Mr. Mak deposits $20000 in a bank account at the beginning of each year for ten years. The interest is compounded yearly and the interest rate is 4.5% per annum.
(a) Find the total amount he will receive at the end of the 5th year.
(b) At the end of the 5th year, Mr. Mak withdraws $45000 for decorating his house. After that, he continues depositing $20000 into his account every year. Find the total amount that he can get at the end of the 10th year.

@sauravshakya @hartnn

Question

Mr. Mak deposits $20000 in a bank account at the beginning of each year for ten years. The interest is compounded yearly and the interest rate is 4.5% per annum.
(a) Find the total amount he will receive at the end of the 5th year.
(b) At the end of the 5th year, Mr. Mak withdraws $45000 for decorating his house. After that, he continues depositing $20000 into his account every year. Find the total amount that he can get at the end of the 10th year.

@sauravshakya @hartnn

tkhunny · Answer

Always, always, solve a problem via Basic Principles.  You can do it after all else fails or you can just jump right in.

i = 0.045 -- Annual Interest
r = 1+i = 1.045 -- Annual Accumulation Factor.

Can we get comfortable with these definitions and variable assignments?

anonymous · Answer

ah well, i am confused by part b

tkhunny · Answer

Trust me on this.  If we can establish a bit of communication, by solving part 'a' together, part 'b' will be simple.  What say you?  Shall we continue with the fundamental approach?  It WILL save you if you become more familiar with it.  You will never have to say, "I am confused by" again.

anonymous · Answer

ok...
part a = 114000 (corr. to 3 sig fig.)

tkhunny · Answer

20000 / year for 10 years and it accumulates to only 114000?  I think not.

20000 * 10 = 200000 > 114000, and that's not even crediting any interest!

Okay, how did you get 114000?  Let's see where it went horribly wrong.

anonymous · Answer

but the answer is 114000 for part a

tkhunny · Answer

I see.  It's only five years in.  Why are we using 3 significant figures?

anonymous · Answer

the questions stated that we should use 3 sig fig. i didn't type that out as i think all the members will know..

anonymous · Answer

$$20000\times(1+0.045)+20000\times(1+0.045)^{2}+...+20000\times(1+0.045)^{5}$$

tkhunny · Answer

Absolutely not.  3 sig fig is quite arbitrary.  Can you think of a single account holder who would agree to that with an actual bank?

Have you drawn a timeline for this?  It often helps to get your thinking straight.

anonymous · Answer

114337.8333

tkhunny · Answer

Okay, since we have established the rather odd 114000 at 5 years, we first simply subtract the 45000 at that time.

What's to stop you from simply calculating it directly?  You already did the 5 year accumulation in part a?

$(114000 - 45000)(1 + 0.045)^{5} + 114000$

anonymous · Answer

i don't think your formula is correct......

anonymous · Answer

@ganeshie8

tkhunny · Answer

Does it make any difference if the accumulation is the first 5 years or the 2nd 5 years?

anonymous · Answer

he also deposits $20000 every year

tkhunny · Answer

All the payments are in there.  Look at it closely.

20000 Payments at times 0, 1, 2, 3, and 4 accumulate to 114000 at time 5
20000 Payments at times 5, 6, 7, 8, and 9 accumulate to 114000 at time 10

tkhunny · Answer

Accumulation at time 5

114000

After withdrawal

114000  - 45000

Accumulation at time 10 is in two pieces

5 years of interest only (114000 - 45000)(1 + 0.045)^5
Accumulation of 2nd five years of payments 114000

anonymous · Answer

201000

anonymous · Answer

thanks