Aptitude - Compound Interest - Discussion

On 1st january customer deposit amount and receive end of year. It means first he deposit for 1 year.

When interest is compounded Half-yearly:

Amount = P * [ { 1 + (1/2) * (R/100) } ^ 2(n) ]
Amount = 1600 * [ { 1 + (1/2) * (5/100) } ^2(1) ] = 1681


But in 1st july he deposit same amount (1600 Rs. ) and receive it after the same date (after 6 month). It means second he deposit for 6 month(1/2 year).

When interest is compounded Annually but time is in fraction, say 1/2 years.

Amount = P * [ 1 + { (1/2)* R/100}]
Amount = 1600 * [ 1 + { (1/2)* 5/100}] = 1640

So Total = 1681 + 1640 = 3321

Will someone explain the first step?

I can't understand

[1600*41/40(41/40+1)]

@Anusha:

You told 2 cases one is half yearly(july) and other is yearly(jan). In yearly case the formula should be p(1+r/100)^n but they used half yearly formula only.

Give the mathemetical formula for compund interest.

Yes, its right , so in first case n (no. of years) is 1 and in second case (for july) n is 1/2 years, so now you can directly put that values in simple formula.

On 1st january customer deposit amount and receive end of year.

But in 1st july he deposit same amount (1600 Rs. ) and receive it after the same date (after 6 month).

It means first he deposit for 1 year and second he deposit for 6 month. And get all amount at the end of year.

@ Anusha: There's nothing. See clearly, 1st january and 1st July are dates of depositions!

And please someone explain why 2nd term is not squared while 1st is?

Customer has deposited twice in a year. Once in the begining i.e on Jan 1st n 2nd time on july 1st i.e after 6 months.

So 1st case n=1(1 yr) & 2nd case n=1/2(half yr).

So you get first and 2nd terms coresponding 2 first and 2nd case.

I didn't understand someone help me.