What is the greatest length $x$ such that $3\dfrac{1}{2}$ m and $8\dfrac{3}{4}$ m are integral multiples of…
What is the greatest length $x$ such that $3\dfrac{1}{2}$ m and $8\dfrac{3}{4}$ m are integral multiples of $x$?
$1\dfrac{1}{2}$ m
$1\dfrac{1}{3}$ m
$1\dfrac{1}{4}$ m
$1\dfrac{3}{4}$ m
Solution
$3\dfrac{1}{2} = \dfrac{7}{2}$ m and $8\dfrac{3}{4} = \dfrac{35}{4}$ m. The greatest length $x$ dividing both is the HCF of $\dfrac{7}{2}$ and $\dfrac{35}{4}$ = $\dfrac{\text{HCF}(7,35)}{\text{LCM}(2,4)} = \dfrac{7}{4} = 1\dfrac{3}{4}$ m.