Question

Find the L.C.M of $\dfrac{2}{3},\dfrac{4}{5}\And \dfrac{5}{7}$ is:
(a) 18
(b) 24
(c) 20
(d) 30

Answer 1

Hint: We are asked to find the L.C.M of $\dfrac{2}{3},\dfrac{4}{5}\And \dfrac{5}{7}$. The L.C.M of fractions is equal to the L.C.M of numerators divided by H.C.F of the denominators of the given dfractions. L.C.M of the numbers written in the numerator of the fractions is $4\times 5$ and then find the H.C.F of the numbers written in the denominator then divide L.C.M of numerators by H.C.F of denominators to get the answer.

Complete step-by-step answer:
We have given three fractions of which we have to find L.C.M are as follows:
$\dfrac{2}{3},\dfrac{4}{5}\And \dfrac{5}{7}$
The formula for finding the L.C.M of the fractions is given below:
$L.C.M\text{ of fractions}=\dfrac{L.C.M\text{ of numerators}}{H.C.F\text{ of denominators}}$ ……….. Eq. (1)
To find the L.C.M of three fractions first of all we have to find the L.C.M of the numbers written in the numerator of three fractions. The numbers written in the numerator are 2, 4 and 5 so we can write 4 as $2\times 2$ and we know that L.C.M is the least common multiple so here we have prime factors 2 appearing two times and 5 appearing once then L.C.M is equal to
 $\begin{align}
  & 2\times 2\times 5 \\
 & =20 \\
\end{align}$
Now, divide the above result with the H.C.F of the denominators of the fraction. The numbers written in the denominator of the fractions are 3, 5 and 7. As all these three numbers are prime to each other so the H.C.F of these three numbers is 1.
From the above, the L.C.M of the numerator of three fractions is equal to 20 and the H.C.F of the denominator of three fractions is equal to 1 substituting these values in eq. (1) we get,
$\begin{align}
  & L.C.M\text{ of fractions}=\dfrac{L.C.M\text{ of numerators}}{H.C.F\text{ of denominators}} \\
 & \Rightarrow L.C.M\text{ of fractions}=\dfrac{20}{1} \\
\end{align}$
Hence, the correct option is (c).

Note: Don’t confuse with the language of the problem as we have to find the L.C.M of the three fractions so you might think that when three fractions are given with the addition sign then we take L.C.M of the denominators which is shown below:
$\dfrac{2}{3}+\dfrac{4}{5}+\dfrac{5}{7}$
In the expression above, we will find the L.C.M of the denominators to solve the addition. In this problem we don’t have to find the L.C.M of the denominators either we have to find the L.C.M of the fractions which are connected with addition or subtraction signs.
Beware of making this blunder.