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: The GCM of any fraction is calculated by the two processes. First, calculate the LCM of the denominators. Then calculate the HCF of the numerators. The fraction will give you the desired result.

Complete step by step Answer:

To find the GCM of any fraction we need to calculate the LCM of the denominators and then calculate the HCF of the numerators. The fraction we will get after that by putting them together will give us our desired result.
So, for the fractions, \[\dfrac{1}{3},\dfrac{1}{6},\dfrac{4}{3},\dfrac{8}{{21}}\],
We try to find the LCM of the denominators first, we have to calculate the LCM of 3, 6, 3, 21,
Using the prime factorization method we can say that
$3 = 3 \times 1$
$6 = 3 \times 2 \times 1$
$21 = 7 \times 3 \times 1$
Therefore the LCM we get, \[3 \times 2 \times 7 = 42\] as our denominator.
Now, We try to find the HCF of the numerator first, we have to calculate the HCF of 1, 1, 4, 8
Using the prime factorization method we can say that
$1 = 1$
$4 = 2 \times 2 \times 1$
$8 = 2 \times 2 \times 2 \times 1$
From the above equation, we can say that HCF(1,4,8)=1
So, the GCM of those fractions, \[\dfrac{1}{3},\dfrac{1}{6},\dfrac{4}{3},\dfrac{8}{{21}}\]is, \[\dfrac{1}{{42}}\] which is an option (b).

Note: Finding the GCM of any fractions can be calculated by first calculating the LCM of the denominators and then calculate the HCF of the numerators, and the respective values are denominator and numerator of the required answer. Note that the LCM of any group of numbers will always be equal or greater than all the numbers individually and HCF of any group of numbers will always be equal or lesser than all the numbers individually.