Question

At the beginning of year 2011, a man had
22,000 in his bank account. He saved some
money by the end of this year and deposited
it in the bank. The bank pays 10% per annum
compound interest and at the end of year
2012 he had 39,820 in his bank account.
Find, what amount of money he had saved
and deposited in his account at the end of
year 2011.

Answer 1

Solution :

Step-by-step explanation:

He had 22000 in 2011. He gets 10% Interest in 1 year = 2200/-

His total value in 2012 is 39820/-

2 years interest of 22000 is = 4400

22000 + 4400 = 26400

39820 - 26400 = 13420 - 1342 = 12078

Answer : Now He deposited 12078/- in the last of the year 2011

Answer 2

Answer:

12,000 /-

Step-by-step explanation:

Its compounded annually thus we need to use A=P(1+r/100)^n

So for the first year he didn't deposit only at then end he did.

thus:

First Year:

==> 22000*(1+10/100)^1

==> 22000*110/100

==> 24,200 {end of 2011}

Now in 2nd year he deposits {start of 2012}

So NOW the money in his account is 24,200/- PLUS x /-

So the amount at the END of 2012 will be 38,820 {GIVEN}

lets us the equation again,

39,820 = {24200+x}(1+10/100)^2

39,820 = {24200+x}*11/10

39,820*10/11 = 24200+x

36200=24200+x

x = 12000 /-

=)