Question

Write down the decimal expansion of the given fraction: $\dfrac{{123}}{{750}}$.

Answer 1

Hint: We will first see which fractions have a terminal expansion. Then, we will modify the given fraction by multiplying and dividing by some number so that we have powers of 10 in the denominator and thus have a fraction.

Complete step-by-step answer:
We know that if we have a fraction then it has a terminal expansion if the denominator is of the form ${2^m} \times {5^n}$, where m and n are non-negative integers.
Now, we see that $\dfrac{{123}}{{750}}$ has a denominator of 750.
$ \Rightarrow 750 = 2 \times 3 \times 5 \times 5 \times 5 = 2 \times 3 \times {5^3}$
$\therefore $ if we compare it to ${2^m} \times {5^n}$, we get m = 1 and n = 3. But we see that there are 3 in the denominator.
We see that 1 + 2 + 3 = 6 which is divisible by 3.
$ \Rightarrow 123 = 3 \times 41$
$\therefore $ we can write $\dfrac{{123}}{{750}} = \dfrac{{3 \times 41}}{{2 \times 3 \times {5^3}}} =\dfrac{{41}}{{2 \times {5^3}}}$
Therefore, the fraction given to us has a terminal expansion.
Now, we have $\dfrac{{41}}{{250}}$.
We can write:-
$ \Rightarrow \dfrac{{41}}{{250}} = \dfrac{{41}}{{2 \times {5^3}}} \times 1$
Now, this can be written as:
$ \Rightarrow \dfrac{{41}}{{250}} = \dfrac{{41}}{{2 \times {5^3}}} \times \dfrac{{{2^2}}}{{{2^2}}}$
Simplifying the RHS, we will get:-
$ \Rightarrow \dfrac{{41}}{{250}} = \dfrac{{41 \times 4}}{{{{\left( {2 \times 5} \right)}^3}}}$
Simplifying the RHS further, we will get:-
$ \Rightarrow \dfrac{{41}}{{250}} = \dfrac{{164}}{{{{10}^3}}}$
$\therefore \dfrac{{41}}{{250}} = \dfrac{{164}}{{1000}} = 0.164$

$\therefore $ the required answer is 0.164.

Note: The students must not forget to check if the fraction has a terminal expansion or not before carrying on with the procedure of division (if preferred by the students).
The students must wonder that they have used the condition for a fraction to be terminal but how does this condition work? Let us ponder over it.
If we have a fraction with a denominator in the form of ${2^m} \times {5^n}$. We can always somehow multiply and divide that fraction with 2 or 5 depending upon whether n > m or m > n to make the denominator a power with base 10 and thus easily replicable into a decimal.