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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.

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    $\begingroup$ You can use the $x^y$ button on your calculator, it it has one. Or you can press the $x^2$ button twice. $\endgroup$ – André Nicolas Sep 7 '13 at 20:00
  • $\begingroup$ if it is compounded annually you get 6050 so the answer is 27.5 $\endgroup$ – MRK Sep 7 '13 at 20:03
  • $\begingroup$ @Jam- it's compounded semi-annually. $\endgroup$ – Ramit Sep 7 '13 at 20:04
  • $\begingroup$ i know but you are trying to compare the two right you said how much more meaning how much more compared to what if not annually $\endgroup$ – MRK Sep 7 '13 at 20:05
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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 $5000(1.05)^4$.

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  • $\begingroup$ edited my question. $\endgroup$ – Ramit Sep 7 '13 at 20:02
  • $\begingroup$ It looks like there is not a quicker way. Because multiplying $105$, $4$ times is as lengthy (or may be lengthier) as $21$, $4$ times. $\endgroup$ – Ramit Sep 7 '13 at 20:10
  • $\begingroup$ You use the $x^y$ button. In one operation it will give you $(1.05)^{47}$, if that's what you need. Or use the Google calculator. Or any number of freely available calculator programs. If you really want to do such calculations by hand, there are shortcuts, not really of much relevance any more. $\endgroup$ – André Nicolas Sep 7 '13 at 20:13
  • $\begingroup$ Sir, calculator is not allowed in exam. $\endgroup$ – Ramit Sep 7 '13 at 20:14
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    $\begingroup$ Sort of quickly: $(1+t)^2=1+2t+t^2$. Use $t=0.1025$. Then $2t=0.2050$, and $t^2$ is basically the same as $(0.1)^2=0.01$. So we get $1.215$. Alternately, view our number as $(1.1+0025)^2$ and only the first two terms matter. $\endgroup$ – André Nicolas Sep 7 '13 at 22:20
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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.

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  • $\begingroup$ that's a pretty nice way to calculate. Really interesting. $\endgroup$ – Ramit Sep 7 '13 at 20:33
  • $\begingroup$ Actually, even quicker would be to go : $$9*9*49*49=9*9*(50*50-50-49)=9*(10*2401-2401)=10*21609-21609=216090-21609=194481$$ And that's truly done in 1 minute, that way. I was sleepy, sorry for not seeing it before. $\endgroup$ – Lery Sep 7 '13 at 21:14
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$10\%$ 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.

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  • $\begingroup$ $5000*1.2= 6000$, which is the same as that in simple interest. So, as per this, answer should be zero. But $5000*1.2155062$ gives me $6077.5$. So, ans is $77.5$ which is what I need. So, though I appreciate your effort and concern, but approximation is not going to help in my case. But binomial can be a great help. Thanks.! $\endgroup$ – Ramit Sep 7 '13 at 20:26
  • $\begingroup$ @Ramit I've edited my answer. Hope it helps. $\endgroup$ – Alraxite Sep 7 '13 at 20:58
  • $\begingroup$ It certainly helps. In the last expression, I guess, we should have $1.2+0.015-1.2$ and not $0.025$? $\endgroup$ – Ramit Sep 7 '13 at 21:03
  • $\begingroup$ @Ramit Yes, I've fixed it. Thanks. $\endgroup$ – Alraxite Sep 7 '13 at 21:07

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