If $1000 is placed into account X, yielding 10% interest compounded an

Is there any way to solve this problem without calculating (1.1)5 ?
Most of the interest questions I've managed to solve it without much calculation. Wonder if this one could be solved in a different way.
Thanks!

Approximate:

In 1st year, both interests will be the same at $100 each.
In 2nd year, CI will be $10 extra (10% interest on previous interest of $100)
In 3rd year, CI will be $21 extra (10% interest on previous interest of $100 + $110 = $210)

It's already obvious that the answer will be 110.51 (= 10 + 21 + approx (30 + a bit extra) + approx (40 + a bit extra)).

else calculate for last 2 years too.

In 4th year, CI will be $33 extra (10% interest on previous interest of $210 + $121 = $331)
In 5th year, CI will be $46 (10% interest on previous interest of $331 + $133 = $464)

Total extra = 10 + 21 + 33 + 46 = $110

I don't fully understand how you get to YR3 interest. I thought the interest payments would be as follows

1st: 100
2nd: 100+10
3rd: 100+10+1

Given: $1000 is placed into account X, yielding 10% interest compounded annually and $1000 is placed into account Y using 10% simple annual interest,

Asked: How much more will be in account X than in account Y at the end of 5 years?  

For simple interest account Y
Interest for 5 years = 1000 * 5 * 10% = $500
Balance in simple interest account Y after 5 years = $1500

For compound interest account X
Balance in compound interest account after 5 years = 1000 (1.1)^5 where 1.1^5 = 1.61051
Balance in compound interest account X after 5 years = 1000 (1.1)^5 = $1610.51

How much more will be in account X than in account Y at the end of 5 years = $1610.51 - $1500 = $110.51

IMO C 

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