Mohit started paying Rs. 800 per month in a 6 year recurring deposit.

1.	Mohit started paying Rs. 800 per month in a 6 year recurring deposit. After 2 years, he started one more R.D. account in which he deposited Rs 1,500 per month. If the bank pays 10% per annum simple interest in both the deposits, find at the end of 6 years which R.D. will give more money and by how much?
Answer» Given : Mohit started paying Rs. 800 PER month in a 6 year recurring deposit. After 2 years, he started ONE more R.D. account in which he deposited Rs 1,500 per month. Interest rate is 10 % in both the deposits . To Find : The end of 6 years which R.D. will give more money and by how much . Solution : We know , interest is Given by : $I=\dfrac{P\times R \times T}{100}$ Here , P is PRINCIPLE amount . R is interest rate . T is total time . Interest of 1st R.D is : $I_1=\dfrac{P_1\times R \times T_1}{100}\\\\I_1=\dfrac{800\times 10\times (6\times 12)}{100}\\\\I_1=5760$ Interest of 2nd R.D is : $I_2=\dfrac{P_2\times R\times T_2}{100}\\\\I_2=\dfrac{1500\times 10\times (4\times 12)}{100}\\\\I_2=7200$ Therefore , second R.D will give more money and by Rs ( 7200-5760) = Rs 1440 . Hence , this is the required solution .

Mohit started paying Rs. 800 per month in a 6 year recurring deposit. After 2 years, he started one more R.D. account in which he deposited Rs 1,500 per month. If the bank pays 10% per annum simple interest in both the deposits, find at the end of 6 years which R.D. will give more money and by how much?

Answer»

Given :

Mohit started paying Rs. 800 PER month in a 6 year recurring deposit.

After 2 years, he started ONE more R.D. account in which he deposited Rs 1,500 per month.

Interest rate is 10 % in both the deposits .

To Find :

The end of 6 years which R.D. will give more money and by how much .

Solution :

We know , interest is Given by :

$I=\dfrac{P\times R \times T}{100}$

Here , P is PRINCIPLE amount .

R is interest rate .

T is total time .

Interest of 1st R.D is :

$I_1=\dfrac{P_1\times R \times T_1}{100}\\\\I_1=\dfrac{800\times 10\times (6\times 12)}{100}\\\\I_1=5760$

Interest of 2nd R.D is :

$I_2=\dfrac{P_2\times R\times T_2}{100}\\\\I_2=\dfrac{1500\times 10\times (4\times 12)}{100}\\\\I_2=7200$

Therefore , second R.D will give more money and by Rs ( 7200-5760) =

Rs 1440 .

Hence , this is the required solution .