How to solve this problem without the assistance of a calculator in under 3 minutes?

Question

I have 3 minutes to solve questions like these:

Someone invested $\$2,000$ in a fund with an interest rate of $1\%$ a month for $24$ months. Consider it to be compounded interest. What will be the accumulated value of the investment after $24$ months?

Remember, I am not allowed to use a calculator or any electronic device. Only pencil and pen.

Anyone have an idea how to solve this very fast within 3 minutes?

Edit [by SN]:

1. Logarithm tables are not allowed either.

2. This is a multiple choice question; the options are:

   • $2,437.53$
   • $2,465.86$
   • $2,539.47$
   • $2,546.68$
   • $2,697.40$

Answer 1

You must compute $C \; (1 + I)^N = 2000 \; (1+0.01)^{24}$ The power is difficult to compute without a calculator. But if the interest is low (more to the point, if $N \; I \ll 1$) , you can approximate it by a first or second order approximation of the binomial formula:

$(1+0.01)^{24} \approx 1 + 24 \times 0.01 = 1.24$ $\hspace{20px} C_f = 2480$ (first order)

$(1+0.01)^{24} \approx 1 + 24 \times 0.01 + \frac{24 \times 23}{2}\times 0.01^2 = 1.2676$ $\hspace{20px} C_f = 2535.20$ (second order)

In this case, the second order appproximation is just decent, if not very precise (real value: 1.2697346..., $C_f=2539.47$)

(Update: and for the multiple-choice decision, it's enough).

Answer 2

Perhaps this is where they (used to--before calculators) use the "Rule of 78s"

Answer 3

If the interest is compounded monthly, which the wording of your question suggests, then you have to calculate $2000 \times 1.01^{24}$, which is not possible in three minutes without log tables. Did the question perhaps specify that the interest was compounded annually?

Edited to add: Now the OP reveals that this was a multiple-choice question. Suddenly it's doable! (As many people have pointed out.)

Answer 4

Develop $2000(1+0.01)^{24}$ using the binomial theorem and start to calculate and add the first terms until you see that the summands are small enough. Four terms should be sufficient.

I don't know if you can do this calculation in 3 minutes. It is certainly doable.

Answer 5

It is fastest to evaluate $(1+0.01)^8$ and cube it: $$ ( 1+ 0.01)^8 \approx 1 + \frac{8}{100} + \frac{8 \times 7}{2} \frac{1}{100^2} = 1.0828 $$ Raising this to the third power $(1+y)^3 \approx 1 + 3y(1+y) = 1 + \frac{3 \times 828}{10,000} \times 1.08$, thus $$ \begin{eqnarray} 2000 \times \left( 1 + 0.01 \right)^{24} &\approx& 2000 + 2000 \times \frac{3 \times 828}{10,000} \times 1.08 = 2000 + \frac{6 \times 828}{10} \times 1.08 \\ &=& 2000 + 496.8 + 496.8 \times 0.08 \approx 2000 + 496.8 + 39.8 = 2536.6 \end{eqnarray} $$ And from the way we did the approximation we should round up, which leaves you with the third choice.

Answer 6

In decimal, 1% of X is just X right shifted two places. Thus, X + 1% of X is just X added with itself shifted two places, which you should easily be able to do in place, starting at the rightmost digit and adding the digit two to its left. The first few months give this:

2000.0000
2020.0000
2040.2000
2060.6020
2081.2080
...

Each if these takes a second or two, so 24 months in 3 minutes should be easy.