please solve this question from the chapter banking.

Dear Student,

Please find below the solution to the asked query:

We know formula to find simple interest when recurring deposit  :

Simple interest = $P \times \frac{n\left(n+1\right)}{2\times 12}\times \frac{r}{100}$ , Here P  =  Recurring deposit amount , n  = Number of months amount is deposited and r  = Rate of interest

Given : Ankita started paying Rs.400 per month in a 3 year recurring deposit and the bank paid 10% p.a. simple interest

Here , Number of months  =  n  =  3 $\times$ 12 =  36 , P  = 400  and r  =  10 %

So,

Bank pays Simple interest to Ankita = $400 \times \frac{36\left(36+1\right)}{2\times 12}\times \frac{10}{100} = 400 \times \frac{36 \times 37}{2\times 12}\times \frac{1}{10}= 20 \times 3 \times 37 =\mathbf{Rs}\mathbf{.}\mathbf{2220}$

Thus,

Ankita gets total money at time of maturity = 400 $\times$ 36 + 1665 = 14400 + 2220 = Rs. 16620

Also given : After six months her brother Anshul started paying Rs.500 per month in a $\left(2\frac{1}{2} = \frac{5}{2}\right)$ years recurring deposit and the bank paid 10% p.a. simple interest

Here , Number of months  =  n  =  $\frac{5}{2}$ $\times$ 12  = 5 $\times$ 6 =  30 , P  = 500  and r  =  10 %
So,

Bank pays Simple interest to Anshul = $500 \times \frac{30\left(30+1\right)}{2\times 12}\times \frac{10}{100} = 500 \times \frac{30 \times 31}{2\times 12}\times \frac{1}{10}= 500 \times \frac{ 31}{2 \times 4}= 62.5 \times 31 =\mathbf{Rs}\mathbf{.}\mathbf{1937}\mathbf{.}\mathbf{5}$

Thus,

Anshul gets total money at time of maturity = 500 $\times$ 30 + 1937.5 = 15000 + 1937.5 = Rs. 16937.5

Therefore,

Anshul gets more money by 16937.5 - 16620 =
Rs.  317.5                                                          ( Ans )


Hope this information will clear your doubts about topic.
Regards