Question

Andrew

Answer 1

Let's say Bailey takes "b" days to paint it.

and Andrew takes "a" days to paint the same house.

now, Andrew is 6 times faster than Bailey, therefore, if Andrew takes "a" days to do it, Bailey takes then "6a" days, or b = 6a.

now, the year they worked together, they finished it in 7 days.

so, after 1 day then, they have only done 1/7 of the whole work.

and Andrew for one day, has done 1/a of the house, whilst Bailey has done 1/b of the house or 1/(6a).

$\bf \stackrel{\textit{Andrew's rate}}{\cfrac{1}{a}}+\stackrel{\textit{Bailey's rate}}{\cfrac{1}{b}}=\stackrel{\textit{1 day of work}}{\cfrac{1}{7}} \\\\\\ \cfrac{1}{a}+\cfrac{1}{6a}=\cfrac{1}{7}\impliedby \begin{array}{llll} \textit{let's multiply all by }\stackrel{LCD}{42a}\textit{ to toss the}\\ denominators \end{array}$

$\bf 42a\left( \cfrac{1}{a}+\cfrac{1}{6a} \right)=42a\left( \cfrac{1}{7} \right)\implies 42+7=6a\implies \cfrac{49}{6}=a \\\\\\ \stackrel{days}{8\frac{1}{6}}=a \\\\\\ \textit{how many days will it take Bailey then?}\quad b=6a \\\\\\ b=6\cdot \cfrac{49}{6}\implies b=\stackrel{days}{49}$