\\\\1840.5= \frac{90000}{R} \times R[\,\frac{R}{10000} + \frac{1}{50}]\\\\1840.5= 90000[\,\frac{R}{10000} + \frac{200}{10000}]\\\\1840.5= 90000[\,\frac{R + 200R}{10000}\\\\1840.5= 9[\,201R]\\\\R = \frac{1840.5}{9 \times 201}\\\\R = \frac{1840.5}{1809}\\\\R = 1.02\%$

Putting the value of R in (1), we get

$P=\frac{90000}{1.02}\\\\\\\\P = 88235.29$

Thus, the sum is Rs 88235.29 and rate of interest is 1.02%