Question

Reduce the fractions to their lowest term by using the HCF method:
a. \[\dfrac{{24}}{{48}}\]
b. \[\dfrac{{42}}{{63}}\]
c. \[\dfrac{{15}}{{27}}\]

Answer 1

Hint: We use the method of HCF to find the highest common factors between the numerator and the denominator. Cancel that highest common factor from numerator and denominator and obtain the lowest term of fraction.
* HCF of two or more numbers is the highest common factor which is given by the multiplication of all common prime numbers that are common in their prime factorization.
* Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.
* Law of exponents states that when the base is same we can add the powers of that element i.e. \[{p^m} \times {p^n} = {p^{m + n}}\]

Complete answer:
We solve for each part separately. Since we have to use the HCF method we calculate the HCF of each term using the prime factorization method and then write the fraction using that method.
a. \[\dfrac{{24}}{{48}}\]
Now we calculate HCF of the three numbers 24 and 48
We write each number in form of its prime factors
\[24 = 2 \times 2 \times 2 \times 3\]
\[48 = 2 \times 2 \times 2 \times 2 \times 3\]
Since, we know \[\underbrace {a \times a \times ..... \times a}_n = {a^n}\]. Therefore, we can write
\[24 = {2^3} \times 3\]
\[48 = {2^4} \times 3\]
HCF is the highest common divisor, so from the prime factorization above we say that \[{2^3} \times 3\]is the highest number which divides both the numbers.
\[ \Rightarrow \]HCF \[ = {2^3} \times 3\]
\[ \Rightarrow \]HCF \[ = 2 \times 2 \times 2 \times 3\]
\[ \Rightarrow \]HCF \[ = 24\]
So, we can write \[\dfrac{{24}}{{48}} = \dfrac{{24 \times 1}}{{24 \times 2}}\]
Cancel same factors i.e. the HCF from numerator and denominator
\[ \Rightarrow \dfrac{{24}}{{48}} = \dfrac{1}{2}\]

\[\therefore \]The lowest term of \[\dfrac{{24}}{{48}}\]is \[\dfrac{1}{2}\].

b. \[\dfrac{{42}}{{63}}\]
Now we calculate HCF of the three numbers 42 and 63
We write each number in form of its prime factors
\[42 = 2 \times 3 \times 7\]
\[63 = 3 \times 3 \times 7\]
Since, we know \[\underbrace {a \times a \times ..... \times a}_n = {a^n}\]. Therefore, we can write
\[42 = 2 \times 3 \times 7\]
\[63 = {3^2} \times 7\]
HCF is the highest common divisor, so from the prime factorization above we say that \[3 \times 7\]is the highest number which divides both the numbers.
\[ \Rightarrow \]HCF \[ = 3 \times 7\]
\[ \Rightarrow \]HCF \[ = 21\]
So, we can write \[\dfrac{{42}}{{63}} = \dfrac{{21 \times 2}}{{21 \times 3}}\]
Cancel same factors i.e. the HCF from numerator and denominator
\[ \Rightarrow \dfrac{{42}}{{63}} = \dfrac{2}{3}\]

\[\therefore \]The lowest term of \[\dfrac{{42}}{{63}}\]is \[\dfrac{2}{3}\].

c. \[\dfrac{{15}}{{27}}\]
Now we calculate HCF of the three numbers 15 and 27
We write each number in form of its prime factors
\[15 = 3 \times 5\]
\[27 = 3 \times 3 \times 3\]
Since, we know \[\underbrace {a \times a \times ..... \times a}_n = {a^n}\]. Therefore, we can write
\[15 = 3 \times 5\]
\[27 = {3^3}\]
HCF is the highest common divisor, so from the prime factorization above we say that \[3\]is the highest number which divides both the numbers.
\[ \Rightarrow \]HCF \[ = 3\]
So, we can write \[\dfrac{{15}}{{27}} = \dfrac{{3 \times 5}}{{3 \times 9}}\]
Cancel same factors i.e. the HCF from numerator and denominator
\[ \Rightarrow \dfrac{{15}}{{27}} = \dfrac{5}{9}\]

\[\therefore \]The lowest term of \[\dfrac{{15}}{{27}}\]is \[\dfrac{5}{9}\].

Note:
Many students get confused while calculating the HCF as they don’t know how to pick the values from prime factorizations of numbers, keep in mind we can pick common factors one-by-one as well and then multiply to get the highest common factor.