Terminating Decimals Application | Number Properties

A recap on Terminating Decimals:

– To figure out whether the fraction is terminating, bring it down to its lowest form.

– Focus on the denominator – if it is of the form $2^a \times 5^b$ , the fraction is terminating, else it is not.

Keeping this in mind, let’s look at a couple of DS questions on terminating decimals.

Question 1: If ${a}$ , ${b}$ , ${c}$ , ${d}$ and ${e}$ are integers and ${m = 2^a \times 3^b}$ and ${n = 2^c \times 3^d \times 5^e}$ , is ${\frac{m}{n}}$ a terminating decimal?

Statement 1: ${a > c}$

Statement 2: ${b > d}$

Solution:

Given: $a$ , $b$ , $c$ , $d$ and $e$ are integers

Question: Is $\frac{m}{n}$ a terminating decimal?

or is $\frac{(2^a \times 3^b)}{(2^c \times 3^d \times 5^e)}$ ?

We know that powers of 2 and 5 in the denominator are acceptable for the decimal to be terminating. If there is a power of 3 in the denominator after reducing the fraction, then the decimal in non- terminating. So our question is basically whether the power of 3 in the denominator gets cancelled by the power of 3 in the numerator. If $b$ is greater than (or equal to) $d$ , after reducing the fraction to lowest terms, it will have no 3 in the denominator which will make it a terminating decimal. If $b$ is less than $d$ , even after reducing the fraction to its lowest terms, it will have some powers of 3 in the dominator which will make it a non-terminating decimal.

Question: Is $b \ge d$ ?

Statement 1: $a > c$

This statement doesn’t tell us anything about the relation between $b$ and $d$ . Hence this statement alone is not sufficient.

Statement 2: $b > d$

This statement tells us that $b$ is greater than $d$ . This means that after we reduce the fraction to its lowest form, there will be no 3 in the denominator and it will be of the form $2^c \times 5^e$ only. Hence it will be a terminating decimal. This statement alone is sufficient.

Answer (B).

Now onto another DS question.

Question 2: If ${0 < x < 1}$ , is it possible to write ${x}$ as a terminating decimal?

Statement 1: ${24x}$ is an integer.

Statement 2: ${28x}$ is an integer.

Solution:

Given: $0 < x < 1$

Question: Is $x$ a terminating decimal?

Again, $x$ will be a terminating decimal if it is of the form $\frac{m}{(2^a \times 5^b)}$

Statement 1: $24x$ is an integer.

$24x = 2^3 \times 3 \times x = m$ (an integer)

$x = \frac{m}{(2^3 \times 3)}$

Is $x$ a terminating decimal? We don’t know. If $m$ has 3 as a factor, $x$ will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Statement 2: $28x$ is an integer.

$28x = 2^2 \times 7 \times x = n$ (an integer)

$x = \frac{n}{(2^2 \times 7)}$

Is $x$ a terminating decimal? We don’t know. If $n$ has 7 as a factor, $x$ will be a terminating decimal. Else it will not be. This statement alone is not sufficient.

Taking both together,

$\frac{m}{24} = \frac{n}{28}$

$\frac{m}{n} = \frac{6}{7}$

Since $m$ and $n$ are integers, $m$ will be a multiple of 6 (and thereby of 3 too) and $n$ will be a multiple of 7. So $x$ will be a terminating decimal.

Answer (C).

I. QUANT · Number Properties

I. QUANT · Number Properties

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