The question asks for the Highest Common Factor (HCF) of the fractions $\frac{2}{3}$, $\frac{3}{4}$, and $\frac{5}{6}$.
To find the HCF of multiple fractions, we use the formula:
$ \text{HCF}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{HCF}(a, c, e)}{\text{LCM}(b, d, f)} $
Here, $a, c, e$ are the numerators and $b, d, f$ are the denominators.
We need to find the HCF of 2, 3, and 5.
Therefore, HCF(2, 3, 5) = 1.
We need to find the Least Common Multiple (LCM) of 3, 4, and 6.
Therefore, LCM(3, 4, 6) = 12.
Using the formula from Step 1:
$ \text{HCF}\left(\frac{2}{3}, \frac{3}{4}, \frac{5}{6}\right) = \frac{\text{HCF}(2, 3, 5)}{\text{LCM}(3, 4, 6)} = \frac{1}{12} $
The HCF of the given fractions is $\frac{1}{12}$.
What is the value of
$\frac{7}{9} - \frac{11}{12} + \frac{13}{16} - \frac{1}{8}$?
5 \(\frac{3}{4}\) + x + 2 \(\frac{1}{2}\) = 10 \(\frac{1}{8}\) Find the value of x.
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
Number 0.232323 can be written in rational form as:
Solve: \(\frac{1}{2}\) [{-2(2 + 3)*20}/2]
Match the following.
Column I | Column II | ||
a. | Equivalent fraction of \(\frac{7}{12}\) is | i. | Proper fraction |
b. | Equivalent fraction of \(\frac{9}{15}\) is | ii. | Improper fraction |
c. | \(\frac{7}{11}\) is | iii. | \(\frac{21}{36}\) |
d. | \(\frac{19}{5}\) is | iv. | \(\frac{3}{5}\) |