Question

The G.C.M of the following fractions \[\dfrac{1}{3},\dfrac{1}{6},\dfrac{4}{3},\dfrac{8}{{21}}\] is
A.\[\dfrac{8}{{42}}\]
B.\[\dfrac{1}{{42}}\]
C.\[\dfrac{4}{{43}}\]
D.\[\dfrac{{11}}{{42}}\]

Answer 1

Hint: First, we will find the GCM of any fraction, we have to calculate by first computing the LCM of denominators and then find the HCF of the numerators. Then the fraction will be the required value.

Complete step-by-step answer:
We are given that the fractions \[\dfrac{1}{3},\dfrac{1}{6},\dfrac{4}{3},\dfrac{8}{{21}}\].
We will find the GCM of any fraction, we have to calculate by first computing the LCM of denominators and then find the HCF of the numerators
So, first we will find the LCM of denominators of given fractions.
Using the prime factorization method in the numbers 3, 6, 21, we get
\[ \Rightarrow 3 = 3 \times 1\]
\[ \Rightarrow 6 = 3 \times 2 \times 1\]
\[ \Rightarrow 21 = 7 \times 3 \times 1\]
Thus, the LCM is \[7 \times 3 \times 2 = 42\] as our denominator.
Now, we will compute the HCF of the numerator first, so we have to calculate the HCF of 1, 1, 4, 8.
Using the prime factorization method in the numbers 1, 4, 8, we get
\[ \Rightarrow 1 = 1\]
\[ \Rightarrow 4 = 2 \times 2 \times 1\]
\[ \Rightarrow 8 = 2 \times 2 \times 2 \times 1\]
Thus, the HCF is only 1 from the above prime factors of 1, 4 and 8.

Therefore, the required value is \[\dfrac{1}{{42}}\].

Note: We know that to find the GCM of any fraction, we have to calculate by first computing the LCM of denominators and then find the HCF of the numerators. Also, we need to remember that the LCM of any group of numbers will be equal or greater than all the numbers and HCF of any group of numbers will be the same or lesser than all numbers individually.