Calculating Interest 
A person deposited $5000\$$ at $10\%$ simple interest for $2$ years. How much more money will she have in her account at the end of two years, if it is compounded semi-annually? $50/40/77.5/85.5$

My attempt: $\dfrac{5000*10*2}{100}=1000,$ so amount$=6000.$ $5000\left(1+\dfrac{10}{2*100}\right)^{2*2}=5000\left(\dfrac{21}{20}\right)^4=6077.5\$$. So, answers is $77.5\$$.
My query: Can we have a quicker way to solve this question? In my approach, the calculation is lengthy. Multiplying $21$, $4$ times is time-consuming.
P.S- It is actually a MCQ and one question should be solved in less than $1$ minute and calculator is not allowed.
 A: Edit: Some time after the answer below, the question was changed to specify no calculator. In that case, one can square $1.05$, getting $1.1025$, and square again. The squaring of $1.1025$ can be done conventionally.  We don't have to pay much attention to the digits on the right. 
You can also use the second order approximation $(1+x)^4\approx 1+4x+6x^2$ for $x$ reasonably close to $0$. This will get you close enough.

Answer before "no calculator" was specified:
You can use the $x^y$ button on your calculator, it it has one. All scientific and financial calculators do.  Or you can press the $x^2$ button twice. 
Note that Google has a nice calculator. If in the search window you type (1.05)^4 it will do the calculation. 
Remark: The answer should be rounded to the nearest cent. It would be good to drop the "percent" notation and think of the interest rate as $0.10$. So when we compound every half-year, the interest rate is $0.05$.
It follows that in $4$ compounding periods our $5000$ grows to $5000(1.05)^4$. 
A: Well, since $21*5=105$ none of the both is quicker... Once you have to multiply $21^4$ with $5$ and once you have to divide it by $20^4$... No really big deal, but maybe the $*5$ way is quicker.
If you have to calculate it by hand the good way to go is with factorization, I think : 
$$21^4=(3*7)^4=3^4*7^4=9*9*49*49=81*49*49=(81*50-81)*49=3969*50-3969=396900/2-3969=198450-3969=194481$$
I just did it entirely by head, so I'm not so sure about the correctness, sorry for that, I'm no calculator ;)
You should check the tips and hints about mental calculation if it's to be solved in less than 1 minute, I did maybe 3 minutes to do the whole while typing it at the same time.
A: $10\%$ interest per year means $5\%$ interest per half-year. So, after each half-year, your principal multiplies by $1.05$. Thus, after four half-years, your amount is $5000(1.05)^4$. For simple interest, your amount is $5000(1+2(0.1))$. So you are seeking $5000(1.05^4- 1.2)$.

You may use the binomial theorem, to calculate $(1.05)^4$:
$(1+5\cdot10^{-2})^4=1+4\cdot(5\cdot10^{-2})+6\cdot(5\cdot10^{-2})^2+4.(5\cdot10^{-2})^3+(5\cdot10^{-2})^4$.

You may also use $f(x+h) \approx f(x) + h\cdot f'(x)$. So for $f(x)=x^4$, $a=1$ and $h=0.05$, we get $f(1.05)\approx 1+0.05\cdot 4(1)^3=1.2$ (the actual value is $1.21550625$).
The Taylor expansion of a function is $f(x+h)=f(x)+h\cdot f'(x)+h^2\dfrac{f''(x)}{2!}+h^3\dfrac{f'''(x)}{3!}+\dots$.
So using more terms will give you a better approximation: Using the next term we get,
$f(1.05)\approx1+0.05\cdot4(1)^3+(0.05)^2\dfrac{12(1)^2}2=1.2+0.025\cdot6\,\,(=1.215)$.
You obtain, $5000(1.05^4- 1.2)\approx 5000(1.2+0.015-1.2)=75$.
Using this you will get an approximate answer and since you're on a multiple choice test, this should be helpful if the values given in the options aren't too close.
