Question 6
The HCF of $$\left(\frac{9}{10},\frac{12}{25},\frac{18}{35},\frac{21}{40} \right)$$ is
Solution
HCF of $$\left(\frac{9}{10},\frac{12}{25},\frac{18}{35},\frac{21}{40} \right)$$ = $$\dfrac{\ HCF\left(9,12,18,21\right)}{LCM\left(10,25,35,40\right)}$$.
Hence, the value = $$\dfrac{3\ }{1400}$$
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