Fractional Approx.

Question

How to approximate $\frac{2*(10^3+10^6+1)}{(10^3 * 1001 *3)}$. In that problem Options are

A) $\frac 35$

B) $\frac{33}{50}$

C) $\frac{333}{500}$

D) $\frac{3333}{5000}$

In that problem I got till above mentioned expression, but question was to find best approximation . According to me, best approximation of this should be Option D as that seems more digits. But answer is option C :

Answer 1

$10^6+10^3+1=1000000+1000+1=1001001=1001\times 1000+1$

$10^3\times 1001\times 3=1001\times 1000\times3$

$\Rightarrow \dfrac{2 \cdot (10^3+10^6+1)}{10^3 \cdot 1001 \cdot 3}=\dfrac{1001\times 1000\times 2}{1001\times 1000\times3}+\dfrac{2}{1001000\times 3}=\dfrac{2}{3}+\dfrac{1}{1501500}$

The expression is a bit larger than $\dfrac{2}{3}$, or a bit greater than about $0.666666666...$

We have (let the answers be $A_1;A_2;A_3;A_4;...$):

• $A_1=3/5=\dfrac{10^1-1}{3\times 5\times 10^0}=0.6$

• $A_2=33/50=\dfrac{10^2-1}{3\times 5\times 10^1}=0.66$

• $A_3=333/500=\dfrac{10^3-1}{3\times 5\times 10^2}=0.666$

• $A_4=3333/5000=\dfrac{10^4-1}{3\times 5\times 10^3}=0.6666$

• $A_n=A_{n-1}\times \dfrac{11}{10}=\dfrac{10^n-1}{3\times 5\times 10^{n-1}}$

We have $A_n=\dfrac{10^n-1}{3\times 5\times 10^{n-1}}<\dfrac{2}{3}$ if and only if

$10^n-1<\dfrac{2}{3}\times 3\times 5\times 10^{n-1}$

$\Leftrightarrow 10^n-1<10^n$, which is always true.

Because $0<A_n<\dfrac{2}{3}$ for all $n\in\mathbb{Z^+}$ and $A_n$ is always equal to $A_{n-1}\times 1.1$, we can conclude that $A_n$ would be closer to the value of the expression (technically, closer to $2/3$) when the value of $n$ is increased.

For this specific problem, the possible choices are $A_1;A_2;A_3;A_4$, so $A_4$ is the closest approximation.

Answer 2

The exact number is clearly $\dfrac23$ and using $\dfrac23\cdot\dfrac32=1$ so you have multiplying by $1$ the given expression, $$\dfrac23\cdot\dfrac32\cdot\dfrac{333333....}{500000...}=\dfrac23\cdot\dfrac{99999.....}{100000.....}$$ The more $3$ and $0$ you add the better the approximation will be because the factors are successively $0.9,0.99,0.999....$ so more and more near of $1$ then a better approach.

Thus the answer is D).