Question

Without actually dividing, find which of the following are terminating decimals.
(This question has multiple correct options)
A) \[\dfrac{3}{{25}}\]
B) \[\dfrac{{11}}{{18}}\]
C) \[\dfrac{{13}}{{20}}\]
D) \[\dfrac{{41}}{{42}}\]

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, let us go through all the options one by one.
Option A: First, we see that \[\dfrac{3}{{25}}\] has a denominator of 25.
$ \Rightarrow 25 = 5 \times 5 = {5^2}$
$\therefore $ if we compare it to ${2^m} \times {5^n}$, we get m = 0 and n = 2.
Therefore, the fraction given to us has a terminal expansion.
Option B: Now, we have \[\dfrac{{11}}{{18}}\]. It has a denominator of 18.
$ \Rightarrow 18 = 2 \times 3 \times 3 = 2 \times {3^2}$
$\therefore $ if we compare it to ${2^m} \times {5^n}$, we get m = 1 and n = 0. But there are 2 3’s in the denominator as well which cannot be eliminated using the numerator also.
Therefore, the fraction given to us does not have a terminal expansion.
Option C: Now, we have \[\dfrac{{13}}{{20}}\]. It has a denominator of 20.
$ \Rightarrow 20 = 2 \times 2 \times 5 = {2^2} \times 5$
$\therefore $ if we compare it to ${2^m} \times {5^n}$, we get m = 2 and n = 1.
Therefore, the fraction given to us has a terminal expansion.
Option D: Now, we have \[\dfrac{{41}}{{42}}\]. It has a denominator of 42.
$ \Rightarrow 42 = 2 \times 3 \times 7$
$\therefore $ if we compare it to ${2^m} \times {5^n}$, we get m = 1 and n = 0. But there is a 3 and a 7 in the denominator as well which cannot be eliminated using the numerator also.
Therefore, the fraction given to us does not have a terminal expansion.

$\therefore $ The required answer is (A) and (C).

Note: 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.
The students must not make the mistake of just multiplying or just dividing some number because that would change the number. If you multiply and divide by the same number, you are basically multiplying the number by 1, which would not result in any change in the number. (If required to convert number into fraction)