Question

Without actual division decide which of the following rational number have trrminating denominator representation

1)33/375
2)15/28
3)16/45
4)12/35
5)80/27
6)123/1250

Answer 1

to determine that you have to do prime factorization of the denominator if it is in the form of 2 m into 5 power n then it's terminating

Answer 2

Answer:

The correct answer is option 6)123/1250.

Step-by-step explanation:

A terminating decimal exists usually described as a decimal number that contains a finite number of digits behind the decimal point.

If x exists a rational number whose simplest form exists $\frac{p}{q}$, where p and q exist integers and $q \neq 0$. Then,

(i) x exists terminating only when q exists of the form $\left(2^{m} \times 5^{n}\right)$ for some non-negative integers m and n.

(ii) x exists non-terminating and repeating, if $q \neq\left(2^{m} \times 5^{n}\right)$.

Given:

1)33/375

2)15/28

3)16/45

4)12/35

5)80/27

6)123/1250

To find:

a rational number has a terminating denominator representation

Step 1

1)  $\frac{33}{375}$

Considering prime factorization of denominator, we get,

$\frac{33}{375}=\frac{33}{3 \times 5^{3}}$

The denominator of the fraction contains factors of 3 and 5. Since the denominator of the given fraction does not contain just factors of 2 and factors of 5. That is denominator exists not of the form $\left(2^{m} \times 5^{n}\right)$, therefore, the given fraction does not contain a terminating decimal representation.

Step 2

5) $\frac{15}{28}$

Considering prime factorization of denominator, we get,

$\frac{15}{28}=\frac{15}{2^{2} \times 7}$

The denominator of the fraction contains factors of 2 and 7. Since the denominator of the given fraction does not contain just factors of 2 and factors of 5. That is denominator exists not of the form $\left(2^{m} \times 5^{n}\right)$,  therefore, the given fraction does not contain a terminating decimal representation.

Step 3

3) $\frac{16}{45}$

Considering prime factorization of denominator, we get,

$\frac{16}{45}=\frac{16}{3^{2} \times 5}$

The denominator of the fraction contains factors of 3 and 5. Since the denominator of the given fraction does not contain just factors of 2 and factors of 5. That is denominator exists not of the form $\left(2^{m} \times 5^{n}\right)$,  therefore, the given fraction does not contain a terminating decimal representation.

Step 4

4) $\frac{12}{35}$

Considering prime factorization of denominator, we get,

$\frac{12}{35}=\frac{12}{5 \times 7}$

The denominator of the fraction contains factors of 5 and 7. Since the denominator of the given fraction does not contain just factors of 2 and factors of 5. That is denominator exists not of the form $\left(2^{m} \times 5^{n}\right)$, therefore, the given fraction does not contain a terminating decimal representation.

Step 5

5) $\frac{80}{27}$

Considering prime factorization of denominator, we get,

$\frac{80}{27}=\frac{80}{3^{3}}$

The denominator of the fraction contains factors of 3. Since the denominator of the given fraction does not contain just factors of 2 and factors of 5. That is denominator exists not of the form $\left(2^{m} \times 5^{n}\right)$, therefore, the given fraction does not contain a terminating decimal representation.

Step 6

6) $\frac{123}{1250}$

Considering prime factorization of denominator, we get,

$\frac{123}{1250}=\frac{123}{2 \times 5^{4}}$

The denominator of the fraction contains factors of 2 and 5. Since the denominator of the given fraction contains just factors of 2 and factors of 5. That is denominator exists of the form $\left(2^{m} \times 5^{n}\right)$, therefore, the given fraction contains a terminating decimal representation.

Therefore, the correct answer is option 6)123/1250.

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